Hartl, Jones - Genetics. Principlers and analysis - 1998 (522927), страница 44
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Assuming that at least one X chromosome is required for viability, what sex ratio is expected amongsurviving progeny from the mating XO × XY ?3.20 In the accompanying pedigree, the shaded symbols represent persons affected with X-linked hemophilia, ablood-clotting disorder.(a) If the woman identified as II-2 has two more children, what is the probability that neither will be affected?(b) What is the probability that the first child of the mating II-4 × II-5 will be affected?3.21 Assume a sex ratio at birth of 1 : 1 and consider two sibships, A and B, each with three children.(a) What is the probability that A consists only of girls and B only of boys?(b) What is the probability that one sibship consists only of girls and the other only of boys?Challenge Problems3.22 A hybrid corn plant with green leaves was testcrossed with a plant having yellow-striped leaves. When 250seedlings were grown, 140 of the seedlings had green leaves and 110 had yellow-striped leaves.
Using the chisquared method, test this result for agreement with the expected 1 : 1 ratio.3.23 A cross was made to produce D. melanogaster flies heterozygous for two pairs of alleles: dp+ and dp, whichdetermine long versus short wings, and e+ and e, which determine gray versus ebony body color. The following F2data were obtained:Long wing, gray body462Long wing, ebony body167Short wing, gray body127Short wing, ebony body44Test these data for agreement with the 9 : 3 : 3 : 1 ratio expected if the two pairs of alleles segregate independently.Further ReadingAllshire, R.
C. 1997. Centromeres, checkpoints and chromatid cohesion. Current Opinion in Genetics-θDevelopment 7: 264.Chandley, A. C. 1988. Meiosis in man. Trends in Genetics 4: 79.Cohen, J. S., and M. E. Hogan. 1994. The new genetic medicine. Scientific American, December.McIntosh, J. R., and K. L. McDonald. 1989. The mitotic spindle. Scientific American, October.McKusick, V.
A. 1965. The royal hemophilia. Scientific American, August.Miller, O. J. 1995. The fifties and the renaissance of human and mammalian genetics. Genetics 139: 484.Page, A. W. and T. L. Orr-Weaver. 1997. Stopping and starting the meiotic cell cycle. Current Opinion in Geneticsθ Development 7: 23.Sokal, R. R., and F. J. Rohlf.
1969. Biometry. New York: Freeman.Sturtevant, A. H. 1965. A Short History of Genetics. Harper & Row.Voeller, B. R., ed. 1968. The Chromosome Theory of Inheritance: Classical Papers in Development and Heredity.New York: Appleton-Century-Crofts.Welsh, M. J., and A. E.
Smith. 1995. Cystic fibrosis. Scientific American, December.Zielenski, J., and L. C. Tsui. 1995. Cystic fibrosis: Genotypic and phenotypic variations. Annual Review ofGenetics 29: 777.Page 122Electron micrograph of the chromosomes in a haploid yeast cell (Saccharomyces cerevisiae) inprophase of mitosis, showing the full complement of 16 chromosomes. The darkly stained mass at thebottom is the nucleus, which is associated with chromosome XII.[Courtesy of Kuei-Shu Tung and Shirleen Roeder.]Page 123Chapter 4—Genetic Linkage and Chromosome MappingCHAPTER OUTLINE4-1 Linkage and Recombination of Genes in a Chromosome4-2 Genetic MappingCrossing-overCrossing-over Takes Place at the Four-Strand Stage of MeiosisMolecular Basis of Crossing-overMultiple Crossing-over4-3 Gene Mapping from Three-Point TestcrossesChromosome Interference in Double Crossing-overGenetic Mapping FunctionsGenetic Distance and Physical Distance4-4 Genetic Mapping in Human Pedigrees4-5 Mapping by Tetrad Analysisof Unordered TetradsThe Analysis of Ordered Tetrads4-6 Mitotic Recombination4-7 Recombination Within Genes4-8 A Closer Look at ComplementationChapter SummaryKey TermsReview the BasicsGuide to Problem SolvingAnalysis and ApplicationsChallenge ProblemsFurther ReadingGeNETics on the webPRINCIPLES• Genes that are located in the same chromosome and that do not show independent assortment are said to belinked.• The alleles of linked genes that are present together in the same chromosome tend to be inherited as a group.• Crossing-over between homologous chromosomes results in recombination that breaks up combinations of linkedalleles.• A genetic map depicts the relative positions of genes along a chromosome.• The map distance between genes in a genetic map is related to the rate of recombination between the genes.• Physical distance along a chromosome is often—but not always—correlated with map distance.• Tetrads are sensitive indicators of linkage because each contains all the products of a single meiosis.• Recombination can also take place between nucleotides within a gene.• The complementation test is the experimental determination of whether two mutations are, or are not, alleles ofthe same gene.CONNECTIONSCONNECTION: Genes All in a RowAlfred H.
Sturtevant 1913The linear arrangement of six sex-linked factors in Drosophila, as shown by their mode of associationCONNECTION: DosXXLilian V. Morgan 1922Non-crisscross inheritance in Drosophila melanogasterPage 124In meiosis, homologous chromosomes form pairs, and the individual members of each pair separate from oneanother. The observation that homologous chromosomes behave as complete units when they separate led to theexpectation that genes located in the same chromosome would not undergo independent assortment but ratherwould be transmitted together with complete linkage. As we shall see, Thomas Hunt Morgan examined this issueusing two genes that he knew were both present in the X chromosome of Drosophila.
One was a mutation for whiteeyes, the other a mutation for miniature wings. Morgan did observe linkage, but it was incomplete. Morgan foundthat the white and miniature alleles present in each X chromosome of a female tended to remain together ininheritance, but he also observed that some X chromosomes were produced that had new combinations of the whiteand miniature alleles.In this chapter, we will see that Morgan's observation of incomplete linkage is the rule for genes present in thesame chromosome.
The reason why linkage is incomplete is that the homologous chromosomes, when they arepaired, can undergo an exchange of segments. An exchange event between homologous chromosomes, crossingover, results in the recombination of genes in the homologous chromosomes. The probability of crossing-overbetween any two genes serves as a measure of genetic distance between the genes and makes possible theconstruction of a genetic map, a diagram of a chromosome showing the relative positions of the genes.
The geneticmapping of linked genes is an important research tool in genetics because it enables a new gene to be assigned to achromosome and often to a precise position relative to other genes within the same chromosome. Genetic mappingis usually a first step in the identification and isolation of a new gene and the determination of its DNA sequence.Genetic mapping is essential in human genetics for the identification of genes associated with hereditary diseases,such as the genes whose presence predisposes women carriers to the development of breast cancer.4.1—Linkage and Recombination of Genes in a ChromosomeAs we saw in Chapter 3, a direct test of independent assortment is to carry out a testcross between an F1 doubleheterozygote (Aa Bb) and the double recessive homozygote (aa bb).
When the genes are on differentchromosomes, the expected gametes from the Aa Bb parent are as shown in Figure 4.1. Because the pairs ofhomologous chromosomes segregate independently of each other in meiosis, the double heterozygote produces allfour possible types of gametes—AB, Ab, aB, and ab—in equal proportions.Independent assortment takes place in the Aa Bb genotype whether the parents were genotypically AA bb and aaBB. The four products of meiosis are still expected in equal proportions. An expected 50 percent of the testcrossprogeny result from gametes with the same combination of alleles present in the parents of the double heterozygote(parental combinations), and 50 percent result from gametes with new combinations of the alleles(recombinants). For example, if the double heterozygote came from the mating AA BB × aa bb, then the AB andab gametes would be parental and the Ab and aB gametes recombinant.
On the other hand, if the doubleheterozygote came from the mating AA bb × aa BB, then the Ab and aB gametes would be parental and the ab andAB gametes recombinant. In either case, with independent assortment, the genotypes of the testcross progeny areexpected in the ratio of 1 : 1 : 1 : 1.
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