Chapter 5Chapter 55.1What are the genetic differences between male- and female-determining sperm in animals with heterogametic males?ANS: The male-determining sperm carries a Y chromosome; the female-determining sperm carries an X chromosome.FEEDBACK: 5.1DIFFICULTY: easy5.2A male with singed bristles appeared in a culture of Drosophila. How would you determine if this unusual phenotype was due to an X-linked mutation?ANS: Cross the singed male to wild-type females, and then intercross the offspring. If the singed bristle phenotype is due to an X-linked mutation, approximately half the F2 males, but none of the F2 females, will show it.FEEDBACK: 5.3DIFFICULTY: medium5.3In grasshoppers, rosy body color is caused by a recessive mutation; the wild-type body color is green. If the gene for body color is on the X chromosome, what kind of progeny would be obtained from a mating between a homozygous rosy female and a homozygous wild-type male? (In grasshoppers, females are XX and males are XO.)ANS: All the daughters will be green and all the sons will be rosy.FEEDBACK: 5.1DIFFICULTY: easy5.4In the mosquito Anopheles culicifacies, golden body (go) is a recessive X-linked mutation, and brown eyes (bw) is a recessive autosomal mutation. A homozygous XX female with golden body is mated to a homozygous XY male with brown eyes. Predict the phenotypes of their offspring. If the progeny are intercrossed, what kinds of progeny will appear in the and in what proportions?ANS: The cross is go/go +/+ female +/Y bw/bw male --> F1: go/+ bw/+ females (wild-type eyes and body) and go/Y bw/+ males (golden body, wild-type eyes). An intercross of the F1 offspring yields the following F2 phenotypes in both sexes.BodyEyesGenotypeProportiongoldenbrowngo/go or Y bw/bw(1/2) (1/4) = 1/8goldenwild-typego/go or Y +/bw or +(1/2) (3/4) = 3/8wild-typebrown+/go or Y bw/bw(1/2) (1/4) = 1/8wild-typewild-type+/go or Y +/bw or +(1/2) (3/4) = 3/8FEEDBACK: 5.3DIFFICULTY: medium5.5What are the sexual phenotypes of the following genotypes in Drosophila: XX, XY, XXY, XXX, XO?ANS: XX is female, XY is male, XXY is female, XXX is female (but barely viable), XO is male (but sterile).FEEDBACK: 5.2DIFFICULTY: easy5.6In human beings, a recessive X-linked mutation, g, causes green-defective color vision; the wild-type allele, G, causes normal color vision. A man (a) and a woman (b), both with normal vision, have three children, all married to people with normal vision: a color-defective son (c), who has a daughter with normal vision (f); a daughter with normal vision (d), who has one color-defective son (g) and two normal sons (h); and a daughter with normal vision (e), who has six normal sons (i). Give the most likely genotypes for the individuals (a to i) in this family.ANS: (a) XGY; (b) XGXg; (c) XgY; (d) XGXg; (e) XGXG; (f) XGXg; (g) XgY; (h) XGY; (i) XGYFEEDBACK: 5.3DIFFICULTY: medium5.7If a father and son both have defective color vision, is it likely that the son inherited the trait from his father?ANS: No. Defective color vision is caused by an X-linked mutation. The sons X chromosome came from his mother, not his father.FEEDBACK: 5.3DIFFICULTY: easy5.8A normal woman, whose father had hemophilia, marries a normal man. What is the chance that their first child will have hemophilia?ANS: The risk for the child is P(woman transmits mutant allele) P(child is male) = (1/2) (1/2) = 1/4.FEEDBACK: 5.3DIFFICULTY: easy5.9A man with X-linked color blindness marries a woman with no history of color blindness in her family. The daughter of this couple marries a normal man, and their daughter also marries a normal man. What is the chance that this last couple will have a child with color blindness? If this couple has already had a child with color blindness, what is the chance that their next child will be color blind?ANS: The risk for the child is P(mother is C/c) P(mother transmits c) P(child is male) = (1/2) (1/2) (1/2) = 1/8; if the couple has already had a child with color blindness, P(mother is C/c) = 1, and the risk for each subsequent child is 1/4.FEEDBACK: 5.3DIFFICULTY: medium5.10A man who has color blindness and type O blood has children with a woman who has normal color vision and type AB blood. The womans father had color blindness. Color blindness is determined by an X-linked gene, and blood type is determined by an autosomal gene.(a) What are the genotypes of the man and the woman?(b) What proportion of their children will have color blindness and type B blood?(c) What proportion of their children will have color blindness and type A blood?(d) What proportion of their children will be color blind and have type AB blood?ANS: (a) The man is XcY ii; the woman is X+ Xc IA IB. (b) Probability color blind = 1/2; probability type B blood = 1/2; combined probability = (1/2) (1/2) = 1/4. (c) Probability color blind = 1/2; probability type A blood = 1/2; combined probability (1/2) (1/2) = 1/4. (d) 0.FEEDBACK: 5.3DIFFICULTY: medium5.11A Drosophila female homozygous for a recessive X-linked mutation that causes vermilion eyes is mated to a wild-type male with red eyes. Among their progeny, all the sons have vermilion eyes, and nearly all the daughters have red eyes; however, a few daughters have vermilion eyes. Explain the origin of these vermilion-eyed daughters.ANS: Each of the rare vermilion daughters must have resulted from the union of an X(v) X(v) egg with a Y-bearing sperm. The diplo-X eggs must have originated through nondisjunction of the X chromosomes during oogenesis in the mother. However, we cannot determine if the nondisjunction occurred in the first or the second meiotic division.FEEDBACK: 5.3DIFFICULTY: medium5.12In Drosophila, vermilion eye color is due to a recessive allele (v) located on the X chromosome. Curved wings is due to a recessive allele (cu) located on one autosome, and ebony body is due to a recessive allele (e) located on another autosome. A vermilion male is mated to a curved, ebony female, and the males are phenotypically wild-type. If these males were backcrossed to curved, ebony females, what proportion of the offspring will be wild-type males?ANS: P(male) = 1/2 ; P(male transmits first WT autosome) = 1/2; P(male transmits other WT autosome) = 1/2; therefore, combined proportion, P(WT male) = 1/8 FEEDBACK: 5.2DIFFICULTY: medium5.13A Drosophila female heterozygous for the recessive X-linked mutation w (for white eyes) and its wild-type allele is mated to a wild-type male with red eyes. Among the sons, half have white eyes and half have red eyes. Among the daughters, nearly all have red eyes; however, a few have white eyes. Explain the origin of these white-eyed daughters.ANS: Each of the rare white-eyed daughters must have resulted from the union of an X(w) X(w) egg with a Y-bearing sperm. The rare diplo-X eggs must have originated through nondisjunction of the X chromosomes during the second meiotic division in the mother.FEEDBACK: 5.2DIFFICULTY: medium5.14In Drosophila, a recessive mutation called chocolate (c) causes the eyes to be darkly pigmented. The mutant phenotype is indistinguishable from that of an autosomal recessive mutation called brown (bw). A cross of chocolate-eyed females to homozygous brown males yielded wild-type females and darkly pigmented males. If the flies are intercrossed, what types of progeny are expected, and in what proportions? (Assume the double mutant combination has the same phenotype as either of the single mutants alone.)ANS: 3/8 wild-type (red), 5/8 brown for both male and female F2 progeny.FEEDBACK: 5.2DIFFICULTY: hard5.15Suppose that a mutation occurred in the SRY gene on the human Y chromosome, knocking out its ability to produce the testis-determining factor. Predict the phenotype of an individual who carried this mutation and a normal X chromosome.ANS: Female.FEEDBACK: 5.4DIFFICULTY: easy5.16A woman carries the testicular feminization mutation (tfm) on one of her X chromosomes; the other X carries the wild-type allele (Tfm). If the woman marries a normal man, what fraction of her children will be phenotypically female? Of these, what fraction will be fertile?ANS: Three-fourths will be phenotypically female (genotypically tfm/Tfm, Tfm/Tfm, or tfm/Y). Among the females, 2/3 (tfm/Tfmm and Tfm/Tfm) will be fertile; the tfm/Y females will be sterile.FEEDBACK: 5.4DIFFICULTY: medium5.17Would a human with two X chromosomes and a Y chromosome be male or female?ANS: Male.FEEDBACK: 5.4DIFFICULTY: easy5.18In Drosophila, the gene for bobbed bristles (recessive allele bb, bobbed bristles; wild-type allele, normal bristles) is located on the X chromosome and on a homologous segment of the Y chromosome. Give the genotypes and phenotypes of the offspring from the following crosses: (a) (b) (c) (d) ANS: (a) 1/2 Xbb Xbb bobbed females, 1/2 Xbb Y+ wild-type males; (b) 1/2 X+ Xbb wild-type females, 1/2 Xbb Ybb bobbed males; (c) 1/4 X+ X+ wild-type females, 1/4 X+ Xbb wild-type females, 1/4 X+ Ybb wild-type males, 1/4 Xbb Ybb bobbed males; (d) 1/4 X+ Xbb wild-type females, 1/4 Xbb Xbb bobbed females, 1/4 X+ Y+ wild-type males, 1/4 Xbb Y+ wild-type males.FEEDBACK: 5.4DIFFICULTY: medium5.19Predict the sex of Drosophila with the following chromosome compositions (Ahaploid set of autosomes):(a) 4X 4A;(b) 3X 4A;(c) 2X 3A;(d) 1X 3A;(e) 2X 2A;(f) 1X 2A. ANS: (a) Female; (b) intersex; (c) intersex; (d) male: (e) female; (f) male.FEEDBACK: 5.4DIFFICULTY: easy5.20In chickens, the absence of barred feathers is due to a recessive allele. A barred rooster was mated with a nonbarred hen, and all the offspring were barred. These chickens were intercrossed to produce progeny, among which all the males were barred; half the females were barred and half were nonbarred. Are these results consistent with the hypothesis that the gene for barred feathers is located on one of the sex chromosomes?ANS: Yes. The gene for feather patterning is on the Z chromosome. If we denote the allele for barred feathers as B and the allele for nonbarred feathers as b, the crosses are: B/B (barred) male b/W (nonbarred) female --> F1: B/b (barred) males and B/W (barred) females. Intercrossing the F1 produces B/B (barred) males, B/b (barred) males, B/W (barred) females, and b/W (nonbarred) females, all in equal proportions.FEEDBACK: 5.4DIFFICULTY: medium5.21A Drosophila male carrying a recessive X-linked mutation for yellow body is mated to a homozygous wild-type female with gray body. The daughters of this mating all have uniformly gray bodies. Why arent their bodies a mosaic of yellow and gray patches?ANS: Drosophila does not achieve dosage compensation by inactivating one of the X chromosomes in females.FEEDBACK: 5.5DIFFICULTY: medium5.22What is the maximum number of Barr bodies in the nuclei of human cells with the following chromosome compositions:(a) XY;(b) XX;(c) XXY;(d) XXX;(e) XXXX;(f) XYY? ANS: (a) Zero; (b) one; (c) one; (d) two; (e) three; (f) zero.FEEDBACK: 5.5DIFFICULTY: easy5.23Males in a certain species of deer have two nonhomologous X chromosomes, denoted X1 and X2, and a Y chromosome. Each X chromosome is about half as large as the Y chromosome, and its centromere is located near one of the ends; the centromere of the Y chromosome is located in the middle. Females in this species have two copies of each of the X chromosomes and lack a Y chromosome. How would you predict the X and Y chromosomes to pair and disjoin during spermato-genesis to produce equal numbers of male- and female-determining sperm?ANS: Since the centromere is at the end of each small X chromosome but in the middle of the larger Y, X1 and X2 both pair at the centromere of the Y chromosome during metaphase so that the two X chromosomes disjoin together and segregate from the Y chromosome during anaphase.FEEDBACK: 5.5DIFFICULTY: medium5.24A breeder of sun conures (a type of bird) has obtained two true-breeding strains, A and B, which have red eyes instead of the normal brown found in natural populations. In Cross 1, a male from strain A was mated to a female from strain B, and the male and female offspring all had brown eyes. In Cross 2, a female from strain A was mated to a male from strain B, and the male offspring had brown eyes and the female offspring had red eyes. When the birds from each cross were mated brother to sister, the breeder obtained the following results:Proportion in Proportion in Phenotypeof Cross 1of Cross 2Brown maleRed maleBrown femaleRed femaleProvide a genetic explanation for these results.ANS: Color is determined by an autosomal gene (alleles A and a) and a sex-linked gene (alleles B and b) on the Z chromosome (females are ZW and males are ZZ) and the recessive alleles are mutually epistaticthat is, aa, bb, or bW birds have red eyes, and A- B- or A- BW birds have brown eyes.Cross 1PStrain A red maleStrain B red femaleaa BBAA bWF1Aa Bb brown malesAa BW brown femalesF2A- Bb brown males (6/16)A- BW brown females (3/16)aa Bb red males (2/16)A- bW red females (4/16)aa BW red females (1/16)Cross 2PStrain A red femaleStrain B red maleaa BWAA bbF1Aa Bb brown malesAa bW red femalesF2A- Bb brown males (3/16)A- bW brown females (3/16)A- bb red males (3/16)A- bW red females (3/16)aa b- red males 2/16aa -W red females (2/16)FEEDBACK: 5.2DIFFICULTY: hard5.25In 1908 F. M. Durham and D.C.E. Marryat reported the results of breeding experiments with canaries. Cinnamon canaries have pink eyes when they first hatch, whereas green canaries have black eyes. Durham and Marryat crossed cinnamon females with green males and observed that all the progeny had black eyes, just like those of the green strain. When the males were crossed to green females, all the male progeny had black eyes, whereas all the female progeny had either black or pink eyes, in about equal proportions. When the males were crossed to cinnamon females, four classes of progeny were obtained: females with black eyes, females with pink eyes, males with black eyes, and males with pink eyesall in approximately equal proportions. Propose an explanation for these findings.ANS: Eye color in canaries is due to a gene on the Z chromosome, which is present in two copies in males and one copy in females. The allele for pink color at hatching (p) is recessive to the allele for black color at hatching (P). There is no eye color gene on the other sex chromosome (W), which is present in one copy in females and absent in males. The parental birds were genotypically p/W (cinnamon females) and P/P (green males). Their F1 sons were genotypically p/P (with black eyes at hatching). When these sons were crossed to green females (genotype P/W), they produced F2 progeny that sorted into three categories: males with black eyes at hatching (P/-, half the total progeny), females with black eyes at hatching (P/W, a fourth of the total progeny), and females with pink eyes at hatching (p/W, a fourth of the total progeny). When these sons were crossed to cinnamon females (genotype p/W), they produced F2 progeny that sorted into four equally frequent categories: males with black eyes at hatching (genotype P/p), males with pink eyes at hatching (genotype p/p), females with black eyes at hatching (genotype P/W), and females with pink eyes at hatching (genotype p/W).FEEDBACK: 5.2DIFFICULTY: hard