Hartl, Jones - Genetics. Principlers and analysis - 1998 (522927), страница 31
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RRpp × rrppRrpp.2. rr PP × rr pprr Pp.3. Rr pp × Rr pp3 4 R— pp : 1 4 rr pp.4. rr Pp × rr Pp3 4 rr P— : 1 4 rr pp.5. RR pp × rr PPRr Pp.6. Rr Pp × Rr Pp (ρ)9 16 R— P— : 3 16 R— pp: 3 16 rr P— : 1 16 rr pp.Page 76(c) The true-breeding genotypes are RR pp (rose), rr PP (pea), rr pp (single), and RR PP (walnut).Problem 5: The pedigree in the accompanying illustration shows the inheritance of coat color in a group of cockerspaniels. The coat colors and genotypes are as follows:BlackA—B—(black symbols)LiveraaB&mdash(pink symbols)RedA—bb(red symbols)Lemonaabb(yellow symbols)(a) Specify in as much detail as possible the genotype of each dog in the pedigree.(b) What are the possible genotypes of the animal III-4, and what is the probability of each genotype?(c) If a single pup is produced from the mating of III-4 × III-7, what is the probability that the pup will be red?Answer:(a) All three matings (I-1 × I-2, II-1 × II-2, and II-5 × II-6) produce lemon-colored offspring aa bb, so each parentmust carry at least one a allele and at least one b allele.
Therefore, in consideration of the phenotypes, thegenotypes must be as follows: I-1 Aa Bb, I-2 Aa Bb, II-1 aa Bb, II-2 Aa Bb, II-5 Aa Bb, and II-6 Aa bb. Thegenotypes of the offspring can be deduced from their own phenotypes and the genotypes of the parents. These areas follows: II-3 aa bb, II-4 aa B—, III-1 Aa B—, III-2 aa bb, III-3 Aa bb, III-4 Aa B—, III-5 aa Bb, III-6 A— Bb,III-7 aa bb.(b) Animal III-4 is either Aa BB or Aa Bb, and the probabilities of these genotypes are 1 3 and 2 3, respectively.(c) If the animal III-4 is Aa BB, then the probability of a red pup is 0; and if the animal III-4 is Aa Bb, then theprobability of a red pup is 1 2 × 1 2 = 1 4 (that is, the probability of an A b gamete from III-4).
Overall, theprobability of a red pup from the mating is 1 3 × 0 + 2 3 × 1 4 = 1 6.Problem 6: From the F2 generation of a cross between mouse genotypes AA × aa, one male progeny of genotypeA— was chosen and mated with an aa female. All of the progeny in the resulting litter were A—. How large a litteris required for you to be able to assert, with 95 percent confidence, that the father's genotype is AA? How large alitter is required for 99 percent confidence?Answer: The a priori ratio of the probabilities that the father is AA versus Aa is 1/3 : 2/3, because the father waschosen at random from among the A— progeny in the F2 generation. With one A— progeny in a testcross, the ratioof probabilities drops to 1/3 : (2/3) × (1/2), because 1/2 of the A— fathers in such a testcross will yield an aaprogeny and so identify themselves as Aa.
Similarly, with n progeny, the ratio of AA : Aa probabilities is 1/3 : (2/3)× (1/2)n, because the probability that an Aa father has n consecutive A— offspring in a testcross is (1/2)n. For 95percent confidence we needorhence, n = 6A— progeny are necessary for 95 percent confidence that the father is AA. For 99 percent confidence,the corresponding formula isso in this case, n = 8 A— progeny are required.Analysis and Applications2.1 With respect to homozygosity and heterozygosity, what can be said about the genotype of a strain or varietythat breeds true for a particular trait?2.2 What gametes can be formed by an individual organism of genotype Aa? Of genotype Bb? Of genotype Aa Bb?2.3 How many different gametes can be formed by an organism with genotype AA Bb Cc Dd Ee and, in general, byan organism that is heterozygous for m genes and homozygous for n genes?2.4 Mendel summarized his conclusions about heredity by describing the gametes produced by the F1 generation inthe following manner: "Pea hybrids form germinal and pollen cells that in their composition correspond in equalnumbers to all the constant forms resulting from the combination of traits united through fertilization." Explain thisstatement in terms of the principles of segregation and independent assortment.2.5 Round pea seeds are planted that were obtained from the F2 generation of a cross between a true-breeding strainwith round seeds and a true-breeding strain with wrinkled seeds.
The pollen was collected and used en masse tofertilize plants from the true-breeding wrinkled strain. What fraction of the progeny is expected to have wrinkledseeds?2.6 If an allele R is dominant over r, how many different phenotypes are present in the progeny of a cross betweenRrPage 77and Rr, and in what ratio? How many phenotypes are there, and in what ratio, if there is no dominance between Rand r?2.7 In genetically self-sterile plants like red clover, why are all plants heterozygous for the self-sterility alleles?2.8 Assuming equal numbers of boys and girls, if a mating has already produced a girl, what is the probability thatthe next child will be a boy? If a mating has already produced two girls, what is the probability that the next childwill be a boy? On what type of probability argument do you base your answers?2.9 Assuming equal numbers of boys and girls, what is the probability that a family that has two children has twogirls? One girl and one boy?2.10 In the following questions, you are asked to deduce the genotype of certain parents in a pedigree.
Thephenotypes are determined by dominant and recessive alleles of a single gene.(a) A homozygous recessive results from the mating of a heterozygote and a parent with the dominant phenotype.What does this tell you about the genotype of the parent with the dominant phenotype?(b) Two parents with the dominant phenotype produce nine offspring. Two have the recessive phenotype. Whatdoes this tell you about the genotype of the parents?(c) One parent has a dominant phenotype and the other has a recessive phenotype.
Two offspring result, and bothhave the dominant phenotype. What genotypes are possible for the parent with the dominant phenotype?2.11 Pedigree analysis tells you that a particular parent may have the genotype AA BB or AA Bb, each with thesame probability. Assuming independent assortment, what is the probability of this parent's producing an Abgamete? What is the probability of the parent's producing an AB gamete?2.12 Assume that the trihybrid cross AA BB rr × aa bb RR is made in a plant species in which A and B aredominant but there is no dominance between R and r. Consider the F2 progeny from this cross, and assumeindependent assortment.(a) How many phenotypic classes are expected?(b) What is the probability of the parental aa bb RR genotype?(c) What proportion would be expected to be homozygous for all three genes?2.13 In the cross Aa Bb Cc Dd × Aa Bb Cc Dd, in which all genes undergo independent assortment, whatproportion of offspring are expected to be heterozygous for all four genes?2.14 The pattern of coat coloration in dogs is determined by the alleles of a single gene, with S (solid) beingdominant over s (spotted).
Black coat color is determined by the dominant allele A of a second gene, tan byhomozygosity for the recessive allele a. A female having a solid tan coat is mated with a male having a solid blackcoat and produces a litter of six pups. The phenotypes of the pups are 2 solid tan, 2 solid black, 1 spotted tan, and 1spotted black. What are the genotypes of the parents?2.15 In the human pedigree shown here, the daughter indicated by the red circle (II-1) has a form of deafnessdetermined by a recessive allele.
What is the probability that the phenotypically normal son (II-3) is heterozygousfor the gene?2.16 Huntington disease is a rare neurodegenerative human disease determined by a dominant allele, HD. Thedisorder is usually manifested after the age of forty-five. A young man has learned that his father has developed thedisease.(a) What is the probability that the young man will later develop the disorder?(b) What is the probability that a child of the young man carries the HD allele?2.17 The Hopi, Zuni, and some other Southwest American Indians have a relatively high frequency of albinism(absence of skin pigment) resulting from homozygosity for a recessive allele, a.
A normally pigmented man andwoman, each of whom has an albino parent, have two children. What is the probability that both children arealbino? What is the probability that at least one of the children is albino?2.18 Which combinations of donor and recipient ABO blood groups are compatible for transfusion? (Consider acombination to be compatible for transfusion if all the antigens in the donor red blood cells are also present in therecipient.)2.19 Red kernel color in wheat results from the presence of at least one dominant allele of each of twoindependently segregating genes (in other words, R— B— genotypes have red kernels).
Kernels on rr bb plants arewhite, and the genotypes R— bb and rr B— result in brown kernel color. Suppose that plants of a variety that istrue breeding for red kernels are crossed with plants true breeding for white kernels.(a) What is the expected phenotype of the F1 plants?(b) What are the expected phenotypic classes in the F2 progeny and their relative proportions?2.20 Heterozygous Cp cp chickens express a condition called creeper, in which the leg and wing bones are shorterthan normal (cp cp). The dominant Cp allele is lethal when homozygous. Two alleles of an independentlysegregating gene determine white (W—) versus yellow (ww) skin color. From matings between chickensheterozygous for both ofPage 78these genes, what phenotypic classes will be represented among the viable progeny, and what are their expectedrelative frequencies?2.21 White Leghorn chickens are homozygous for a dominant allele, C, of a gene responsible for colored feathers,and also for a dominant allele, I, of an independently segregating gene that prevents the expression of C.
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