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Study Guide and Workbook for GENETICS By Irwin H. l;1erskowifz ASSOCIATE PROFESSOR'· OF BIOLOGY SAINT LOUIS UNIVERSITY _~;I McGRAW-HILL BOOK COMPANY New York Toronto London 1960 ro THE STUDENT AND INSTRUCTOR This Study Guide and Workbook has been prepared for use with the recorded audiovisual series on 'Principles of Genetics". Near the beginning of the book you will find introductions to the distinguished -;eneticists who lectured in this series. For each of their lectures I have provided a chapter containing a ist of readings, a set of notes, and questions for discussion. At intervals a number of tests also are in~luded, each covering the material in a group of chapters. While these contents could fulfill a variety of needs of student and instructor, it should be emphasized that the Guide and the films were designed only to supplement and not to supplant a textbook and an in3tructor. It is hoped all of these used together will provide a fund of useful and interesting organized mowledge, will serve to stimulate clear thinking and independent learning on the part of the student, and the desire to remam informed of advances in genetics. Finally, as a geneticist, I would like especially to encourage students who find this subject fascinatlng, as I do, and who wish to contribute to the growth of genetlc knowledge via research. Remember that vast areas remain to be investigated, in which innumerable questions are still to be either answered, or ~ven asked. ACKNOWLEDGEMENTS This work was made possible by a grant from the Fund for the Advancement of Education to Educational Televlsion Station KETC and to Saint Louis University, both of St. Louis, Missouri. I wish to thank The Calvin Company of Kansas City, Missouri, for their cooperation in preparing most of the visual_ aid materIal. Special thanks are due my wife, Reida Postrel Herskowitz, for her efforts during every phase of preparation of this work. I should like also to express to the lecturers my appreciation for the opportunity of working with such an illustrious group of scientists, for their cooperation in numerous ways, and for the pleasure of their company. Irwin H. Herskowitz STUDY GUIDE AND WORKBOOK FOR GENETICS Copyright © 1960 by the McGraw-Hill, Inc. Printed in the United States of America. All rights reserved. ThIS book, or parts thereof, may not be reproduced in any form without permission of the pubhshers. 28390 Acknowledgment is made to: Henry Holt & Co., Inc., for the figures, used in questions 13.20, 16.6, and 16.7, from OUTLINES OF MODERN BIOLOGY by Plunkett, Charles Robert, 1930, copyright 1958 by C. R. Plunkett. McGraw-Hill Book Co., Inc., for the figures, used in questions 4.1, 4.6, 22 on p. 53, 36 on p. 145, from INTRODUCTION TO CYTOLOGY by Sharp, L. W., 1934; 2.20, 3.3, 6.17, 6.24, 6.25, 7.11, 7.15, from PRINCIPLES OF GENETICS, 3rd Ed., by Sinnott, E. W., and Dunn, L. C., 1939; and Fig. 19-3, from PRINCIPLES OF GENETICS, 5th Ed., by Sinnott, E. W .• Dunn, L. C., and Dobzhansky, Th., 1958. Georg Thieme Verlag, Stuttgart, for the fIgures, used in Fig. 33-1 and Fig. 33-2, from LETALFAKTOREN by Hadorn, E., 1955. Yale University Press, for the figure, used in question 26 on p. 142, from THE THEORY OF THE GENE by Morgan, T. H., 1926. STUDENT USE OF STUDY GUIDE - WORKBOOK For each chapter there is a Pre-Lecture Assignment, which includes a list of suggested readmgs from a number of recently published textbooks on general genetics. Reference to special books or SCIentific journals also may be included. You will learn from your instructor which book or books to purchase, and in what respects he wishes you to follow the readings suggested. Please note that listing a text does not necessarily mean it is recommended. The general texts referred to are: Altenburg, E. 1957. "GenetIcs" Revised EditIOn. 496 pp. New York: Henry Holt & Co. ColIn, E. C. 1956. "Elements of genetics" 3rd Edition. 498 pp. New York: McGraw-Hill Book Co., Inc. Dodson, E. O. 1956. "Genetics, the modern science of heredIty" 329 pp. Philadelphia: W. B. Saunders Co. Goldschmidt, R. B. 1952. "Understanding heredity, an introduction to genetics" 228 pp. New York: John WIley & Sons, Inc. Snyder, L. H., and David, P. R. 1957. "The principles of heredity" 5th EdItion. 507 pp. Boston: D. C. Heath & Co. Srb, A. M., and Owen, R. D. 1953. "General genetlcs" 561 pp. San Francisco: W. H. Freeman & Co. Sinnott, E. W., Dunn, L. C., and Dobzhansky, Th. 1958. "Prmciples of genetics" 5th Edition. 459 pp. New York: McGraw-Hill Book Co., Inc. Stern, C. 1960. "Principles of human genetics" 2nd Edition. San Francisco: W. H. Freeman & Co. Winchester, A. M. 1958. "Genetics, a survey of the principles of heredity" 2nd EdItion. 414 pp. Boston: Houghton MifflIn Co. For the sake of brevity, only the names of the authors of the general textbooks are given in the reading lIStS, in the case of additIOnal references SUitably complete bibliographic cItation is provided. Students who are interested in reading some of the key papers in genetics can fmd a number in Peters, J. A. (EdItor) 1959. "Classic papers III genetics" 282 pp. Englewood Cliffs, N. J. : Prentice-Hall, Inc. If more information is sought on chemical or microbIal genetics the student may refer to McElroy, W. D. , and Glass, B. (EdItors) 1957. 'The chemical basis for heredIty" 848 pp. Baltimore: The Johns Hopkins Press. References which will help answer many other specific questions can be obtained from the subject indexes of Parts II and III of "BIbliography on the genetics of Drosophila". Part I is by H: J. Muller (1939, 132 pp., Edinburgh: Oliver and Boyd); Part IT (includes a subject index for Part I) and Part III are by 1. H. Herskowitz (1953, 212 pp., Oxford: Alden Press, and 1958, 296 pp. Bloomington, Ind.: Indiana University Press, respectively). The Pre-Lecture Assignment is followed by the Lecture Notes. These Notes cover only the material in the recorded lectures. They were made as complete as possIble, yet, in each case, can be read in about 20 mIllutes. Space is left for additional notes you may wish to make from readings or from su_pplementary lecture and/or discussion sessions. Items of interest III the lectures omitted from the Notes are often referred to in the section QuestIOns for DiSCUSSIOn. After the Notes a Post-Lecture ASSignment is given. This may be modified by your instructor, who will assign specIfIC readings as well as partlcular questions, for which he wishes you to prepare answers orally or III writing, from the textbook and from Questions for Discussion and Examinations III the Study Guide. The Questions for DISCUSSIon are supplementary to those III the textbooks listed, deal largely with the material of the lecture they follow, and usually involve the subject matter in the' saie order as did the lecture. Accordingly, successive questions may vary in difficulty. After different groups of chapters you will find Examinations covering their subject matter as a whole. The questions, of both short answer and problem type, are designed to test not so much your memory but your ability to utilize, synthesize, extrapolate, and generally understand the contents of these chapters and their relation to the principles of genetics. CONTENTS Page iii To the Student and Instructor Student Use of Study Guide and Workbook v INTRODUCTION TO THE LECTURERS 1 Chapter 1. HEREDITY AND VARIATION (Lecturer-- L. C. Dunn) 17 Chapter 2. MENDEL'S PROOF OF THE EXISTENCE OF GENES (Lecturer--L. C. Dunn) 21 Chapter 3. MITOSIS (Lecturer-- R. E. Cleland) 26 Chapter 4. MEIOSIS (Lecturer--R. E. Cleland) 31 Chapter 5. HUMAN TRAITS SHOWING SIMPLE MENDELIAN INHERITANCE (Lecturer--L. C. Dunn) 38 Chapter 6. INDEPENDENT SEGREGATION (Lecturer--E. Altenburg) 43 EXAMINA TION I 50 Chapter 7. EXPRESSION AND INTERACTION OF GENES (Lecturer--Jack Schultz) 55 Chapter 8. MULTIPLE FACTOR INHERITANCE (Lecturer--J. F. Crow) 60 Chapter 9. ALLELISM, AND LETHALS (Lecturer--C. Stern) 66 Chapter 10. PLEIOTROPISM, PENETRANCE AND EXPRESSIVITY (Lecturer--C. Stern) 69 Chapter 11. TWINS, NATURE AND (Lecturer--C. Stern) Chapter 12. SEX-LINKED INHERITANCE (Lecturer--C. Stern) 77 Chapter 13. SEX DETERMINATION I (Lecturer--C. Stern) 82 Chapter 14. SEX DETERMINATION II (Lecturer--C. Stern) 86 EXAMINA TION II ~ruRTURE 73 90 vii Chapter 15. LINKAGE (Lecturer--G. W. Beadle) 95 Chapter 16. CROSSINGOVER IN TERMS OF MEIOSIS (Lecturer--G. W. Beadle) 99 Chapter 17. CROSSINGOVER, CHIASMATA, AND GENETIC MAPS (Lecturer--G. W. Beadle) 106 Chapter 18. CHANGES IN GENOME NUMBER (Lecturer--H. J. Muller) 112 Chapter 19. CHROMOSOME ADDITION AND SUBTRACTION (Lecturer--H. J. Muller) 116 Chapter 20. STRUCTURAL CHANGES IN CHROMOSOMES I (Lecturer--H. J. Muller) 120 Chapter 21. STRUCTURAL CHANGES IN CHROMOSOMES II (Lecturer--H. J. Muller) Chapter 22. "SPONTANEOUS" GENE MUTATION (Lecturer--H. J. Muller) 130 Chapter 23. MUTAGEN-INDUCED GENE MUTATION (Lecturer--H. J. Muller) 134 EXAMINA TION III 138 Chapter 24. GENETICS OF MENDELIAN POPULATIONS (Lecturer--Th. Dobzhansky) 149 Chapter 25. GENETIC LOADS IN MENDELIAN POPULATIONS (Lecturer--Th. Dobzhansky) 154 Chapter 26. SELECTION, GENETIC DEATH AND GENETIC RADIATION DAMAGE (Lecturer--Th. Dobzhansky) 158 Chapter 27. . THE GENETICS OF RACE (Lecturer--Th. Dobzhansky) 162 Chapter 28. THE ORIGIN OF SPECIES (Lecturer--G. L. Stebbins) 166 Chapter 29. INTERSPECIFIC HYBRIDIZATION AND ITS CONSEQUENCES (Lecturer--G. L. Stebbins) 170 Chapter 30. INBREEDING AND HETEROSIS (Lecturer--J. F. Crow) 175 Chapter 31. CYTOGENETICS OF OENOTHERA (Lecturer--R. E. Cleland) 180 EXAMINA TION IV viii 186 Chapter 32. DEVELOPMENTAL GENETICS I (Lecturer-- L. C. Dunn) 191 Chapter 33. DEVELOPMENTAL GENETICS II (Lecturer-- L. C. Dunn) 195 Chapter 34. CYTOPLASMIC HEREDITY (Lecturer--T. M. Sonneborn) 200 Chapter 35.- NUCLEO-CYTOPLASMIC RELATIONS IN PARAMECIUM (Lecturer--T. M. Sonneborn) 205 Chapter 36. PSEUDOALLELISM (Lecturer--E. B. LewIs) 210 Chapter 37. VARIEGA TED PERICARP AN UNSTABLE ALLELE IN MAIZE (Lecturer--R. A. Brink) 215 Chapter 38. DNA STRUCTURE AND REPLICATION (Lecturer--J. D. Watson) 220 Chapter 39. BIOCHEMICAL GENETICS I (Lecturer--G. W. Beadle) 225 Chapter 40. BIOCHEMICAL GENETICS II (Lecturer--G. W. Beadle) 229 Chapter 41. GENE STRUCTURE AND GENE ACTION (Lecturer--G. W. Beadle) 233 Chapter 42. CHROMOSOME CHEMISTRY AND GENETIC ACTIVITY (Lecturer--J ack Schul tz) 237 EXAMINA TION V 241 Chapter 43. BACTERIAL GENETICS: CLONES (Lecturer--J. Lederberg) 247 Chapter 44. BACTERIAL GENETICS: SEXUAL REPRODUCTION (Lecturer--J. Lederberg) 251 Chapter 45. BACTERIAL GENETICS: GENETIC TRANSDUCTION (Lecturer--J. Lederberg) 256 Chapter 46. VIRUS GENETICS: BACTERIOPHAGE (Lecturer--J. Lederberg 260 Chapter 47. VIRUS GENETICS: PLANT AND ANIMAL VIRUSES (Lecturer--J. Lederberg 264 Chapter 48. BIOCHEMICAL ORIGIN OF TERRESTRIAL LIFE (Lecturer--J. Lederberg 268 EXAMINA TION VI 272 ix INTRODUCTION TO THE LECTURERS THE REDUCTION DIVISION IN THE F,HYBRID ~0 - - ~ RY(rcm:fyellow) ([iJ) ® OR ~@) --: c@@ r Y(wnnkled gr.) __ C» ~ I), ~ ,(round green\ ..._:-:;.~ @)(2) -- <.ED0 EDGAR ~ . ALTENBURG Dr. Altenburg was born in Jersey City in 1888. He received all his academic degrees - the A. B. in 1911, A. M. in 1912, and Ph. D. in 1916 - from Columbia Uinversity, where he was a student of T. H. Margan.. It was there that he and H. J. Muller met and became friends and research collaborators, and in both respects the intensity and fruitfulness of their association continues to the present. Since obtaining his doctorate Professor Altenburg has been in t11e Department of Biology at The Rice Institute in Houston, Texas. His continued wide range of interest in genetic matters is reflected and expressed in his textbook "Genetics" (1957, revised edition), and his articles in scientific journals. Hisresearch uses Drosophila for studies of natural and induced mutation rates. Dr. Altenburg was the first to demonstrate that heritable changes are produced by ultraviolet light. 2 GEORGE WELLS BEADLE Dr. Beadle had already made significant contributions to the cytology and genetics of maize and to the physiological genetics of Drosophila before the experiments were performed on the chemical genetics of Neurospora, for which he and his collaborator Dr. E. L. Tatum were awarded a portion of the Nobel Prize in physiology and medicine in 1958. He was born in Wahoo, Nebraska, in 1903, and went to the University of Nebraska for the B. S. (1926), and M. S. (1927) degrees, and Cornell University for the Ph. D. (1931). He has received honorary D. Sc. degrees from Yale University, University of Nebraska, Northwestern University, Rutgers University, Kenyon College, Wesleyan University, University of Birmingham, and Oxford University. Dr . Beadle is a member of the National Academy of Sciences and the American Philosophical Society. He has been President of the Genetics Society of America (1946), the Western Society of Naturalists (1947), and the American Association for the Advancement of Science (1955). Dr. Beadle's teaching and research career includes positions at Cornell University (1928-1931), Harvard University (1936-1937), and Stanford University (1937 -1946). In 1946 he assumed his present position as Professor and Chairman, Division of Biology, California Institute of Technology in Pasadena, California. 3 ROYAL ALEXANDER BRINK Dr. Brink has been Professor of Genetics in the Department of Genetics at the University of Wisconsin, at Madison, since -;_931. He came to Wisconsin as Assit>Lant Professor in 1922, becoming Associate Professor in 1927, and from 1939-1951 was Chairman of the Department. Dr. Brink was born near Woodstock, Ontario, Canada, in 1897, and enrolled at the Ontario Agricultural College (University of Toronto) where he received the B. S. A. degree in 1919. His M. S., obtained from the University of illinois in 1921 was followed, in 1923, by the Sc. D. degree in genetics from Harvard University. Dr. Brink was a National Research Fellow, in Berlin and Birmingham, in 19251926. He is a member of the National Academy of Sciences. Dr. Brink served as managing editor of "Genetics" from 1952-1957, as Treasurer of the American Society of Naturalists (1936-1939), and as Vice President (1953) and President (1957) of the Genetics Society of America. His research publications include work done on pollen physiology, reciprocal translocations in maize, the endosperm in seed development, plant breeding, and gene action and mutation. Most recently Dr. Brink has been analyzing unstable alleles in maize, in particular, variegated pericarp, and a phenomenon he has termed paramutation. 4 RALPH ERSKINE CLELAND Until recently, Dr. Cleland was simultaneously Professor of Botany, Chairman of this Department, and Dean of the Graduate School at Indiana University, Bloomington, Indiana. He is now Distinguished Service Professor of Botany at that institution where he teaches cytology and carries on research on the genetics, cytology and evolution of the evening primroses, Oenothera. He was born in LeClaire, Iowa in 1892 and attended the University of Pennsylvania where between 1915 and 1919 he received the A. B., M. S., and Ph. D. degrees. Starting as Instructor in 1919 he became Professor and Chairman of the Biology Department at Goucher College by 1937. A year later he joined the faculty at Indiana University. Dr. Cleland was a Guggenheim Fellow in 1927-1928, received the Lewis Award of the American Philosophical Society in 1937, and from 1950-1953 was Chairman of the Division of Biology and Agriculture of the National Research Council. He was editor in chief of the "American Journal of Botany" (1940-1946), and starting in 1925 he continues to be editor of the section on plant cytology of "Biological Abstracts ". Dr. Cleland is a member of the National Academy of Sciences, the American Academy of Arts and Sciences, the Botanical Society of America (President in 1947), the American Society of Naturalists (Secretary from 1938-1940, and President in 1942), the Genetics Society of America (Vice-President in 1954, President in 1956), the American Philosophical Society, and the International Union of Biological Sciences (Vice-President since 1953). 5 CHROMOSOME PfRC£NT SURVIVAL x 2301510204060 80 2_== 3=_= 4== __ 1=== " -.. . = =_ .... B- __ .... g---- • ('1_=-: 7 10 - - 11- - _ 12-13 -_ 14 1516 - - • JAMES FRANKLIN CROW Dr. Crow was born in Proenixvi1le, Pennsylvania, in 19U6. He received his A. B. degree from Friends University in 1937 and his Ph. D. degree in genetics at the University of Texas in 1941. After this and until 1948 he was first Instructor and then Assistant Professor at Dartmouth College. He then went to the Department of Genetics at the University of Wisconsin where he advanced by 1954 to the rank of Professor of Zoology and Genetics. Since 1958 he has been Chairman of the Department of Medical Genetics. Dr. Crow was Vice President of the Genetics Society of America in 1958, and President in 1959. While Dr. Crow uses Drosophila in experimental studies of the mechanism of transmission (classical) and population genetics, he has been increasingly active in applying the principles of both these branches of genetics to the study of human populations. He has been concerned recently with the problem of estimating the genetic consequences of exposure of populations to penetrating radiations. Dr. Crow, author of more than 25 scientific papers, and of TlGenetics Notes Tl , is writing, in collaboration with Dr. M. Kimura, a book on the mathematical theory of population genetics. 6 THEODOSIUS DOBZHANSKY Dr. Dobzhansky, Professor of Zoology at Columbia University in New York City, was born in Russia in 1900. Educated at the University of Kiev, he began teaching and research at Kiev Agricultural Institute and the University of Leningrad. He came to the United States in 1927, worked with T. H. Morgan and his group at Columbia University and the California Institute of Technology from 1929-1940, where he was Professor, before accepting his present post. He received the Kimber Genetics Award in 1958. Dr. Dobzhansky is a member of the National Academy of Sciences, the American Philosophical Soc iety, the American Academy of Arts and Sciences, the Royal Danish, Royal Swedish, and Brazilian Academies of Sciences, and has received honorary D. Sc. degrees from the College of Wooster, the University of Mttnster, the University of Montreal, and the University of Sft'o Paulo. He has been President of the American Society of Naturalists, the Genetics Society of America, and the Society for the Study of Evolution. He is author of "Genetics and the Origin of Species tl (1951, 3rd edition), tlHeredity, Race, and Society" (1957, 3rd edition, with L. C. Dunn), ttprinciples of Genetics If (1958, 5th edition, with E. W. Sinnott and L. C. Dunn), and ttEvolution, Genetics, and Man" (1955). In recent years Dr. Dobzhansky's research has been concerned chiefly with population genetics and evolutionary problems, using laboratory and wild Drosophila as experimental material. 7 .. LESLIE CLARENCE DUNN Dr. Dunn is Professor of Zoology at Columbia University in New York City. Born in Buffalo, New York in 1893, he rec e~ ve d t:18 B. S. degree at Dartmouth College and the M. S. and Sc. D. degrees from Harvard University. He was then geneticist at The Storrs Agricultural Experiment Station in Connecticut from 1920-1928, after which he assumed his professorship at Columbia. Dr. Dunn is a member of the National Academy of Sciences, the American Philosophical Society, the Norwegian Academy of Sciences, and Accademia Pativina (Italy). In 1932 he was President of the Genetic: Society of America, and was Vice-President of the American Society of Naturalists in 1942. Besides editorial work with "Genetics ", "The American Naturalist", "Journal of Experimental Zoology", "Encyclopedia Britannica", and "Columbia Biological Series" Dr. Dunn is the editor of the book "Genetics in the Twentieth Century" (1951). He has written several books: "Heredity and Variation" (1932), "Biology and Race" (1951), "Principles of Genetics" (1958, 5th edition, with E. W. Sinnott and Th. Dobzhansky), "Heredity, Race, and Society" (1957, 3rd edition, with Th. Dobzhansky), and "Heredity and Evolution in Human Populations" (1959) . For many years his research has been devoted to a study of the ways in which heredity affects development, using as experimental material Drosophila, poultry, mice, and men. Among his current interests are human genetics, and population and developmental genetics of the mouse. 8 JOSHUA LEDERBERG Born in New Jersey in 1925, Dr. Lederberg received the Ph. D. degree in microbiology from Yale University in 1947, having earlier obtained the B. A. degree from Columbia University in 1944. He then went to the University of Wisconsin as Assistant Professor of Genetics, became Associate Professor in 1950, and Professor in 1954. In 1959 he went to Stanford University, Palo Alto, California, where he heads a new Department of Genetics in the School of Medlcine. Dr. Lederberg shared the Nobel Prize in physiology and medicine in 1958 with G. W. Beadle and E. L. Tatum. The award was based on his "discoveries concerning genetic recombination and the organization of the genetic material of bacteria". He is a member of the National Academy of Sciences. His wife, associate and collaborator, Dr. Esther Marilyn Lederberg, is also an active researcher in problems dealing with the genetics of microorganisms, lysogenicity, and bacterial recombination. 9 EDWARD B. LEWIS Born in 1918 in Wilkes-Barre, Pennsylvania, Dr. Lewis attended the University of Minnesota where in 1939 he received the B. A. degree. He obtained the Ph. D. degree in Genetics in 1942 from the California Institute of Technol G ~,y in Pasadena, California. Dr. Lewis was a Rockefeller Foundation Fellow at the School of Botany at Cambridge, England, in 1948-1949, and served in the Army Air Force from 19421945. Since 1946 he has been at the California Institute of Technology as Instructor until 1948, Assistant Professor (1948-1949), Associate Professor (1949-1956) and Professor in the Department of Biology since 1956. His studies on the nature of position effect and of pseudoallelism with Drosophila have contributed significantly to our knowledge of transmission and physiological genetics. Recently he has been active, in addition to continuing his past research interests, in exploring the genetic effects of radiation, and particularly the relationship between radiation exposure and incidence of leukemia. 10 HERMANN JOSEPH MULLER After receiving his doctorate at Columbia University in 1916, where he studied with T. H. Morgan and E. B. Wilson, Dr. Mul.ler taught and did research at The Rice Institute, Columbia University, and the University of Texas. It was in Texas, in 1926, when Dr. Mul.ler was 35 years old, that the experiments were begun, proving the ability of X-rays to produce mutations, which earned him the Nobel Prize in physiology and medicine in 1946. Since 1953 he has been Distinguished Service Professor in the Department of Zoology at Indiana University., Bloomington, Indiana. His scientific works, numbering more than 300, have contributed to such basic and diverse aspects of genetics as; linkage and crossingover; multiple factor analysis; theory of the gene; evolution; gene mutation and chromosome change and their artificial production; biological effects of radiation; speciation; human genetic s; properties of chromosome parts. Dr. Muller has received additional degrees from Columbia University and the University of Edinburgh' and the Kimber Genetics Award. He is a member of the National Academy of Sciences, the American Society of Naturalists (President in 1943), the American Philosophical Society, the American Academy of Arts and Sciences, the Genetics Society of America (President in 1947), and the American Society of Human Genetics (President in 1949). He was President of the 8th International Congress of Genetics (Stockholm, 1948). Dr. Mul.ler is an honorary or foreign member of the British Genetical Soc., the Genetics Soc. of Japan, the Mendelian Soc. of Lund, the Royal Swedish Acad., the Royal Danish Acad., the Nat. Inst. of Sci. of India, the Royal Soc. of London, the Accad. Naz. dei Lincei (Rome), and the Acad. Sci. et Litt. Moguntina (Mainz). JACK SCHULTZ Dr. Schultz condnuc.s to investigate a large variety of genetic and cytological problems at the Institute for Cancer Research in Philadelphia, Pennsylvania, where he is Chairman of the Division of Biology and Head of the Department of Genetics and Cytochemistry. His research work has contributed to an understanding of the nature and function of the gene, of chemical genetics as studied in Drosophila by genetic, cytochemical, and nutritional techniques, of the cytochemistry of growth, and of the pattern of human chromosomes. Dr. Schultz was born in 1904 in New York City and his academic background includes receipt of the A. B. (1924), A. M. (1925), and Ph. D. (1929) degrees from Columbia University. He has been a National Research Council Fellow (1927-1929) and Visiting Professor at the University of Missouri (1942-1943). During the years 1937-1939 he was an International Fellow of the Rockefeller Foundation, associated with T. Caspersson at the Karolinska Institute in Stockholm, Sweden. Before accepting his present position in 1943 he was, for a number of years, a plember of T. H. Morgan's research group. One of Dr. Schultz's Research Associates and collaborators is his wife Dr. Helen Redfield, who is especially interested in the study of intra- and inter-chromosomal effects on crossingover. \ 12 or o----~ '..t' I ~" \. o CLONE I " TRACY 0 CLONE ~ MORTON SONNEBORN Dr. Sonneborn's research interests lie primarily in the area of the genetics of microorganisms, especiaJ.ly Paramecium. He is, since 1953, Distinguished Service Professor in the Department of Zoology, Indiana University, Bloomington, Indiana. He came to Indiana in 1939 from The Johns Hopkins University where he had received the B. A. degree in 1925, the Ph. D. degree in 1928, and had remained first as assistant, then Research Associate, and later Associate in Zoology. In 1951 he was Visiting Professor at the University of Chile. Dr. Sonneborn is a member of the NationaJ. Academy of Sciences, The American Philosophical Society, and the American Academy of Arts and Sciences. He has held office in such scientific organizations as the American Society of Naturalists (Treasurer 1944-1947, Vice President 1948, and President 1949), the Genetics Society of America (Vice President 1948, and President 1949), and the American Society of Zoologists (President 1956). In 1959 Dr. Sonneborn received the Kimber Genetics Award of the National Academy of Sciences. 13 GEORGE LEDYARD STEBBINS Born in Lawrence, New York, in 1906, Dr. Stebbins attended Harvard University where he received the A. B. (1928), A. M. (1928), and Ph. D. (1931) degrees. After a position at Colgate University he went to the University of Califorr.ia a t Davia where he has been Professor in the Department of Genetics since 1947. Dr. Stebbins is a member of the National Academy of Sciences and was Vice-President of the California Botanical Society in 1948, and Vice-President (1947) and then President (1948) of the Society for thE Study of Evolution. He is author of "Variation and Evolution in Plants lt (1950) based upon a series of Jesup Lectures delivered at Columbia University. Dr. Stebbins is a botanical geneticist whose research interests have included studies on the cytogenetics of parthenogenesis in the higher plants, cytogenetics of Paeonia, grasses, and other plants, production of hybrid and polyploid types of forage grasses, and natural selection in the higher plants. CURT STERN Dr. Stern, born in Hamburg, Germany, received his Ph. D. degree from the University of Berlin in 1923 when he was 21 years old. From 1924-1926 he was a Fellow of the Rockefeller Foundation working in the famous Drosophila laboratory at Columbia University under T. H. Morgan, A. H. Sturtevant and C. B. Bridges. Some years later he left Germany permanently for a position at the University of Rochester, where in 1946 he became Professor of Experimental Zoology and Chairman of the Division of Biological Sciences. In 1947 he assumed his present post as Professor of Zoology at the University of California at Berkeley. Among the societies to which Dr. Stern belongs are the National Academy of Sciences, the Genetics Society of America (of which he was President in 1950), the American Society of Human Genetics (President in 1957), and the American Philosophical Society. Dr. Stern is distinguished not only for his researches with Drosophila on the cytogenetic basis of independent assortment of chromosomes, somatic crossingover, position effect, developmental and radiation genetics, but for his contributions to human genetics as exemplified by his book I'Principles of Human Genetics" (1960, 2nd edition). 15 JAMES DEWEY WATSON Native of Chicago, where he was born in 1928, Dr. Watson went to the University of Chicago where he received the B. S. degree in 1947. He attended Indi;:u;,<:;. University and received the Ph. D. there in 1950 Dr. Watson was a National Research Council Fellow in Copenhagen, Denmark (1950-1951) and in Cambridge, England (1951-1952), and a National Foundation for Infantile Paralysis Fellow (1951-1953) at Cambridge. He was a Senior Research Fellow in Biology at the California Institute of Technology from 1953-1955. In 1955 he assumed his present post as Associate Professor in the Department of Biology at Harvard University in Cambridge, Massachusetts. Dr. Watson was active in studying bacteriophage reproduction and bacterial genetics before he and Dr. F. H. C. Crick published their now-classical research on the chemical organization of nucleic acids in 1953. 16 Chapter 1 HEREDITY AND VARIATION Lecturer-L. C. PRE-LECTURE ASSIGNMENT 1. Suggested readings from textbooks: Altenburg: Chap. 1, pp. 18-23. Colin; Chap. 11, pp. 181-193, 208-212. Dodson: Chap. 1, pp. 1-7; Chap. 5, pp. 46-50. Goldschmidt: Chap. 1, pp. 3-17, 24-27. Sinnott, Dunn, and Dobzhansky: Chap. 1, pp. 1-12; Chap. 2, pp. 17-30. Snyder and David: Chap. 1, pp. 3-10; Chap. 16, pp. 222-224; Chap. 26, pp. 380381; Chap. 28, pp. 415-417. Srb and Owen: Chap. 1, pp. 1-14; Chap. 2, pp. 16-17. Stern: Introduction. Winchester: Chap. 1, pp. 1-18; Chap. 2, pp. 19-33. 2. Read the lecture notes through the portion headed "Chief branches of genetics". LEC TURE NOTES The remarkable capacity of living things to convert non-living material into living stuff results in bodily growth or in the production of new individuals. The formation of new individuals, or reproduction, provides the continuity of living things from one generation to the next. This continuity is the subject of the study of heredity, while the resemblances and differences between individUals of the same or different generations is the concern of the study of variation. These two processes of continuity and of change, which we refer to as heredity and variation, are universal and primary consequences of all organisms. Their organized study is the scope of the science of genetics. A. Heredity 1. Continuity of life is accomplished via reproduction which, except for viruses, al ways involves the separation of cells from parents to form new individuals. In asexual DUNN reproduction the single cells, or (less frequently) the group of them involved, come from a single parent, while in sexual reproduction two single reproductive cells (gametes) unite in a process called fertilization to start the new life as a single cell, the zygote. In higher animals and in flowering plants the gametes are designated as ~ (female) and sperm (male). 2. Continuity's narrow bridge It is clear that even though organisms undergoing reproduction may be themselves composed of billions of cells, the hereditary contribution to offspring must be contained in the single cell by which the new life is initiated. We know that the essential elements which form the living link between generations reside in the nucleus of the cell. In sexual reproduction the gametic nuclei join to form the zygotic nucleus. 3. Inheritance of codes Since nuclei do not contain bodily features (limbs, eyes, blood) they must possess an organization or pattern which carries in very condensed form, like a code, the instructions transmitted from parents to offspring. These instructions guide the metabolism, which is also influenced by environment, so that the final result is the appearance of specific characters. 4. Chief branches of genetics a. Transmission genetics aims to analyze the content of nuclei into the structural and functional units which transmit the nucleic code. Study of mitosis and meiosis (Chaps. 3 and 4) will help. b. Physiological or developmental genetics attempts to discover how the instructions received from parents (1) guide the course of development, and (2) are carried out in such specific ways that the 17 5. 6. 7. 8. new individual repeats in its living activities the same pattern of metabolism and growth which the parent showed. c. Population or evolutionary genetics Since descendants sometimes differ from ancestors one has to determine the origin of hereditary variations, how they are distributed in populations, whether the populations are rendered more fit by them, and in what respect these variations explain the ways that populations change. Misunderstandings regarding biological inheritance a. Not to be confused with inheritance of property or culture. b. Non-heritability of acquired characters. The only biological variations which are heritable are mutations -- changes in the hereditary material itself. c. Not a miscible fluid like blood. There is no blending between what is inherited from different parents, for this remains discrete, emerging from mixtures intact and unaltered (Mendel). d. Need to appreciate the relative contribution of heredity and environment. Johannsen's experiments with the self-fertilizing bean pl:.L.'1t showed the relative role of heredity and environment in determining seed weight. Seed weight could be effectively selected for when samples were drawn from a mixed population but not when applied to progeny from self-fertilization, for these comprise a pure line. Members of pure lines have the same inherited constitution or _genotype. The expression of the genotype in traits or characters is called the phenotype and is dependent upon the environment. Variation within pure lines is entirely phenotypic, barring the rare event of mutation. The genotype is composed of genes which are the units of the inherited code which together with the environment produce the phenotype. Himalayan albinism in rabbits is a phenotype determined jointly by heredity and environment. What is inherited is the ability of the pigment formation system to respond to environmental temperature in a specific way. The possible variation in coat color pattern in such rabbits is the phenotypic range or norm or range of reaction of their genotype. The black rabbit has a genotype B. with a narrow range oJ phenotypic expression. Variation · The lack of identical phenotypes in members 0 cross-fertilizing species living in similar environments is largely due to the existence of a variety of genotypes each with a different nor of reaction. The amount and source of the variety which exists was first illustrated using 9-omesticated species. 1. Darwin's example of variation among genotypically different breeds or fanciers' varieties derived from the rock pigeon (Columbia li vi a) was shown. 2. Hereditary variations in the house mouse, Mus musculus, may involve not orily color, pattern, structure, and chemical differences, but behavioral muta~ions . (like the waltzing mice shown) also exist. When the different variations are combined with ~ach other in all possible ways the total variety possible is staggering. 3. Were there no heritable variety it would not be possible to study transmission genetics. Mendel recognized this and took advantage of genotypiC variation in analyzing the process of inheritance. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture, or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. QUESTIONS FOR DISCUSSION 1. 1. What are the genetic implications of the biblical statement that organisms reproduce each after its own kind? 1. 2. A relation is sometimes traced between the Inquisition, which was particularly active in Spain, and the decline of Spain as a great power which took place not long afterward. VVhat genetic basis do you think there may be for this? 1. 3. How might a knowledge of genetics help solve agricultural, industrial, medical, poli- tical, or social problems? Even though cultural acquisitions like language are not biologically heritable, describe possible heritable variations which could prevent the learning of a language, that is, the ability to use symbols. rabbits. 1. 9. How could you get a rabbit to grow your name on its back a) in black on a white background, and b) in white on a black background? 1. 10. How could temperature be used to mimic phenotypically (phenocopy) a genetically black by a genetically Himalayan rabbit? Would this influence the transmission behavior? Why? By means of temperature changes, could you make a phenocopy of the Rimal ayan using a genetically black rabbit? Why? 1. 4. 1. 5. The practice of cutting off the tail in puppies is commonly used in certain breeds of dogs. In these breeds some puppies are born with short tails or without tails, and this is transmitted to their offspring. Would you consider this valid evidence of inheritance of an acquired trait? Explain. Suppose, accidentally, a group of beans is found whose lengths range from 5 to 20 mm. Upon pI anting these and I ater obtaining the beans produced by self-fertilization it is found that 5 mm. parents each produced 5-7 mm. beans, while the 20 mm. parents each produced 12-20 mm. beans. What conclusions could you come to regarding a) the relative contributions of heredi ty and environment to bean length, b) the pure line status of the beans originally found and of the offspring later obtained, c) norms of reaction? How would your conclusions have been affected if the range of 5-20 mm. was obtained from each of the beans pI anted '? 1. 11. Normal (''wild type") strains of Drosophila melanogaster have grayish-brown bodies if developed on food media free of silver salts, but have straw-colored bodies if certain silver sal ts are added to the food on which the larvae develop (Rapoport, DiStefano). Strains of the mutant straw in the same species have straw body color regardless of whether they develop on food with or without silver salts. Suppose you have a straw fly and the food on which it has developed is unknown. Row would you find out whether this fly is a mutant or a phenocopy? 1. 6. 1. 7. How would you explain the occurrence of a large number of eminent people within a single family? 1. 12. "When hair from a dark region of a Siamese cat is removed and this region is kept warm the new hair is dark. " In what respect is this statement true or false? 1. 13. How could you demonstrate a wide range of reaction for the same genotype using cuttings from a single plant? If members of a white-skinned race are exposed to bright sunlight, their skin is darkened, or "tanned". Races native to regions of bright sunlight, like Negroes in the tropics, are genetically dark-skinned. How would Lamarck explain the dark skin of such races? Row would you? 1. 14. 1. 15. 1. 8. Explain why it is or is not possible to change the eye color of adult Himalayan albino "What is insulin's effect upon the diabetic's norm of reaction? 19 1. 16. 'What has been the effect of human variability on the development of human societies? 1. 17. How do the varieties of the pigeon, Columba, differ phenotypically? 'What evidence can you offer that these differences have a genotypic basis? 1.18. Name 6 cultivated plants or domesticated animals which show abundant genotypic variety. Why have pedigrees been kept much more carefully for our domestic animals than for our cultivated pl.ants? 1. 19. AI though the number of different kinds of mice which are possible is practically infinite, how many different kinds would be possible if the total number of traits recognized (each having two alternative states) was 1? 2? 3? 47 5? n? 1. 20. 1. 21. To what do you attribute the great variability within the human species? Would you expect the present population of Israel to be more variable than that of most European countries? Why? 1. 22 . Can you provide an explanation why measles and chicken pox killed thousands of natives of the South Sea Islands at their first contact with white men, although these diseases are rarely fatal to Europeans? 1. 23. What could you possibly learn about genetics if all the organisms studied possessed identical genotypes? What advantage does genotypic variation provide? 1. 24. What effect do you think the Napoleonic wars had on the average stature of the French people? To what would you attribute this? 20 Chapter 2 MENDEL'S PROOF OF THE EXISTENCE OF GENES Lecturer__:l.. C. PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings from textbooks: Altenburg: Chap. 2, pp. 25-36. Colin: Chap. 1, pp. 1-10; Chap. 2, pp. 11-21. Dodson: Chap. 2, pp. 8-18. Goldschmidt: Chap. 3, pp. 55-71. Sinnott, Dunn, and Dobzhansky: Chap. 3, pp. 32-42. Snyder and David: Chap. 2, pp. 11-19; Chap. 3, pp. 22-25. Srb and Owen: Chap. 2, pp. 17-29. Stern: Introduction. Winchester: Chap. 5, pp. 63-71. LECTURE NOTES A. Mendel's law of the splitting of hybrids is of primary importance in the study of transmission genetics. 1. His success was the consequence of a brilliant experimental design, which was then revolutionary in plant hybridization work. It required that: a. the experimental' organisms provide heritable variations producing sharp, visible differences; b. all crosses between varieties be possible and fully fertile; c. all matings be controlled by the experimenter. 2. The pea plant was selected, after preliminary tests, because it satisfied these requirements. Peas normally self-fertilize, but controlled cross-fertilization can be done by hand, by the method described. 3. For the first in a series of matings Mendel made cross-fertilizations between parental (P1) pure lines differing in one character. In all 7 different crosses of this type the offspring (first filial generation, or Fl) were Uniform and phenotypically like one of the Pl. .. DUNN He then permitted each of the F1 to self-fertilize (hence serve as a P2) and collected the F2 progeny in large numbers. These were not uniform, for some were like one grandparent in the P1 and some like the other, the phenotypes being, on the average, roughly 75% like one and 25% like the other. 4. Each F2 plant was then permitted to self-fertilize (serve as a P3)' In F3, he obtained three kinds of results: a. Each one of the 25% of P3 plants phenotypically like one P1 again produced only the same phenotype . They bred true and were just as pure as the variety which was used as Pl' The character had not been changed even though it had passed through a generation (Fl or P2) in which it had not shown itself. At the time it was recovered in F2 it was uncontaminated. b. There were two categories of P3 plants among the 75% which looked like the other Pl' 1) Of these, approximately 1/3 (or 25% of all F 2 ) bred true like their PI type, the observations made in "a" applying here also. 2) The remainder, about 2/3 (or 50% of all F2) bred like their parent (P2) did. 5. From this it was clear that among the F2, what was important was not the phenotypic ratio of 3:1, but the fact that it was composed of 25% of individuals which bred pure like one of the PI, 25% of individuals which bred pure like the other P1' and 50% which split upon self-fertilization, like the P2 did. 6. These results were obtained regardless of which one of the 7 pairs of contrasting characters Mendel bred in this manner. B. Mendel's hypothesis of unitary factors 1. 'tTnitary factors (now called genes) were assumed to occur in alternative states (or 21 1. 16. What has been· the effect of human variability on the development of human societies? 1. 17. . How do the varieties of the pigeon, Columba, differ phenotypically? What evidence can you offer that these differences have a genotypic basis? 1. 18. Name 6 cultIvated plants or domesticated animals which show abundant genotypic variety. 1. 19. Why have pedigrees been kept much more carefully for our domestic animals than for our cultivated plants? 1. 20. AI though the number of different kinds of mice which are pOSSIble is practically infmite, how many different kinds would be possible if the total number of traits recognized (each having two alternative states) was 1? 2? 3? 4? 5? n? 1. 21. To what do you attribute the great variability within the human species? Would you expect the present population of Israel to be more variable than that of most European countries? Why'? 1. 22. Can you provide an explanation why measles and chicken pox killed thousands of natives of the South Sea Islands at their first contact with white men, although these diseases are rarely fatal to Europeans? 1. 23. What could you possibly learn about genetics if all the organisms studied possessed identical genotypes? What advantage does genotypiC varIation provide? 1. 24. What effect do you think the Napoleonic wars had on the average stature of the French people? To what would you attribute this ? 20 Chapter 2 MENDEL'S PROOF OF THE EXISTENCE OF GENES Lecturer-L. C. DUNN PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings from textbooks: Altenburg: Chap. 2, pp. 25-36. Colin: Chap. 1, pp. 1-10; Chap. 2, pp. 11-21. Dodson: Chap. 2, pp. 8-18 Goldschmidt: Chap. 3, pp. 55-71. Sinnott, Dunn, and Dobzhansky: Chap. 3, pp. 32-42. Snyder and David: Chap. 2, pp 11-19; Chap. 3, pp. 22-25. Srb and Owen: Chap. 2, pp. 17-29. Stern: Introduction. Winchester: Chap. 5, pp. 63-71. LECTURE NOTES A. Mendel's law of the splitting of hybrids is of primary importance in the study of transmission genetics 1. His success was the consequence of a brilliant experimental design, which was then revolutionary in plant hybridization work. It required that: a. the experimental organisms provide heritable variations producing sharp, visible differences; b. all crosses between varieties be possible and fully fertile; c. all matings be controlled by the experimenter. 2. The pea plant was selected, after preliminary tests, because it satisfied these requirements. Peas normally self-fertilize, but controlled cross-fertilization can be done by hand, by the method described. 3. For the first in a series of matings Mendel made cross-fertilizations between parental (PI) pure lines differing in one character. In all 7 different crosses of this type the offspring (first filial generation, or F1) were uniform and phenotypically like one of the Pl. He then permitted each of the F1 to self-fertilize (hence serve as a P2) and collected the F2 progeny in large numbers. These were not uruform, for some were like one grandparent in the PI and some like the other, the phenotypes being, on the average, roughly 75% like one and 25% like the other. 4. Each F2 plant was ~en permitted to self-fertilIze (serve as a P3)' In F3, he obtained three kinds of resul ts: a. Each one of the 25% of P3 plants phenotypically like one PI again produced only the same phenotype. They bred true and were just as pure as the variety which was used as Pl' The character ,had not been changed even though it had passed through a generation (F 1 or P2) in which it had not sh_own itself. At the time it was recovered in F2 it was uncontaminated. b. There were two categories of P3 plants among the 75% which looked like the other Pl' 1) Of these, approximately 1/3 (or 25% of all F 2 ) bred true like their PI type, the observations made in "a" applying here also. 2) The remainder, about 2/3 (or 50% of all F2) bred like their parent (P2) did. 5. From this it was clear that among the F2, what was important was not the phenotypic ratlO of 3:1, but the fact that it was composed of 25% of individuals which bred pure like one of the PI, 25% of individuals which bred pure like the other PI, and 50% which split upon self-fertilization, like the P2 did. 6. These results were obtamed regardless of which one of the 7 pairs of contrasting characters Mendel bred in this manner. B. Mendel's hypothesis of unitary factors 1. Unitary factors (now called genes) were assumed to occur in alternative states (or 21 alleles} which were responsible for the production of the alternative characters. 2. Gametes each possess only one gene for a particular trait and are said to be haploId, so that the fertilized egg must carry a pair of this kind of gene and is said to be diploid. (This is paralleled by the behavior of chromosomes of sexually reproducing organisms which are paired or diploid in the fertilized egg and in adults, and single or haploid in gametes. ) C. The PrinCIple of Segregation 1. If adults possess paired genes and their gametes have these unpaired, the members of a pair of genes must segregate, or separate, from each other upon entering the gamete. 2. The gene present in a gamete is uncontaminated (uninfluenced) by the gene, whatever the allele, whICh was its partner in the adult. 3 Hybrids, formed by crossing two pure lines dlffering in one trait, must possess one gene contributed by each parent, and Mendel's statement that "one or other" of these two factors enters the gametes produced by the hybrid is essentially the principle of segregation as just described in 1 and 2. D. Abstract model validates experiments 1. Consider the trait flower color, and its alternatives colored and white. Let Q = gene for colored, ~ = gene for white. The results of crosses described in section A may be represented as shown below in Fig. 2-1, upon making the addltional assumptions a) that the members of a gene pair are each present in 50% of the gametes, and b) that union of gametes in fertilization is random. 2. The facts of experimentation were in this way supported by the statistical proportlOns expected - hence we speak of the statistical gene. 3. Testing the hypothetical 1:2:1 ratio can be done via a series of tosses of two coins, or by a series of WIthdrawals and replacements of two items from a large population composed. of equal numbers of be. and .!!. .. x CC all c all G G1 (gametes) (cross-fertilization) cc all Cc x P2 Cc G2 1/2C, 1/2~ Cc (self-fertilIzation of FI) 1/2Q, 1/2~ F2 Sperm 1/2Q 1/2~ 1/2Q 1/4CC 1/4Cc 1/2~ 1/4Cc 1/4cc (random fertilization) Eggs OR P3 1/4CC; t breeds like PICC when selfed 1/2Cc; t breeds like P 2 Cc when selfed Figure 2-1 22 1/4cc t breeds like PlcC when selfed E. Experimental test of segregation in the hybrid 1. A cross of Fl hybrId colored (Cc) WIth white (cc) should produce offspring of whIch half are Cc and half cc, colored and white, respectively, anci it does. 2. In all of Mendel's 7 crosses the Fl hybnd was phenotypically indistinguishable from one PI (for example, Cc looked colored), the gene received from that parent (f.l being called dominant and the one from the other PI called recessive (E.). Dominance refers to the phenotYPIC expression of genes in diploids and has no relation to the transnnssion mechanism of segregatIOn. 3 In chickens black X white produces blue-grey F l' Mating between two blue-grey F 1 produces in 'F2 1/4 black, 1/2 blue-grey, and 1/4 white. F. Corollaries of Segregation 1. A consequence of self-fertilization is to increase the proportIOn of homozygotes (pure indiVIduals, like CC and ccl and reduce the proportion of heterozygotes (hybnds, lIke Ccl. 2. Hardy-Weinberg equilIbrium principle In a large cross-breeding population the proportions of genotypes present in adults mating at randum will remam unchanged, in equilibrium, if there IS no discrimination agamst the transmission of any genotype to the next generation. 3. Item 1 IS especially important m the theory of inbreeding and in plant breeding, item 2 in animal breeding. Both corollarIes are important in understandmg populabon genetics and evolution. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, makmg additions to them as desired. 2 Review the readmg assignment. 3. Be able to discuss or define orally or in wnting the items underlined in the lecture notes. 4. Complete any additional assignment. 23 QUESTIONS FOR DISCUSSION 2. 1. What advantages and disadvantages have plants over animals as material for the study of heredity? 2. 2. Mendel criticized all previous work on plant hybridization as having failed "to make it possible to determine the number of different forms under which the offspring of hybrids appear, or to arrange these forms with certainty accorchng to their separate generations, or definitely to ascertain their statistical relations". In what ways did Mendel assure his experiments were satisfactory in these respects? 2. 3. Do you think that the principle of segregatIon means essentially that the transmission of heredity is by particles? Explain. 2. 4. Does Mendelian segregation take place in asexual reproduction? Does segregation occur in homozygotes? Explain. 2. 5. List.all the assumption:::; which were involved in Mendel's symbolic representation of the facts observed in the splitting of hybrids. 2. 6. Describe how the process of controlled cross-fertilization is performed in peas. What precautions need be talc en against self-fertilization, or foreign cross-pollination? 2. 7. If one of the 7 pairs of alternative traits studied by Mendel showed no dominance in the Fl hybrid in what way would this have affected Mendel's thinking concerning segregatIOn? 2. 8. AI though these terms were unknown to Mendel in what respect did his experiments involve a) norms of reaction, b) pure lines, and c) emphasize genotypes rather than phenotypes. 2. 9. What evidence is there that genetic factors occur in the body cells as well as in the gametes? 2. 10. Do you think Mendel's study of the mode of inheritance of 7 different traits was a large 24 enough sample of all traits to be the basis of the general proposition of segregation? Explain. 2.11. Give an example of another biological p'rinciple which has been found to apply equally well to animals .and plants. 2. 12. Suppose you have two bags each containing 60 black and 60 white marbles. Without looking, one marble is withdrawn from each bag to make a pair, this procedure continuing untIl all 120 pairs are obtained. What is the ide~ expectation of the number of pairs containing 2 blacks, 2 wh~tes, and 1 black and 1 white? What would you expect the results to be in actual practice? What, if anything, in this model could represent parents, gametes, genes, segregation, fertilization, hybrid, pure, heterozygotes, homozygotes, genotype, and phenotype? 2.13. How do we know that one gene does not chalig6 the character of its partner in the diploid state? 2.14. If you had only 1 pea plant how could you determine whether it was a heterozygote or homozygote? 2.15. Explain how it can be that individuals which look very much alIke breed very chfferently. 2.16. Suppose from a single pair of flies 78 longwinged and 22 short-winged Fl are obtained. What can you hypothesize concerning the heritability of wmg length? ' Represent the genotypes and phenotypes of the parents and offspring, so far as is possible. What is the phenotypIC relationship between the two alleles involved? 2. 17. Is a test cross necessary to determine genotypes in cases of no donunance? Why? 2. 18. What is the phenotypic ratio for cases of dominance and of no dominance, respectIvely, for genotypic ratios of 1:2:1 and 1:1? By what cross can the latter ratio be obtained? 2. 19. If the gene 1:. for tall is dominant to its al- lele !. for short, and if .I and !. should have visible effects on the gametes within which they are supposed to occur (as happens in some cases) what should you expect to observe in-gametes of TT, Tt, and tt parents, respectively? 2.20. StudY,the diagram (after ntis) and then select appropriate genetic symbols and show the genotypes of parents, Fl, and F2 individuals. PARENTS themselves? Note: In summer squashes the gene for white fruit color (!Y) is completely dominant over the gene for yellow (~. 2.25. What fruit colors will appear in the offspring of the following crosses? Ww X ww WW X Ww ww X WW Ww X Ww 2.26. A white-fruited squash plant crossed to a yellow-fruited one produces 12 white and 10 yellow offspring. What are the genotypes of the parents? If the white-fruited parent had been selffertilized what would be the expected fruit color of the offspring? P, Colored ilowers P, and seed coats uncolored flowers and seed coats 2.27. Two white-fruited squash plants when crossed produce 1 yellow offspring. What are the genotypes of the parents? What would each have produced in F1 if crossed with a yellow, fruited plant? F. GENERATION 2. 28. If in a species like peas which reproduces by self-fertilization, a heterozygote Aa and all its progeny are equally viable and fertile, what will be the proportions of the genotypes in the progeny after 4 generations of self-fertilization? What would you predict about this population after 10 generations of self-fertilization? , , 705 plants colored (3151 3 F2 GENERATION 224 plants whlte(I), I 2.21. In short horn cattle red X white produces roan in Fl. Matings between F1 roans produce in F2 approximately 1/4 red, 1/2 roan, 1/4 white. How can you explam these results? 2. 22. Which do you think would be easier to handle in breeding, a trait which shows complete dominance or one which does not? Why? 2.23. Under what specific circumstances may phenotypes represent genotypes? 2.24. Why is the F1 between two homozygous parents typlCally as uniform as the parents 2. 29. What consequence does the Hardy-Weinberg equilibrium have in populations where it applies? Chapter 3 MITOSIS lecturer--R. E. CLELAND PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings from fextbooks: Altenburg; Chap. 1, pp. 1-5. Colin; Chap. 5, pp. 63-69. Dodson: Chap. 3, pp. 20-27. Goldschmidt: Chap. 2, pp . 28-41. Sinnott, Dunn, and Dobzhansky: Chap. 1, pp. 13-16. Snyder and David: Chap. 4, pp. 35-42. Srb and Owen: Chap. 7, pp. 103-115. Stern: Chap. 2. Winchester: Chap. 3, pp. 35-45. LECTURE NOTES A. Mendel knew from his studies of transmission genetics that 1. an individual, reproduced sexually, possesses two sets of genes, one received from the father, and another, homologous, one from the mother; 2. the two sets of genes contained in the fertilized egg become characteristic of the individual as a whole; 3. when such an individual itself forms sex cells, the members of gene pairs separate during gametogenesis. B. Mendel did not know the mechanism whereby these results were achieved, for chromosomes were not discovered until about 10 years after his work was published. C. ReI ation between behavior of genes and of chromosomes 1. We know now the former is a reflection of the " latter. 2. The rules for the distribution of genes to different cells are the consequence of the distribution of chromosomes to these ceUs. 3. Therefore, an understanding of genetical phenomena requires a familiarity with the essentials of chromosome behavior. 26 D. The chromosomal content of ~uclei 1. Gametes. Mature eggs or sperms each contain one complete set of chromosomes, "the number in a set being characteristic for the species. In peas the chromosome number per set (!!) is 7, and in man g is 23. 2. Zygote. The fertilized egg has 2g or 2 complete sets of chromosomes, one set received from each parent. 3. After fertilization, in most cases, the individual develops into a multicellular organism, in which as a rule each cell has 2n chromosomes (g paternally derived and g maternally derived). This is especially true for the cells ultimately giving rise to gametes. E. Each cell of the adult body typically has the 2g number because as the cells have divided, so have the chromosomes. This chromosome constancy is accomplished by mitosis, the process by which the nucleus divides. F. Early cytologists suspected that the chromosomes were intimately associated with the determination of hereditary traits. They noted that when a cell divides 1 .. no special mechanism assures that the cytoplasmic constituents are divided equally among daughter cells, while, 2. in contrast, the chromosomes are very precisely and accurately divided and separated. By 1900, cytologists had worked out the details of chromosome behavior throughout the life cycle and were formulating the theoretical rules which govern heredity, on the assumption that the hereditary determiners were chromosome-borne. Rediscovery of Mendel's work showed that the rules of heredity being postulated by cytolOgists had long before been proved by Mendel's breeding experiments. G. The nucleus of a non-dividing cell is metabolic, not "resting". As seen in killed and stained preparations it contains a network of fine threads with which one or more nucleoli are associated (Fig. 3-1). Figure 3-3 Figure 3-1 H. Prophase. Mitosis begins with prophase, the stage in which the chromosomes first become clearly visible and then are transformed into thick rods while the nucleoli and nuclear membrane disappear. 1. Early prophase (Fig. 3-2). Each chromosome consists of two delicate threads irregularly coiled about each other. These are called half chromosomes or chromatids. Figure 3- 4 --~ Figure 3-2 2. Mid prophase (Fig. 3-3). The chromatids become shorter and thicker and untwist from each other. 3. Late prophase (Fig. 3-4). The chromatids become thick rods while the nuclear membrane and nucleoli disappear. The latter may contribute some of the material involved in the thickening of the chromosomes. Figure 3-5 27 1. Metaphase (Fig. 3-5). Just before the next main stage, the chromosomes, quiescent while the nuclear membrane remained intact, suddenly begin to move about. This movement is centered not in the whole chromosome but in the centromere. At the same time, the medium in which the chromosomes lie becomes reorganized into a spindle. The developed spindle helps move the chromosomes so that their centromeres come to lie in a plane across the middle of the polar axis of the· . spindle. This arrangement is the metaphase of mitosis. J .. Anaphase (Fig. 3-6). Until now the two chromatids of a chromosome are still attached to each other at or near the centromere, although elsewhere they are largely free. They next separate at the centromere also, forming two centromeres which suddenly move apart, one going toward one pole of the spindle, the other toward the other pole, while the rest of the chromatids are dragged passively. The stage during which the chromatids separate and move toward the poles is called anaphase. reverse of those observed in prophase - a new nuclear membrane and new nucleoli are formed. Material is drained away from the chromosomes and each chromosome then appears to consist of two delicate threads wound about one another (as in early prophase), so that each 'chromosome is. already in duplicate in preparation for the next division. Finally, the chromosomes achieve the appearance they have in the metabolic condition. • Telophas(!l is the stage of mitosis which begins with nuclear membrane formation and ends with the chromosomes losing their visible identity. Figure 3-7 M. Cytoplasmic or cell division (cytokinesis) usually '" Figure 3-6 K. At each pole there will be as many chromatids as there were whole chromosomes in the parent nucleus. 'What has happened is that each original chromosome had previously become longitudinally double (composed of two chromatids), and the halves were separated to opposite poles of the cell. . In this way the chromosomal material has been precisely and completely duplicated and divided, so that both qualitatively and quantitatively the two groups of chromosomes at the poles are identical in every way. L. Telophase (Fig. 3-7). Once the chromosomes are at the poles, the events which follow are the 28 follows mitotic division. The cytoplasm is divided in higher plants via cell plate formation and in animals via furrowing (the former process starts the separation from the inside of the cell and continues towards the perifery; the latter process works in the reverse direction). In no case is there a mechanism to insure exact division of the extra-nuclear cell components. N. Consequences of mitosis 1. So far as chromosomal content is concerned, the two nuclei formed by mitosis are identical with each other and with the nucleus from which they arose. 2. It follows then that the fertilized egg and all the cells produced from it mitotically are also identical in this respect, all such cells having two complete sets of chromosomes, one derived from the mother, one from the father. Mendel discovered that the genes in an organism were present in double dose. It is understandable why this should be so once it is recognized that the genes are located within the chromosomes. 3. Yet if every cell has two full sets of chromo- somes and of genes, how does it come about that the sperm and egg each have only one set? This is explained by the process of meiosis (Chap. 4 ). O. TIle work of Dr. Bajer, using the phase contrast microscope and time lapse photography, showing mitosis in living cells was demonstrated. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. ·Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 29 QUESTIONS FOR DISCUSSION 3. 1. The genes contained in the fertilized egg determine the characteristics of the organism as a whole. What is your opinion of this statement as a basic principle of transmission genetics? 3. 7. What might be expected to happen to a chromosome 1. that lost its centromere? 2. that contained two centromeres ? 3. in which one arm is extremely long? 3. 2. 'Which stage would be rarest among a group of cells killed while undergoing mitosis? How could prophase be distinguished from telophase? 3. 8. "What mig~lt be expected to happen chromosomally and genetically to an Ul....fertilized egg that undergoes mitosis, but fails to undergo cytokinesis? 3. 3. Name the process illustrated below. Start with the stage of longest duration and fill in the fonowing chart. I I Iflj. A B 3.10. Describe briefly various reasons for the failure of a cell division to produce daughter nuclei which are chromosomally' qualitatively and quantitatively identical to each other and to the parent nucleus. 3.11. Can you recall seeing il\ Bajer's study of . mitosis a feature of chromosomal behavior neither mentioned previously in the lecture nor described in the lecture notes? E G Letter c ::~~r~,~t o Stage H Characteristic features 1. 2. 3. 4. 5. 6. 7. 8. 3. 4. Design an experiment to test the idea that the chromosomes maintain their individuality during the metabolic stage. 3. 5. Would Mendel's principle of segregation hold true were there no such phenomenon as mitosis? Explain. 3. 6. Do mitotic divisions occur in the human germ line? Explain. 30 3. 9. What might be expected to follow if a cell formed a tripolar spindle (one with three poles) ~ 3.12. In what respects do you think that the diagrams (of the mitotic phases) and'the photographs (of killed and stained cells which served as their models) are or are not satisfactory , "still s II of the process of mitosis as it occurs in living cells? Chapter 4 MEIOSIS Lecturer-R. E. CLELAND PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings from textbooks: Altenburg: Chap. 1, pp. 5-13. Colin: Chap. 5 ., pp. 69-81. Dodson: Chap. 3, pp. 27-33. Goldschmidt: Chap. 2, pp. 42-54. Sinnott, Dunn, and Dobzhansky: Chap. 4, pp. 45-58; Chap. 6, p. 82. Snyder and David: Chap. 4, pp. 43-48. Srb and Owen: Chap. 7, pp. 116-123. Stern: Chap. 4. Winchester: Chap. 4, pp. 46-62. LECTURE NOTES A. The problem How is it that sperm and eggs have only a single set of chromosomes and of genes? For were they produced by mitosis they, like other body cells, would contain two sets. Reduction from two sets to one is brought about by a peculiar type of mitosis, called meiosis, which actually requires two successive nuclear divisions to accomplish this result. B. Meiosis is characteristic of sexually reproducing organisms, and always occurs sometime in their life history. 1. In most animals meiosis comprises the last two nuclear and cell divisions before the maturation of sperm or egg. 2. In different plants it occurs at different points in the life history, but almost never just before the formation of gametes. C. Chromosome behavior in meiosis and mitosis 1. In mitosis each chromosome, of the two sets present, behaves independently of all others, and its halves separate independently of all others in anaphase. 2. In meiosis each chromosome seeks out and pairs with its corresponding (homologous) chromosome in the other set. This occurs early in prophase of tlle first of the two divisions and whatever these chromosomes do throughout the rest of this prophase and metaphase they do as a pair. D. Each of the two meiotic divisions has prophase, metaphase, anaphase, and telophase stages. E. Prophase I Prophase of the first meiotic division is of long duration, as compared with mitotic prophase, and is divided into several SUb-stages. 1. Leptonema - thin thread (Fig. 4-1). The chromosomes as they emerge from the metabolic stage are long and thin, more so than in earliest prophase of mitosis. t . • r. I ., .. ...... ~'.,;._ .' -~,Ji,. -,,<,> . Figure 4-1 2. Zygonema - joining thread (Fig. 4-2). Corresponding (homologous) chromosomes pair with each other. The pairing process, known as synapsis, is very exact, being not merely between corresponding chromosomes but between corresponding individual points. Synapsis proceeds zipper-wise until the two homologs are completely apposed. 3. Pachynema - thick thread (Fig. 4-3). 31 The paired threads are so intimately associated it is difficul t to see that they are composed of two chromosomes. d. The paired chromosomes do not separate from each other completely since they appear attached to each other at various points along their length; each such point '/ is called a chiasma (cross; plural form is chiasmata). A chiasma represents a place where two chromatids associated on one side of a point of contact appear to sepa::: rate and become associated with different-~~ chromatids on the other side of the contact point (see Figs. 4-4, 4-5). Figure 4-2 Figure 4-4 Figure 4-3 4. Diplonema - double thread (Fig. 4-4). a. The threads separate from each other here and there. b. It then can be seen clearly that each pair of synapsed chromosomes contains four threads because the paired chromosomes had themsel yes doubled. It is unknown just how much earlier the duplication first seen in diplonema actually took place. c. A pair of synapsed chromosomes is known as a bivalent if one thinks in terms of chromosomes, but is called a tetrad if one is thinking of chromatids or half chromosomes. 32 Figure 4-5 A chiasma is thought to involve an actual exchange of corresponding chromatid segments between homologous chromosomes. The chiasma is important since it is the physical basis for the genetic phenomenon of crossingover (discussed in detail in Chap. 16). e. By the end of diplonema the chromosomes have become shorter and thicker than they ever become in mitosis. f. Especially in the developing oocytes of some aJ).imals, there now follows a diffuse or growth stage, in which the nucleus takes on the appearance of that in a non-dividing cell. This stage, as in humans, may last for decades until the oocytes are to mature and be ovul ated. :). Diakinesis (Fig. 4 - 6) a. is reached when the bivalents contract (or, if there was a diffuse stage, recontract) maximally, each lying separately in the nucleus; b. ends prophase I when the nucleoli and nuclear membrane disappears . G. Anaphase I (Fig. 4-8). The centromeres of the bivalent separate and move toward the poles as univalents, each composed of two chromatids (making now a diad). This results in the segregation of maternal centromeres from paternal. However, each diad usually contains a chromatid composed of segments of both maternal and paternal origin, because of crossingover associated with a chiasma. • Figure 4-8 H. In telophase I the daughter nuclei are formed. Figure 4-6 F. Metaphase I (Fig. 4-7). Movement of chromosomes to the mid-spindle takes place, as in mitosis, except that they move as bivalents, made up of a tetrad of chromatids still connected by some chiasmata. This is followed by interphase, (Fig. 4-9), (a term used to describe the period between successive nuclear divisions), whose length varies with rlifferent orgaJ'lisms. Prophase II is as expected from mitosis. " Figure 4-9 I. In metaphase II (Fig. 4-10) diads line up mid- Figure 4-7 spindle, and in anaphase II the members of a diad separate and proceed to the poles as single chromatids (or monads). 33 Meiosis produces nuclei containing one set of chromosomes, whose maternal or paternal composition is a matter of chance because the way bivalents line up at metaphase I relative to each other is a matter of chance (see also Chap. 6). Thus, meiosis provides the physical basis for Mendel's principle of independent segregation (Chap. 6). " Figure 4-10 Telophase II (Fig. 4-11) shows the formation of the nuclei in the daughter cells. MEIOSIS --/ \...~/ MITOSIS ~ / I \ "- /I\~ I' ''' ... """, ~\I/ \/"... ." v _8lB _~ '/ Figure 4-12 Figure 4-11 J. Comparison of meiosis and mitosis 1. with regard to chromosome number (Fig. 412). Mitosis involves duplication and separation in regul ar al ternation. Meiosis involves only one duplication by the chromosomes but two separations during two nuclear divisions. Result is maintenance of two sets of chromosomes after mitosis, and reduction from two sets to one in meiosis. Meiosis, then, provides the physical basis for Mendel's principle of segregation. 2. with regard to the maternal-paternal composition of chromosomes (ignoring chiasmata for simplicity) (Fig. 4-13). Mitosis results in daughter nuclei each with the same chromosomes as the parent cell. 34 ~\JO aa bb ~ bb 7-r m, ~" aa Be. Figure 4-13 K. Therefore, both of Mendel's basic principles of transmission genetics may be explained as the resul t of chromosome behavior and distribution in meiosis. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possibl e, making additions to them as desired. 2 . Review the reading assignment. 3 .. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 35 QUESTIONS FOR DISCUSSION 4. 1. Examination of the female reproductive tract in Ascaris revealed the stages shown below. List the correct order of these stages. In 250 words or less describe the events in process. 2 4 3 5 Assuming normal development, what would be seen at metaphase I? 4. 6. Number in order the diagrams below showing megasporogenesis and development of th,e female gametophyte in an angiosperm. "What diagram shows the last meiotic cUvision? 6 4. 2. What processes are shown in the diagrams below? Which ki ngdom is involved? Why? Considering diagram 5 . how ruany arrangements are possible in diagram 6? From what is shown in 6 in what respect is 7 inaccurate? 6 7 4. 3. Would you consider the evidence so far presented as proof that the genes are in the chromosomes? Justify your opinion. 4. 4. What number of sperm and eggs will be produced following ~ male and ~ female cells undergoing meiosis in humans? 4. 5. What would be found at mitotic metaphase of a zygote produced from the union of a sperm containing a haplOid set of two chromosomes and an egg with an homologous set, plus an extra non-homologous chromosome? 36 4. 7. Could you, in theory, simplify the meiotic process so that only one nuclear division was necessary, yet obtain, per nucleus, the same chromosomal and genetical end resul ts? Explain. 4. 8. If the suffix "-tene" means thread, what word might be used instead of "diplonema"? 4. 9. What would happen to the chromosomal constitution of the nucleus if the reduction division cUd not take place? 4 . 10. If the germ cells were formed by direct division of the nucleus (with neither mitosis nor meiosis), how would this affect the character of the gametes and the process of inheritance? 4. 11. Do you think processes like mitosis or meiosis would be requirements for higher forms of life on other planets? Explain. 4.12. For each of the following pairs explain your opinion as to which came first in the course of evolution. mitosis and cytokinesis mitosis and meiosis unpaired and paired chromosomes 4.13. What do you think happens in gametogenesis in a species with an odd number of chromosomes? 4.14. What explanation can you suggest for the presence of polar bodies in the development of the animal egg? 4. 15. If a character is found to be transmitted only by the mother, in what part of the gamete is it probably transmitted? 4.16. At the time of synapsis preceding the reduction division, the homologous chromosomes align themselves in pairs, and one member of each pair passes to each of the daughter nuclei. Assume that in an animal with four pairs of chromosomes centromeres A, B, C, and D have come from the faL_lter, and A', B', 'c t, and Dr have come from the mother. In what proportion of the germ cells of this animal will all of the paternal centromeres be present together? all of the maternal ? 4.20. If one gametophyte displays characters A and B and is crossed with another which displays ~ and Q, what will the next gametophyte generation look like with respect to these two characters? 4.21. If a given character A pertains to the gametophyte and the gametophyte of one plant shows it while that of another plant shows its allele ~ and if gametes from these two gametophytes unite, what will be the appearance of tile succeeding generation of gametophytes with respect to this character? 4.22. Given 2g_ = 2, draw diagrams showing chromosome appearance in: 1. pachynema 2. diakinesis 3. metaphase I 4. mitotic metaphase 5. metaphase II 6. anaphase II 7. mitotic anaphase 4.23. Drawing maternal chromatids as broken lines and paternal ones as unbroken, illustrate a chiasma resulting from chromatid exchange (breakage followed by cross union). 4.17. Of what advantage in plant reproduction is the complicated system of accessory structures such as the calyx and corolla? 4.18. In maize there are 10 pairs of chromosomes in normal sporophyte tissues. What number would you expect to find in a. endosperm b. poll en - tube nucl eus c. embryo sac d. leaf e. root tip f. embryo of seed? 4. 19. Can you determine from those photographs included in the lecture notes which are Dr. Rhoades r photographs of maize (2!!_ = 20)? 37 Chapter 5 HUMAN TRAITS SHOWING SIMPLE MENDELIAN INHERITANCE Lecturer-L. C. PRE-LECTURE ASSIGNMENT 1. QUIckly review notes for the previous lecture. 2. Suggested readings from textbooks: a. General genetics textbooks Altenburg: Chap. 12, pp. 214-217, 218221. Colin: Chap. 2, pp 25-32; Chap. 7, pp. 108-116. Dodson: Chap. 21, pp. 249-258; Chap. 8, pp. 90-94. Goldschmidt: Chap. 3, pp. 71-76; Chap. 11, pp. 188-191. Sinnott, Dunn, and Dobzhansky: Chap. 5, pp. 59-68; Chap. 9, pp. 116-117. Snyder and David: Chap. 13, pp. 180-186. Srb and Owen: Chap. 19, pp. 405-411. Stern: Chap. 6. Winchester: Chap. 5, pp. 71-76; Chap. 12, pp. 161-166. b. Additional references Mohr, O. L. 1932. Woolly hair a dominant mutant character in man. J. Hered., 23: 345-352. Neel, J. V., and Schull, W. J. 1954. Human heredity. pp. 83-86, 89-91, 240-241. Chicago· The University of Chicago Press. LECTURE NOTES A. Human heredity 1. The genic basis of heredity applies universally to all organisms, including man. 2. Absence of experimental breeding in man necessitates the development and use of special methods for genetic studIes. B. PedIgree method uses family trees extendIng over several generations. In pedigrees, squares represent males and circles females. Filled in symbols represent persons affected by the anomaly under discussion. 1. Albimsm, lack of pigment, occurs rarely (in 38 DUNN about 1 birth per 20,000). The assumption it occurs in homozygotes for receSSIVe gene is supported by: ' I a. Both parents of albinos may be non-albino. b. SInce the albino gene IS rare, the trait should appear most frequently in progeny· sharing a common ancestor. This IS true in SwedIsh and Japanese populations where the percent of cousin marriages is less than 5% in the general population, but is 20-50% amongst parents of albinos. c. Non-albino parents of albino progeny , should be heterozygott::s, so that among 2 child families (see photo on p. 8 ) the phenotypic ratio should be non-albino: albino = 4:3. The actual numbers agree well with the numbers predicted in families of 2 or of more children. d. Albino X albino produces only albino. _.../' e. MonozygotIc (identical) twInS are either both albino or non-albino. 2. Hypotrichosis, lack of hair, is caused by a rare recessive gene when homozygous. 3. Woolly hair is caused by a rare dominant gene. Here one affected parent (heterozygote) directly produces children half of which are expected to be woolly. 4. Ataxia, lack of neuromuscular coordination, appeared in some cases In progeny of first cousins, and in other cases from unrelated parents. The assumption that the former cases were due to a recessive and the latter due to a different, dominant gene was supported by clinical tests shOWIng dIfferent symptoms for the two groups. C. Family method uses only parents and their children. 1. This method is used extensively in the study of the genetIc basis for blood group type. 2. When human red blood cells are injected into a rabbit, the rabbit's blood forms antiserum 'a - serum containing antibodies against the specific blood injected. If the specific human red cells are then placed mto this antiserum they are agglutmated (clumped). 3. MN blood groups worked out by Landsteiner and Levine in 1927. a. Two kinds of antisera were obtaIned. There are three types of people: those whose blood agglutinates (1) only in one of the two sera, (2) only m the other of the two sera, (3) in·both. Phenotypically, all people are arbItrarily designated as having, respectively, M, N, and MN blood group types. b. Distribution of phenotypes in families Parents (1) (2) (3) (4) M N M MN (5) MN (6) MN X X X X X X M N M All N All N N M MN Children MN 1/2 1/4 All 1/2 1/2 1/2 1/2 1/4 c. Genetic explanation involves one segregating pair of genes. Let M = gene for blood group type M Let m = gene for blood group type N Mating (6) must be, genotypically, Mm X Mm. Note that dominance is absent. -- -4. ABO blood group system a. Landsteiner in 1900 obtained two test antisera, called anti-A and anti-B. In this case there are 4 kinds of blood: 1) clumped in the former (blood group A), 2) in the latter (B), 3) in both (AB), and 4) in neither (0). b. Distribution of phenotypes in families Parents AB X AB (8) AB X 0 (9*) A X 0 (10*) A X 0 (11 *) B X 0 (12*) B X 0 (13) 0 X 0 *in some families (7) A 1/4 1/2 1/2 All Children AB B 1/2 1/4 1/2 0 1/2 1/2 All 1/2 All c. Genetic explanation (7) and (8) show AB phenotype is heterozygote for 1 pair of alleles, each dominant to O. (9-12) show 0 and A, and 0 and B dif- fer in a single pair, one of which IS not the same as A or B. Assume that a gene can occur in three alternative or allelic forms, rather than two as in the case of Mm alleles. Let: LA = gene for blood group A LB = gene for blood group B !._ = gene for neither A nor B blood group Genotype Phenotype LALB AB or LAI A LB i)3 B - - or LBI I1 0 Note that each person is diploid, and carries only 2 genes for this blood group type. 5. People can be described according to the different blood group systems simultaneously. With just the MN and ABO systems 12 (3X4) different phenotypes are possible. 18 (3X6) different genotypes are possible assuming the MN and ABO systems are determined by different pairs of genes. 6. The DiPorto family is large, providing examples of segregation in a number of different blood group systems. The phenotypic results observed are summarized: LA LA (14) (15) (16) (17) (18) Father B M r P Leb Lea Mother 0 M R1 R2 P Leb Lea Children 2 B, 70 9M 3 Rl; 6 R2 9P 6 Leb , 3 Lea D. Population method can reveal the distribution and frequency of different genes in human populations. 1. Mediterranean anemia or thalassemia Among Italians living in Italy or those who have emigrated, there may be an anemia of these two types: a severe form which is fatal, usually in childhood, called Cooley's Anemia or thalassemia major, and a moderate form called microcytemia or thalassemia minor. 2. Pedigree and family studies show that t. maJor children are homozygotes for a single pair of genes. Their parents both have t. minor and are heterozygotes. 3. Drs. Silvestroni and Bianco have examined more than 100,000 people in Italy and classified them for the presence of one or two doses of this gene. E. The list of human traits known now to exhIbit segregation grows every day. 39 POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, makmg additions to them as desired. 2. RevIew the reading aSSIgnment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 40 QUESTIONS FOR DISCUSSION 5. 1. As at least partial compensation for the disadvantages of humans as material for genetic study, can you describe some advantages they provide? 5. 2. To conclude that a human anomaly was hereditary and due to the presence of a rare recessive gene when homozygous what criteria could you apply? How would these criteria have to change to permit you to conclude the anomaly was due to the presence of a dominant gene? 5. 3. How c an you expl ain that the chfferent phenotypic effects of a gene in heterozygous and homozygous condition are often given names which bear no relation to each other? 5. 4. "If two people heterozygous for brown eye color have three children with brown eyes the fourth child will have to be blue-eyed. " In what respect is this statement true or false? 5. 5. In man, why are parents of homozygotes for rare recessives likely to be related? 5. 6. In marriages between two heterozygotes for albinism what is the chance that the first child is non-albino? is albmo? What are these chances for the second child born, when the first-born IS albino? is non-albino? Explain. 5. 7. In marriages as described in question 5. 6, what is the chance that of three successive chIldren all will be albino? all non-albino? two non-albino and one albino? two albino and one non-albino? 5. 8. all albino children? 5.11. In marriages as described in question 5. 6 what proportion of all 3-child families will fail to have an albino child, thereby escapmg detection when the family method of analysis is employed? 5. 12. In question 5.11 what will be the relative frequencies of non-albinos and albinos in families detected (having at least one albino child) ? 5. 13. What conclusions would you draw if a case was found in which two of three identical triplets were hairless? Why? 5.14. What conclusions could you draw if one parent had brittle nails, three children did, and one child did not? Explain. 5.15. Can one always determine from the study of pedigrees whether a gene is dominant or recessive? Explain. 5.16. Under what circumstances would you suspect that a case of albinism was due to the transmission of a dominant gene whose effects were similar to that of the recessive gene for albinism? 5.17. Give the genotypes of parents and children from marriages (1)-(6) in the lecture notes. 5.18. What genotypes are possible for the parents of a child whose blood group is MN? M? N? 5. 19. Give the genotypes of parents and children from marriages (7)-(13) in the lecture notes. In human families traits are often observed to "skip" a generation or two. How do you explain this? 5. 9. When the parents are an albino and a heterozygote for albinism what proportion of one child families will have a non-albino child? an albino child? 5.10. In marriages as described in question 5. 9 what proportion of 3-child families will have 5.20. What genotypes are possible for the parents of a child whose blood group is AB? A? B? O? 5.21. From the phenotypes given in (14)-(18) of the lecture notes decide for different blood gro~up systems whether the mother or father was heterozygous. 5.22. Justify your opinion of the statement: "Each 41 person is genetically unique". 5.23. How many phenotypes are possible in each of the following genetically-determined traits: albinism hypotrichosis woolly hair MN blood group ABO blood group Mediterranean anemia 5.24. What is the total number of different phenotypes and of different genotypes possible when all the alternatives for the traits in question 5.23 (each assumed to be determined by different pairs of genes) are considered simultaneously? Note: In the following pedigrees calculate the probability that the trait in question will appear in the offspring of the various matings indicated. Assume that these individuals have had no children and that the only indication as to their genotype is the occurrence of the trait in the pedigree. Assume further (unless there is evidence to the contrary) that individuals who have mal'ried into these families and who do not show the trait in question do not carry the recessive factors for 1t. 5.28. Completely dominant trait 1 X 5 2 X 4 ¥~~ Pfi •• 5. 25. In what respects are the followmg statements true or false? 1 a. Persons with thalassemia major or minor have parents with a common ancestor fewer generations back than have non-anenuc persons. h. A person genetically albino is more likely to have hypotrichosls than is a non-albino. c. The albino phenotype gives a person a greater likelihood than the non-albino one of having hypotrichosis or diabetes, but not woolly hair. Note: Determine for each pedigree the method of inheritance of the trait in question (whether dominant or recessive); and, as far as possible, determine for that trait the genotype of each individual m the pedigree. 5.26. 5.29. 2 3 4 5 6 Completely recessive trait 1 X 6 1 X 2 X 4 3 X 10 7 6 X 11 h\ 6t~~ff66 2 5.30. 3 4 5 6 7 8 9 10 11 Completely dominant trait 1 X 3 2 X 4 Feeble-mindedness 234 5.31. 5.27. Completely recessive trait 1X7 2X4 Short-sightedness ffi6 1 42 2 3 6X8 Chapter 6 INDEPENDENT SEGREGATION Lecturer-E. ALTENBURG PRE-LECTURE ASSIGNMENT 1. Quickly review notes of the previous lecture. 2. Suggested readings from textbooks: Altenburg: Chap. 3, pp. 40-55. Colin: Chap. 3, pp. 37-46. Dodson: Chap. 4, pp. 34-42. Goldschmidt: Chap. 4, pp. 77 -88. Sinnott, Dunn, and Dobzhansky: Chap. 6, pp. 71-82. Snyder and David: Chap. 5, pp. 51-57. Srb and Owen: Chap. 3, pp. 33-40. Stern: Chap. 4. Wmchester: Chap 6, pp. 78-81. LECTURE NOTES A. Review of Mendel's pea crosses showing segregatIOn 1. When round hybrids (Rr) were self-fertilized, the phenotypes of the offspring were in the approxImate ratio 3 round:l wrinkled. 2. Self-fertIlizatIOn of yellow hybrids (IY_) produc~'d 3 yellow:l green. B. Mendel also mated two pure lines differing SImultaneously both in seed shape and seed color (see FIg. 6-1) From the PI cross of round yellow by wrinkled green (RR YY X rr IT) all Fl were round yellow (Rr .rr), as expected. Inbreeding (self-fertllizatlOn) of the Fl dlhybrids (Rr W produced the results shown in the figure. C. How can we explam the Fl results Mendel obtamed? Assume that two pairs of genes are involved, each pair located in a different pair of chromosomes (see Fig. 6-2). Note that the gametes of the parents contain only one chromosome of each pair, and only one gene of each pair, as expected according to segregation (Mendel's first principle). D. Mendel's results in F2 can be explained after considenng the ways that the reduction divislOn may take place during gametogenesis in the MENDEL'S Ol-HYBRID CROSS cf} P WRINKLED GREEN ROUND YEllOW ~ I ROUND / YEllOW c {) r2~~0· 9 RD. YEL 300. GR. 3WR. YEL I WR. GR Actual Nos. 315 101 [08 32 Actual 9.06 2.9 3./ 0.9 RatiO Figure 6-1 EXPLANATION OF Ol-HYBRID CROSS -- PARENTS(P,) ROUND e~(Y) YELLOW~Ql (~~) ~ ~ ~ "~" Gametes ~/ WR'NKLED~(r»<® --~ GREE N ~r7li"t\ (rr!l) ~ ~~,~ @ FI (lLY) HYBRID r ~ ~ ROUND YELLOW Figure 6-2 43 THE REDUCTION DIVISION IN THE F,HYBRID __ ~ RY(rourdyello~ ~ y ~ -~ ry(wnnkled gr) OR " __ ~ Ry(round green) ~ -8 rY(wnnKled ~eI) GAMETES (re-arranged) RY, Ry,rY.ry y Figure 6-3 (EGGS) } OLD COMBINATIONS R~ r Y r I NEW COMBINATIONS J(RFCOMBINATIONS) R Y R ~ R Y roo ~el. R Y R Y roo yel. roo R Y R ~ r RY RY '(j) ....J ..r._ Y _[_ lL ~el. roo yel. roo gr. ~ ~rY !::!2 roo roo ~el. Y J:_ JL ucd R~ If-g-ff-g-RY er: ~ ~el. R Y roo gr. RY.RlLL..Y..r.lL r y r y r y rY roo ~el. rD. ~el. wr: yel. wr. yel. R.y.B_lt_ 1: Y r Y dihybrid. Fig. 6-3 shows the two ways that the r Ij r y r 'j r ':J two pairs of chromosomes can be arranged on r~ roo ~el. roo gr. wr. yel. wr. gr.· the spindle with reference to each other (see also Chap. 4, on meiosis). When the upper alignment occurs the gametes formed after reduction divIsion contain genes in the old combinations, 1. e., m the combinations present in the gametes which formed the Fl' The lower Figure 6-4 alignment produces ga:r::lete" contaming the genes in new combinations. On the assumption G. To get the 9:3:3:1 ratio we have used two printhat there is equal chance for these two alignciples: segregation for each gene pair, and indements to occur in different cells undergoing rependent segregation for different gene pairs. The ductIOn, the dihybrid will produce all four expected ratio could have been more simply obclasses of gametes in equal proportions. ThIS tained as shown in Fig. 6-5. means that at the reductlOn division the yellowgreen pair may be turned in either the one direction or the other one with respect to the round-wrinkled pair. Alignment is followed by nuclear division resul ting in segregation. Accordingly, one pair of genes has no influence on which way the other pair segregates. E. Independent segregation of genes (or of chromo3: I RATIOS somes) may be defined as the separatIOn of the members of one pair of genes (or chromosomes) ROUND without reference to the separation of any other pair. F. Assuming independent segregation has occurred, .the offspring expected from self-fertilizatIOn of the dihybrid may now be derived by random ferYELLOW ~ GREEN tilization of gametes as shown in Fig. 6-4. Note in F2 that 12/16 are expected to be round and 4/16 green. And when mdividuals are classified phenotypically both as to shape and color of RD. YEL RD. GR. seed simultaneously the 9:3:3:1 ratIO IS preFigure 6-5 dicted. 9 Ro. YEL. : -3 Ro.GR : 3 WR YEL: IWR. GR. COMBINING TWO INDEPENDENT %. %. W6 44 /\ t 1 1(6 H. Mendel postulated that his Fl dihybrid formed four types of gametes in equal proportions because each pair of genes segregated independently of other pairs. He made this analysis without knowing the relationship between genes and chromoso_mes. Independent Segregation is Mendel's second principle. Segregation and independent segregation are basic to an understanding of transmission genetlCs. 1. Independent segregation can be tested by making a back cross with a dihybrid (Fig. 6-6). Mendel TEST CROSS (OR BACK CROSS) FI HYBRID (~ ~) X WRINKLED GREENU ;) Gamet es R Y R ~ r Y r y y r -~ _r_1:_ - -~ -Iir - -R r ~ r ~ r y ~ Rn YEL. Ro. GR. WR.YEl. WR.GR. Figure 6-6 crossed the round yellow dihybrid to the double recessive plant (wrinkled green). Since the wrinkled green parent produces only IT gametes, this cross should enable one to determine di~ectly from the phenotypes of the offspring that the dihybrid formed four classes of gametes in equal proportions, because the phenotypes of the offspring will correspond to the genotypes received from the gametes of the hybrid parent. And, in fact, Mendel found a 1:1:1:1 ratio for the four phenotypes in the offspring. J. Since a chromosome contains hundreds or thousands of genes, sometimes an individual is dihybrid for two pairs of genes in the same pair of chromosomes (see Fig. 6-7). When such a hybrid forms gametes each chromosome tends to pass as a whole when it goes into a gamete. Thus, there is a tendency for genes in the same chromosome to go together from parents to offspring. This is referred to as linkage. In cases of linkage, therefore, new gene combinations are less frequent than old gene combinations among the gametes produced by a dihybrid (linkage is discussed in more detail in Chap. 15). LINKAGE Figure 6-7 K. AI though Mendel studied the breeding behavior of many dihybrids, in each case he observed independent segregation between the different gene palrs. It was learned later that each of the gene pairs Mendel studied was in a different palr of chromosomes. Since Mendel's time numerous cases of linkage have been found. L. Mendel's discovery of independent segregation was the first proof that the germ plasm was not composed of a single hereditary unit but consisted of a number of units (genes) separable in heredity. M. Discovery of genetic recombination by separable units of inheritance has contributed to our understanding of evolution in species reproducing sexually. Mutations are relatIVely rare events. Because of this the joint occurrence of two mutatIOns in the same line of descent would be extremely rare. However, by genetic recombination mutations arising in different lines of descent can enter a single line of descent. If the new combination of genes is advantageous it might produce a line which in time displ aces the old lines in competition with them. In this way genetic recombmation can speed up evolutionary processes. Therefore, when Mendel demonstrated genetic recombination he helped us understand how evolution takes place. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, maldng additions to them as desired. 2. Review the reading assignment. 3. Be able to chscuss orally or III writing the concepts presented in the lecture. 4. Complete any additional assignment. 46 QUESTIONS FOR DISCUSSION cessive alleles and if these two indiVIduals are crossed, what proportion of the F2 from this cross will resemble each parent, respectively, in appearance? Note: In thlS series of questions, unless otherwise specified, assume that different pairs of genes are in different pairs of chromosomes. 6. 1. Should Mendel's second principle be called "independent assortment" or "independent segregation"? Why? 6. 2. Does independent segregation take place in homozygotes? III asexual reproduction? Explain. 6. 3. Expl aJh the statement: "Independent segregation of different gene pairs is determined prior to the time segregatlOn actually takes place. " 6. 4. How many different ways can the chromosomes of a trihybrid be arranged on the spindle at the maturation division? 6. 5. What are the genotypic and phenotypic requirements for obtaining the phenotyplc ratio 9:3:3:1? 1:1:1:1? 6. 6. How many dlfferent genotypes produce the fo1lowmg phenotypes, m each case: round yellow wrlnkled green round green wrinkled yellow 6. 7. How many kinds of gametes, and in what proportions, would be produced by hybrids for 3, for 4, for 5, and for n pairs of genes. 6. 8. How many dlfferent Fl genotypes and phenotypes are possible from self-fertillzation of a trihybrid? a tetrahybrid? 6. 9. Describe two procedures for determining the genotype of a round yellow seeded pea. 6.10. A man of blood types AB, MN marries a woman of 0, M blood types. What is the chance this couple will have a child phenotypically like the father? the mother? A, MN? 6.11. If one individual is homozygous for four dominant factors and another for their four re- Note: In Jimson weeds, purple flower color (R) is dominant over white (g) i and spiny pods @) over smooth ~). 6.12 Make the following crosses in Jimson weeds: (1) homozygous purple, spiny WIth white, smooth (2) homozygous purple, smooth Wlth whIte, smooth. Cross the Fl of cross (1) with the Fl of cross (2). What phenotypes are expected in the offspring and in what reI ati ve amounts? 6.13. A purple, smooth Jimson weed plant crossed with a white, spmy one gives 320 purple, spiny and 312 purple, smooth. If these two types of offspring are bred together, what will their offspring be phenotypically and genotypically, and in what proportions? Note: .In cattle the polled condition (hornless) (R) is dominant over the horned (g), and the heterozygous condition of red coat (B:) and white coat (.!:) IS roan. 6.14. If a homozygous polled, white animal IS bred to a horned, red one, what will be the appearance of the Fl? of the F2? of the offspring of a cross of the F1 with the polled, white parent? Wl th the horned, red parent? 6.15. A po1led, roan bull bred to a horned, white cow produces a horned, roan daughter. If thls daughter is bred to her father what phenotypes may be expected in the offsprmg as to horns and coat color, and in what proportlOns? Note: In summer squash, white fruit (}Y) is domlnant over yellow 0.'{_)i and dlsk fruit shape (Q) is dominant over sphere shape (Q). 6.16. What are the gametes formed by the following squash plants, of which the genotypes for fruIt color and shape are given; and what will be the phenotypic ratlOs of the offspring from each cross? 47 WWdd WwDD ---WwDd ---WwDd ---Ww - -Dd WwDd ---- X X X X X X 6.20. White, disk crossed with white, sphere gives 3/8 white, disk; 3/8 white, sphere; 1/8 yellow, disk; and 1/8 yellow, sphere. wwDD - wwdd WwDD ---Wwdd wwdd WwDd _--- 6.21. White, disk crossed with yellow, sphere' gives 1/4 white, disk; 1/4 white, sphere; 1/4 yellow disk; and 1/4 yellow, sphere. 6.17. From the diagram which follows, what proportion of the F2 are genotypically different from either P1 ? How could you determine which F2 squash were genotypically like the F1 ? 6.22. White, sphere crossed with white, sphere gives roughly 3/4 white, sphere; and 1/4 yellow, sphere. 6.23. White, dIsk crossed WIth yellow, sphere gives approximately 1/2 white, disk; and 1/2 white, sphere. P, Yellow Sphere 6.24. What phenotypes, in what proportions, would you expect from crosses of F1 ~th (1) a white peloric (2) a pink peloric (3) another F1 ? White Disc r~~ !~~ WhIte, DI~c i White Disc. P, Rod,Normal RR NN White Disc F, , White, Pelonc rr nn Plnk,Normal Rr Nn 6.25. From the 4 offspring shown below, which are all from the same mating, ~ve all the possible genotypes of their parents. Wlute DIsc. I F~< ! ,\\ , i~ ! WhI~:t:S~ Pmk, Normal White DISC White DISC White Sphere \ i l Y,lId" Yellow DISC Ye\low Disc Yellow Sphere WhIte, Pelonc Pmk, Pelonc Red, Normal Note: In certain breeds of sheep both sexes bear horns, but in others horns are absent from both. In crosses between the two breeds, the horned condition is dominant in males and re,cessive in females. White fleece is dominant over black in both sexes and in all breeds. The lnhentance of two paIrS of characters In summer squashes 6.18. White, disk crossed with white, disk gives 28 white, disk plants; 9 white, sphere plants; 10 yellow, disk plants; and 3 yellow, sphere plants. 6.26. A horned, black ram bred to a hornless, white ewe has the following offspring: Of the males, 1/4 are horned, white; 1/4 horned, black; 1/4 hornless, white; and 1/4 hornless, black. Of the females, 1/2 are hornless, black; and 1/2 hornless, white. What are the genotypes of the parents? 6.19. Yellow, dIsk crossed with white, sphere gives all white, disks. 6.27. If a homozygous horned, white ram is bred to a homozygous hornless, black ewe, what Note: In the following questions the appearance of parents and offspring is stated. Determine in each case the genotypes of the parents. 48 will be the appearance of the F1 and the F2 generations as to horns and color? Note: In man assume that brown eyes @) are dominant over blue ~), and right-handedness (B) over left-handedness _(!). 6. 28. A right-handed, blue-eyed man marries a right-handed, brown-eyed woman. They have 2 children, ·one left-handed and brown-eyed, and the other right-handed and blue eyed. By a later marriage with another woman who is also right-handed and brown-eyed, this man has 9 c!hildren, all of whom are right-handed and brown-eyed. What are the genotypes of this man and his 2 wives? 6.29. A brown-eyed, right handed man marries a blue-eyed, right-handed woman. Their first child is blue-eyed and left-handed. If other children are born to this couple in what combinations and proportions will their appearance be as to these two traits? 6.30. A right-handed, blue-eyed man whose father was left-handed marries a left-handed, brown-eyed woman from a family in which all members have been brown-eyed for several generations. What offspring may be expected from this marriage as to the two traits mentioned? Note: Assume, in man, that baldness (§) is dominant over nonbaldness ~ in males, but is recessive in females. 6. 31. A brown-eyed, bald ImIll tvhose father was nonbald and blue-eyed marries a blue-eyed, nonbald woman whose father was bald and all of whose brothers were also bald. What will be the probable appearance of their children as to eye color and baldness? gene, what fraction of the population would you expect to be heterozygous for both albinism and taste-blindness? 6.33. In swine, white coat is dominant over black and the "mule-footed" condition over that with normal feet. A white, mule-footed boar, A, always produces white, mule-footed offspring, no matter to what sow he is bred. Another boar, B, however, also white and mule-footed, when bred to black sows produces about half white and half black offspring, and when bred to normal-footed sows, about half mule-footed and half normal offspring. Explain this difference between these two animals by comparing their genotypes for these two traits. Note: In the following questions, which deal with flower color and plant surface in stocks, find the genotypes of the parents. 6.34. A violet, smooth plant crossed with a white, smooth one produces offspring of which 1/16 are VIOlet, hairy; 1/16 violet, smooth; 1/16 red, hairy; 1/16 red, smooth; 1/4 cream, smooth; and 1/2 white, smooth. 6.35. A violet, hairy plant crossed with a red, smooth one produces one white, smooth plant, one cream, smooth plant, and one red, smooth plant. What are the chances vi getting a violet, hairy plant out of thIS cross? 6.36. A cream, smooth plant crossed with a white, smooth one produces offsprmg of whlCh 3/8 are violet, smooth, 1/8 red, smooth, and 1/2 white, smooth. 6.32. In a human population breeding at random an albino indIvidual appears about once in 20, 000 births. Taste-blind persons appear about 3 times per 10 births. If the genes for taste-blindness and albinism show independent assortment in the same population, what should be the frequency of taste-blmd albinos? If taste-blindness is due to a recessive 49 EXAMINA liON UNDERLINE ALL THE ALTERNATIVES WHICH ARE CORRECT. 1. a. shows that genes are uncontaminatecl by theIr neighbors. b. is the same as the principle of segregation. c. applies to any species possessing suitable contrastmg inherited characters. d. provIdes a mechanism for genetic recombination. e. is not involved in his second law of heredity. 2. both mvolve nuclear division. cytokinesis must follow. a chromosome's identity is maintained. after completion more functional cells have been produced than were present initially. e. the spindle has an important role in both cases. production of two cells by one VIa mitosis. parthenogenesis. conjugation in Paramecium. the divisions which produce gametes. runners of strawberrIes. 50 6. The number of chromosomes per cell is usually a. b. c. d. e. 7. haploid in somatic tissue. haploid in the gametes. diploid in somatIc tissue and oogonia. diploid m all cells except sperm and egg. constant because of mitosis and meiosis. Meiosis a. reqUIres two nuclear divisions for completion. b. results in haploidy from diploIdy. c. is restricted to organisms reproducing sexually. d. precedes gamete formation. e. is essential for maintaining constancy of chromosome number in present-day organisms reproducing sexually. A man of 0 blood group type marries a woman of AB blood type. a. None of their children will have the same blood type as their parents. b. They could have grandchildren of A, B, AB, or 0 types. c. This couple carries three dIfferent alleles of a gene. The chromosomes found in a sperm are a. haploid in number. b. produced by segregation. c. reduced in number because of synapsis earlier. a. all have a paternal origin. e. are uncontaminated after fertilization. Examples of asexual reproduction include a. b. c. d. e. 4. 5. Meiosis and mitosIs are sIIDllar in that a. b. c. d. 3. d. A gIVen child cannot, If it marries only once, have all four types among its children. ' e. Their children will show the results of segregation and theIl' grandchil~ren the consequences of independent segregation. Mendel's law of the "splittmg of hybrIds II 8. Joh::lnnqpn'q pxnprimpntq with hp::ln nl::lntq short, pink offspring, then, assuming two pairs of genes are involved and Mendel's experimental requirements were met, a. studied the effects of selectIOn on seed weIght. b. distinguished between genotype and phenotype. c. indicated that in normally self-fertilizing speci~s most members are homozygotes. d. proved segregatIOn takes place even in homozygotes. e. showed that pure lines gIve only non-genetic variabilIty, barring mutatIOn. 9. Assume red (RR) x white (!D produces pink @!). a. this proves no dominance for color. b. one parent is heterozygous for tall. c. short IS dominant to tall or the reverse is true. d. no dominance could exist for the alleles determining height. e. the parents are either pure or monohybrids. 13. a. is a monohybrid or heterozygote. b. will produce gametes half of which contain a. This is an example of a test or backcross. b. These results prove color has a genetic basis. c. Only pink mdividuals are essential in provmg segregatIOn. d. This shows how genes blend in the Fl' e. It would be expected that PInkS do not breed true. 10. A. c. wi,ll be phenotypically A even without defining the gene symbols m terms of each other. d. will produce offsprmg phenotypically unlike itself upon self-fertilization. e. undergoes segregation for this gene pair durIng gametogenesis. Reduction from the diploid number of chromosomes to the haploid 14. a. IS unnecessary in asexual reproduction. b. is essential to avoid doubling chromosome number each generation in sexually reproducing organisms. c. must precede mitosis. d. results in chromosomes becoming unpaired. e. has nothing to do Wlth the constancy of chromosome number among the somatic cells of any particular individual. 11. A trait is primarily the result of action of a single gene pair In a dihybrid Aa Bb reproducing sexually, a. four types of equally frequent gametes are produced. b. the gametes are pure. c. will produce a 9:3:3:1 phenotypic ratio when selfed. d. one can test segregatIOn and Independent segregation by means of a test cross. e. mayor may not look AB depending upon dominance. 15. When an animal genotypICally Aa Bb Cc is crossed with another genotypically AA bb Cc a. no trihybrid offspring are possible. b. the number of different gametes each parent produces is different. c. homo zygotes for 2 of the 3 gene pairs are pOSSIble. d. more than 8 offsprIng are needed to obtain all possible genotypes. both undergo independent segregation even e. though the second parent is a monohybrid. a. if from a cross of two different pure lines, the Flare phenotypICally uniform. b. when there is dominance of one alternative for the trait over another. c. because we always assume the simplest explanation. d. when test crosses produce offspring which are either all uniform or in a 1:1 ratio. e. when matings produce a 3:1 or a 1:2:1 phenotypic ratio. 16. 12. A plant genetICally Aa Chromosomes and genes are similar In that If tall, red x short, white produces one 51 a. both maintain their integrity mitosis after mitosis. b. their diploid number IS restored by fertilization. 17. c. both are uncontaminated by their neighbors. d. both are paired in somatic cells. e. both undergo independent segregation. Fill in blanks with words or statements which best complete the following paragraphs. The book "Chromosome Numbers in Animals", written by S.\Makino, states that the diploid number of chromosomes in the house mouse, Mus musculus, is 40. Knowing this, one can say that there are pairs of homologous chromosomes. As a result of mitosis two nuclei are produced, each containing chromosomes, qualitatively like those in the parent nucleus. At metaphase I one would find that the chromosomes have not only but that homologous ones have to form tetrads. Whereas an oogonium is , a secondary oocyte is and fertilization of an egg by a sperm restores the condition. On the assumption that homologous chromosomes are similar but not identical, one would expect to be able to prove that a study of gametogenesis provides two important facts concerning, chromosome distribution. These principles of chromosome distribution are and 18. Assume, in humans, that the gene for brown eyes @ is completely dominant to the one for blue eyes (Q). Two brown-eyed people marry and have a blue-eyed child. This couple plans to have two more children. What is the chance: a. both of the next two children WIll be blue-eyed? b. both of the next two children will be brown-eyed? c. that of the next two children one will be blue-eyed and one brown-eyed? 19. In Andalusian fowls the heterozygous condition Bb gives blue plumage, while black (BB) and white ~ feathers are shown by homozygotes, representing a case of no dominance. A blue feathered fowl is bred to birds of the following plumage colors (1) black, (2) blue, (3) white. Only one offspring is obtained from each mating. What chance has the offspring from matmg (1) to be blue feathered? What chance has the offspring from mating (2) to be blue feathered? What chance has the offspring from mating (3) to be blue feathered? What is the chance all three of the offspring will be blue feathered? 20. In summer squash white frult color is completely dominant over yellow. Two white fruited squash plants were crossed and produced a single offspring. This was yellow fruited. a. What are the genotypes of the parents and the offspring? b. What is the theoretical genotypic ratio for such a cross? c. What is the theoretical phenotypic ratio for such a cross? d. If the same mating was made many times, and from each mating two offspring were obtained, how often would both be yellow? how often woul d both be white? 21. The gene for tall CD is completely dominant to the gene for short (!), while the genes for red @) and white (!:) show no dominance, the hybrid @!:) being pink. From a cross between the dihybrid and a short, white individual give the kinds and frequencies of genotypes and phenotypes expected in the Fl' 52 22. In what respect is the series of diagrams below incorrect? How many different combinations of maternal and paternal chromosomes are possible in the haploid egg? .. ' b e . . ._... . ,\._1,.'~f· . DIagram of t ... o maturatlOn divislOns of the egg. The first polar spindle is shown III a. The separahon of the paternal and maternal chromosomes (reductIon) lS shown in b. The first polar body has been given off III c. The second polar spindle is formed ln d; each chromosome has spltt lengthwise mto daughter halves (equatIOnal diVIsIOn). The second polar body IS belllg gn'en off in e. The egg· nucleus is left m f WIth the half (haploid) number of chromo· somes. 23. 24. Assuming the facts in columns (1) and (2) as to the blood character of mother and child, draw circles around the correct alternatives for the father in columns (3) and (4). (1) (2) Mother Child 0 0 A M N AB M MN (3) Father (4) Could not be Could be A;B;AB;O , A;B;AB;O A;B;AB;O A;B;AB;O M;MN;N M;MN;N M; MN; N M; MN; N In man, woolly hair ('!!) is dominant over non-woolly ® and normal pigmentation ® is dominant over albino~. These two pairs of genes are independently segregated. A woman of genotype Ww aa marries a man of genotype ww Aa. Answer the following questions. 1. TIle phenotype of the woman is: 2. The phenotype of the man is: 3. The woman produces eggs of the following gene combinations and relative frequencies: 4. The man produces sperm of the following gene combinations and relative frequencies: 5. Make a checkerboard showing the genotypes and their proportions of the offspring producible by this man and woman. 25. Give the different types of gametes and their relative frequencies produced by an individual of the genotype Aa BB Cc dd Ee. 53 26. How many sequences are possible in a family of 5 children in which four children show dominant trait -A and one child shows the recessive trait -a? 27. In the cross of Aa Bb by Aa bb, what is the chance that three and only three children in a family of four children will be aa bb? 28. Suppose there are only two kinds of chromosomes in an organism and that they look like this: 1st kind: 2nd kind: For this organism, show by diagram the kinds, numbers, and arrangements of chromosomes found in each of the following: a. Metaphase of the first cleavage division: b. In the primary spermatocyte, at the stage when synapsis is in progress: c. In the primary spermatocyte, during anaphase: d. In the secondary s];:erms.tocyte , during anaphase: 29. 30. Two parents are heterozygous for 4 pairs of alleles in different chromosomes. Aa Bb Cc Dd. What proportion of their offspring would be AA bb CC DD? Each parent is A mother belonging to blood group B has a child of blood group O. The The The The genotype of the mother is _ _ _ __ genotype of the child is ----three possible genotypes of the father are _ _ _ _ _ _ _ _ _~ three genotypes which the father could not have are _ _ _ __ 31. In a cross of aa Bb Cc by Aa Bb cc, what proportion of the families with three children will have two, and only two, of genotype Aa BB cc? 32. Define Hrange of reaction" as the term is used in genetics and give an example of it, maldng clear how the example illustrates the term. 54 Chapter 7 EXPRESSION AND INTERACTION OF GENES lecturer-JACK SCHULTZ PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 3, pp. 55-63. Colin: Chap. 6, pp. 83-86, 88-95. Dodson: Chap. 9, pp. 97-103. Goldschmidt: Chap. 9, pp. 159-170. Sinnott, Dunn, and Dobzhansky: Chap. 7, pp. 85-96. Snyder and David: Chap. 6, pp. 61-73. Srb and Owen: Chap. 4, pp. 54-62. Stern: Chap. 3; Chap. 16. Winchester: Chap. 6, pp. 81-87. b. Additional references Bateson, W. 1909. Mendel's principles of heredity. 413 pp. Cambridge: Ca.mbridge University Press. Bridges, C. B., and Brehme, K. S. 1944. The mutants of DrOsophila melanogaster. 257 pp. Washington, D. C. : Carnegie Institution of Washington. which presumed that the hereditary materials blended with each other and were not identity-retaining particles. Figure 7-1 . L EC TURE NOTES A. The particulate theory of inheritance l. Fig. 7-1, showing the genotypic results of a dihybrid x dihybrid cross, condenses Mendel's contributions to transmission genetics: a. the concept of self-reproducing, untaintable, hereditary particles, b. segregation, and c. independent segregation (also called independent assortment). 2. This theory can be tested by mathematical methods. 3, Not everyone accepted this theory early in the 1900s. Karl Pearson believed in a law of ancestral heredity, first proposed by Galton, 4. William Bateson studied transmission gene- B. tics in a wide variety of animals and plants. The results he obtained, supporting Mendel's principles, are summarized in his book, whose introduction is quoted. Genic interaction and phenotypes 1. Mendel studied independent segregation of gene pairs affecting different organs, so that absence of an effect of one gene pair on the phenotype produced by another gene pair gave unambiguous results. Thus, the mathematical relationships between genotypes, as in the progeny in Fig. 7-1, could be easily determined from the phenotypes or by further breeding tests. 2. But soon cases were found where the pheno55 c. D. typic resul ts did not appear, at first, to conform to the genotypic expectation. 3. This was due to different gene pairs affecting the same character. 4. It was Bateson and his associates who showed that these cases of genic interaction also followed Mendelian rules. Such cases provided information relative to gene action --the relationship of genes to physiology. 5. Interaction between pairs of genes in Drosophila is discussed. Drosophila as material for studies of genic interaction 1. T. H. Morgan began work with Drosophila about the time Bateson's book was published. 2. He looked for new hereditary variants (mutants) of this fly. 3. Drosophila is a good organism for this work because it a. is easy and inexpensive to raise, and furnishes large numbers of progeny, and b. has many easily observed morphological characteristics. 4. Though mutants are rare, Morgan found many. 5. Morgan's student, Calvin E. Bridges a. found more Drosophila mutants than any other worker, b. started a compil ation of all Drosophil a mutants which was completed and published after his death by Brehme Warren, c. made a worksheet of the external morphological traits of the normal fly. 6. Not all phenotypic variants from normal (wild-type) are heritable. Only those proven to be mutants are discussed. New ones are still being found. Mutants affecting wings of Drosophila (Fig. c) :. "RED. WING" Figure 7-2 2. A "Maxy" male, although genetically con- miniature, vestigial). 56 bcnv8 Gla/Cy rb ex 2. Other mutants affect wings by causing them E. min ~ r 7-2) 1. Some mutants affect wing size (reduced, to curve, e. g., to tilt up (Curly, curlex, wavy, Turned-up ). 3. Still others, like dumpY, change wing shape. 4. At least 200 non-allelic mutants affecting the wings have been found. 5. What happens in crosses involving two or more of these wing mutants? Epistasis and hyPostasis 1. Recall that dominance refers to the phenotypic interaction between alleles of a gene pair. Our present question asks about the phenotypic relationship between different pairs of genes. WILD TYP~ F. stituted to show the effects of both the miniature and the Turned-up mutants, actually shows only the miniatur:e phenotype. 3. This phenomenon of obliteration of one mutant trait by another non-allelic one was \ discovered by Bateson and co-workers. a. The miniature gene is epistatic to the Turned-up gene. b. Turned-up is hypostatic to miniature. 4. A "Maxy" male also contains eight different genes affecting eye color. Singly, each would produce a phenotypic effect. Seven of these would change the normal red eye to various shades of pink or purple. But the eighth gene is white, which removes all eye pigment. So white is epistatic to all the other seven eye color genes present, which are therefore hypostatic, and the eye is phenotypically white. 5. This phenomenon is really an example of dominance between non-allelic genes affecting the same trait. Other Drosophil a mutants affecting wings (Fig. 7-3) 1. Some affect the wing margins (cut, Notch, serrate, Xasta). 2. Others change wing venation (Confluens, crossveinless, cubitus interruptus, radius ct6 N-8 WinS ser k~ ~. ~ ~ Co.-WinS ~1 'w; ,. ( cv C3 Xa Ax WinS .r: ~r ~, Ci ri ~ r~ ".. __..... '" r I. . 2. The expression of genes affecting a given character is diverse. 3. Combinations of non-allelic genes affecting the same trait may produce phenotypic effects which a. superpose, b. show epistasis, c. are new to the series of crosses. 4. Breeding experiments showed that abnormal ratios resulted from phenotypic interactions, not from genetic exceptions to Mendel's principles. The analysis by Bateson of gene expression and interaction 1. showed that Mendel 's principles obtained even in apparently exceptional cases, 2. cleared the way for the analysis of cases of inheritance involving multiple factors (Chap. 8), Figure 7-3 G. H. incompletus, Abruptex). 3. Combinations of some of these mutants provide other examples of eipstasis and hypostasiS'. 4. However, some combinations of mutants affecting the same organ produce no phenotypic conflict, their effects being individually re cognizable' because they are superposed. Combinations producing new phenotypic classes 1. Eye color in .Drosophila a. Normal eyes contain both red and brown pigments. b. The brown mutant lacks the red pigment; the cinnabar mutant 1acks the brown pigment. c. The homozygous recessive for both cinnabar and brown has white eyes. d. This combination produces a new phenotypic class resulting from the additive subtraction of eye pigments. 2. Comb shape in domestiC fowl a. This was studied by Bateson. b. There are comb shapes like rose, pea, and single. Each breeds true. Rose and pea are dominant to single. Rose or pea x single gives a 3 :1 ratio in F2. c. Yet when rose is crossed to pea all F1 are walnut - a new comb type (seen before only in the Malay fowl). d. Because of this combination effect, crosses between walnut combed fowl give progeny with a distorted ratio. Principles of gene expression and interaction 1. Many genes can affect the same trait. 3. aided our understanding of the relationships between gene, morphology, and physiology, 4. stimulated Wbeldale, Onslow, and also Garrod to lay the basis for modern biochemical genetics (Chaps. 39, 40). POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lec- ture or as soon thereafter as possible. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. ;::'7 QUESTIONS FOR DISCUSSION 7. 1. Explain specifically in what respects Fig. 7-1 embodies Mendel's principles of transmission genetics. 7. 2. Are our present questions in genetics the same or different from those asked in Bateson's time? 7. 3. How many different phenotypes would the offspring in Fig. 7-1 show if for the two gene pairs there was no dominance for either pair and the traits they affected were entirely different? 7.11. In the following figure make a key of genes and their phenotypic effects and relationships and fill in the genotypes of all individuals shown. What proportion of F2 are genotypically like one of the PI? like the F1 ? P, r, 7. 4. "Dominance of one allele over another may depend upon factors not inherent III the gene. " In what respect is this statement true or false? 7. 5. Using Fig. 7-2, give four different genotypes which would show epistasis. 7. 6. Using Fig. 7-2, give the genotypes of two individuals which when mated would give progeny in a 9:7 ratio of normally long wings: not normally long wings. 7. 7. "Identical phenotypes may result from different genotypes, and vice versa. " In what respect is this statement true or false? 7. 8. A strain of black, solid-colored mice has bred true for 10 generations, except for one albino which appeared in the second generation. In the sixth generation a black- and White-spotted mouse is born in this strain. The breeder's explanation is that the blacks have been carrying white as a latent character and that, through long association, the two traits have affected or "contaminated" each other, resulting in a black-and-white animal. Criticize this explanation and offer an alternative one. 7. 9. The difference between dark andJ;iue eyes in man is probably determined by a single factor, but there are many shades of brown and of blue eyes. How would you explain these minor differences? 7.10. Using Fig. 7-3, give two genotypes which would show hypostasis. 58 F, W W W \l7 \l7 W W. \l7 \V 17 W 17 W \l7 17 ¢ , -The 15 1 ratIO Checkerboard showlllg the expected composItion of the F2 from & cross between a type of shepherd'a-purse (Bursa) wIth trIangular capsules (homozygous for two duplicate factors) and a type wIth top-shaped capsules (After G H Shull) 7. 12. With the help of Fig. 7-3, draw a wing expected in a fly homozygous for a. crossveinless ~ and radius incompletus <ED. b. cubitus interruptus ~ and radius incompletus (ri). . What phenotypic relationship is shown between non-allelic genes in these cases? 7.13. With the help of Fig. 7-3, draw a wing expected in a fly heterozygous for Abruptex and Notch. What phenotypic relationships are shown between the genes in this case? 7. 14. In each of two different strains of maize, plants have been found which, when selfed, produce about three-fourths normal green and one-fourth lethal white (albino) seecllings. If two such albino-producing plants, one from each strain, are crossed, the F1 is found to be all green, but certain of the F2 popula- tions are approximately nine-sixteenths green and seven-sixteenths white. Explain, giving genotypes. three-fourths are white, three-sixteenths yellow, and one-sixteenth green. Find the genotypes ofille parents when a white plant crossed with a green one produces offspring of which about one-half are white and one-half yellow. 7.18. List all the assumptions one would need to make in order to explain the inheritance of the .traits involved as shown in the following figure. 7.15. ~ P, 7. 19. Find the genotypes of the parents when a white plant crossed WIth a yellow one produces offspring of which roughly one-half are white, three-eighths yellow, and one-eighth green. ~ 10.,. ~ Pea rr PP R05e RR pp ~Walnut F, Rr Pp ~ ~_p~~~_p~ ~_rp~ ~_rp~ __ __ __ ®~ ~ ~ ~ RR PP RR Pp Rr PP Rr Pp Walnut Walnut Walnut Walnut Note: In maize, factors Q and B:_ are both necessary for the production of red aleurone color, the absence of either resulting in white aleurone. If factor R is present in addItion to Q and B:_, the aleurone is purple, but R has no effect in the absence of either C or R or both. The factor W prevents the appearance of any color in the aleurone at all, and its presence thus results in white aleurone. Its recessive allele w allows the development of color. What will be the aleurone color of the offspring of the following crosses, the genotypes of the parents being given? 7.20. ~~ ~~ RR Pp Walnut RRpp Rr?p Rrpp R05e Walnut Rose R. PP R. Pp Walnut rr PP Peo rr Pp Walnut Rr Pp Walnut Pea rA~fi\ Rr pp rr Pp rr pp Rose Pea Single. Note: In summer squashes the factor for white fruit color, W, is epistatic to that for yellow, Y; WYand Y:!:i_ plants are white, wY plants yellow, and '!!)L plants green. What is the color of the fruit in the offspring of the following crosses, the genotypes of the parents being given? Ww"X;t_ x Ww IT ww YY x Ww IT Ww IT x ww"X;t_ 7.16. Find the genotypes of the parents when a white plant crossed with another white one produces offspring of which approximately 7. 17. Ww Cc Rr E2. Ww cc Rr 22. ww Cc rr E2. ww cc Rr 22. x x x x ww cc rr 22. Ww cc Rr 22. WW cc rr E12. Ww - -Cc - -rr-PP - Note: In stocks, factor C, in the absence of factor produces cream-colored flowers; s:. produces whi te ones; .Q with B:_, red ones; .Q with B:_ and y, violet ones, but Y has no effect in the absence of either C or R or both. Factor!! causes the plant to be hairy, but it is operative only in the presence of both .Q and B:_. Its recessive allele, Q, causes a smooth condition. White-flowered and cream-flowered plants are thus always smooth, and red-flowered and violetflowered ones may be either hairy or smooth. _g, In stocks, what is the appearance of the offspring of the following crosses, the genotypes of the parents for flower color and plant surface being given? 7.21. CC Rr VV Hh Cc Rr Vv Hh Cc rr vv Hh Cc Rr Vv Hh x x x x Cc rr Vv - HH Cc Rr Vv Hh - cc RR Vv hh cc rr vv hh 59 Chapter 8 MULTIPLE FACTOR INHERITANCE Lecturer-J. F. CROW PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings from textbooks: Altenburg: Chap. 4, pp. 76-88. Colin: Chap. 6, pp. 86-89; pp. 470-475. Dodson: Chap. 10, pp. 112-118. Goldschmidt: Chap. 10, pp. 174-184. Sinnott, Dunn, and Dobzhansky: Chap. 8, pp. 99-107. Snyder and David: Chap. 14, pp. 197-208; Chap. 15, pp. 211-214. Srb and Owen: Chap. 15, pp. 304-309, 312-322. Stern: Chap. 18. Winchester: Chap. 13, pp. 173-182. LECTURE NOTES A. Discontinuous and continuous traits 1. Discontinuous traits include such phenotypes as flower or seed color in peas, albinism vs. pigmentation, or blood group types in humans. In each case an individual belongs clearly to one class or another. 2. Contmuous traits do not permIt sharp classifification of individuals as to type - as in heIght of corn ears, or height or intelligence in humans. B. Characteristics of continuous traits 1. The continuous range of phenotypes requires each individual be measured. Hence these are often called quantitative traits. 2. They are determined by a large number of genes, and are, accordingly, often called also multiple factor traits, multi-factorial or polygenic. Usually so many gene pairs are involved that the effect of anyone is difficul t to distinguish. 3. They are rather strongly influenced by the environment, which has a large effect as compared with one of the polygenes. (Ex60 amples include the environmental effect of fertilizer on corn ear size, and of diet on height in humans.) C. To understand multiple factor inheritance and its properties it is desirable to consider first simpler cases involving one or a few pairs of genes. D. Traits involving'one pair of factors showing no dominance . 1. Example is coat color in cattle. 2. The monohybrid AA', from mating red(A'A') by white (~, is roan. 3. Matings between roans produce offspring of ',,'hich 1/4 are red, 1/2 roan, and 1/4 white. E. Two pairs of factors showing no dominance (Fig. 8-1) TWO PAIRS Of FACTORS - NO DOMINANCE GRAIN COLOR IN WHEAT GENOTYPES AA BB A'A BB A'A B'B AW B'B NA' B'8' AA B'B AW BB A'A B1 81 AA B'B' COLOR WHITE LlG"T MEDIUM DARK VERY DARK COLOR GRADE 0 1 2 3 4 1 AA BB x A' A' B8' WHITE VERY DARK NA B'B MEDIUM VERY DARK DARK MEDIUM LIGHT WHITE 1/16 4/16 6ft6 4AG 1/16 Figure 8-1 F. Expected distributions from cases of no dominance involving 1, 2, 3, or many factors (Fig. 8-2) 1. In all cases the F1 are uniform and phenotypically intermediate between the two Pl. 2. F2 results, from matings between F1 (by cross- or self-fertilization), are shown when different numbers of gene pairs are involved. 3. As the number of gene pairs involved increases, a. the classes of offspring increase until the number is so great that environmental action causes loss of discrete classes. b". the fraction of all F2 resembling either PI becomes smaller. c. the narrower becomes the shape of the F2 curve, especially when, for example, 20 rather than 10 pairs of genes are involved. • • • • I • _111F2 WITH ONf PAIR OF FACTORS Fz WITH 00 mlRS OF F~TORS •• 111 •• F2WITH MANY FACTORS Figure 8-2 }. The nature of DDT resistance in Drosophila (see Fig. 8-3) 1. Two strains were developed, one genetically resistant (#1) of which more than 60% of individuals survived a treatment 'With DDT that permitted less than 1 % of the other genetically susceptible strain (#16) to survive. 2. At the same time, it was possible to identify genetlcally any particular chromosome (X, 2, or 3) with its stock, because it carried different alleles in the two stocks, that is, because marker genes were used for the dJ.fferent chromosomes. 3. From a series of crosses it was possible to obtain offspring having a number of different, and identifiable, combinations of chromosomes from the two stocks. 4. The greater the number of chromosomes from the resistant strain, the greater was the resistance (proceeding upward in the figure). This trait, then, appears to be a continuous one. 5. Flies of types 13, 14, and 15 as compared with 1 and 16 prove that the genes making 1 resistant (or 16 susceptible) are located on all three chromosome pairs tested. 6. There is no way of knowing, from these experiments, how many genes are involved, but there is a minimum of one per chromosome, and, undoubtedly, many more than this. CHROMOSOME PtRC[NT SURVIVAL X 2 3 01 5 10 20 40 60 80 ----3 -=4- ----= --=5 = --= -= 6 ---== 7 --=== 8 --1 2 =-- 9~~~----· 10ES~~--- 11 ~B~--. 12~~B13g8~- 14-B~B - 15~BB- 16 §5BB. = CHROMOSOME !=ROM CONTROL STRAIN - CHROMOSOME FROM RESISTANT STRAIN Figure 8-3 H. Unless special breeding techniques are available, as in Drosophila, it may be illfficult to distinguish, especially from F2 data, whether 2 or 3 gene pairs cause a quantltative character, and almost impossible to know just how many are invol ved whenever more than 3 are concerned. This necessitates use of statistical treatment in order to make predictions. 1. It is desirable to measure the variability of a trait. This involves calculating 1. the mean, m, or the simple arithmetic average, and 2. the variance, y_. For a group of measurements y_ is obtained by determining the difference between each measurement and the mean, squaring this difference in each case, adding all the values obtained, and dividing the total by 1 less than the number of measurements involved. 3. An example of these determinations was shown. 4. The square root of y_ is called the standard deviation. J. The effect of dominance in polygenic inheritance 1. Fig. 8-4 deals with two pairs of genes, each showing dominance and having opposite effects on a trait. a. As before the Fl are uniform and intermediate to the Pl. b. However, the F2 ratio might be, in practice, difficult to distinguish from a 1:2:1 61 ratio. c. So two pairs of genes with dominance can give much the same result as one pair of genes with no dominance. Thus, if there is dominance the number of genes involved in a polygenic trait will be underestImated. And a certain amount of dominance is the rule in genetics. d. The average of the F2 (1. 75 units) is less than the average of their parents (2 units). This is an example of regression. TWO PAIRS OF FACTORS- COMPlETE DOMINANC£ GENOTYPES A-bb A-B- P~ENOTYPE 2 UNITS AA bb 1UNIT Pl 2 aa bb A-bb 3/16 1 a aa 8- 10/16 3/16 CROSS BHWfEN TWO HtTEROZYGOTfS, EAC~ 2 UNITS PARENTS Aa bb x Aa bb OFFSPRING A- bb aa bb 3/4 1/4 AVERAGE 2 UNITS AVERAGE 1.75 UNITS Figure 8-4 2. Regression means that, as a consequence of dominance, an individual phenotypically extreme in either direction will have progeny less extreme. Loss of extreme individuals generation after generation because of regression does not make the entire population more and more homogeneous, since there is an exactly counterbalancing tendency from the average members of the population (see Fig. 8-5). ILLUSTRATION OF REGRESSION PARENTS orrSPRI~ FIgure 8-5 62 PARENT G:~ ~~ruECTEOA5 ~NTS t oUNITS A-B- aa bb SELECTION FOR AQUANTITATIVE CHARACTm aa BaaBB Aa Bb F1 K. Regression is involved in selection for a quantitative character (FIg. 8-6), for the average of the offspring is not that of the selected parents but less than this; yet somewhat larger than the original mean. By continuing selection for a number of generations the offspring approach the size of the parents. 1', MEAN OFFSPRING GENERATION MEAN or SELECTED GROUP ~ .,. MEAN Figure 8-6 L. The concept of heritability and its use in breeding 1. Heritability is the ratio of additive variance diVIded by total variance. . 2. Additive variance is the variance if all of the genes were without dominance. Total variance is obtained as described in I. 3. Hentability permits accurate prediction of traits in offspring from those of the parents. 4. Predicted offspring average = population average + (heritability x the difference between parents' average and' popul ation average). 5. Exampl e of weight in swme: population average - 100 pounds heritability = 1/3 parents' average = 130 pounds (father was 135, mother 125) therefore, 100 + (1/3 x 30) =110 pounds. 6. Although actual animal or plant breeding procedures are considerably more complex than this, all are based on this simple principle of quantitative inheritance. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture· or as soon thereafter as possible, making additlOns to them as desired. 2. Review the reading assignment. - 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 63 QUESTIONS FOR DISCUSSION 8. 1. "Genes for quantitative characters show clearcut Mendelian ratios in F2'" In what respect is thIS statement true or false? 8. 2. Why is it more difficult to .study the inheritance of such characters as size, yield, and intelligence than of color? 8. 3. How might a "qualitative" trait become a quantitative one, even though the genotypes remamed unchanged? Under what circumstance might the reverse occur? 8. 4. Even when the genes for a polygenic trait cannot be studied individually, would you expect them to undergo segregation or independent segregation? Expl ain. 8. 5. Is coat color in cattle or grain color in wheat an example of polygenic inheritance? What function did a discussion of these serve? 8. 6. Can wheat plants of medium grain color breed true? Explain. 8. 7. What proportions of F2 resemble PI when there is no dominance and the number of factors mvolved is I, 2, 3, 4, n? 8. 8. What relationships exist between the genes for DDT resistance and those for DDT susceptibility? 8. 9. Compare the variances of populations A and B. A 103 100 104 97 96 100 B 45 50 50 50 55 50 8.10. Why does dominance tend to make qualitative what would in its absence have been a quantitative trait? 8.11. Other things being equat, what happens to the variance as the number of factors invol ved in a quantitative character increases? 64 8.12. With reference to Fig. 8-4, how many genotypes produce a "I unit" phenotype? Of these, which can produce "0 unit" F1 upon self-fertilization? 8.13. Why is regression absent in cases of no dominance? 8.14. Are quantitative traits restricted to sexually reproducing organisms? What advantage does sexuality provide for quantitative characters? 8. 15. If the average I ength of corn ears in a population was 8 inches and all 10 inch ears were selected as parents for the next generation, what would the heritability be if the new ears produced were, on the average, 9 inches long? 8.16. A breeder has three squash plants, each of which bears 4-pound fruits. Plant 1, when selfed, breeds true to 4-pound fruits. So does plant 2. In plant 3 the offspring range from 3 .... to 5 pounds. Plant 1 crossed willi plant 2 gives offspring all of 4 pounds, but their offspring when inbred range from 3 to 5 pounds, and selection cannot increase this above 5 pounds. Plant 1 crossed with plant" 3 gives offspring which range from 3-1/2 to 4-1/2 pounds, and selection among their offspring can raise llie fruit weight to 6 pounds. Plant 2 crossed with plant 3 gives offspring which also range from 3-1/2 to 4-1/2 pounds, but selection among their offspring is able to raise fruit weight only to 5 pounds. Give genotypes for these three parent plants which will expl ain these resul ts. 8. 17. A breeder has a race of pI ants which has been self-fertilized for 10 generations. He has repeatedly tried to increase the flower SIze of this race by selection, but to no avail. Explain why this is so. Finally, he crosses this race With anollier which is exactly similar to it in flower size. The hybrids resemble their parents, but by selection among the offspring of the hybrids the breeder is able in a few generations to increase the flower size considerably. Explain. 8.18. Suppose that the difference between a 26- decimeter and a 10-decimeter corn plant is caused by four pairs of multiple factors and that the difference between a one-stalked and a nine-stalked corn is also due to four other pairs of cumulative multiple factors. A breeder has a nine-stalked 10-decimeter race and a one-stalked 26-decimeter race. He wants a pure race 26 decimeters high, with nine stalks. If he wants it in two generations, how many plants should he raise in the F2? By spending more time how can he get it more easily? 8.19. A breeder has a number of plants which are 14 decimeters high. He crosses some of these together, selfs their offspring, and selects among their offspring for increased tallness for several generations. His results are as follows: Two throw all 14-decimeter offspring, and selection fails to raise their height. Two others throw offspring varying from 10 to 18 decimeters, and selection among these fails to raise the height above 18 decimeters. Two others throw offspring varying from 12 to 16 deCimeters, and selection is able to raise the limit to 22 decimeters. Two others throw offspring varying from 10 to 18 decimeters, and selection is able to raise the limit to 22 decimeters. Two others throw offspring varying from 10 to 18 decimeters, and selection is able to raise the limit to 26 decimeters. Explain, by giving parents' genotypes for height, why these results obtain. 8.20. What would you consider the advantages and disadvantages of qualitative traits for sexually reproducing organisms? Chapter 9 ALLELISM, AND LETHALS Lecturer-c. STERN PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 11, pp. 203-211; Chap. 12, pp. 214-219; Chap. 17, pp. 297-298. Colin: Chap. 7, pp. 104-114; Chap. 6, pp. 97-99. Dodson: Chap. 8, pp. 88-94; Chap. 7, pp. 82-87. Goldschmidt: Chap. 11, pp. 185-191, Chap. 7, p. 1.42. Sinnott, Dunn, and Dobzhansky: Chap. 9, pp. 112-116, 119-120; Chap. 10, pp. 125126. Snyder and David: Chap. 13, pp. 174-182; Chap. 3, pp. 27-29. Srb and Owen: Chap. 12, pp. 239-241; Chap. 5, pp. 74-75. Stern: Chaps. 11, 7. Winchester: Chap. 12, pp. 159-164; Chap. 10, p. 136. b. Additional references Hadorn, E. 1955. Letalfaktoren. 338 pp. Stuttgart: Thieme Verlag. Robertson, G. G. 1942. An analysis of the development of homozygous yellow mouse embryos. J. expo Zoo!., 89: 197231. Race, R. R., and Sanger, R. 1959. Blood groups in man. 3rd Ed. 377 pp. SpringfIeld, 111.: C. C. Thomas Publ. Wiener, A. S., and Wexler, 1. B. 1958. Heredity of the blood groups. 150 pp. New York: Grune & Stratton. LECTURE NOTES A. Allelism The two partners of a gene pair are called alleles and may be alike, as in a homozygote, or different, as in a heterozygote. Many cases are known 66 where alleles can be of more than one different al ternatIve type, forming a series of mul tiple alleles. 1. ABO blood group in man (see. also Chap. 5) Indians In South America all have a blood group, those of certain tribes in North America have a and A_types. Other peoples have B and AB types, in addition. These four main types are due to three allelic genes, whIch may be symbolized lA, I B , 10, although any one person may carry, of course, only any two. It is known also that "A" blood type is really comprised of three different sub-types resulting from slightly different forms (al-' " leles) of IA called IA1, IA2, I A3 , and three alternatives of IB also exist to produce subtypes within liB" blood group. Thus, there are 3 main alleles, or (at least) 7 alleles If finer blood tests are employed. 2. Eye color in Drosophila Morgan discovered a series of multiple alleles producing a graded series of eye colors from red to white which includes: red (~, blood (wbl ), coral (w co ), apricot (wa) , buff (wbf ) , and whi te (~. 3. Coat color in rabbits As in 2, there is also a graded series of coat colors In rabbits due to multiple alleles: agouti (9, chinchilla (cch), and albino (s:_). However, another allele, c h , produces not a simple dilutIOn in color, but a change in coat color pattern called Himalayan (see also Chaps. 1 and 10). 4. Sel£sterility in Nicotiana (Fig. 9-1) A pollen grain cannot grow down a style if the S allele it carries is also present in the style. Some species have 50 or more different S alleles responsible for sel£sterility, group sterility, or group incompabIlity. 5. Wing venation in Drosophila In different wild populations the wing veins are SELF ST[RILITY PISTI L STYLt '" OVARY -.it OVARY '" OVARY Figure 9-1 normally uninterrupted, and the genes responsible may be called CI+1 , ci+2 , ci+3 , even though normally they are phenotypically indistinguishable. But when a heterozygote is made by crossing wild flies with those of a strain homozygous for the recessive allele CI, which causes the cubitus vein to be interrupted (ci = cubitus interruptus), ci+1 ci may be normal, but ci+ 3 ci may show the cubitus vein interrupted. Thus, two alleles which at first seem alike prove to be different when tested further. Such alleles are said to be Isoalleles. The greater the fllleness and the number of different tests made, the greater is the chance for demonstrating isoallelism. Besides these Wlld type isoalleles, there are also mutant isoalleles, such as those producing white eye in different strains of Drosophila. B. How genes act to produce phenotypes Study of the phenotypic effects of mul tiple alleles furnishes some clues. 1. In Drosophila different eye color alleles may produce their effects because an enzyme, involved in red pigment production, is less and less abundant or effIcient as one proceeds in the allelic series from red toward white, the white allele being completely inefficient in this respect. This suggests that intermediate alleles act in the same way as the normal one but to a lesser degree. 2. This interpretation is supported by Stern's study of bristle length in Drosophila. While the wild type, normal, fly has long bnstles due to the dominant gene bb+, there is a strain with shorter, thinner brIstles due to the recessive allele bb (bobbed). Making use of certain abnormalities in chromosome distribution it was possible to obtain a fly which carried three bb genes, in which case the bristle form produced was almost like that of wild type. C. G€nes may be lethal when homozygous 1. Different alleles, when homozygous, may affect viability to different degrees. Those which kill before their carriers can reproduce are called recessive lethal genes. 2. Color in snapdragons There are two types of adult plants - green and auria (yellow golden). Auria x auria produces seedlings of which 1/4 are green (AA), 1/2 auria ~), and 1/4 white (~. The latter die because of lack of chlorophyll, so that among the grown plants one finds 1/3 green: 2/3 auria. 3. Yellow coat color in mice CuEinot proved this first case of lethality. Yellow x yellow produces 2 yellow (11) :1 nonyellow (rr). Others later showed that from such matings 1/4 of all fertilized eggs fail to develop and abort early. 4. A horse called "Superb", imported by Japan for breeding purposes, proved to be heterozygous for a lethal gene which, when homozygous, caused colonic obstruction and death. D. Lethals may act, therefore, very early or very late in development, or at any stage in between. Sometimes lethality is produced by the coming together in offsprlllg of several non-allelic genes which were separate in the normal parents. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 67 QUESTIONS FOR DISCUSSION 9. 1. How are multiple alleles proved if anyone individual carries but a sIngle pair of them? Do multiple alleles exist in asexual organisms? Explain. 9. 2. How many different genotypes are possible when there are 5 different alleles? 9. 3. "Marriages of 0 blood type with AB blood type individuals may produce offspring of four different blood types. " In what respect is this statement true or false? 9. 4. If both parents belong to blood group AB, wh:it proportion of their children would be expected to be of such a type as to be able to give blood to their parents? 9. 5. Discuss the statement: "Genetic studies are confused by the fact that under certain conditions there are 3, and under other conditions 7, alleles for ABO blood group". 9. 6. Can you present evider,ce at,ainst t..'le view that the members of a multiple allelic series invariably represent a series of progressive degradations of the "top dominant" (wild-type) gene? 9. 7. Why is it normally impossible to have a homozygote for a selfsterility gene? 9. 8. Fill in the following chart involving selfsterility genes in Nicotiana with the expected percentages of aborted pollen tubes. Male Parent S1 S3 S1 S4 S3 S4 S1 S2 Sl S3 Female Sl S4 Parent S2 S3 S2 S4 S3 S4 9. 9. When can a mating between two selfsterile plants be completely sterile? completely fertile? partially fertile? 9.10. Discuss the statement: "One cannot ever be certain that two genes are the same". 9.11. 68 Can pure stocks of the following be pro- duced: yellow mice; walnut-combed fowls; blue Andalusian fowls? ExplaIn. 9. 12. Describe one way to test whether the genes for white eye in two different populatIOns of Drosophila are isoalleles. I 9.13. What phenotype would you expect from the presence of 4 bobbed (~ genes in an otherwise diploid fly? Expl ain. 9.14. The dominant gene for Brachyury causes a considerable shortening of the tail in the European house mouse. When such mice are crossed with Oriental house mice, hybrids with the Brachyury gene have tails of normal or nearly normal length. Suggest an explanation. 9.15. "Multiple alleles, like multiple factors, show Independent segregation. " In what respect is this statement true or false? 9.16. In mice how would you recognize a recessive trait which 'has al so a lethal effect? 9.17. What proportion of F1 would survive following self":fertilization of a plant heterozygous for 4 recessive lethals belonging to pairs each of which segregates independently? 9.18. Two plants when self-fertilized each produced some albino seedlings, but when crossed to each other produced offspring all of which were green. Explain. 9.19. How might one distinguish whether a plant was heterozygous for selfsterility or for a recessive lethal gene? 9.20. If aa is a receSSIve lethal genotype, what would be one which is a dominant lethal ? Is it easier to prove death was due to a dominant than to a recessive lethal? Why? 9.21. Invent a pedigree for "Superb" which would prove this horse carried a recessive lethal gene. 9.22. Pure line A crossed with pure line B produces F1, all of which die as embryos. Yet matings between these lines and pure line C produce fully viable Fl. What is the simplest genetic explanation? Chapter 10 PLEIOTROPISM, PENETRANCE AND EXPRESSIVITY lecturer-Co STERN PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks ,Altenburg: Chap. 1, pp. 15-16; Chap. 21, pp. 374-375. Colin: Chap. 6, pp. 95-101. Dodson: Chap. 9, pp. 97, 99-100; Chap. 21, pp. 250-252. Sinnott, Dunn, and Dobzhansky: Chap. 9, pp. 121-122; Chap. 10, pp. 129-131. Snyder and David: Chap. 28, pp. 426-428. Srb and Owen: Chap. 5, pp. 76-82; Chap. 15, p. 303. Stern: Chaps. 3, 16. Winchester: Chap. 10, p. 135; Chap. 21, pp. 290-298. b. Additional references Dobzhansky, Th., and Holz, A. M. 1943. A re-examination of manifold effects of genes in Drosophila melanogaster. , Genetics, 28: 295-303. Goldschmidt, R. B. 1938. Physiological genetics. 375 pp. New York: McGrawHill Book Co., Inc. Hadorn, E. 1956. Patterns of developmental and biochemical pleiotropy. Cold Spring Harbor Symp. on Quant. Biol., 21: 363-374. GrUneberg, H. 1938. An analysis of the "pleiotropic" effects of a new lethal mutation in the rat (Mus norvegicus). Proc. roy. Soc. Lond. B, 125: 123-144. LECTURE NOTES Pleiotropism refers to the multiple effects on the phenotype which a gene may have. 1. Yellow coat color in mice At least two effects are involved, one on coat color and one on viability. One allele is dominant for viability but recessive for color A. (non-yellow), the other is a recessive lethal but dominant for yellow. 2. Eye color and spermatheca shape in Drosophila Dobzhansky studied whether the eye color genes for red (~ and white (~ would have an effect upon a sperm storage organ in females called a spermatheca. For this organ the ratio of diameter divided by height was Significantly different in strains of red as . compared with that of white. These strains were so constructed that they differed genetically almost only for these alleles, being therefore otherwise isogenic. 3. Lethal-translucida gene and biochemical pleiotropism in Drosophila Hadorn has studied this recessive lethal, which causes pupae to become translucent and die, with regard to the chemical changes it produces in the blood fluid. In the photograph (see p.15) the white column gives the amount of a substance in the normal organism, the black one the relative amount in lethal-translucida. In the sample of results shown, some substances are unchanged in amount (peptide ill), others are more abundant in the lethal than in the normal (peptide I, peptide II, and proline), still others are less abundant (glutamine) or absent (cystine) in the lethal. 4. Himalayan rabbits ~ ~ have black coat color at the extremities and white elsewhere (see Chaps. 1 and 9). Here the mul tiple effects were proven to be due to the effect of temperature upon the action (at less than 340 C) or inaction (at more than 340 C) of an enzyme which transforms nonpigmented into pigmented material. 5. Sickle cell anemia in man a. Certain individuals may have anemia, spleen enlargement, heart failure, brain damage causing paralysis, kidney damage, 69 and .skin lesions, either singly or III any combination, so that often they die as adolescents or young adults. b. It was found that the red blood cells of these individuals may become sickle shaped instead of being circular. Such defective cells are destroyed (causing anemia) but before this they may clump and clog blood vessels in various parts of the body causIng the various other ailments mentioned in 5a. c. As shown by Pauling and co-workers, sickling of red blood cells is the result of the abnormal type of hemoglobin they contain.' d. The abnormal hemoglobin is produced in individuals homozygous for a particular gene. e. For this gene, then, we have a pedigree of causes for multIple effects. The first cause is the gene, the second is the abnormal hemoglobin it produces, the third is the Sickling that follows, the fourth is the subsequent red cell clumping and destruction WhICh, in turn, produces the multipfe effects listed in 5a. 6. It may be concluded that many genes are pI eiotropic. 7. It may be true for many genes that the multiple effects each produces are all due to a single, unitary activity on the part of the gene. This activity would be of a biochemical, perhaps enzymatic, nature which would then affect many varied chemical reactions involved in the production of chfferent, at first apparently unrelated, phenotypIc traits. However, it is not proven that all genie action is unitary from the beginning, for there are, possibly, some genes which produce more 13 than one primary effect. . R__enetrance refers to the abIlity of a gene to express itself in one way or another. A gene so expressed is said to be penetrant. Those genes studied up to now were fully penetrant, but there are others, now discussed, which are sometimes Penetrant and other times not penetrant. R_olydactyly (Fig. 10-1) refers to a rare heritable Condition in WhICh individuals may have more than five fingers or toes on a limb. 1. The top female in the figure had five fingers on each hand, but six toes on each foot. Her husband was normal. They had five children, of whom three were affected. These results would suggest fuat polydactyly is due to a single dOminant gene, p, so that the parents would be, then, mothe;-RP_, father!>p'. 70 POLYDACTYLY Figure 10-1 2. As expected on this hypothesis, an affected daughter had ~wp sons, of whom one was affected, and this affected son, in turn, had some affected children. 3. But shown on the other side of the pedigree is the first born son, who was unaffected, yethad an affected daughter. ThIS is best expI amed by the assumption that the son was R:Q but that the R was not penetrant in him. " 4. This interpretation IS supported by the penetrance observed witllln affected individuals, who may have normal fingers but extra toes, the reverse, or extra fingers on one hand only and different numbers of extra toes on the two feet. Sinae there may be penetrance in one limb and not the other in the same Individual, it is reasonable that R:Q indlviduals may sometimes occur who are not penetrant in either the hands or the feet. e. ExpresslVity refers to the kind of effect a gene produces when it is penetrant. 1. In polydactyly, expressivity may Involve one or two extra digits. Moreover, the extra digits are sometimes developed to different degrees. 2. A dominant gene is known whIch has the following phenotypic effects: blue sclera, brittle bones, and deafness. Different individuals may have one or more of these effects penetrant, those effects which are penetrant having variable expressivity. D. Factors involved in variable penetrance 1. Polydactyly in guinea PIgS has been found by S. Wright to be more frequent in litters from younger than from older mothers. 2. Abnormal abdomen in Drosophila, studied by Morgan, is more penetrant if moisture during development is abundant than if it IS scarce. 3. Gene transmission is constant. But gene action in different individuals or within an individual may be variable, producing different amounts of penetrance and expressivity, because of the environment. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additIonal aSSIgnment. 71 QUESTIONS FOR DISCUSSION 10. 1. In Drosophila matings between plum eye colored flies produce offspring of which 2/3 have plum and 1/3 red eyes. Explain these resul ts genetically. Are the genes involved pleiotropic? Explain. 10. 2. Why did Dobzhansky choose spermatheca shape as a test of pleiotropism for eye cEllor genes? What resul ts would you expect he obtamed after red and white eyed flies were compared as to testis pigmentation and general viability? 10. 3. Is it the gene for red eye color which has been proven pleIOtropic, or the gene for white eye color? Expl ain. 10. 4. It has been shown in primroses that genes which result in deeply cut and lobed leaves tend to affect the petals of the flower in somewhat the same way. Explain. 10. 5. In Radorn's study of lethal-translucida pleiotropism do you think the translucid condition produced the changes in blood chemistry, or that the reverse was .rue, or neither? Explain. 10. 6. Give an example of the pleiotropic effects produced by three different environmental factors. 10. 7. What can you conclude about the multiple effects of the normal allele C from a study of c h c h rabbits? Explain. c h pleiotropic in a Q c h rabbit? Explain. - Is 10. 8. Do Himalayan rabbits provide evidence that a gene can have more than one primary effect? Explain. 10. 9. Can one allele affect the penetrance or pleiotropIsm of another? Explain. 10.10. Could there be different degrees of expressivity in the absence of pleiotropism? Explain. 10.11. Distinguish multiple effects from multiple factors. 10.12. 72 Is penetrance and no penetrance of a gene an example of multiple effects? Explain. 10.13. Discuss the statement: "The chance of correctly ascribing an effect to a particular gene is reduced the further the phenotypic criteria used are from the place and time of gene action. " 10.14. Describe two instances in man where the expression of an inherited defect has been modified by training or environment. 10.15. Why is it so difficult to prove or disprove the "one gene, one primary action" hypothesis in even a single case? 10.16. What criteria are required for a gene ,to be considered fully penetrant? Had Mendel knowingly used fully penetrant genes in his studies? Explain. 10.17. If a gene located in a constant genetic background shows variable expressivity what would you predict regarding its penetrance? 10.18. What evidence can you present that polydactyly is not due to a recessive gene? • 10.19. What genotypes are possible for a person who is non-polydactylous? polydactylous? 10.20. How many different explanations can you offer for the following results? Two red-eyed wild-type strains of fly are mated to a pure line of white-eyed flies in your laboratory. One of these strains, collected m the tropics, produces some whiteeyed F1' the other one, collected from a temperate climate, produces only red-eyed offspring. 10.21. Choose one explanation for the results of questIOn 10.20 and describe how you would proceed in order to demonstrate it was incorrect. 10.22. Penetrance of rickets depends upon genotype and upon the lack of vitamin D. How could you test the validity of this statement? Chapter 11 TWINS, NATURE AND NURTURE Lecturer-c. PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 24, pp. 426-439. Colin: Chap. 9, pp. 148-156; Chap. 11, pp. 194-202; Chap. 16. Dodson: Chap. 21, pp. 253-256. Goldschmidt: Chap. 3, pp. 66-68. Sinnott, Dunn, and Dobzhansky: Chap. 11, pp. 133-141. Snyder and David: Chap. 29, pp. 466-472. Srb and Owen: Chap. 24, pp. 522-530. Stern: Chaps. 25, 26, 27. Winchester: Chap. 3, pp. 42-44; Chap. 13, pp. 183-184; Chap. 21, pp. 302-307. b. Additional references Kallmann, F. J. 1953. Heredity in health and mental disorder. 315 pp. New York: Norton & Co. Newman, H. H. 1940. Multiple human births. 214 pp. New York: Doubleday, Doran & Co. 'Osborn, F. 1951. Preface to eugenics. Rev. Ed. New York: Harper & Bros. LECTURE NOTES A. How much of a trait may be attnbuted to genetics (nature) and how much to environment (nurture) ? In humans, twins provide favorable material for investigating this problem, which is stated better as: how much of the variation in a trait has a genetic and an environmental basis? 1. Different eye color or blood group types are due entirely to different genetics, the environment causing none of the variations observed. 2. In the case of body weight, it is clear that diet and genetics both may have important roles. STERN 3. In organisms other than man, it is possible to experimentally standardize conditions. A standard genotype in different environments and a standard environment for different genotypes show to what extent environment and genetics, respectively, are responsible for phenotypic variability. 4. Twins provide natural experiments in humans a. Identical twins raised in the same family have identical genotypes and rather similar environments before and after birth. b. Non-identical twins raised in the same family have dissimilar genotypes, but similar environments - like those of identicals. c. Identical twins raised in separate families have identical genotypes but varied environments. B. Embryology of twinning 1. Identical or monozygotic twins start as a single fertilized egg which divides repeatedly. At some time, however, the cells produced fail to adhere to each other to form a single developing unit, but separate into two masses, each of which becomes a complete individual genetically identical to its twin. 2. Non-identical or dizygotic twins start with the fertilization of two separate eggs each fertllized by a different sperm. These twins are genetically different, being, lD this respect, no more similar than siblings conceived at different times. C. Distinguishing identlcal from non-identical twins 1. The number of pI acentas or birth membranes (chorions) are not good criteria. 2. The best way is to compare the twins with regard to a large number of genetic traits known to have complete penetrance, such as sex, eye color, ABO, MN, Rh, and other blood group types, etc. Anyone such difference between twins proves them non-identical, 73 whereas if both are the same in all these respects the probability becomes very large that they are monozygotic in origin. D. Basis of physical traits of twins (Fig. 11-1) Comparisons are made of % concordance (cases where glVen one twin to be affected the other one is also, as indIcated by ++ in the Figure) between Identical and non-identical twins of the same sex. IDENTICAL - %++ ABO NON IDENTICAL %++ BLOOD GROUPS~~~~~ CLUB mOT TUBERCULOSIS PARALYTIC ~~~"_--1 POLIOMYELITIS ....,_,,_,,"""-"'-"-L_ _--' form throughout the population from which the twin data were obtained .. E. Basis of mental traits in twIns 1. Personal tempo (Fig. 11-2) For metronome speed there is substantially the same % deviation in preference by the same indiVIdual at different times as between identical twins. In comparison, this deviation is almost double for non-identical twins or for sibs, and is still greater for unrelated mdividuals. There is, then, a component of thIs personality trait WhICh has a genetic basis. , \ • O[VIATION 20 19.5 15 %10 5 INTRA Figure 11-1 1. ABO blood group is 100% ,concordant for identicals, but only 64% for non-identicals since 36% of the time the latter receive different genotypes from their parents. Here heredity explains all of the dIfference between the two kmds of twins, environment being roughly equal for each paIr. 2. Club foot data for identicals show that the genes are penetrant only 32% of the time, whereas the 3% concordance for non-IdentIcals (or for SIbs born at different times) is still lower because these do not have identical genotypes. 3. TuberculoSIS concordance in identIcals usually involves the same form of disease, attacking the same place, with the same severity, whereas these similarIties are less frequent among non-identlCals. 4. Paralytic polIomyelitis, like tuberculosis, has a basis in genes whose penetrance depends not upon the infective organisms, because most humans are exposed to these normally, but upon the rest of the enVIronment. S. Measles shows very high concordance among both types of twins. This means that the gene basis for measles susceptibility is quite uni74 ID. PERSON NON- SIBS ID. NOT RELATED PtRSONAL TEMPO FIgure 11-2 2. Schizophrenia twin studies (Fig. 11-3) support the view there is an hereditary predisposition to this mental disease, although it is likely that more discordance was produced by differences in social environment in the case of non-identical than in that of Identical twins. In support of an hereditary contribution to thIS disease are two cases of identical tWIns who were separated, grew up in different environments, yet were concordant at about the same age. IDENTICAL %++ CRIMINALITY Figure 11-3 NON IDENTICAL %++ 3. Crnninality is a trait for whIch, even though concordance is different for twin types (Fig. 11-3), it IS v_ery difficult to evaluate the relative contributions of nature and nurture. 4. Test intelligence Listed below are the apprmamate differences in test performance found between identical twins, 3.1; between non-identical tWlns, S.5; between identicals reared apart, 6. There is clearly an hereditary and an enVIronmental component to IQ performance. Dr. Stern points out: "You IDlght say that we are really all born with different endowments for intellIgence performance. These may be compared to different lengths of rubber bands, where environment stretches the rubber band into performance. Some short rubber band may be stretched into a long performance, and some long rubber band may be stretched very littie. Human beings are biologIcally different, they are not alike. But there is no end to what, with different endowments, can be accomplished by applymg the rIght kind of environment. " POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, makmg additions to them as desired. 2. ReVlew the readlllg assignment. 3. Be able to discuss or define orally or in writing the Items underlined in the lecture I notes. 4. Complete any additional assignment. 75 QUESTIONS FOR DISCUSSION 11. 1. Give three examples of the nature-nurture problem. 11. 2. What disadvantages to the study of human genetics are overcome througb the use of the twin method? What does monozygotIc twinning indicate about the genetic consequences of cell division? 11. 3. 11. 4. If the babies in a hospital nursery were ll1ixed up, could you prove two were identical tWins? fraternal twins? Explain. 11. 5. What are "Siamese" twins? What causes them? Can they be dizygotic in origin? 11. 6. What genetic information, otherwise unobtainable, may be furnished by the study of hu.man twins? 11. 7. Wben are dizygotic twins not useful for the study of the relative importance of nature and nu.rture? 11. 8. Under what conditions are traits, even When due to genes that are 100% penetrant, no belp in distinguishing between monozygotic and dizygotic twins? 11. 9. Why is it necessary to determine concordance between twins of the same sex? 11. 10. Why may concordance be less than 100% for identlCal twins? 11. 11. Describe two ways in which triplets may be formed. 11.12. In what respects are triplets less or more Useful than twins in studying human inheritance? 11.13. How might concordance of tuberculosis in identical twins vary for different economic groups? 11.14. What would you conclude if concordance Was the same for dizygotic twins as for siblings? 11.15. 76 Is tbere a gene for personal tempo? Wby? 11.16. How can two children be as similar genetically as siblings, yet have no parents in common? 11. 17. Interpret the following data: Number of twin pairs Bqth members Only one memconVicted ber convicted of crime of crime Identical twins 45 21 Fraternal twins (of sa,me sex) 32 52 11. 18. If monozygotic twins are identical genetically, how can they be mirror'images, e.g. one left-banded and the other right-handed, one developing warts only on the right and the other only on the left side of the body? ' 11. 19. What would you conclude if concordance for married identic'al twins was lower than for unmarried ones? 11. 20. What effect might the improved nutrition of modern times have upon concordance? Explain. 11.21. Do you agree with the statement (D2) that the penetrance of club foot in identical twins is 32%? Why? 11.22. Is penetrance in identical twins estimated as the square root of concordance? Explain. Chapter 12 SEX-LINKED INHERITANCE Lecturer-c. STERN PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 5, pp. 99-107. Colin: Chap. 10, pp. 165-175. Dodson: Chap. 6, pp. 63-72. Goldschmidt: Chap. 8, pp. 153-156. Sinnott, Dunn, and Dobzhansky: Chap. 12, pp. 143-158. Snyder and David: Chap. 8, pp. 90-108. Srb and Owen: Chap. 6, pp. 85-99. Stern: Chap. 13. Winchester: Chap. 9, pp. 121-127; Chap. 7, p. 97; Chap. 15, p. 211. b. Additional references Bridges, C. B. 1916. Non-disjunction as proof of the chromosome theory of heredity. Genetics, 1: 1-52, 107-163. Hutt, F. B., Rickard, C. G., and Field, R. A. 1948. Sex-linked hemophilia III dogs. J. Hered., 39: 2-9. Morgan, L. V. 1922. Non-criss-cross inheritance in Drosophila melanogaster. Biol. Bull., 42: 267-274. LECTURE NOTES 1» A. Typical Mendelian inheritance occurs when reciprocal crosses between different pure lines produce F1 which are genotypically and phenotypically uniform, that is, when there is no relation between the traits which appear and the sex of the individuals. B. ExceptlOnal inheritance of colorblindness in man 1. Affected cf x normal 9 produces all normal children. 2. Affected 9 x normal cf produces both affected and normal children. All sons are affected, like the mother; all daughters are normal, like the father. 3. These phenotypic rules were empirical ones already formulated in the 19th century. 4. The cytogenetic explanation, by Morgan (geneticist) and Wilson (cytologist) in the early 1900's, was based on work with Drosophila (see C and D). C. Exceptional inheritance of white eye in Drosophila 1. Red 9 x white cf produces cfcf and 99, all red, red being dominant. 2. White 9 x red cf produces only white cfcf and red 9? 3. The crosses in B2 and C2 show a crisscross pattern of inheritance, sons resembling mothers, daughters resembling fathers. 4. It. was assumed all chromosomes are al ways paired except one, called the X chromosome pair. In females the XS are paired, but in males there is only one X. It was further assumed that in Drosophila the genes for red (~ and white (~ eyes are located in the X chromosome. 5. The cross in C1 is, then: 6. The cross m C2 is, then: PI ~~9 x ~cf F1 ~ cfcf and Xw+)@ 99 D. Regular Mendelian inheritance is expected for chromosomes other than the X (called autosomes), while the exceptional heredity can be explained on the basis of X-linked (sex-linked) inheritance. The same explanation as for eye color heredity in Drosophila was applied to the case of colorblindness in man by Morgan and Wilson. 77 E. Sex-linked inheritance in birds and moths 1. In chickens, non-barred feather Cj' x barred r:J produces offspring, all barred, barred (£) being dominant to non-barred (Q). 2. Barred Cj' x non-barred r:J produces all sons barred, all daughters non-barred. 3. Such results, opposite of those in Drosophila, can be explained if in birds (and moths, too) the male has the X paired and the female has it singly. This proved true upon cytological examination. 4. The cross in El is, then: PI ~~ x X£X£r:J Fl X£ ~ r:Jr:J and x£ ~~ 5. The cross in E2 is, then: PI X£~ x ~xQr:J Fl x£ xQ r:Jr:J and )@ '?~ 6. Sexmg of chIcks becomes easy if the matings made are between parents as in E2 (E5). F. In humans, the classical bleeder's disease (hemophilia type A) is due to an X-linked r:ecessive. It is a rare disease usually occurring in males, but recently a few caS8S of hemophilic women have been discovered in England. These homozygotes are viable, but are so infrequent because they must have for parents a hemophilic father and a mother heterozygous for this recessive allele. G. The chromosome theory of inheritance is based upon the parallellsms between genes and chromosomes. 1. Both usually occur as pairs in somatic cells. 2. Both show segregation of the members of a pair during gametogenesis. 3. Both retain their 1I1dividuallty regardless of the nature of their partners. 4. Both show independent segregation of different pairs. 5. When exceptions to the paired condition were found in transmission genetics, there were corresponding exceptions found cytologICally in chromosome type. These exceptions were sex-lInkage 111 Drosophila and in birds and moths. While these parallelisms might have resulted from chance, the finding, of an exception to the exception of sex-linked inheritance and the demonstration cytologically that there was a correspondmg exception to the exceptIOn of the number of X chromosomes present (see H), 78 proved that the chromosomes were the physical basis of inheritance and carried the genes. H. Non-disjunction of X chromosomes in Drosophila 1. Bridges, a student of Morgan's, also found that because of sex-linkage crosses of white Cj'Cj' x red d'r:J produced white sons and red daughters. 2. But, in addition, he found among several \ thousand F 1 one or two offspring which were exceptions to sex-linkage, namely, white daughters and red sons. 3. He reasoned that a. since whIte daughters possess two X , chromosomes each carrying Y:!_, both Xs must have come from the mother. b. red sons, carrying wt must have received their single X from their father. 4. Normally segregation produces a haploid egg contaming one X, the other Xlof the pair going into a pol ar body. 5. Bridges assumed that segregation sometimes fails, so that the members of the X pair fail to disjoin, i. e. -undergo non-disjunction, and both enter the egg or a polar body. In the latter case the egg formed carries no sex chromosome and is deSIgnated as a "0" (zero) egg. 6. Males normally have one X plus a partner chromosome called a Y which does not carry any gene for eye color. 7. From the cross described in HI and H2, then, non-disjunctional eggs fertilized by sperm would produce four types of offspring: , Sperm Offspring ~ (1) ~ XY:!_ XY:!_ Y (2) xY:!_ xY:!_ Y o ~ (3) Xw+O o Y (4) YO Eggs 8. Bridges assumed that type (4) flies, with no X, and type (1), with 3 Xs, rue. We know now that the former type always dies, and that only occaSionally does the latter survive. 9. Type (2), with 2 Xs, should be a white female, and type (3), with one X, shoul d be a red male - thus accounting genetlcally for the exceptional flies expected to survive. 10. This genetic explanation by Bridges predicted that exceptional white females would have diploid cells containing an additional Y chromosome, and that exceptIOnal red males would have one X but no Y chromosome in cells otherwise diploid. And when these cells were examined cytologically Bridges' predictions proved true. 11. This work of Bridges, published in 1916, (see Pre-Lecture references), represents a great feat of the human mind. I. The study of sex-lirlked inheritance, an extension of Mendelian inheritance in general, demonstrates how the architecture of human knowledge is often created out of painstaking work with very small objects. POST-LECTURE ASSIGNMENT 1. Read the notes immedIately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading aSSIgnment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 79 QUESTIONS FOR DISCUSSION 12. 1. In Drosophila, pure line, straight-bristled <¥<jl x forked-bristled eM produce offspring which are all straight-bristled. The reciprocal cross produces forked-bristled sons and straight-bristled daughters. Explain. 12. 2. Colorblind mothers always produce sons who are colorblind, but colorblind fathers rarely have colorblind sons. Explain. 12. 3. Whether or not the father is hemophilic is normally of no phenotypic concern for his children. Explain. 12. 4. The son of a normal mother whose father was normlli may be hemophilic. Explain. 12. 5. In Drosophil a, if a white-eyed female is crossed with a red-eyed male and the F2 allowed to interbreed freely, what will be the appearance of the F 3 as to eye color? 12. 6. In Drosophila, if a homozygous red-eyed female is crossed with ::. whlie-eyed male and the F2 allowed to interbreed freely, what will be the appearance of the F3 as to eye color? 12. 7. A hemophilic man marries a woman who is colorblind. She has no hemophilic ancestors. What IS the expectation for these traits in their children? 12. S. An albino, colorblind, hemophilic man with 0, M blood group types marries a normal woman with AB, N blood group types. What phenotypes are expected in their children? 12. 9. A genel_ III Drosophila is recessive, lethal, and sex-linked. If female Ll is crossed with a normal male, what should be the sex ratio of the progeny? 12.10. Assume that right-handedness is dominant over left-handedness and brown eye color over blue, and that both these traits undergo segregation independently of each other and of sex. The mother of a right-handed, brown-eyed woman of normal vision is right-handed, blueeyed, and of normal vision, and her father is left-handed, brown-eyed, and colorblind. 80 This woman marries a man who is lefthanded, brown-eyed, and of normal vision, and whose father was blue-eyed. What chance will ,the sons of this couple have of resembling their father phenotypicllily? 12.11. What offspring would you expect from a mating of one moth with another that carried a sex-linked recessive lethal gene in heterozygous condition? 12.12. Distinguish between and gi~e an example of sex-linked and sex-li'mited characters. Note: In Drosophila, reciprocal matings between vestigial-winged and long-winged flies produce only long-winged offspring. 12.13. A white-eyed, vestigial male is crossed to a red-eyed, long female from a pure line. The Fl female is crossed to her father. What will be the genotypic and phenotypic expectation in F2? 12.14. A white-eyed, vestIgial female is crossed to a red-eyed, long male from a pure line. The Fl male is crossed to a female like his mother. What phenotypes are expected from this last cross? 12.15. In blrds, how would you proceed in order to prove that a particular dommant gene was located in the X chromosome? 12.16. How could you prove that the Y chromosome in Drosophila carries genes for male fertility which are absent from the X chtomosome? 12. 17. Do you agree wi th the definitlOn of autosomes given in D in the Lecture Notes? 12.18. Can a gene, like bobbed in Drosophila, which has a locus in both the X and the Y chromosomes, show crisscross inheritance? Explain. 12.19. How could you prove that the Y chromosome, like the X, carries a locus for bobbed? 12.20. Can you explain how Stern (Chap. 9) might have obtained and identified flies containing three bobbed genes? 12.21. In Drosophila, what would be the consequence of a non-disjunction of sex chromosomes during spermatogenesis? Devise a genetic method for identifying the produc_ts of such a non-disjunction. 12.22. How could you test specifically whether non-disjunction can occur in somatic cells? 12.23. Mrs. Morgan (see Pre-Lecture references) found a case in Drosophila in which yellow-bodied females bred to gray-bodied (wild-type) males always produced yellow daughters and wild-type sons. How can this reversal of the ordinary course of sex-linked inheritance be explained? 12. 24. Has evidence been presented that a chromosome contains more than one gene? Explain. 12. 25. Do you agree that all genes linked to sex are linked to e~ch other? Why? 12.26. What evidence would be required to prove that a trait in man was due to a gene in the Y chromosome which normally had no allelic partner in the X? 12.27. Except for a few genes, the Y chromosome in Drosophila is said to be blank for the genes present in the X chromosome. Can you justify this explanation rather than one whlCh states that, with few exceptions, the Y carries the least efficient alleles of the same gen~s as the X carries? 81 Chapter 13 SEX DETERMINATION I \ Lecturer-c. STERN PRE- LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General ~enetlcs textbooks Altenburg: Chap. 5, pp. 97-101, 113-117, 124-127. Colin: Chap. 9, pp. 159-160, 139-145. Dodson: Chap. 15, pp. 187-190; Chap. 6, pp. 63-65, 79-80. Goldschmidt: Chap. 8, pp. 144-148. Sinnott, Dunn, and Dobzhansky: Chap. 23, pp. 324-325; Chap. 22, pp. 303-305. Snyder and DaVId: Chap. 23, pp. 332-333, 341-345. Srb and OWen: Chap. 6, pp. 85-89; Chap. 14, pp. 287-288, 295-298. Stern: Chap. 20, 21. Winchester: Chap. 7, pp. 90-96, 99-100; Chap. 8, pp. 113-116. b. Additional references' Warmke, H. E. 1946. Sex determination and sex balance in Melandrium. Amer. J. Bot., 33: 648-660. Welshons, W. J., and Russell, L. B. 1959. The Y-chromosome as the bearer of male determining factors in the mouse. Proc. nat. Acad. SCI., U. S., 45: 560-566. Whiting, P. W. 1943. Multiple alleles in complementary sex determination of Habrobracon. Genetics, 28: 365-382. LECTURE NOTES A. Sex determination deals basically with the question of how it is that eggs and sperm are produced. Normal sex determination is discussed III this chapter, whIle abnormalIties which help explain the normal situatIOn are discussed prlllclpally in Chap. 14. B. Sex cell formation in hermaphrodites An hermaphrodite is an animal that produces both eggs and sperm in the same indlvidual. 82 C. D. E. F. G. 1. Helix, the snall, has a gonad which produces eggs and sperm from cells very close to each other. How is It that genetically' identical cells so close to each other form different, gametes? The answer to this particular question requires an unders.tanding of the process of differentlatIOn in general. 2. Earthworms produce eggs and sperm in separate organs located in different body segments. Sex cell formatIOn in monoecious mosses MonoeclOus refers to plants which carry male and female sex organs in the same individual. In these mosses, egg and sperm-like gametes are produced in separate sex organs located on the haploid gametophyte. Gamete formation in Ophryotrocha In this marine annelld size IS correlated with sex determination. For when the individual is small, because of youth or of amputation, sperm are formed, whereas larger individuals produce eggs. Sexual differentiation in Bonellia This marine worm has walnut-sized females with a long proboscis, and microscopic, ciliated males which live as parasites in the body of the female. Fertilized eggs develop into females when grown in the absence of adult females; they become males when grown in the presence either of adult females or of an extract of the proboscis. All cases so far described (B-E) are examples of non-genetic, phenotypic sex determination. In these cases, different sexes are determined not by genetic differences between cells, organs, or individuals, but by environmental differences which cause differentiation to switch toward maleness or femaleness. Of course, even here genes playa role, for it IS they that make possible different responses to different environments. Genotypic sex determlllation Cases where there are separate sexes, because genes determine that they are separate, are dis- cussed next. 1. Chlamydomonas is a haploid umcellular plant with two flagella and a chlorophyll-containing chr'omatophore. Though it divides by mitosis, individuals from cultures of illfferent origlll may fuse in pairs to form diplOId zygotes. Each zygote immediately undergoes meiosis to produce four haploid cells. These four cells, grown separately, produce cultures within which there is no mating, all individuals in a culture being + or -. Upon testing these cul tures two prove to be - and two +, matlllg being between + and - individuals. The meiotic segregation of + from -, producing the 1 : 1 ratio of sexual types, shows that sex determination in Chl amy do monas is genetic, even though the sexes are not visibly dlfferent. 2. Grasshoppers a. McClung, and WIlson, and Stevens showed there are sex chromosomes; in certam species females have 14 and males 13 chromosomes. b. Thus, females have two X chromosomes and males one X chromosome, both sexes having 12 auto somes (12 A), as six pairs (6 AA). c. After melOsis, eggs contain 6 A + 1 X, half of sperm are 6 A + 1 X and half 6 A + OX. d. Sex determination is obvious, zygotes with IX forming males, those with 2X forming females. e. The X chromosome produces a tendency towards femaleness, since adding IX to 12A + IX makes a change from male to female. 3. Droso'plllla melanogaster a. Females are 3AA + XX, while males are 3AA + IX + lY. b. From this, one cannot decide the chromosomal basis for sex determination, since there are two varIables, the Xs and the Y. c. Bridges obtained flies contaimng XXY (Chap. 12) or XXYY besides 3AA. These were females. He obtained also 3AA + XO flIes (Chap. 12) which were males. The Y, therefore, is not sex determimng in this species. d. However, the Y is necessary in males for sperm motility. 4. Gynandromorphs or gynanders are inillviduals abnormally part male and part female. a. In moths and butterflies males may have H. large, beautifully colored wmgs, females small stumps of wings. Gynanders may have wings like a male on one side and those like a female on the other side. b. This could be explained only after the chromosome theory of inheritance was established. c. Then it was found, in Drosophila, that a gynander starts as a zygote containing 3AA + 2X, that is, as a female. The zygote nucleus divides mitotically to produce two nuclei. But occasionally the nntosis IS abnormal so that while one daughter nucleus is normal, containing 3AA + XX, the other daughter nucleus is defective, containing 3AA + X, because one X which failed to be included in the nucleus degenerates and IS lost. Cells produced following mitOSIS of the XX nucleus WIll produce female tIssue, while those derived from the X nucleus will produce male parts. d. Sometimes front and hind halves and other times right and left sides are of different sex in gynanders produced this way. e. This theoretical explanation was proven by Morgan and Bridges. Zygotes were obtained with one X carrying the dominant gene for normal bristles, sn+, and another X carrying the receSSIve allele for singed, short bristles, sn. Gynanders produced by loss of th;)c carryi'ng sn+ have smged bristles everywhere they are male, but normal bristles wherever they are female. f. Slllce hormones play almost no role in insect differentiation, each body part forms largely according to the genotype it contains, usually resulting in a sharp borderline between male and female parts m gynanders. g. Gynanders may al so occur after fertillzation of an abnormal egg containing two haploid nuclei. Because of polyspermy in insects (more than one sperm normally enters each egg), one egg nucleus may be fertilized by an X-carrying sperm, the other by a Y-carrying one. Resulting individuals may be simultaneously mosaic for sex and autosomal traits if the two sperm fertilizing differed genetically III the autosomes each carried. Sex determination in man and mouse 83 1. 1. Males are usually XY, females XX. 2. In 1959 (see pp. 709-716, #7075, vol. 1, of Lancet), certain underdeveloped human males (having testicular hypoplasIa as part of Klinefelter's syndrome) were found to be XXY, and certain underdeveloped human females (having Turner's syndrome) were found to be XO. XO in the mouse is also female. 3. In these forms, then, unlike DrosophIla, the male is determined by the presence of the Y chromosome. There is a variety of non-genetic and of genetic sex determining mechanisms. It is important to study the general phenomenon of determination as well as a variety of species, since one cannot necessarily generalize from one case to another. POST-LECTURE ASSIGNMENT 1. Read the notes Immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional ;}sf'ignment. 84 QUESTIONS FOR DISCUSSION Differentiate between sex determination and sex differentiation. 13. -1. 13. 2. How do hermaphroditic and monoecious individuals differ? 13. 3. How can you change egg producing Ophryotrocha into sperm producers? How might you do the reverse experimentally? 13. 4. Why would the Bonellia worm probably become extinct if suddenly males were produced independently of the presence of fe_males? What results are possible following loss of a sex chromosome from one of the first two nuclei formed by an XY-containing zygote of Drosophila? 13.16. 13. 17. Could you detect the loss of a Y chromosome from one of the early cleavage nuclei formed from division of an XY-containing zygote of Drosophil a? Expl am. How can you explain a moth which is approximately 1/4 female and 3/4 male? 13.18. What do gynanders of moths and Drosophila tell about the role of hormones in sex determination in these forms? sex differentiation in these forms? 13.19. 13. 5. Is sex differentiation restricted to diploid or to multicellular organisms? Explain. 13. 6. Discuss the basis for sex determination in Melandrium. 13. 7. What procedure would you follow to show that of the four cells produced by meiosis in Chl amydomonas two are + and two - in sexual behavior? In what respects would you say fly A and fly B are abnormal? Can you offer a genetic explanation for these phenotypes? 13.20. 13. 8. How do we know that mating pairs are not ++ or -- in Chlamydomonas? 13. 9. What reason can you give that sex deternunation has a genetic basis in Chlamydomonas? 13. 10. What is a sex chromosome? Is the Y in Drosophila an example of one? Explain. 13.11. . From the evidence so far presented, is it necessary to include autosomes in describing the chromosomal basis for sex determination? Why? Describe how one might obtain an XXYY individual if the strains of Drosophila provided consisted of XX and XY individuals. 13.12. Since the Y chromosome in Drosophila is not sex determining, why has it not been lost in the course of evolution? 13.13. 13.14. Differentiate between hermaphrodite, gynandromorph, and mosaic. What is the evidence that sex determination in Drosophila is genetic rather than nongenetic? 13.15. How can you explain a female Drosophila which has one eye red and one white? 13.21. 13.22. If, in Drosophila, the parents were how could you explain an offspring which was phenotypic ally a. half f+ and half f? b. half male and half female, but allr? c. male half f and b+, female half rand b? In man, what sex would you expect in the offspring following maternal non-disjunction of Xs? paternal non-disjunction of sex chromosomes? What would be the sex chromosome content of the offspring in each case? 13.23. 85 Chapter 14 SEX DETERMINATION II lecturer-Co STERN PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 5, pp. 108-111, 117124, 126-127. ColIn: Chap. 9, pp. 156-159. Dodson: Chap. 15, pp. 179-183. Goldschmidt: Chap. 12, pp. 203-204. Sinnott, Dunn, and Dobzhansky: Chap. 22, pp. 305-314. Snyder and David: Chap. 23, pp. 333338. Srb and OWen: Chap. 14, pp. 282-287, 289-292. / Stern: Chaps. 20, 21. Winchester: Chap. 7, pp. 97-99; Chap. 8, pp. 106-113, 116-120. b. AdditlOnal references Bridges, C. B. 1925. Sex in relation to chromosomes and genes. Amer. Nat., 59: 127-137. Goldschmidt, R. 1934. Lymantria. Bibliograph. Genet., 11: 1-186. LECTURE NOTES A. How does the XX vs. X, or the Y vs. no Y, chromosome mechanism accomplish its developmental effect of making for femaleness or maleness? In Drosophila, if the X leads towards femaleness (addition of one X to one already present produces female), then why are not IX individuals already females instead of males? There must be an inherited tendency towards maleness, due to genes not in the X, which is stronger than the female tendency produced by IX, so that a IX mdividual is male. However, the same inherited male tendency must be overpowered by the strength of femaleness of 86 B. C. 2Xs, so that in this circumstance a female is produced. Autosomal genes and sex m Drosophila 1. Sturtevant showed that when a specific autosome carnes the dominant gene.:!: III homozygous conditlOn, XX flies are female and X flies male, as expected. 2. But if flies are homozygous for tlie recesSlY'? allele transformer (~, though X flies "remain males, XX llldividuals develop as males. 3. This sex transformation proves autosomal genes also are concerned with sex determination. Sex determination in Lymantria dispar 1. This gypsy moth, studied by R. Goldschmidt, is found, in many countries, to have males with large, white wings and wide, featherlike antennae, and females with smaller, darker wings and thin, threadlike antennae. 2. Crosses between llldivlduals from different countries frequently produ~ed sexually intermediate offspring, which he called inter- , sexes. Included among the organs intermediate in intersexes are WlllgS, antennae, and gonads. 3. Ignoring the Y chromosome, in moths normally XX is male and X is female. 4. Goldschmidt reasoned that each X produces a tendency for maleness, M, which within a race is so balanced against the tendency for femaleness, F, present elsewhere, that F + M produces normal females and F + MM produces normal mal es. 5. He assumed that there are different alleles for F and M, some stronger, others weaker, but that within a race the F and M alleles are of the same strength. 6. Consequently, within a race sex determination would be normal, but if the strengths of two races differed and were mixed by cross- D. E. 4. Among the viable offspring are normal males, normal females, and triploid females. Others have three of each autosome, but only two Xs; these appear as sterile intersexes. Still others have 3 Xs and are otherwise diploid (superfemales), or are XY and otherwise triploid (supermales). 5. An intersex, a female with regard to Xs, has its development shifted towards maleness by the extra autosomes present. 6. In additional work Bridges found still other sex types (Fig. 14-2). One of each kmd of autosome comprises a set of autosomes. ing, mtersexes could be produced. Balance theory of sex determination 1. Sex is determined by the balance of genes located on different chromosomes. 2. This idea was formul ated first on the basis of studie_s of the gypsy moth. 3. Additional evidence was obtained by Sturtevant when crosses between Drosophil a repleta and neorepleta produced intersexes. 4. Proven conclusively by Bridges' study of abnormal sex types in Drosophila melanogaster. Sexual types in Drosophila melanogaster 1. Bridges found individuals, called triploids, having each chromosome threefold (FIg. 14-1). In the Figure, Xs are filled in while auto somes are not, and the Y IS represented by a broken line. TrIploids are females which are a httle larger than normal females. SEXUAL TYPtS IN OROSOP~ILA MtLANOGASTER (AFTER X SETS OF SEX INDEX CHROMOSOMfS AUTOSOMES (RATIO XIA) (A) SEX ~,~ ,(;- ~oo~ 11\ \~\ SUPERFEMALf TRIPlOI~F[MALE) SUPERMALE ~ ~ \ FtMALE INT~RSEX SUPERFEMAlE TETRAPLOID NORMAL FEMALE { TRIPLOID DIPLOID ~APLOID INHRStx NORMAL MALE SUPERMALE 3 4 3 2 4 3 1.5 1.0 1.0 2. 1 2 1 1 2 1 3 2 1.0 1.0 0.67 0.50 3 0.33 Figure 14-2 MALE FIgure 14-1 2. When triploids undergo meiosis, bundles of three homologous chromosomes (trivalents) are formed at synapsis, and after segregation eggs are produced containing anyone or any two of the chromosomes of a trIvalent. Since different trivalents undergo segregation independently, eggs may be produced which have each type of chromosome singly, or contain two of each type, or have any combination in which some chromosomes are represented once and others twice. 3. When such eggs are fertilized by normal sperm some zygotic combinations are lethal but others are viable. BRIDGES J F. a. When the sex indices were tabulated it became clear that there was a spectrum of values from 1.5 (superfemales, or ultrafemales) down to 0.33 (supermales, or inframales). b. Note that even though the chromosome content varies, so long as the balance between Xs and autosomal sets is such that the sex index is 1. 0, essentially normal females are produced. c. The correspondence between visible numbers of chromosomes and sexual types was proof of the balance theory of sex determination. Sex determination and differentiation in mammals 1. The basis of Y, or no Y, chromosome determining sex in humans is discussed elsewhere (Chap. 13). 87 G. H. 88 2. The early gonad is neutral, consisting of an outer cortex and an inner medulla. In males (persons with a Y) the cortex degenerates and the medulla forms testis, while in females the medulla degenerates and the cortex becomes ovary. 3. Once ovary and testis are formed they produce hormones which direct the development or degeneration of various sexual ducts, the formation of gemtalia, and other sexual characters. 4. In cattle, male sex hormones from the male twin may enter the female co-twin which then becomes an intersex, called a freemartin. 5. Therefore, though sex determination is genetic, sex differentiation is due largely to sex hormones produced by the gonads. The sex ratio refers to the relative numbers of males and females produced. 1. In humans, there should be a 1 : 1 ratio of girls to boys, since half the sperm should carry Y and half not. 2. While at birth 106 boys are born for every 100 girls, at conception the numbers may be equal. 3. One family is reported to have had only boys in 47 births. Another, well substantiated, family, has had 72 births, all girls. This is too rare to be merely a matter of chance! 4. Work in Drosophila suggests some possible explanations. a. CertaIn males with a gene for "sex tatio" produce daughters almost exclusively. These males are XY. When they form sperm almost all Y chromosomes degenerate so that nearly all sperm carry an X. b. Certain females transmit a virus to their offspring through their eggs. When such females mate with normal males the zygotes produced start development. But very early in embryology the virus kills XY individuals, so that almost all survivors are females. Predetermination of sex 1. When there are chromosomally two kinds of male or of female gametes it is theoretically possible these can be separated, thereby leading to control of the sex of progeny. 2. A Russian worker reported some success in separating X and Y sperm of rabbits by electrIc current. Suggestive results were also obtained this way by a U. S. worker. 1. A Swedish worker, using centrifugation, has reported some success in separating cattle sperm into two types. 3. These results are not yet definite, but someday a method for doing this will be discovered> It will be of great use in animal husbandry. Will it be used in man? "We have always made use of knowledge, and although we might hesitate at the moment to think about the consequences of such a separatIon, it is pretty likely that such a method will be used also in our case." "The study of sex determination shows again that one can be interested in kn~wledge itself, and that such studies have contributed great things to such knowledge, but that it also, at I the same time, involves giving power to mankind over nature, including over himself. " POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading aSSIgnment. 3. Be able to discuss or define orally or in wrlt~r.b the items underlined in the lecture notes. 4. Complete any additional assignment. QUESTIONS FOR DISCUSSION 14. 1. If the male tendency in Drosophila is due to genes not on the X chromosome, must these genes be numerically equal in females and males? Explain. 14. 2. What sex ratio is expected from each of the following crosses? a. b. 14. 3. + tra cJ x .:!:. tra <;> XY, tra tra x XX, + tra How do intersexes differ from gynanders? 14. 4. List all the different chromosome combmations which a trIploid female of Drosophila may produce in her gametes. 14. 5. Would triploid Drosophila maintain themselves in nature? Explain. 14. 6. Why are the terms ''ultrafemale'' and "inframale" preferable to "superfemale" and "supermale" in Drosophila? 14. 7. What are the chromosomal makeups of parents of intersexes m Drosophila? 14. 8. Can you tell the sex of an early human embryo? Expl ain. 14. 9. and Y sperm have been successfully separated artificially? 14.16. Are humans with Turner's or Klinefelter's syndrome (Chap. 13) inters exes ? Explain. 14. 17. If a hen which undergoes sex reversal, and thus becomes a functional mal"e, produces gametes of the same chromosomal constitution as before (al though they are now sperms instead of eggs), what will be the sex of her offspring when she is mated with a normal hen? 14.18. If a sex-reversed hen as in 14.17 was barred, what would be the appearance of her offspring when bred to a non-barred hen? 14.19. Why can or cannot a situation in which the chromosome number is 2 in females and 1 in males be explained on the chromosome balance theory of sex determination? 14.20. Is either the control of the external environment or the determmation of chromosome number enough to tell whether sex is environmentally or genetically determined? Explain. List evidences that the sex chromosomes are not the only ones having to do with sex. 14.10. Hqw might you explain the occurrence of inters exes in humans? 14.11. Can humans be gynanders? hermaphrodites? Explain. mosaics? !4.12. If equal numbers of XY and XX zygotes are formed, what is responsible for the greater number of boys born than of girls? 14.13. What is the likely explanation of 8-child familes in which the offspring are all boys or all gIrls? 14.14. From the results with Drosophila, can you suggest an explanation of the more than 70 consecutive female births in a French family? 14.15. How does one proceed to test whether X 89 EXAMINA liON II UNDERLINE ALL THE ALTERNATIVES WHICH ARE CORRECT. 1. fail. b. provides a poor test of the chromosome theory of inheritance. c. can produce phenotypic exceptions tq the exception of sex-linked inheritance. d. is a rare phenomenon. Non-disjunction . can occur at any nuclear division but not byautosomes. e. does not serve as an exception to independent segregation. Absence of pigment in a bean plant is lethal a. since the plant is homozygous for a recessive gene. b. because the plant cannot make food via photosynthesis. c. but albinism is viable in fungI and animals. d. and, ignormg mutations, the genes causing this came from parents both of which were green. e. because all traits are the result of interaction of genotype and environment. 5. 2. Characters due to rare recessive genes confined to the X chromosome a. generally appear in men only. b. may appear in some of the grandsons produced by a daughter of a man with the trait. c. appear more often in men than in women. d. appear in the daughters, but not the sons of an affected man. e. have been studied, and the genes for these characters are mostly in men. 3. 4. 6. if the twins are identIcal. if the parents have the same blood group. if the parents have different blood groups. a. plus d. b. plus c. Non-disjunction of sex chromosomes in Drosophila Hypostasis a. cannot be observed in the progeny of a backcrossed dihybrid. b. is the reclprocal of epistasis. c. and dominance played an important role in Mendel's work. d. shows that phenotypes may not follow the rules of transmiSSIOn genetIcs. e. involves, at some level, interaction of non-'allelic gene pairs in the production of visible traits. if the twIllS are fraternal. a. shows that segregatIOn sometimes can 90 a. in differences in birth weight of identical twins. b. in identical twins reared together. c. in identical twins reared apart. d. in every indlvidual, although the amount may be obscure. e. least clearly III fraternal twins reared together. Blood transfusions can always be made safely between twins of the same sex a. b. c. d. e. f. The effect of environment on genotypic expression is shown 7. Characters which are polygenic ally determined a. require a constant environment for their . b. c. d. e. s. expression . make unsatisfactory examples for demonstrating Mendel's principles. obey the ordinary rules of Mendelian inheritance. will show regression only if there is no dominance between alleles. require measurement and use of statistical procedures in order to determine certain of their transmissive properties in successive generations. d. in the frequency with which polydactyly occurs in guinea pigs whose mothers were or were not aged. e. by genes producing either quantitative or qualitative characters. 11. a. is a phenotype based upon the presence of a number of different wild-type alleles. b. can be produced sometimes by the presence of one gene, and sometimes by the presence of two. c. may occur in a female heterozygous for the normal allele of vermilion. d. and red eye provided Morgan with material to prove sex-linked inheritance. e. in females can result in sons all of which are white eyed, even though the father has brown eyes. Two genes are alleles if they a. always segregate. b. both affect the same trait in the same way, but to dIfferent degrees. c. are isoalleles. d. are both allelic to a third gene. e. exist in a single individual -- one'in one chromosome and the other in the homologous chromosome. 12. 9. Variable penetrance is shown a. in the case of the Himalayan rabbit grown under different temperature conditions. b. when the phenotypic effect of three bobbed genes in Drosophila is compared with the effect of two. S. in the case of abnormal abdomen, studied by Morgan, which depends upon the amount of moisture during the adult stage. 14. You can be sure a trait has an hereditary basis Genes which produce a lethal phenotype a. must be recessive and homozygous to produce their effect. b. may cause death at any stage in development but prior to reproduction. c. occur only in sexually reproducing organisms. d. include those producing albinism in some plants, but not in others. e. follow Mendel's rules even if they give a I 2:1 phenotypic ratio. 10. In Drosophila, white eye , a. only if it occurs in two or more alternatives. b. if it is a trait like polydactyly. c. when there are many qualitatively different alternatives. d. if it shows dominance in a hybrid. e. if a change in just one gene produces an effect on it. 13. Sickle cell anemia in man a. is not to be confused with thalassemia minor which also occurs in homo zygotes for a recessive gene. b. shows variable expressivity at the biochemical level. c. sometimes acts as a recessive lethal. d. is completely non-penetrant in heterozygous condition. e. shows multiple phenotypic effects all of which are indIrectly attributed to a single effect. What kinds of adult offspring could be produced from mating these exceptional Drosophila? a. ~ YY d x W~ <jl b. ~ Yd x W~ Y <jl 91 15. Each time the number preceding the organism applies to a statement write it in the space provided. 1. 2. 3. 4. 5. Drosophila Ophryotrocha Grasshopper Man Bonelli a Chlamydomonas Chickens Helix Mouse 10~ Lymantria 6. 7. 8. 9. - - - - - Individuals can be intersexes. Males are XY in constitution. ----_ _ _ _ _ Females are XY in constitution. _ _ _ _ _ Sex is not determined by chromosomal differences. - - - - - Diffusable chemicals are known to be involved in sex determination or differentiation. - - - - - The chromosome balance theory of sex applies. _ _ _ _ _ Non-motile gametes are produced. _ _ _ _ _ True hermaphroditism is shown. 16. Before each of the following statements write T if complete true, or write F If not completely true. Give a reason or example for your decision in each case. a. Almost all males of a species have the same chromosome number. b. One may obtain, among adult offspring, a ratio of 4 dormnant individuals to 1 recessive individual following a mating between monohybrids. c. Each human contains only about half of the total genetic variability possessed by his parents. d. The environment can sometimes produce as much phenotypic change as a mutation in an individual. e. The ability to reproduce itself and some of its modifications are properties unique to genes. f. Backcrossing a dihybrid produces 4 types of progeny only in cases of complete dominance. g. Barring mutation, normal males cannot have hemophilic daughters. h. Phenotypic interaction between different pairs of alleles is suggested by such ratios as 9:7, 9:3:3:1, 13:3, 63:1, but not by 1:2:1, or a 2:1 ratio. i. If one fraternal twin is an albino genetically there is a 25% chance or greater chance for concordance in the other twin. j. In all mammals the Y chromosome is male-determining. 92 17. Discuss the following: A rare autosomal dominant gene with low penetrance can be distinguished from a rare autosomal recessive gene in either of two ways: a. the relative frequency with which the trait appears in siblings, grandchildren, and first cousins of individuals that have the trait, and b. the relative frequency of first cousin marriages between the parents of individuals that show the trait. 18. The coat color of mice is controlled in part by the following three genes: a is a recessive for black. b is a recessive for cinnamon. c is a recessive for albino. Individuals that are homozygous recessive for both ~ and Q are brown; those carrying all three of the dominant genes~, ~, and Q are agouti. A black mouse is crossed to an agouti mouse and they produce offspring of the following types: Agouti Albino Black Brown Cinnamon In the space before each of the phenotypes listed write a number representing the proportion of the offspring that theoretically should be of that type. 19. Deaf mutism in humans IS due to the presence of either or both of the completely recessive nonallelic genes ~ and Qin homozygous condition. These genes are located in different autosomes. a. Give the genotypes of a deaf mute man and a deaf mute woman who can produce only normal children. b. A deaf mute of genotype aa bb has normal parents. for his parents. 20. Give the genotypes possible In poultry, barred feathers (dominant) and black feathers (recessive) are due to sex linked genes @' Q). Split comb (dominant) and single comb (recessive) are due to a pair of autosomal alleles (~, ~). What is the expected phenotypic ratio in the offspring from the cross of Ss 21. ~y '? by Ss Bb d'? In corn, genes ~ and ~ are non-allelic, and both are required for purple seed. If either a or b is present in homozygous condition the seeds are white. What phenotypic ratio is expected from a cross of Aa Bb by Aa bb? 22. When were identical twins not twins? 93 23. Ten people chosen at random from a population have heights (in inches) of: 64 66 68 68 68 68 70 70 70 72 - - - - is the population mean. - - - - is its variance. deviation. - - - - is its standard , 24. In humans assume brown eyes @) is completely dominant to blue (!:!), and is not sex-linked. Red-green colorblindness is due to a completely recessive gene (0, normal vision to (9, and is X-linked. A blue-eyed, red-green colorblind woman marries a brown-eyed man with normal vision whose mother was blue-eyed. a. What are the genotypes of the man and woman whose marriage we are considering? b. What genotypes and phenotypes are expected in the sons and daughters from this marri ag-e? 25. In Drosophila, crosses of Curly winged flies by Curly always give approximately two-thirds Curly to one-third non-Curly (normal) winged offspring. Curly x normal gives approximately one-half Curly and one-half normal. a. How would you explain these results? b. Make up an appropriate key of gene symbols. Give the genotype of one fly that is phenotypICally Curly and one that is phenotypically non-Curly. 26. Assume that in man the difference in skin color is due to two pairs of factors: that AA BB ,is black ' and aa bb white; and that any three of the genes for black produce dark skin'; any two medium skin; and anyone light skin color. A marriage of medium by light yields a large family of which 3/8 are medium, 3/8 light, 1/8 dark, and 1/8 white. Give the genotypes for the individuals listed below. Note: There is more than one possible answer to part c. Choose anyone correct answer and make the other answers consistent with this. Even after you select one correct answer for part c there remains more than one genotype for two of the other parts. In these cases, give all the possible genotypes. d. a. _ _ _ _ _ _ _ _ _ white children medium parent e. ___________ medium children b. _ _ _ _ _ _ _ _ _ dark children c. _ _ _ _ _ _ _ _ _ light parent f. _ _ _ _ _ _ _ _ _ light children ----------- 94 Chapter 15 LINKAGE Lecturer-G. W. BEADLE PRE-LECTURE ASSIGJ\TMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings from textbooks: Altenburg: Chap. 9, pp. 160-166. Colin: Chap. 8, pp. 118-124. Dodson: Chap. 11, pp. 126-132. Goldschmidt: Chap. 6, pp. 105-122. Sinnott, Dunn, and Dobzhansky: Chap. 13, pp. 160-164. Snyder and David: Chap. 10, pp. 125-137. Srb and Owen: Chap. 9, pp. 150-156. Stern: Chap. 14; Chap. 15. Winchester: Chap. 14, pp. 189-190. B. LECTURE NOTES A. Symbols for genotypes have evolved (Fig. 15-1) 1. The sYmbol contaming + always refers to the normal gene, the other symbol of the pair representing the mutant form. C. 2. _+_ may also be written _+_ or +/w or I ~ W + w. (Except for the two preceding sentences, gene symbols in the text of this book are underlined. Thus, the heterozygote under discussion may be represented as :!:./'!!_.) S'fMBOLS FOR ALLELES IN ROUND - WRINKLtD HYBRID R W w+ +w + === OR === OR=== OR~OR == r w w w w Figure 15-1 Mendel and independent segregation 1. While the garden pea has seven pairs of chromosomes, more than seven pairs of genes are known. 2. There must be more than one gene per chromosome, therefore, the genes being transmitted in a bundle, linked, corresponding to the chromosome in which they lie. 3. However, the seven pairs of genes Mendel studied all showed independent segregation, so each gene pair was on a different pair of chromosomes. 4. Because of this Mendel did not discover linkage. Linkage in the garden pea (Fig. 15-2) 1. This was studied by deVilmorin and Bateson. 2. The acacia variety has no tendrils due to a recessive mutant. 3. A double recessive pea plant (wrinkled, no tendrils) was crossed to a pure double dominant plant (round, tendrils). 4. The F 1 was self-fertilized and produced the F2 shown. 5. Were the two pairs of genes inherited independently, there would have been in F2 a 9:3:3:1 ratio since the F1 is a dihybrid. 6. Each gene pair, in fact, showed segregation in F 2 . a. Round:wrinkled gave a 3:1 (323:126) ratio. b. Tendrils:no tendrils gave a 3:1 (322:127) ratio. 7. But these two 3:1 ratios were not produced independently, for the four phenotypic classes do not form a 9:3:3:1 ratio. 8. There are in F2 relatively too many plants phenotypically like the PI parents (wrinkled, no tendrils; round, tendrils) and relatively too few new, or recombinational types 95 (round, no tendrils; wrinkled, tendrils). 9. The non-allelic genes entering the cross in P1 are not inherited independently, but are inherited together in a package -- linked. 10. Only seven recombinational F2 plants are exceptions to complete linkage. P1 WRINKLf~N~ HNDRILS X ROU~D.]ENDRILS F1 E. wt ++ ROUN~ ~NDRllS wt P2 F1 ROUN~T~NDRILS) Sf IF -1[~'LlZm) wt r2 ROUND,TENDRILS ROUND, NO HNDRIlS WRINKL[D, HNDRILS WRINKLm, NOHNDRILS wt ++ 319 ? ? +t 4 ?t w+ 3 W? wt wt 123 Figure 15-2 D. Linkage in the sweet pea 1. This was the first example of linkage found, reported by Bateson and Punnett. 2. Purple flowered, long pollen pea plants were crossed to red flowered, round pollen plants. The genotypes are, respectively, ~ ~/~ ~ x .!:. ro/.!:. roo The F1 H.!:. .!:2) was self-fertilized. 3. As in C, the F2 contained relatively too many P1 phenotypes and too few new recombinational, non-parental types (purple, round; red, long). 4. While there was not independent segregation, linkage was not complete (as in C). 5. The frequency of recombinational F 2 progeny can be explained if the P2 plant forms (haploid) spores* in the relative proportions 10.:t. .:t.:10 .!:. ro:1 .:t. ro:1.!:. ±. 6. On this view, the parental chromosome types come out from meiosis 10 times as frequently as the recombinational types, the former being called non-crossover chromosomes, the latter crossover chromosomes. F. e 96 G. 7. The frequency of recombinational, crossover F2 progeny in C can be explained if linkage obtains in 63 of each 64 spores*. *Note that in higher plants the meiotic products are haploid spores which produce haploid gametophyte plants which form haploid gametes. This is essentially like the situation in most animals Where meiosis is followed directly by formation of haploid gametes. Crossingover as the exception to linkage 1. Linkage is the result of genes being carried in the same chromosome. 2. The failure of linkage to be comp~ete is explained by a process of crossingdver between homologous chromosomes (Chap. 16). 3. Bateson and Punnett did not accept the chromosome explanation of incomplete link:" age. a. They proposed that the excess of parental types of gametes in cases of linkage was due to a preferential multiplication of these gametes after meiosis. b. The P 1 cross can be made two ways with reference to the two recessives. When they enter the cross together, in thesame parent, they tend to be transmitted together (coupling), but if they enter separately they tend to remain apart when transmitted (repulsion). c. Coupling and repulsion are terms acceptable today even though the preferent~al reduplication hypothesis has proved incorrect. Mendel's chance of not discovering linkage 1. Assume the chance for a gene to be on any chromosome is equal. 2. In the garden pea recall that n is seven. 3. The chance, after the first gene, that the next six each will be on a different chromosome is 6/7·5/7·4/7' 3/7 ·2/7' 1/7, which equals approximately 1/163. 4. Thus, the chance for Mendel finding seven genes all showing independent segregation (or fo,r his not discovering linkage) is about 1 chance in 163. • Crossingover involving sex-linked genes 1. Morgan crossed a White-eyed fly (~ with one having miniature wings (gD. Both genes are X-linked. 2. The F1 female, having two Xs, was dihybrid @±/.:t..!!!). 3. The sons of this F1 female will receive an X from their mother and a Y, genetically empty with respect to the loci under consideration, from their father. . H. 1. ~, 4. So, examination of the sons of the Fl female will show directly what kind of an X they received. 5. If w and m were segregating independently, four types of males would be equally frequent, i. e., 50% of sons would have the old combinations @ ~ or ~ ~ and 50% recombinations 0: ~ or w ~. 6. But actually Morgan observed linkage, the recombinational sons comprising about 33% of all sons. 7. Morgan assumed correctly that the 33% of recombinations was due to crossingover whlch involved an actual physical exchange of corresponding segments by the two homologs . Linkage and crossing over in other organisms 1. Both are found in every organism adequately studied. 2. In man a. Colorblindness (£) and hemophilia ® are recessive X-linked genes (absent from the Y). b. Although rare, there are women with the genotype ~ .!!/£~, i. e., having one of these mutants on each X. c. Such mothers produce sons of which about 10% are recombinations. The proportion of old: new combinations in linkage cases 1. Garden pea (C) 63: 1 2. Sweet pea (D) 10: 1 3. Drosophila (G) 2: 1 9:1 4. ,Man (H2) 5. The variation observed is due to variation in crossingover, which is discussed in Chap. 16. ' POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 97 QUESTIONS FOR DISCUSSION 15. 1. Discuss the statement: "Linkage violates both of Mendel's laws. " 15. 11. In maize if a gene increases the rate of pollen-tube growth, how would you determine the linkage group to which it belongs? 15. 2. Suggest one or more ways to test the now disproven reduplication hypothesis of Bateson and Punnett. \ 15.12. In Drosophila why is X-linkage easier to study than autosomal linkage? 15. 3. Do you believe Mendel was lucky or unlucky in not discovering linkage? Explain. 15.13. Would you now require that Mendel's second 1aw be restated? Why? 15. 4. What was the chance for Mendel discovering linkage in the garden pea? 15. 5. For the cases discussed, in which two pairs of genes fail to segregate independently, the best explanation includes two new assumptions. What are these? 15. 6. Why are there question marks in the F2 in Fig. 15-2? By what could they be replaced? 15. 7. Can one detect linkage in homozygotes ? Why? 15. 8. In sweet peas a cr05S between a homozygous bright-flowered, tendril-leaved plant and a dull-flowered, acacia-leaved plant produced an F 1 which was all bright, tendril. The F2 from this cross was as follows: 424 bright, tendril 99 dull, tendril 102 bright, acacia 91 dull, acacia The cross of bright, acacia on dull, tendril also gave an F 1 which was all bright, tendril, but the F 2 in this case was as follows: 847 bright, tendril 298 dull, tendril 300 bright, acacia 49 dull, acacia Why do you consIder that the two loci are linked? What is the percentage of crossingover between these genes? 15. 9. What are the advantages and disadvantages of linkage? 15.10. Does linkage occur in asexually reproducing organisms? In such forms what is the consequence of the absence or presence of linkage? 98 '11~.14. , How would you prove that purple flower color and dark stem color in Datura, which ,occur together.. are due to a single gene rather than to two linked genes? 15.15. What do you suppose is meant by the ''law of the limitation of linkage groups"? 15. 16. From the .roatings described in D in the Notes draw a checkerboard showing the genotypes and frequencies of F 1 gametes and of F2 zygotes. 15.17. From the information in H2 in the Notes give the genotypes of mother and father and the expected frequenci.e,s of different geno': types among their sons and daughters. • 15.18. Assume that an individual homozygous for .::.:: is crossed with one homozygous for ~ Q and that the F2 from this cross is as fol.lows: 334 + +, 37 + b, 38 + a, and 87 a b. Is this result different from that which you would expect if segregation of ~ and Q were independent? If so, what is the recombination rate due to crossingover? 15.19. In Drosophila y_ and'!!.. are X-linked. A female genetically.:: .::/y y:!... produces 1. 5% of sons which carry either.:: y:!... or Give the relative frequencies of gametes which the mother produces. Is the father's genotype important? Why? y.::. Chapter 16 CROSSINGOVER IN TERMS OF MEIOSIS Lecturer-G. W. BEADLE PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings from textbooks: Altenburg: Chap. 10, pp. 186-202. Colin: Chap. 8, pp. 124-131. Dodson: Chap. 11, p. 132; Chap. 3, pp. 27-29. Goldschmidt: Chap. 6, pp. 122-125. Sinnott, Dunn, and Dobzhansky: Chap. 13, pp. 164-168, 176-178. Snyder and David: Chap. 11, pp. 141-154. Srb and Owen: Chap. 9, pp. 156-164. Stern: Chap. 4. Wmchester: Chap. 14, pp. 190-193. 3. Read the lecture notes through section C. LECTURE NOTES A. Complete linkage in Drosophila males (Fig. 16-1) 1. Assume two mutants, ~ and _Q_, are linked autosomally. 2. From the mating shown, the F1 are dihybrids. These are crossed to provide the F2· 3. The Figure assumes that the new, crossover, combinations are 20% of the total. 4. This means that the P 2 female produces eggs of which 80% carry parental combinations (40% 2: _Q_; 40% ~ 2) and 20% recombinations (10% .:!: 2:; 10% ~ _Q_). 5. Because of complete linkage the male produces only parental sperm, 2: _Q_ and ~ 2:, in equal frequency. No recombinations occur because there is no crossingover in the male. 6. The F2 genotypes, when collected, give phenotypes as follows: + +: a: b : a b = 10: 5: 5: 0, or 2 : 1 : 1 : o. 7. No double recessive F2 are produced because no ~ ~ sperm are formed. a+ a+ P2 0 t ~1 ~ ~ ~ x x F1 + b d' + b ~~ d GENOTYPES SPERMS + b (1) a + (1) ,F2 ++(1) EGGS +b(4) ! ~ (1) ! g(4) ; ~ (1) ; ~ (4) a+(4-) ~; (4) ~! (4) ab(1) ~ ~ (1) g~ (1) Figure 16-1 8. No matter what may be the actual recombination rate from crosslllgover. the ratio 2:1:1:0 will be obtained for autosomal genes if only one sex undergoes crossingover! 9. Linkage is easy to detect under these conditions; genes are not linked and linked when results fit the 9:3:3:1 and 2:1:1:0 ratios, respectively. 10. In animals, in general, the heterogametic sex has reduced or no crossingover. 11. Complete linkage in male Drosophila is useful in studying the genetic contribution of whole chromosomes to quantitative characters (like DDT resistance, in Chap. 8). B. Variation in the strength of linkage 99 1. This has already been shown to occur in dif- exchange of exactly equivalent sections between two non-homologous strands of a tetrad. 3. Shows the diads present after the first meiotic division. The top nucleus contains one .± .± non-crossover strand and one .± Q_ crossover strand, the bottom nucleus contains the crossover strand a + and the non-crossover ferent organisms (Chap. 15). 2. It also occurs for different genes in the same organism. 3. Fig. 16-2 gives the results for different sex-linked genes in Drosophila. MOTHER ~ + + SC 8 + + CV %CROSSOVER + + ct ~ ~ abo CHROMOSOMES + -- o 13 20 + m 34 + + f 48 ~ ' 1 Figure 16-2 C. 100 a. The mother's genotype and the frequency of crossover combinalib'lS in her sons are shown. b. The recombination rates are given between the gene for yellow body color and other genes (scute bristles absent, crossveinless wings, cut wings, miniature wings, and forked bristles). c. Linkage between y_ and sc is complete, since no recombinants occur. d. A female dihybrid for y_ and cv produces 13 eggs of each 100 which carry crossover recombinants (:!:.± or y_ ~. 4. What do these different linkage values mean in terms of meiosis? Chiasma and its theoretical genetic consequences (Fig. 16-3) Schematically, meiosis is shown to produce four haploid products. 1. Shows a pair of homologous chromosomes; one member (broken line) carries the recessives .§: and Q_, the other (unbroken line) carries their normal alleles. The black dots represent centromeres. 2. Shows the appearance after synapsis and chromosome replication (except of the centromere), where a chiasma has formed between the two non-alleles. It is supposed that a chiasma is the result of a physical 4 Figure 16-3 D. 4. Shows the four haploid nuclei produced after the centro meres replicate and the second meiotic division is completed. a. Note that following one chiasma two gametes are produced containing non-crossover, parental chromosomes while the two others contain crossover, recombinational chromosomes. b. Proof of this is ordinarily difficult to obtain either because only one meiotic product is functional (in females the others often form polar body nuclei) or because the four products are not identifiable (in animals the four sperm produced following a given chiasma separate and mix with other sperm). Proof of 4-strand stage crossingover 1. That crossingover involves two of the four strands in a tetrad cannot be proven when only one product of meiosis is observed at a time. 2. Bridges obtained favorable evidence using Drosophila with attached-Xs whose eggs receive two of the four meiotic products at one time. E. 3. This is easily proven in Neurospora where all of the meiotic products of the same mother cell can be recovered and tested. Meiosis in Neurospora (Fig. 16-4) 1. Neurospora (meaning nerve spore) is a red bread mold. 2. The organism as observed is haploid and comes in-two different sexes. iP TWO HAPLOID NUCLEI F. directly. 9. This enables one to follow all the products of meiosis from a single diploid nucleus. Chiasma and crossingover in Neurospora (Fig. 16-5) 1. Using the same symbols as in Fig. 16-3, a chiasma is shown in the one of the seven pairs of chromosomes represented. 2. Fig. 16-5 is correct, showing the linear arrangement of nuclei vertically instead of horizontally as in Fig. 16-4. + + DIPLOID NUCLEUS -·a--b"- ~ DIPLOID NUCLEUS + + ~+ -6.' ............ l DIPLOID NUCLEUS 0) ~ rlRST MflOTIC DIVISION :& o:;v S~COND M[IOTIC DIVISION MITOTIC DIVISION & SPORE FORMATION Figure 16-4 3. In the sexual process fruiting bodies are formed composed of cells each containing two haploid nuclei, derived originally from one nucl eus from each parent. 4. 'f,wo such haploid nuclei fuse to form a diploid nucleus containing seven pairs of chromosomes, and the cell elongates to form a sac. 5'. The diploid nucleus immediately undergoes meiosis, in the manner shown, so that at completion the four haploid products are arranged in tandem, i. e., the rightmost two nuclei come from one first division nucleus, the leftmost two from the other first division nucleus. 6. Each haploid nucleus then divides once mitotically, in tandem, so that each meiotic product is in duplicate. 7. Each nucleus becomes encased in a football-shaped spore contained in the sac. S. Each haploid spore can be removed, grown individually, and its genotype determined -a DIPLONEMA 0' + + =--=---- __]1[~____ __i__ AnfR FIRST DIVISION a b + + -4-~- rOUR M[IOTIC PRODUCTS - • -8 --'b""" -+----+ + + + tIG~T SPORES Figure 16-5 3. The four meiotic products contain two noncrossovers and two crossovers. 4. Note that if crossingover occurred in the 2-strand stage (in the topmost nucleus), all meiotic products would be recombinants. 5. Crossingover between genes located close together regularly produces two recombinants and two parental types, proving crossingover occurs in the 4-strand stage. 6. Suppose that in SO% of spore sacs no chiasma occurs in the genetically marked region. This would produce SO% of spores of parental genotype. From the 20% of spore sacs with such a chiasma one would obtain 10% spores that were recombinational and 10% parental. So, 20% chiasma frequency would produce 10% recombination. 7. The distance between loci a and b would be 101 G. H. 1. 102 10 map units. A map unit is that distance which gives one crossover per hundred. S. For the simple case, crossover frequency is just one-half chiasma frequency. Ways to measure crossingover frequency in Neurospora Assume 20% is the chance for a chiasma in the region tested. 1. Determine from a large number of spore sacs how many sacs have crossover spores and how many do not. 20% of sacs would have crossovers, SO% would not. Since each sac containing crossovers has four spores that are crossovers and four that are not, crossingover frequency would be 10%. 2. All the spores from many sacs are mixed, then a random sample is taken. This also would give 10% recombmation. This is like the procedure for testing crossover frequency in human sperm. 3. If from each sac one spore is taken and the rest discarded, 10% recombmation would still be obtained. This is like the situation in the human female where one random product of a meiosis enters the egg, the others bemg lost. ' 4. Note, for statistical purposes, that the sample size is number of sacs in G1 and number of spores m G2 and G3. Normality of chiasmata 1. Chiasma formation is a normal part of meiosis. 2. A chiasma prevents premature separation of chads, holding them together as a tetrad until anaphase 1. 3. Thus the consequence of the chiasma, the crossover, IS a normal part of meiosis. 4. There is usually at least one chiasma, and as many as six chiasmata, per tetrad. 5. The relation that the crossovers of one chiasma have to those of the next chiasma in the tetrad is chscussed in Chap. 17. Cytologlcal demonstration of crossingover 1. It is possible to obtam individuals with a pair of homologous chromosomes whose members dlffer from each other cytologically in two regions. It is also possible to have such individuals simultaneously dihybrid for genes located between the two cytolOgically dlfferent regions. 2. In this way it becomes possible to show that non-crossovers retain their original chromosome arrangement while crossovers have a new chromosome arrangement which re- sul ts from exchange of segments between the homologs. 3. In such a way, genetic crossovers were correlated exactly with cytological crossovers by Stern (1931) using Drosophila, and by Creighton and McClintock (1931) using maize. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible. 2. Review the reading assignment. 3. Be able to discuss or, define orally or in writing the items underlined in the lecture notes. • 4. Complete any additional assignment. QUESTIONS FOR DISC USSION 16. 1. What are the conditions under which a 2:1:1:0 ratio is obtamed? Are these conditions-suitable for measuring the strength of linkage? Explain. 16. 2. Using the information in the Notes and Fig. 16-1, indicate in detail your method for constructing a fly homozygous for the recessives a and b. 16. 3. To what use could the fly required in 16. 2 be put? 16. 4. Using Fig. 16-2 give the genotypes and their relative frequencies of sons produced by each of the mothers indicated. 16. 5. Explain the lack of recombination between the y_ and sc loci in Drosophila females. 16. 6. II! the following diagrams an F 1 female, from a cross of gray vestigial and black long, is backcrossed to a black vestigial male. What can you determine about the inheritance of body color and wing size from the results obtained? Select symbols for genes and fill in circles with genotypes which fit your explanation. 0/ \ 1)I( '±l1o 18% 91. 82"!o 103 16. 7. In the following diagrams an F 1 male, from a cross of black vestigial and gray long, is backcrossed to a black vestigial female. What can you determine about the inheritance of body color and wing size from the results obtained? \ Select symbols for genes and fill in circles with genotypes which fit your explanation. IO 0 :«/\ 10/\ 0/\ ~ /\ 0000 0000 \ I 9 \ 090\1' 00 . \XI o 0 44 50~ 50% 16. 8. What will be the genotypes and phenotypes of the daughters produced by the mothers in Fig. 16-2 if the father is phenotypically scute, cut, and forked? 16. 9. Using Drosophila with attached-Xs heterozygous for y and 1, mdicate the kinds of recombinants which would support the VIew that crossingover occurs ~n the 4-strand stage. 16.10. What does the study of crossingover tell us about meiosis? 16.11. Compare the fate of the zygotic nucleus in man and in Neurospora. 104 I \ .Q Q~ 0·0 00 \X/·· 00 4- 501. 5010 16.12. What advantage does Neurospora offer for the study of transmission genetics? 16.13. Compare the meaning of "crossingover" with that of "chiasma" and "crossover". 16.14. Why should it be true that if crossingover is between chromatids, the percentage of chiasmata in a gIVen region will be twice as great as the percentage of crossovers there? 16.15. In what percentage of the germ cells should a chiasma have been formed between two linked genes If the number of crossover units betweenthemis2? 4? 7? .3? 16.16. How can one tell that the last division in a Neurospora spore sac is mitotic? 16.17. In Neurospora how many spores from a sac need be tested before it is certain whether or not a crossover has occurred between two closely linked genes? 16.18. In the fowl assume that ~ (early feathering) and ~ (barring) are sex-linked and show 20% of crossingover (in the male only). If a male from a cross of late feathered, barred male x early, black female is mated with an early, black female, what will be the frequencies of phenotypes in their offspring with regard to feathering and barring? Predict the chromosomal constitutions of the gametes and the progeny of this plant produced by self-fertilization. 16.24. Invent chromosomal diagrams and gene symbols which could be used in correlating genetic and cytological crossingover. 16.25. Using the answer to 16.24, describe the procedure you would follow and the results you would have to obtain to prove conclusively the simultaneity of genetical and cytological crossingover. 16. 19. How would you determine whether characters which show no crossingover were due to alleles or to closely linked genes? In tomatoes Jones has found that tall vine Note: is dominant over dwarf and spherical fruit shape over pear. Vine height and fruit shape are linked, with a crossover percentage of 20%. 16.20. If a homozygous tall, pear-fruited tomato is crossed with a homozygous dwarf, spherical-fruited one, what will be the appearance of the F1? of the F1 crossed with a dwarf, pear? of the F2? 16.21. What genotypically different types will there be in the F 2 of the preceding cross? What offspring will each of these produce if selfed? I 16.22. A certain tall, spherical-fruited tomato plant crossed with a dwarf, pear-fruited one produces 81 tall, spherical; 79 dwarf, pear; 22 tall, pear; and 17 dwarf, spherical. Another tall, spherical plant crossed with a dwarf, pear produces 21 tall, pear; 18 dwarf, spherical; 5 tall, spherical; and 4 dwarf, pear. What are the genotypes of the two parental tall, spherical plants? If they were crossed to each other, what would their offspring be phenotypically? 16.23. In a species with 10 pairs of chromosomes, a plant is found in which one of the homologs of pair 9 has a knob while the other has none; "the members of pair 5 are also heteromorphic, one of them having a terminal satellite. 105 Chapter 17 CROSSINGOYER, CHIASMATA, AND GENETIC MAPS lecturer.:.-G. W. BEADLE wild-type ones. 3. Crossover frequency changes with ~ of female and with temperature. It depends also upon what other genes are present. 4. Even when genetic and environmental conditions are standardized, the exact standard map distances are not obtained experimentally for large distances, as will be explained. PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings from textbooks: Altenburg: Chap. 9, pp. 166-176, 178-180. Colin: Chap. 8, pp. 131-137. Dodson: Chap. 11, pp. 132-142. Goldschmidt: Chap. 6, pp. 125-129. SInnott, Dunn, and Dobzhansky: Chap. 13, pp. 168-176. Snyder and David: Chap. 12, pp. 155-165, 169-171. Srb and Owen: Chap. 9, pp. 164-176. Stern: Chap. 15. Winchester: Chap. 14, pp. 193-201. 3. Read the lecture notes through section C. LECTURE NOTES A. The standard genetic map 1. Using crossover frequency as a measure of distance between genes, it is possible to place linked genes in a linear sequence. 2. Such a crossover map for the Drosophila X chromosome would have the genes y_, cv, ct, m, andl lined up at the respective positions: 0, 13, 21, 36, and 52. 3. From this standard map ct and mare 15 map units apart ( 36 - 21). . 4. Since one map unit equals one crossover per hundred gametes, the dihybrid for ct and m (Fig. 17-1) should produce 15% crossovers (7.5% 2:.2: and 7.5% ct m). 5. However, this result is obtained only when certain conditions are met. B. Factors influencing crossover rates 1. The larger the size of sample the closer the observed value approaches the standard map distance. 2. Different phenotypic classes may show differential viability. For example, phenotypically ct m males are not as viable as 106 +m C? X ct + SONS~ +m 42.5% ct+ 42.5%' ++ 7.5% '7.5%, ctm Figure 17-1 C. ANY d Crossingover and single chiasma in a chromosome model (Fig. 17 -2) 1. Assume that a chromosome has five equal segments marked by six genes and that each tetrad has one chiasma which occurs at random among the segments. 2. Therefore, the chance the chiasma lies between _§: and Q is 20%. 3. Only _§: and 12. are followed in the F 1 hybrid shown. a. From 25 tetrads 100 meiotic products will be produced. b. Of each 25 tetrads 20%, or 5 tetrads, will have the chiasma in the _§:-Q region. + + + -r a b c d I a 20 + + a b 5 + + :=x::: a b + + b + + a f b + + a PRODUCTS e + + F1 ~YBRID T~TRADS + -I- b a 40% 40% b + + 57.. a b 57.. + b 570 a + 5% TOTAL O~ ALL PRODUCTS(IN7a) + + a b + b a + 45% 4570 570 5% Figure 17-2 D. These wIll produce 10 crossover and 10 non-crossover strands. The latter added to the 80 non-crossover strands from the other 20 tetrads give 90 non-crossovers. c. So 20% chiasmata gives 10% crossovers, as explained before (Chap. 16). 4. Following now only the ~-.2_ region, 40% of the time the chiasma will fall there, and 20% of gametes will be crossovers between a and c. (The rule for finding the probability of mutually exclUSIVe events is the sum of their separate probabilities. ) 5. Thus, for example, the chance of a chiasma between ~ andl is 100 (20+20+20+20+20)%, recombination is 50%, and the number of map units 50. 6. But a tetrad usually has more than one chiasma. 7. How will a second chiasma be related to the first? Positions possible for two chiasmata 1. Consider region ~-.2_ in the model, and suppose in the tetrad stage the four strands are numbered 1, 2, 3, 4, where 1 and 2 are +++, and 3 and 4 are abc. 2. If one chiasma occurs between strands 2 and 3 in the ~-Q region, there are six combinations of strands possible for a second chiasma in the Q-.2_ region, namely, 1 with E. 2, 3 with 4, 2 with 3, 2 with 4, 1 with 3, and 1 with 4. a. The first two combinations given involve sist~r strands which are genetically identical. Sister strand crossingover would have an effect only under other, very special, circumstances, so for our purposes these need not be considered further. b. The last four are non-sister combinations which actually produce genetic recombination. 3. What are the crossover consequences when each of these four non-sister types of chiasma in the Q-.2_ regIon is present with one in the ~-Q regIon? Crossovers consequent to double chiasmata (Fig. 17-3) 1. Diagrams showing the four non-sister types of double chiasmata are shown at the left in the Figure from top to bottom, in the order mentioned in D2. 2. Also, the correct genotypes of their meiotic products are shown in the middle column. + + + 1++ + --~,,\,-,~ a.'bl\e ]A_"t2' ~£= + + + + b + ~ .... ---~--- 2-STRAND ...abc ------- + + + +++ 1±\ + 2 DOUBLES a + c.. -- 2 NON-CROSSOVfRS ....±..l2__<;_ 1DOUBLE a'rt;.rc a + c 2 SINGLES l a"6-:~-c- ...------3-STRAND ...~t.Q_±.. 1NON-CROSSOVm + _ _ _ :I _ _ _ + + + I+,;,N + a .. b '•• c. 1ft1{:~= 3-STRAND • + + __<;,....±J?_-±... 1DOUBLE .. 9... + + 2SINGL[S .....9_9__<;,- 1NON{ROSSOV[R .+ + -~+ b c -+------- ... 51 + + 4-STRAND 4 SINGL~S ......a_ P....±... Figure 17-3 3. The top left diagram shows a 2-strand double chiasmata, in which the same two strands are involved in both chiasmata. 4. The next two diagrams below show the two kinds of 3-strand doubles, while the bottom 107 F. G. H. 108 one shows a 4-strand double (involving all strands). 5. The last column classifies the meiotic products of double chiasmata according to whether the strands are, for the ~-~ region, noncrossovers, single crossovers, or double crossovers. 6. Notice that a double crossover has the middle gene switched; single crossovers have one end gene switched. 7. What types of double chiasmata actually occur, and in what relative frequencies? Frequency of double chiasmata types 1. The occurrence of all four types with equal frequency would mean that the strands forming one chiasma are uninfluenced by those which form another adjacent chiasma. 2. Information on this can be obtamed by studying crossovers, since these differ following 2-, 3-, and 4-strand double chiasmata. 3. This has been studied using Neurospora where all products of meiosis are available for analysIs. 4. The Neurospora work shows all four types do occur. Since, in some experiments, the four types occur with equal frequency, we shall assume, for our p'lrposes, that this al so is proven. 1. Recombination rate between end genes 1. What is the recombination rate between ~ and c in Fig. 17-3? 2. Eight of 16 equally-frequent meiotic products shift one end gene. 3. So, even if every tetrad had two chiasmata between a and c there would be 50% recombination,-but n;t more, for ~ and ~. 4. If four genes were studied and three chiasJ. mata occurred in each tetrad, one in each region, it would turn out that, of the 64 meiotic products, 32 would be crossovers for the end genes and 32 would not. 5. In other words, 50% is the maximum amount of recombination for end genes (see also J). Interference in chiasma formation 1. Does the occurrence of one chiasma influence the probability of another one occurring (regardless which strands are involved) ? 2. If, in Fig. 17-3, .3 is the probability of a chiasma between ~ and Q or between Q and c, the probability of chiasmata occurring ;imul taneously both places is . 3 x. 3, or .09, if these events are independent. (The probability for the simultaneous occurrence of two independent events is equal to the product of their separate probabilities. ) 3. Suppose an experiment actually yields not the expected 9% double chiasmata but 4.5%. This would indicate interference of one chiasma on the formation' of another. 4. The d~gree of interference is expressed by the fraction double chiasmata observed = .045 = .5. double chiasmata expected . 09 This is called the coefficient of coincidence. A coefficient of coincidence of 0 would mean one chiasma completely prevented the other one from occurring. 5. The same coefficient of coincidence can be obtained from the crossover values: double crossovers observed = .01125 double crossovers expected .5* *The double crossovers expected are. 15 x .15, or .0225. (Since each region has. 3 chance for a chiasma, each has . 15 chance for a single crossover.) The frequency of observed double chiasmata is .045. Since only 4 of the 16 products from double chiasmata are double crossovers (Fig. 17-3), the observed double crossovers will be .045/4, or .01125. Map distance and coefficient of coincidence in Drosophila 1. For distances between genes up to 10-15 map units, the coefficient is O. This meap.s that no double chiasmata (hence no double crossovers) occur withm such distances. 2. As the distance increases above 15 map units the coefficient increases gradually to 1, at which time there is no interference. Map distance and recombination 1. For a single, double, or ~riple chiasmata one can get no more than 50% recombination for the end genes (see G). 2. When there are more than three chiasmata per tetrad there is still only a maximum of 50% recombination for the end genes. a. One can work out the fact that, for cases where there are four or more chiasmata between end genes, the number of crossover strands obtained bearing odd numbers of crossovers (1, 3, 5, etc.) is 50%. These will shift one end gene relative to the other producing 50% recombination. b. However, the remaining strands contain either even numbers of crossovers (which do not cause end genes to shift relative to each other) or no crossovers. 3. This explains ' K. a. why the crossover rates observed for large distances are less than standard map distance (reread B4). b. why standard linkage maps are constructed by summation of short distances within which only a single chiasma can occur. 4. Map length will always be the mean number of chiasmata per tetrad x 50. Determination of gene order 1. Given a trihybrid as shown on p. 3 one can determine gene order from the results of a test cross. 2. The frequencies of phenotypes obtained from such a cross are indicated in the photograph. These represent the frequencies of corresponding genotypes in the gametes produced by the trihybrid. 3. The middle gene would be the one which switches least often from the original gene order, for only it requires two chiasmata to switch. 4. This is gene .Q_ , so the actual gene order is .!!: .£ 2_ (or 2_ .£ .!!:). POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lec- ture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 109 QUESTIONS FOR DISCUSSION 17. 1. What is the map distance between the follOwing genes? y and cv ct and f y andJ m and f cv and f ct and m 17. 2. Which would require a larger sample of tested gametes in order to obtain equal reliabilIty in map dlstance, genes moderately close or very close together? Explain. 17. 3. In Fig. 17-1 what percentage of daughters look hke their father if the father is wildtype? miniature? cut? cut mimature? 17. 4. In one experiment the crossover rate between two genes was 8% while in another it was 11%. List the possible causes for this difference. 17. 5. Using Fig. 17-2, calculate the expected chance of a a. single chiasm:.!. between i! and 2. or between e and f. b. double chiasmata m the ~-Q region. c. double chiasmata in the ~-2. region. d. triple chiasmata in the i!-Q region. 17. 6. What two relations between adjacent chiasmata were discussed? Give the conclusions reached m each case. 17. 7. How many gene pairs are needed to detect single crossovers? double crossovers? Explain. 17. 8. Draw an ascus (spore sac) of Neurospora and the eight spores it contains for each of the following cases. Using as a parental genotype ~ :!=i::!:: 2., fill in genotypes for the spores which would demonstrate a. the last division was clearly mitotic. b. only non-crossovers. c. a crossover between a and b. d. a crossover between the centromere and the nearest of the marker genes ~ or 2.). 17. 9. Can you tell from the spores in a single ascus whether two pairs of genes are on the same or different chromosomes? Explain. 17.10. Why should the coefficient of coincidence be the same when calculated from chiasmata 110 data as from crossover data? \ 17. 11. What may be the order of the genes mentIoned if double chiasmata can occur between Q and ~ and between !'i and Q, but not between Q and Q? 17. 12. From the data in the photograph on p. 3 determine the coefficient of coincidence for the ~-2. region. 17. 13. What is the coefficient of coincidence if the observed double crossovers have a frequency of .05 and the double chiasmata were expected at a frequency of . 8 ? 17.14. Suggest a possible physical basis for interference over small distances. 17.15. Describe how you would go-about-demonstrating that genes are arranged linearly on a crossover map. 17.16. Would a Drosophila female trihybnd for three hnked genes spaced evenly over a distance of, 30 map units give a linear crossover map for these genes when testcrossed? Explain. 17.17. In Chinese primroses short style is dominant over long, magenta flower over red, and green stigma over red. When from the cross of homozygous short, magenta flower, green stigma by long, red flower, red stigma, the y~ was crossed with long, red flower, red stigma, the following \ offspring were obtained: Style Flower Stigma Long Red Red 1,032 Short Magenta Green 1,063 Short Magenta Red 634 Long Red Green 526 Short Red Red· 156 Long Magenta Green 180 Short Red Green 39 Long Magenta Red 54 Map the chromosome in which these genes lie. 17.18. A breeder of Chinese primroses has three plants, each of which has short styles, magenta flowers, and green stigmas. TIle offspring of each, when crossed on a triple re- cessive plant, are presented below, symbols being used instead of words (L short style, I long, R magenta flower, r red, S green stigma, s red). Plant 1 x Irs 290 LRs 151 LRS 288 IrS 147 Irs 37 LrS 20 Lrs 39 IRs 21 IRS Plant 2 x Irs 21Irs 19 LRS 37 IrS 40 LRs 289 LrS 150 Lrs 291lRs 148 IRS Plant 3 x Irs 221 LRS 218 IrS 57 LrS 60 lRS fruited plant. The" F 1 were backcrossed to dwarf yellow-fruited plants. The results were as follows: tall red-fruited 51 tall yellow-fruited 49 dwarf red-fruited 50 dwarf yellow-fruited 49 Are these factors linl(ed? Would you expect to find g linked with E? Explam. What are the genotypes of these 3 plants? Note: The three questions following deal with tomatoes. 17.19. The following 3 pairs of factors are linked and show complete dommance: Q, Q (tall, dwarf); E, 2. (smooth epidernns, pubescent epidermis); Q, Q (oblate fruit shape, ovate fruit shape). The crossover value between D and P is 3.5%; between Q and Q is 17.4%, and between R. and Q is 13.9%. In what order do these genes occur on the chromosome? 17.20., The genes ~, ~ (simple inflorescence, compound inflorescence) are found to be linked with Q, Q (see 17. 19) showing 12.8% crossingover. Where could you place ~ on the map? What further data would you need in order to place it accurately? The crossover percent between ~ and Q is found to be 24.6. With this information where would you place ~ on the map? The sum of the crossover percents between Q and Q, and Q and~, is 30.2%. Yet the crossover percent actually found between D and ~ is only 24. 6%. How do you account for this? Approximately what crossover percent would you predict between ~ and R.? 17.21. Another pair of factors consists of g, a factor for red fruit color, and I_, its recessive allele for yellow fruit color. A homozygous tall plant with yellow fruits was crossed to a homozygous dwarf red- 111 Chapter 18 CHANGES IN GENOME NUMBER Lecturer-H. 1. MUl.LER PRE-LEC TURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 16, pp. 274-283. Colin: Chap. 12, pp. 236-241; Chap. 14, pp. 300-301. Dodson: Chap. 14, pp. 172-174. Goldschmidt: Chap. 7, pp. 140-142. Sinnott, Dunn, and Dobzhansky: Chap. 21, pp. 296-302. Snyder and David: Chap. 20, pp. 297302. Srb and Owen: Chap. 11, pp. 214-219, 229-230. Winchester: Chap. 15, pp. 213-217. b. Additional references Dobzhansky, Th. 1941. Chap. 7 in: Genetics and the origin of species. 2nd Ed. New York: Columbia University Press. C. D. LECTURE NOTES A. The study of transmission genetics is based upon the existence of heritable differences. How do these differences originate? B. Origin of hereditary differences 1. The genetic differences found today were not always present. 2. Some hereditary variation is provided by new combinations of old genes via independent segregation and crossingover. 3. Before Johannsen's work it was not possible to readily recognize truly new hereditary types. 4. However, as in Johannsen's work with beans, once homozygous lines were studied, sudden phenotypic changes were occasionally found which represented the origination of really new genetic materials. 112 E. 5. Such really new, novel, hereditary types were called mutants by DeVries, these having arisen by the process of mutation. There are four general categories of mutations: . 1. changes in the number of whole sets of chromosomes (genomes), or ploidy; 2. changes involving addition or subtraction of singl e" whol e chromosomes; 3. structural changes involving parts of one or more chromosomes; 4. gene mutation. 5. This chapter and the next five discuss these changes in the order given, from the largest al terations of the genetic material to the smallest one. 6. Paradoxically, the smaller the genetic change the more important it is apt to be in the end, for nature usually proceeds better by small steps. . 7. Note that a given mutant may not always clearly belong to one category even when the boundaries between categories are sharp. Ploidy 1. Most organisms are diploid, having two genames - one derived from the mother and one from the father, whereas gametes are haploid, having one genome. 2. In Oenothera, DeVries' ~ type was shown to have three genomes - to be triploid. 3. Some diploid plants have relatives that are tetraploid, i. e., have four genomes. 4. Other plants have six sets (hexaplOidy) or eight (octaploidy). Ploidy in Datura 1. This Jimson weed was studied by Blakeslee and Belling. 2. In Fig. 18-1, line B shows, from left to right, haploid, diploid, triploid, and tetraploid flowers; their seed capsules are shown above, in line A. 3. While flower size increases with ploidy, the H. fertilization. 3. Incomplete mitosis in haploids can form completely homozygous diploid cells. Ploidy in animal.s 1. Triploid and diploid Drosophila are compared (Fig. 18-2) (see also Chap. 14). B Figure 18-1 triploid capsule is smaller than the diploid because it does not set as many seeds. Tetraploids also set fewer seeds than diploids. 4. If each chromosome is to have a partner at meiosis, genome number must be even, not odd. a. Triploids produce gametes which usually have, in addition to one genome, some, but not al.l, chromosomes of another set. Offspring from such gametes have an overdose of some genes and are so unbalanced physiologically that often they die during development. Hence, many fewer seeds are produced by triploids than by diploids. b. Yet in tetraploids, where set number is even, the four chromosomes of a kind sometimes segregate abnormally (3 and 1) so that defective seeds are produced. F. How ploidy is increased (polyploidy) 1. AutopolyPloidy refers to addition of genomes of the same kind, as discussed in Datura. '2. Amphiploidy, or allopolyploidy (see Chap. 29) refers to combinations of two, or more, ,d iploid genomes derived from different species, as in cultivated wheat. As expected, amphiploids often combine characteristics of their different parent species. G. Haploids 1. These usually do not produce gametes containing a whole genome. 2. Occasionally a haploid gamete is formed, which produces the diploid condition after Figure 18-2 2. Bridges found tetraploid females; these breed more normally than triploid ones. 3. Tetraploids cannot establish a race. 1. a. Since a "tetraploid" male would have 2Xs and 2Ys, usually the XS would synapse with each other as would the Ys, so that each sperm would carry IX and 1 Y. b. Fertilization of an egg carrying 2Xs by such sperm would produce zygotes with 3Xs and a Y. c. These zygotes would produce intersexes. 4. This difficulty probably explains why polyploidy is rare among animals which have two different kinds of homologous sex chromosomes. 5. Polyploid larvae of salamanders and frogs have been obtained by Fankhauser. 6. Polyploid races exist of the water shrimp, Artemia, and of the moth, Solenobia, in which femal es are parthenogenetic. Here meiosis produces haploid eggs which start development as haploids, then nuclei fuse in pairs to reestablish the diploid (female) condition. Mechanisms for producing autopolyploidy 1. Failure of mitotic anaphase can be followed by formatIon of a nucleus with genome number doubled. 113 2. Fusion of haploid nuclei produced by meiosis can resul t in a diploid gamete which forms a triploid zygote. 3. Artificial induction of parthenogenesis initiates haploid development. Such development is abnormal probably due to the change in surface-volume reI ation haploidy produces. If, however, chromosome doubling occurs early in development so that diploid cells are produced, a normal diploid embryo may resul t. Such chromosome doubling has produced parthenogenetic salamanders (in Fankhauser's work) and parthenogenetic rabbits (in work of Pincus done more than 20 years ago), which are female. . 4. Drugs like colchicine and environmental stresses like starvation (as Muller found in Drosophila) can artificially induce polyploidy by interfering with mitosis. 5. Polyploidy occurs normally in certain somatic cells, as in human liver or in the larval salivary gland of Drosophila. POST- LEC TURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items uDde ri ;.led in the lecture notes. 4. Complete any additional assignment. 114 QUESTIONS FOR DISCUSSION 18. 1. Distinguish between hereditary variation and mutation. 18. 2. Discuss the adequacy of the hypothesis of special creation of novel hereditary types. 18. 3. It has been suggested that so-called mutations are really the resul ts of segregation from remote hybrid ancestry. Of what significance in this question is the fact that such variations are al so found among the offspring of diploids which have had their origin through the (rarely occurring) self-fertilization of haploids? 18.14. Name at least one important practical resul t which may be expected to follow from our ability to double the number of chromosomes in a plant. 18. 15. Why cannot a sexually-reproducing haploid race ordinarily be established? 18.16. What would be the chromosome constitution of tetraploid males and females in Drosophila? What would be the consequence of breeding these to normal females and males? to each other? 18. 4. What contributions did Johannsen make to the study of mutation? 18.17. How does Muller explain the scarcity of polyploid series in animals? 18. 5. How can mutations be detected in asexually reproducing organisms? 18.18. Describe three mechanisms by which autopolyploidy is produced. 18. 6. How can one determine whether the appearance of a new phenotype is due to crossingover, to independent segregation of chromosomes, or to mutation? 18.19. What would be the expected sex of parthenogenetically produced mammals? birds? Why? 18. 7. Which category of mutation has the most far-reaching consequences? Explain. 18.20. How could one distinguish a diploid individual from an autopolyploid derived from the same strain, cytologically? genetically? 18. 8. Why are the mutations most easily detected those which are probably least important in nature? 18. 9. To what category of mutation would an individual belong that possessed two genomes, one from each of two different diploid species? Have you any additional conclusions? 18.10. In Datura, as compared with diploids, in what way are tetraploids abnormal although less so than triploids? Explain why this is so. 18.11. Can a triploid be an amphiploid? Can this be true for a pentaploid? Explain. 18. 12. Are allopolyploids or amphiploids more likely to be reproductively successful ? Explain. 18. 13. What are the advantages and disadvantages of changes in ploidy? 115 Chapter 19 CHROMOSOME ADDITION AND SUBTRACTION Lecturer-H. J. MULLER PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 15, pp. 258-264; Chap. 16, pp. 283-289. Colin: Chap. 12, pp. 241-246. Dodson: Chap. 12, pp. 149-154; Chap. 14, pp. 168-172. Goldschmidt: Chap. 12, pp. 197-200. Sinnott, Dunn, and Dobzhansky: Chap. 14, pp. 189-192; Chap. 15, pp. 195-198. Snyder and David: C:!:tap. 41, pp. 309313; Chap. 20, pp. 294-296. Srb and Owen: Chap. 10, pp. 184-185; Chap. 11, pp. 207-214. Winchester : Chap. 16, pp. 218-223; Chap. 15, pp. 210-213. b. Additional references Blakeslee, A. F. 1934. New Jimson B. weeds from old chromosomes. J. Hered., 25: 80-108. LECTURE NOTES A. Polyploidy in somatic cells 1. Larval salivary glands of Drosophila a. In ordinary cells at metaphase the chromosomes are sausage-shaped, containing threads coiled tight in a series of spirals like a lamp filament. b. In salivary cells chromosomes are in interphase condition, largely uncoiled. Their thickness is due to two factors. Each chromosome has doubled (as many as) nine times, producing 512 chromosome threads, and all strands, instead of separating, remain in contact with homologous loci apposed. This appears as a many threaded, polytene, cable. The members of a homologous pair of chro- 116 C. D. mosomes also plllr at homologous points, by what is called somatic synapsis, to form a double cable with 1024 strands. c. Differences in stainability along the length of the chromosome fiber give the salivary "cables" a cross-banded appearance. d. Within and between chromosomes the banding pattern is so characteristic that it is useful in correlating genetical and cytological results. e. Painter and Heitz, independently, were the first to recognize these structures as chromosomes. 2. Giant human and other mammalian cells, thousands of times polyploid, were produced by Puck by irradiation of cell cultures, While such cells no longer undergo cell division, their chromosomes can divide· repeatedly. Polyploidy and evolution 1. Polyploidy is not often established in animal evolution (see Chap. 18). 2. While many plants are polyploid, this type of mutation cannot and does not occur many times in succession, for chromosome numbers would become unwieldy. 3. Other types of mutations have difficul ty expressing themsel ves in polyploids. Euploid and aneuploid changes 1. Changes in genome number preserve the same ratios that genes or chromosomes have to each other in the normal diploid. Such changes are euploid, or right-fold. 2. Addition or subtraction of single, whole chromosomes upsets this normal relationship to produce aneuploid, or not-right-fold, genetic constitutions. These mutations comprise the next category to be discussed. Non-disjunction-produced aneuploidy 1. E. B. Wilson found a specimen of the bug E. Metapodius which was diploid except for one chromosome present thrice. He exrlained this as a mitotic mistake in which a daughter chromosome went into the same cell as the other daughter chromosome daughter chromosomes having undergone non-disjunction. 2. Bridges then found, genetically, parallel cases in Drosophila involving non-disjunction of X and Y (see Chap. 12). Recall that in humans, unlike Drosophila, XXY individuals are intersexes with Klinefelterrs syndrome. Haplo-IV and triplo-IV Drosophila (Figs. 19-:4 19-2) F. 1. These were discovered by Bridges. 2. Addition or subtraction of a IV chromosome made visible phenotypic changes from the normal diploid condition, but was viable. 3. He soon found that adding or subtracting one of the larger autosomes to diploids was lethal before adulthood. Trisomies in Datura (Fig. 19-3) Normal ROiled Elongate M IcrocarplC Glossy Buckling Ech in us Cocklebur Reduced • POinsettia Figure 19-3 Normal Figure 19-1 G. H. Q b Tnplo-TIl Haplo-N Figure 19-2 1. Blakeslee and Belling obtained 12 different abnormal types of seed capsule, each produced by a different chromosome, of the 12 that make up a haploid set, when present in addition to the normal two. 2. Each of these is a trisomic (as is also the Drosophila triplo-IV), and each was given a separate name. 3. Diploids with one chromosome missing are monosomics or haplosomics, those with two extra chromosomes of the same type are tetrasomics. Mongolian idiocy in man. is due to the presence in trisomic condition of the smallest chromosome except the Y. Tetraploids and tetrasomics (Fig. 19-4) 1. Datura seed capsules for 2N +2 (tetrasomic) individuals depart from normal (2N) even more than do 2N+1 (HOlobe n trisomic) individuals. 2. Thus even though a chromosome when te- 117 trasomic is genetically more stable than when trisomic, the tetrasomic phenotype is too abnormal to establish a race. dance, that these mutations all involve one or more chromosome breakages. Consequences of a singl e chromosome break (Fig. 19-5) L. I .] 2N+l ,C,LOBE1 2N+2 4N-+ 1 4Nt2 • 4N+3 Figure 19,-4 1. J. K. 118 5 4 3. The tetraploid which has this same chromosome extra once (maldng it a pentasomic tetraploid) deviates from the tetraploid in the same di.rection as does 2N + 1 from 2N, but less extremely. 4. The deviation of hexasomic tetraploids from 4N is similar to that of 2N + 1 from 2N. 5. Polyploids can stand whole chromosome additions or subtraction better than can diploids. Induction of non-disjunction 1. Mohr working with the locust, Decticus, found radium induces a high frequency of non-disjunction (also called mis-division or mitotic dislocation). This was extended by Mavor, at Bridges' suggestion, to Drosophila Xs. 2. Patterson found carbon dioxide increases non-disjunction in Drosophila. 3. Older women are more apt to have mongolian idiot children, probably because of some metabolic defect that occurs with age. Addition or subtraction of whole chromosomes invol ves too drastic a change to be very useful in evolution. Structural changes in chromosomes 1. TIllS is the third category of mutations, involving aneuploidy of part of one or more chron10somes in some cases. 2. It was found, especi31ly after X-radiation was shown to produce them in great abun- Figure 19-5 1. Shows the normal chromosome. 2. 3. 4. 5. The cen-. tromere is indicated by a black spot. Shows the chromosome broken. Shows thte broken chromosome has reproduced two broken chromosomes. Broken ends are sticky, and can join in pairs. Note that the ends closest together are the corresponding ends of the sister (or motherdaughter) strands. . Shows that when corresponding broken ends of sister strands unite, 'one strand has no centromere, and is called acentric, .while the other has two, and is called dicentric. As the cell divides the acentric piece is lost, since it is not pulled to the poles, while the dicentric one is pulled to both poles, forming a bridge between the daughter nuclei. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. QUESTIONS FOR DISCUSSION 19. 1. Why does Muller not consider polyploidy a main source for the increase in genetic material which has occurred in the course of evolution? 19. 2. How did Muller use cake making as an analogy to explain the phenotypic consequences of euploid and aneuploid mutation? 19.15. What other alternatives for joining are possible following the single break produced in Fig. 19-5? What would be the genetic consequences of these alternatives? 19.16. In an essay of about 200 words describe the events depicted in the foHowing figure. 19. 3. What phenotypic differences exist between normal, haplo-IV, and triplo-IV Drosophila (Figs. 19-1, 19-2)? f 19. 4. Why do trisomies affect many different characters of an individual ? 19. 5. How many chromosome types of gametes are produced by monosomics ? trisomies? How many zygotes are possible when each of these is crossed with a normal diploid? 19 ...6. In what respect is a tetrasomic more stable than a trisomic? Explain. 4 ) 19. 7. How can a tetrasomic diploid be distinguished cytologically from a doubly trisomic diploid? 19. 8. Why does one chromosome added to a diploid produce more phenotypic effect than does adding the same one to a tetraploid? 19. 9. How many extra chromosomes would be needed in 4N and 6N individlUlls to get a deviation approximately equal to that of 2N + 1 from 2N? Explain your answer. 19.10. What explains the variety of phenotypes shown by the trisomic seed capsules in Fig. 19-3? 19.11. Do you think a double trisomic is viable in man? Explain. 19.12. Explain how a trisomic and a monosomic may be formed simultaneously. 19 . 13. What factors do you think delayed the discovery of the basis for mongolian idiocy until 1959 ? 19.14. How would you define aneucentric? 119 Chapter 20 STRUCTURAL CHANGES IN CHROMOSOMES I lecturer-H; J. MULLER PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 14, pp. 232-236, 240244, 247-249. Colin: Chap. 12, pp. 230-231. Dodson: Chap. 12, pp. 144-149. Goldschmidt: Chap. 12, pp. 196-197. Sinnott, Dunn, and Dobzhansky: Chap. 15, pp. 198-209. Snyder and David: Chap. 21, pp. 304308, 314-318, 323. Srb and Owen: Chap. 10, pp. 185-198. Winchester: Chap. 15, pp. 204-209. b. Additional references Dobzhansky, Th. 1941. Chap. 4 in: Genetics and the origin of species. 2nd Ed. New York: Columbia University Press. Muller, H. J. 1950. Radiation damage to the genetic material. Amer. Scientist, 38: 33-59, 126, 399-425. LECTURE NOTES A. Chromosome breakage 1. Ends produced by breaking chromosomes are "sticky" and are able to join permanently in pairs. 2. Originally free ends of chromosomes are not sticky, having genes called telomeres that serve to seal them off. 3. Usually the ends produced by a break restitute, i. e., join to restore the original linear order of the chromosome. 4. The union of ends, produced by one or more breaks, to form a gene arrangement other than the original, is called non-restitutional or exchange union. 5. Exchange unions produce structural changes 120 B. C. D. in chromosomes. 6. Hereafter, only non-restituting breaks will be discussed in this chapter. Consequences of a single break (see Chap. 19 and Fig. 19-5) 1. A single break, followed by chromosome division, then by union between the two ' centric fragments, produces a dicentric chromosome. 2. The dicentric is pulled to both poles at once during anaphase, and may a. fail to enter either daughter nucleus. b. form a bridge between daughter nuclei. 3. The acentric pieces are lost whether or not they join together. 4. Both daughter cells may die a. after B2a, since each is deficient for a . whole chromosome. b. after B2b, due either to the loss of genes, or to the entanglement of the nuclei which frustrates further attempts at cell division. 5. Such aneuploid genomes can be transmitted to the next generation, in gametes of animals, where they may produce a dominant lethal effect. 6. An isochromosome is composed lengthwise of identical halves (like the dicentric mentioned). Two breaks in the same chromosome These can occur in two positions: 1. paracentric, where both breaks are to one side of the centromere, or 2. pericentric, where the centromere is between the breaks. The conseque·nces of these are discussed. Deficiency 1. Paracentric position The end pieces unite to produce a centric chromosome deficient for the acentric interstitial piece. The latter is lost whether E. or not its ends join to form a ring. Chromosomes with small deficiencies may act like recessive lethals; those with large ones usually act, in the next cell generation, as dominant lethals. 2. Pericentric position Here tlle end pieces are lost, being acentric, whether or not they join together. The centric middle piece can survive if its ends join to form a ring, and the deficient sections are not extensive. Paracentric inversion 1. Suppose a chromosome is linearly differentiated by genes in the order ABCDEFG. H, the period representing the position of the centromere. 2. If this chromosome is broken between A and B, and between F and G, and the middle piece undergoes exchange unions with the end pieces, the gene order becomes AFEDCBG.H. 3. The result is that, relative to the ends,. the interstitial segment is inverted, or vice versa. 4. Suppose, in an individual carrying the inverted chromosome and also a non-inverted homolog, that in the tetrad formed during meiosis a single crossingover occurs within the inverted region. 5. The result would be a. two normal strands (one inverted, one . not inverted), b. one acentric strand (with some regions duplicated, others deficient), c. one dicentric (with duplicated and deficient regions the reverse of the acentric's) . 6. In males of most species such crossovers occur and they produce defective sperm, this disadvantage leading to the elimination of the inversion in nature. 7. In Drosophila, however, there is no crossingover in the male. In the female, as found by Sturtevant and Beadle, the dicentric is formed. But the meiotic divisions occur in such a way that it is shunted away from the egg nucleus, which therefore receives one of the two eucentric, non-crossover strands. 8. Therefore, paracentric inversions when heterozygous are not disadvantageous in either sex of Drosophila and can become established in nature. 9. Very small paracentric inversions can survive in any species. F. G. H. Pericentric inversion 1. If the chromosome had its centromere between C and D (ABC. DEFGH) and breaks occurred, as before, between A and E, and between F and G, the middle section can become inverted to form a chromosome AFED.CBGH. 2. A single crossover within the inverted region, in an individual heterozygous for the rearrangement, produces four eucentric strands: a. two are non-crossovers, b. two are crossovers having duplicated and deficient sections. 3. Such crossovers cause trouble both in males, if crossingover occurs in the male, and females, since in this case there is no shunting of non-crossover strands into the haploid egg. 4. Very small inversions may survive, however. Two breaks in two chromosomes 1. which are homologous. If the breaks are in different positions, between C and D in one, and D and E :in the homolog, exchange union can produce eucentric chromosomes in which D is deficient in one and duplicated in the other. 2. which are non-homologous .. The consequences of this are discussed next . Translocation 1. Aneucentric type a. If the two centric pieces unite a dicentric is formed. The two acentric pieces are lost in the next division, whether or not they join. b. This type often acts as a dominant lethal in a subsequent division, particularly when the dicentric is pulled to both poles at once. 2. Eucentric type a. This is as likely to happen as the aneucentric type. b. Here the centric piece of one chromosome joins the acentric fragment of the non-homolog, and, vice versa, the centric piece of the second joins the acentric piece of the first. c. This is a reciprocal exchange of pieces by non-homologs. d. In translocation heterozygotes, gametes may be formed with deficiencies and duplications because segregation caused them to receive only one of the two I. J. translocation chromosomes. e. Because of this translocations tend to be eliminated from the population. 3. Oenothera (see Chap. 31) is commonly heterozygous for many translocations. Special mechanisms for chromosome disjunction assure production of euploid gametes. 4. Whole arm translocation a. These can occur when both chromosomes are broken close to their centromere. b. When an H4a translocation is heterozygous in Drosophila, and probably most other species, chromosome synapsis and disjunction are so regular that euploid gametes are usually formed. 5. Half-translocation a. In oocytes, and probably other cells, too, usually only one exchange union occurs (the other two ends failing to join) so that only half of a reciprocal translocation is produced. b. Usually, descendant cells either die or produce visible abnormality. c. Lejeune has ascribed a vertebral anomaly in humans to a half-translocation. Consequences of three breaks 1. Three breaks in one chromosome can cause the two interstitial pieces to ex~hanbe positions, resulting in what is called a shift. 2. Two breaks in one chromosome and one in a non-homolog can result in the interstitial piece of the first chromosome being inserted into the second. This result is called transposition. Increasing gene number following breakage 1. If a chromosome containing a shift crosses over in the region of the shift with a normal chromosome, one of the crossover strands has a section in duplicate. 2. In subsequent generations a chromosome containing a transposition may come to be present not with the non-homologous chromosome from which the piece was transposed, but with a normal one. 3. Suppose two breaks occur in a single chromosome. If joining is delayed until the chromosome reproduces, and then both interstitial pieces join so as to become inserted into the same daughter strand, duplication in situ is produced. 4. Only small duplications are likely to survive in nature. 5. The larger the number of genes, the more complicated the organism that can be formed. 122 K. Evidence for duplicated regions in Drosophila was 1. found by Muller in genetic studies. 2. independently discovered cytologically by Bridges, in repeats of banding pattern observed in salivary gland chromosomes. 3. In evolution small duplications are more important than changes involving one or more whole chromosomes in increasing the amount of genetic material. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. QUESTIONS FOR DISCUSSION 20. 1. Describe all possible chromosomal consequences following production of a single break. 20. 2. Define and give an example of a eutelomeric and all aneutelomeric chromosome. 20. 3. Why do you suppose restitutional union is more common than non-restitutional ? nucleus of the gamete. How does this information explain the freedom of eggs from dicentrics produced by crossingover in inversion heterozygotes ? 20.15. Why are m.ore paracentric than pericentric inversions found in wild Drosophila popul ations ? 20. 7. Why is a deficiency more likely to be lethal in males than in females in such an organism as Drosophila? 20.16. Suppose a cell contains the non-homologous chromosomes ABCDEF. GH and JKL. MN. Using two non-restituting breaks diagram the formation of a. a eucentric ring chromosome b. an aneucentric ring chromosome c. a deficient V-shaped chromosome d. a paracentric inversion e. a pericentric inversion f. a eucentric reciprocal translocation g. an aneucentric reciprocal translocation h. a eucentric half-translocation i. an aneucentric half-translocation 20. 8. What is the consequence of a single chiasma in a. a ring homozygote? b. an inversion homozygote? 20.17. Cytological examination of the maturation process in a plant reveals that some chromosomes are joined to form circles at diakinesis. How can this be explained? 20. 9. What is the consequence of a 2-strand double chiasma within the inversion of an inversion heterozygote? 20.18. Crossover suppressors were at first regarded as genes whose specific effect was to prevent crossingover. It was found, however, that they produced this effect only when heterozygous. Show how the modern conception of inversion as a cause of "crossover suppression" explains this fact. 20. 4. Is a cell killed by having undergone a mutation in which a 1 arge section of a chronl0some is deleted? Explain. 20. 5. Draw a diagram of synapsis in a tetrad containing homologs differing by a a. paracentric inversion. b. pericentric inversion. 20. 6. Do all chromosomes have telomeres? Explain. 20.10. Why are small deficiencies not transmitted by pollen while even large ones are in animal sperm? 20.11. Draw a diagram showing synapsis between a ring and a homologous non-ring chromosome in Drosophila. What happens following a single chiasma? 20.12. Is possession of a ring chromosome disadvantageous to Drosophila females? males? Explain. 20.19. In a species of the roundworm Ascaris, both the fertilized egg and cells of the germ line carry a small number of large chromosomes, while somatic cells have a large number of small chromosomes. What can be concluded from this with regard to centromeres and telomeres ? 20.13. Do inversions suppress crossingover? Explain. 20.20. Why can very small paracentric inversions survive in any species? 20.14. As a result of oogenesis in Drosophila, four haploid nuclei are produced, arranged linearly. One of the end nuclei becomes the 20.21. How does a shift differ from transposition? Can both occur in haploid cells? Why? 123 20.22. Diagram how a chromosome with a shift can produce, after crossingover, a chromosome with a duplication. 20.23. Diagram how an in situ duplication can occur after a chromosome is broken twice. 20.24. How can repeats within Drosophila chromosomes be recognized? 124 Chapter 21 STRUCTURAL CHANGES li N CHROMOSOMES II, , Lecturer-H. J. different from that in mel anogaster. 6. Such changes involve whole-arm translocation. PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 14, pp. 249-252. Colin: Chap. 12, pp. 224-225,. Dodson: Chap. 14, pp. 175-178. Sinnott, Dunn, and Dobzhansky: Chap. 15, pp. 209-212; Chap. 28, p. 383. Snyder and David: Chap. 21, pp. 318321. Srb and Owen: Chap. 10, pp. 198. 201- ~~ >j' >,<.. >;< 203. Winchester: Chap . 16, pp. 224-225; Chap. 15, pp. 209-210. b. Additional references Lewis, E. B. 1950. The phenomenon of position effect. Adv. in Genet., 3: ',L "r 75-115 . Muller, H. J. 1950, Radiation damage to the genetic material. Amer. Scientist, 38: 33-59, 126, 399-425. Muller, H. J.. 1958. General survey of mutational effects of radiation. Chap. 6, pp. 145 -177 in: Radiation biology and medicine. Ed. W. D. Claus. Reading, Mass. : Addison-Wesley Publ. Co. LECTURE NOTES A. Chromosome configuration in Drosophila species (Fig. 21-1) 1. Studied by Metz and others. 2. Haploid configurations are shown. 3, The chromosomes of the melanogaster species group are shown in row 2, column 1. 4. Row 2, colunm 2 shows, in a related species group, a change from a V to two rods. (Note also the smallest chromosome is missing. ) 5. Row 2, column 3 shows two rods form a V MULLER Figure 21-1 B. '" ,,'r *- Breakage and chromosome configuration 1. Forming a V from two rods a. One of the arms of a rod chromosome is very short. b. Suppose two rods are broken near their centromeres, one:in the long arm, the other in the short. c. If the acentric long arm joins to the centric piece of the other broken chromosome a V is formed. d. The remaining pieces, which mayor may not join to form a small chromosome, are composed largely of heterochromatin and may be lost without producing much phenotypic detriment. 2. Forming two rods from a V a. Another centromere is needed. 125 C. D. 126 b. The Y in Drosophila, being almost entirely heterochromatic, can furnish it. c. A eucentric reciprocal translocation then occurs between the Y and a V broken near its centromere. d. This produces two chromosomes which later, by paracentric deletions of Y portions, become rods derived from the original V. 3. Pericentric inversions change the relative length of arms when the two breaks are different distances from the centromere. Discoveries made possible by structural changes 1. Structural changes of parts of chromosomes a. involved groups of genes already known to be linked from crossover studies. b. Analysis showed that the gene arrangement in the chromosome was the same as in the genetic crossover map. 2. The centromere has a genetic basis, and crossingover near it is reduced. 3. The telomere has a genetic basis. 4. Heterochromatin a. normally occurs near the centromere and, to a lesser extent, the telomeres. b. contains genes. These produce blocks of chromatin ::n mit otic chromosomes, so that the relative contribution to chromosome size during mitosis is greater per gene for those in heterochromatic than in other (euchromatic) regions. 5. Cooper discovered collocho res, genetic el ements near the centromere important for synapsis. 6. Gene dosage changes became feasible and their effects better understood. 7. Position effect became better understood. Position effect 1. Sturtevant, studying the Drosophila eye mutant Bar, was the first to show that a change merely in a gene's neighbors can produce an hereditary change in phenotype. 2. Many structural changes cause genes near the points of breakage to be somewhat changed in their functioning, i. e., to show position effects. 3. Whenever the same rearrangement occurs, the same position effect is produced. 4. Position effects can occur also for genes located some small distance from a breakage point. 5. The original functioning of a gene is restored either by its removal to a normal chromosome via crossingover, or by reversal of the structural change. E. F. 6. A gene from a normal chromosome shows position effect when placed in the abnormal position by crossingover. 7. Position effects are thought to be due often to the products of one gene reacting with those of its linear neighbors. 8. Placing a gene near or in heterochromatin often produces a special, wavering, position effect, which is expressed in a mosaic or variegated phenotype. Such a position effect is reduced when, by breeding, chromosomes are added which are largely or entirely heterochromatic. Mechanism of induced structural changes 1. These mutations are induced both by radiations and chemical substances. 2. High-energy "radiation produces ions (see Chap. 23) which cause breaks. 3. The number of ions and of the breaks they produce are proportional to the radiation dose. 4. Since break frequency is independent of radiation dose rate, so also are all onebreak structural changes. Rearrangements involving two brea1\:s very close together also increase linearly with dose. 5. Following two breaks located in widely separated positions, structural changes are proportional to a power of the dose greater than 1. Such cases depend upon dose rate, si?-ce the second breal<: must be produced soon enough after the first, or the first might meanwhile have restituted. 6. More structural changes are produced a. in heterochromatic than in euchromatic regions. b. when condensed chromosomes are irradiated. 7. Joining of broken ends requires energy, being enhanced by oxygen and prevented by nitrogen when present after irradiation, as shown first in plants (Wolff) and then in Drosophil a (Abrahamson). 8. Oxygen present during irradiation increases the number of breaks. Radiation effects in different tissues 1. Gonia are often killed by induced structural changes. 2. Animal gametes function even when carrying structural changes that kill after fertilization. 3. Somatic cells most harmed are those most actively dividing, as expected if the detriment is caused by structural changes lead- G. ing to aneuploidy after cell division. 4. There is an ageing effect, approximately proportional to dose, expressed as a lesser ability to survive at all ages. It is suggested this is due to the loss or impairment of somatic cells in which structural changes have been produced. Much somatic damage can be produced even from single breaks. "We must be careful for our own selves, not only for our children, even of small amounts of radiation, if they are repeatedly or continuously received so as to total up to a sizable amount. II POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 127 QUESTIONS FOR DISCUSSION 21. 1. Sometimes one species has a small number of large chromosomes, while a closely related species may have a large number of small chromosomes. Suggest an expl anation for this. 21. 2. From Fig. 21-1 can the direction of evolution within column 1 or row 4 be determined? Explain. 21. 3. Explain in detail how attached-Xs in Drosophila may become detached. 21. 4. How might a small chromosome, like the fourth of D. melanogaster, be used in forming a V from two rods? 21. 5. How does breakage lead to changes in genome configuration in the course of evolution? 21. 6. Referring to Fig.. 21-1, explain in each of the following pairs how the first chromosome configuration listed might give rise to the second. a. b. c. column 1, row 4; column 1, row 3 column 4, row 5; collann 5, row 5 column 1, row 4; column 1, row 5 21. 7. Describe three ways of demonstrating that position effects due to structural changes are not due to mutation at a point of breakage. 21. 8. From an examination of the chromosome maps of Drosophila do you think that mutations are equally likely to occur in all regions of the chromosome? Explain. 21. 9. In what ways have chromosomal aberrations helped to strengthen the hypothesis that hereditary factors are carried in the chromosomes? 21. 10. How did the use of radiation aid the discoveries made possible by structural changes? 21. 11. How is it possible to identify a given gene locus with a given band on the salivary chromosome? 21. 12. Using the same scale, draw a curve showing an effect which increases as (dose)l, and 128 one increasing at a power of the dose greater than 1. 21. 13. If translocations increased as (dose)2 and small deletions as (dose)l, how many times would each be increased by changing the dose from 100r to 400r? 21. 14. What happens to the frequency of translocations when a dose is given in a protracted rather than a concentrated manner? Explain. 21. 15. Suggest two reasons why more gross rearrangements are produced by irradiating spermatozoa, spermatids, metaphases, or chromosomes halted in division by colchicine, than by irradiating interphase stages. 21. 16. Why is it that certain two break rearrangements increase linearly with radiation dose while others increase faster than the first power of the dose? 21.17. In Drosophila what possible factors expI ain the greater number of .rearrangements which involve heterochromatin than euchromatin? 21. 18. How may oxygen be used to obtain in spermatids the maximum number of struc-: tura! changes for a given dose of radiation? 21.19. To what does Muller attribute radiation sickness and other immediate and postponed effects of radiation? What evidence supports his view? 21. 20. What genetic reasons can be given for the use of radiation to arrest or cure malignant growths? 21. 21. Compare the effects in man of a given qose of radiation upon a nerve cell, a red blood cell, and a white blood cell, at the genetical and physiological levels. 21. 22. What was Quastler's analysis of the effect of large doses of radiation upon the mammalian intestinal tract? 21. 23. What sample of evidence did Muller present that radiation-induced ageing may be due to structural changes? Design an experiment to test this hypothesis. 21. 24. Gametic lethals are unknown in animals, but pollen lethals are of common occurrence in plants. Explain. 21. 25. Would it be appropriate to call this chapter ''What structural changes have taught us "? Why? 129 Chapter 22 "SPONTANEOUS" GENE MUTATION Lecturer-H. J . MULLER 2. Before 1909 it was difficult to demonstrate PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 17, pp. 292-304. Colin: Chap. 12, pp. 214-215', 225-227, 234-236. Dodson: Chap. 13, pp. 157-162; Chap. 16, pp. 198-200. Goldschmidt: Chap. 7, pp. 130-136. Sinnott, Dunn, and Dobzhansky: Chap. 16, pp. 214-220, 223-224, 226-229. Snyder and David: Ghap. 24, pp. 348- B. 354. Srb and Owen: Chap. 12, pp. 236-240. Stern: Chap. 22. Winchester: Chap. 18, pp. 243-251, 255-259; Chap. 19, p. 273. b. Additional references Muller, H. J. 1923. Mutation. Republished 1925 in Newman's "Readings in evol ution, genetics, and eugenics", pp. 495-502. Muller, H. J. 1950. Radiation damage to the genetic material. Amer. Scientist, 38: 33-59, 126, 399-425. Muller, H. J. 1958. General survey of mutational effects of radiation. Chap. 6, pp. 145-177 in: Radiation biology and medicine. Ed. W. D. Claus. Reading, Mass. : Addison-Wesley Publ. Co. 3. Read the lecture notes through section D. LEe TURE NOTES A. Gene mutations involve additions, subtractions or other changes in individual or very small groups of genes. (Stadler has called these point mutations.) 1. These supply the building blocks for evolution's edifices. 130 C. D. their origination because a, since many are recessive they could have been long present in the population in heterozygotes. b, gross structural changes might ~xplain the new heritable phenotypes, 3. Yet some were known, especially those which were dominant (Ancon sheep) or occurred in inbred lines (as in Johannsen's beans). Spontaneous gene mutations 1. A number were found, beginning in 1909, by Morgan studying Drosophila and Bauer studying the snapdragon, Antirrhinum. 2. Shadow views of Drosophila exhibiting some early-found mutations, vestigial, Curly, or truncated wings, were shown. Characteristics of gene mutations , 1. A given type is rare. 2. They occur singly. 3. Incidence bears no seeming relation to environment. Other characteristics (later collated by Muller, 1914-1923) 1. In a diploid cell only one gene of a pair mu- tates, suggesting the process is a sub-microscopic accident. 2. Any single mutation may be pleiotropic, yet mutations in different genes can affect the same character. Environment affects penetrance and expressivity, but not the gene itself. "* 3. Mutations with small phenotypic effects are more frequent than those with large effects. The smaller the effect the more important it is in evolution. 4. The vast majority of mutations are detrimental, as expected of accidents in a complex organization. a. Per 100 detectable mutations about 25 are lethal. Of the non-Iethals only 3-4 produce some readily visible detrimental effect. The remainder are invisible, yet detrimental to some degree. b. The majority of mutations affecting a trait or organ cause its degeneration. c. Increasing the detectability of small-effect mutations is higlUy desirable, since each detrimental mutation, regardless of how small its effect may be, is eventually eliminated from the population by a genetic death (see Chap. 26). 5. Most mutants are recessive rather than dominant a. However, mutants are seldom, if ever, completely recessive. b. Experiments of both Stern's and Muller's groups showed that Drosophila heterozygous for a recessive lethal have, on the average, 4% less chance to reach maturity than have normal homozygotes. c. In Drosophila, about one mutation in 10 is visibly expressed in the heterozygous individual, which, however, resembles the normal homozygote more than the mutant homozygote (which is frequently lethal) . d. The phenotypic effect in heterozygous condition is relatively so large that most mutants suffer genetic death as heterozygotes before persisting long enough to appear homozygously. 6. Dosage changes reveal how mutants act to produce phenotypes. ~. Phenotypic effects and gene dosage Radiation-induced structural changes aided this study. 1. Hypomorphs are mutants having a simil ar but lesser effect than the normal gene. a. Most mutants are of this type. b. Addition of further doses of such mutants causes the phenotype to become more normal. c. Bobbed is a hypomorphic mutant (see Chap. 9). 2. Amorphs are mutants having no phenotypic effect, even when present in extra dose. 'White eye in Drosophila is an example. 3. Neomorphs produce a new effect. Adding more doses of the mutant causes more departure from normal, while adding more doses of the normal allele has no effect. F. How mutants act to produce phenotypes (see P.ll) 1. A single.:!: allele almost produces the wild- G. H. type phenotypic effect. Two + alleles reaches this level. 2. In the case of hypomorphic mutants, even three doses may not reach the phenotypic level reached by one.:!: allele. 3. Note that genetic modifiers or environmental factors would shift the phenotype less and less as one proceeds from individuals carrying one dose of mutant toward those containing two ;:. alleles. 4. Natural selection favors alleles producing effects near the curve's plateau, for this insures phenotypic stability. 5. The fact, then, that the heterozygote with one;:. and one mutant gene has practic ally the same effect as the normal homozygote best explains dominance. Spontaneous mutation frequencies 1. At specific loci a. In Drosophila, Schalet obtained an average of one mutation in a given gene per 200,000 germ cells tested. b. In mice, Russell found this rate to be about one in 100, 000. c. In man, dominants per gene occur one time per 50,000-100,000 germ cells per generation. d. For these species, mutation rate differences are approximately proportional to differences in number of cell divisions per generation (i. e., from gamete to gamete). 2. Total mutation rate a. In Drosophila, one gamete in 20 (or one zygote in 10) contains a new detectable mutation which arose that generation. b. This total gametic frequency is 10,000 times tl1le frequency obtained for individual loci in Drosophila (Gla). c. In mice, multiplying the gametic per gene rate of 1/100, 000 by this minimum factor of 10,000 yields a total mutation frequency of one in 10 for gametes, or one in five for zygotes. d. In humans, 10, 000x(1/50, 000) would give one germ cell in five, or two zygotes of each five, with a new spontaneous mutation. This agrees with the estimate of Morton, Crow, and Muller based upon a study of abnormalities and mortality in offspring of first cousins. e. For each five humans, then, at least two are eliminated by genetic death. Conditions for spontaneous mutations 131 1. They occur in males and females, at about comparable rates in Drosophila. 2. They occur in germ cells at various stages, and in somatic cells. 3. Ageing effects on mutation rate have been studied in Drosophila by Muller. a. Prolongation of larval or adult stages has little effect on mutation rate, showing few mutations occur during these stages. b. Peri-fertilization stages (later stages in gametogenesis and very early development) are relatively rich in mutations, as decided by ageing spermatids and spermatozoa and by a process of elimination. c. Flies, mice, and men are not very different in length of time occupied by perifertilization stages, so that their mutation rates would tend to be alike if most mutations occur then. 4. Higher organisms, containing more genetic material, probably have been selected for genotypes that reduce spontaneous mutation rate in order to avoid over-mutation. 5. Genetic control of mutability is demonstrated, in Drosophila, by mutator genes which increase mutation rate lO - fold or more. 6. Environmental conditions a. Temperature Altenburg and Muller obtained about a 5fold increase in mutations for a rise of 10 0 C in the normal temperature range. Violent temperature changes in either direction produce a much greater effect. b. Detrimental conditions of almost any kind increase mutation rate (Demerec; Novick and Szil ard) . 7. Mutations in the old gene a. Mutation rate is substantially the same whether or not colon bacilli are actively dividing (Novick and Szilard). This indicates mutations are not usually caused by a mistake in the reproduction of the new gene by the old one. b. In Drosophila, ageing sperm or spermatids increases the number of mutations which occur, not in a quarter, but in half or more of the body of the individuals formed. Since the gene is two-stranded and a new gene is composed of one old and one new strand (see Chaps. 38, 4i), error in the making of one of the two new strands would later involve one quarter of the 132 body. The absence of such mutations indicates mutations occur in the old gene. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3 .. Be able to discuss or define orally or in writing the items underlined in the lecture notes. ' 4. Complete any additional' assignment. QUESTIONS FOR DISCUSSION ·22 . . 1. Specify the conditions required for concluding that a suddenly appearing heritable change is due to a point mutation. 22. 2. What evidence is there that point mutations are chemical accidents at the sub-microscopic level? 22. 3. Are point mutations with slight effects more or less easy to recognize in man than in Drosophila? Explain. 22. 4. Explain which would be expected to have greater significance in evolution, point mutations with large or small phenotypic effects. 22. 5. In what respect are mutants producing large and small detrimental effects the same? 22. 6. How many generations, on the average, would a mutant persist in a large POPul ation of constant size if in heterozygous condition it produced a detriment of 100%? 50%? 5%? 1%? to dominant alleles in man? Explain. 22.16. How can a minimum estimate of gene number be made in Drosophila? Make the calculation and list the assumptions involved. 22.17. How may mutations which occur in the mature gamete be distinguished from those occurring in gonial stages? 22.18. Give evidence for increase and decrease of mutability being under genetic control. 22.19. Why is it expected that a rise in temperature in the normal range will increase mutation rate? 22.20. Explain, using a series of diagrams, how mosaic formation in Drosophila illustrates that the old gene has mutated. 22. 7. Of what consequence is the fact that mutants are not completely recessive? 22. 8. Can a mutant be proven hypomorphic from a study of euploid individuals? Explain. 22. 9. Which are more important in evolution hypomorphs or neomorphs? Explain. 22.10. How can it be explained that the variability in penetrance and expressivity is greater for a mutant than for its normal allele? 22.11. How is dominance related to natural selection? 22.12. Suggest another way by which dominance may arise. 22.13. Design a procedure for detecting mutations at the miniature locus in Drosophila. 22.14. What factors would tend to make the total cell mutation rate not very dissimilar in man, mouse, and fly? ~erm 22.15. Is it valid to compare the mutation rate to recessive alleles in fly and mouse with that 133 Chapter 23 MUTAGEN-INDUCED GENE MUTATION ,L ecturer-H. J. MULl.ER PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General. genetics textbooks Altenburg: Chap. 18, pp. 308-318, 322325. Colin: Chap. 12, pp. 227-234. Dodson: Chap. 13, pp. 162-166. Goldschmidt: Chap. 7, pp. 136-140; Chap. 12, pp. 195-196. Sinnott, Dunn, and Dobzhansky: Chap. 17, pp. 231-240. Snyder and David: Chap. 24, pp. 354363. Srb and OWen: Chap. 12, pp. 245-248, 250-255. Stern: Chaps. 23, 24. Winchester: Chap. 19, pp. 260-262, 264-273; Chap. 20, pp. 278-287. b. Additional. references Auerbach, C. 1956. Genetics in the atomic age. 106 pp. Fair Lawn, N. J. : Essential Books. Beadle, G. W. 1959. Ionizing radiation and the citizen. Scient. Amer., 201: 219-232. The biological. effects of atomic radiation. Summary reports. 1956. See report of Genetics Committee, pp. 3-31. Washington, D. C. : National Academy of Sciences - National. Research Council. Crow, J. F. 1959. Ionizing radiation and evolution. Scient. Amer., 201: 138-160. Muller, H. J. 1927. Artificial transmutation of the gene. Science, 66: 8487. Reprinted in "Great experiments in biology", 1955, Eds. Gabriel, M. L., and Fogel, S. Englewood Cliffs, N. J.: Prentice-Hall, Inc. Also reprinted in 134 "Classic papers in genetics", 1959, Ed. Peters, J. A. Englewood Cliffs, N. J. : Prentice-Hall, Inc. Muller, H. J. 1950. Radiation dal1:}age to the genetic material. Amer. Scientist, 38: 33-59, 126, 399-425. Muller, H. J. 1955. Radiation and human mutation. Scient. Amer., 193: 58-68. Muller,. H. J. 1958. General survey of mutational effects of radiation. Chap. 6, pp. 145-177 in: Radiation biology and medicine. Ed. W. D. Claus. Reading, Ma.ss. : Addison-Wesley Publ. Co. Platzman, R. L. 1959. What is ionizing radiation? Scient. Amer., 201: 7483. Report of the United Nations Scientific Committee on the effects of atomic radiation. 1958. See Chaps. 5-6, annexes G-I. New York: General Assembly Official Records: 13th Session, Suppl. 17 (A/3838). 3. Read the lecture notes through section D. LECTURE NOTES A. Spontaneous mutation Naturally occurring mutations are believed to involve chemical modification of nuc1eotides (see Chaps. 38, 41). Mistakes in the synthesis of new nucleotide chains, i. e., new genes, may be caused by a variety of agents. B. Mutagen-induced mutation 1. Mutagens are specific agents which increase the mutation rate enormously. 2. High-energy radiations are mutagenic. a. X-rays were the first known mutagen (Muller 1927, and independently, Stadler 1928). b. Acting simil arly are gamma, beta, and alpha rays. C. Chemical effects of high-energy radiations 1. Although X and gamma rays are electromagnetic waves, like visible light, they have relatively shorter wavelengths, more energy and greater penetrating ability. 2. When these rays are stopped by an atom, their energy is absorbed by the atom which is caused to lose an orbital electron. This electron shoots off at great speed (as a fast electron) which in turn causes otller atoms to lose electrons. 3. All atoms losing electrons become positively charged ions. 4. The free electrons are finally captured by other atoms which become negatively charged, or negative ions. 5. Since each electron lost from one atom is gained by another atom, ions occur as pairs. 6. A track of ionizations is produced which often has smaller side branches. 7. Alpha rays are particulate, being composed of helium nuclei. D. Ions and mutation 1. Ions undergo chemical reaction to neutralize their charge. In doing so they can cause mutation a. directly, if they occur in an atom of a gene, or b. indirectly, by changing oxygen-carrying molecules that in turn react with genes. 2. Increasing oxygen or poisoning a cell's reducing substances during irradiation produces more mutations. During irradiation, replacement of oxygen by nitrogen is protective. 3. The mutated gene is about as stable as it was previously. 4. Point mutations and chromosomal breakages are produced by individual or small groups of ions occurring in the course of a track of ionizations. Accordingly, such mutation rates are simply proportional to the total number af ionizations produced. 5. Thus there is no dose without risk of mutation. 6. Tightly spiralized chromosomes appear to' be especially radiasensitive. E. X-ray dose and ionizations 1. One roentgen (r) praduces about 1. 8 x 109 ion pairs/cm 3 of air. 2. 1r produces about 1. 5 ion pairs/fl 3 of tissue. 3. Drosophila sperm heads are about 0.5 1-1 3 so that 1r produces less than ane ion pair in each. 4. 400-500r in a concentrated total-body expo- F. G. H. sure will kill 50% of humans. 5. Chest X-rays deliver about O. 1r internally; fluoroscopic examinations may deliver 100 times more. Mutation-doubling dose for X-rays 1. The dose of radiation which doubles the spontaneous rate of mutations per generation is called the doubling dose. 2. Drosophila doubling dose is a. 60r for mature spermatozoa, b. 20r for spermatids, c. 250r for gonia. 3. As shown by Russell, mice, as compared with Drosophila, have twice the spontaneous mutation rate and 12 times the mutatian rate following acute X-ray exposure. Since such X-rays are six times more effective relatively, daubling dose for mouse gonia is only about 401'. Nature of radiation-induced mutations 1. They are the same kinds as occur spontaneously_ 2. They are localized events, involv:i.ng one gene when a pair is present. 3. Most are detrimental, recessive (but not completely so), and hypomorphic (otherwise usu.a lly amorphic). 4. Sometimes nO' structural change is associated with them; they can subsequently undergo reverse mutation (reversion) to or near the normal gene. Such mutatians are not readily explained as mere losses of genetic material. 5. Point mutatians, such as minute deletions and inversions, simul ate intragenic changes. 6. Most are changes in the old gene, not in its reproduction. 7. Somatic and germinal ones are produced at about tl1e same rate. 8. Leukemia is thought by some to be due to samatic mutations. Permissible doses of high-energy radiation 1. A committee of the United States National Academy of Sciences recommends that per reproductive generation (or 30 years) the general population receive not more than lOr. Already 3r are received from natural sources, 5r from medical uses of radiation, leaving only 2r before reaching the permissible level. 2. Simil ar recommendations were made in 1958 by the International Commission on Radiation Protection. 3. Both graups make exceptions of people oc- 135 1. 136 cupationally exposed, allowing them to receive 50r each 10 years, so long as the population average is not raised above lOr. 4. In a population of 100 million, about 1. 25 million would contain mutations induced by lOr of permitted radiation, of which about 6%, or 75,000 would show effects in the first generation following exposure. The rest of the detriment would be spread over many future generations, until, eventually, all the mutations were eliminated by genetic death (Chap. 26). 5. "Thus, the words permissible dose only mean the least dose that we can manage without giving up too many of the benefits that the use of radiation can afford us, but by no means implies a harmJ.ess dose. " Other mutagenic agents 1. Ultraviolet light, though less energetic than X-rays, is mutagenic, as shown first by Altenburg using Drosophila. Since ultraviolet is not deeply penetrating, it does not produce mutations in the human germ line. 2. Chemical substances a. Mustard gas and its derivatives are about as mutagenic as X-rays, as discovered by Auerbach and Robson in 1941 working with Drosophil a. b. Peroxides, epvxides, and carbamates (including formaldehyde) are mutagenic in Drosophil a as shown by the Russian worker, Rapoport. c. Many other highly reactive substances, including triethylenemelamine, are mutagenic. The Pull mans maintain all chemical mutagens have a high concentration of electrons at a particular point in the molecule. d. Unlike radiation, chemicals often have delayed effects, so that mutations may occur even several generations after treatment. e. Like radiation, they produce breakages and point mutations. £. Nucleotide analogs (wrongly made nucleotides) have been used to produce mutations. g. Mutational specificities exist. Chemicals tend to act more on one set of loci or alleles than on others. Different chemicals act differently from each other and from natural or radiation mutation agents. h. We do not know, but we may be getting today more mutagenic effect from chem- J. icals than froIn radiation. Counteracting mutations 1. Despite mutation::ll specificities of different mubgens, it is unlikely that we shall be able for a long time to cause only particu1ar genes to mutate in particul ar ways. 2. Avoidance of genotypic degeneration due to mutation requires some type of selection. a. Saving lives, by medical or other improved conditions, reduces selection and increases our load of detrimentnl mutations more than does radiation. b. More social conscience is needed in reproduction, in which the good of future children is considered. "Remember that it is these tiny gene mutations, not the great genome or whole chromosome mutations or even structurnl changes of chromosomes, that have managed, as a result of rigorous sel ection of them in the past, to result in our own wonderful organization, but that, if unchecked, they can result in misery. Which the outcome will be must depeno upon the decisions that we humans make concerning these gene mutations. II POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. QUESTIONS FOR DISCUSSION 23. 1. Design an experiment to test whether mutations sometimes occur when the old gene reproduces. 23. 2. Describe how a track of ionizations is produced after X-ray energy is absorbed. 23. 3. One end of a track of ionizations initiated by X-rays usually appears quite different from the other end. Describe how the ends could differ, and explain the basis for this. 23. 4. Can one obtain an odd number of ions following X-ray absorption? Explain. 23. 5. Compare the spacings between clusters of ionizations produced by X-rays and cosmic rays. 23. 6. Are alpha rays or ordinary X-rays more likely to produce simultaneously mutations in adjacent loci? Explain. 23. 7. If point mutation rate for Drosophila sperm is linearly proportional to dose in the range 25r-100r, what would you expect the relationship to be between 11' and 25r? Explain. 23. 8. What difficulties exist in mutation rate experiments using low dosages that are relatively less important in studies employing higher dosages? 23. 9. Describe the cell stage, the environmental and other conditions which would be least mutagenic when delivering 1 OOr. 23. 10. Discuss the view that condensed, closely spiralized chromosomes are especially susceptible to X-ray-induced mutation. 23.11. Does each ion pair have a probability of 1 of producing mutation? Expl ain. 23.12. What precautionary measures may be taken to reduce radiation damage to future generations a. when a man is exposed to an acute gonadal dose? b. when a woman needs pelvic X-ray treatment? c. when selecting employees for indus- tries employing atomic energy? 23.13. Shoul.d the doubling dose be higher or lower for spermatozoa than for spermatogonia? Explain. 23.14. How may the higher per locus X-ray mutation rate for mouse than for Drosophila genes be expl ained ? 23.15. What evidence is there that ionizations need not be in the gene itself in order to cause mutation? 23.16. Should radiation-induced and spontaneous mutations have similar mutagenic specificities? Explain. 23.17. Why are not all X-ray-induced intrachromosomal mutations explicable as loss of genetic material? 23.18. Would it be expected that certain chemical substances are useful in cancer therapy? Why? 23.19. Do you believe that peroxides are likely to cause somatic or germinal mutations in humans? Expl ain. 23.20. Is it possible to determine what chemical substance is actually producing mutations? Explain. 23.21. Design an experiment using Drosophila to compare the mutagenic specificities of mustard gas and triethylenemelamine (TEM). Design such an experiment using Neurospora. 23.22. Discuss the statement: "Intragenic mutations can be differentiated from intergenic ones by the relations each has to chromosome breakage. II 23.23. Which would be expected to produce a more specific effect on the phenotype - point mutations or chromosomal rearrangements? Why? 1 37 • EXAMINATION III \ UNDERLINE ALL THE ALTERNATIVES WHICH ARE CORRECT. 1. Linkage 4. a. provides exception to independent segregation by different gene pairs. b. would be complete for a chromosome which was monosomic in both sexes. c. is incomplete because of chiasmata. d. can explain how different genes fail to be inheri ted together. e. is directly associated with the production of crossovers. 2. Point mutations a. b. c. d. is a prime consequence of sexuality. is produced by gene mutation. made Mendel's discoveries possible. occurs in asexual organisms via chromosome rearrangement. e. is produced by crossingover which provides the advantage of placing non-allelic genes from non-homologous chromosomes into the same chromosome. 5. a. are the most important type of mutation in evolution, yet produce the least effect per cell of any mutational type, ignoring position effect. b. have no radiation threshold and are also induced by chemical mutagens. c. can be detected cytologically. d. which are recessive are more likely to be advantageous than those which are dominant. e. which are hypomorphs are less damaging than are neomorphs, one would expect. A chromosome which has Gene order a. can be determined only when more than two pairs of genes are studied. b. cannot be determined by backcrossing trihybrid Drosophila males. c. has to be considered in two dimensions when a ring chromosome is involved. d. rarely is changed by crossingover, but always is changed after a chromosome break fails to restitute. e. is presumably linear in the chromosome because it is linear in the crossover map. 138 Aneuploidy a. can be produced only following chromosome breakage. b. not only occurs spontaneously, but is induced by mutagenic agents. c. is more likely to contribute to evolutionary progress thai! is ploidy or point mutation. d. of heterochromatic regions produces less genic imbalance than would be produced by equally long euchromatic regions. e. can provide material for studying the effects of gene dosage changes. 6. 3. Genetic recombination a. lost its centromere may cause little trouble if it lies in a neuron. b. lost its telomeres will be a cause of cell damage or death within a relatively few cell divisions. c. exchanged equal segments with another chromosome may have done so by crossingover. d. once been broken by X-rays is more likely to break at the same place next time rather than at some other randomly chosen site, not because it has been weakened by the first break but because it has an in- b. due to point mutation increase linearly with X-ray dosage. c. may arise from chromosomal rearrangements, and so may require two breakages in chromosomes for their production. d. may be due to position effects consequent to chromosome rearrangements not involving loss of genes. e. are genes that have lost their power to self-reproduce. trinsic chemical makeup which makes it prone to brealcage. e. no homologous partner is heteromorphic, like the Y chromosome in birds. 7. Pericentric inversions a. do not produce aneucentric chromosomes upon crossingover with similarly inverted or with non-inverted chromosomes. b. in Drosophila can be detected when homozygous by the change produced in banding pattern of the salivary cell chromosomes. c. always form a loop when seen in salivary nuclei of Drosophila. d. are usually lost from the population unless they are not small. e. almost always modify the shape of a chromosome. 8. a. results in the total absence of genetic recombination under normal conditions. b. results in the absence of crossovers. c. is modified by the presence of heterozygous inversions. d. means that sperm produced are genetically of only two types. e. does not involve absence of segregation. 12. Lethal mutants a. may be completely or partially recessive, but never completely dominant. Mutants a. are relatively more important to the individual receiving them the smaller the number of genes that are involved. b. involve changes, iE number, nature, or relative positions of genes. c. whether spontaneous or induced, are usually partly dominant. d. were first produced by X-rays in Muller's studies. e. are the raw materials of evolution and therefore the majority of those which occurred billions of years ago must have been advantageous. The "maximum permissible dose" of X-rays a. refers only to exposures for non- medical purposes. b. reduces the mutation rate to zero. c. is subject to revision as more information on the genetic consequence of irradiating humans is obtained. d. produces the minimum induced mutation rate. e. is harmful but we accept this damage for certain reasons. 10. Complete linkage, as in Drosophila males, Variation in crossover frequency a. is dependent upon genetic but not environmental factors. b. occurs between different gene pairs both in different organisms and in the same organism. c. cannot be attributed to a change in distance between two pairs of genes. d. is directly correlated with variation in chiasma frequency. e. provides the raw material for the making of genetic maps of chromosomes. 9. 11. 13. Since crossingover occurs in the tetrad stage a. only two strands of the four will be found to be crossovers. b. an ascus can contain either eight crossovers or eight non-crossovers following a double chiasmata. c. a single chiasma produces half of the mei oUc products of its tetrad without crossovers. d. a double chiasmata involving only two different strands produces only 50% recombinant meiotic products. e. sister-strand crossingover is possible, 139 whereas this would have been impossible if crossingover occurred in the two-strand stage. 14. 19. 140 17. Mutations which occur spontaneously a. are rare for a given gene and ·are rarely beneficial when homozygous. b. differ qualitatively from those induced by X - rays. c. are usually amorphs or hypomorphs. d. are based only upon changes of the old gene. e. have probably arisen many times during the previous history of the species. The crossover rates, used for making genetic maps of chromosomes, a. must have sister-strand crossover rates subtracted from the total crossover rate in order to be utilized in map making. b. are directly proportional to the physical distances separaEl1g genes if the distances are so short as to permit no double chiasmata within them . c. are never more than 50% between any two genes. d. have their physical basis in chiasmata each of which involves but two strands. e. may be lower than the standard expected rate because of interference. Translocations a. always involve two or more non-homologous chromosomes which have broken and undergone non-restitutional unions. b. change linkage relationships of genes. c. of the reciprocal type make synapsis during meiosis difficult when they are heterozygous, so that crossingover is somewnat reduced. d. of the non-reciprocal type are called halftranslocations, and are produced by means of a single breakage. e. when reciprocal and aneucentric often lead to death of the cell line. Changes in number of genomes, or ploidy mutations, a. are such gross mutations that they are never the basis for new species formation. b. were involved in the production of Oenothera gigas. c. can occur in some tissues and not in others of the same individual. d. are expected to be more commonly found in asexually reproducing plants than in sexually reproducing animals. e. are rarely found in higher animals which have a genetic mechanism for sex determination. 15. 16. 18. Interference a. measures reduction in number of double crossovers. b. is complete when distances are long. c. plays no role in a tetrad forming only single crossovers. d. is expressed also by the coefficient of coincidence. A student claims that the parents of the offspring shown above were Wild-type flies. a. Can you suggest an explanation for the results obtained? Outline your explanation. b. How would you test your explanation experimentally? 20. The underlined part makes each statement true or false. On the line supplied write TRUE or write a substitute for the underlined part which will make the statement true. a. Tissues which divide rapidly, such as epithelium, embryonic and germinal tissues, are most susceptible to damage by ionizing radiation. b. Position effects increase linearly with X-ray dose. c. The presence of a higher than normal oxygen concentration in dividing cells results in an increase in mutational damage from radiation. d. Cell death produced by ionizing radiations is due principally to aneucentric rearrangements. e. When the average number of chiasmata per tetrad is 1/2, the maximum lengtll of the genetic map for this chromosome is 50 map units. L The point mutations induced by certain chemical mutagens are qualitatively different from those which occur spontaneously. g. Other things being equal, the smaller the detrimental effect of a mutation the fewer the number of generations through which will persist before being eliminated from the population. h. The genetic effects of radiation are cumulative in the individual organism and its hereditary material. i. More damage is done genetically to the U. S. population by present diagnostic, therapeutic, and industrial X-rays than by the present rate of fallout from H-bomb tests. j. Different stages of spermatogenesis show identical induced mutation rates for the same dose of penetrating radiation. 21. Give in the accompanying chart the probability for a Drosophila male deriving a. a gene in his X chromosome from his b. his Y chromosome from his a. b. mother's mother mother's father father's father father's mother 22. A test cross indicates that gametes of various genotypes were produced with the following frequencies: ~ RS Rs rS rs 23. % 32.4 16.9 17.0 33.7 a. The genotype of the tested individual is b. Recombination frequency between genes at the two loci is %. Male Drosophila whose sperm contain the normal X chromosome genes ABCDEF, in this order, were irradiated with a large dose of X-rays. They were mated to females carrying the recessive alleles abcdef homozygously. All progeny were phenotypically normal except two, one of which was phenotypically ABcdeF, and the other phenotypically ABCDef. Investigation of the salivary gland chromosomes of these two exceptions revealed the presence of a buckle in similar regions of the X chromosome. One exception showed a buckle involving bands 4451, inclusive, the other showed a buckle of bands 32-44, inclusive. a. The gene best localized by this method is _ _ _ __ b. The band or bands associated with this gene would be _ _ _ __ 141 24. Explain how it is possible, by studying the behavior of two genes close to the Bar locus, to prove that as a result of crossingover there may be two Bar "genes" in one X chromosome of Drosophila and none in the other X chromosome. 25. The FI females from a cross of ++ ++ by QQ vv were test crossed with the following results: Offspring phenotypes ++ pv +v p+ Total Observed numbers 450 450 50 50 1000 a b 1) Under 'a' above fill in the numbers most likely to occur among 1000 test cross offspring if the two pairs of genes had been segregating independently. 2) Under 'b' above fill in the numbers most likely to occur among 1000 test cross offspring if the genes involved occupied the same map loci, but if the tested Fl had corne from parents whose genotypes were PQ ++ and ++ vv. 26. 27. 142 Describe as completely as possible the genetics of the crosses and their results shown in the diagram below. Assume 20% chiasmata occurs between genes S! and £. What would be the genotypes and the probability of each, among offspring of the following cross: dR/Dr x dr/dr? In a sexually reproducing animal assume it has been established that genes short @), red (B), and straight IT) are completely dominant to those for long @_), white (D, and bent (.0. respectively. a. Place the correct genotypes on the lines provided. x (pure recessive) (homozygous dominant for all genes involved) x (F 1 offspring) (pure recessive) b. If each of the 3 pairs of genes was located on a different pair of chromosomes, how many different combinations of characters (not including sex) would be expected among a large F2 population? c. What would be their relative frequencies? Suppose the F2 obtained from the P 2 above actually were: Phenotypes short, red, straight long, white, bent long, red, straight short, white, bent short, white, straight long, red, bent short, red, bent long, white, str.aight lio . 279 257 128 136 70 64 34 32 1000 total F2 d. Circle the numbers of F2 which are non-crossovers. e. Calculate in the spaces provided below the percentage of F2 which are single crossovers between the gene pairs indicated. Size gene and color gene Size gene and posture gene Posture gene. and color gene f. The genes for which traits are at the ends of a crossover map one would construct from these data? ----------------------------g. Having determined the order of the three genes, calculate the percentage of F2 offspring of all the offspring which are double crossovers. h. Calculate the total percentage of observed crossingover which had actually taken place between the two genes furthest apart. i. Show the calculations of the amount of double crossingover expected from the amount of single crossingover which was observed. j. The coefficient of coincidence in the present case is ___________ 143 29. A gene in a primary oocyte of Drosophila finds itself connected to a gene different from the one to which it was joined 10 minutes previously. List the different types of events which could have produced this change. 30. Three linked genes, G, H, 1, and their recessives, g, !!_, and 1, are being investigated for their distances and location with respect to one another. In a cross made between GH1/ghi x ghi/ghi the following phenotypic results were obtained: GH1 71; GHi 3; Ghi 17; Gh1 14; gHl 18; gHi 11; ghI 2; ghi 64. a. The middle gene of the three genes involved is _ _ _ __ b. Briefly justify your choice. c. What is the observed map distance between loci G and H? d. What is the percent of observed crossingover between loci G and 1? 31. In Drosophila the chromosomes in the largest larval salivary gland cell differ from those in a diplOid cell just before mitotic metaphase in the fol1f'lwing respects: a. b. c. d. e. f. 32. In Drosophila, ~, g, and s:. are recessive genes. Females are mated to wild-type males. The progeny are phenotypically as follows: all daughters were ABC; there were 1000 sons, and they were: AbC 0 ABc 45 Abc 427 ABC 23 aBc 1 aBC 424 abc 26 abC 54 a. In what chromosomes are the genes located? _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ b. Reconstruct the genotype of the female parent, showing the correct gene order and the alleles on each homolog. 33. Assume ~ and g are sex-linked genes located seven map units apart in the X chromosome of Drosophila. A female of genotype +b/a+ is mated with a wild-type male. a. What is the probability that one of her sons will be either a+ or +b in phenotype? h. What is the probability that one of her daughters will be ++ in phenotype? 144 ----- 34. Assume a normal chromosome bears the genes ABC~EF, in this order, and with the centromere between C and D. Males appearing abcdef are crossed with normal virgins, and the F1 females are backcrossed to males like their father. The backcross offspring obtained were numerous and only of the following 8 phenotypes. ABCDEF aBCDEF ABCDEf aBCDEf abcdef Abcdef abcdeF AbcdeF Explain these results. 35. A cross of a trihybrid for the genes ABC is made with an individual homozygous for the complete recessives abc. The following phenotypes, in the amounts indicated, were obtained in 1000 Fl' Abc aBC ABC 202 abc 198 49 51 ABc 200 abC 200 AbC 50 aBc 50 a. Are A and B on different pairs of homologous chromosomes? Explain. b. Are Band C on different pairs of homologous chromosomes? Explain. c. What do you conclude concerning the location of A and C from your answers to the preceding portions of this question? d. How many different phenotypes would you expect in this F1 were each of the different pairs of genes located on non-homologous chromosomes? e. Assuming that genes.!! and Q_ are in the same linkage group, 1) what percentage of the F1 retained the parental gene order for these genes? _ _ _ __ 2) what is the percentage of single crossovers between the loci '~' and 'Q_'? 3) describe how you would proceed in order to detect double crossovers between the 'a' and 'b' loci. 36. Shown below are key diagrams of selected portions of the chromosome complement in meiotic prophase of Zea mays. Describe the chromosomal events depicted in each in the space provided. A. B. C. D D. E. F. Alter .M cClintock 145 37. Suppose you found among the F1 shown to the right one additional fly (A or B or C or D). What explanation would you offer in each case? A. , B. C. D. A 38. Use the following crossover data between gene pairs to map the genes concerned. in map units, using lengths of the shortest intervals. yg - Q sh - c 39. 146 c B 16.9% 3.1% bz - -wx -c - -bz sh - yg - 23.7% 5.1% 19.5% wx - -sh - 8h - bz D Show the order 24.5% 2.0% Suppose you are a geneticist working with the banana fly, Drosophila melanogaster. In a pure wild type strain, raised under constant environmental conditions, you notice one male fly whose body color is purple, all the other flies in the culture having their normal grey body color. Purple body color has never been observed before in this species. The purple male is mated and a pure line of purple flies (both males and females) is eventually obtained which breeds true. all types of flies are available to you for maldng various tests. Assume stocks of a. With what chromosome is the mutant probably not connected? ___________ b. Describe how you would determine whether purple was dominant to grey. c. What would you predict concerning the fitness of the mutant relative to wild-type 1) when the mutant is heterozygous? 2) when the mutant is homozygous? Suppose next it was proven that purple arose as a phenotype associated with a reciprocal translocation that occurred in the mutant fly but not in its normal brothers or sisters. d. Describe one method you could use to detect such a translocation. e. What tests would you make and what results would you require before conduding purple was due to a position effect of the translocation? 40. Given a Neurospora tetrad of the following constitution, indicate the position of an exchange or exchanges that would result in: a. first division segregation of alleles at the a locus and second division segregation of alleles at the b locus. B B o a b a b. second division segregation of alleles at both loci, o -0 A B A B a b a b c. first division segregation of alleles at the b locus and second division segregation of alleles at the a lOCUS. o A B A B a b a b 147 41. Suppose two non-homologous chromosomes are linearly differentiated, as shown, where the circle represents the centromere. __~A~__-=B~__C~o D E 1) Using these chromosomes draw the following types of chromosome rearrangement: a. aneucentric reciprocal translocation e. deficiency b. tandem duplication f. eucentric reciprocal translocation c. paracentric inversion g. eucentric half-translocation d. pericentric inversion h. centric ring chromosome 2) Each time one of the above types of chromosome rearrangement in heterozygous condition could be responsible for the following observations, place its number in the space provided. a. salivary chromosome shows ''buckle'' b. salivary chromosome shows "loop" c. salivary chromosome shows "cross lt configuration d. dicentrics can be formed e. the number of crossovers is reduced 42. 148 Define the gene. Chapter 24 GENETICS OF MENDELIAN POPULATIONS Lecturer-Th. DOBZHANSKY PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks AJ.tenburg: Chap. 26, pp. 449-453; Chap. 27, pp. 473-474, 481-483. Colin: Chap. 14, pp. 285-287. Dodson: Chap. 19, pp. 229-233. Goldschmidt: Chap. 12, pp. 206-207. Sinnott, Dunn, and Dobzhansky: Chap. 18, pp. 241-246. Snyder and David: Chap. 28, pp. 429431. Srb and Owen: Chap. 20, pp. 426-435. Stern: Chap. 10. Winchester; Chap. 24, pp. 330-334, 337-338. b. Additional references Dobzhansky, Th. 1955. Evolution, genetics, and man. 398 pp. New York: Wiley & Sons, Inc. Dobzhansky, Th. 1951. Genetics and the origin of species. 3rd Ed. 364 pp. New York: Columbia University Press. LECTURE NOTES A. If the pre-Mendelian idea was true, that heredity is transmitted from parents to offspring through blood, 1. a child's heredity would always be a mixture or solution of its parents' heredity, 2. successive generations would become more and more uniform, 3. until, eventually, the population would become a pure race, i. e., composed of individuals uniform in their heredity. B. The genetic basis of heredity 1. Mendel showed that heredity is transmitted not by blood but by genes in the gametes, and that these genes do not mix, fuse, or C. D. contaminate each other. 2. Mendel's laws permit us to study and to predict the distribution of hereditary traits in families and in individuals. The genetics of populations 1. What will happen to the percentages of blood group types shown on the left in Fig. 24-1 a. 1 or 30 generations from now? b. if immigration or emigration occurs of people whose blood types are of a different proportion? 2. What will be the effect on society of the mutations produced by the use and mis-use of atomic energy? by the increasing exposure to medical X-rays? 3. These questions are concerned with the genetics of populations. 4. Population genetics is important in organic evolution, including the present and future evolution of man. The Hardy-Weinberg equilibrium principle 1. This is the basic law of population genetics. 2. Suppose a South Sea island, previously uninhabited, received a population in which 80% of people had blue (bb) eyes and 20% brown (BB), since presumably their ancestors were all brown-eyed), B being dominant to Q. Assume also that marriages are at random with respect to eye color. 3. The frequencies of the two eye colors in the children making up the next generation are calculated in the middle section of Fig. 241. People with the same, or different, eye colors will marry each other with frequencies proportional to their incidence in the popul ation. (Note that 32% of offspring will be Bb because 16% come from BB c!d' x bb nand 16% from the reciprocal marriages. ) 4. The same results are obtained more easily (shown to the right in Fig. 24-1) if, instead of marriages, the gene pool of the population 149 Figure 24-1 is considered, supposing each individunl provides an equnl number of gametes for the next generation. Therefore, 20% of all eggs and 20% of all sperm will carry B, 80% of all eggs and 80% of ~ 1 sperm will carry Q_. Combining the gametes at random produces the same results as was obtained from random marriages between parents. 5. What distribution of eye colors will be observed in generations following? In the first generation offspring, 4% were BB, 32% Bb, and 64% bb. When these offspring produce gametes 20% will carry B (all 4% of gametes from BB individuals, plus half of all gametes, or 16%, from Bb individuals) while the 80% remaining will carry g. 6. In other words, the gene pool of the next and all generations following will contain 20% B gametes and 80% g. This means that the frequencies of genes and of people with brown and blue eyes will henceforth remain constant. 7. This will be true regardless of what the initial frequencies of the two genes in the popul ation may be. a. Let p equal tlle fraction of male and of female gametes in the population gene pool which carry B, and q the fraction of male and of female gametes which carry g. Naturally, for eggs p + q 1, as it is also for sperm. These sex cells are now combined at random (Fig. 24-2). b. The expression which summarizes the resul tant offspring population, p2 BB + 2 pq Bb + q2 bb, = 150 is called the Hardy-Weinberg equilibrium formula. c. p2 + 2 pq is the fraction of all individuals who will be brown-eyed, while q2 is the bl ue-eyed fraction. Figure 24-2 d. In the next and all generations following these frequencies will remain unchanged. For among the gametes of the population obtained in 7b the frequencies of B and g will be p and q, respectively, as it was among the gametes of the previous generation. B = p2 + pq =. p(p+q) = p g = q2 + pq :; q(q+p) =q into the fit - adapted to the environment e. Note that neither dominance nor recesgenotypes of new races and species), random siveness will change the gene frequengenetic drift (which can produce rapidly cies in subsequent generations. changes in gene frequency in small populaE. Importance of the Hardy-Weinberg law tions) , and migration (interchange of indiviUnder this law: duals between different populations). 1. Mendelian populations preserve their genetic variability indefinitely. POST-LECTURE ASSIGNMENT 2. Sexual reproduction does not 1ead to pure 1. Read the notes immediately after the lecraces. ture or as soon thereafter as possible, 3. The population remains static. making additions to them as desired. F. Factors disturbing the Hardy-Weinberg 2. Review the reading assignment. equilibrium 3. Be able to discuss or define orally or in 1. If this law held indefinitely,. evolution could writing the items underlined in the lecture not occur, for evolution's basis in Mendelian notes. populations is change in gene frequencies. 4. Complete any additional assignment. 2. There are four main factors which upset the equilibrium, which change gene frequencies, . and therefore can be called jointly the causes of evolution. a. Mutation Gene frequencies will change if the mutation rates to and from an allele are different. b. Selection If individuals of a certain genetic endowment produce more surviving offspring than do those with a different genetic endowment, the genes which confer this higher biological fitness will tend to increase their frequency in the population while tl10se genes with lower fitness will tend to decrease. c. Random genetic drift When populations are very large, oscillations in the number of children produced by individuals with different genotypes do not matter. In small popul ations such oscillations can change gene frequencies. For example: Suppose the Paterfamilias family produces a large number of children for several generations, so that a large number of individuals called Paterfamilias appear. In a city with a large population the percentage of all people with this name will be very small, whereas in a small town the number of people with this name will be relatively frequent. d. Migration can change gene frequencies if, for example, natives interbreed with immigrants who are genetically different from the natives. G. The principal causes of evolution are mutation (which supplies the raw material of evolution), selection (which shapes these raw materials 151 QUESTIONS FOR DISCUSSION 24. 1. What is meant by a "Mendelian 'l population? 24. 2. How do the predictions with regard to populations differ for the blood theory as compared with the genetic theory of inheritance? 24. 3. What evidence would you require before conel uding that the frequency of gene A in the population pool was at Hardy-Weinberg equilibrium with its allele ~? 24. 4. Does the Hardy-Weinberg rule apply only for cases of complete dominance? Explain. 24. 5. What would be the consequences with regard to genotypes, phenotypes, and gene frequencies should a cross-fertilizing population obeying the Hardy-Weinberg rule suddenly become exclusively self-fertilizing? 24. 6. What can you conclude about a population which, with respect to one gene locus, obeys the Hardy-Weinberg law? 24. 7. If a population obeys the Hardy-Weinberg law for one pair of alleles is this probably or automatically true for all others? for any others? EArpl ain. 24. 8. Suppose a South Sea island population started with 10% blue-eyed and 90% homozygous brown-eyed people. Under the HardyWeinberg rule what will be at equilibrium the percentages of heterozygotes? of each type of homozygote? 24. 9. Suppose the population in 24. 8 started with 100% of people with blood group type AB. What will be the blood group situation at equilibrium, assuming the Hardy-Weinberg law obtains? 24.10. What will be the relative proportions of different genotypes at equilibrium, when both traits described for the population in questions 24. 8 and 24. 9 are considered simultaneously? 24.11. The proportion of recessives aa in a large crossbreeding population in equilibrium is .04. Assuming aa to be normal in viability and fertility, what should be the proportion 152 of Aa in the popul ation ? 24.12. Under the Hardy-Weinberg law, what is the frequency of gene P if its only allele .2. is homozygous in 49% of the popul ation ? 24.13. What are the frequencies of genes R and r if 50% of the people in a population at Hardy=Weinberg equilibrium are heterozygotes ? What could you say if the heterozygotes were 42% of the population? 24.14. The frequency of blood types in a sample of 300 persons is as follows: Type M, 42.7%; Type MN, 46.7%; Type N, 10.7% . Does this fit the assumptions of segregation and random mating? 24.15. In an isolated population of poultry which has gone wild and is mating at random, three color types are found as follows: black 490, gray 420, white 90. Suggest an expl anation to account f~)l' these proportions. 24.16. What would be the Hardy-Weinberg expectation for the frequency of Q if 99% of people were phenotypically like the dominant allele B? t 24. 17. Several investigators have each found that about 70% of Americans get a bitter taste from the drug phenyl thiocarbamide (PTC) and 30% get no bitter taste from it, that is, the latter are "taste blind" to PTC. Assume that taste blindness is recessive to normal. What proportion of marriages between normal and taste blind persons have no chance of producing a taste blind child? 24.18. Snyder found 278 taste blind, tt, offspring out of 761 tested offspring from marriages of taster by taste blind. On the basis of the gene frequencies of.!. being. 55 and T being. 45, calculate the expected frequencies of taster and taste blind children from the matings indicated.. What conclusions can you draw? 24.19. What frequencies of p and q give the greatest proportion of heterozygotes ? 24.20. Assume that an animal population is made up of the following individuals, which inter- breed and multiply in large numbers: AA, Aa, AA, AA, and Aa. If there is no selective factor in operation, what will be the proportion of AA, Aa, and aa individuals in the third generation? 24.21. Make the same assumptions as in the preceding problem, except that the population has the following initial composition: Aa, AA, aa, AA, AA, .t\a, aa, and AA. What will be the genetic constitution of the second generation? 24.22. Compare the importance or usefulness of the Hardy-Weinberg law when it does and does not obtain? 24.23. What are the causes of evolution? does each operate? How 24.24. If organisms reproduce asexually, is the Hardy-Weinberg law unable to operate? Explain. 24.25. May the Hardy-Weinberg rule be applied to loci which have multiple alleles? Explain. 24.26. Compare the effects of mutation, of selection, and of migration in large versus small Mendelian populations. 15~ Chapter 25 GENETIC LOADS IN MENDELIAN POPULATIONS ., lecturer-Tho DOBZHANSKY PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 26, pp. 453-458. Colin: Chap. 15, pp. 333-334; Chap. 16, p. 378. Dodson: Chap. 21, pp. 258-264. Sinnott, Dunn, and Dobzhansky: Chap. 18, pp. 248-253. Snyder and David: Chap. 28, pp. 432440. 8rb and Owen: Chap. 21, pp. 452-457. Stern: Chap. 10. Winchester: Chap. 24, pp. 334-335; Chap. 18, pp. 248-250. b. Additional references Dobzhansky, Th. 1951. Genetics and the origin of species. 3rd Ed. 364 pp. New York: Columbia University Press. Dobzhansky, Th. 1955. Evolution, genetics, and man. 398 pp. New York: Wil ey & Sons, Inc. LECTURE NOTES A. Heredity VB. mutation in evolution 1. So far as evolution is concerned, heredity is conservative, making offspring like their parents, while mutation is progressive, resulting in changes in the genetic composition of populations. 2. Mutants (the great majority are harmful) furnish the raw materials which natural selection constructs into evolutionary changes. 3. What happens to mutants in the population gene pool? B. Dominant lethal mutations 1. These are lethal when heterozygous. 2. Since they are eliminated from the gene pool the generation they arise, the muta154 C. tion frequency or rate (u) must be equal. to the frequency of affected individual s divided by 2. 3. This kind of lethal situation obtains for a mutant causing the eye cancer retinoblastoma, except for afflicted children receiving recently invented medical treatment. Dominant detrimental mutations 1. Achondroplastic or chondrodystrophic dwarfism is due to a gene which when heterozygous produces short legs and arms. 2. Such dwarfs (AI A2) have a lower Darwinian fitness or adaptive-value of the genotype than have normals (AI AI). 3. In populations fitness is measured by the reproductive rate. 4. Dwarfs have a fitness 20% that of normals, since the former have only 20 children for every 100 produced by the latter (Fig. 25-1). Figure 25-1 5. The selection coefficient (s) by which the heterozygote is discriminated against is) therefore, 1-.2, or .8. D. 6. M¢'rch determined u from the frequency of such dwarf children born to normal parents. 7. The expected frequency, p, of AZ in the popillation is u/s. As shown, thisagrees well with the actual p observed by M¢'rch. 8. The fact that for dwarfism p is not very much larger than u illustrates the efficiency of natural selection in eliminating such mutants. Recessive lethal mutations 1. The gene for j,uvenile amaurotic idiocy has no apparent effect when heterozygous (AI a2) but causes homo zygotes (a2 a2) to die as children (Fig. 25-2). -- E. Figure 25-2 Figure 25-3 2. Neel and co-workers have estimated u from Al to a2 to be 1/100,000. 3. Calculations for the frequency of a2 and the frequencies of Al AI, Al a2, and~ a2 individuals in a populationatequilibrium are shown in the Figure. 4. Note that heterozygous carriers are 600 times more frequent than afflicted individuals. Genetic load in Drosophila pseudoobscura populations (Fig. 25-3) 1. This fly is common in western United States and Mexico. 2. It has five pairs of chromosomes (see diagram at the left in the Figure), including three large, rod-shaped autosomes. 3. In nature almost all the individuals are alike phenotypically. Yet it was possible to show, through the use of special genetic techniques involving a series of crosses, that individuals in the popillation carry an astounding number of recessive mutant genes in heterozygous condition which if homozygous woilld be deleterious. 4. The percentages of the three autosomes studied which, when made homozygous, showed detrimental mutations of various types is shown at the right in the Figure. 5. Natural popillations, therefore, carry a tremendous load of detrimental mutations. 6. How is this genetic load distributed in the fly popill ation ? a. Consider first one pair of chromosomes. Each member has a 25% chance of carrying a lethal or semi-lethal and a 75% F. G. chance of being free of such a gene (see central part of the Figure). b. In the population (D.25)2 or 6-1/4% of the time both members of a chromosome pair will carry (non-allelic) lethals or semilethals. c. What portion of the popul ation would have one second, or one third, or one fourth chromosome with such a mutant? This can be calculated as the chance thDt one chromosome will have this mutant (0.25) times the chance that its homolog will not (0.75) times 2 (to account for the equal likelihood this happens in the reverse order). d. Both members of a given chromosome pair would be free of lethals or semi-lethals in 56% of individuals. e. When all three autosomes and all types of detrimental mutations are considered together, very few, if any, flies in natural populations are free of a load of detrinlental mutations. The genetic load in man is as large as, or greater than, that in flies. This is expected because, 1. like Drosophila, man is a sexual, outbreeding species. 2. like Drosophila, man's genes mutate to deleterious variants. 3. defective progeny more often are produced from marriages bet\veen related individuals than from marriages in the general population. If most mutants are detrimental, how can they provide the raw material for evolution? 1. Mutants are almost always detrimental in the environment in which the species normally lives. 2. But under abnormal environmental conditions some mutants prove useful. 3. In insects, for example, mutations to DDT resistance are useful in a DDT-containing environment, but are useless, or even harmful, in a DDT-free environment. 4. Other examples are found in microorg:lllisms where mutants occur that confer resistance to antibiotics. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lec- ture or as soon thereafter as pOSSible, mak.ing additions to them as desired. 2. Review the reading assignment. 3. Be ahle to discuss or define orally or in writing the items underlined in the lecture 156 notes. 4. Complete any additional assignment. QUESTIONS FOR DISCUSSION 25. 1. What is expected to happen to the frequency of the gene for retinoblastoma in future generations? Expl ain. 25. 2. Why is the mutation rate, u, for retinoblastoma estimated by the frequency of affected individuals divided by 2 ? 25. 3. What percentage of all achondroplastic babies born in a given generation have an achondroplastic parent? grandparent? 25. 4. Compare the chances for establishment of a new viable mutation in a self-fertilized and in a cross-fertilized population. 25. 5. The frequency of juvenile amaurotic idiots at birth in Sweden is .000083. By means of the Hardy-Weinberg equilibrium formula calculate the probable frequency of Swedish persons who are heterozygous for this gene. 25. 6. What would happen to the frequency in the population of a completely recessive gene if its selection coefficient changed from 1 to 1/4? Explain. 25. 7. Suppose that an autosomal and a sex-linked recessive mutant gene are each equally deleterious to their carriers. With equal mutation rates, which of these mutants will be more frequently encountered in a sexual cross-fertilizing population? 25. 8. Is selection against a dominant gene, other things being equal, more or less effective than selection against a recessive gene? Why? 25. 9. Why is the frequency (q) of a recessive lethal in a population at equilibrium equal to the square root of u? 25.10. What happens to the frequency of heterozygotes relative to that of mutant homozygotes when the gene frequency of a rare mutant is reduced by a change in mutation rate? 25. 11. What is the expected frequency at equilibrium of a completely recessive gene which has a selective coefficient (s) of .09 and a mutation rate (u) of . 000036 ? 25.12. What adaptive value would a completely recessive gene have whose mutation rate was 1 per million and equilibrium frequency in the population was. 001 ? .01 ? How frequent would be the mutant homozygotes in each case? 25.13. What would be the mutation rate of a completely recessive. gene which affects 1. 44% of a population at equilibrium and has the following adaptive values? a. 1 b. 0.25 c. 0.36 25.14. In a population observing the Hardy-Weinberg rule how frequent would homozygotes be if the mutant allele had a frequency of 20%? 25.15. Referring to Figure 25-2, how many homozygous normal people would there be in a population of 900,000? 25.16. Describe how you would proceed to determine whether a female Drosophila pseudoobscura collected in nature possessed a chromosome 2 carrying a recessive lethal. Assume males of the same species are available having one chromosome 2 wild-type, and one containing both a mutant dominant to its Wild-type allele and inversions which completely prevent the appearance of crossovers between these homologs. 25.17. From Figure 25-3 and the Notes calculate the chance that a fly has a. no lethal or semi-lethal on any of the three pairs of autosomes tested. b. one sub-vital mutant on a third and another one on a fourth chromosome. c. both a female sterile mutant and a male sterile mutant on the same third chromosome. d. all of the above simultaneously. 25. 18. Did genes for DDT resistance arise in response to exposure to an environment containing DDT? Expl ain. 25. 19. What assumptions are made when referring to the adaptive value of a gene in a population? 25.20. Can the adaptive value of a gene differ in males and females? Explain. 157 Chapter 26 SELECTION, GENETIC DEATH AND GENETIC RADIATION DAMAGE lecturer-Tho DOBZHANSKY PRE-LECTURE ASSIGr-..TMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Colin: Chap. 14, pp. 288-291. Sinnott, Dunn, and Dobzhansky: Chap. 18, pp. 246-247; Chap. 19, pp. 259263. Snyder and David: Chap. 24, pp. 359362. Srb and Owen: Chap. 12. pp. 253-254. Stern: Chap. 2"1. Winchester: Chap. 20, pp. 278-287; Chap. 24, pp. 330-333. b. Additional references Allison, A. C. 1956. Sickle cells and evol ution. Scient. Amer., 195: 87 -94. Dobzhansky, Th. 1951. Genetics and the origin of species. 3rd Ed. 364 pp. New York: Columbia University Press. Dobzhansky, Th. 1955. Evolution, genetics, and man. 398 pp. New York: Wiley & Sons, Inc. LECTURE NOTES A. Natural selection 1. Most mutants are harmful and contribute to the genetic load which Mendelian populations carry. 2. Mutation is the price paid by a population for possession of raw materials from which adaptations to environmental changes can be built by natural. selection in the process of evolution. 3. Harmful mutants are eliminated by natural selection, reducing the genetic load. 4. But varying numbers of generations elapse between the time a mutation occurs and the mutant is eliminated. 5. If, per generation, more mutations arise 158 B. C. than are eliminated the genetic load increases, and vice versa. 6. However, an .equilibrium is reached when, per generation, origination and elimination of mutants are equal. Genetic death 1. Harmful mutants are eliminated by genetic death. 2. Such death is not necessarily synonymous with death of an individual. 3. Genetic death refers only to the inability of a genotype to produce descendants. 4. A person with retinoblastoma (Chap. 25) who dies of this cancer suffers genetic death just as does an achondroplastic dwarf (Chap. 25) or a normnl person who fails to marry. 5. In a population at equilibrium the genes producing achondroplastic dwarfism and retinoblastoma, though different in the severity of the damage they do, both produce genetic deaths in numbers equal to their respective mutation rates. 6. However, the number of individuals in the population carrying a particular mutant depends also upon how damaging the mutant is. Genetic damage can be produced by exposure t@ high energy radiations or other mutagenic agents (see Chap. 23). 1. The top half of Fig. 26-1 shows the genetic load in a population initially at equilibrium which at some point is exposed to additional high energy radiation generation after generation. The genetic load increases until a new equilibrium is reached. Then each generation the number of genetic deaths equals the number of new mutations. 2. The bottom half of Fig. 26-1 shows the genetic load of a population receiving additional radiation for a limited number of generations. The genetic load increases and when the additional radiation exposure stops the load gradually returns to the old equilibrium, but only via genetic deaths. Figure 26-1 3. Regardless of their number, mutants add to D. the genetic load and are removed only by genetic deaths. 4. Mutants usually persist through a number of generations before they are eliminated. Amount of r::l(liatiol1-induced genetic damage 1. It is desirable to know a. how many mutants of various sorts are produced by a given dosage, b. how many genetic deaths will be produced,. c. how many generations will be required to accomplish this. 2. Precise estimates meet with clifficul ties. a. For hUmans, little is known regarding G. Dl. E. F. b. Mutants are usually harmful when homozygous and most, if not all, show also some effect when in heterozygous condition (Chap. 23). However, the relative frequencies of mutants that are harmful and beneficial when heterozygous are unknown. Genetic variants showing heterosis 1. In natural popul ations of Drosophil a there are a number of mutants which, though harmful when homozygous, are heterotic, better than the normal homozygote, when heterozygous. 2. There is, so far, one well-established case of a heterotic mutant in man. Sickle cell anemia and heterosis 1. Humans who are homozygous for the sick- H. ling gene (A2 A2) usually die from anemia before adolescence (see Chap. 10). 2. Al Al individuals are normal. 3. Heterozygotes (AI AZ) are phenotypically either normal or have slight anemia. 4. But Allison has discovered that a. heterozygotes are resistant to certain forms of malaria; b. A2 is particularly frequent in those regions having the forms of malaria to which the heterozygote is relatively immune. 5. Suppose the fitness of heterozygotes is 1 in malarial countries. There, normal homozygotes have a lower fitness, say 1- SI, because they are not resistant to malaria. Mutant homozygotes have a fitness of 1- s2. Here s2 equals 1, fitness being zero since they all die. 6. What natural selection does in such a case is to maintain both Al and A2, A2 having a frequency equal to S'ldivided bYSl + s2. 7. When the heterozygote is more adaptive than either homozygote natural selection will maintain a gene in the popul ation even though lethal when homozygous. 8. Note, however, that in countries where malaria is absent or less frequent, Al A2 has no advantage and may even be slightly disadvantageous as compared with Al AI. 9. This is supported by the fact the sicklecell anemia is rare or absent in most of the world where certain forms of malaria are absent. Balanced polymorphism 1. The hereditary material in the population is maintained in more than one form, is ~ morphic, if heterozygotes are superior to homozygotes. 2. Ambivllient, heterotic mutants are maintained in a popul ation by the reI ati ve fi tness of the heterozygote and the mutant homozygote. 3. There are, therefore, two kinds of genetic load: a. mutational load maintained by recurrent mutation and eliminated by genetic death; b. balanced load maintained not so much by mutation as by selection pressure. Such a load is retained because natur31 selection keeps mutants at such a frequency that the average fitness of the whole population is maximized. It is only after we 811al1 learn the relative importance of mutation:il and bal anced loads in 159 human and other popul ations that we shall be able to make more precise estimates of genetic damage which these populations are likely to suffer. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 160 QUESTIONS FOR DISCUSSION 26. 1. Compare the effects of mutation and natural selection on the genetic load. 26. 2. If most mutants are harmful how can they furnish the raw materials for evolution? 26. 3. Give specific examples in three different species of genetic deaths which actually produce cadavers. 26. 4. What is the relation between Darwinian fitness and genetic death? 26. 5. How would the curves in Fig. 26-1 be changed, if, coincident with the first generation exposed to extra radiation, a medicine was discovered which made each person 20% healthier? 26. 6. Assume that a sexually reproducing population remains stationary in numbers generation after generation, and that every pair of parents produces two surviving children. In a certain generation there appears a single mutant individual which is neither superior nor inferior in adaptive value to the prevailing condition. What is the chance that this mutant will be present in one individual in the next generation? that it will be lost? that two mutant individuals will appear? 26. 7. Suppose that the popul ation is like that described in the preceding problem, but that some parents leave no surviving progeny at all and other parents leave one, two, or several offspring. . What are the chances of loss, of retention, and of increase in frequency of an adaptively neutral mutant in such a population? 26. 8. The frequency of a Race A of wheat rust increased manyfold in the United States during a period when the frequency of a related Race B of the same species declined and became nearly extinct. Suggest explanations which could be tested. 26. 9. How would you explain the following situations? a. The frequency of a rare gene in a large popul ation is greater than expected on the Hardy-Weinberg rule. b. The frequency of a rare gene in a small population is lower than expected but increases a certain amount as the population size rises in subsequent generations. c. A gene, lethal when homozygous, has a frequency in the popul ation equal to the square root of its mutation rate. d. The frequency of a recessive detrimental gene in a large population gradually increases in successive generations. 26.10. With regard to sickle cell anemia in malarial countries, how could the frequency of the normal gene Al be expressed in terms of selection coefficients? 26.11. Melanic (darker colored) variants have appeared in several species of moths in England and Germany, nearly always first near large industrial cities, where in some cases they have suppl anted the normal type of the species. Assuming that the dark and light types differ in one or a few genes, outline an explanation for the above facts, together with suggestions for testing the hypothesis experimentally. 26.12. What information would be desirable in order to correctly evaluate the consequences of raising the mutation rate in man? 26.13. What are the criteria for balanced polymorphism? 26.14. Compare very detrimental and slightly detrimental genes with regard to their relative roles in mutational and in balancedloads-; 26.15. Do geneticists disagree on the harmfulness of most mutants? Explain. 26.16. Explain which would be expected to suffer more from genetic damage caused by a large continued increase in radiation exposure through fallout: a. rapiclly or slowly dividing tissues in the same organism; b. a species with a long or short life cycle; c. species reproducing sexually or asexually; d. germinal or somatic tissue; e. diploids or polyploids; f. adults of mammals or of insects. 161 Chapter 27 THE GENETICS OF RACE Lecturer~ Th. PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. GeneraJl genetics textbooks Altenburg: Chap. 26, pp. 462-463, 458460. Colin: Chap. 7, pp. 113-114; Chap. 17, pp. 389-395. Sinnott, Dunn, and Dobzhansky: Chap. 20, pp. 277-285. Srb and Owen: Chap. 24, pp. 537-540. Stern: Chap . 81 ; C'l-tap. 32. b. AdditionaJl references Dobzhansky, Th. 1951. Genetics and the origin of species. 3rd Ed. 364 pp. New York: Columbia University Press. Dobzhansky, Th. 1955. Evolution, genetics, and man. 398 pp. New York: Wiley & Sons, Inc. Dunn, L. C., and Dobzhansky, Th. 1957. Heredity, race, and society. 3rd Ed. New York; The New Amer. Libr. of World Lit., Inc. LECTURE NOTES A. Number of human races 1. Racial variations exist. 2. In the 1700s people were divided into races according to skin color. 3. Since people differ also in other traits, anthropologists later divided mankind into greater and greater numbers of races to accommodate these differences. 4. There was confusion, however, even when 200 races were proposed. B. Use of averages to define races 1. Numerous individuals are studied with regard to a number of traits. 2. The data collected are treated statistically to obtain mean vaJlues for the characteris162 DOBZHANSKY tics of a race, thereby prOviding a race standard. 3. While this is a satisfactory procedure for a preliminary orientation in classification, it leads to trouble when studying the basis for races. C. Racial mean values and heredity 1. Race averages describe the ideal racial type. 2. Does this average type have any hereditary basis? 3. According to the blood theory of inheritance it would represent the end product produced by heredity as it blended generation after generation until finally the population became a genetically uniform race. 4. But, in fact, Mendelian heredity occurs, and sexually reproducing popul ations never become genetically uniform, pure races. There is no such thing as a mean genotype characteristic of a race. 5. Phenotypic criteria are often unsatisfactory for describing populations or races. D. Characterizing populations genetically 1. Since people are either A, B, AB, or 0 blood type and there are no intermediates, populations have to be described in terms not of mean blood types, but of relative frequencies of different blood types. These types are inherited in a simple Mendelian manner (see Chap. 5). 2. As shown in Dr. Mourant's map, 3/4 of the gene pool has 0 blood type in western Europe, Iceland, Ireland and parts of Spain. Eastward the frequency decreases. 3. The opposite is true for the B blood group gene. 4. In a world map the B gene is most frequent in central Asia and some populations in India, and becomes less and less frequent as one proceeds away from this center. E. F. G. 5. Among American Indians B is rare or absent. 6. In this way racial variations can be described by the relative frequencies of different genes in the gene pool, blood type serving as an appropriate example since it is the genetic trait studied in most detail in man. The reality of races 1. Are racial differences real ? a. Yes, as can be seen by simple observation of phenotypes. b. Yes, genetically, since one can objectively compare two populations with regard to gene frequencies. If these are the same the two groups are racially alike, and vice versa. 2. How many races are there? a. The number is arbitrary, a matter of convenience, depending upon the purpose. b. Races are not divided by sharp lines; for example, the frequency of the B blood group gene changes gradually in going from one population to an adjacent one. c. Anthropologists Coon, Garn, and Bird. sell recognize about half a dozen basic races of m:inkind, or define about 30 races when finer details of some populations are to be considered. Races of Drosophila pseudoobscura 1. This fly is common in southwestern United States (see p. 7 ). 2. Here flies are very similar phenotypically al though they may differ with respect to their chromosome structure. 3. Different populations can be described in terms of the relative frequencies of the different gene arrangements they carry. a. California populations are rich in a chromosomal type called Standard. b. Populations of Arizona and New Mexico are nearly uniform with respect to the Arrowhead chromosome configuration. c. Pikes Peak arrangement is most common in Texas. Adaptedness of different populations 1. In D. pseudoobscura different chromosome types are adaptively different in different environments, as shown in 1aboratory experiments. 2. There is reasonable certainty that in nature, too, these gene arrangements are adaptive, the differences between populations being the result of natural selection. 3. Similar results were obtained by Clausen, Keck, and Hiesey in studies of three races of a plant species. H. r. J. a. One race lives at sea level, another at mid-elevation in mountains, the third in the alpine zone. b. The sea level race is killed when grown in the alpine environment. c. The alpine race grown at lower elevations proves less resistant to rust fungi than the lower elevation races. d. In this way, each race, or population, was shown to be adapted to the conditionS of its habitat. Adaptedness of a species 1. Each species occupies a certain territory. 2. Different parts of the territory usually have different inorganic and organic environments. 3. How does a species reach maximum fitnesS in all parts of the territory it occupies? a. Not by a single genotype, since no one genetic endowment would be equally adapted to all the ,d ifferent environments encountered. b. One method is through the differentiation of the species into geographic populationS or races, differing from each other genetically in such ways as to produce maximum fitness of the species as a whole. Allopatric and sympatric races 1. Races of cross-fertilizing species are usually allopatric, occupying geographically separate territories. 2. Different species may be allopatric or they may be sympatric, i. e., occupying the same territory. In the latter case the two specieS are kept separate by reproductive isolating mechanisms (Chap. 28). 3. Races are not always allopatric but are sometimes sympatric. a. Man Several thousand years ago mankind was differentiated into allopatric races. Development of civilization and improved communications have made such races sympatric in part. Gene exchange in the now-sympatric races may be prevented by social and economic forces. b. Domesticated plants and animals There are dozens of different breeds of dogs living close together in any sizable city. These races are sympatric yet do not exchange genes to form one mongrel breed because their reproduction is controlled by man. Genetic diversity yet individual equality Dr. Dobzhansky points out: "All men are cre- 163 ated equal. But they are not alike genetically. Genetic diversity is however compatible with equality. The diversity is a biological, the equality is an ethical and a religious concept. People are genetically unlike, but they are equal before the law and before God. " POST-LEe TURE ASSIGl\TMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desir~d. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 164 I QUESTIONS FOR DISCUSSION 27. 1. What is the relationship between a race and a popul ation ? 27. 2. Write a paragraph of 100 words or less on the meaning of the title of this chapter. loid races? Explain. 27.13. Are races permanent entities? Give an example to support your view. 27. 3. Is it possible for a species to consist of just one race? Explain. 27.14. Is genetic study of populations required before decisions regarding their racial classification can be made? Explain. 27. 4. Why is the method of race averages inadequate to describe races of cross-fertilizing species? 27.15.. Are overlapping inversions useful in deducing the relationship between two interfertile races? Explain. 27. 5. 0, A, B, and AB are different in human popu- 27.16. Is a knowledge of human genetics useful to anthropologists? Explain. lations of different parts of the world. What working hypotheses can you suggest to account for these different frequencies? 27.17. What would have to happen to races before they could not lose their identity? The relative frequencies of the blood groups 27. 6. In what respect is the study of blood group types less satisfactory for understanding racial origin than is the study of gene arrangernentin D. pseudoobscura? 27.18. What genetic explanation can you give for the lack of clearcut boundaries between races? 27. 7. Explain why the frequencies of the following are or are not expected to conform to the Hardy-Weinberg formula. a. ABO blood group genes in man b. Gene arrangement types in D. pseudoobscura c. Genes for size in dogs d. Genes for presence and absence of chlorophyll in a self-fertilizing plant 27. 8. In a population mating at random (as in man) what will be the relationship between effective size of the interbreeding population and the frequency of cousin marriage? 27. 9. Discuss the genetic differences between races. 27.10. Give two reasons why animals and plants under domestication are more variable than corresponding wild species. 27. 11. Discuss the statement: "From the standpoint of modern genetics races are real entities existing in nature, not merely the inventions of some anthropologists and biologists." 27. 12. Do you support the view that all humans belong to the Europoid, Negroid., or Mongo165 Chapter 28 THE ORIGIN OF SPECIES ledurE!r-G. L STEBBINS PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 27, pp. 483-484. Colin: Chap. 14, pp .. 193-194. Dodson: Chap. 19, pp. 233-236. Sinnott, Dunn, and Dobzhansky: Chap. 21, pp. 286-289, 290-296 . Srb and Owen: Chap. 21, pp. 462-464. Stern: Chap. 33. b. Additional references Dobzhansky, Th. 1951. Genetics and the origin of species. 3rd Ed. 364 pp. New York: Columbia U'liversity Press. Dobzhansky, Th. 1955. Evolution, genetics, and man. 398 pp. New York: Wiley & Sons, Inc. Stebbins, G. L. 1950. Variation and evolution in plants. 643 pp. New York: Columbia University Press. LECTURE NOTES A. The nature of species 1. In sexually reproducing organisms a species consists of a number of races adapted to the regions they occupy. These races are Mendelian populations kept in genetic continuity by intermediates. 2. Different species are genetically discontinuous, brought about by their inability to hybridize, or if they do, the hybrids are more or less sterile. B. Morphology alone may not be adequate for defining species. 1. As was shown, dogs of different breeds appear so different that an Irish wolfhound and a Boston terrier might be classed as different species. But all breeds belong to one species because they can interbreed and 166 C. D. form fertile mongrel or hybrid offspring. 2. Although the coyote may at first glance look like some dogs, it has certain characters which do not appear in dogs, and for which there are no intermediates. Dog and coyote are separate species because their hybrids often die before maturity, so that the two forms remain reproductively isolated. The nature of reproductive isolating mechanisms 1. They include several ,d ifferent kinds of barriers. 2. Each is produced by a 1arge number of genes and chromosome segments. 3. Any two species are separated by several different ones. 4. The amount of morphological difference is not well correlated with degree of reproductive isolation between related species. These four items are discussed in turn. Kinds of reproduCtive isolating barriers 1. Ecological In California the Monterey cypress grows along the coast on the rocks (Fig. 28-1). The Gowen cypress grows two miles inland in the sand barrens. The hybrid between them would have no chance in nature, although it can be raised in the experimental garden. 2. Seasonal In the same area is Evolution Hill where two species of pine live near each other but show no intermediates. The Monterey pine sheds its pollen before March, the bishop pine sometime later. However, hybrids can be obtained experimentally. 3. Sexual or ethological a. Drosophila pseudoobscura and D. persimilis show mating preferences for their own species, due apparently to females rejecting males of other species MAPS SHOWING 1lI[ LOCATHW OF HW RELATED AND INTERFfRTll[ SPECIES OF CYPRESS (CUPR[SSUS MACROORPA AND CGOVENlANA)IN CAUFORNIA . Figure 28-1 (Fig. 28-2). b. Color in fishes and feather color in birds are also a basis for mating preferences. In the case of pheasants, females of different species are quite similar, whereas the males are very different. Hybrids are rare because, apparently, the female is stimulated only by males of her own species. MATING PRErtRtNCtS AS REPRODUCTIVE ISOLATING MECHANISMS IN DROSOPI-II LA FEMALES PSEUOOOBSCURA MAlf:S I PERSIMILIS INSfMINATtO VIRGIN INSEMINATED VIRGIN PSEUDOOBS(URA 84.3 15.7 7.0 93.0 PERSIMILIS 22.5 775 79.2 20.8 Figure 28-2 E. 4. Hybrid inviability This was already mentioned in B2 for the case of dog x coyote. 5. Hybrid sterility a. The mule is a vigorous hybrid, but sterile. b. Sterility can be caused in two different ways. c. Genic sterility results from disharmony in development, often of the gonads and/ or the sex hormones, so that the hybrid may not show secondary sex characters. This was illustrated by the hybrid of turkey and pheasant which shows neither the wattle of the tom turkey nor the ring or long tail of the cock pheasant. d. Chromosomal sterility is due to differences in the arrangement of the chromosome segments in the two species (because of translocations, inversions, etc.), so that the hybrid produces irregular pairing at meiosis. This results in irregular chromosome distribution to the gametes, which are therefore rendered non-functional. This was shown in hybrids between Crepis negleda (n=4) and C. fuliginosa (n=3) which had, in meiosis, chromosomes which were unpaired, paired, and in groups of 3 and 4, resulting in pollen sterility. The multiple genic or chromosomal basis for any particul ar barrier 1. Drosophil a pseudoobscura x D. persimilis produces sterile males but partly fertile females. The hybrid female can be backcrossed to pseudoobscura males, and, with additional crosses, offspring may be obtained which possess various combinations of the chromosomes of the two species (Fig. 28-3). These combinations were detectable because different chromosomes carried suitable genetic markers. These combinations were tested for fertility, as reflected in testis length. When the X is from pseudoobscura the testis is essentially normal in length. When the X is from persimilis the testis is shorter, and becomes even more abnormal as more and more of the autosomes come from pseudoobscura. 2. Simil ar results of Mitntzing show a multichromosomal basis for sterility of hybrids between Gal eopsis species. 167 S[GR[GATION FOR CHROMOSOMAL COMPOSITION & TESTIS 51Z£ IN BACKCROSS PROGENY Of HYBRIDS (DROSOP~ILA PS EUDOOBSC URA~x PERSIMJLlSd')~)( PS[uDOOBS(URAd' = - CHRONOSOM£S OF D. PSEUDOOBSCURA CHROMOSOMfS OF I), PERSIMllIS llN6JU OrTESnS IN MICRA CHROMOSOMES 0 1 2 X [:::=J [:::=J 3c:::::::::J 4c:::::::::J 5c:::::::::J 6 [:::=J 7c:::::::::J 2 3 4 ----- -- --, -- -- --- -= - -. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = - . ~. ~~". - " ~ ...._ . I ....!"........ ,;- • ~', .. ~ \" • '1: • • ·w H. -~ = = = = = 8c::= 9_ = = = = 1 0= = 11 __ = = = = = 12_ = = = 13_ = = = = = 14_ = = = = = 15- = = = = 16_ 25 50 100 200 400 700 - = = = = = = ... ~ ' _ 0- ,:.- ~4; , .. 1> , ..... # •:of" .., .. ~ ~.!J ... o· ~. J>. : N . .. '.1' :.t' . "" ~ o 2S 50 100 200 400 700 Figure 28-3 F. G. 168 Any two species are separated via a number of reproductive barriers. In the case of D. pseudoobscura and D. persimilis these barriers include ecology (the former lives in a dryer and warmer habitat), mating preferences, mating behaviors (usually pseudoobscura mates in the evening, persimilis in the morning). genic sterility, hybrid inviability, and hybrid breakdown in future generations. Each barrier is not a complete one, but taken together they cause complete reproductive isolation so there is no gene exchange in nature. Morphology is not well correlated with reproductive isolation between related species. 1. Genera which fail to produce hybrids when crossed usually differ morphologically. This is true for cottontail rabbit x jack rabbit or hare, and for spruce tree x pine or fir. But European cattle and the Tibetan yak (usually placed in different genera) can be crossed, and in Tibet many cattle have yak-like traits. 2. Species of one genus may look so similar they cannot be told apart by simple inspection. For example, D. pseudoobscura and D. persimilis were once considered races of the same species. But they are genetically 1. discontinuous, and, recently, careful measurements showed their genitalia differ. Such morphologically similar species are called sibling species, and have been found among other Drosophila, mosquitoes, and other insects. 3. Sibling species are a1 so found in pI ants, in the tarweeds of the aster family, and in the blue wild rye, Elymus glaucus which Stebbins found to consist of a large number of Sibling species. The nature and origin of reproductive isolating mechanisms tells the way that species originate . 1. Valid species do not arise by a simple mutation, but as the result of many different, independently occurring, genetic changes. 2. Species do not arise automatical1y by accumulating the same kinds of changes that separate races. Special kinds of effects, each contributing to reproductive isolation, are needed. A major unsolved problem of organic evolution is just how these special kinds of genetic effects originate and become established. Some of the factors are known, others inferred. 1. Populations must usually be separated geographically while reproductive barriers are being buH t up, otherwise hybridization would break down barriers as they were being built up by selection. Plants with a high rate of self-fertilization provide an exception. In Elymus glaucus Stebbins found many microspecies seem to have arisen in the same general area, if not side by side. Another exception is the case of aphids which might change their host suddenly and form a population that would not have a chance to exchange genes with one living on the old host. 2. Accumulation of genetic factors promoting reproductive isolation is furthered by natural selection, acting both directly and indirectly (see Chap. 29). POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. QUESTIONS FOR DISCUSSION 28. 1. Distinguish between races and species in cross-fertilizing organisms. 28. 2. In order to decide whether a new type of plant or animal which arises under observation is a new species or not, what criteria would you employ? 28. 3. Is morphology a good criterion for distinguishing species? Expl ain. 28. 4. What criteria would you suggest to distinguish species among asexual organisms? 28. 5. Do you believe that different kinds of reproductive isolating barriers between races first arise and only later acquire a genetic basis? Explain. 28. 6. Design an experiment to test mating preferences between two morphologically identical species of Drosophila. 28. 7. If chromosomal sterility has a genetic basis, how does it differ from genic sterility? such restricted distributions? 28.14. Do the mechanisms for the formation of species apply also to the formation of higher taxonomic categories? Explain. 28.15. Discuss Dobzhansky's contributions to our understanding of reproductive isolation. 28.16. Darwin observed that species belonging to large genera tended to be more variable than species of small genera. What genetic explanation can you offer for this difference in variability? 28.17. What genetic interpretation can you suggest for the fact that species have apparently evol ved much more rapidly among flowering plants than among gymnosperms (coniferous trees and their allies) ? 28.18. Are Drosophila triploids reproductively isolated from diploids? Explain. Has this any bearing on the origin of species? Why? 28. 8. From Fig. 28-3 give evidence as to which pseudoobscura chromosome has the least effect on testis length when the X is from persimilis. 28. 9. Is the ability or inability to successfully cross two individuals in the laboratory or garden a good criterion for classifying them as the same or different species? Explain. 28.10. What do reproductive isolating mechanisms tell about the mechanism of speciation? 28.11. Describe briefly one cause of reproductive isolation in each of the following cases: 1. Monterey and bishop pine 2. Turkey and pheasant 3. Crepis neglecta and C. fuliginosa 4. Species of Galeopsis 5. Dog and coyote 28.12. Can sympatric races become reproductively isolated? Explain. 28.13. Certain species, or genera, are endemic, occupying a range restricted to a certain region. What explanations can you offer for 169 Chapter 29 INTERSPECIFIC HYBRIDIZATIION AND ITS CONSEQUENCES lecturer-G. L STEBBINS PRE-LECTURE ASSIGNMENT 1. QUickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 16, pp. 278-281. Colin: Chap. 14, pp. 296-301. Dodson: Chap. 14, pp. 172-174; Chap. 19, pp. 239-240. Sinnott, Dunn, and Dobzhanslcy: Chap. 21, pp. 289-290, 296-302. Snyder and David: Chap. 20, pp. 297302. Srb and Owen: Ch.ap. H , pp. 217-219, 224-231. Winchester: Chap. 15, pp. 213-217; Chap. 24, pp. 340-341. b. Additional references Dobzhansky, Th. 1951. Genetics and the origin of species. 3rd Ed. 364 pp. New York: Columbia University Press. Dobzhansky, Th. 1955. Evolution, genetics, and man. 398 pp. New York: Wil ey fir Sons, Inc. Stebbins, G. L. 1950. Variation and evolution in plants. 643 pp. New York: Columbia University Press. LECTURE NOTES A. Accumulation of genetic factors responsible for reproductive isolation can be aided by natural selection in two ways. 1.~ This explains seasonal isolation of Monterey pine from the more northern bishop pine. 2. Inclirectl.l a. In Moore'S studies of leopard frog races (Fig. 29-1) northern females crossed to males from progressively more southern regions produce offspring whose developruent is progressively slower, and head 170 size progressively larger. The reciprocal crosses, of southern females with more and more northern males, produce hybrids which have a still slower rate of development and whose head becomes progressively smaller. INCR~A~ING INVIABILITY O~ lfOPARO mOG (RANA PJPJENS) HYBRIDS AS GEOGRAPHIC DISTANCE BETWHN PARENTAL RACES B[(OMfS GREATER ~ V[RMONT NEW Jt:RSEY VERMONT 45°N RATE LHfJID (::lcoNTRltS (::1 ::!: NORMAL NOItMAL + - OCALA. ---FLORIDA TEXAS NEW JERSEY OCALA. flORIDA 40°1'1 ±OR- :!:OR+ --- +++ (~)cIlNTRllS(~) - -- --- -- ----------- ... TEXAS 2S'N RATE I HEAD RATE IHEAD + (~}CONTIIIlS!1 !) _--- ++ + 32'N RATE ---~ - - .... = -- 1 HEAD ++++1" +++ ++ (!ICONTROLS( ~I Figure 29-1 Al though these races are still continu0us and can exchange genes through intermediate forms, this type of hybrid inviability is similar to that separating different species of frogs. This suggests natural selection has sorted out different rates of development with different temperature optima, producing developmental disharmony in hybrids. b. Reinforcement of reproductive isolating barriers B. If strains of Drosophil a pseudoobscura and D. persimilis are placed in a population cage, some hybrids are formed because isolation by mating preference is not complete. Using genetic markers for identification, Koopman aided selection by removing all Fl hybrids, thereby penalizing females which accepted males of the wrong species. After several generations the two groups were even more strongly isolated by mating preference. Reproductive isolating mechanisms may originate through interspecific hybridization. In pI ants, particul arly, interspecific hybrids may be converted into stable intermediate types, that are isolated from both parental species, by several methods. 1. Amphiploidy (doubled hybrids, a term similar to allopolyploidy) (Fig. 29-2). a. The theoretical Fl hybrid is sterile because of irregular chromosome pairing at meiosis. If, however, the chromosomes present are doubled, artificially by colchicine or spontaneously, the chromosomes, now paired, behave normally in meiosis and produce fertile haploid gametes. The amphiploid is phenotypically intermediate to, and isolated from, both parental species. Radish (2n:;18) x cabbage (2n:;18) pro- . duced an Fl hybrid with 18 unpaired chromosomes and an intermediate type of pod. If the hybrid's chromosomes become doubled an amphiploid is produced with 36 chromosomes (9 pair each from radish and cabbage). ' d. The New World common cultivated cotton (Fig. 29-3) Gossypium hirsutum and sea island cotton G. barbadense are tetraploids with 2n=52. These are phenotypically intermediate between the diploid species from Africa, G. herbaceum, and from Peru, G. raimondii, which each have 2n=26. Cytological studies of triploid hybrids showed in meiosis: 1) barbadense x raimondii gave 13 pairs + 13 singles 2) barbadense x herbaceum gave 13 pairs + 13 singles 3) raimondii x herbaceum gave 26 singles This can be explained by the behavior of the following chromosome sets, respectively: 1) AA DD x DD produces DD + A 2) AA DD x AA produces AA + D 3) DD x AA produces AD Therefore barbadense is an amphiploid of raimondii and herbaceum. ANCESTRY OF NEW WORLD COTTONS SPECIES LEAF Co. HERBACEUM AFRICAt+UM 2:n-2.6 . AA . 'V..... BRACT CAPSULE ~ W Q \:1b 6 / G . TOMENTOSUM 2n- 52 . AADO . ~J 6 Q C£] n Q 4J G - G. BARBAOE: NSE. 2" .. '52. AAOD. Figure 29-2 b. Of present flowering plant species 20% to 25% have originated this way, and as many or more have done so in the past and then diverged to form different genera. c. Example of artificially produced amphiploidy (Karpechenko) C. H1RSUTUM YA" . PUN(.TATUM 21"1 · 52.A"'DD. G. RAIMONDI! Zn.26. DO . " I 6 Figure 29-3 171 e. But these two parental species live on different continents. How could they have produced the amphiploid barbadense? 1). It is unlikely man brought these two species together. 2). More likely is that there was once present in the New World a species like herbaceum that hybridized with raimondii's ancestors. This is supported by the fact that, although there is no fossil record for cotton because its leaves are too fragile, there is fossil evidence for the presence in the New World of leaves of woody plants belonging to the Old World tropics. These species, living about 60 million years ago in the New World, are now extinct. Moreover, G. tormentosum, native of Hawaii, has 2n=52 (AADD), and probably arrived there by long distance dispersal. f. Three species of goatsbeard Tragopogon (po rrifol ius , pratensis, and dubius), native to Europe, were introduced to eastern Washington about 40 years ago. All have 2n=12, forming sterile interspecific hybrids. But about 10 years abo Ownbey found T. mirus which was phenotypically intermediate between dubius and porrifolius, and which was shown to be an amphiploid of these species; he also found T. miscellus to be the amphiploid of dubius and pratensis. This illustrates species origin in recent times. 2. Introgression can also stabilize the products of interspecific hybridization. This process, described by Dr. Edgar Anderson, involves the establishment of new variant types as a result of backcrossing of an interspecific hybrid to one of its parents, followed by selection for favorable backcross recombinant types. a. Artificial introgression has been demonstrated in tobacco, Nicotiana. N. alata x N. langsdorffii (both 2n=18) produce hybrid F1 which can produce an F2 containing a rich array of recombinant types. If, for example, alata-like F2 are backcrossed to alata, a true breeding type can be obtained having the large flowers of alata but the color and some rounding of the corolla lobes as in langsdorffii. b. Natural introgression is of two types: Sympatric, which occurs between two species having the same geographic 172 range and increases the variation of one of those species through backcrossing in the original range. Allopatric, which occurs when products of backcrossing have a different enough adaptive ability to colonize new habitats. c. In nature both types are demonstrated in two genera of shrubs of the rose family, the bitterbrush Purshia tridentata and the cliff rose Cowania mexicana, in western United States (see p.14). The appearance of the two is very different. In Utah, bitterbrush seedlings vary in the direction of cliff rose; elsewhere this is not so. Moreover, several partly fertile hybrids have been found in this, utah area. This, then, is evidence of hybridization and sympatric introgression in Utah. P. tridentata variety glandulosa has apparently acquired heat resistant genes from Cowania and has extended the range of Purshia to new, more southerly, hotter habitats. d. Introgression in cultivated plants Mangelsdorf has shown, using archeological and experiment31 breeding evidences, that evolution of corn through artificial selection was aided by genes incorporated from teosinte, Zea mexicana. In a brilliant application of theoretical genetics to practical plant breeding, Sears was able by artificial introgression to introduce into wheat, segments of goat grass chromosomes containing genes for rust resistance (Fig. 29-4). 3. Direct recombination of the products of crossing and the establishment of recombinations offers another way products of hybridization may be stabilized. This is demonstrated in the work of Lewis and Epling on the larkspur genus Delphinium (Fig. 29-5). a. D. gypsophilum is morphologically intermediate between D. recurvatum and D. hesperium. All three have 2n;;16. b. Crosses of recurvatum and hesperium produces Fl. F1 hybrid x _gypsophilum gives offspring which are more fertile and more regular than those produced either (a) by the Fl hybrid backcrossed to either parent or (b) by crossing gypsophilum with either of the two other species. c. It is very likely that gypsophilum, TRANSFER OF LEAF-RUST RESISTANCE FROM AEGILOPS UMBELLULATA TO WHEAT (Sears, E. R. 1956. Brookhaven Sympos. BioI., 9: 1-22.) T. dicoccoides AABB - 14 II x A. umbell uJ. ata CUCU - 7 II Triticum aestivum var. "Chinese Spring'_'_'- - ---, AABBDD - 21 II~ '*-,::-- - -_1 F 1 - AABBDCu - 14 II, 14 I BC 1 -<2 ~91~ ~~~. AABBDD + (lor 2CU) 21 II + 1 I or 22 II o~ BC 2 AABBDD - 21 II Explanation of symbols: Plants enclosed in rectangles carry the rust resistance gene. II indicates bivalents, I indicates univalents. Gametic sets of 7 chromosomes: A: wheat, derived from T. macrococcum B: wheat, derived from A. speltoides. D: wheat, derived from Aegilops squarrosa. Cu: Aegilops umbelluJ.ata. Figure 29-4 *~, , ~. ~ D.G PSOPHILUM I I I~I I I 'I [ D. R£CURVATUM D. HESPfRlUM PAlLESCENS 't;q C. which lives in a new habitat probably made available only since the pluvial period of the ice age, arose by hybridization between the older,. more widespread recurvatum and hesperium. Hybridization can be a powerful stimulus in species formation and an experimental tool for analyzing the genetic basis for differences between species. In this way it helps to solve the problem of the origin of species. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. Figure 29-5 173 QUESTIONS FOR DISCUSSION 29. 1. To what can the differences obtained from reciprocal matings in Moore's experiments be attributed (Fig. 29-1)? 29. 2. In some groups of animals and plants the chromosome number of one species is a multiple of that in another. Thus in wheat some species have 14, some 28, and some 42 chromosomes. What does this suggest as to the evolutionary history of these species? 29. 3. Draw a figure of mitotic metaphase for each of the two theoretical parental species whose diploid hybrid is shown in Fig. 29-2. 29. 4. Describe in words what happens in meiosis of "diploid hybrids" and of "amphiploids" as illustrated in Fig. 29-2. 29 .. 5. Is amphiploidy a common method of species formation in animals? in plants? Explain. 29. 6. How does amphiploidy differ from aneuploidy? polyploidy? 29. 7. What reason can you suggest for the fact that it is apparently impossible to increase chromosome number experimentally beyond a certain point? 29. 8. What are "mitotic poisons!!? Have they any role in evolution? Explain. 29. 9. Why does polyploidy tend to prevent the phenotypic expression of point mutations? 29.10. Give two examples of highly polyploid cells that no longer divide. 29.11. Referring to the Notes and Fig. 29-3, explain how, in Gossypium, interspecific crosses indicate barbadense to be an amphiploid of raimondii and herbaceum. 29.12. Using a case from modern times, illustrate how interspecific hybridization has led to speciation via amphiploidy. 29.13. What relationship can you suggest between the occurrence of amphiploids and the occurrence of duplicate factors? 174 29.14. What factor affecting size might be operative in amphiploids but not in autopolyploids? 29.15. In polyploid series of species in nature (such as the Wheats, barleys, and many others) the differences in body size and in cell size between members of the series are often very much less than in polyploid series which have been developed experimentally. Explain. 29.16. What are the similarities and differences between amphiploidy and introgression? 29.17. How might an amphiploid arise other than by doubling the chromosome number of a hybrid? 29.18. Describe in words the transfer of leafrust resistance from Aegilops to Triticum as outlined in Fig. 29-4. 29.19. What have studies of larkspurs contributed to our understanding of speciation? 29.20. Is gene mutation the only source of raw material for evolution? Explain. Chapter 30 INBREEDING AND HETEROSIS lecturer-J. F. CROW PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 7, pp. 136-147 . Colin: Chap. 13, pp. 248-270. Dodson: Chap. 21, pp. 255-256; Chap. 10, pp. 118-124. Sinnott, Dunn, and Dobzhansky: Chap. 19, pp. 254:"'258. Snyder and David: Chap. 16, pp. 229242. Srb and Owen:. Chap. 16, pp. 330-333, 336-340. Winchester: Chap. 25, pp. 346-35l. b. Additional references Crow, J. F. 1955. Genetics notes. 2nd Ed. 124 pp. Minneapolis: Burgess Pub!. Co. Gowen, J. W. (Editor) 1952. Heterosis. 552 pp. Ames, Iowa: Iowa State College Press. LECTURE NOTES A. Review of the Hardy-Weinberg principle (see Chaps. 2 and 24) 1. Applied in studying phenylketonuria a. This type of feeblemindedness occurs in individuals, homozygous for a recessive gene, who are unable to metabolize phenyl al anine. b. The frequency in the popul ation of the normal gene (A) is .99, and of the abnormal gene (~) is .01. c. In the population, therefore, AA: Aa: aa individuals will have frequencies of 9801/10,000:198/10,000:1/10,000, respecti vely. d. Notice that Aa individuals are 198 times more frequent than aa, so that even if B. C. every aa did not reproduce only 1% of ~ genes present would be eliminated from the population per generation. 2. Principle assumes mating is random a. This is true for traits such as blood group type in humans. b. Although feebleminded people do not marry in the population at random, this has little effect on the assumption since aa people have so few of all ~ genes. Only marriages between Aa individuals are consequential since they are the major source of aa offspring. AA and Aa apparently marry with each other at random. Departures from random mating 1. Assortive mating is the tendency for phenotypically similar individuals to mate. This is generally true in animals and in human marriages also. 2. Inbreeding is the tendency for mates to be more closely related to one another than they would be were selection of mates made at random from the popul ation. a. Self-fertilization, in plants, is the closest form of inbreeding possible. b. Other examples include brother-sister matings in animals and cousin marriages in man. The effect of inbreeding upon the distribution of heterozygous genes. 1. Self-fertilization Each generation 1/2 of the heterozygosity in the parents becomes homozygosity in the offspring. 2. Brother-sister (sib) matings These reduce heterozygosity, or increase homozygosity, by 1/4. 3. Half-sib matings These involve individuals who have one parent in common. The frequency a given al175 lele in the common parent shall pass to the male half-sib is 1/2, and the frequency an offspring of a male half-sib having this gene shall receive this gene is 1/2. The frequency of both events occurring is, therefore, 1/2 x 1/2, or 1/4. The frequency is also 1/4 for these events to occur through the female half-sib, so that the frequency of a given allele becoming homozygous in an offspring from a halfsib mating is 1/4 x 1/4, or 1/16. Since the other allele could, in this way, also become homozygous 1/16 of the time, the combined chance for homozygosity from half-sib matings is 1/8. In other words, 1/8 of heterozygous genes would be made homozygous by this process. 4. Cousin marriage This type would reduce heterozygosity by 1/16. 5. The inbreeding coefficient (F) designates the amount by which heterozygosity is reduced. The way F is computed for more elaborate pedigrees is given following the notes of this chapter. 6. Inbreeding produces homozygosity, which in turn produces greater phenotypic uniformity in the p·'pul a-' 1011. This is :lccompanied by a reduction in vigor, expressed in various ways. Loss in vigor due to inbreeding is attributed to the production of homozygotes 1. one of which is inferior to the heterozygote; 2. in which recessive genes with harmful effects are unmasked that were previously masked in the heterozygote. Disease risk from consanguineous marriages 1. Phenylketonuria (Fig. 30-1) D. E. P~OIGRH SHOWING A RfCESSIVE OISfASE (PHENYLKETONURIA) IN CHILDRfN fROM COUSIN MARRIAGES ¥Id a. In both of the cousin marriages affected children were produced. b. It can be calculated that there is a 7-fold greater chance for affected children from cousin marriages as from m_arriages between unrelated parents. 2. Data from Japan (Fig. 30-2) were collected by the Atomic Bomb Casualty Commission (and are unassociated with the atomic explosions) . INCREASED RISK Of GfNfTIC DEr[CT WITH COUSIN MARRIAG[S (DATA fROM HIROSHIMA AND NAGASAKI) UNRELATED INCREAS[ WITH PfRCENT PARENTS (OOSIN MARRIAG£ INCREASE CONGENITAL 011 MALFORMATION . .005 48 PfRCENT STILLBIRTHS .025 INFAN r DEATHS .023 .006 24 PERCENT 34 PERCfNT .008 I Figure 30-2 3. The death rate values in Fig. 30-3 include fetal death, all childhood, and very early adul t deaths. How much "genetic death" was carried in this French population? COMPUTATION OF THE NUMBfR OF LETHAL fiOUIVAlENT5 CARRIED AS HfTIROIYGOOS RECESSIVfS BY AN AVERAGE PERSON DEATH RATE FROM COUSIN MARRIAGf .25 DEATH RA1E FROM UNRELATfD PARENTS .12 E~(ESS FROM COUSIN MARRIAGE .1 3 0.13 ~16 ~ 2= APPROXIMATELY 4 LETHAL EQUIVAllNTS PER PERSON (DATA FROM RURAL FRANCE) Figure 30-1 176 Figure 30-3 Because of cousin marriages 13% more people died than would have died normally. Assuming this excess has a genetic basis, death must have been due to increased homozygosity from cousin marriages. a. The amount of lethal -effect present in heterozygous condition can be calculated if it is recalled 1/16 of heterozygous genes become homozygous in offspring of cousin marriages. It must be, therefore, that 16 x . 13 x 2 is the chance of having abnormal genes in heterozygous condition. The factor of 2 corrects for heterozygotes whose offspring also became homozygous, but for the normal alleles. b. Four lethal equivalents means that each person would carry four times the amount of detriment it would take to kill were the genes involved made homozygous. The actual number of genes responsible is unJ.mown. c. Recent improvement in environment makes this an overestimate of our present load of detrimental genes. d. These lethal equivalents produce some detrimental effect even when the genes responsible are heterozygous since genes are not usually completely recessive. Increase in heterozygosity increases vigor. This is called heterosis, or hybrid vigor. 1. Self-fertilization produces homozygosity, phenotypic uniformity, and decline in vigor. Crossing two pure lines homozygous for different detrimental recessives can produce F1 heterozygotes which will be uniform yet more vigorous than the parents because the dominants hide the effects of the detrimental recessives. 2. Hybrid corn a. Normal corn is too variable. b. To obtain uniformity and vigor in F1, hybrids are made between two selected inbred lines. c. Although the seeds produced in F1 are hybrid,. they are not plentiful because the ears in which they are located have grown on one of the non-vigorous inbred lines. d. This is overcome by making two different hybrids by crossing four different selected inbred lines. The two hybrids are then crossed. Seed produced by this double cross is plentiful, since it is formed on a single cross hybrid plant, and can be sold inexpensively to farmers for planting. 3. Hybrid vigor is applied in the breeding of many economically important plants and animals, and from the financial standpoint is the most valuable contribution genetics has made. COMPUTATION OF F FROM PEDIGREES In the diagram above, suppose that gamete gl carries a certain gene. The probability that gamete g2 carries a descendant of this same gene is 1/2, for the gene in g2 has an equal chance of having come from the egg gl or the sperm g5' In a similar way the two gametes gl and g3 have a probability of 1/2 of both containing the same gene from the ancestor A. However, if A is also inbred, her two genes may also be identical by having descended from a more remote ancestor. The probability of this is FA' the coefficient of inbreeding of A. Hence the probability that gl and g3 contain descendants of the same gene is 1/2 + F A/2 or 1/2(1 + FA)' Finally the probability that gamete g4 contains the same gene as g3 is 1/2. We can get the probability that g2 and g4 both contain descendants of the same gene, which is the inbreeding coefficient of I, by multiplying all these together. Hence F = 1/2 x 1/2 (1 + FA) x 1/2 = (1/2)3(1 + FA) By extension of this principle we can say that the contribution of a common ancestor to the inbreeding coefficient of an individual, I, is (1/2)n (1 + FA) where n is the number of individuals in the path leading to the common ancestor and back through the other parent. Finally, since there may be more than one common ancestor, it is necessary to add together the effect of each of them so the 177 general formul a for determining the inbreeding coefficient of an individual is F = S «1/2)n (1 + FA) ) where n is the number of individuals in a particular path, FA is the inbreeding coefficient of the common ancestor in this path, and S means to add up all the terms, one for each common ancestor. If the common ancestor is not inbred, the formula is simplified to F S (1/2)n. If the genes being considered are sex-linl<:ed, only the females in the path are counted. = (This section is from Dr. Crow's "Genetics Notes" -- see Pre-Lecture references.) POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 178 , QUESTIONS FOR DISCUSSION 30. 1. What would be the relative frequencies at equilibrium of tile heterozygotes and each homozygote type if radiation doubled the frequency of the gene for phenylketonuria? 30. 2. Under the circumstances described in 30. 1, how would the risk of homozygosis for disease be affected in cousin marriages from what it was originally? 30. 3. What is the relation between inbreeding and assortive mating? 30. 4. In the ninth generation of the offspring of a self-fertilized plant heterozygous for Aa, what proportion would tilere be of heterozygous individuals? 30. 5. Starting with a population composed entirely of heterozygotes (Aa), what proportion of it will be homo zygotes after four generations of 1. self-fertilization? 2. sib mafings ? 3. half-sib matings ? 4. cousin marriages? 30. 6. If the frequency of A is .1, and of £\: is .9, what are the zygotic proportions expected according to the Hardy-Weinberg Principle at equilibrium? If, then, suddenly all future matings were entirely brother-sister, what would be the frequency of heterozygotes in the second generation? What would you expect to happen to the frequency of A ? 30. 7. If PI is .02 for the recessive gene £\:' compare the frequency of aa among offspring from cousin and from half-sib matings after a single generation of either type of inbreeding. What would happen after this generation if random mating was resumed? 30. 8. How many lethal equivalents would be present heterozygously if it were found that halfsib matings produced 20% more deaths than occur in the normal popul ation ? 30. 9. The inbreeding of normally cross-fertilized animals and plants is usually followed by a reduction in vigor. How is it, then, that many species of plants in nature are almost always self-fertilized and thus closely inbred but still continue to thrive and maintain themsel ves successfully? 30.10. Will inbreeding result in the reduction of a heterozygous to a homozygous condition more rapidly in a species with a large number of chromosomes or in one with a smaller number? Explain. 30.11. Does crossingover affect the rate of homozygosis from inbreeding? Explain. 30.12. Name four aspects of modern society which have reduced inbreeding in humans, and give a brief explanation for each choice. 30.13. If heterozygous individuals are more vigorous than homozygous ones, why is so much emphasis laid by animal breeders on the desirability of 'pure bred animals, which are relatively much more homozygous than ordinary stock? 30. 14. Why is selection for vigor in inbred lines likely to delay the attainment of complete homozygosity? 30.15. Why may persistent inbreeding in a number of lines, later followed by crosses between them, result in more vigorous individuals than are produced by crosses between members of lines which have not been inbred? 30.16. W4en a farmer procures from his experiment station some "crossbred" maize seed which gives him a high yield through heterosis, he is tempted to grow his own seed for the next season, hoping to perpetuate these favorable traits. Why is such a procedure doomed to failure~ 30.17. Can you suggest a case in which the double cross method would be unnecessary in order to economically obtain the benefits of hybrid vigor? 30.18. How might specific chromosomal rearrangements be used in retaining heterotic combinations? 179 Chapter 31 CYTOGENETICS OF OENOTHERA Lecturer-R. E. CLELAND PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 20, pp. 348-363. Dodson: Chap. 19, pp. 236-238. Sinnott, Dunn, and Dobzhansky: Chap. 19, pp. 266-267. Srb and Owen: Chap. 10, pp. 190-193; Chap. 16, PI' . 341-342. Winchester: Chap. 2, pp. 28-30. b. Additional references Cleland, R. E.. 1936. Some aspects of cyto-genetics of Oenothera. Bot. Rev., 2: 316-348. LECTURE NOTES A. Oenothera, the evening primrose (see p. 5 ), is a weed found along roadsides, railwayembankments, and in abandoned fields. 1. Studied first by DeVries, who found most plants were exactly alike; but an occasional one, derived from the rest, was different and bred true. 2. He took this as evidence of evolution proceeding through large steps, contrary to Darwin's view that such steps were so minute as to be individually unobservable. 3, DeVries thought the new plants were new species formed all at once, and formulated his mutation theory of evolution on this basis. 4. In the 1880s and 1890s DeVries studied inheritance in other plants, his conclusions being essentially the same as Mendel's. However, these results were so out of line relative to those from Oenothera that they were not published. 5. When in 1900 he rediscovered Mendel's paper DeVries immediately published a short 180 B. summary of his work on other plants. Exceptional breeding behavior of Oenothera 1. In Jleas, the F 1 monohybrid between two pure-breeding lines is uniform but splits in F2 to give several phenotypes (Fig. 31-1). OENOTHERA PEAS TALL DWARr TALL '" DWARr '" -TALL -lDWARF '" F1 ~ TALL / . LAMARCKIANA lAMARCKIANA lAMAR~KIANA I \ Fe 3TALL: 1DWAR~ ~ BIENNIS ~ BlfNNIS BIEN"'NIS / ABC !ABC !1 Figure 31-1 2. In Oenothera, DeVries found that two purebreeding lines (races) when mated produced an F1 which was split into several types, each of which when self-fertilized bred true in F2. 3. Renner's analysis showed that Oenothera normally bred as though all its genes were in a single linkage group. 4. Whereas self-fertilization of a monohybrid would, in peas, give 50% hybrids and 50% homo zygotes , self-fertilization of Oenothera gave only hybrids. 5. This absence of homozygotes in Oenothera was due to zygotic and gametophytic lethals, as Renner showed (Fig. 31-2). Zygotic lethals are recessive lethals which kill homozygous zygotes. Gametic lethals act in haploid condition t ,o render gametes non-functional. LtT~ALS fo p ZYGOTIC @@ @)@ @@j @@ GAMETOP~YTIC C. ft I------I~@~I ~~ Figure 31-2 R m P B Sp Cu FlAVENS {URVANS .PERCURVANS ·FlECTENS " .VELANS RUBtNS· FlAVENS '" -CURVANS CURVANS· V[LANS II RUBENS· VElANS 6. Oenothera is normally self-fertilized. A race like biennis produces only two kinds of gametes (Fig .. 31-3), each composed of a complex of genes, all linked. A gene complex is carried intact generation after generation. The two gene complexes in biennis are called albicans and rubens. Lamarckiana's complexes are gaudens and ve1ans. BIE NNIS~ ~ LAMARCKIANA ~~ (G) (V)~ GAMETES\.S)A\.BK:AWS RUBENS@;@GAUDENS VELANS@J PLANT ~@V~@~ GftIJEns (A) ®A (R) . . J® y@~ -]@~ ~@ GAMETES ~ ~@ @~ PLANT V@)D. @) @( GPI.4[T[s ®z )@®f. )@ PLANT Figure 31-3 7. These races are true-breeding heterozygotes because of gene complexes plus balanced 1ethals (see Chap. 33). Linkage relations in artificial hybrids 1. Renner crossed different races of Oenothera to obtain hybrids containing the complexes indicated in Fig. 31-4. Figure 31-4 2. The genes shown were originally in one linkage group. 3. When these hybrids were bred, it was found that they formed more than one linkage group. 4. For instance, tests of the hybrid containing flavens·curvens showed ill and P still linked, but separate from B, which was, tn turn, separate from §Q and Cu, so that there were now at least three linkage groups. 5. Different hybrids had these genes in different linkage arrangements, although some (curvans·velans) still showed one linkage group. 6. The same hybrid combination always produced the same linkage groups. 7. The explanation of how the genes sometimes could be linked and other times not linked came only after cytological investigation. The cytology of Oenothera 1. Cleland, beginning in 1919, found that the evening primrose, from the Rocky Mountains eastward to the Atlantic seaboard, gave an unusual meiotic picture. 2. At meiosis the 7 pairs of chromosomes do not form 7 pairs, but a closed circle of 14 chromosomes arranged end-to-end (Fig. 181 31-5, the number is clear at the left, where the circle has broken open). gene linkage would be explained by this type of chromosome linkage and separation procedure. 6. This would produce gametes identical with the parental gametes (Fig. 31-7) . .-.- .... _--PARENTAL GAMETES GAMfTtS PRODUCfD Figure 31-5 3. All chromosomes have median centromeres and are the same size. 4. Adj_acent chromosomes go to opposite poles, so that at the start of separation chromo- somes assume a zigzag arrangement (Fig. 31-6). Figure 31-7 E. Cytology of artificial hybrids 1. The more than one linkage group formed in race hybrids should be paralleled by finding cytologically more than one chromosome group. 2. There are 15 different ways to arrange 14 chromosomes in circles (which always have an even chromosome number) and pairs (Fig. 31-8). 014 01004 I 0806 , 060404, , 012,1 PAIR Figure 31-6 5. If it is assumed that paternal and maternal chromosomes alternate in the circle, then all paternal chromosomes go to one pole, all maternal ones to the other. Normal 182 o 8,04. 1PAIR 06,06,1 PAIR 04- 0404-1 PAIR I J , Figure 31-8 010,2 PAIRS o 6,04, 2 PAIRS 08,3 PAIRS 04043PAIRS , , 06, 4 PAIRS 04, 5 PAIRS 7 PAIRS ( 0 = CIRCLE) F. G. 3. Various race hybrids were made, and, indeed, all 15 types were found. 4. Any particular hybrid, whenever formed, always gave the same configuration. Cytogenetic investigation of race hybrids 1. To test critically the relation between genetic and chromosomal linkage, many hybrids were made, their chromosome configurations were determined, and tests were made to find out the number of genetic linkage groups. 2. Cleland did the cytology, Renner and Oehlkers the genetic tests. 3. The number of linkage groups turned out to be the same as the number of chromosome groups. Basis for Oenothera chromosome behavior 1. Belling had found, after crossing two races of Datura, a circle of 4 in Fl. 2. He explained the circle as resulting from the presence of a reciprocal translocation (then called segmental interchange) in heterozygous condition. 3. Fig. 31-9 (upper left) shows how this could come about. Two pairs of non-homologous chromosomes exchange approximately equal segments so tllat the ends are no longer 1-2 and 3-4, but 1-3 and 2-4. be due to presence of heterozygous reciprocal translocations. 6, Larger circles can be formed by successive interchanges of this type (the right side of Fig. 31-9 shows how a circle of 6 may be formed from a circle of 4 and a pair). 7. Cleland working with Blakeslee followed up this idea and at the same time Sturtevant and Emerson also began studying the situation from this viewpoint. Segmental arrangements in gene complexes 1. The chromosome ends in a standard complex were numbered 1-2, 3-4, ... , 14-15. 2. By making a series of hybrids containing the standard plus different complexes, and other crosses, it was possible to define the ends present in other complexes. 3. The ends in one hybrid, forming a circle of 4 and 5 pairs, are defined in Fig. 31-10 (top); complexes differing by six interchanges in different chromosome pairs would form a circle of 14 (Fig. 31-10, middle and bottom). H. COMPLEXES DIFFER BY C»JE INTERCHANff UHEOOET/CAL) 1·2 3-4 \ I 1-2 ---4 1 2 2 4 3-4 /\/ 7-8 9·10 I I I I I I 11·12 13-14 I I I I 5·6 \/ \/ 7-8 9'10 \/ 11·12 13·14 \/ \/ \/ 10·11 12·13 14-1 AN ACTU4L EXAMPLE OF RACE WIT~ 014 (MURICATA) 1-2 3-4- 6-5 13·12 7-11 10-9 8-14 2·3 3 3 5'6 2-3 4·1 5-6 7-8 9·10 11-12 13-14 ~PLEXES DFFER BY SIX INTERCHANGES (THEORETICAL) 1_ _ 2 3 4 --~-123 4 1 \ / y~ 1) \~ 2\~ 25''5 \. 4. The heterozygote for this reciprocal translocation (left middle) will synapse to form an X-shaped structure (lower left) which will open out to a circle of 4 (upper right). 5. Belling suggested Oenothera circles might \/ \ 2·3 4·6 / 6-7 8-9 \/ \ / 5-13 12-7 \ / 11-10 \ / 9-8 \ 14-1 Figure 31-10 ft) Figure 31-9 4·5 r. 4_ This interpretation was tested by predicting the chromosome arrangement to be found in a hybrid not yet formed. 5. All such predictions came true. Reciprocal translocations in nature 1. During evolution, the ends of Oenothera chromosomes have been shuffled by reciprocal translocation. 2. The 14 ends can be arranged in seven groups of two in 135,135 different ways. 3. In Cleland's laboratory 350 complexes have 183 / J. been analyzed, and more than 160 different segmental arrangements were found. 4. There must be hundreds or thousands of arrangements in nature. 5. Most races form a circle of 14 due to a long history of interchanges. 6. In nature they breed true because a. only two kinds of gametes are produced, b. there is self-pollination, c. lethals prevent homozygotes from surviving. 7. Evening primroses in nature are constant hybrids. Oenothera is a genetic exception which, like all exceptions, should be treasured. Since the genes behave abnormally and the chromosomes behave in a corresponding abnormal manner, this genus provides an outstanding example of the validity of the chromosome theory of inheritance. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing thE" iter<1::: underlined in the lecture notes. 4. Complete any additional assignment. 184 QUESTIONS FOR DISCUSSION 31. 1. If a single explanation fits almost all the observations, should one be interested in the rare observations which do not fit? Explain. 31. 2. In what way was DeVries unfortunate? 31. 3. Has the term mutati.on changed from its original meaning? Explain. 31. 4. Was DeVries studying mutation? Explain. 31. 5. In what respects are peas and evening primrose genetically similar? different? 31. 6. Write a description of Fig. 31-2 in less than 150 words. 31. 7. How could one test whether gametophytic and zygotic lethals occur in Oenothera? 31. 8. Is Renner's concept of gene complexes valuable in the light of our present knowledge of Oenothera? Expl ain. 31. 9. Of what consequence is the fact that all Oenothera circles contain an even number of chromosomes? 31. 10. Differentiate between "circle" and "ring" as used with regard to chromosomes. evening primroses. 31.17. Would you expect to find all of the 135,135 different possible segmental rearrangements in a complete sampling of Oenothera in nature? Explain. 31. 18. Draw a diagram of a circle of 4 with chromosome ends numbered. Place on the chromosomes genes constituting a balanced lethal system which, after self-fertilization, would perpetuate only the circle configuration. Be sure to expl ain the meaning of the gene symbols. 31. 19. Why is 6 rather than 7 the minimum number of interchanges by which two complexes in a cell must differ in order that they form a circle of 14? 31. 20. What cytological phenomenon is still unsol ved in Oenothera? 31. 21. What advantages might Oenothera have because of its anomalous cytogenetics? 31. 22. Discuss the statement: The principles of genetics are based on studies of organisms which are the exception rather than the rule," 31. 11. Can circles of chromosomes occur in meiosis involving odd numbers of chromosomes? Explain. 31. 12. How could one obtain a chain of five chromosomes at meiosis? 31.13. Explain how genes can be linked genetically yet be located on different chromosomes. 31. 14. For each hybrid in Fig. 31-4. give the minimum number of chromosomes included in one or more circles at meiosis. 31. 15. All the Oenothera plants in a certain locality show a circle of 14 at meiosis. Does this mean they have the same constitution genetically? chromosomally? Explain. 31. 16. Compare the expected role of point mutations in the future evolution of peas and of 185 EXAMINATION IV UNDERLINE ALL THE ALTERNATIVES WHICH ARE CORRECT. 1. A study of the races that ma}.;:e up a particular species a. can be made only if the species reproduces sexually. b. usually will show that the races are allopatric. c. shows that each race is best described by an average genotype. d. shows they are adapted only to their current ecological niche. e. does not provide clues as to the factors responsible for species formation. 2. 4. 5. a. most individuals showing the trait occur in families where only one parent shows it. 186 Natural selection a. operates on phenotypes through genotypes. b. is a conservative force, in that it tends to eliminate from populations mutations which are the raw materials for species formation. c. can aid the formation of reproductive isolating barriers. d. tends to preserve the biologically fit by having them leave more descendants than the less fit. e. would be meaningless were it not for Mendel's discovery of the gene. In the case of a human trait due to a fully penetrant, rare, autosomal, recessive gene, b. in collections of many family pedigrees, the siblings, of individuals showing the trait, include approximately one-fourth that also show the trait. c. the great majority of the genes for the trait are in individuals that do not show The distribution of ABO blood group genes a. is clearly correlated with their adaptive value. b. in the world is so variable that it indicates the concept of race is misleading. c. can be used in a study of migration and racial intermarriage rates. d. can be used to characterize populations genetic ally. e. shows that a race cannot ever be pure in this respect .. Interspecific hybrHlization a. cannot cause two species to merge into one. b. can lead to new species formation via amphiploidy and introgression. c. of two plants, one with n= 10 and the other with n= 12, will produce amphiploids with 22 chromosomes in the nucleus of a root tip cell. d. permits an almost instantaneous creation of new species. e. is the main method of species formation in plants if not in animals . 3. the trait. d. the probability that the gene is carried by a normal first cousin of an individual showing the trait is much greater when the trait occurs in the general population with a frequency of . 0009 than when it occurs with a frequency of .000009. e. the trait appears one-fourth as often among first cousins, as among siblings, of individuals with the trait. 6. In the case of a human trait due to a fully penetrant, rare, autosomal, dominant gene, a. the great majority of the recessives occur in heterozygotes. b. normals never have children with the abnormality. c. abnormal individuals commonly have one normal parent. d. all the children of an abnormal individual are abnormal. e. the normal first cousin of an abnormal individual has one chance in eight of carrying the gene for the abnormality. £. the character must appear among the children and grandchildren of an affected individual. g. in random mating, abnormal individuals practically always marry homozygous recessives. 7. The frequency of a mutant in the gene pool of a large Mendelian population a. depends not only upon mutation, but upon selection pressure. b. can be estimated at equilibrium only if it has some dominant effect in heterozygous Since heterozygotes for the gene causing sickle cell anemia are resistant to certain forms of malaria, a. it is reasonable to suppose most heterozygotes are better than either homozygote in at least one respect. b. their anemia is more or less severe than it would be in non-malarial countries. c. this is a case of heterosis, or balanced polymorphism in man. d. such individuals, in some countries, are more biologically fit than are the normal inhabitants. e. the frequency of the sickle cell allele in the gene pool is expected to be 50% or more in certain malarial regions. 11. Natural populations carry a tremendous load of detrimental mutations a. derived primarily from the action of natural and man-made penetrating radiations. b. derived from mutations most of which arose in preceding generations. c. some of which, under different environmental circumstances, probably would be adaptive. d. some of which mayor may not confer an adaptive advantage when heterozygous. e. which is reflected in the great variety of phenotypes shown by Drosophila pseudoobscura caught in nature. Species formation is speeded up by a. mutation, genetic drift, selection, and random mating. b. rapid changes in the environment in part of a species territory. c. races being kept allopatric by geographical barriers. d. exposure to ionizing radiations. e. maintaining a gene pool which violates the Hardy-Weinberg rule in a number of ways. 9. 10. In the process of forming different species from sexually reproducing races of the same species, a. one race usually differs from another by many physiological gene-based differences. b. the races cannot be continuously sympatric. c. different reproductive preferences must first become established genetically. d. the races will usually show different adaptive values in the same environment. e. a combination of factors usually causes reproductive isolation between the races. 8. condition. c. is not subject to radical change through the action of genetic drift. d. is estimated, in some cases, to be onehalf the frequency of affected individuals. e. is not very much larger than the frequency of affected individuals in the case oJ a dominant semi-lethal mutant. 12. Genetic death a. may not require a cadaver, but each cadaver represents a genetic death. b. occurs when a family has only one child. c. is the only way by which the mutant form is removed from a population. d. rate will increase as exposure of reproductive organs to mutagens increases. e. is delayed by medical treatment, but its 187 occurrence usually is not prevented from happening eventually. 13. Inbreeding is a departure from random mating a. as a result of which heterozygosity is increased and homozygosity decreased. b. while assortive mating is not. c. and is exemplified by self-fertilization, which is the closest form of inbreeding. d. and has one consequence in the loss of heterotic effects. e. which is invariably undesirable from the genetic point of view. 17. The tion. Hybrid corn a. is the most significant contribution genetics has so far made. b. has earned so much money that were the funds made available they would be enough to support genetic research for decades. c. requires two generations for its production from commercial seed. d. is highly heterozygous, yet grows uniformly and vigorously. e. grown by farmers shows heterosis, yet the farmer should not use the seed from his crop for the next generation. In Oenothera, circle formation a. may be absent from some hybrids produced experimentally. b. indicates the presence of reciprocal translocations in homozygous condition. c. is perpetuated in part by the presence of a balanced lethal system. d. means that the genes in non-homologous chromosomes will be distributed during meiosis as though linked. e. indicates most plants of this genus in nature are constant complex hybrids. 14. 15. 16. Use of Oenothera in experimental studies of inheritance a. is a poor choice because of its unorthodox genetic behavior. b. was unfortunate for DeVries. c. provides a means of testing the basic principles of transmission genetics. d. provided a better understanding of the population genetics of this genus. ;:;. showed that sometimes exceptions need be made to the principle of linkage. f. shows how cytology and genetics are both useful in the solution of biological problems. law describes a static frequency of genotypes in a populaFour factors which operate to shift gene frequencies in populations are ------------------________________________________ , and __________________________ __ As a result of the operation of these factors usually the most of the available genotypes are retained in the population. Before different populations can become separate species there is usually isolation leading eventually to _-,__________ isolation, which arises through the following five types of barriers: 1. 2. 3. 4. 5. 18. In the present century, the Congress of the United States has a number of times debated and acted upon the question of immigration of various foreign peoples into this country. Among others, geneticists have been asked to give expert testimony on this subject. Discuss the purely genetic aspects of the problem, citing as fully as you can the various genetic facts and principles bearing on it and explaining how they do bear on it. 19. Discuss the requirements from the genetic standpOint for converting two races of elephants into two different species. 188 20. What bearing do the factors which upset the Hardy-Weinberg equilibrium have on the formation of different races of a cross-fertilizing population? 21. other things being equal, would you expect to find a larger number of recessive lethals in a large or a small Mendelian population? Explain. 22. Would autopolyploidy or amphiploidy ever be expected to produce heterosis? Explain. 23. What are the characteristics of a biologically fit Mendelian population? 24. Cite the genetic facts and principles bearing on each of the two following statements and discuss their implications. a. "All men are created equal. " b. The Oriental race is superior to other human races and its gene pool should be kept free from contamination by the genes in other races. 25. In a large population of range cattle, the following frequencies are observed: 49% red (RR) ; 42% roan (lg:_) ; 9% white (rr). The percentage of all gametes that give rise to the next generation of cattle in this population containing the allele R is _ _ _ _ _ __ In another population, only 1% of animals are white, and 99% are either red or roan. The percentage of I._ alleles in this case is 26. In a human population which had been marrying at random for many generations, the four ABO blood types were found to have the following frequencies: AB 12%; A 72%; B 7%; 0 9%. The estimated frequencies of the three alleles in the population are: Show calculations or method followed in arriving at these estimates. 27. In a large population of Drosophila which had been mating at random for many generations, 72% of the males were red eyed, and 28% white. White eyes are due, in this case, to a recessive gene in the X chromosome. 189 What percentage of the females in the population would you expect to be red eyed? white eyed? _ _ _ _ _ _ _ _ __ 28. The two members of a pair of X-linked genes in Drosophila had initial frequencies of 90% A: 10% Assume random mating has occurred for many generations. ~. According to the Hardy-Weinberg principle, the genotypic frequency expected at equilibrium is --------------------------------------------------- 29. 30. How is genic interaction related to the operation of natural selection? Two pairs of linked genes show a crossover value of 40%. The initial allelic frequencies were 20%~, and 7 0% B, 30% 12.: These genes are located autosomally. 80% A, After several hU!ldred generations of random mating in large populations and the absence of mutation, what percentage of aa bb individuals would you expect? Outline the reasons for your answer. 31. Assuming the Hardy-W2inberg principle obtains, what proportion of the children will be phenotypically recessive from matings between parents, one of whom is phenotypically dominant and the other phenotypically recessive, in each of the following cases? a. One marriage in 10, 000 is between phenotypically recessive individuals. b. 16 children in 10, 000 show the recessive trait, born of parents both of whom show the recessive trait. c. 4% of the children produced are phenotypically recessive when both parents are phenotypically dominant. d. 51% of the population consists of individuals showing the dominant trait. 32. How doyou account for the fact that most of the known mutations in man are "dominant", whereas in other organisms most mutations are "recessive"? 33. Suppose a new mutation appears in a human being. this gene in later generations? 34. Marriage between brothers and sisters was the custom among ancient Egyptian royalty, and the recent royal families of Europe were closely interrelated. In some parts of the United States, on the other hand, marriage between first cousins has been prohibited by law. What will happen concerning the frequency of Discuss the pros and cons of marriage between relatives from the genetic standpoint. 190 Chapter 32 DEVELOPMENTAL GENETICS I lecturer-L. C. DUNN PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 25, pp. 441-443. Dodson: Chap. 16, pp. 193-194, 197. Goldschmidt: Chap. 12, pp. 204-205. Sinnott, Dunn, and Dobzhansky: Chap. 24, pp. 32(]-328; Chap. 25, pp. 339341, 346-349. Snyder and David: Chap. 26, pp. 380381. Srb and Owen: Chap. 18, pp. 372, 380381. b. Additional references Goldschmidt, R. B. 1955. Theoretical genetics. 563 pp. Berkeley: University of California Press. Hutt, F. B. 1949. The genetics of the fowl. 590 pp. New York: McGraw-Hill Book Co., Inc. Landauer, W. 1954. On the chemical production of developmental abnormalities. J. cell. compo Physio!., 43 (Suppl.): 261-305. Wright, S. 1941. The physiology of the gene. Physiol. Rev., 41: 487-527. LECTURE NOTES A. Genes and characters 1. Since characters themselves are not inherited, how are they produced by the action of genes? 2. What happens between the time alleles enter the zygote and the time the individual shows the character from which the gene's presence is inferred? 3. Such questions are dealt with in developmental genetics. 4. The procedure is to learn how development B. is changed in individuals with different genotypes. Gene activity and levels of organization (Fig. 32-1) 1. Knowing what genes do helps to understand their developmental effects. -----, r----- ·, :------------?" ~: GENE , __ ~-~ GENE I~ t _____ J Gene duplication 1 Catalysis I ,-----1 -~'k--7'l)~ ENZYME I --T-- J Catalysis --~)~ ,-------- * ) 1 CELL : 1__ , _____1 Cellular metabolism External actions of cytoplasm I ) I _ _ ___,*'--_~~I Morphogenesis 71I ORGANIC ---,---~ Organic response 1 ) I STRUCTURE II r------, ~ I EXTRA I _ _ _ _ _ _..It_)~1 ORGANIC I Behavior of individual L.STR~C_!I!.Ri I Figure 32-1 2. The gene's most important action is to reproduce itself from surrounding materials. 3. Genes also regulate metabolism without being themselves modified, i. e., they act as catalysts, usually through the enzymes which are formed (see Chaps. 39 and 40). 4. The effects produced may be internal or external to the cell in which the gene acts. 5. External actions involve, for example, nerve impulse transmission, muscular contraction, and action at a distance by hormones. 6. As a result of such external actions, cells in different parts become specialized as tis191 C. sues and organs in a process called morphogenesis. 7. It is usually at the morphogenetic level that observations in developmental genetics are made first. 8. The question becomes how do genes accomplish a morphogenetic result? 9. The study of how phenotypes come into being via gene action is the subject of phenogenetics. Genetics of Creeper domestic fowl 1. This is a case very thoroughly worked out by Landauer with contributions by Hamburger, Rudnick, and Dunn. 2. Fig. 32-2 shows roosters of normal (right) and Creeper (left) phenotypes. Figure 32-2 D. 3. Reciprocal crosses of Creeper by normal gave a 1:1 ratio of Creeper: normal chicks. 4. Creeper x Creeper gave 775:388 birds as Creeper:normal .. This is a 2:1 ratio. 5. Creeper, therefore, is heterozygous for a single pair of segregating genes. The Creeper gene (.Q.2) is dominant to that for normal (::). 6. The 2:1 ratio, like that obtained from crosses between yellow mice, suggests QE_ fE. is lethal. 7. This was supported by finding that about 25% of embryos produced by Creeper x Creeper died about or before the third day of incubation. 8. What is the connection between _QQ _Qp_ which acts as a recessive lethal, _Qp_ ~ which produces Creeper, and 2: 2: which produces normals? The phenogenetics of the QQ locus 1. The techniques used involved studying a. embryology b. histology c. tissue transplantation d. tissue culture e. changing embryonic chemical environment f. experimental induction, by chemical and other treatments, of the same kind of phenotype as the gene in question produces. 2. At 48 hours of incubation (Fig. 32-3) a Qp_.:t embryo (left) is smaller, less developed, and does not have the head flexure already present in a ~ ~ embryo (right). Differences lil<e this can be seen even 12 hours earlier. Figure 32-3 3. Rarely, a.92.Q.p embryo survives 19 days, just before hatching. Fig. 32-4 illustrates at that time the normal (right), and the Creeper homozygote (left) which shows the Figure 32-4 192 E. F. following pleiotropic effects (syndrome): a. no eyelids, microthalmia (small eyes), misshapen head and small body. b. only toes are formed (as seen on top of the black paper background). c. the skeleton is not ossified. 4. Differentiation of cartilage is one system primarily affected by _QQ. a . .Q2.:. has a cartilage disease called chondrodystrophy. h. This disease in _QQ 92_ is called phokomelia. c. Both diseases had been recognized in human families by Vir chow 100 years ago. Families showing chondrodystrophy (then called achondroplasia) sometimes had individuals whose fingers protruded from the shoulders and toes came from a deficient hind limb. This can be explained now as being due to a gene, like .92., in single and double dose, respectively. .QQ and developmental rate 1. As compared with normals one or two doses of f.E. retards early development. a. The leg buds which appear at 7 days are shorter in heterozygotes than in normals. b. This must be based upon events occurring still earlier. 2. Suppose one of the main effects of QQ relative to .:. is to slow down development at a stage when certain parts are at their peak growth rate. Those structures should be most affected by the slowdown which are growing most rapidly at the time. a. This is true of the legs. b. The long bones of fore and hind limbs are affected to the greatest extent. Transplantation experiments 1. elucidate the origin of some pleiotropic effects of .QQ. 2. Prospective hind leg tissue, a. from a normal chick embryo, transplanted to a more forward position in another normal mouse embryo, grows out as a normal limb. b. from a homozygous Creeper embryo, transplanted to a more forward position in a normal embryo, grows out as a Creeper type leg. c. Thus, even at a very early stage before there is any hind limb, tissue from G. Creeper is already permanently determined to develop as Creeper limb. 3. Early eye anlage, a. from a normal embryo, transplanted to an abnormal locus in a normal embryo, grows to be an eye which is smaller and split, just like the eye in homozygous Creeper. b. from a homozygous Creeper embryo, transplanted to the eye-forming region of a normal embryo, grows into a normal eye. c. Thus, the Creeper eye abnormality i,a; " due to abnormal surroundings. In the Creeper homozygote the eye is probably undergoing a kind of starvation due t9 the bad circulation the genotype prodiices. "Morphological characters by which we judge the outcome of development are determined by metabolic steps which arise much earlier, probably presided over by enzymes, and it is these which are first affected by the genes, have their later consequences in the morphological features, which we see as the outcome of the developmental processes. 11 POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 193 QUESTIONS FOR DISCUSSION 32. 1. At what period in the life of a cell do you think that the genes would be most likely to affect the cytoplasm? 32. 2. What are the chief reasons for supposing that genes exert their effects through intermediate products in the cytoplasm? 32. 3. Can one study the developmental effects of a particular dominant lethal mutation? Explain. 32. 4. How can a gene affecting egg characters be detected in a male? 32. 5. Discuss the relationship Fig. 32-1 has to pleiotropism. 32. 6. Give two examples of possible gene effects which would be restricted to the cell in which they are produced. Give an example of an effect which would influence other cells in addition to the one in which it is produced. 32. 7. How can one decid", which of a gene's multiple effects are secondary consequences of a more primary effect? 32. 8. Do you suppose that the gene for Creeper in fowl and the gene for chondrodystrophy in humans are alleles? Explain. 32. 9. How does genetics help us understand developmental processes? 32.10. In what way did experiments utilizing transplantation of embryonic tissues between normal and Creeper homozygotes aid our understanding of how genes influence morphogenesis? 32.11. In poultry a bit of tissue from an early homozygous Creeper embryo, when grafted into a normal egg, may live beyond the time when such a Creeper embryo usually dies. What conclusions can you draw from this fact? 32. 12. 194 What would you expect to happen a. when limb rudiments from normal embryos are grown in dilute culture media with low nutritive content? b. when embryos of normal genotype are injected with insulin, which interferes with carbohydrate metabolism? 32.13. In what way can changes in gene dosage be of assistance in studying developmental processes? Chapter 33 DEVELOPMENTAL GENETICS II Lecturer-L. C. DUNN PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 25, pp. 443-446. Colin: Chap. 12, pp. 220-221. Dodson: Chap. 16, pp. 194-197, 202203. Goldschmidt: Chap. 9, pp. 159-160; Chap. 12, pp. 204-205. Sinnott, Dunn, and Dobzhansky; Chap. 25, pp. 355-357, 349-353, 358, 344-346. Snyder and David: Chap. 26, pp. 382384. Srb and Owen: Chap. 18, pp. 381-392, 396-397. Winchester: Chap. 17, pp. 235-236. b. Additional references Gluecksohn-Waelsch, S. 1951. Physiological genetics of the mouse. Adv. in Genet., 4: 2-49. Grtlneberg, H. 1952. The genetics of the mouse. 2nd Ed. 650 pp. The Hague: M. Nijhoff. Hadorn, E. 1955. Letalfaktoren. 338 pp. Stuttgart: G. Thieme Verlag. Waddington, C. H. 1947. Organizers and genes. 160 pp. London: Cambridge University Press. B. gers. b. upon tissue induction systems. Pituitary dwarfism in the house mouse 1. has been studied by Snell, Smith and MacDowell, and Francis. 2. Genetics Mice homozygous for a completely recessive gene have their whole size reduced so that they are proportionate dwarfs. 3. Development a. Dwarf and normal mice grow equally fast at first. b. Then the dwarf stops growing suddenly, remaining immature sexually. c. The anterior pituitary gland in the dwarf is much smaller than in the normal mouse. 1) Certain large cells are absent from dwarf pituitaries. 2) These cells secrete growth hormone. 4. Effect of pituitary growth hormone upon dwarfs (Fig. 33-1) 77 t gr 75 a 73 77 LECTURE NOTES A. Studies in developmental genetics show 1. genes produce phenotypic differences by physiological, perhaps chemical, actions originating very early in development (see also Chap. 32). 2. the effects of genes in individual cells may spread to other cells. 3. The present chapter deals with genes acting a. at a distance through chemical messen- 9 7 ~~ __ <r_ •• "'.cr----()--.--~---~----o-------.Q------------~---O----- --Q ]9 Y9 Figure 33-1 195 C. a. Two dwarf Ii tter mates 34 days old were used. b. 30 days later the control dwarf still weighed about the same number of grams (b). d. During this time the other mouse, injected daily with pituitary glands ta}.(:en from normal mice, became virtually normal (a). 5. The gene controls growth hormone production by the anterior pituitary gland. Lethal giant larva in Drosophila 1. Fig. 33-2 shows diagrammatically the effect of the homozygous recessive gene l_ethal giant larva (!_g!_) on larval structures (right half) as compared with the anatomy of a normal larva (left half). +/+ /gl//9/ RING GLAND-' ' . RING GLAND BRAIN ' .... ' D. E. these cells. The other larval effects followed secondarily. a. Early implantation of ring gland from normals into !..g!_ l@ larvae gives normal development. b. Early injury of the ring gland of a normal larva produces abnormal development elsewhere. 6. The!.g!_ gene has an early effect in ring gland cells. This effect reverberates through development because the hormonal substances these cells produce are altered. Induction refers to the developmental influence of one part upon another. This is another mechanism for organization in complex vertebrates. Brachyury in the house mouse 1. The normal mouse (.::::_.::::_) has a long tail containing about half the spinal cord. Other mice inJ.-rt¥rit shortened tails (Brachyury). 2. Brachy x Br::why gives 2/3 Brachy : 1/3 normal offspring, suggesting the gene Brachy CD is dominant for shorttailness and recessive for lethality. 3. Chesley studied embryologically the offspring of T .: : _ x T.:!:. (Fig. 33-3). IMAGINAL DISCS-.', SALIVARY GLAND·· ··INTESTINAL TRACT x , fAT BODY MALPIGHIAN TUBE" BP,,",CHY T, BR"'CHY T· ,--·GERM CELLS _. - . -- SOMATIC CELLS GENITAL DISC -- ..... - - .-.---- GENITAL DISC Figure 33-2 AT 2. The mutant gene acts as a recessive lethal by preventing the metamorphosis of the larva into a pupa. 3. Affected larvae are defective particularly in their imaginal discs (anlagen) from which adult legs, wings, eyes, etc., are developed. 4. The norma] larva has a ring gland, near the brain, containing cells which secrete a pupation (metamorphosis) hormone. 5. Experiments by Hadorn showed that a primary effect of the mutant was to affect 196 I MONSTER TT \ 5RACHY TT BIRTH NORMAL"'+ Figure 33-3 a. 25% were normal (.:!:_:!:) (shown at birth at a different magnification than the other embryos). b. 50% developed as Brachys IT _:!:), tail degeneration starting at 11 days of incubation. c. 25% of embryos IT T) had posterior limb buds misdirected dorsally and zigzag F. neural tubes. Since their whole posterior part is not developed they cannot form a placental connection and die between 10 and 11 days. These monsters have no notochord (chorda dorsalis). 4. Bennett studied notochord induction in tissue culture. a. (Notochordal tissue normally induces the surrounding mesoderm to form cartilage and vertebral segments.) b. Mesoderm from normal embryos will develop into cartilage and vertebral segments when surrounding presumptive (future) notochord tissue from young T T embryos. c. T T mesoderm does not form cartilage or vertebrae when surrounding presumptive notochord from normal embryos. d. In the case of T T, then, an inductive relationship has been altered, in which the mesoderm is not able to respond to the normal inductive stimuli of notochordal tissue. Taillessness in the house mouse (Fig. 33-4) 1. A mutant occurred which was tailless. G. ment -- at about 5 days, just after implantation. These individuals had ectodermal hypertrophy and no mesoderm. 4. This turns out to be a case of balanced lethals in which T to mice are tailless, and T T and to to individuals (F3a and F3c, respectively) die. In balanced lethal systems recombinants are prevented by the lethals being alleles or by inversions. 5. A whole series of alleles of to have been found. Another allele stops development even earlier than does to -- at the morula stage. There is some evidence that this mutant stops production of ribonucleic acid, illustrating the principle stated in AI. Development of eye color in Drosophila 1. This has been studied by Beadle and Ephrussi by transplantation of an eye imaginal disc from one larva into the body cavity of another larva. The eye later developed is dug out of the adult and its color observed. 2. The eye anlage of most mutants when transplanted into wild-type larvae develop the mutant eye color. 3. One exception to this is illuminating (Fig. 33-5) . x TAILLESS n' TAILLESS SUMMARY or Tj" + T~t R£SUlTS O~ TRANSPlANTATION [XPfRIMfNTS vOI~cn v en V bW V bW QQQQQO t i iii i TT DiE. DjE AT Ii DAYS AT 5 D.AYS IOXOAYS 16 DAYS "0% BORN NO MESODERM TA'LLE,>S Figure 33-4 2. Tailless x tailless produced only tailless progeny. 3. A study of embryos showed that a. about 25% which died between days 10 and 11 were phenotypically like T T individuals (E3c). b. 50% became tailless mice. c. about 25% died very early in develop- Figure 33-5 a. (The wild-type, dull red, eye contains both red and brown pigments. ) b. Vermilion~) and cinnabar (~I1) produce bright red eye colors because they lack the brown pigment. c. Either type of mutant disc transplanted 197 into a wild-type larva (B into A in the Figure) develops wild-type eye color. d. Thus, the wild-type fly supplies something which the transplanted anlage lacks. e. Cinnabar anlage transplanted into vermilion larva (D into C) remains cinnabar. So vermilion cannot supply what cmnabar misses in order to develop wild-type eye color. f. Vermilion anlage transplanted into cinnabar larva (C into D) develops wild-type eye color. Cinnabar can supply what vermilion lacks to produce wild-type eye color. 4. Chain of chemical reactlOns for production of WIld-type eye color (Fig. 33-6) SUBSTANC[S BROWN 1- GENES PIGM~NT WILD TYPE 3-0XYKYNURfNINE= ll-U~RE ( J KYNURENINE t ERE ( 1-1 PHENOTYPE of SUBSTANCE eN BREAKS C"AIN CINNABAR V+ SUBSTANCf V BRfAKS CWAIN VERMILION TRYPTOP~AN Figure 33-6 a. In the presence of v+, kynurenine (v+ substance) IS produced. b. In the presence of cn+, 3-oxykynurenme (cn+ substance) IS produced. c. The presence of both v+ and cn+ genes permi ts the cham of reactIons to go to completion to Wild-type eye color. d. The absence of either one breaks the reaction chain and no brown pigment is produced. 5. This chainwisc orgamzatlOn of chenncal processes may be considered a general model for the ways other genes affect the dIfferent kinds of developmental processes which have been discussed. 198 POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writmg the Items underlined in the lecture notes: 4. Complete any additional assignment. QUESTIONS FOR DISCUSSION 33. 1. How might you show that an operation which destroys the ring gland of a Drosophila larva and produces a phenotype like lethal giant larva does this due to the lack of pupation hormone rather than to the injury of the operation itself? 33. 2. Discuss each of the following, citing evidence to support your opinion w~enever possible. a. Some genes produce early and others late morphogenetic results. b. A mutant may change one tissue's ability to respond yet may have no internal effect on another tissue of the same individual. c. External effects of genes can be produced by means other than hormones. type can be brought about by non-allelic genes acting differently. 33. 9. What will be the eye color expected in a mostly female gynandromorph of Drosophila which is ~ y.. in one eye and y.. in the other? Explain. 33.10. Referring to Fig. 33-5, explain why it is that the homozygote for y.. bw in E has some pigment in its eyes, whereas in F it does not. .r 33.11. In four-o 'clocks there are two factors, and _!!, which affect flower color. Neither is completely dominant, and the two interact on each other to produce seven different flower colors, as follows: Discuss the different kinds of effects gencontrol of growth rate can have in vertebrates. YY-RR - = crimson yY_ Rr = magenta-rose YY Rr = orange-red yY_ RR = magenta 33. 3. Name three substances in Drosophila which can act at a distance. In each case give one genotype which does and one which does not produce the substance or its normal amount. = yellow yY_ rr = pale yellow IT RR and IT rr = white YY!E. 33. 4. lC 33. 5. Both of the dominant, non-allelic, mutants !!: and ~ are lethal when homozygous and are linked autosomally. Construct a balanced lethal system including these mutants. 33. 6. What would you conclude from the results of each of the following experiments with Drosophila? a. When wild-type larval fat bodies are transplanted into yermilion larvae, the host develops wild-type eye color as an adult. h. When wild-type larval salivary glands are transplanted into cinnabar larvae, the adult host has cinnabar eyes. What will be the flower color in the offspring of the following four-o 'clock crosses? a. b. IT RR x IT Rr YY Rr x rr rr 33.12. What assumptions can you make regarding the chief chemical effects of the genes and B in the preceding problem? How would you test the assumptions? .r 33. 13. What are the main generalizations of developmental genetics? 33.14. How are developmental, physiological, and biochemical genetics related to each other? 33. 7. Glve an example of a developmental effect of a speciflc mutant which can be produced . also by specific modification of the environment acting in the presence of the normal gene. 33. 8. Give an example of h~w the same pheno- 199 Chapter 34 CYTOPLASMIC HEREDITY Lecturer-T. M. SONNEBORN PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 23, pp. 406-409, 415421. Dodson: Chap. 18, pp. 215, 218-220. Goldschmidt: Chap. 12, pp. 201-202. Sinnott, Dunn, and Dobzhansky: Chap. 26, pp: 359-360, 364-366. Snyder and DavId: Chap. 25, pp. 372373. Srb and Owen: Chap. 13, pp. 261, 268272. Wmchester: Chap. 22, pp. 309-310. b. Additional references Rhoades, M. M. 1946. Plastid mutations. Cold Spring Harbor Sympos. Quant. BioI., 11: 202-207. Rhoades, M. M. 1955. Interaction of genic and non-genic hereditary units and the physiology of non-genic inheritance. pp. 19-57 in Vol. 1 of "Handbook of plant physiology", W. Ruhland, Editor. Berlin: Springer Verlag. TRAIT U)O ~ :z LJ..J e::::: ~ S2 F, 200 AUTOSOMES LECTURE NOTES A. Chromosomal and cytoplasmic heredity 1. The vast majority of heredi.tary traits that have been carefully analyzed are determined by genes contained in ,chromosomes. 2. In animals and plants a con.siderable llumber of hereditary traIts are determined by cytoplasmic factors. 3. Chromosomal heredity a. obeys the law of segregation, b. gives regular and predictable ratios, c. IS transmitted to offspring equally from both parents, except in the case of sex chromosomes, d. requires hereditarily different individuals possess differences in genotype (chromosomal gene make-up). B. Characteristics of cytoplasmic heredIty (Fig. 34-1) 1. Typically, the trait in F1 is that of the mother. 2. Such transmission cannot be attributed to chromosomal genes. a. If it was, such traits would follow the rules of transmission for autosomes, or for X or Y chromosomes. They do not. X e 1\(\ I Y - 0 ~rrt ~ ~ 0 1\ r1\) ~- ~ I a 9 - r - - - Figure 34-1 r cj - - ~ CYTOPLASM ANIMALS ~ PLANTS l 0 0 EB PLANTS AND ANIMALS b. Even when all chromosomes have been derived from the paternal line (accomplished by repeated crosses back to the paternal type), the maternal trait still is produced. 3. Such traits may be transmitted through the cytoplasm. a. In the simple case (see Fig. 34-1) little or no relevant cytoplasm comes in from the male parent with the sperm o'r pollen nucleus, so the F 1 resembles the mother. b. In exceptional cases some cytoplasm does come from the father and then some paternal transmission also occurs: In that case: 1) transmission is irregular, not following segregation, 2) mtermediate traits may be produced, 3) some F 1 parts may resemble the mother, others the father. c. Only examples of the simpler case are presented in this chapter. The inheritance of chlorophyll production 1. When chlorophyll production fails a. the area involved is non-green (white). b. there is no sugar synthesized. c. in a seed, the seedling dies once its stored nutrients are exhausted. 2. Failure of chlorophyll production often is inherited, a. usually due to a mutant nuclear gene which acts as a recessive lethal. b. sometimes due to cytoplasmic factors, in which event the offspring are_ phenotypically like the mother. 3. Inheritance can be studied by using mosaics -- part green and part white plants. The green part provides nourishment for the wHite part. a. Normal (all green) and mosaic (green and white striped) corn plants were illustrated. b. Striping can run through the reproductive organs, so pollen and eggs can be obtained both from green and from white • '. E. LEUCOPLASTS YOUNG GROWING CHlOF lEAr C~LOROPLASTS L[UCOPLASTS lEAF CEll mOM lEAF CELl mOM GREEN PLANT WHITE PLANT PRIMORDIUM Figure 34-2 pa~ts. c. Striping occurs in the corn cob. This is detected by the groups of green and of albino seedlings obtained from growing kernels in rows corresponding to their positions in the cob. Cytoplasmic inheritance of chlorophyll production 1. In this case it makes no difference from which part of the plant the pollen comes. 2. Only the place where the seed is set, i. e. , where the ovum comes from, is important. Eggs from green tissue develop later as green seedlings; eggs from white tissue produce white seedlings. 3. Occasionally, a seed at the border between green and white seeds again will produce a striped plant, and again the eggs alone will determine the chlorophyll character of the offspring produced. 4. This trait is the only one in plants for which the physical basis of cytoplasmic inheritance is known. Plastids and chlorophyll production 1. Chlorophyll is contained in plastids called chloroplasts (Fig. 34-2, center). . F. 2. A corresponding cell from a white plant contains white plastids, called leucoplasts (Fig. 34-2, right). 3. In normal, young, growing cells one finds only colorless plastids of various sizes (Fig. 34-2, left). In a normal plant leucoplasts develop chlorophyll on exposure to light, but in an hereditarily non-green plant they do not . 4. Occasionally a striped plant produces an egg whose cytoplasm contains both leucoplasts and chloroplasts. If during development of the fertilized egg the plastid types become segregated striping will occur. Plastids· reproduce true to type. Iojap and plastid mutation 1. Marcus Rhoades contributed greatly to our understanding of this case and of cytoplasmic inheritance in general. 2. He investigated a striped corn plant in 201 which both genic (chromosomal) and cytoplasmic factors were involved in chlorophyll production. 3. The recessive chromosomal gene iojap ill) when homozygous causes plastids to mutate so they can no longer produce chlorophyll. 4. The mutant plastids reproduce their kind even after !i is substituted by its normal allele. 5. A similar case, in which a nuclear gene controls chlorophyll production by mutating the plastids, is known in the catnip, Nepeda. 6. In other cases, gene control of chlorophyll production may be accomplished by some other mechanism. Inheritance of male sterIlity in plants 1. The fact that male sterile corn plants occasionally produce some functional pollen permitted the demonstration that both nuclear genes and cytoplasmic factors operate also in the case of male fertility. 2. The iojap gene can cause a different mutation in the cytoplasm. This results in male sterilIty that is, thereafter, cytoplasmically inherited. 3. The physical baSIS in the cytoplasm for the sterility factor is unknown. 4. In this case nucleaI genes have a function additional to mutating the cytoplasmic factor (Fig. 34-3). 0- GENOTYPE + C'lI~~t~~MIC~p~~NOTYPE 1 S5 2 ss 6 - ST[RILf 0' - STERILE 6 - ~~RTILE 0- rfRTlLE 3. SS (OR S5) d - FtRTILt 0 - rfRTlLE 4. SS(ORSS) 6-STERILE a-FERTIlE Figure 34-3 a. The male sterile phenotype requires both the o"-sterile cytoplasmic factor and the recessive gene ~ in homozygous condItion. 202 b. If ~ or the o"-fertile cytoplasmic factor is present (types 4 and 2, respectively), or both are (type 3), the phenotype is male fertile. 5. The cytoplasmic factor self-reproduces regardless which "~" ,genes are present. (PI) ss, cytoplasmically o"-sterile 9 x SS, cytoplasmically o"-sterile 0" (P2 ) Ss, cytoplasmically o"-sterile F1 is seHed. (F 2 ) 25% are ss, cytoplasmically o"-sterile and produce the malelsterile phenotype. Thus the o"-sterile cytoplasmic factor: of the P generation, self-reproduced in the P1 l which was fertile, and appeared again in F2' 6. Other examples-of autonomous cytoplasmic factors, which come to expression only in the proper genotype, are known in onion, in sugar beets, and m two other cases of male sterility in corn (each involving still other cytoplasmic factors and other nuclear. g~lles). H. Cytoplasmic inherItance and cell organization Certain cells have an elaborate organization which is faithfully reproduced at successive divisions. But mutations occur which change cell organization, and the mutant form is thereafter reproduced. 1. Cell form in Paramecium (demonstrated with motion pIctures) a. Paramecium is usually single, reproducing this form by fission. b. Double paramecia also occur and fission reproduces the double condition. c. Double paramecia have two complete sets of the structures necessary for conjugation; single paramecia have only one set., d. Occasionally doubles revert to singles, which thereafter produce only singles. 2. Self-perpetuation of s,ingle and double cell form in Paramecium a. is not due to nuclear genes. As seen, conjugation can occur between smgles and doubles. When conjugation is completed the ex-conjugants are genetically identical (Chap. 35). Yet the single ex-conjugant produces only singles, and the double ex-conjugant only doubles. b. is due to cytoplasmic factors. r. These are self-perpetuating and their expression is independent of nuclear genotype. SUmmary of main pOints 1. Cytoplasmic inheritance occurs and differs from nuclear inherItance in several ways, particularly in being transmitted usually by one parent- only. 2. The clearest, normal, regular physical basis of cytoplasmic inheri'tance is the plant plastid (chlorophyll production). 3. Certain factors for organization self-perpetuate in the cytoplasm independently of the nucleus (Paramecium cell form). 4. Nucleocytoplasmic interactions occur a. Nuclear genes (iojap) can have mutagenic effects on cytoplasmic factors (plastids and male fertility factor in corn). b. Nuclear and cytoplasmic factors interact in the expression of traits (male sterIlity in corn). POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, makmg additlOns to them as desired. 2. Review the reading assIgnment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 203 QUESTIONS FOR DISCUSSION 34. 1. Compare the number and importance of the hereditary traits determined by chromosomal genes and cytoplasmic factors. 34. 2. In what respects do cases of cytoplasmic inheritance resemble those of sex-linked inheritance? In what respects do they. differ? 34. 3. How IS the trait in question in Fig. 34-1 explained as not being determined by chromosomal heredity? 34. 4. Compare the characteristics of chromosomal and cytoplasmic heredity. 34. 5. In what way is the use of plants striped with green and white like the use of heterozygotes for a recessive lethal ? 34. 6. What kinds of plastids are found in a white portion and in a non-whIte portion of a striped corn plant? Starting with the zygote, explain how this could come about. 34. 7. What evidence can you give that plastids regularly reproduce true to type? mice is due in part to factors not in the chromosomes. State two sources of evidence which could be used to test the truth of this hypothesis. 34.13. How did Sonneborn prove that double paramecia hav~ two complete sets of structures needed for conjugation? How C'aD one distinguish double paramecia from singles that are in the process of fission or of conjugation? 34.14. Cases like that of variegation in Pel argonium zonale are inherited through both parents, but show no clear segregation as in Mendelian inheritance. ' How might such cases be explained? 34.15. If an infectious disease were to be transmitted from mother to offspring through the egg, could you distinguish this from cytoplasmic inheritance?- Explain. 34.16. How do studIes of cytoplasm-free nuclei and of' nucleus-free cytoplasm bear on the view that the cell as a whole depends for its complex of hereditary characters and potenfialities upon both nucleus and cytoplasm? 34. 8. How would it be possible to determine 34. 17. How might it be possible to distinguish bewhether the plastid differences in cases of tween true cytoplasmic inheritance and the invariegation are due to two types of plastids or duction of a cytoplasmic change by a gene just to a diseased condition of some of the plastids? before the meiotic divisions? 34. 9. Can o"-sterility and o"-fertility in corn be 34.18. What is "maternal" inheritance? How explained by the presence and absence of a would you distinguish between it and "cytoplassingle cytoplasmic factor? Explain. mic" inheritance? 34.10. Give all the hereditary constitutions pos34.19. Discuss the bearing that present knowledge sible for maize plants which of cytoplasmic inheritance has on the speCUlation that the cytoplasm orders the sequence, a. upon self-fertilization produce some form, pattern, or arrangement of the cytoplasmale sterile and some male fertile mic particles carrying enzymes or other spephenotypes. cific substances, while the nucleus functions to b. when used as the male parent always produce these enzymes or specific substances. produce the male fertile phenotype. c. could show no effect of :U.:U. upon male fertility. 34.11. Using Fig. 34-3, give the hereditary constitutions and phenotypes among the offspring expected from reciprocal matings between individuals of types 2 and 4. 34.12. 204 Castle has maintained that body size in Chapter 35 NUClEO-CYTOPlASMIC RELATIONS IN PARAMECIUM Lecturer-T. PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 23, pp. 412-415, 421424. Dodson: Chap. 18, pp. 221-222. Goldschmidt: Chap. 12, pp. 201-202. Sinnott, Dunn, and Dobzhansky: Chap. 26, pp. 363-364, 366-367. Snyder and David: Chap. 25, pp. 374377. Srb and Owen: Chap. 13, pp. 272-277. Winchester: Chap. 22, pp. 310-315. b. Additional references Btlale, G. H. 1954. The genetics of Paramecium aurelia. 178 pp. CambrIdge: Cambridge University Press. Sonneborn, T. M. 1937. Sex, sex inheritance and sex determination in P. aurelia. Proc. nat. Acad. Sci., U. S., 23: 378-395. Sonneborn, T. M. 1943. Gene and cytoplasm. 1. The determination of inheritance of the killer character in variety 4 of P. aurelia. Proc. nat. Acad. Sci., U. S., 29: 329-338. Sonneborn, T. M. 1951. The role of the genes in cytoplasmic inheritance. Chap. 14, pp. 291-314, in "Genetics in the 20th century", L. C. Dunn, Editor. Sonneborn, T. M. 1957. Breeding systems, reproductive methods, and species problems in Protozoa. pp. 155324, in "The species problem", E. Mayr, Editor. Washington, D. C.: Amer. Assoc. Adv. Sci. Publ. Sonneborn, T. M. 1959. Kappa and related particles in Paramecium. Adv. Virus Res., 6: 229-356. M. SONNEBORN 3. Read the lecture notes through section D. LECTURE NOTES A. Paramecium is remarkably favorable for the study of nUeleo-cytoplasmic relations, as will become elear after considering certain of its cytogenetic actiVIties. B. Asexual reproduction (Fig. 35-1) 1. A typical Paramecium contains a diploid micronucleus and a highly polyploid macronucleus (meganucleus). MACRONUCLtUS (ABOUT 1)000 N) MICRONUCLEUS (2 N) Figure 35-1 C. 2. During fission both types of nuclei divide so that the two daughter cells produced are genetically identical. 3. A group of genetically identical individuals derived from one ancestral parent is called a clone. Sexual activity (Fig. 35-2) 1. Suppose two individuals, which differ in 205 micronuc13ar genotyPe as shown, are brought together and allowed to mate in a process called conjugation. 2. During conjugation each mate undergoes meiosis and produces four haploid nuclei, of which three dismtegrate. The remairung nucleus divides mitotically once, producing two th nuclei in one conjugant and two ±. nuclei in the other. --.... E. F. : :MEIOSISI ! " :' "e':'''~'''; f \t "::;. _ ":'" ~-..- __ \. I ! Figure 35-2 3. One th and one +- nucleus then migrate from D. 206 one conjugant t;- the other, and nuclear fusion produces a ±.Ith diploid nucleus in each llldividual. 4. The two individuals separate and each produces a clone. 5. When two Fl ~/th indivIduals are mated, 75% of the time both ex-conjugants will produce phenotypically normal clones and 25% of the time both will produce thin clones. 6. The macronucleus diSintegrates during conjugation. However, the diploid fertilization nucleus diVIdes mitotIcally, one product forming a new macronucleus, the other remaining as the new micronucleus. Consequences of conjugatIOn 1. MICronuclear genes show typical Mendelian heredity. 2. All members of both ex-conjugant clones are identical with respect to chromosomal genes. G. 3. If chromosomal genes determine a trait, all members of both ex-conjugant clones will show the same expression of that trait. DistrIbution of cytoplasm 1. after fission is apprOximately equal III daughter cells. 2. during conjugation (demonstrated with motion pictures) a. Usually the interiors of conjugants are kypt apart by a boundary which temporarily is penetrated by_ the migrant nuclei. As a result, little or no cytoplasm IS exchanged between conjugants. b. Under certain expe~imental conditions conjugants form a wide bridge through which the cytoplasmic contents of both mates can mix. The extent of mixing can be controlled e:kperimentally. Control of cytoplasmIc mixing during conjugation is a powerful tool for analyzing the role of cytoplasm in heredity in P~rameciutn. 1. For traits determined chromosomally (genically) tYPICal MendelIan rules are followed, as described in C, whether or not cytoplasms mix. 2. It may be found for a given trait that, a. when there has been no cytoplasmic exchange, the two ex-conjugant clones are different, each resembling"in its tr:ut its . cytoplasmic parent. , b. when there has been cytoplasmic mixing, the two ex-conjugant clones are the sQ.me phenotypically (see p.13). c. Such a trait IS clearly determmed by cytoplasmic factors. d. Three dIfferent traits inherited this way will be dIscussed in detail in this chapter. Killers and sensitives (demonstrated with motIon pictures) 1. Animal-free flUid, from cultures of killer paramecia, will kIll sensitIve paramecia. 2. The killer and sensitive traits are hereditary in fisslOn. 3. All conjugants are resistant to killing ac- \ tion, so that the cross can be made between killer and sensitive; when cytoplasmic mixIng occurs m this case, both ex-conjugants, are killers. 4. Killers contain hundreds of small cytoplasmic particles kalled kappa particles. SenSItives contain no kappa. 5. Kappa a. can be seen stamed and unstained. b. is self-reproducing, and was shown dividing. nricronuc13ar genotype as shown, are brought together and allowed to mate in a process called conjugation. 2. During conjugation each mate undergoes meiosis and produces four haploid nuclei, of which three disintegrate. The remaining nucleus divides mitotically once, producing two th nuclei in one conjugant and two 2: nuclei in the other. Ol /II~+/----+~t !cJi _th~/th_ _ 1 00e@<[ I th/+ >CTJQ)(lXDI th/+ t _CLQNEJ " ") MEIOSIS: ( :MEIOSISI C_Dl Figure 35-2 3. One th and one.:!: nucleus then migrate from D. 206 one conjugant to the other, and nuclear fusion produces a 2:/th diploid nucleus in each individual. 4. The two individuals separate and each produces a clone. 5. When two Fl 21th individuals are mated, 75% of the time both ex-conjugants will produce phenotypically normal clones and 25% of the time both will produce thin clones. 6. The macronucleus disintegrates during conjugation. However, the diploid fertilization nucleus divides mitotically, one product forming a new macronucleus, the other remaining as the new micronucleus. Consequences of conjugation 1. Micronuclear genes show typical Mendelian heredity. 2. All members of both ex-conjugant clones are identical with respect to chromosomal genes. 3. If chromosomal genes determine a trait, all members of both ex-conjugant clones will show the same expression of that trait. E. Distribution of cytoplasm 1. after fission is approximately equal in daughter cells. 2. during conjugation (demonstrated with motion pictures) a. Usually the interiors of conjugants are kept apart by a boundary which temporarily is penetrated by_ the migrant nuclei. As a result, little or no cytoplasm is exchanged between conjugants. b. Under certain experimental conditions conjugants form a wide bridge through which the cytoplasmic contents of both mates can mix. The extent of mixing can be controlled experimentally. F. Control of cytoplasmic mixing during conjugation is a powerful tool for analyzing the role of cytoplasm in heredity in Paramecium. 1. For traits determined chromosomally (genically) typical Mendelian rules are followed, as described in C, whether or not cytoplasms mix. 2. It may be found for a given trait that, a. when there has been no cytoplasmic exchange, the two ex-conjugant clones are different, each resembling in its trait its cytoplasmic parent. b. when there has been cytoplasmic mixing, the two ex-conjugant clones are the same phenotypically (see p.13 ). c. Such a trait is clearly determined by cytoplasmic factors. d. T11ree different traits inherited this way will be discussed in detail in this chapter. G. Killers and sensitives (demonstrated with motion pictures) 1. Animal-free fluid, from cultures of killer paramecia, will kill sensitive paramecia. 2. The killer and sensitive traits are hereditary in fission. 3. All conjugants are resistant to killing action, so that the cross can be made between killer and sensitive; when cytoplasmic mixing occurs in this case, both ex-conjugants are killers. 4. Killers contain hundreds of small cytoplasnric particles kalled kappa particles. Sensitives contain no kappa. 5, Kappa a. can be seen stained and unstained. b. is self-reproducing, and was shown dividing. H. L c. is mutable, mutant kappas producing different poisons. d. is liberated into the medium once it deyelops a highly refractile granule, appearing sometimes as a "bright spot". One "bright spot" kappa particle is enough to kill a sensitive individual. Foreign par:ticles in the cytoplasm 1. Kappa a. resembles a bacterium in size and shape, but differs from a bacterium in certain staining reactions. b. (in comparisons made under the electron microscope, Fig. 35-3), differs from a bacterium in internal morphology (left and center photographs, respectively), particularly when kappa develops the refractile granule (right photograph). c. resembles a foreign organism because it is infective and not found in all paramecia. 2. The hereditary significance of kappa is that it furnishes a model of how a parasitic or symbiotic microorganism could become so well adapted to its host as to a. become a part of the host's hereditary system, and b. determine hereditary traits of the host. 3. The rickettsia causing Rocky Mountain spotted fever, .like kappa., is visible and inherited through the cytoplasm. 4. Sensitivity to carbon dioxide in Drosophila is due to an invisible, infectious, cytoplasmic factor. In Drosophila, W10 other traits having these properties are sensitivity to ether and a tendency for early death of male zygotes. 5. Much of cytoplasmic inheritance can be attributed to probably foreign particles, visible or invisible, carried in the cytoplasm. Serotypes (demonstrated with motion pictures) 1. Anti-A serum is obtained from a rabbit previously injected with paramecia of type A. Since type A parafbecia are immobilized when placed into anti-A serum, they are said to have serotype A. 2. Type B paramecia, immobilized by anti-B serum obtained similarly, have serotype B. 3. Serotype A paramecia in anti-B serum and serotype B individuals in anti-A serum are unaffected. 4. Ex-conjugants from crossing A and B individuals give rise to serotypically similar clones only when cytoplasm has been exchanged. Figure 35-3 J. 5. The cytoplasmic basis for the serotype trait is invisible and non-infectious. 6. It has been shown that every Paramecium is homozygous for two loci, one for serotype A and the other for serotype B. 7. Yet a Paramecium is serotype A, or B, but never AB. 8. This is so because the cytoplasmic factor determines which one of the two gene loci will come to phenotypic expression. It is not known, however, whether the cytoplasmic element a. operates on the gene products once they get into the cytoplasm, or, b. acts directly on the gene loci, determining which gene will be active or inactive. Mating types (demonstrated with motion pictures) 1. When cultures of different mating types are mixed together, there is a mating reaction in which individuals of different mating types stick together forming larger and larger clumps. This leads to conjugation in pairs whose members are always of different mating types. 2. Call one mating type alpha and the opposite type beta. Ex-conjugants, from a mating of alpha by beta, form clones of identical mating type only if the mates mixed cytoplasms. 3. Besides the hereditary cytoplasmic factor there is a genetic factor in the macronucleus; both factors interact in the control of mating types. 4. The role of the cytoplasmic factor 207 K. a. (When the fertIlization nucleus dIvides, one daughter nucleus forms a young macronucleus which develops into the adult macronucleus. This thereafter divIdes at every fission and goes to all daughter cells of a clone. ) b. If a young macronucleus is placed in alpha cytoplasm, then as an adult macronucleus it comes to determine the alpha mating type; a genetically identical young macronucleus in beta cytoplasm becomes an adult macronucleus determining beta mating type. c. ThIS fIxation of the macronucleus is irreversible. d. Thus, the cytoplasmic factor causes macronuclear dIfferentiation. S. The role of the macronucleus a. Once the macronucleus is determined and persIsts in this conditIOn, it produces or determmes cytoplasmic factor of the same kind. b. For example, an alpha adult macronucleus causes alpha cytoplasm to be produced which, in turn, is ready to determine young macronuclel in the next sexual generation. Serotype and matn.g tYP2 !'lystems 1. Unlike the kappa system, these are normal systems involving invisible cytoplasmic particles. 2. In both systems, the cytoplasmic factors regulate or control nuclear action by deciding whlCh nuclear potential in a set of alternatives will come to phenotypic expression. In the mating type case, at least, the nuclear change is irreversible. 3. This sort of SItuation is probably comparable to many others encountered In higher organisms, as in the case of nuclear differentiation in amphibia studied by Briggs and King. 4. Dr. Sonneborn concludes: "The whole problem of nuclear differentiatIOn and developmental differentiation seems to be here in a nutshell, waiting to be cracked in the near future. " POST-LECTURE ASSIGNMENT 1. Read the notes immedlately after the lecture or as soon thereafter as possible. 2. Review the reading assignment. 3. Be able to discuss or define orally or in wrIting the items underlined in the lecture notes. 4. Complete any additIOnal aSSIgnment. 208 QUESTIONS FOR DISCUSSION 35. 1. Describe in detail how a 3:1 phenotypic ratio IS obtained from segregating nncronuclear genes In Paramecium. 35. 2. Describe how one can obtain a 1:1 genotypic ratio in Paramecium. 35. 3. Why is the macronucleus of Paramecium of no consequence in obtaining typical Mendelian segregatIOn ratios? 35. 4. Describe a procedure for determining whether a trait m Paramecium is due to nuclear or cytoplasnnc heredity. 35. 5. Describe the physical changes which occur in sensitive paramecia exposed to killers. 35. 6. Discuss the view that the macronucleus and micronucleus in Paramecium are like the nuclei of somatic and germinal tissues, respectively. pha macronucleus was placed in a cell carrying the beta cytoplasmic factor? 35. 15. Describe how you would proceed expel'imentally in order to decide whether a given strain of Paramecium was of alpha, beta, or of some other mating type. 35.16. Does describmg an hereditary cytoplasmic factor as a plasmagene, cytogene, viroid, or virus have any advantage or disadvantage? Explain. 35.17. Of the cytoplasmic factors mentioned in this chapter and Chap. 34, a. which are and whIch are not aSSOCIated with visible cytoplasmic particles? b. which are known to be affected and which are known to be unaffected by nuclear genes? c. which are known to affect nuclear genes or their products? 35. 7. Design a hypothetical experiment, not based upon cytological observatIOn, which would prove that there was a reciprocal exchange of cytoplasm by conjugants. 35. 8. Is conjugation a method of reproduction? Explain. 35. 9. DeSCrIbe the experIment which showed, in the moving pictures, the effects of anti-sera on serotypes. 35.10. How can it proven, when cultures of different mating types are combined, that contacts are always made between two animals of different mating type? 35.11. What do you consider the most useful discovery about Paramecium from the standpoint of genetics? Defend your choice. 35.12. What do you consider as Paramecium'S most unique technical contribution to genetics? Why? 35.13. Excluding phenotype, describe one feature unique to each of three cases of cytoplasmic heredity In Paramecium. 35.14. What would you expect to happen if an al- 209 Chapter 36 PSEU DOAllELISM Lecturer-E. B. LEWIS PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 11, pp. 205-206, 208211. Sinnott, Dt!nn, and Dobzhansky: Chap. 27, pp. 370-372, Chap. 28, pp. 383387. Snyder and DaVId: Chap. 26, pp. 393397. Srb and Owen: Chap. 19, pp. 415-417. b. Additional references Carlson, E. A. 1959. Comparative genetics of complex loci. Quart. Rev. BioI., 34: 33-67. Dunn, L. C., and Cg_spari, E. 1945. A case of neighboring loci with similar effects. Genetics, 30: 543-568. Green, M. M. 1954. Pseudoallelism at the vermilion locus in Drosophila melanogaster. Proc. nat. Acad. Sci., U. S., 40: 92-97. Green, M. M., and Green, K. C. 1949. Crossing-over between alleles at the lozenge locus in Drosophila melanogaster. Proc. nat. Acad. Sci., U. S., 35: 586-591. Lewis, E. B. 1951. Pseudoallelism and gene evolution. Cold Spr. Harb. Sympos. Quant. Biol., 16: 159-174. Lewis, E. B. 1952. The pseudoallelism of white and apricot in Drosophila melanogaster. Proc. nat. Acad. Sci., U. S., 38: 953-961. Pontecorvo, G. 1958. Trends in Genetic analysis. 145 pp. New York: Columbia University Press. "Pseudoalleles and the theory of the gene", a symposium by M. M. Green, 210 E. B. Lewis, J. R. Laughnan, Clyde Stormont, and S. G. Stephens. 1955. Amer. Nat., 89: 65-122. LECTURE NOTES A. A gene 1. occupies a fixed locus in the chromosolI\e, 2. exists in many forms, i. e., lias multiple alleles, 3. can be considered a unit of function, 4. can be studied with regard to its transmission properties and its function. B. The gene as a unit of function 1. The functlon of producing normal, red eye color pigment in Drosophila is perf_gr_med by several genes (see also Chap. 33). These include the following: a. White (~), located at position 1. 5 on the X chromosome crossover map, removes all eye pigment. b. Vermilion (y), located at 33.0 on the X chromosome map, removes one of the eye color pigments. c. Cinnabar (£!!), located on chromosome 2 at 57. 5, removes the same pigment as does v. 2. These eye color genes are similar in function yet are located at widely different places in the same chromosome, or in different chromosomes. 3. The meaning of the gene as a functional unit is cl arified by breeding tests. C. Allelism as determined by breedmg tests 1. In Drosophil a the heterozygote for white and vermilion ~ :::_/:::_~) has red (wild-type) eye color. These genes work on different processes in eye color formation so that the two :::_ genes In the dIhybrid form full pigmentation. So these two genes, located at very different positions in the genome, are said to be non-allellc genes. D. 2. A heterozygote containing white and cherry ~/~ch) has a dilute cherry eye color. These genes work on the same process, both have the same locus on the X chromosome (the hybrid produces only gametes with one or the other allele); they are called alleles. 3. -The mutant apricot ~ produces a phenotype like ~ch does. By ordinary crossover tests it is located at 1. 5 on the X chromosome. a. The heterozygote for white ~) and apricot ~ has a dilute apricot (cherry) phenotype, indicating the two mutants work on the same process. In this respect they behave like alleles. b. But the heterozygote produces not only gametes containing either ~ or apr, but, on rare occasions, some carrying an X chromosome which produces the full red eye color. In this way w and apr act non-allelically. c. 'The exceptions in C3b are not reverse mutations to .:!: because, though rare, they are too frequent, and because they always occur in association with crossingover. d. At ''locus 1. 5" there are really two separate loci which are said to carry pseudoallelic genes. Techniques for demonstrating pseudo allelism 1. Closely linked marker genes are introduced on either side o.f the pseudoallelic series under investigation. In the example discussed, the pseudoallelism of ~ and apr, yellow body color (y) and split bristles ~ were used as markers (Fig. 36-1). CQOSSING DVm BHWffN PS[uOOAllWC HE COLOR GENES '~PRICOT' (APR) AND "w~ln" (WlIN "ATTAC"W X" ffMALfS 0.0 I ~ / 1.5 +'\ '\ '3.0 E. J W SPL / I .Y + + + \ ~ / 1 __ ,3:> I-I I \ I + APR WSPL Figure 36-1 F. 2. A genetic plan was used by which both complementary crossovers between the pseudoalleles are recoverable. Attached-X chromosomes permit the simul taneous recovery of two strands of the four involved in a crossingover. 3. Fig. 36-1 (to the left in figure) shows schematically a portion of an attached-X as it would appear at meiosis at the time of crossingover. 4. The attached-X had y ~ ~ on one arm and L apr spl+ on the other. The female carrying this had a pale (dilute) apricot phenotype. 5. The daughter progeny of such a female are usually pale apricot, or white, or apricot. When large numbers of daughters were examined some of Wild-type phenotype also were found. 6. Six such exceptional red-eyed daughters were shown, by detaching the arms of the attached-X and identifying the genes each contained, to have one arm with y, the normal alleles of apr and~, and spl +; the other arm carried L apr w ~ (right side of Fig. 36-1). 7. The genes found in the detached arms of these exceptional attached-X females proved apr lies to the left of ~ on the X chromosome. 8. The presence of the double mutant combination ~ ~ was proven by breeding the six exceptional red-eyed attached-X females, scoring numerous daughters, and obtaining some which were pale apricot. These new exceptional daughters were shown to contain the original gene arrangement restored by crossingover. 9. Crossingover between different pseudoalleles occurs with frequencies usually ranging from 0.1% to 0.01%, the value 0.03% being typical. The cis-trans position effect (see photo on p.10) 1. The trans heterozygote (.:!: ~/apr .:!:) and cis heterozygote (.:!: ::/apr ~ produce different phenotypes. 2. This phenotypic difference is the resul t of a pOSition effect, since the only difference between cis and trans is the arrangement which the same genetic material takes. Other cases of pseudoallelism involve 1. color in cotton (Stephens), 2. taillessness in mice (Dunn and Caspari), 3. lozenge and vermilion eyes in Drosophila (the Greens), 211 G. 4. traits in Aspergillus (Pontecorvo), other microorganisms, and corn. 5. The effects of w and apr are so similar that another example from Drosophila is discussed to illustrate how pseudo alleles differ functionally. Pseudoalleles involving the halteres (balancers) 1. Lewis has shown a locus in Drosophila to be composed of five pseudoallelic genes. 2. The normal fly (Fig. 36-2) has small clubshaped balancers located on the posterior part of the thorax. Figure 36-",1: 4. Close examination reveals these two recessive pseudoalleles really do different things. Bithorax converts the front portion, and postbithorax the back portion of the haltere into wing-like structure. 5. This is verified by taking these mutants in trans form, obtaining the double mutant combination (cis form) by crossingover (at a rate of . 02%), and observing the phenotype of flies made homozygous for the double mutant combination. Such flies (Fig. 36-5) have a fully developed second pair of wings. Figure 36-2 3. One of the pseudoalleles, bithorax (~ converts the hal ter'C) into il i argo wing-like structure (Fig. 36-3), another called postbithorax (pbx) at first appears to do much the same thing (Fig. 36-4). Figure 36-5 H. Figure 36-3 212 Cis-trans effects for bx and pbx 1. The cis form <.:::: .:!::_/bx pbx) has normal balancers, while the trans form (bx .:!::/.:!:: pbx) shows a slight postbithorax effect. 2. This is another example of cis-trans posi- 1. J. K. tion effect. Models explaining cis-trans position effects 1. One model compares the two homologous chromosomes to assembly lines, each making products independently. a. The cis form can make all the products in turn in the strand containing the two normal alleles. (The strand with both mutants makes less or no end product.) Much end product is produced. b. Since each strand of the trans form contains a mutant (defective machine) the total end product produced is zero or relatively little. c. This model offers clues as to how genes act and applies well to Drosophila pseudoallelism. 2. Other model s have been suggested which may apply better in other cases. Chromosome morphology of pseudoallelic regions 1. The "white" loci-are associated with a double band (doublet) in the salivary gland chromosome (Bridges and Metz); apr may be in one band, w in the other. 2. Vermilion is associated with a doublet; the five loci of the bithorax series are connected wi th two doubl ets. 3. Many doublets appear in salivary chromosomes. Origin and evolution of pseudoalleles 1. Some present pseudoalleles may have arisen a. by selection of mutants of genes which were origimilly different but adjacent; b. by selection of rearrangements which brought together non-adjacent but similar genes, 2. Lewis finds the simplest assumption to be that pseudoalleles represent duplications of an ancestral gene which occurred on one occasion or more (as in "bithorax l l ). Following duplication, mutation would advance evolution by leading to functional differellPes between adjacent pseudoalleles. 3. Pseudoalleles may provide real clues regarding action and origin of genes. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 213 QUESTIONS FOR DISCUSSION 36. 1. Give the sexes of the flies shown on p.lO and in the lecture notes of this chapter. 36. 2. Does Modu1.ator in maize (Chap. 38) violate Lewis I statement that genes occupy a fixed position in a chromosome? Why? 36. 3. What two tests should be performed before concluding two mutant genes are allelic? 36. 4. In what respects are pseudoalleles similar to and different from alleles and non-alleles? 36. 5. Why is it necessary in proving pseudoallelism to border the locus under test with marker genes? 36. 6. What difference would it make in question 5 if the marker genes were each 10 crossover map units from the pseudoallelic locus? 36. 7. Is it possible for w and apr both to be located at 1. 5 on an X chromosome crossover map? Expl ain. 36. 8. Using Fig. 36-1, show how the maternal genotype could produce female progeny of white and of apricot phenotype. 36. 9. How might one obtain detachment of attached-Xls? How could one identify an individual which received a detached arm from its mother? 36.15. What is the relation between bands in the salivary gland chromosomes and the pseudoallelic series for "white" and ''hithorax''? 36.16. Wllat did Bridges and Metz conclude regarding the Significance of doublets in salivary chromosomes? 36.17. How would you go about deCiding which locus to study for pseudoallelism in Drosophila? 36.18. What are the requirements for proving a cis-trans position effect? 36.19. Show diagrammatically 110W an ancestral gene might have duplicated to form two genes side by side. 36.20. Is it possible, from present knowledge, to determine which is the primary mechanism by which most pseudo alleles have arisen? Explain. 36.21. How has pseudoallelism influenced your concept of the gene? 36.22. Can pseudoalleles be considered to be parts of one gene (sub-genes) rather than separate genes? Expl ain. 36.10. Can one prove pseudoallelism using P.v'o mutants dominant to wild-type? Explain. • 36.11. What gene arrangement in red-eyed exceptional females of Fig. 36-1 would have proved the gene order was y w apr ~? Expl ain. 36.12. How does one proceed to show that the rare, pale apricot daughters from red-eyed attached-X mothers of Fig. 36-1 have the original gene order restored? 36.13. How many differences are there between cis and trans heterozygotes? To which of these genetic differences can a position effect be attributed? 36.14. In what respects are the pseudoallelic loci for "white" and "bithorax" different? 214 Chapter 37 VARIE 'G ATED PERICARp· AN UNSTABLE AllELE IN MAIZE Lecturer-R. A. BRINK PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General gerietics textbook Altenburg: Chap. 19, pp. 344-345. b. Additional references Brink, R. A. 1954. Very light variegated pericarp in maize. Genetics, 39: 724-740.' Brink, R. A. , and Nilan, R. A. 1952. The relation between light variegated and medium variegated pericarp in maize. Genetics, 37: 519-544. McClintock, B. 1951. Chromosome organization and genic expression. Cold Spr. Harb. Symp. Quant. BioI., 16: 13-47 . . van Schaik, N. W., and Brink, R. A. Transpositions of Modulator, a modifier of variegated pericarp in maize. Genetics, (in press). LECTURE NOTES A. Comparison of mutation rates 1. In maize the gene for yellow endosperm (Y) mutates to color! ess (y) about 1. 3 times per million genes .. 2. From a medium variegated pericarp parent about 10% of offspring are mutant. B. Medium variegated pericarp (Fig. 37-1) 1. The pericarp of a kernel encloses the seed which contains the embryo. 2. The embryo comes from the generation following that during which pericarp was formed. 3. Note the variable pattern of red pigment in the pericarp. C. Some earlier conclusions concerning variegated pedcarp R. A. Emerson pioneered this study. He RANDOM SAMPL~ O~ KfRNfLS ~ROM A MtDIUM· VARIEGATED PERICARP tAR Figure 37-1 D. concluded: 1. Variegated is an unstable allele at the P locus. Later P was located on the longest maize chromosome, chromosome 1. 2. Variegated normally produces colorless pericarp (when heterozygous with pW, a stable colorless allele), but in somatic cells frequently mutates to a red-producing allele. 3. The size of the mutant stripe or sector depends upon when in ear shoot development the mutation occurs. Early mutations make large sectors red, late mutations produce small red sectors. Studies in Wisconsin, begun about 1950, showed that medium variegateds produce not only red mutants but also light variegated ones (Fig. 37-2). Lights have about half as many sectors 215 mutant as have mediums. F. MEDIUM VARIEGATED ~I PERICARP 6. Lights x pW pW produce about equal. numbers of lights and mediums among the colored progeny plus a few reds (new mutations from variegated). Occurrence of twin spots (Fig. 37 -4) 1. Occasionally, medium variegated ears shov.r the two mutants, light and red, as twin patches of kernels. 2. Red and lights are not merely related in origin (E4) but these mutations appear complementary -- i. e., in the mutation process one gained something the other had lost. (PARENTAL TYPE) ,SEH-RED PERICARP " (MUTANT) Figure 37-2 E. Results of test crosses (Fig. 37-3) 1. The gene for medium variegated pericarp is designated P V. 2. Half of offspring are, as expected, colorless (pW pW). 3. The remaining half are colored, of which a. 90% are mediums (pV pW), b. about 6% are reds, c. about 4% lights. 4. The similar frequency of reds and lights suggested that the origins of these two mutants were reI <lted in some way. 5. Reds x pW pW produce offspring which, if colored, are all red, red being stable. TESTCROSS - INBRED MI:DIUM VARlfGATED PflRICARP INBRED TWIN MUTANTS tROM MEDIUM VAR1EGATED PER'CARP Figure 37-4 G. New genetic hypothesis for pericarp variegation (Fig. 37-5) Note that gene symbols are changed. G~NOTYP~S [P~ENOTYPt pw pw d' MfDIUM VARIEGATED PERICARP RtD 50% ~MEOI\jM VARIEGATfU r-----l 90% l 50 % : + RED 670 LC_9.LO_R'!JS~ COLORED MlITANT[tIGHT VARIEGATED (DISCARDtD) o 47.] Figure 37-3 216 I GENOTYPE I I COLORLESS X pr Mp/pw ::; prjpw MUTANTS [ LIGHT VARIEGATED pr M'pjPW + TRANSPOSED Figure 37-5 ~p/- 1. 1. All genotypes shown are heterozygous for ?W, the stable gene for colorless pericarp. 2. The variegated allele is a dual structure, containing pr, the top dominant allele for red, and Mp, Modulator, which suppresses pigment production. 3. pr Mp together suppresses pigmentation, so that ll1.ediums are produced. 4. ~ alone produces · stable red. 5. pr Mp plus an additional Mp somewhere else (transposed Modulator) produces lights. 6. If the transposed Modulator in a light is returned to a red at its position near pr, both resultants would be expected to give medium phenotypes. TranspOSition mechanism for instability of mediums 1. In studies of other unstable alleles in maize, McClintock found transposition occurs, whereby an accessory element (like Modulator) is removed from one position in the genome and inserted in it elsewhere. 2. This hypothesis is applied to the present case (Fig. 37 -6). 1. J. K. OIFFtRENTIAL MITOSIS OIVlDfD PARENT CHROMOSOME DAUGI·n~R CO-TWIN CWROMOSOMES P8fNOTYPES .If- pr Mp - "" 0 ~ Pr M. -1 MEDIUM VAQ I[GATED PH [NOTYP[ /"-pw~ f_ - ........... ,\ I ~J\f\l'() ~ /,f~'" ~. , - -'V\('\() ~ pr Mp ,,, \ - --- ---~~Mp I RED 1-----1 LI GI-IT VARlfGAT~D .... -t-.J..}J pw~ '- - - / (NOT ILLUSTRATED) Figure 37-6 a. Parent chromosomes (pr Mp and PW) are shown already divided, daughter strands still connected at the centromere. b. Normal mitosis would produce two daughter cells, each carrying pr Mp/pw. c. In a differential mitosis, however, Mp is lost from one chromosome and inserted into a non-homologous (wavy line) chromosome which is received by the sister cell carrying pI Mp. ~ d. Subsequent mitosis, of the cell containing pI' will produce reds, and of the sister cell lights, these cells becoming adjacent mutant patches in a medium background. Expectations from differential mitosis 1. Colored progeny from red should be red only. They are. 2. Reds lack Mp. This is often the case (however, see K6). 3. Lights should produce l:i.ght and medium offspring, and they do. Excess lights among the progeny indicates transposed Mp has been inserted somewhere in the same chromosome as P. Modulator and gene instability in general 1. Factors like Modulator are quite common in maize. 2. Unstable loci in other organisms might be due to similar factors, but this is not yet proven. Other properties of Modulator 1. Transposability of Mp is under genetic control. If Mp transposes from pr Mp 100 times, the presence of one transposed Mp elsewhere reduces this frequency to about 60, while presence of two transposed Mp further reduces the value to about 5. 2. Mp may become fixed at the P locus, so that a medium becomes a stable colorless mutation . 3. Transposed Mp may make the recipient locus unstable. Mp transposed near the starchy gene in chromosome 9 caused a mutation to the waxy allele which was unstable thereafter, frequently mutating back to starchy. 4. Transposed Mp may occupy a variety of sites. a. In 57 of 87 cases treated Mp was still linked to chromosome 1, having been transposed less than 50 crossover units from P. b. In the remaining cases Mp was found transposed to one of five different nonhomologous chromosomes. 5. Linked sites for transposed Mp are clustered near the P locus. 37 of 57 transposed, but still P-linked, Modulators were within five crossover units of P, 10 were between five and fifteen units, and the remainder were further away. Hence Mp tends to move from the P locus by a short, rather than a long, jump. Contact between old ar..d new sites may be re- 217 quired for transposition. 6. Rarely, reverse mutations occur from red back to medium. a. In each case M.l:? had been transposed to chromosome 1. b. The frequency of reversions from red to variegated is greater the closer the transposedMPis to pr (Fig. 37-7). PERCENT RECOMBINATION p- Mp VARlfGATfV SfCTORS PfR 1000 KERNELS 2.6 4.3 15 7.6 12.0 B 3 42.0 0.2 11 Figure 37-7 L. 7. If genetic elements like MP are vlldespread among organisms, these facts may be of basic significance for mutation theory in general. In summary, medium variegated mutates to red by loss of MP from the E locus. In this process complementary lights are produced with an extra, transposed MI!_. The medium type is reconstItuted by transposition of a transposed M.l:? back near the E locus. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 218 QUESTIONS FOR DISCUSSION 37. 1. Approximately how many more times more mutable IS medium variegated pericarp than is yellow endosperm color? 37.16. What is the expected effect of the presence of extra transposed Modulators upon the frequency of mutation to lights? to reds? 37. 2. How can one explain the fact that somewhat more reds than light variegateds are observed as mutants from medium variegateds? 37. 17. Do you think the mutation of starchy to waxy is due to position effect? Explain. 37. 3. How do you suppose Emerson showed the locus for pericarp variegation was due to a factor on chromosome 1 ? 37. 4. Why is medium variegated kept in heterozygous and in inbred condition? 37. 5. Compare the detectability of light variegateds m a medium individual when the mutation occurs early and late in development of the ear shoot. 37. 6. What facts connect hght variegated and full colored (red) mutants? 37. 7. Are the twin mutations, lights and reds, restricted to adjacent patches of kernels on an ear? Explain. 37... 8. How can you explain the origination of whole ears of light and of red type from a medium parent? 37. 18. What genetic results would support the idea that it was Modulator which caused the unstability of waxy? 37.19. How would you proceed to show that Modulator was transposed to a new site on chromosome I? chromosome 9? 37.20. Is pericarp variegation an example of position effect? Explain. 37.21. Might other Modulators exist in corn whose effects have so far escaped detection? Explain. '37.22. Describe different mechanisms which produce what at first appears to be reverse gene mutation. 37.23. Does Modulator violate Lewis' statement (Chap. 36) that genes occupy a fixed position in a chromosome? Explain. 37. 9. What effect is expected from transposition of a transposed Modulator to a position adjacent to Pw? I 37.10. Have you any criticism of the chromosomes diagrammed in Fig. 37 -6? Explain. 37. 11. If transposition is a breakage event, how many breaks are required to transpose Modul ator? Why? 37.12. How might a red without Modulator acquire it? 37.13. What IS expected from a red which acquires one Modulator? 37.14. How can you explain a red that produces a. all reds? b. varIegated sectors only occasionally? c. variegated sectors frequently? 37.15. Is pr or .ME unstable? Explain. 219 Chapter 38 DNA STRUCTURE AND REPLICATION lecturer-J. D. WATSON PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 22, pp. 393-396. Colin: Chap. 12, pp. 216-220. Dodson: Chap. 17, pp. 205-207. Sinnott, Dunn, and Dobzhansky: Chap. 27, pp. 374-378. Snyder and David: Chap. 24, pp. 363369. Winchester:, Chap. 17, pp. 232-234. b. Additional references Crick, F. H. C. 1954. The structure of the hereditary material. Scient. Amer., 191 (Oct.): 54-61. CrIck, F. H. C. 1957. Nucleic acids. Scient. Amer., 197: 188-200. Watson, J. D., and Crick, F. H. C. 1953. The structure of DNA. Cold Spring Harbor Sympos. Quant. BioI., 18: 123-131. LECTURE NOTES A. Nucleic acids and genetic information 1. Earlier work favored proteins as the primary genetic material. 2. We are certain now that nucleic acids are the carriers of much, if not most, genetic information. 3. There are two forms of nucleic acid, deoxyribonucleic acid (DNA) which is apparently the prime genetic material of chromosomes, and ribonucleic acid (RNA). B. RNA 1. This is less well known than DNA although similar to it chemically. 2. It serves a function in protein synthesis, perhaps as a carrier of genetic information from chromosomes to the protein-making 220 C. D. E. cytoplasm. 3. RNA appears to carry, all the genetic information in certain viruses, like tobacco mosaic and poliomyelitis. DNA is discussed with regard to its 1. chemical structure as related to its biological function, and 2. exact replication. Electron microscope photograph of DNA, taken by Dr. Cecil Hall, shows a long (40,.000 A, or 4 microns), thin (20 A) molecule of high molecular weight (about 8 million). 1. DNA samples from the same species have a surprising uniformity in molecular we~ght. 2. DNA is the largest uniform molecule of biological origin. DNA is a polymer 1. DNA consists of a repetition of a series of simple chemical units called nucleotides. 2. Each nucleotide has three parts, a phos:" phoric group, a 5 carbon sugar (deoxyribose), and a nitrogenous base. In Fig. 38-1 the base is adenine. Deoxyribonucleotide (e.g. 3'-deoxyadenylic acid) Figure 38-1 3. There are four main nucleotides; these differ only In their bases (Fig. 38-2). F. PURINES N~~(' ~NAN/ HN-!yN~ AW,l!.._N/ H NH2 AD[NINE NH GUANINE PYRIMIDINES' z 0 HN:JCH I NG16 1 51 HO H '~ 34 N ~ HO CYTOS INE 3 N THYMINE Figure 38-2 BASt COMPOSITION SERRATIA o Punne 01". I pyrlmldme 5' C~HZ 0 base H H" 3' 2' o I -O-p=o H G. H I o I C~2 0,,--- Base H H H H o H I -O-p=O I o I CH z Figure 38-3 20.7 20. 1 212 CYTOSINE 21.5 20.4 20.7 17. 7 17.420.3 35.4 31.9 Figure 38-4 ~ 4 or DNA mOM VARIOUS ORGANISMS ADENINE THYMINE GUANINE 28.8 29.2 20.5 CI·HCKfN SALMON 29.7 29.1 20.8 LOCUST 29.3 29.3 20.5 17. 7 Sf A URC~IN 32.8 32.1 'lEAST 31.7 32.6 18.8 VACCINIA VIRUS 29.5 29.9 20.6 TUBERCl[ BA(l[RIA 15.1 14.6 34.9 4. The sugar and phosphate groups form the backbone of the molecule (Fig. 38-3). The phosphate unites carbon 3 of one sugar to ~arbon 5 of the next sugar, in a very regular way. 5. The result is a linear, unbranched molecule. 6. The sequence of bases is less regular, the H molecular specifIcity of DNA molecules being attributed to the particular sequences of bases they contain. Base content of DNA (Fig. 38-4) 1. Different species have different base contents. 2. Evidence is good that within a species DNA is a mixture of molecules differing in base content and sequence. 3. There are, however, two striking regularities. a. Total purines always equal total pyrimidines. b. Adenine always equals thymine, and guanine equals cytosine. 4. There must be some structural feature of DNA requiring these equivalences. The three dImensional organization of DNA 1. Through the use of X-ray diffraction it is possible to determine the three-dimensional arrangement of parts of molecules. 2. Workers in England, particularly Watson and Crick, found a. all DNA molecules gave the same X-ray diffraction pattern (despite F1 and 2). This indicated a configuration common to all DNA molecules. b. Moreover, this unit consisted of two chains twisted about each other (Fig. 38 -5). 3. Specifically, sugar and phosphates are on the outside, bases inside. The backbone completes a turn each 34 A, and there is a nucleotide every 3.4 A. Bases lie perpen- 221 dicul ar to the fiber axis and those in different chains are held together in pairs by H bonds. H. . 34 }\ 1. 5. These are the only possibilities to allow the regular double helical configuration. 6. At any level it is possible to face the base pairs two ways (A-T or T-A, or, G-C or C-G). 7. The base sequences on the two chains are complementary, their sequence on one chain specifying that on the other. Templates or molds 1. Molds have been used to copy complex surfaces. 2. If a mold was made of the genetic surface, and this mold used to make a new mold, the second mold would be an exact copy of the original surface. 3. The two complementary strands of the DNA molecule can be viewed as templates for each other! Either strand might act as a mold on which the complementary strand could be synthesized. Hypothesis of DNA replication (Fig. 38-7) Figure 38-5 4. H bonding between bases is very specific, adenine (A) with thymine (T), guanine (G) with cytosine (C) (Fig 38-6), 2 i Figure 38-7 Figure 38-6 222 1. The two strands come apart. 2. The two free chains exist in the presence of nucleotides or of their precursors. 3, When the correct free nucleotide approaches the single chain its base would be H bonded. When several free nucleotides have bonded to the single chain, perhaps an enzyme would link them to start the new complementary chain. J. 4. Preliminary results of experimentl? carried out to test this hypothesis are encouraging. The chemical basis for spontaneous mutations 1. DNA specificity resides in base sequence. 2. It is a reasonable hypothesis that mutation results when the wrong base gets incorporated during the formation of a new chain. 3, The probability such a mistake will occur is calculated, according to physical-chemical theory, as 1 in 10 10 bon dings between base pairs. 4. Faul ty incorporation is rare enough, then, for the usually orderly transmission of hereditary information to be attributed to correct hydrogen bonding. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, maldng additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 223 QUESTIONS FOR DISCUSSION 38. 1. Why is DNA not considered the only substance capable of carrying genetic information from one organism to its progeny? 38. 2. Why is it surprising that DNA samples from the same species show a uniformity in molecular weight? 38. 3. What fundamental differences exist between purines and pyrimidines? 38. 4. What is the relationship in the DNA from various organisms between a. different purines? b. different pyrimidines? c. total purines and total pyrimidines? d. adenine and thymine? e. guanine and cytosine? f. A+T/G+C? 38. 5. Give the base sequence of the chain complementary to the one indicated below. A G C T T 38.13. How are the two Watson-Crick DNA chains positioned relative to each other a. chemically? b. physically? 38.14. What would you consider the major unsolved problem of Watson's hypothesis of DNA replication by chain separation? 38.15. How does Watson explain spontaneous mutation? What bearing has this hypothesis on the evidence that mutations can occur in the old gene (see Chap. 22)? 38.16. At the chemical level, describe how ions might induce mutations. 38.17. Has the Watson-Crick structure of DNA any bearing on the basic principles of transmission genetics? Explain. G C 38. 6. In Fig. 38-5b, what do the ribbons represent? the crossbars? What is the diameter (1 the double helix? How does this compare with DNA diameter as seen in the electron microscope? 38. 7. In what condition are DNA and RNA found in living organisms? 38. 8. In Figure 38-7 which parts represent sugar, phosphate, purine, and pyrimidine? 38. 9. Do you believe that the number of possible base pair combinations is too small for DNA to be the prime genetic material? Explain. 38.10. Is it critical for genetic theory that DNA forms a linear, unbranched molecule? Explain. What X-ray diffraction finding directly supported results of chemical analysis of DNA? 38.11. 38.12. Previously, DNA had been called "thymonucleic" acid, "nuclear" nucleic acid, and "animal" nucleic acid. Can you suggest a reason why each of 224 these names was given? Give a reason for rejecting each of these alternative names. 38.18. Presuming life exists on other planets in the universe, what chemical basis for hereditary material would you expect there? • Chapter 39 BIOCHEMICAL GENETICS I lecturer-G. W. BEADLE PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 21, pp. 366-373. Dodson: Chap. 16, pp. 200-202. Sinnott, Dunn, and Dobzhansky: Chap. 24, pp. 331-338. Snyder and David: Chap. 26, pp. 389393. Srb and Owen: Chap. 17, pp. 350-358. Winchester: Chap. 17, pp. 239-242.. b. Additional references Bearn, A. G. 1956. The chemistry of hereditary disease. Scient. Amer., 195: 126-136. Hsia, D. Y. - Y. 1959. Inborn errors of metabolism. 358 pp. Chicago: The Year Book Publishers, Inc. McElroy, W. D., and Glass, B. (Editors) 1957. The chemical basis of heredity. 848 pp. Baltimore: The Johns Hopkins Press. Peters, J. A. (Editor) 1959. Classical papers in genetics. 282 pp. Englewood Cliffs, N. J. : Prentice-Hall, Inc. Wagner, R. P., and Mitchell, H. K. 1955. Genetics and metabolism. 444 pp. New York: John Wiley & Sons, Inc. LECTURE NOTES A. Biochemical genetics 1. The specifications for making a complete individual must be contained in the fertilized egg. 2. In humans, it is estimated that these specifications, in the form of words, would fill about a thousand standard library volumes. a. Every mitotic division must reprint this thousand volume recipe for a human. B. b. Replication of the specifications must be highly accurate for only occasionally are errors made (mutation) in the 20-40 divisions occurring between gamete production by consecutive generations. 3. The genes, containing the specifications, function in the presence of cytoplasm and other environmental factors. It takes about ten tons of food to make an adult from the human fertilized egg. 4. Biochemical genetics, then, deals with how these specifications are written and how they are translated during development in the functioning of the individual. 5. This branch of biology started very soon after 1900 with Garrod's study of heritable diseases due to inborn errors of metabolism. Alcaptonuria (Fig. 39-1) , -~, ,~(, C:_- -c ,'J~,- c~ ')'iE:NYLAlAMINE: yrt 01-\ _/~- II I '-...//' c- -¢-'lJH.t. C~~ T'l'RDSINE. (}I-f -/~/ I /1 'V~,.:-. ~ r r -/~/ )()-CCCOH , OH C'__:;Q CGrrH ~HfN,(~ ALCAPTo~ f\,R_lWK ("OfIIOCiTNTKIC AClb, (')ell:'» P-OH Figure 39-1 1. In this rare disease urine blackens upon exposure to air. a. The symptom appears at birth and persists thereafter. 225 b. Affected individuals are otherwise llttle inconvemenced. 2. The chemical substance responstble for the blackening is alcapton (homogentisic aCid). (The FIgure omits C atoms in tho' ring, and also H atoms whose positions are indIcated by short lines. ) 3. Alcaptonuria occurs in homozygotes for a recessive gene. 4. Garrod thought alcapton was an intermediate substance in a series of normal chemical reactlOns. a. In normal people, alcapton is rapIdly metabolized to another form. b. In alcaptonurics, alcapton accumulates. 5. He postulated that there was a serIes of chemical precursors for alcapton, including two amino acids (essential in our diet) -phenylalanine and tyrosine. a. Phenylalanine is normally converted to tyrosine (by addItion of one oxygen atom). b. Tyrosine is normally converted to p-OH (para hydroxy-) phenylpyruvic aCid (by replacing the amino group by an oxygen atom). c. p-OH phenylpyruvic acid is normally converted to alcapton. 6. He also postulated that alcaptonurics are defective in an enzyme which catalyzes the conversion of alcapton to some other form, and that the enzymatic defect was the result of the genetIC defect. 7. Garrod's views, like Mendel's, were correct but ahead of their time. C. New approaches to metabolic investigations 1. The gene for alcaptonuria a. made it relatively easy to identify alcapton as an intermediate substance in metabolism. b. made fruitful the use of techniques that identify metabolic precursors of alcapton. 2. Alcapton fed to a. alcaptonurics is accumulated, then excreted. b. normals is metabolized and not excreted. 3. p-OH phenylpyruvic acid (or tyrosIne, or phenylalanine) fed to a. alcaptonurics increases excretion of alcapton. b. normals is not excreted as alcapton. 3. Chemical pathways in metabolism can be worked out usmg genetIC defects as tools. D. Albinism (see also Chap. 5) 1. ThIS disease, also lllvestigated biochemi226 E. F. G. cally by Garrod, is characterized by absence or decreased amount of the pigment melanin. 2. Tyrosine, by a different chemical pathway, is also a precursor of melanin. 3. Albinism can be caused genetically by the defective production of an enzyme necessary in the conversion of tyrosine to melanin. Phenylketonuria (see also Chap. 30) 1. In thIf disease phenylalanine IS not converted to tyrosine because a gene produces a defect in the enzyme reqUIred for this process. 2. What happens instead IS that phenylalanine , is converted to phenylpyruVic acid (by replaCing the amine group by an oxygen atom) which accumulates and is excreted In the urine. 3. Phenylpyruvic acid IS toxic, producing feeblemindedness or lower mental abIlity. 4. The disease can be par~i31ly alleviated, or circumvented, if dietary phenylalanine is reduced to an amount enough for protein synthesis b_ut not enough to accumulate as phenylpyruvic acid. Galactosemia l. In humans, the 12-carbon sugar lactose, found in milk, is spilt into the 6-carbon sugars galactose and glucose. 2. Normal humans then convert galactose int!? "utilizable galactose" by means of an enzyme. 3. Homozygotes for a recessive gene cannot convert galactose this way because the enzyme is inactive. 4. Such galactosemic people accumulate galactose which IS toxIC, causeS degeneration of the central nervous system, and finally death. 5. The disease is alleviated by eliminating galactose from the dIet. No lactose is fed. 6. While "utilizable galactose" is an essential (non-toxic) nutrient, it can be made by the body from glucose. Curing gene-caused diseases l. The gene defect is not corrected when the disease it produces is cured or alleVIated. 2. If treatment permits more genetically defective people to reproduce, tbe frequency of defectives in the popUlation will gradually increase (see Chaps. 23 and 24). 3. This will become an increasmgly important medical problem as more and more genetic diseaseS are treated. 4. Society may have to prOVide some correc- H. I. J. tion for the buildup of defective genes. The one gene - one enzyme hypothesis 1. Flower color pigmentation and eye color pigmentation m Drosophila (see Chap. 33) are examples of many other cases in which a series of chemical reactions have been shown to be controlled by genes. 2. This work led to the concept that, in many c~ses, the gene functions through the intermediary of an enzyme. 3. The one gene - one eJ1.,Zyme hypothesis says, m its sImplest form, that the total specifiCIty of an enzyme is determined by a single gene. 4. Although modified m certain respects (see Chap. 40), this view is essentially correct. S. Neurospora has been used to test certam predictions from this hypothesis. Neurospora as a tool in biochemical genetics 1. Certain aspects of Neurospora genetics have been discussed already (Chap. 16). 2. Its basIc medium for growth may conSIst solely of water, simple inorganic salts, a carbon and energy source (cane sugar), and the vitamin biotin. 3. From these raw materials It can synthesize some 20 amino aCIds, all essentIal vitamins (but biotin), purines and pyrimidines -everythmg else it needs for its total actIVIty. 4. On the hypotheSIS described in H, it should be possible to induce mutations In Neurospora which would block varIOUS of its chenncal syntheses. Vltannn·B 1 (thiamin) production in Neurospora 1. B1 is known to be produced by the action of an enzyme which combines a particular thiazole and a particular pyrimidme. 2. If enzymes owe theIr specifiCIties to genes, it should be pOSSIble to mduce a mutation m the gene that normally specifies this B1forming enzyme. If a mutant no longer specifIes active B1-forming enzyme, no Bl WIll be made, and the mold will reqUIre B1 m its dIet m order to grow. 3. Asexual spores are treated with a mutagenic agent and grown on the baSIC medium supplemented with vitamin B 1 . 4. The spores whIch grow mto cultures WIll include those whIch make B1 themselves and those whICh get it from the medIUm. 5. A large number of such cultures' are then tested for their abIlity to grow on basic medIUm containing no B1. 6. Those now falling to grow presumably carry mutants mvolving faIlure to malce B 1 . K. Detection and localization of B1 -reqUIring mutants (see Chap. 16) 1. A Bl-requiring strain is crossed WIth the normal strain. 2. After the hybrid undergoes meIOsis, a sac containing 8 haploid ascospores is produced. 3. When all 8 spore~ grow on B1 -containing medium, but 4 of them do not when transplants are placed in B1-free medium, segregation is proven and the B1-requiring form is mutant. 4. Mutants can be located relative to the centromere using heterozygotes. a. When no chiasma occurs between the mutant ® and the centromere, segregation occurs in the first meiotic division, and the 8 ascospores WIll be in the relative order +++ th-th _+-th_ - tho b. When a chiasma occurs between the mutant and the centromere, segregation occurs in the second meiotic division, and the ascospores will be m the relative order +_ I- th th + + th tho -----c. If 20% of sacs showed second division segregation the mutant gene would be located 10 map units from the centromere. 5. Many instances are known, in several different organisms, of this type of relation between gene and chemical reaction. POST-LECTURE ASSIGNMENT 1. Read the notes immedIately after the lecture or as soon thereafter as poSSIble, malcmg additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the Items underlined III the lecture notes. 4. Complete any additional assignment. 227 QUESTIONS FOR DISCUSSION 39. 1. What percentage of human bIrths do you suppose suffer from inborn errors of metabolism? What support can you give for your answer? 39.13. Why is it necessary in Neurospora to precorrect an enzymatic defect by providing suitably enriched medIUm before the defect may be detected? 39. 2. What would you predict about the genotypes and kinship of parents of alcaptonurics ? 39.14. Explain how fIrst and second meiotic division segregation is detected in Neurospora from ascospores in the case of a monohybrid. 39. 3. Would you expect that classifying all the siblings in alcaptonuric families would give a 3:1 ratio as normal: affected? Explam. 39. 4. Of what significance is each of the facts that Garrod was a physician, knew biochemistry, and was a friend of a student of heredity like Bateson? 39. 5. Using rabbits for your example, explain, on the basis of enzymes, how under different environmental condItions a gene can produce different phenotypes, whereas its allele produces only one phenotype. 39. 6. What is the specific functIon of each of the normal alleles of the genes producing alcaptonuria, albinism, phenylketonuria, and galactosemia? 39. 7. What evidence is there that alcapton is not an anomalous substance but a normal intermedIate in metabolism? 39. 8. Why were Garrod's contributions ignored by biochemists and geneticists for almost a third of a century? 39. 9. How do the diseases discussed in this chapter illustrate the interrelations of genes, enzymes and chemical reactions? 39.10. In cases where a gene produces a disruption of metabolism, can one learn anything about the biochemistry of the reactIOns which would normally follow those disrupted? Explain. 39.11. "A mutant gene can change the rate of development by changing the specifIcity of a single enzyme." Explain. 39.12. What are the advantages and dIsadvantages to the somatic cure of genetic diseases? 228 \ 39.15. What kinds and numbers of spore sacs of Neurospora would you most likely fmd among 100 sacs from ~ e_/_:1_: _:1_:, if ~ and e_ were located paracentrically 3 and 4 map units, respectively from the centromere. 39.16. What are the two functI~ns of a gene according to the one gene - one enzyme hypothesis? 39.17. Discuss whether or not the one gene - one enzyme hypothesis IS subject to direct test. 39.18. To what factors may the present great interest in biochemical genetics be attributed? 39.19. In what way does biochemical genetics differ from developmental genetics? Chapter 40 BIOCHEMICAL GENETICS II lecturer-G. W. BEADLE PRE- LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 21, pp. 373-376; Chap. 22, pp. 393-398. Dodson: Chap. 17, pp. 205-210. Sinnott, Dunn, and Dobzhansky: Chap. 24, pp. 330-331. Snyder and David: Chap. 26, pp. 385389; Chap. 24, pp. 363-368. Srb and Owen: Chap. 17, pp. 358-368. Winchester: Chap. 17, pp. 237-239. b. Additional references Anfinsen, C. B. 1959. The molecular basis of evolutIOn. 230 pp. New York: John Wiley & Sons, Inc. Beadle, G. W. 1957. The physical and chemical basIS of inheritance. Condon Lectures. 47 pp. Eugene, Ore.: Oregon State System of Higher Education. Crick, F. H. C. 1957. Nucleic acids. Scient. Amer., 197: 188-200. Hsia, D.Y.-Y. 1959. Inborn errors of metabolism. 358 pp. Chicago: The Year Book Publishers, Inc. Ingram, V. M. 1958. How do genes act? Scient. Amer., 198: 68-74. McElroy, W. D., and Glass, B. (EdItors) 1957. The chemical basis of heredity. 848 pp. Baltimore: The Johns Hopkins Press. Peters, J. A. (Editor) 1959. Classical papers III genetics. 282 pp. Englewood Chffs, N. J. : Prentice-Hall, Inc. Structure and functIOn of genetic elements. 1959. No. 12, Brookhaven Sympos. BioI. Washington, D. C.: Office of Technical Services, Dept. of Commerce. Wagner, R. P., and Mitchell, H. K. 1955. Genetics and metabolism. 444 pp. New York: John Wiley & Sons, Inc. Zirkle, R. E. (EdItor) 1959. A symposium on molecular biology. 348 pp. Chicago: The University of Chicago Press. LECTURE NOTES A. Efficient detection of biochemical mutants in Neurospora 1. Potentially mutant spores are grown on a medium supplemented with all substances which might conceivably be involved in biochemical mutation. 2. Growing cultures are then transferred to basic medium (see Chap. 39) containing no additions, where failure to grow indIcates a mutant culture which had lost the ability to synthesize some component added to the basic medium. 3. The specific ability lost is determined by testing for growth in basic medium supplemented in turn with the individual enriching components of the complete medium. 4. Techmques which selectively eliminate nonmutant strams also are employed In work with Neurospora and other organisms. B. Tests of the one gene - one enzyme hypothesis 1. If an enzyme's total speCIfiCity is determined by one gene, mutants of independent origlll, defIcient for a particular enzyme, all should be allelic (see Chap. 36). 2. Tryptophan a. The final step in synthesizing this essential amino aCId is the catalyzed union of indol and the 3-carbon amino acid serine by the enzyme tryptophan synthetase. b. Separately occurring tryptophan-reqmring mutants were obtained. Those tested 1) were blocked at the final synthesis 229 c. 2) and lacked activity for tryptophan synthetase. c. 25 qualifying mutants tested all proved allelic. 3. Adenine a. The last step in the synthesis of this essential purine is catalyzed by adenylosuccinase which splits succinic acid off adenylosuccinic acid to leave adenine. b. More than a hundred independently occurring mutants, each lacking adenylosuccinase activity, were all allelic. 4. Were two or more genes specifying one enzyme, some of the mutations changing its specificity should have been non-allelic. 5. That a gene specifying an enzyme has no other activity is suggested by the fact that mutants lacking tryptophan synthetase or adenylosuccinase require nothing to becQme normal but tryptophan and adenine, respectively. 6. Many other cases gave similar results. Biosynthetic pathways 1. Neurospora has been used extensively to identify the pathways for synthesis of many amino acids and vitamins. 2. The pathways for tryptophan synthesis is summarized by the diagram following, in which X is an unidentified SUbstance. -----7 Anthranilic ~ X -----7 Indol -----7 Tryptophan (a) acid (b) + Serine D. E. 230 a. Anthranilic acid was identified as a precursor of tryptophan by tryptophan-requiring mutants (a) and (b). b. Mutant (b) makes, but accumulates, anthranilic acid. c. Mutant (a) cannot make anthranilic acid, but once supplied with this it can make tryptophan. 3. In general all organisms carry out metabolic activities of this type by the same chemical reactions. Genic control of protein specificity 1. The protein part of an enzyme determines its specificity. 2. If a gene speCifies an enzyme by specifying the amino acids in its protein part, one should find proteins which are specified genetically. 3. This has been found. Genetic determination of protein structure 1. Hemoglobin consists of four separate chains F. G. of amino aCids (polypeptides) and four ironcontaining ring compounds (heme groups), one for each protein chain. a. The amino acids occur as two pairs of chains called alpha and beta. The members of one pair are identIcal but differ from those in the other pair. b. Each pair contains about 150 amino acids. 2. Homozygotes for the gene causing sickle cell disease have abnormal 8 hemoglobin. a. At a particular pomt in both beta chains glutamic acid is replaced by another amino acid, valine. , b. Glutamic acid has one free carboxyl group (giving a total of two negatIVe ions) while valine is neutral. 80 the net positive charge of normal adult (A) hemoglobin is increased in S he'moglobin by two (or is decreased in net negative charge by two). I c. This was the first case ':worked out, by Ingram, of a specific gene-caused change in protein. 3. C hemoglobin- a. This is produced 'by an allele of the sickle cell gene. b. In this case, lysine, which has a free amino group, replaces the same glutamic acid in the beta chains. c. Thus changing from A to C hemoglobin increases the net positive charge by four. 4. I hemoglobin and Hopkins-2 hemoglobin invol ve changes in the charge of the amino acids of the alpha chains, and are caused by genes non-allelic to the one producing 8 or C type. 5. Thus, the two kinds of polypeptide chains in hemoglobin are specified by two different genes. 6. Tryptophan synthetase has been shown to have two dis associable and reassociable polypeptide chains, each <?ontrolled in a bacterium by different, but adjacent, genes. 7. Other molecules, like insulin, also are composed of two scparate polypeptide chains \ which conceivably might be also under the control of separate genes. The hypothesis, one gene - one enzyme, might be restated one gene - one polypeptide chain. "In other words, the information to put the amino acids together in a partIcular protein chain is carried in a genetic unit which we define as a gene. " To understand how the genetic material makes speCifications one needs to H. 1. 1. identify the genetIc material, 2. learn its chemical composition and organization, 3. hypothesize how this knowledge relates to a. gene replication b. nature of mutation c. gene action 4. test these hypotheses. These topics are discussed now and in Chap. 41. Evidence nucleic acid is genetic material (see also Chap. 42). 1. The effectiveness of ultraviolet light in inducing mutations varies according to wavelength in a way paralleling the absorption by nucleic acid (not by protein). 2. PurifIed DNA from one bacterium can cause a heritable change when introduced into another bacterium, as though a gene in the latter was replaced by one from the former (Avery, et al). 3. Bacterial virus _a. conSIsts of a tadpole-shaped umt whose head contains a core of DNA. b. The VIrus attaches to the bacterium by its taIl, injects the DNA into the host, leavmg the protem coat mostly outside. c. After injection, removing the protein coat (using a Waring Blendor) has no effect on virus reproduction in the bacterium. d. Using radioactive. labels (P-32 for DNA and radioactive S for the protein) it was found 1) the DNA entering the bacterium is conserved for the next generation of viruses, while 2) the small amount of protein entering IS not so conserved. " e. Viruses are mutable; theIr genes have been studIed and found to be arranged linearly (see Chap. 43 and those following). f. Since only DNA carries over from one virus generation to another, DNA must be their primary genetic material. 4. In cellular forms DNA also is probably the primary genetic material. 5. RNA comprises the core of certain viruses and is clearly genetic material in these forms. DNA structure (see also Chap. 38) 1. DNA conSIsts of units called nucleotides. 2. Each nucleotide contains a deoxYribose (5-carbon) sugar jomed at one place to a phosphate and at another place to a N-con- 3. 4. 5. 6. taining ring compound (a purine or pyrimidine base). The purine may be adenine (A) or guanine (G); the pyrimidine may be thymine (T) or cytosine (C). Nucleobdes can string together, forming a polynucleotide chain with a phosphate-sugar backbone. One polynucleotide chain pairs with another in a complementary way. a. By means of H-bondmgs, A is always paired with T, and G with C. b. Thus, If part of one cham is A-A-T-G-C, the pairing chain will necessarily be T-T-A-C-G. The double chain is wound in a helix as shown in a three dImensional model. "This is the so-called Watson-Crick structure of DNA and, because of its significance to genetics, can be SaId to represent one of the most important advances in modern biology of the present century. " POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possIble, making additions to them as desired. 2. Review the reading aSSIgnment. 3. Be able to discuss or defme orally or in writmg the items underlmed in the lecture notes. 4. Complete any additIOnal aSSIgnment. 231 QUESTIONS FOR DISCUSSION 40. 1. Describe a way to enrich the harvest of mutants, obtained after X-raying asexual spores of Neurospora, through the use of a. filters. b. antibiotics that kill actively metabolizing cells. 40. 2. Using Neurospora, how would you proceed to find mutants for unknown growth factors? 40. 3. How would you use Neurospora mutants requiring unknown growth factors in the isolation and identifIcation of such substances? 40. 4. Some so-called biochemical mutants are not made essentIally normal by supplying one chemical substance. Can these be used to test the one gene one enzyme hypothesis? Explain your decision. 40. 5. In what ways are biochemical mutants used as tools in biochemistry? 40. 6. How many mutants are needed to identify a substance and its precursor when normally both are rapidly metabolized? Explain. 40. 7. What kinds or kind of hemoglobin would you expect to find in the red blood cells of mdividuals heterozygous for the gene for sickle cell anemia? Explain. 40. 8. Describe how you could separate A, S, and C hemoglobins from a mixture. 40. 9. What evidence can you cite that one gene determines amino aCld sequence in a polypeptide? 40.10. What would you conclude was the genotype of an individual who had hemoglobin molecules of all the following types? a. b. c. d. normal alpha, normal beta normal alpha, S beta Hopkins-2 alpha, normal beta Hopkins-2 alpha, S beta. What else would you conclude? 40.1l. From what source may evidence be obtained as to whether enzyme proteins and 232 other proteins are frequently or only occasionally made up of separable protem chains ~ Explain. 40.12. What evidence is there that some quantitativ\ely minor component of a chromosome is not the genetic material ? 40.13. What evidence was presented that protein is not the pnme genetic material ? 40. 14. What subsequent observations would test the view that, in bacterial "transformation", donor DNA replaces tl?-e DNA of the recipient cell ? 40.15. Why were radioactive P and S used in labeling the DNA and protein of the bacterial virus? Why were labeled H and C not used? - 40.16. Do the experiments described show that the gene is pure DNA? Explain. 40.17. What are the major aims of biochemical genetics? Chapter 41 GENE STRUCTURE AND GENE ACTION Lecturer-G. W. BEADLE PRE- LECTURE ASSIGNMENT of Technical Services, Dept. of Com1. Quickly review notes for the previous lecmerce. ture. Wagner, R. P., and Mitchell, H. K. '2. Suggested readings: 1955. Genetics and metabolism. 444 pp. New York: John Wiley g, Sons, Inc. a. General genetlcs textbooks Altenburg: Chap. 22, pp. 393-398. Zirkle, R. E. (Editor) 1959. A symposium on molecular biology. 348 pp. Dodson: Chap. 17, pp. 205-210. Chicago: The University of Chicago Sinnott, Dunn, and Dobzhansky: Chap. 27, pp. 373-378. Press. Snyder and David: Chap. 24, pp. 363368. LEC TURE NOTES A. Questions relative to DNA Winchester: Chap. 17, pp. 228-235. b. AdditlOnal references 1. What is its chemical and physical organizaAnfinsen, C. B. 1959. The molecular tion? This has been elucidated by the Watsonbasis of evolution. 230 pp. New York: Crick model (see Chaps. 38 and 40). John Wiley & Sons, Inc. 2. What in it contains the code for making Beadle, G. W. 1957. The physical and specIfIcations? chemical basis of inheritance. Condon a. The base pairs A-T, T-A, C-G, and Lectures. 47 pp. Eugene, Ore.: G-C can be thought of as the symbols of Oregon State System of Higher Education. a code. Crick, F. H. C. 1957. NucleIC acids. b. This is a four symbol code, whereas our SCIent. Amer., 197: 188-200. alphabet is a code With 26 symbols and Exchange of genetic material. 1958. the Morse code uses only two. Vol. 23, 435 pp., Cold Spring Harb. c. The sequence of these four kinds of base Sympos. Quant. BioI. Long Island, N. Y.: pairs can spell messages. L. I. BIOI. Assoc., Inc. d. This code could carry 1000 volumes of McElroy, W. D., and Glass, B. (Edimessage within the DNA of one human tors) 1957. The chemical basis of heredity. 848 pp. Baltimore: The cell. 3. How does it replicate? (See also Chap. 38) Johns Hopkins Press. a. The Watson-Crick model suggests the Peters, J, A. (Editor) 1959. Classical -"" two chains separate by the opening of the papers in genetics. 282 pp. Englewood hydrogen bonds holding base pairs toCliffs, N. J. : Prentice-Hall, Inc. Pontecorvo, G. 1958. Trends in genegether. tic analysis. 145 pp. New York: b. Each single chain then collects its complementary nucleotides from the surColumbia University Press. roundmg materials to make a complemenReVIews of Modern Physics, 31. 191268. 1959. tary chain. 4. How IS it related to mutation? (See also Structure and function of genetIC eleChap. 38) ments. 1959. No. 12, Brookhaven a. Mutation could be a change m the code. Sympos. BioI. Washington, D. C. : Office 233 B. 234 b. The code could change by addition, subtraction, substitution, and transpositlOn of symbols. c. Such errors could be made during DNA replication. 5. What function does it perform? The results described in Chaps. 39 and 40 suggest DNA orders or sp~cIfies the amino acids in protem molecules. Tests of chain separation hypothesis for DNA replIcation 1. Experiments WIth heavy DNA a. Bacteria are grown in a medium containing heavy nitrogen (N-15). b. Smce each base in DNA contains several N atoms, their normal light N-14 is replaced by N-15. c. DNAs of dIfferent densities can be separated by physical methods. d. When heavy DNA is permitted to replicate in a medium containing only light N, the expectation is: 1) after the first replication the density of the DNA molecules should be exactly intermediate between light and heavy. (The two heavy chains separate,and each synthesizes a light complementary cham.) 2) after a second replicatIon half of the DNA molecules are light and half are intermediate between light and heavy. (The two chains of the hybrid DNA separate, and light complementary chains are made by both the light and the heavy chains. ) e. These predictions were fulfilled completely in experiments done first with bacteria (Meselson and Stahl) and later with other organIsms. 2. ExperIments with radioactive chromosomes a. Herbert Taylor and associates made chromosomes radioactive by feeding tritiated thymidine which becomes incorporated into DNA as radioactive thymine. b. When a chromosome so labeled was allowed to replicate in unlabeled medIUm, 1) after one cell division, both daughter chromosomes were stIll radioactive. 2) after a second cell dIViSIOn, radlOactiVlty was present in one daughter chromosome and not in the other. c. Such results are consistent with the replication hypothesis and have been obtained in several different plant species. FOlD[D UNfOLDED R[PLICATION Figure 41-;1 d. Fig. 41-1 shows a model of a chromosome, suggested b_yl Freese, ,in which old DNA is separated from new after two chromosome diVIsions. (The_ black ovals are rather hypothetical protein links between the DNA segments in a chain. ) 3. Replication of DNA in vitro a. This has been performed by Nobel Laureate Arthur Kornberg,ill2d his aSSOCIates. b. DNA is syntheSIzed rapidly in a test 'tube contmning 1) a polymerizmg enzyme (obtained frbm bacteria), 2) a buffered solution of nucleotides (with A, T, C, and G bases), each of whICh carries two extra phosphate groups (to provide energy for polymerization), 3) a piece of already formed DNA, which serves as a primer. c. The DNA synthesized is lIke the prImer in so many SIgnificant respects (for example, It has the same ratio of C + G : A + T) that it is difficult to belIeve that it is not being synthesized by essentially the same mechanism as that of the living cell. , d. When all the ingredients of B3b are present, except for the primer DNA, a polymer forms slowly which contains A-T base pairs and excludes C-G pairs. e. When such an A-T polymer is used as the primer under the conditions m B3b, more DNA is made containmg A-T pairs but not C-G pairs. C. D. f. This is a verificatIOn of the Watson-Crick structure. 4. All these results are consistent with, though they do not prove, the hypothesis of DNA replication by chain separation. Carrying out DNA's specifications 1. The evidence indicates that DNA encodes the Information necessary to make polypeptides. 2. How does DNA's four symbol code translate, for example, into an alpha chain of hemoglobin that is synthesized in the cytoplasm? 3. A popular working hypothesis is: a. the coded information in DNA is transferred to RNA which also has a four symbol code (being composed similarly of four nucleotIdes in a, perhaps smgle, chain). b. The RNA then migrates into the cytoplasm, where it is incorporated into microsomes -- organelles which are the site of protein synthesis. c. The RNA serves as a template against which amino aCIds collect m the right sequence to make an alpha chain. d. Once complete, the alpha chain peels off so a new alpha chain may be formed similarly. The unit in DNA's code 1. This cannot be a single base pair, since this would provide only four "words" for specifymg 20 kinds of amino acids. 2. SImilarly, the unit cannot be two base pairs, for this would provide only 16 (4x4) different combinations (words). 3. If sequences of three base pairs made up the unit, 64 (4 x 4 x 4) different combmations would be possible. 4. However, there are other requHements for the DNA code. a. It must be a "comma-free" code, SInce there is no separation between words. If in the following sequence 1 A-T 2 T-A 3 C-G 4 A-T 5 T-A 6 T-A amino aCIds A and B are specified by 123 and 456, respectively, errors would be possible if 234 or 345 encoded different ammo acids. These last two triplets, therefore, would have to be eliminated as words m the code. When, of the 64 E. F. possible triplets, all overlapping ones are eliminated, 20 non-overlappmg triplets (words) remain. b. A DNA chain is oriented or polarized by the linkages made by the number 3 and number 5 carbon ·atoms in each sugar (see Fig. 38-3), so that a chain has direction. But in the Watson-Crick model complementary chains run in opposite directions, and it is possible to read the symbols either way. c. One way to insure the double helix is not read the wrong way is to have all triplets read in one dIrection make nonsense. This reduces the number of legitimate triplets to 10, which is too few. 5. A umt consisting of four nucleotides (four letter words) provides 27 words satisfyin~ the requirements in D4. But whether this is the way the code is constructed is unknown. 6. Solution of the coding problem will be a very important contributIOn to genetIcs and to biology. Mutational sites within the gene 1. If the sequence C?f amino acids m a protem is determined by a segment of DNA, called a gene, the DNA segment must be several hundred base pairs long. 2. It ought to be pOSSIble, then, to make mutations a. at many sites within the gene, and b. these sites should be arranged linearly. 3. Hybrids can be made between Neurospora mutants which affect adenylosuccinase's protein in different ways. a. RecombinatIOns occur which restore the enzyme's activity. b. Recombination frequencIes from many dIfferent such hybrIds show the mutational sites are numerous and linearly arranged. 4. Such results have been obtained also for bacterial and viral genes. "For the first time in the history of biology It is now possible for physicists, chemists, and bIOlogists to talk about these problems in the common terms of molecular biology. " POST-LECTURE ASSIGNMENT 1. Read the notes immedIately after the lecture or as soon thereafter as possible. 2. Review the reading assignment. 3. Be able to dISCUSS orally or in writmg the items underlined In the lecture notes. 4. Complete any additional assignment. 235 QUESTIONS FOR DISCUSSION 41. 1. What value has the Watson-Crick model other than to describe DNA structure and organization? 41. 2. At the biochemical level, in what way may mutations be like typographical errors in language? 41. 3. How might unnatural purines or pyrimidines act as mutagens? 41. 4. State two kinds of coding changes whICh nught result from the substitution of one single base pair by another. What might be the resul ts upon the polypeptide chain specified? 41. 5. Describe how Fig. 41-1 is consistent with DNA replication by chain separation. 41. 6. In following up the work of Kornberg, what additional information would you seek experimentally? 41. 7. Suppose sequences of four numbers make up words in a code. How many sequences must be nOllrense if 12343421 is to encode just two words when read in either direction? 41. 8. In what respect is the devising of specific codmg systems for DNA more than just an amusing game? 41. 9. Should the one gene - one polypeptide hypothesis be restated as "one base pair one amino acid"? Why? 41. 10. In Neurospora, a gene concerned with a specific reaction in the synthesis of the vitamin pantothenic acid is known to mutate in several different ways. Explain how you would use conventional mapping techniques to test whether the mutational sites of this gene were linearly arranged. 41. 11. Do you suppose that only one gene is responsible for "malting" a polypeptide? Explain. 41.12. Do you suppose there are genes which act in the synthesis of amino acids? Give evidence for your opinion. 236 41. 13. Wh,at would you consider the most important unsolved problem in biochemical genetics? Why? 41. 14. Can epistasis occur at the biochemical Explain. l~vel? 41. 15. Has the biochemical study of gene action influenced our ideas concerning the nature of the gene itself? Explain. 41. 16. Has biochemical knowledge about the genetic material aided our understanding as to how it produces 'phenotypes? Explain. Chapter 42 CHROMOSOME CHEMISTRY AND GENETIC ACTIVITY Lecturer-JACK SCHULTZ PRE- LECTURE aSSIGNMENT 1. Quickly review notes for the prevIOUS lecture. 2. Suggested readings: Allfrey, V. G., Mirsky, A. E., and Stern, H. 1955. The chemistry of the cell nucleus. Adv. in Enzymol., 16: 411-500. Beermann, W. 1959. Chromosomal differentiation in insects. Chap. 5 in "Developmental cytology", D. Rudnick, Editor. New York: The Ronald Press Co. Brachet, J. 1957. The nucleus of the resting cell. Chap. 4 in "Biochemical cytology". New York: Academic Press, Inc. Gall, J. G. 1958. Chromosomal differentiation. pp. 103-135 in "The chemical basis of development", W. D. McElroy and B. Glass, Editors. Baltimore: The Johns Hopkins Press. Pavan, C. 1958. Morphological and physiological aspects of chromosomal activities. Proc. lOth intern. congr. Genet., 1: 321336. Rudkin, G. T., and Woods, P. S. 1959. Incorporation of H3 cytidine and H3 thymidine into giant chromosomes of Drosophila during puff formation. Proc. nat. Acad. Sci., U. S., 45: 997-1003. LECTURE NOTES A. Aspects of modern genetics started about 1870: 1. studies of the rules of transmission, initiated by Mendel; 2. studies of the physical basis of heredity, initiated by the Hertwigs' demonstration that the sperm contributes a nucleus at fertilization; 3. studies of the chemistry of the hereditary material, initiated by Miescher. B. Miescher's contributions 1. He studied the chemical composition of cell C. D. components. 2. The chemical similarities found between the nuclei of ordinary cells and the sperm head led him to suggest that the sperm contributes a nucleus to the egg. 3. He showed the nucleus contains nucleic acid and other substances. Cytological distribution of nucleic acids 1. DNA is restricted to the nucleus This can be seen in a total mount of a larval salivary gland of Drosophila which shows that only the nuclei have been stained by the Feulgen procedure. (The Feulgen staining reaction is specific for DNA. ) 2. Nucleic acids within chromosomes a. In the nucleus, DNA is restricted to the chromosomes. b. In photographs taken in ultraviolet light, nucleic acid appears to have a uniform distribution in the 40 tiny chromosomes of a mouse tumor cell, but, as seen in the giant salivary gland chromosomes of insect larvae, it has an uneven, but characteristic, distribution as seen by bands, interbands, and puffed regions. c. By cytogenetic study of a series of different rearrangements involving the same chromosome region, it is possible to show that a particular gene resides in a particular nucleic-acid-containing band. 1) A band can contain more than one gene. 2) What appears as one band often may be actually a doublet. d. What is the detailed chemical composition of bands, mterbands, and puffs? Chemical composition of chromosomes 1. in groups of isolated nuclei a. Refined histochemical techniques have been developed by A. E. Mirsky and his collaborators. b. The gross amount of DNA extracted from 237 E. 238 different tissues of an organism is proportional, within 10%, to the number of chromosome sets contained. c. This regularity is not true for the protein or RNA in chromosomes, for their amounts can vary quite widely. d. The 10% variability for DNA could allow for an enormous number of differences in hereditary code in different tissues. 2. in a single nucleus a. Caspersson adapted physical-chemical methods to the microscope in order to study the detailed composition of cell structures. b. He used absorption of ultraviolet light by microscopic structures to measure their distribution and chemical composition. The ultraviolet microscope was used because its lenses transmit such wavelengths. c. The wavelength of ultraviolet that is especially favorable to use (257 ml-l) happens to be one which is highly absorbed by nucleic acids. d. His refined cytochemical techniques made it possible to measure the amount of nucleic acid in a single band of a salivary gland chromosome. Chemical content of salivary chromosomes and their bands 1. using enzyn:atic <.nd chemical treatments. a. This was illustrated with Rudkin's experiments on Rhynchosciara giant salivary chromosomes. b. The same chromosome was shown photographed successively with ultraviolet light: 1) as it was originally, then 2) after treatment with ribonuclease, which removes all RNA, then 3) after treatment with a deoxyribonuclease, which removes a good part of the DNA, then 4) after treatment with hot trichloracetic acid, which removes all nucleic acids. c. The faint object remaining after these treatments was the protein of the chromosome. d. Ultraviolet measurements made at such successive stages showed that in a single band there is a lot of DNA, protein, and some RNA. 2. in a chromosome with and without a mutant (see p. 12 ). a. Female larvae of Drosophila were made heterozygous for the mutant white and for a cytologically visible difference which in salivary cells distinguished the X chromosome bearing w from the homolog bearing w+. b. When the amounts of ultraviolet absorbing material in the region containing w and w+ were compared, there was nodifference to within about the 5% error of the method. c. Calculations showed that, in the sperm, w and w + could differ in amount by no more than about five nucleotides. d. This maximum nucleotide difference between w and w+ is so small as to support the view these alleles differ not in nucleotide number but in nucleotide arrangement (sequence or code). 3. Nucleic acid content of different bands a. The ultraviolet absorbance of different bands can be shown in a frequency distribution (Fig. 42-1; the vertical axis gives the number of bands, the horizontal axis shows their absorbance). Figure 42-1 b. The bands have different amounts of absorbing material, as expected from their appearance. c. From the smallest detected amount of ultraviolet absorbed by a hand, one can calculate the amount of material tlus would involve in a sperm cell. d. This material would have a molecular weight of the same order of magnitude, 8 million, as has genetically active DNA in transformation experiments with microorganisms (see subsequent Chapters). F. G. H. Cell activity and chromosomal chemicals 1. Goldschmidt long ago proposed that chromatin was of two types -- one was the hereditary material, the other the working material of the cell. 2. This view, once discarded, was revived in terms of DNA and RNA, respectively. 3. Caspersson and Schultz, and Brachet have shown RNA is important in cell growth. Recently, others have shown biochemically that RNA has to do with protein synthesis. 4. What is the relation of chromosome activity to DNA and RNA production? Activities of larval giant chromosomes 1. Beermann, working with Chironomus, found that in different tissues (malpighian tubules, salivary glands, intestinal caeca, and rectal glands) the same chromosome region shows different degrees of local swelling, puffing, and different chromosome regions have different puffing patterns in different tissues. 2. At the same time Pavan and Breuer, working with Rhynchosciara, showed that in the same cell cllfferent puffs appear and regress at different times of development. 3. The different behavior of puffs in the same and in different tissues leads to the supposition that the puffs have a specific activity, that the chromosome takes on particular functions with specific stages of development. Gene action and the chemistry of puffing 1. Pavan and Breuer believed that there was a disproportionate increase in the amount of DNA associated with puffing. 2. Rudkin then studied the matter quantitatively (Fig. 42-2). a. The upper part of the figure shows the different chromosome segments whose DNA ultraviolet absorbance was determined at different stages of puffing. b. The lower part summarizes the values obtained. c. In non-pUff segments (I, IV, V) twice as much DNA was present after puffing as there was before puffing (center column). d. Puff a (segment II) had increased its DNA fourfold in this period, so that it had made twice as much DNA as did non-pUff segments (right column). 3. Thus DNA is synthesized disproportionately in chromosome regions which seem to be concerned with developmental changes. 4. Using autoradiographic techniques, RNA and DNA synthesis can be studied separately. In some cases it is found that the puffs are local centers for the synthesis of RNA, even where the DNA is not synthesized disproportionately. Figure 42-2 5. There are then two modes of gene action: a. disproportionate increase in amount of coding substance (DNA), and b. increase in the amount of metabolic RNA. 6. This is a general principle for gene action. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 239 QUESTIONS FOR DISCUSSION 42. 1. Why do you suppose progress was so slow in the chenucal study of the hereditary material for the half century following Miescher's dIscoveries? 42. 2. Do you accept the statement in the Notes "DNA is restricted to the nucleus"? Explain your answer. 42. 3. Do you suppose that the DNA in mouse chromosomes is, in fact; distributed evenly along the long axis? Explain. the chromosome? 42.13. How might you label the RNA and not the DNA in the same cell? How could you do the reverse? 42.14. What in the present chapter is evidence for identifying the hereditary material chemically? 42.15. Would you expect the principles of gene action presented in this Chapter to apply to all eplls? 42. 4. How do the techniques of histochemistry and cytochemistry differ? 42. 5. How could you prove that a band contained more than one gene? 42. 6. DescrIbe how you could prove that the Feulgen procedure stains DNA speCIfically. 42. 7. Describe how you could, if it is possible, localize a gene in a particular band using only a. b. c. d. deficiencies inversions translocations point mutatIOns 42. 8. What advantage or disadvantage does microphotometric chemical analysis of chromosome regions have over gross chemical analysis of isolated nuclei? 42. 9. If ultraviolet light is damaging to living cells can it properly be used in cytochemical investigations? Explain. 42.10. In what way do you suppose the chromosomes in Fig. 42-2 may have been treated pnor to measurement for ultraviolet absorbance? Why? 42.11. What do you suppose is the cytologlcal marker which distlllguishes the X chromosome bearlllg ~ from the homolog bearlllg w+ in the photograph on p.12 ? Explain. 42.12. USlllg ultraviolet light, how would you proceed to measure the separate amounts of RNA and DNA present in the same region of 240 Rxnlmn EXAMINATION V UNDERLINE ALL THE ALTERNATIVES WHICH ARE CORRECT. 1. When complete culture medium is supplied to Neurospora a. one cannot observe a bIOchemical mutan:t which fails to grow. b. a culture which grows mayor may not be mutant. c. nothing can be learned about the number 01 growth requiring mutants unless additional tests are made using basic or minimal medium. d. it is not possible to detect the presence of morphologIcal mutants whICh grow under these conditions. e. it is not possible to detect mutations requiring a component supplied by this medium unless transplants are made to minimal medium. 2. 5. Cytoplasmic inheritance a. refers to self-duplicating factors which are not in the chromosomes. b. is sometimes the cause of failure of chlorophyll production. c. is normally never associated with known cytoplasmic organelles. d. is rarely transmitted by factors contained in the male gametes. e. is based upon RNA while nuclear factors are associated with DNA. 6. The Watson-Crick double helix configuration for DNA a. does not require that nuclear RNA be double-stranded. b. does not require that the protein component of deoxyribonucleoprotein molecules be double-stranded. c. does not require all nuclear DNA to be double-stranded. d. requires precisely as much adenine as guanine. e. does not require the total purines to equal the total pyrImidines in the molecules. Inborn errors of metabolism a. are the basis of many morphological abnormalities. b. involve production of enzymes which are defective quantitatively or qualitatively. c. furnishes a tool for assaying biologically certain chemical substances. d. comprise all of our load of detrimental mutations, except for those arising in the current generation. e. are of no consequence to Neurospora grown on synthetic basic medium, plus biotin. The Creeper gene is due to a mutant which is lethal when homozygous a. but does not comprise a balanced lethal system as does a single gene for taillessness in the mouse. b. but this is not true for the gene for pituitary dwarfism in mice. c. and is the cause of developmental delay even when heterozygous. d. in which case the embryo's eye is defective for intrinsic reasons. e. has an effect similar to that of a gene known in man. Developmental genetics a. is concerned with the study of how phenotypes come into being. b. deals with gene action both on the intraand inter- cellular levels. c. studies how a particular allele acts upon morphogenetic processes. d. is restricted to a study of multi-cellular forms. e. is based upon biochemical reactions controlled by genes. 3. 4. 7. The presence of the mutant iojap in homozygous condition 241 a, makes the plant non-green and male sterile. b. makes the plant striped and male fertile, c. makes more than one cytoplasmic factor mutable. d. has nothing special to do with the phenotypic expression of cytoplasmic factors. e. has no effect upon the mutation of ~ to ~, although in the presence of ~ the plant mayor may not be male sterile. 8. Position effect a. does not require pseudo allelism for its expression. b. can occur only in the diploid condition. c. cannot be demonstrated for adjacent loci which are kept in this position and are identical and homozygous. d. can be increased or decreased by heterochromatin. e. in pseudoallelic series requires that the cis form show the wild-type phenotype and the trans form the mutant one. 9. 242 Nucleocytoplasmic interactions The loci in a pseudo allelic series a. are located very close to each other in the chromosome. b. are detectable only when they produce different phenotypes in cis and trans positions. c. are in trans pOSition when the normal genes of the series are not all in the same member of a chromosome pair. d. mayor may not have originated as duplicated genes which l:lter underwent mutation. e. normally exist in heterozygous condition. 13. Modulator a. presumably transposes by some multibreak chromosomal rearrangement. b. affects both its alleles and non-alleles. c. does not cause mutation of adjacent loci, but may cause them to have their action suppressed. d. need not be absent from a completely nonred pr homozygote. e. causes fewer spots the more transposed Modulators are present in addition. DNA a. is the only polymeric material which is genetic. b. has its specificity determined only by its base content. c. differs from RNA in two or more ways. d. is probably genetic material on some other planets also. e. has sugar molecules which are attached at three places to other molecules. 11. 12. A corn plant having a medium variegated pericarp a. would have been solid red if Modulator was absent. b. would have beer, ligU variegated were one transposed Modulator added to the genotype. c. is probably showing a wavering type of position effect. d. cannot be homozygous for pw. e. shows a mutation has occurred each time a red spot occurs. 10. a. include effects of cytoplasmic factors upon the mutability of nuclear genes. h. include effects of nuclear genes upon the coming to expression of cytoplasmic factors. c. involve effects of cytoplasmic factors upon the coming to expression of nuclear genes. d. refers to all of the interactions between nuclear and cytoplasmic components, whether or not these components are se1£replicating. e. may provide important clues to our understanding both of normal and abnormal cell differentiation. 14. Ultraviolet light of wavelength 2570A a. is more highly absorbed by RNA than by protein. b. is more effective in producing mutations than is ordinary visible light. c. is transmitted by quartz but not by glass lenses. d. would be less highly absorbed after a nucleus was treated with RNAase than after treatment with hot trichloracetic acid. e. is absorbed by prec.isely those chromosomal components which are stained by the Feulgen procedure. cate by a method consistent with the hypothesis of chain separation. 17. - 15. The information encoded in the hereditary material In a given region of a salivary gland chromosome a. viable mutations can occur which probably remove all of the DNA. b. mutations have occurred which produce no detectable change in DNA. c. each band may contain more than one gene 31ld hundreds or thousands of DNA double helices which are identical. d. the amount of DNA present can be determined in one measurement. e. appearance and regreSSion of puffs is correlated with the disproportionate manufacture of RNA or DNA or both. a. always resides in the base pairs of DNA. b. must be in comma-free form because the code can be read in more than one direction. c. of a human sperm would fill a thousand books even if a sequence of four base pairs were required to form a word. d. is in no way transmitted just by the process of gene replication itself. e. may be conveyed to protoplasm by another coded system like RNA, or by some other mechanism. 18. If the gene is defined as a unit which specifies amino acid sequence in a polypeptide 16. In chromosomes, the nucleic acid polymers a. non-polypeptide substances cannot be specified by genes directly. b. each gene can have only one primary action besides self-replication. c. one can discard the older definitions based upon mutation and crossingover. d. protein specification may involve action by more than one gene. e. there should be within a gene many mutational sites. These sites must be arranged linearly whenever DNA is the primary genetic material. a. found are always of both DNA and RNA types. b. of DNA type are practically constant in total amount, although some fluctuations are found in local chromosome regions. c. are always combined with basic proteins in the form of nucleoproteins. d. are clearly responsible for the linear organization of the chromosome as seen with the light microscope. e. comprising the hereditary material repli- x 25% 25% .F, 19. What can you decide from the results, consistently obtained, summarized in the figure above? 243 20. Suppose in Neurospora gene 1 controls production of-indol, gene §. controls production of serine, and gene I controls the combination of indol and serine mto tryptophan. All are located on dIfferent paIrS of chromosomes. The formatlOn of tryptophan is, of course, blocked if an organism is homozygous for anyone (or more) of the three recessive alleles (!_, §_, or!). Such an organism is called tryptophanless. If all three dominants are present, the organism is wild-type. What ratio of wild-type to tryptophanless would result from the cross of Ii Ss Tt to ii ss tt? 21. In Neurospora, mutant 30300 will grow on minimal (basic) medium plus citrulline, or on minimal medIUm plus arginine, but not on minimal plus ornithine. Mutant 21502 will grow on minimal medium plus ormthine or citrulline or arginine. Mutant 36703 requires arginine for growth, and will not grow on minimal medium plus either 6f the other substances. None of the mutants grow on minimal medium without supplement. Each of them segregates as a single gene difference when crossed to the Wild-type. Construct a directed chain of reactions and show where each mutant acts. 22. Many traits, including sexual characteristics, body build, metabolic characteristics, and others, are determined by the prevailing balance of varIOUS hormones, yet these same traits are found to be heredl tary. How do you reconcile these endocrinological facts with the facts and principles of genetics? 23. Describe how pseudoalleles demonstrate position effect. 24. Usmg no more than a few sentences, descnbe briefly 6 eVIdences that DNA behavior parallels genetic behavior. 1. 2. 3. 4. 5. 6. 244 25. In a species of animals aa normally produces "a" phenotype, bb normally produces "b" phenotype, and ++ normally produces "+" phenotype. Transplantations were made between individuals which were isogenic except as indicated below .• Genotype of implanted embryonic tissue aa bb ++ ++ aa bb Genotype of host ++ ++ aa bb bb aa Phenotype developed by implanted tissue + + + + + b What conclusions can you draw from these results? 26. In the nucleus, _ _ _ _ _ _-:-_ _--:-_ _ _ _ _ _ are always found in chemical combination wlth _ _ _ _ _ _ _ _ _ in the form of _ _ _ _ _ _ _ _ _ _ _ _ _ . The acids are of two types called and The basic unit of either acid is the _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ , which is composed of a _________ , a _ _ _ _ _ _ _ _ __ and an ------------------------- The prime genetic material in the tobacco mosaic virus is _____________ , while in most cells It is The latter contams the purines and _ _ _ _ _ _ _ _ _ _ _ _ and the pyrimidines and _ _ _ _ _ _ __ _ _ _ _ _ _ , all of which have a structure. In the Watson-Crick model a is always _ _ _ _ _ _ _ _ bonded to a and the backbone of a chain is made up of _ _ _ _ _ _ _ _ __ linkages. The base complementary to is _ _ _ _ _ _ _ _ _ _ _ _ , and the reverse. while the complement to is and the reverse. The worle of _ _ _ _ _ _ _ _ _ _ _ showed that to obtain rapld _ _ _ _ _ _ _ _ _ _ _ __ synthesis m the test tube one needs _ _ _ _ _ _ _ _ _ _ __ _______________ , and _ _ _ _ _ _ _ _ _ _ _ _ _ _ . In the absence of just the _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ polymerization slowly occurs but contains only _ _ _ _ _ _ __ 27. Describe how you would obtam, from a wild-type strain of Neurospora, a mutant strain which requires pantothenic acid for its growth. 245 28. Would a human that was half XY and half XO appear as it would in Drosophila, half male and half female? Explain. 29. The followmg statements should refer to various aspects of the phenomenon of position effect, as discussed at different tImes durmg this course. If the statement IS relevant and true, write the word true in the space provided following each statement; if it is irrelevant, write the word irre-1 - levant. If it IS false, change the underlIned word or phrase to an approprIate form, usmg thc space provided, whether or not the statement is relevant. 1. One possible explanation of position effect suggests that rearrangements might result in an altered chemical environment for the genes WhICh have been separated from one another. 2. If kynurenine is supplIed artificially to vermilion flies, wild-type eye color will result. 3. The presence of an extra Y chromosome m a male fruit fly contaimng a mottl!ld white eyed mutant associated with an inversion causes the mutant effect to be suppressed. 4. When associated with heterochromatic rearrangements the mutant effect is expressed uniformly. 5. The ClS form of two recessive adjacent alleles lS usually wlld-type. ____________ 6. The frequent reversions of homozygous Bar eyed flies to the wild or round eyed form may be interpreted as back mutations at the Bar locus. 7. Pseudoallelism is a widespread genetic phenomenon throughout the plant and animal kingdom"!.. 8. When four doses of the Bar region in Drosophlla are present in the two X chromosomes of a female, it makes no difference whether two doses are on each chromosome or whether three doses are pre-sent on one chromosome and one dose on the other. 9. Position effects assocIated with rearrangements are rare in most animal and plant groups. 10. Rearrangements resultmg in yellow, achaete, or scute mutations when analyzed for salivary structure m Drosophila show good evidence that these three mutants are associated with one salivary band or less. 246 Chapter 43 BACTERIAL GENETICS: CLONES lecturer-J. LEDER BERG PRE-LECTURE ASSIGNMENT 1. Qmckly reVlew notes for the prevIOUS lecture. 2. Suggested readings: a. General genetics textbooks Snyder and David: Chap. 26, pp. 401407. 8rb and Owen: Chap. 24, pp. 534-537. Winchester: Chap. 23, pp. 318-321. b. AddltlOnal references Bryson, V., and Szybalski, W. 1955. Microblal drug resistance. Adv. in Genet., 7: 1-46. Lederberg, J., and Lederberg, E. M. 1952. Replica plating and indirect selection of bacterial mutants. J. Bact., 63: 399-406. Luria, S. E., and Delbrttck, M. 1943. Mutations of bactena from virus sensitivity to VlruS resistance. GenetIcs, 28: 491-511. Reprinted III "Papers in microbial genetlcs", selected by J. Lederberg. 1951. Madlson: Uruversity of WlsconSlll Press. B. C. I LECTURE NOTES A. Motivations for research with bacteria are: 1. their importance in general ecology, agrIculture, and disease; 2. the very large populations which are easily handled in the laboratory; 3. their simple cell structure. a. Escherichla colI contains two or lour nuclei, chemlcally defined as masses of DNA, each con taming about six million nucleotide units. b. DNA <..:ontent of a mouse cell lS about five billion units. c. Although the morpholOgIcal mechanism of nuclear division in bacteria is still controversial, the exact replication of D. DNA occurs each cell dlvision. Vegetative (asexual) reproduction 1. This is the most important means for increasing bacterial numbers. 2. A clone (see also Chap. 35) is a population of llldividuals all derived from a single cell by vegetative reproduction. 3. Barring mutation or genetic recombination, all clonal members are genetically identICal. 4. Bacteria multiply rapidly. a. E. coli divides about each half hour. b. One such cell, in suitable nutrient medium, will produce a population of N = 22t = 2 n individuals in t hours, or n generations. c. Thus, from a single ancestor, 30 generations (requirmg 15 hours)' would produce about 10 billIon organisms. Methods for lsolating a single bacterium 1. Directly, by the tedious but exact procedure of micromanipulation 2. Indlrectly, by dilution a. When a fluid suspenslOn of bacteria is sufficiently diluted, a sample spread on agar will contain relatlvely few bacterIa. b. Each such cell will be located on the agar at random and givc rise to a vlsible clonal colony (Flg. 43-1, top left plate). 3. Indirectly, using the simple innoculating loop a. A sample of a broth culture IS streaked upon fresh agar. b. At some places single cells wlll have been deposited some dIstance apart, YIeldmg separate colonies (Fig. 43-1, top right plate). Typing bacteria by clonal phenotype 1. Individual cells show few morpholOgical variations -- like presence or absence of flagella. 2. Because there is not enough material per 247 .. . . ' ' / , .. ~... ~.., " I .' : "! " .. ,,'" ". .. .' ... ." '" II .. •..... ,,' .::.: ": .. It ... I" ':. ",. .. ....... .. . ... '". -: ~.'.,; '. ..~ ~ '. 0' .:"~. ~:.: ... .' ~: <" ",,: "--~>,,7- ,,0::.:. . •• • ===._ -. ... Fh :::···· ." Figure 43-1 E. 248 cell, a bacterium is typed from the physiological and biochemical behavior of the clone to which it gives rise. 3. The top right plate in Fig. 43-1 has an eosin, methylene blue, agar medium containing lactose. a. The top half was streaked with wild-type organisms, which ferment lactose via beta galactosidase, generating colored, dark, clones. b. The bottom half was streaked with an ultraviolet induced lactose-negative mutant which produces light clones. Origin of bacterial mutants 1. Bacteria are intimately exposed to their chemical environment. 2. Mutations from lactose-negative to lactosepositive could be detected by appearance of dark colonies in the lower half of the top right plate in Fig. 43-l. 3. Would the medium in that plate have induced the mutations, or would these have occurred anyway? 4. The same question may be asked also whenever a mutant form has a selective advan- F. tage on the culture medium employed for its detection. 5. For example, streptomycin-sensitive cells do not form colonies when plated on agar containing streptomycin. But a streptomycin-resistant mutant which occurs among them will form a visible colony. 6. Are the mutants pre-adapted or post-{ldapted with respect to the detecting medium? Are they spontaneous or medium-induced? Pre-adaptiveness and spontaneous nature of bacterial mutations 1. Fluctuation test (Luria and Delbrtick) a. The number of mutants within a clone will depend upon the amount of time there is for multiplication before the test for them is made. b. On the post-adaptive view, the number of mutants in different samples tested will fluctuate and form a normal distribution because of the random occurrence of mutations in the final generations exposed to the testing medium. c. On the pre-adaptive view, the number of mutants in different samples should form a skewed distribution. For a very few samples should contain a very large number of mutants because mutation had occurred early in clonal life, long before exposure to the testing medium. d. The frequency of cultures containing jackpots of mutants demonstrated these were pre-adaptive and spontaneous in origin. 2. Three clone-sampling procedures are available. a. A single streptomycin-sensitive clone is plated on agar to produce a large number of colonies. Each colony is individually streaked across a streptomycincontaining line in the agar. All clones will grow except in the streptomycin region, but if enough clones are tested one will grow there also -- being a spontaneous, pre-adaptive, streptomycin-resistant mutant (see Fig. 43-1, top center plate). This method is too laborious to test the pre-adaptation hypothesis. b. Replica plating of separate colonies An agar plate containing up to a thousand separate colonies is pressed on velvet so that a sample of each colony is left on it. The velvet is then used as a master to plant a corresponding pattern of growth on a series of additional agar plates. The three lower plates in Fig. 43-1 show the master (left) and two of its subsidiary plates prepared this way. A masfer plate not containing streptomycin can be used to make a first copy, also on drug-free agar, and then additional copies on plates containing streptomycin. On the streptomycin plates only the resistant colonies will grow. This also is too laborious for testing the pre-adaptation hypothesis. c. Replica plating of unseparated colonies A billion or so organisms plated on agar will form small clones so closely spaced as to show continuous growth. Replicas can be made as already described. Subsidiary streptomycin-containing plates will show growth where there are drug-resistant mutants. On the preadaptive view one can return to the corresponding site on the master plate and obtain a sample which is richer in drugresistant mutants than is a sample taken G. H. from another part of the plate. This result has been found. d. In all of these methods only a sample of each colony is exposed to the medium that tests for mutants, making it possible to prove the testing medium has not played a direct role in producing the mutants. Detection of bacterial mutants 1. Spontaneous mutants a. can be selected for, using deleterious agents. b. to nutritional independence can be detected easily among nutritional mutants plated on media lacking the required nutrients. 2. Very low mutation rates can be measured wi th these techniques. The lowest rate so far detected is one per one billion divisions for mutation from streptomycin sensitivity to resistance in E. coli. 3. Mutagen-induced mutants also can be detected through the use of these techniques. Induced mutation 1. X-rays and many other agents are mutagenic in bacteria. 2. Novick and Szilard showed that purines like caffeine, adenine, and guanine increase, while their ribosides decrease, bacterial mutation rate. 3. No mutagen is known at present which produces a given mutation at will. 4. This reflects the fact that each gene must contain all four of the nucleotides in DNA, different genes having these in different arrangements. 5. A specific mutagen would have to recognize specific assemblages of nucleotides, being itself as complicated, chemically and structurally, as the gene it mutates. 6. Spontaneous mutation is in many respects an incident of the normal metabolism of the cell. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 249 QUESTIONS FOR DISCUSSION 43. 1. What advantages do bacteria have as material for genetic study? 43. 2. What disadvantages do bacteria have as genetic material ? 43. 3. If division occurred once an hour, how many bacteria would be produced :=t. after 4 hours, starting with one bacter- ium? b. after 3 hours, starting with four bacteria? c. after n-1 hours, if on the nth hour there were 2n ? 43. 4. What proportion of a clone would be mutant if one cell produced by the third division underwent a mutation, but was adaptively unchanged? What would you expect to fmd in this clone if there was s"election for or If there was selection against the mutant? 43. 5. Discuss the advantages and disadvantages of various techniques for obtaining a clone from a single bacterium. 43. 6. Bacterial clones have been compared to the soma produced by zygotes of multicellular organisms. Discuss whether or not this view is justified or potentially fruitful. 43. 7. What is the virtue of the necessity of using biochemical traits in most mutation studIes with bacteria? 43. 8. What morphological traits of clones would be of use in bacterial genetics? 43. 9. What disadvantage has the use of the clone for typing the parental cell ? 43.10. Does the post-adaptive view of the origin of bacterial mutations ever apply? Explain. 43.11. Suppose from a single clone of streptomycin-sensitive E. coli approximately 100 bacteria are placed in each of 100 test tubes containing drug-free broth. When each test tube contains about one billion individuals its contents are poured on the surface of nutrient agar medium containing streptomycin. 250 a. DescrIbe the kind of result which would prove mutations to streptomycin resistance were pre-adaptive. b. What kind of result would prove neither the pre-adaptive nor the post-adaptive hypothesis of mutant origin? 43.12. Ho'Y could you show that the streptomycin resistance seen in the two central streaks in the top center plate of Fig. 43-1 was not induced by the exposure to streptomycin? 43.13. Explain how you would proceed to detect and isolate independent mutations from methionine-requiring to methionine-mdependence using the techniqu'es of replica plating a. separated colonies, arid b. unseparated colonies. I 43.14. How would you proceed to detect bacterial mutations from wild-type to threonine-requiring? from threonine-requiring to threonine-independence? 43.15. Design a specific experiment which would test whether X-rays induce mutations in bac, teria. 43.16. Design an experiment using semi-solid culture medium to detect and collect mutations to motility. 43. 17. Discuss Lederberg's statement that spontaneous mutation IS in many respects an incident of the cell's normal metabolism. Chapter 44 BACTERIAL GENETICS: SEXUAL REPRODUCTION lecturer-J. LEDERBERG PRE- LEC TURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 22, pp. 377-385, 392393. Sinnott, Dunn, and Dobzhansky: Chap. 23, pp. 315-318. Snyder and David: Chap. 26, pp. 407409. Winchester: Chap. 23, pp. 326-327. b. Additional references Jacob, F., and Wollman, E. L. 1958. Genetic and physical determination of , chromosomal segments in Escherichia coli. Sympos. Soc. exper. BioI., 12: 75-92. Lederberg, J. 1947. Gene recombination and linked segregations in Escherichia coli. Genetics, 32: 505-525. Reprinted in "Papers in microbial genetics", selected by J. Lederberg. 1951. Madison: University of Wisconsin Press. Lederberg, J. 1956. Conjugal pairing in Escherichia coli. J. Bact., 71: 497498. Lederberg, J. 1959. Bacterial reproduction. Harvey Lect., 53: 69-82. Wollman, E. L., and Jacob, F. 1956. Sexuality in bacteria. Scient. Amer., 195: 109-118. LECTURE NOTES A. Genetic recombination in bacteria 1. Genetic analysis based only upon mutations in' asexually reproducing lines is severely limited (see also Chap. 6). 2. The infrequency of sexual processes shows their detection under the microscope is ordinarily improbable. B. 3. The search for sexuality is greatly expedited by the selective isolation of specific genotypes, using mutants for different nutritional defects. a. The wild-type Escherichia coli strain K-12 is nutritionally sufficient, i. e., is a prototroph (~ T+) (see central part of Fig. 44-1). b. Two nutritionally-dependent mutants, or auxotrophs, were obtained -- one requiring the amino acid methionine (M- T+), and the other requiring the amino acid threonine ~ D. c. Neither auxotroph can produce colonies when plated separately on a basal, minimal culture medium containing neither amino acid. d. If the two mutant strains are mixed and genetic recombination produces M+ T+ prototrophs, these will be the only cells forming colonies when plated on agar containing the minimal medium. e. Such evidence for genetic recombination was first obtained by Lederberg and Tatum (1946). 4. The test for prototrophs derived from different auxotrophs is very efficient for detecting genetic recombination. The fertilization process 1. Early experiments used mating type F+, and gave a very low frequency of recombination. 2. Later, other strains were found giving very !!_igh gequencies of .!:_ecombination, hence called Hfr strains, which were useful in learning the details of fertilization. 3. As demonstrated, a r:J (Hfr) cell is seen under the microscope to form a conjugation bridge with a 9 (F-) cell once these make a random contact (see also left side of Fig. 44-1). 251 Figure 44-1 4. When Wollman and Jacob used a Waring blender to separate the cells at different times after mating started, they found a. bacteria so separated were viable. b. there was a progressive linear transfer of genetic markers from d to <il. 1) This transfer required 100 minutes for completion. (The total DNA length transferred is about 221-1, or 5 to 10 times the bacteria's length. ) 2) The DNA to be transferred must have unwound. 3) One end of the DNA string was preferentially transferred first. c. Our most detailed map of bacterial genes was made this way. 5. The rate of transfer is 1, 000 nucleotide units, or nrits, per second (6 x 10 6 n'its in 100 minutes). 6, The genetic unit (cistron) specifying a typical protein is about 2, 000 n'its. a. Levinthal and Garen studied a variety of mutants defective for alkaline phosphatase, using procedures of this sort. b. They found it required about 2, 000 n'its to code this enzyme, which contains about 400 amino acids. c. Such evidence supports a 4: 1 (n'H: amino acid) coding ratio (see Chap. 41). 7. The injected d DNA synapses with that in the C. <jl. 8. Recombination between the d and 9 DNA oc- curs, producing recombinant strands with markers from both parents. The mechan252 D. ism for this is not known, and could result from a. breakage and cross-unions between parental strands. b. a copy-choice mechanism, in which the daughter DNA alternates in using maternal and paternal DNA as a template. Markers for genetic recombination 1. usually 6 to 12 markers are studied in a single cross. 2. The top center plate in Fig. 43-1 shows, using two markers, how recombinants typically are detected (see also Chap. 43 F2a). a. A loop of streptomycin was brushed vertically on the agar, then four clones were streaked horizontally across this region. b. Suppose one parental type (top colony -in Fig. 43-1 top center plate) was lactosenegative (light colored) and streptomycin-sensitive (interrupted streak). c. Suppose the otller parental type (third colony down on this plate) was lactosepositive (dark colored) streptomycinresistant (uninterrupted streak). d. The other two colonies would be recombinants -- lactose-positive streptomycin-sensitive (bottom one), and lactosenegative streptomycin-resistant (colony next to top). 3. Traits of mutants may involve their nutrition, bacteriophages, specific antigens, motility, etc. Basis for sexual differentiation 1. Male sexuality is an infectious phenomenon a. One F+ (cf) cell can rapidly convert all F(?) cells in a culture to F+. b. The new F+ cells transmit this trait to their progeny. c. The infective factor must multiply at least twice as fast as the typical cell. d. This factor, called F, is extra-chromosomal and not isolable as a cell-free virus particle (see right side of Fig. 441, where the chromosome is diagrammed as a singl e line). 2. Properties of the F particle a, It is transferred from d to ? in a transient mating. b. Such matings are more unstable but more frequent than matings which involve chromosome transfer. c. The dye, acridine orange, inhibits reproduction of F but has nQ appa~f ect on chromosomal genes. Treatment with this dye results in converting F+ to F- cells. Mating types 1. F+ and F- types have been described already. 2. The F particle a. must modify the d cell wall so as to recognize and react with a '? cell it contacts, b. must form a bridge between d and ?, c. probably confers motility to the F+ cf chromosome by attaching to it at least temporarily (Fig. 44-2, small circle attached to chromosome in F+ cell). 3. Hfr mating type carries F on the chromosome end transmitted last in fertilization (Fig, 44-2). - E. F. c. Many Hir strains are unstable and revert to F+ type, simultaneously losing their high fertility and gaining the property of infectious F particles. d. The d character of Hir is not inhibited by '\ acridine orange. Episomal cycle refers to the facultative participation of a factor as an extra-chromosomal or as a chromosomal element. 1. F is an episome a. In F+ cells, F is normally extra-chromosomal. It is suggested that the low fertility of F+ is due to a transient connection of F with the chromosome. b. In Hfr cells F has a chromosomal locus. 2. The episome lambda (see also Chaps. 45 and 46) a. E. coli §train K::J2 normally carries a s~mbiotic bacteriophage, lambda, to which it is relatively insensitive. b. A sensitive mutant, which had lost lambda, was found. This strain is used to test for lambda. c. In the normal strain, lambda is temperate since it does not cause conspicuous lysis. d. Since the normal insensitive strain can potentially produce infective lambda, it is said to be lysogenic. e. Crosses between lysogenic and sensitive bacteria showed that lambda, as a prophage, has a definite locus on the chromosome, .l£._ closely linked to...Gal.- a locus responsible for galactose fermentation. f. This close linkage was observed in the haploid segregants from diploid exceptional clones of genotype Gal + Lp+ / Gal- Lps. (That is, ability to use galactose, Gal+, was closely linked to capacity to produce bacteriophage, Lp+, as was Gal- to~. g . Occasionally the Lp-+- chromosomal factor enters a cycle of vegetative multiplication in the cytoplasm, after which it matures as intact virus. This occurs quite frequently when Lp+ cells are treated with ultraviolet light (Lwoff). h. Once Lp+ forms mature virus the cell lyses and free virus is liberated. i. The free lambda can enter other ceBs either to multiply as a parasitiC virus, or to reenter the chromosome to again produce the lysogenic state . J U () Figure 44-2 a. Hfr strains do not transmit F particles contagiously. b. All Hfr strains are derived from F+ cells (which may have come from Fcells) . . 253 3. Other particles in E. coli also act as episomes. 4. It is possible that cytoplasmic factors in other organisms are due to genic factors which can enter the cytoplasm. POST-LEe TURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 254 QUESTIONS FOR DISCUSSION b. Mutant A x C never gave prototrophs when the pairs were separated before 1,000 seconds, but gave increasing percentages of prototrophs up to 1,002 seconds, after which this maximum frequency remained unchanged. 44. 1. What kinds of genetic information obtained from sexually reproducing organisms cannot be obtained from organisms reproducing only asexually? 44. 2. Why is it futile to search cytologically for mating couples in an ordinary culture of E. coli strain K -12 in which F+ fertilizes F':once per million bacteria? 44. 3. How do you suppose it was possible to prove that the results from mixing auxotrophs were not the consequence of mutation but rather of genetic recombination? 44. 4. Suppose an Hfr clone has the normal genes r- c. Mutant C x D gave no prototrophs before 1,001 seconds, but reached its maximum frequency beginning 1, 002 seconds after mating. 44. 7. Draw a suitably labeled diagram showing a daughter DNA strand and the maternal and paternal DNA strands from which it was produced by a copy-choice mechanism. ROT S V while an F- clone is mutant for ----- these markers. The clones are mixed at lOa. m. At the times specIfied below the "happy couples" were separated, and analysis showed the normal genes indicated there had been transported into the F- cells. 10:02 10:05 10:15 ]10:25 10:35 a. a. a. a. a. m. m. m. m. m. ------ none T 0 T SOT V -_-_ Y §. Q I Make a genetic map for the marker genes which is as complete as the data allow. 44. 5. If the DNA in a bacterial nucleus which is transferred is 22J.,l10ng, approximately how much of thIS is transferred 10 minutes after fertilization begins? , Approximately how long does it take for an average gene to be transferred in a bacterIal fertilization? 44. 6. Several independently-arising, auxotro-' phic, mutants for the same trait were obtamed and crossed to each other in pairs. What conclusions could you reach from the results following, from whlCh mutational events have been excluded, on the supposition that bacteria can be mated and can be separated instantaneously? a. Mutant A x B never gave prototrophs, even when fertilization was permitted to be completed. 44. 8. List as many differences as you can between F+ and F- cells. 44. 9. List as many differences as you can between F+ and Hfr cells. 44.10. Specify the kinds of particulate genetic matter whieh may be transferred from one bacterium to another. 44.11. In bacteria, is sex type determined genetically? chromosomally? Explaip. 44.12. Describe how one might obtain experimental evidence that F IS transferred extra-chromosomally from F+ to F- cells. 1-' 44.13. What would you predict about the chemical composition of the F particle? Justify your answer. 44.14. DeSCrIbe how you would proceed to obtain a Gal+ Lp+ stram from a Gal- Lp+ strain. How would you test to show the desired genotype was obtained? 44.15. What are the sImilaritIes and differences between F and lambda? 44.16. What experimental evidence can you give in support of the schematic diagram at the left of Fig. 44-1? 44.17. WhIch cytoplasmic factors discussed in Chapters 34 and 35 might be episomes? 255 Chapter 45 BACTERIAL GENETICS: GENETIC TRANSDUCTION lecturer-J. LEDERBERG PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 22, pp. 392-393, 396397, 398-399. Sinnott, Dunn, and Dobzhansky: Chap. 23, pp. 318-319; Chap. 28, pp. 384386. Snyder and David: Chap. 26, pp. 410413. Winchester: Chap. 23, pp. 324-326, 322-324. b. Additional references Lederberg, E. M., and Lederberg, J. 1953. Genetic studies of lysogenicity in Escherichia coli. Genetics, 38: 51-64. Lederberg, J. 1956. Genetlc transduction. Amer. Scient., 44: 264-280. Lederberg, J. 1959. A view of genetics. Stanford Med. Bull., 17: 120-132. This Nobel Prize lecture is published also in "Science" . Lederberg, J. 1959. Bacterial reproduction. Harvey Lect., 53: 69-82. Zinder, N. D. 1958. "Transduction" in bacteria. Scient. Amer., 199: 38-43. LECTURE NOTES A. Chap. 44 described a sexual mechanism in Escherichia coli involving the mediation of F which, like fertilization in higher forms, 1. involved intact cells as participants, and 2. had an entire genome as the unit of transfer. Relatives of E. coli and some filamentous bacteria also show sexual processes. B. Genetic transductions refer to processes of fragmentary genetic exchange. One type, discussed at length here, involves bacteriophage. C. Salmonella typhimurium 256 D. 1. causes mouse typhoid and is an agent in human food poisoning. 2. Like its close relative E. coli, it is cultured on simple medium. 3. Zinder and Lederberg found genetic recombination, of erratic pattern, between certaiIl strains. a. Mixing an M+ T- strrun with an M- T+ strain produced a filterable, heat-resistant agent with M+ activity. b. This agent made prototrophs when added to an M- indicator strain. c. The agent, smaller th[J.n a bacterium, could not be isolated from a pure M+ Tculture although this must have donated the M+ factor when mixed with the M- T+ strain. d. Yet a drop of filtrate from the IDlxed cul~ ture added to fresh W~arrying cells evoked more filterable W factor. e. Two activitles were involved -- evoking and transferring W. Bacteriophage as the evoking agent 1. The phage P22 is lysogenic in the M- T+ ----=--~ strain. 2. A stock suspension of P22 grown on the M+ T- strain yields particles with M+ achvity. 3. The indicator strain M- was derived,from the M- T+ strain and is, therefore, lysogenic for P22. Bacteriophage as the transferring agent 1. P22 acquires fragments of genetic material from the host on which it is grown. 2. Evidences for the association of bacteriophage with the genetic transferring capacity of the phage suspension -- the transductionaJ capacity -- include: a. both show the same temperature inactivation pattern; b. both have the same susceptibility to an -- E. F. antiserum that blocks phage attachment to cells; c. both attach to susceptible cells sImultaneously; d. size and mass of both are the same, as determined by filtration and sedimentation tests. 3. Jt is strongly suspected that phage that carries part of a bacterial genome is defective for virus genome -- that there is a replacement of the latter by the former. Example of transduction experiment in Salmonella 1. Phage 22 is grown on bacteria genetically M+ T+ X+y- Z-. --_2. Part of the crop of phage harvested is then tested on suitable indIcator strains (M-, T-, X-, Y-, B one at a time. 3. This is done to show that the phage filtrate has the same range of activity as the bacteria on whlCh it was grown. 4. Another part of the crop is now grown on a new bacterial strain, for example, M+ T- x+y+ Z-. -- a. The new crop of phage harvested has now lost T+ and gained Y+. b. The phage is passive with respect to the content of the genes it transduces. 5. To harvest the phage, the liquid culture is centrifuged and the supernate heated at 600 C for 20 to 30 minutes (to kill any re.maining bacteria). 6. To detect transduction of M+, phage is grown on M+ bacteria, harvested as described, mixed with M- bacteria and plated on agar containing methionine-deficient medium. a. Phage attaches and injects its DNA into the M- bacterium (see Chap. 46). b. If the bacterium survives this attack and if it acquires the M+ fragment from the phage a clone will be formed. c. This transduction can by symbolized: G. Genetic scope of transduced material 1. Usually a single bacterial marker is transduced. a. Wr+ x+ b. The latter bacteria are grown on different media -- one which selects for M+, another for T+, and a third for X+.c. When the W clones are further typed H. they are still T- X-. Similarly, T+ clones are still M- X-, and x+ clones are still M- T-. 2. In contrast, in sexual recombination, large blocks of genes are transmitted together from Hfr to F- cells. 3. Several markers may be transduced together in linked transduction or co-transduction. a. Demerec has shown, using transduction, that the genes for the biosynthetic sequence: anthranilic acid to indol to tryptophan (see Chap. 40), are closely linked to each other. b. Different mutants involved in defecting a particular enzyme are even more closely linked. c. Histidine biosynthesis is also controlled by a cluster of genes. d. This correspondence between biosynthetic and genetic association, though it does not apply to some higher organisms (e. g. Neurospora) may be adaptive in providing a mechanism for turning on or off a whole series of enzymes. 4. Any locus in Salmonella is transducable by P22. Co-transduction in E. coli 1. From bacterial crosses, ~ and Gal are known to be closely linked (Chap. 44). 2. When lambda is harvested from Gal+ Lp+ prophage cells, with the aid of ultraviolet light (Chap. 44), it has Gal+ transducing activity. 3. Gal is the only marker known to be transduced by lambda. The typical rate is one transduction per 100,000 phage particles. 4. The transduced Gal- strain is a heterogenote (partial heterozygote), having one complete and unchanged chromosome of the host (Gal- ~ and a fragment (perhaps attached to the chromosome), carried over with lambda, containing Gal+ Lp+. 5. Heterogenotes can undergo reduction during which the Gal+ may exchange places with Gal-. 6. Nearly every lambda obtained from a heterogenote contains Gal. This was demonstrated by obtaining lambda from a Gal+ heterogenote and cross-brushing it over a Gal- clone streaked on galactose-deficient nutrient agar. The Gal+-carrying lambda particles can be counted by the number of Gal+ colonies that grow at the zone of intersection. 257 r. 7. About one hundredth of the bacterial DNA is transduced at one time. Recombination mechanisms in bacteria may 1. involve whole nuclei (sexuality). a. Heterokaryosis, in certain filamentous fungi, involves concurrence of nuclei in common cytoplasm and leads to b. heterozygosis, as in E. coli K-12. 2. involve fragments of genomes (transduction) via a. bacteriophages, like P22, b. episomes (see Chap. 44). like F and lambda, and c. purified DNA, in bacterial transformation (see Chap. 40). POST- LEC TURE ASSIGNMENT 1. Read the notes immediately after the lec- ture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional assignment. 258 QUESTIONS FOR DISCUSSION 45. 1. Is the infective F particle (Chap. 44) appropriately placed under the heading of genetic transduction? Explam. 45. 2. What evidence can you cite that the genetic recombination observed in Salmonella is not accomplished by a sexual process? 45. 3. Describe how you would perform an experiment to transduce the X- Z- 10Cl present in a given strain of Salmonella. 45. 4. What evidence would you accept as proof that phage P22 is passive with respect to the genes it transduces? 45. 5. How would the results in G of the lecture notes be changed if co-transduction occurred a. between M and T only? b. between T and X only? 45. 6. Should lambda be called a virus or a segment of a bacterial chromosome? Explain. 45. 7. Compare P22 with lambda, as to similarities and differences. 45. 8. How is it possible to estimate the proportion of the total bacterial chromosome which can be carried in a transducing phage? 45. 9. Of all the types of transduction, what is unique to bacterial transformation? 45.10. Compare the genetic behavior of E. coli and S. typhlmurium. 45. 11. Learn what colic ins are from the suggested readings. Have they any bearing upon genetic recombination in bacteria? Explain. 45. 12. Do you suppose transduction occurs also in higher organisms? Explain. What are the possible advantages and disadvantages of transduction as compared with sexuality? 45.13. 45.14. Do bacteria obey Mendel's laws of inhentance? Justify your answer. Chapter 46 VIRUS GENETICS: BACTERIOPHAGE Lecturer-l. LEDERBERG PRE- LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 22, pp. 386-393. Colin: Chap. 12, pp. 216-220. Sinnott, Dunn, and Dobzhansky: Chap. 16, pp. 225-226; Chap. 23, pp. 317318; Chap. 27, p. 373. Snyder and David: Chap. 27, pp. 412413. Srb and Owen: Chap. 14, pp. 282-283. Winchest~r: Chap. 23, pp. 327-328. b. Additional references Anfinsen, C. B. 1959. The molecular basis of evolution. 230 pp. New York: John Wiley & Sons, Inc. Brenner, S. 1959. Physiological aspects of bacteriophage genetics. Adv. Virus Res., 6: 138-158. Burnet, F. M., and Stanley, W. M. Editors, 1959. The viruses. Vol. 1, General virology. 609 pp. New York: Academic Press. Luria, S. E. 1953. General virology. 427 pp. New York: John Wiley & Sons, Inc. LECTURE NOTES A. Nucleotide pairs (n'its) in different organisms Mouse 5 billion per cell Drosophila 80 million per cell Bacterium 6 million per cell If 2, 000 n'its specify an average protem (see Chap. 44), bacteria could have about 3,000 different enzymes specified by its DNA content. It is surprising they could be free-living with so Iowa number. B. Bacteriophages like T2 or lambda contain about 100,000 n'its, which, on this view, per260 C. D. E. mi t the manufacture Qf only about 50 different proteins. 1. It is not surprising, then, these are not free-living organisms. 2. The host cell provides most ~f the accessory metabolIsm for synthesizing new virus. 3. The virus, in adding elements unique to its structure, a. competes with the metabolic systems of its host, and b. produces the protective structures enclosing it when outside its host. Smaller viruses 1. These include animal viruses causing encephalitIc diseases and the exceptional phage 0'X174. 2. These have only about 5,000 n'its, which could encompass only one or a few genes. 3. Their nucleic acid is one-stranded. a. Most small viruses of plants and animals contain RNA. b. 0'X174 is exceptional in containing onestranded (otherwise typical) DNA. 4. It is remarkable that by probably producing but a few distinctive proteins these viruses can divert the host's metabolism so that virus nucleic acid and protein are made rather than the host's normal metabolic products. Bacteriophage structure 1. These bacteria-attacking viruses are. 1 to .2j.llong, about one-tenth the bacterial diameter. 2. They have a somewhat-crystalline protein coat within which is packed double-stranded DNA (about 34).l10ng when extended). 3. The protein coat has a tail, covered by a spiral protein, with tilll tip fibers. Lytic cycle of a bacteriophage 1. Phage attaches to the bacterium tail first. 2. The DNA enters the bacterium leaving a shell of protein outside. a. Hershey and Chase labeled the DNA by P32 and the protein by 835. b. The empty shells, containing all or almost all the phage protein, can be sheared off the bacteria without changing the fate of the infected cells. c. The bacterium was innoculated, therefore, not with whole phage but with phage's genetic element. 3. An eclipse period follows during which no infective phage can be demonstrated in the recently infected bacterium. a. During the first several minutes there is replication of the phage DNA. b. This vegetati'Ve phage forms a pool of DNA units. 4. From time to time this pool is sampled, in that a fractlOn of it undergoes condensation and is surrounded by a new skin (head and tail) formed from a cycle of protein synthesis and organization; as a consequence infective phage is produced. 5. Phage-infected bacteria produce endolysins which, 20-40 minutes after infection, lyse the cell wall and liberate infective phage in the medium. 6. The free infective phage can attach and penetrate a sensitive bacterium, then a. repeat its lytic cycle, b. or, as in the case of lambda (Chaps. 44 and 45), establish a relationship with the bacterial chromosome, so that the bacterium is lysogenic and has the property of having some of its descendents produce some lytic phage. F. Methods for assa;ying phage I 1. One method is to determine the time required for complete lysis of a liquid culture of senSItive bacteria. 2. If a few phage are added on top of an agar medium recently seeded heavily with sensitive bacteria, a. the bacterial clones will form a continuous and somewhat opaque lawn. b. each phage particle that enters a bacterium will lyse it and release up to several hundred daughter particles which will attack bacteria near the original burst. This cycle will result in a progressively increasing zone of lysis wluch appears as a clearing or plaque in the bacterial lawn. c. each plaque will be a phage colony de- G. H. rived from one ancestral particle. d. Genetically different phages may produce different types of plaques. What is unique to phage DNA? 1. It must be able to stop bacteria from malcing their own protein and DNA. 2. Even-numbered phages, T2, T4, and T6, have hydroxymethyl-cytosine in their DNA instead of cytosine. 3. But in T1 or lambda there is no such peculiarity in base composition, a. yet T1 is virulent (causes lysis), b. and lambda can act either temperately (in a lysogenic bacterium) or lytic ally (upon ultraviolet induction). 4. The complete answer is still unknown. Genetic recombination in phage 1. Given one phage strain mutant for both host range, !!_, and plaque type, .!:_, and one strain that is wild-type. 2. Sensitive bacteria are exposed to a mixture of such large numbers of the two mutants that some will have been multiply-infected. 3. Some doubly-infected cells wIll carry both phage types. 4. The daughter phage from such mixedly-infected cells may be of parental (!!..!:_ and h+ r+) and recombinational ~.!:_ and!!_ types. Fine structure analysis of T4 1. Since T4 contains about 100,000 n'its, there could be that number of mutational sites. 2. One segment, .!:_, is about 1% of the total map length, or about 1,000 n'its long. 3. The wild-type, r+, grows on both strains B and K-12 of E. coli, while the .!:_ mutants grow on B but not K-12. 4. Benzer made mixed infectIOns with different r mutants, then easily detected the rare "(i in 106 or 108 ) recombinant particles capable of growing on K -12. 5. Of 1,500 spontaneous .!:_ mutants Benzer tested, 100 were different. Thus there are at least 100 sites for mutation in the .!:_ segment. 6. Certain mutants must have recurred frequently, indicating certain sites in .!:_ were "hot spots" of spontaneous mutation. 7. The frequency of least recombination observed between two mutants was . 0001. Since there are about 10 exchanges (morgans) per phage per generation (lytic cycle) over its whole linkage map, the total number of crossover sites in phage should be D I. 261 10/.0001 or 100,000. 8. Thus, crossover site number equals n'it number; the genetic unit of recombination equals the smallest meaningful chemical unit -- the interval between two successive n'its. 9. Benzer also found phenotyPic interaction between different!. mutants growing in the same bacterium. a. By Itself neither!. mutant can lyse K-12. b. The vegetative phage of both together, in the same cell, can cause lysis. c. Since the phage liberated were usually of parental type, this phenotypic cooperation did not require genetic recombination. d. The!. region can be dIvided into A and B sections or cistrons. Only a mixed infection including one mutant from A and one from B results in phenotypic cooperation. e. It is concluded, therefore, !. governs the production of two separate protein elements. f. Mutants which behave as deletions for different regions in !. have been obtained. It is possible to arrange the spontaneous point mutations In only one order which will give a compatible superposition of the various deletion types. g. Even in the 'J.tmost details of its fine structure the genetlc material of phage T4 is organized in a linear fashion. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making addItIons to them as desired. 2. Review the reading assignment. 3. Be able to dISCUSS or define orally or in writing the items underlined in the lecture notes. 4. Complete any addItional assignment. 262 QUESTIONS FOR DISCUSSION , 46. 1. What functions are attributed to the spiral protein and tip fibers of the tail of a bacteriophage? 46. 2. 46. 14. What hypotheses can you offer to explain the occurrence of ''hot spots" for spontancous mutation in the .!:_ chromosome segment of E. coli? Distinguish a virus from a bacteriophage. 46. 3. Defend the statement: "Viruses are not genes. " 46. 4. What do you consIder the most remarkable and important feature of phage jl{X174? Why? 46. 5. What evidence proves or suggests that a transducing virus cannot a. carry a whole bacterial chromosome? b. simultaneously carry a viral genome in addition? 46. 6. Design an experiment which would demonstrate the "eclipse period" of phage. 46. 7. Discuss the phenotypic interactions between viral and bacterial genes. 46.15. Of what significance IS the fact that crossover percentages lower than. 01% were not found between!. mutants, even though lower values would have been detected readily? 46.16. What bearing has Benzer's work to our understanding of the gene as a umt? 46.17. Briefly describe the eVIdence that phage DNA is organized linearly. 46.18. What attributes are common to all viruses? 46.19. Which do you suppose came first in evolution, bacterial-like or viral-like organisms? Explain. 46.20. Compare viruses and antibiotics. 46. 8. Are phages harmful to a bacterial species as a whole? Explain. 46. 9. Discuss the statement: "The holes in a bacterial lawn are contagious. " 46.10. Why is it so important to learn the chemical nature of the proteins specified by phage? 46.11. From what is known about the chemical nature of the hereditary material in different organisms, discuss the evolutionary changes which the gene itself may have undergone. 46.12. What preliminary experiments had to be performed and what results obtained, do you suppose, before Benzer accepted an r+ genotype as the result of phage recombination? 46.13. Using different.!:_ mutants of phage T4, indicate the requirements and procedures necessary, in each case, to obtain a. genetic recombination without phenotyplC cooperation. b. both phenotypic cooperation and genetiC recombination. c. spontaneous reverse mutations to r+. 263 Chapter 47 VIRUS GENETICS: PLANT AND ANIMAL VIRUSES Lecturer-J. LEDERBERG PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings' a. General genetics textbooks Smnott, Dunn, and Dobzhansky: Chap. 27, pp. 377-378. Srb and Owen: Chap. 19, pp. 400-402. b. Addltional references Burnet, F. M., and Stanley, W. M. Editors, 1959. The viruses. Vol. 1, General virology. 609 pp. Vol. 2, Plant and bacterial viruses. 408 pp. Vol. 3, Animal viruses. 428 pp. New York: Academic Press. Fraenkel-Conrat, H., and Williams, R. C. 1955. Reconstitution of active tobacco mosaic virus from its inactive protein and nucleic acid components. Proc. nat. Acad. Sci., U. S., 41: 690698. Reprinted in: "Classic papers in genetics", J. A. Peters, Ed. 1959. Englewood Cliffs, N. J. : Prentice-Hall, Inc. Luria, S. E. 1953. General vlrology. 427 pp. New York: John Wiley & Sons, Inc. LECTURE NOTES A. Plant and animal viruses are 1. of great economic and medical significance. 2. difficult to study for technical reasons. B. Determination of an infective umt 1. This is a chief difficulty. 2. Plant virus a. To tItrate a virus attacking leaves, a sample IS rubbed on the leaf surface. b. Only a small fraction of the virus particles find and penetrate susceptible cells and give a demonstrable lesion. 3. Poliomyelitis virus 264 c. a. When propagated on intact host animals, quantitation of particles is expensive and time-consuming. b. Samples to be titrated may be plated onto agar layers seeded with ~usceptible tissue culture cells. Clearing plaques are produced as by bacteriophage. This kind of technique is very useful. 4. Unfortunately, many viruses (like influenza virus) do not produce sufficient cytopathic effect to produce detectable plaques on such agar plates. For these viruses, the techniques of lImit-dilution must still be used. 5. Influenza VIrus a. This has been adapted to grow on cells lining the fluid cavities of the chick embryo. b. A sample of virus to be tested is sufficiently diluted, and then aliquots innocuIated into a series of eggs. c. After 48 hours or so, the eggs are harvested to determme the fraction which contained a virus particle. d. If near-limit dilutions are used (so the probabIlity IS low that an aliquot contains a virus particle), one can estimate the virus content of the entire sample. e. At near-limit dilutions, the virus partlcles harvested from an egg are probably from one clone. Life cycle of animal viruses 1. Influenza virus cycle IS used as an example, whose general features may apply also to other ammal and, to some extent, plant viruses. 2. The mammalian host cell (FIg. 47-1) has a. a flexible shape, b. an ambiguous margin, and c. an outermost mucoid coat which acts as D. a substrate for an enzyme located on the virus surface. This coat constitutes, therefore, a virus receptor. 3. The influenza virus has a. an inner core of RNA genetic material, llnd b. an outer protein coat containing the mucin-reacting enzyme. 4. The virus cannot attach if the mucoid coat is stripped by specific enzyme or periodate treatment. 5. After attachment, the particle enters the cell, perhaps by being engulfed via the cell's normal pseudopodial activity. 6. Once inside the cell, the particle enters an eclipse phase (see Chap. 46) and multiplIes vegetatIvely, at which time the cell's RNA content increases. 7. After some hours intact particles are gradually liberated. a. Evidence indicates that the influenza viral coat is added during emergence from the host. b. This coat contains some material made (by the host -- therefore host cell-specific) before infection and some made after infection (by host and virus together). Consequences of mixed infections 1. Burnet and others used MEL and WSJ:: strains of influenza virus which differ in their markers. a. Serologically, WSE is W and MEL is ~. b. WSE is inactivated by ovomucin ~ while MEL is not (9. c. WSE is pathogenic when placed on the egg's chorio-allantoic membrane (~ while MEL is not @. 2. Egg membranes are multiply-infected with mixtures of the two strains. 3. Phenotypic mixing (Burnet) a. From such mixed infections, daughter particles are neutralized almost perfectly efficiently by antiserum for either strain. b. The coats show this phenotypiC mixing even though the genomes witl:p.n them are either WSE or MEL. c. This effect, then, is not due to genetic recombination. 4. Heterozygosls (Burnet) a. The virus from mixed infections is harvested and clones obtained via limit-dilution (see B5). b. Some clones contain more than one genetic type. Figure 47-1 E. F. c. This heterozygosis may be explained either by adhesion of two, whole, genetically-different particles which act as a unit on hmit-dilution, or by one particle containing two different genomes. d. Since there has been no exchange of parts of the genetic material in forming a stable clone, this also is not a complete phenomenon of genetic recombination. 5. Genetic recombination a. Influenza virus Mixed infections give particles which yield pure clones that are stable recombinants ~ ~ or W .Q). b. Vaccinia virus (Fenner) Similar evidence was obtained, for this more complex virus, in experiments involving markers for hemagglutinins, heat resistance, virulence, and pock character. Poliomyelitis viruses 1. No evidence has been obtained so far for genetic recombination in these viruses. 2. If fa"und, recombination would greatly accelerate the deletion of the specific adaptation to the human host which these viruses possess. 3. Present work is limited to mutational studies. Chemical separation of viral components 1. Aqueous phenol solution destroys the protein but leaves the nucleic acid intact (Schramm, using animal viruses). 2. Tobacco mosaic virus protein is removed, 265 almost unchanged, by moderate alkalinity. G . .;I InfectlOn by nucleic acid 1. RNA, minus demonstrable protein, has been shown to be capable of infecting tobacco and mammalIan cells. 2. After vegetative multiplication, the mature particles formed have the protein coats that would be specified by this RNA when introduced by VIrUS. 3. Thus virus protein plays no part in replIcation either of the genetic material or of itself. 4. As compared to nucleic acid in virus, purified nucleIc acids have infectivities up to several percent. 5. Isolated RNA is more susceptible to rIbonuclease, temperature, and pH, and less susceptible to detergents, than is intact virus. 6. Recent evidence suggests isolated DNA from bacteriophage IS infective. H. Reconstitution experIments with tobacco mosaic virus (TMV) 1. TMV is a long linear particle; the outside is protein, built up of stacks of monomeric blocks; the inside is spiralled RNA. 2. Fraenkel-Conrat has reconstituted essentially the origmal virus by mixing, under certain conditlOns, its separate protein and RNA. 3. Using two strains of TMV, the standard (TMV) and Holmes ribgrass (HR), he was able to construct virus with TMV RNA and HR protem. a. The behavior of these particles was somewhat like that of animal viruses showing phenotypic mixmg. b. However, particles have such low infectivity in plants that mixed infections, WhICh could produce phenotypic mixing, do not occur. c. Such a reconstituted particle is mactivated not by anti-TMV serum but by anti - HR serum. The progeny of such a particle are typical TMV. d. Corresponding results were obtained from reconstituted particles havmg HR RNA and TMV coats. 1. Biologically-actIve nucleic acids have not yet been synthesized in the laboratory. 1. The success of systems USing DNA as a prImer for synthesizing more DNA (Kornberg; see Chap. 41) IS a gIant step m this direction. 2. The genetic behavlOr of RNA motIvates attempts to replIcate it III vitro. 266 J. -.!..!It should be stressed that DNA and RNA are not unique materials. They have the same relationship to the information contained in them as carbon black does to the words in a dictionary." POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Revie:w the reading assignment. 3. Be able to discuss or define orally or in writmg the items underlined in the lecture notes. 4. Complete any additional assignment. QUESTIONS FOR DISCUSSION 47. 1. What are the comparative advantages and disadvantages of influenza and poliomyelitis viruses as material for genetic investIgation? cludmg statement, quoted in J in the lecture notes. 47. 2. Why is the WSE strain of tobacco mosaic virus called an egg-adapted strain? 47. 3. Is the technique of limit-dilution used m assaying tobacco mosaic virus? Explain. 47. 4. In what respects do the host cells of bacterIOphage, on the one hand, and of plant and anImal viruses, on the other hand, dIffer from each other wIth respect to virus? 47. 5. What evidence was presented that genetically recombinant clones of influenza virus are stable and pure? 47. 6. How can the dIfferent consequences of mixed infections with influenza VIruS be dlstmguished from each other? 47. 7. Specifically what would you do experImentally, in order to benefIt mankind, if the phenomenon of genetic recombination was dIscovered III the,poliomyelitis viruses? 47. 8. What evidence does Fraenkel-Conrat's experiments furnish that- RNf\ is the sole or primary determinant of the coat protein of the tobacco mosaic virus? 47. 9. Cite evidences for the exact replication of RNA. 47.10. What genetic attributes does RNA share with DNA? 47. 11. In what respects have genetic investIgations using plant and arumal viruses been more fruitful, to date, than those using bacteriophages ? 47.12. How would you proceed to determine whether a nucleic acid synthesized in vitro was biologically active? 47. 13. In what respects has our understanding of the genetIcs of hIgher organisms been aided by the genetIc study of microorganisms? 47.14. Discuss the meaning of Lederberg's con- 267 Chapter 48 B,IOCHEMICAL ORIGIN OF TERRESTRIAL LIFE Lecturer-J. LEDERBERG PRE-LECTURE ASSIGNMENT 1. Quickly review notes for the previous lecture. 2. Suggested readings: a. General genetics textbooks Altenburg: Chap. 27. pp. 487-488. Srb and Owen: Chap. 1, pp. 1-7. b. Additional references Lederberg, J. 1959. A view of genetics. Stanford Med. Bull., 17: 120-132. This Nobel Prize lecture is published also in IIScience" . Lederberg, J., and Cowie. D. B. 1958. Moondust. SciencG, 127: 1473-1475. Miller, S. L., and Urey, H. C. 1959. Organic compound synthesis on the primitive earth. Science, 130: 245-251. Oparin, A. I. 1957. The origin of life on the earth. 3rd Ed. New York: Academic Press. Sinton, W. M. 1959'. Further evidence of vegetation on Mars. Science, 130: 1234-1237. LECTURE NOTES A. Terrestrial life and its origin 1. The universe is about 10 billion years old. 2. The earth is about five billion years old, and has a fossil record only for the past one billion years or so. 3. The absence of fossils before this time requires a synthetic attempt to reconstruct a. the features of the earth when it originated. b. the most likely features from which organisms developed. B. Common plan of present organisms 1. Comparative biochemistry of present higher plants and animals, bacteria, and many viruses shows that a. the same amino acids are found univer268 C. D. sally. b. DNA is the carrier of genetic information. 2. At a higher level of organization, . the chromosome and the remarkably uniform process of mitosis are found both in plant and animal cells. 3. It is concluded, then, life has had a common plan ever since plants and animals became separate nearly a billion years ago. 4 . Nucleic acid and protein are, perhaps, the most durable geochemical features of the parth. Minimum requirements for the first organism 1. It must be self-reproducing, and capable of mutation (heritable changes). 2. The mutants must be subjected to a natural selection that retains the fitter forms. 3. The free-living bacterium, too complex to have arisen spontaneously, must be the result of a previous evolution of complexity. 4. The replication of DNA in vitro (Kornberg) would be one of the simplest systems subject to evolution. However, complex nucleoside triphosphates and other accessories are required for this synthesis. 5. Accordingly, the primary living material may have been DNA synthesized by some mechanism even simpler than is presently known, or may have been even simpler than DNA. 6. Attempts should be made, aided by the polymer industry, to construct linear polymers which show some degree of self-replication. Organic synthesis prior to organisms 1. Not very much is learned about the evolutionary steps leading to the first organism from our present habitat, for this is, in large measure, the result of the metabolism of living forms. Figure 48-1 a. Living things have been responsible, for example, for large deposits of carbon in coal sediments and for the release of oxygen to the atmosphere consequent to photosynthesis . b. Organisms tend to destroy large accumulations of organic compounds. c. For these reasons the distribution of organic compounds might have been more complex before life started than it is now. 2. Organic compounds a. Such compounds of carbon are not made only by the metabolism of organisms. b. Wohler synthesized the organic compound, urea, by heating the inorganic compound, ammonium cyanate (Fig. 481, center). 3, Oparin, about 1928, proposed that certain organic syntheses could occur naturally in the absence of life (Fig. 48-1, right top portion). a. Carbon or methane reacting with metals produces carbides (CaC2)' which upon hydrolysis yields acetylene (C2H2), which when hydrolyzed yields acetaldehyde (CH3' CHO). b. Nitrogen can produce ammonia by reacting with hydrogen. c. Other reactions, between aldehydes and ammonia, would produce high molecular weight polymers of considerable interest for biological development. 4. :Miller and Urey subjected mixtures of E. gases, predicated as being in the primitive earth atmosphere, to ultraviolet light and spark discharges. In this way, they have produced large amounts and large varieties of amino acids. a. The synthesis of glycine, indicated in Fig. 48 -1 (right lower po rtion), is an example. b. The detailed chemical steps in these syntheses are unknown, but supposedly involve the production of highly-reactive free-radical intermediates. Comparative chemistry of earth and universe (Fig. 48-1, left) 1. Most of the universe is hydrogen and helium. 2. If all the other atoms are totaled, the elements of unique importance in organic compounds (0, N, C) a. comprise 80% of this total in the case of the universe, while b. the earth has, relatively, only traces of carbon. 3. Whereas the universe is a good place, the earth is a very poor place for starting an organic chemistry which would be of biological interest. 4. Yet, despite the unlikelihood, life did develop on earth! 5. Comets have been shown to contain CH, CN, CC, and CO radicals. Such radicals are common in organic compounds. 6. The primitive earth could have accumulated large amounts of different complex organic 269 F. 270 materials WhICh remained undegraded until the advent of organisms. u. As the first organisms used up these resources, there would be a selection in favor of mutants capable of synthesizing these organic materials from simpler or from inorgamc components. b. In this way organisms would acquire synthetic capabilities. Search for organic compounds and hfe on other planets 1. Mars a. Astronomers have reported variations in apparent color and texture of its surface. b. Using the Palomar 200-inch telescope, Sinton found infra-red spectroscopic evidence for the presence there of orgamc molecules of an asymmetric type. c. While there is, therefo~e, evidence for appreciable quantities of organic material on Mars, these mayor may not be of organismic origin. d. MiSSIOns to or near Mars will be necessary in order to determine definitely its organic contents, the presence of DNA, and the presence of life. 2. Accidental transplantation of terrestrial genotypes to other planets must be avoided. a. If a single bacterium, like Escherichia coh, were placod on suitable medium it would occupy a volume the size of the earth in about 48 hours. b. Such a premature transplantation would be disastrous for our study of 1) the indigenous forms of life, or, 2) in the absence of organisms, the preorgamsmal evolution of organic compounds. 3. Venus a. WhIle estimates of its temperature vary widely, some are compatible with the existence of life. b. Its surface IS unlmown, being hidden completely by an opaque highly-reflectmg cloud layer. c. It cannot be assumed biologlCal activity is impossIble there. 4. Moon a. Havlllg no atmosphere and probably no water, the presence there of earth-lIke life is out of the question. b. It has been suggested that the moon mIght act as a gravitational trap for fosSIl spores which may have drifted between planets. G. c. Although improbable, the very possibility of an interplanetary gene flow is too important to ignore in our plans to exploit and explore space. The genetics of the minutest organisms and the rephcative powers of DNA have cosmic importance. POST-LECTURE ASSIGNMENT 1. Read the notes immediately after the lecture or as soon thereafter as possible, making additions to them as desired. 2. Review the reading assignment. 3. Be able to discuss or define orally or in writing the items underlined in the lecture notes. 4. Complete any additional~assignment. QUESTIONS FOR DISCUSSION 48. 1. Discuss the geochemIcal evolution of the earth from Its origin to the time the first gene was formed. 48. 2. Discuss the gene as the basis of life. 48. 3. What properties would you predict for genes present on other planets? 48. 4. What evolutionary processes do you imagine took place on earth between the origin of the first gene and the occurrence of the first free-living organism? 48. 5. Do you suppose other planets have forms of life superior to ours? Explain your answer. 48. 6. What mformation might we obtain about life on other planets without leaving our own ? 48. 7. What genetic predictions would you make for marriages between earth humans and planetary ·"humans"? 48. 8. What specific suggestions did Lederberg make with regard to future research on the origin of life? 48. 9. Do you believe planetobiological research should be supported regardless of cost? Explain. 48.10. Would superhumans on some other planet be likely to beam radio signals specIfically to the earth? Why? I 48.11. Are our present human genotypes adapted for living on Mars? Explain. 48.12. What would you predict about an orgarusm whlCh drifted to the moon from another planet? 48.13. Is RNA likely to be a basic component of the first organism on our own or on other planets? Explain. 48. 14. In what respects do the earth's minutest organisms and DNA have cosmic importance? 271 EXAMINA liON VI UNDERLINE ALL THE ALTERNATIVES WHICH ARE CORRECT. 1. a. separate colonies, one should make the first copy on the same medium as the master plate. b. continuous colonies, the first copy can usually be used in place of the master plate. c. continuous colonies, it is necessary to label the positions the plates have taken on the velvet. d. separate clones, spontaneous mutations within a clone can be disregarded. e. one can transfer phage as well as bacteria. 2. The fertilization process in bacteria always a. produces recombination. b. involves two whole cells. c. involves two different mating types, of 272 Salmonella typhimurium a. undergoes sexual recdmbination as does its relative E. coli. b. undergoes transduction,via phage P22. c. is attacked by T4, which can carry more than one bacterIal marker at a time. d. has n lits, each of which is capable of being carried by a transducing phage. e. has some of its genes arranged in the same order as the chemical processes occur which these genes control. 6. "£1 becoming infected, a bacterium a. can change its mating type. b. can shift its manufacture of DNA and protein from one type to another. c. runs the risk of lysis because all the virulent virus usually enters the cell. d. may become a prototroph even if it was previously auxotrophic. e. may give rise to progeny which have lost a specific allele present before infection. Escherichia coli strain K-12 a. usually reproduces sexually, each organism bemg one of three mating types. b. produces one clone from a single individual. c. is attacked by more than one type of virus. d. is best isolated by the use of the replicaplating technique. e. was discovered by Lederberg and Tatum in 1946. 4. 5. DNA a. is sometimes one-stranded, like RNA must be. b. synthesized in vitro has many of the properties of the prime" DNA but is not biologically active. c. is probably self-replicating in many places in the universe at the present time. d. is itself a linear polymer. e. can normally occur uncombined with other substances. 3. which one must be F-. d. invdlves two of the nuclei in an E. coli cell. e. includes transfer of DNA, but excludes transfer of RNA. In replica plating 7. Smaller plant and animal viruses a. are always composed of RNA which is usually single-stranded. b. may require several hours after infection to produce daughter particles. c. which specify only a few proteins may not be so virulent for this reason. d. do not have the protein tail so characteristic of bacteriophage. e. are often difficult to assay because of their size. 8. Mixed infections involving different strains of a virus b. a. do not occur wIth tobacco mosaic virus although they do occur with influenza virus. b. can give rise to phenotypic mixing and heterozygosis, but not to true genetic recombinatIOn. c. occur with vaccima virus, leading to recombination of its genes. d. do not occur for viruses containing RNA. e. do not occur if the technique of limit-dilution is employed. 9. c. d. e. 12. a. is greater for free-living than for parasitic or symbiotic organisms. b. can give an estimate of the number of genes' therein, simply by dividing by 2 x 10 3 . 10. b. c. d. e. 11. 13. no appredable growth will occur on the plate if the clone was prototrophic and the medium minimal. its growth, everywhere but on the streptoI,llycin, shows it is streptomycIn-resistant. you can be sure no sexual processes are taking place to confuse the results. mutation to streptomycin-resistance would be indicated if a very small amount of growth occurred in the streptomycin-containing region, but much growth occurred in the drug-free region. uniform growth along the streak indicates the clone is streptomycIn-resistant, although It may be auxotrophIc. Genetic recombination In bacteria a. cannot be visualIzed as an asexual pro- Punfied nucleic acids can act genetically a. when introduced Into tobacco, mammalIan, or bacterial cells. b. when they are either RNA or DNA. c. when these have been obtained from organisms other than viruses. d. and when taken from virus can produce complete daughter virus particles. e. even after they are surrounded by a protein coat unlike their OrIginal one. If a clone is streaked across agar including a section containing sufficient streptomycin, u. The fluctuation test of Luria and DelbrUck a. showed that the medium used did not select mutants preferentially. b. gave a normal dIstribution for the number of mutants in dIfferent samples tested. c. was the fIrst demonstration with bacteria of the occurrence of mutations without regard to the specific medium upon whlCh the mutants are detected. d. showed that mutations can occur any tIme in clonal growth. e. gave results easily reproduced by using techniques of replica platmg. The number of n'its in a genome c. is a good estimate of the total number of crossover sites. d. is approximately equal to the number of mutatIOnal sites. e. refers to the number of linearly arranged organic bases in a set of genes, whether in RNA or DNA. cess except through the use of the electron microscope. proved that gene exchange must have occurred by a sexual process. can only occur between auxotrophs for different nutritIOnal requirements. may be the cause of new virulence. can be induced by man in a way that ordinarily does not occur in nature. 14. F- cells of E. colI a. cannot contribute any of their heredItary material to other cells. b. can become Hfr cells only by first becoming F+. c. are sites for synapsis and crossingover. d. can be heterogenotes or heterozygotes. e. cannot be lysed by lambda unless exposed to ultraviolet light. 15. BacterIophage a. does not contain phosphorous in Its protem. 273 b. may be infective even when its coat is punctured or removed. c. contains no RNA in Its coat, tail, or spiral fiber. d. is not infective when multiplying. e. could not be detected if it multiplIed slower than its host. 17. The !_ region of T4 a. contains 1% of the total n'its present in that phage. b. is composed of about 100 clstrons, as detected from spontaneous mutation experiments. c. is especially suitable for studies of genetic recombination rates. d. cannot lyse strain K-12 when it is mutant. e. can only demonstrate recombination when hosts are mixedly-infected. Whenever the research ot the workers lIsted in column A can be associated with an item in column B place the appropria_te' number III the space provided. . 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 274 16. A. Sinton Benzer Hershey and Chase Schramm Lederberg and Tatum Demerec Fraenkel-Conrat LUrIa and DelbrUck Wollman and Jacob Zmder Levinthal Miller and Urey Burnet B. Plant virus Ongm of hfe Protein Salmonella Phage RNA F+ m~ting type DNA NOTES