Chapter 4

1. In the XX-XY system, females are homogametic (XX) and males are heterogametic (XY). In the ZZ-ZW system, females are heterogametic (ZW) and males are homogametic (ZZ).

14. (a) Yes; (b) yes; (c) no; (d) no.

15. 1. F₁: ½ X⁺Y (gray males), ½ X⁺X^y (gray females); F₂: ¼ X⁺Y (gray males), ¼ X^yY (yellow males), ¼ X⁺X^y (gray females), ¼ X⁺X⁺ (gray females)

    2. F₁: ½ X^yY (yellow males), ½ X⁺X^y (gray females); F₂: ¼ X⁺Y (gray males), ¼ X^yY (yellow males), ¼ X⁺X^y (gray females), ¼ X^yX^y (yellow females)

19. Because Bob must have inherited the Y chromosome from his father, and his father has normal color vision, a nondisjunction event in the paternal lineage cannot account for Bob’s genotype. Bob’s mother must be heterozygous X⁺X^c because she has normal color vision, and she must have inherited an X^c chromosome from her color-blind father. For Bob to inherit two X^c chromosomes from his mother, the egg must have arisen from a nondisjunction in meiosis II. In meiosis I, the homologous X chromosomes separate, and so one cell has the X⁺ chromosome and the other has X^c. The failure of sister chromatids to separate in meiosis II would then result in an egg with two copies of X^c.

25. 1. F₁: All males have miniature wings and red eyes (X^mY s⁺s), and all females have long wings and red eyes (X⁺X^m s⁺s). F₂: male, normal, red; male, normal, sepia; male, miniature, red; male, miniature, sepia; female, normal, red; female, normal, sepia; female, miniature, red; female, miniature, sepia.

    2. F₁: All females have long wings and red eyes (X⁺X^m s⁺s), and all males have long wings and red eyes (X⁺Y s⁺s). F₂: male, long wings, red eyes; male, long wings, sepia eyes; male, miniature wings, red eyes; male, miniature wings, sepia eyes; ⅜ female, long wings, red eyes; ⅛ female, long wings, sepia eyes.

26. 1. The results of the crosses indicate that cremello and chestnut are pure-breeding traits (homozygous). Palomino is a hybrid trait (heterozygous) that produces a 2:1:1 ratio when palominos are crossed with each other. The simplest hypothesis consistent with these results is incomplete dominance, with palomino as the phenotype of the heterozygotes resulting from chestnuts crossed with cremellos.

    2. Let C^B = chestnut, C^W = cremello. The parents and offspring of these crosses have the following genotypes: chestnut = C^BC^B; cremello = C^WC^W; palomino = C^BC^W.

28. To have long earlobes, the child must inherit the dominant allele and also express it. The probability of inheriting the dominant allele is 50%; the probability of expressing it is 30%. The combined probability of both is 0.5 × 0.3 = 0.15, or 15%.

29. 1. The 2:1 ratio in the progeny of two spotted hamsters suggests lethality, and the 1:1 ratio in the progeny of a spotted hamster and a hamster without spots indicates that spotted is a heterozygous phenotype. If S and s represent alleles at the locus for white spotting, spotted hamsters are Ss and solid-colored hamsters are ss. One-fourth of the zygotes expected from a mating of two spotted hamsters are SS—embryonic lethal—and missing from the progeny, resulting in the 2:1 ratio of spotted to solid progeny.

    2. Because spotting is a heterozygous phenotype, obtaining Chinese hamsters that breed true for spotting is impossible.

32. 1. B or AB.

    2. No. Many other men have these blood types. The results would have meant only that Chaplin could not be eliminated as a possible father of the child.

33. 1. All walnut (Rr Pp)

    2. ¼ walnut (Rr Pp), ¼ rose (Rr pp), ¼ pea (rr Pp), ¼ single (rr pp)

    3. walnut (R_ P_), rose (R_ pp), pea (rr P_), single (rr pp)

    4. ¾ rose (R_ pp), ¼ single (rr pp)

    5. ¼ walnut (Rr Pp), ¼ rose (Rr pp), ¼ pea (rr Pp), ¼ single (rr pp)

    6. ½ rose (Rr pp), ½ single (rr pp)

36. Let A and B represent the two loci. The F₁ heterozygotes are Aa Bb. The F₂ are A_ B_ disc-shaped, A_ bb spherical, aa B_ spherical, aa bb long.

38. In genetic maternal effects, the genotype of the mother determines the phenotype of the offspring. Because Martha is sinistral, we know her mother must be genotype ss. If Martha’s mother is ss, Martha must carry at least one s allele. We have no information about Martha’s father.

    1. False. Martha might have inherited an s⁺ from her father and therefore could be s⁺s.

    2. True. Martha must have inherited an s allele from her mother and therefore cannot be s⁺s⁺.

    3. False. The phenotype of Martha’s offspring will be determined by Martha’s genotype, which we do not know. Martha might have inherited an s⁺ allele from her father, in which case her genotype would be s⁺s and she would produce all dextral offspring.

    4. False. Martha’s genotype could be s⁺s, in which case all her offspring would be dextral.

    5. False. Her mother’s phenotype is determined by her own mother’s genotype, which could have been s⁺s.

    6. True. Because Martha is sinistral, her mother’s genotype is ss and all her offspring should be sinistral like Martha.