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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
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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
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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.
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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