Chapter 15Slide 2Slide 3Slide 4Slide 5Recombination of Unlinked Genes: Independent Assortment of ChromosomesSlide 7Slide 8Slide 9Slide 10Slide 11Slide 12Inheritance of Sex-Linked GenesX inactivation in Female MammalsAbnormal Chromosome NumberSlide 16Slide 17Human Disorders Due to Chromosomal AlterationsDisorders Caused by Structurally Altered ChromosomesYou should now be able to:Slide 21PowerPoint Lectures for Biology, Eighth EditionNeil Campbell and Jane ReeceChapter 15Chapter 15The Chromosomal Basis of InheritanceFig. 15-2P GenerationYellow-roundseeds (YYRR)YF1 GenerationYRRRYrrryyyMeiosisFertilizationGametesGreen-wrinkledseeds ( yyrr)All F1 plants produceyellow-round seeds (YyRr)RRYYrryyMeiosisRRYYrryyMetaphase IYYR RrryyAnaphase IrryYMetaphase IIRYRyyyyRRYYrrrryYYRRyRYryrYR1/41/41/41/4F2 GenerationGametesAn F1 F1 cross-fertilization9: 3: 3: 1LAW OF INDEPENDENTASSORTMENT Alleles of geneson nonhomologouschromosomes assortindependently during gameteformation.LAW OF SEGREGATIONThe two alleles for each geneseparate during gameteformation.123321•Morgan first observed and noted–Wild type, or normal, phenotypes that were common in the fly populations•Traits alternative to the wild type–Are called mutant phenotypesFigure 15.3Figure 15.4 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.Morgan then bred an F1 red-eyed female to an F1 red-eyed male toproduce the F2 generation. RESULTSPGenerationF1GenerationXF2Generation Morgan mated a wild-type (red-eyed) female with a mutant white-eyed male. The F1 offspring all had red eyes.EXPERIMENT•Morgan determined –That the white-eye mutant allele must be located on the X chromosomeCONCLUSION Since all F1 offspring had red eyes, the mutant 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.PGenerationF1GenerationF2GenerationOva(eggs)Ova(eggs)SpermSpermXXXXYWW+W+WW+W+W+W+W+W+W+W+W W+W W WRecombination of Unlinked Genes: Independent Assortment of Chromosomes•When Mendel followed the inheritance of two characters–He observed that some offspring have combinations of traits that do not match either parent in the P generationGametes from green-wrinkled homozygousrecessive parent (yyrr)Gametes from yellow-roundheterozygous parent (YyRr)Parental-type offspringRecombinantoffspringYyRr yyrr Yyrr yyRrYRyrYryRyr•Morgan crossed flies–That differed in traits of two different charactersDouble mutant(black body,vestigial wings)Double mutant(black body,vestigial wings)Wild type(gray body,normal wings)P Generation(homozygous)b+ b+ vg+ vg+xb b vg vgF1 dihybrid(wild type)(gray body, normal wings)b+ b vg+ vgb b vg vgTESTCROSSxb+vg+b vgb+ vgb vg+b vgb+ b vg+ vg b b vg vgb+ b vg vgb b vg+ vg965Wild type(gray-normal)944Black-vestigial206Gray-vestigial185Black-normalSpermParental-typeoffspringRecombinant (nonparental-type)offspringRESULTSEXPERIMENT Morgan first mated true-breedingwild-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).CONCLUSION If these two genes were on different chromosomes, the alleles from the F1 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, onlyoffspring with parental phenotypes would be produced. Since most offspring had a parental phenotype, Morganconcluded that the genes for body color and wing sizeare 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.Figure 15.5Double mutant(black body,vestigial wings)Double mutant(black body,vestigial wings)Figure 15.6TestcrossparentsGray body,normal wings(F1 dihybrid)b+vg+b vgReplication ofchromosomesb+vgb+vg+bvgvgMeiosis I: Crossingover between b and vgloci produces new allelecombinations.Meiosis II: Segregationof chromatids producesrecombinant gameteswith the new allelecombinations.Recombinantchromosomeb+vg+b vgb+ vgb vg+b vgSpermb vgb vgReplication ofchromosomesvgvgbbbvgb vgMeiosis I and II:Even if crossing overoccurs, no new allelecombinations areproduced.OvaGametesTestcrossoffspringSpermb+ vg+b vgb+ vg b vg+965Wild type(gray-normal)b+ vg+b vg b vg b vg b vgb vg+b+ vg+b vg+944Black-vestigial206Gray-vestigial185Black-normalRecombinationfrequency=391 recombinants2,300 total offspring100 = 17%Parental-type offspringRecombinant offspringOvab vgBlack body,vestigial wings(double mutant)b•Linked genes–Exhibit recombination frequencies less than 50%•A linkage map–Is the actual map of a chromosome based on recombination frequenciesRecombinationfrequencies9%9.5%17%b cnvgChromosomeThe b–vg recombination frequency is slightly less than the sum of the b–cn and cn–vg frequencies because double crossovers are fairly likely to occur between b and vg in matings tracking these two genes. A second crossoverwould “cancel out” the first and thus reduce the observed b–vg recombination frequency. In this example, the observed recombination frequencies between three Drosophila gene pairs (b–cn 9%, cn–vg 9.5%, and b–vg 17%) best fit a linear order in which cn is positioned about halfway between the other two genes:RESULTS A linkage map shows the relative locations of genes along a chromosome.APPLICATIONTECHNIQUE A linkage map is based on the assumption that the probability of a crossover between twogenetic loci is proportional to the distance separating the loci. The recombination frequencies used to constructa linkage map for a particular chromosome are obtained from experimental crosses, such as the cross depictedin Figure 15.6. The
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