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Chapter 15: The Chromosomal Basis of Inheritance Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Figure 15.1 Chromosomes tagged to reveal a specific gene (yellow) Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Figure 15.2 The chromosomal basis of Mendel’s laws P Generation Starting with two true-breeding pea plants, we follow two genes through the F 1 and F2 generations. The two genes specify seed color (allele Y for yellow and allele y for Y green) and seed shape (allele R for round and allele r for wrinkled). These two genes are on different chromosomes. (Peas have seven chromosome pairs, but only two pairs are illustrated here.) Yellow-round seeds (YYRR) Green-wrinkled seeds (yyrr) Y R r R y r y Meiosis Fertilization y R Y Gametes r All F1 plants produce yellow-round seeds (YyRr) R R y F1 Generation y r r Y Y Meiosis LAW OF SEGREGATION r R Y 1 The R and r alleles segregate R at anaphase I, yielding two types of daughter cells for this locus. Y y R Y y LAW OF INDEPENDENT ASSORTMENT r Anaphase I y Y r 1 Alleles at both loci segregate in anaphase I, yielding four types of daughter cells R depending on the chromosome arrangement at metaphase I. Compare the arrangement of the R and r alleles in the cells y on the left and right r R Metaphase II Y y R R r 1 YR 4 Y r r 1 yr 4 F2 Generation 3 Fertilization recombines the R and r alleles at random. Y Y r 1 yr 4 2 Each gamete gets a long and a short chromosome in one of four allele combinations. y y Y Y Y Gametes r r R 2 Each gamete gets one long chromosome with either the R or r allele. Two equally probable arrangements of chromosomes at metaphase I y y R R 1 yR 4 Fertilization among the F1 plants 9 :3 :3 Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings :1 3 Fertilization results in the 9:3:3:1 phenotypic ratio in the F2 generation. Figure 15.3 Morgan’s first mutant Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Figure 15.4 In a cross between a wild-type female fruit fly and a mutant white-eyed male, what color eyes will the F1 and F2 offspring have? EXPERIMENT Morgan mated a wild-type (red-eyed) female with a mutant white-eyed male. The F1 offspring all had red eyes. P Generation X F1 Generation Morgan then bred an F1 red-eyed female to an F1 red-eyed male to Produce the F2 generation. RESULTS The F2 generation showed a typical Mendelian 3:1 ratio of red eyes to white eyes. However, no females displayed the white-eye trait; they all had red eyes. Half the males had white eyes, and half had red eyes. F2 Generation Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings CONCLUSION Since all F offspring had red eyes, the mutant 1 white-eye trait (w) must be recessive to the wild-type red-eye trait (w+). Since the recessive trait—white eyes—was expressed only in males in the F2 generation, Morgan hypothesized that the eye-color gene is located on the X chromosome and that there is no corresponding locus on the Y chromosome, as diagrammed here. P Generation W+ X X X X Y W+ W+ W W+ W Ova (eggs) F1 Generation Sperm W+ W W+ Ova (eggs) F2 Generation Sperm W+ W W+ W+ W+ W W W+ Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Unnumbered Figure p. 278 b+ vg+ b vg X Parents in testcross b vg b vg Most offspring b+ vg+ b vg or b vg Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings b vg Unnumbered Figure p. 278 Gametes from yellow-round heterozygous parent (YyRr) YR Gametes from greenwrinkled homozygous recessive parent (yyrr) yr Yr yR Yyrr yyRr yr YyRr Parentaltype offspring Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings yyrr Recombinant offspring Figure 15.5 Are the genes for body color and wing size in fruit flies located on the same chromosome or different chromosomes? EXPERIMENT Morgan first mated true-breeding wild-type flies with black, vestigial-winged flies to produce heterozygous F1 dihybrids, all of which are wild-type in appearance. He then mated wild-type F1 dihybrid females with black, vestigial-winged males, producing 2,300 F2 offspring, which he “scored” (classified according to phenotype). P Generation (homozygous) b b vg vg F1 dihybrid (wild type) (gray body, normal wings) b+ b vg+ Double mutant (black body, vestigial wings) x Wild type (gray body, normal wings) b+ b+ vg+ vg+ Double mutant (black body, vestigial wings) TESTCROSS x b b vg vg vg RESULTS b+vg+ b vg 965 944 Wild type Black(gray-normal) vestigial b+ vg b vg+ 206 Grayvestigial 185 Blacknormal b vg Sperm b+ b vg+ vg b b vg vg b+ b vg vgb b vg+ vg Parental-type offspring Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Recombinant (nonparental-type) offspring CONCLUSION If these two genes were on different chromosomes, the alleles from the F 1 dihybrid would sort into gametes independently, and we would expect to see equal numbers of the four types of offspring. If these two genes were on the same chromosome, we would expect each allele combination, B+ vg+ and b vg, to stay together as gametes formed. In this case, only offspring with parental phenotypes would be produced. Since most offspring had a parental phenotype, Morgan concluded that the genes for body color and wing size are located on the same chromosome. However, the production of a small number of offspring with nonparental phenotypes indicated that some mechanism occasionally breaks the linkage between genes on the same chromosome. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Figure 15.6 Chromosomal basis for recombination of linked genes Testcross parents b+ vg+ Gray body, normal wings b vg (F1 dihybrid) Replication of chromosomes b+ Meiosis I: Crossing over between b and vg loci produces new allele combinations. Black body, vestigial wings b vg (double mutant) Replication of chromosomes vg b vg b+ vg+ vg b b vg vg b b vg b vg Meiosis II: Segregation of chromatids produces recombinant gametes with the new allele combinations. Gametes b vg Meiosis I and II: Even if crossing over occurs, no new allele combinations are produced. Recombinant chromosome Ova Sperm b+vg+ b vg b+ vg b vg+ b vg b+ vg+ Testcross offspring Sperm b vg 965 Wild type (gray-normal) b vg 944 Blackvestigial b+ vg b vg+ 206 Grayvestigial 185 Blacknormal Ova b+ vg+ b vg+ b+ vg+ Recombination 391 recombinants = b vg+ frequency 2,300 total offspring b vg b vg b vg b vg Parental-type offspring Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Recombinant offspring 100 = 17% Figure 15.7 Constructing a Linkage Map APPLICATION A linkage map shows the relative locations of genes along a chromosome. TECHNIQUE A linkage map is based on the assumption that the probability of a crossover between two genetic loci is proportional to the distance separating the loci. The recombination frequencies used to construct a linkage map for a particular chromosome are obtained from experimental crosses, such as the cross depicted in Figure 15.6. The distances between genes are expressed as map units (centimorgans), with one map unit equivalent to a 1% recombination frequency. Genes are arranged on the chromosome in the order that best fits the data. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings RESULTS In this example, the observed recombination frequencies between three Drosophila gene pairs (bcn 9%, cnvg 9.5%, and bvg 17%) best fit a linear order in which cn is positioned about halfway between the other two genes: Recombination frequencies 9.5% 9% 17% Chromosome b cn vg The bvg recombination frequency is slightly less than the sum of the bcn and cnvg frequencies because double crossovers are fairly likely to occur between b and vg in matings tracking these two genes. A second crossover would “cancel out” the first and thus reduce the observed bvg recombination frequency. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Figure 15.8 A partial genetic (linkage) map of a Drosophila chromosome I Y II X IV III Mutant phenotypes Short aristae Black body 0 Long aristae (appendages on head) Cinnabar eyes Vestigial wings 48.5 57.5 67.0 Gray body Red eyes Normal wings Wild-type phenotypes Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Brown eyes 104.5 Red eyes Figure 15.9 Some chromosomal systems of sex determination 44 + XY 22 + X Sperm 44 + XX 44 + XX Parents 22 + Y Zygotes (offspring) Ova 22 + XY 44 + XY (a) The X-Y system 22 + XX 22 + X 76 + ZW 76 + ZZ 32 (Diploid) 16 (Haploid) (b) The X-0 system (c) The Z-W system (d) The haplo-diploid system Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Figure 15.10 The transmission of sex-linked recessive traits XAXA (a) A father with the disorder will transmit the mutant allele to all daughters but to no sons. When the mother is a dominant homozygote, the daughters will have the normal phenotype but will be carriers of the mutation. Ova Xa (c) If a carrier mates with a male who has the disorder, there is a 50% chance that each child born to them will have the disorder, regardless of sex. Daughters who do not have the disorder will be carriers, where as males without the disorder will be completely free of the recessive allele. Y XAXa XAY XA XAYa XAY XA Ova XaY XA XAXa (b) If a carrier mates with a male of normal phenotype, there is a 50% chance that each daughter will be a carrier like her mother, and a 50% chance that each son will have the disorder. XAY Y XA XAXA XAY Xa XaYA XaY XAXa Sperm Sperm XaY Sperm Xa Ova Y XA XAXa XAY Xa XaYa XaY Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Figure 15.11 X inactivation and the tortoiseshell cat Two cell populations in adult cat: Active X Early embryo: X chromosomes Cell division Inactive X and X chromosome Inactive X inactivation Allele for orange fur Allele for black fur Orange fur Black fur Active X Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Figure 15.12 Meiotic nondisjunction Meiosis I Nondisjunction Meiosis II Nondisjunction Gametes n+1 n+1 n1 n+1 n –1 n–1 Number of chromosomes (a) Nondisjunction of homologous chromosomes in meiosis I Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings n n (b) Nondisjunction of sister chromatids in meiosis II Figure 15.14 Alterations of chromosome structure (a) A deletion removes a chromosomal segment. (b) A duplication repeats a segment. (c) An inversion reverses a segment within a chromosome. (d) A translocation moves a segment from one chromosome to another, nonhomologous one. In a reciprocal translocation, the most common type, nonhomologous chromosomes exchange fragments. Nonreciprocal translocations also occur, in which a chromosome transfers a fragment without receiving a fragment in return. A B C D E F G H A B C D E F G H A B C D E F G H A B C D E F G H Deletion Duplication Inversion A B C E F G H A B C B C D E A D C B E F G H M N O C D E Reciprocal translocation M N O P Q Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings R A B P Q F G H R F G H Figure 15.15 Down syndrome Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings