Application Questions and Problems

Introduction

Question 4.14

As described in the introduction to this chapter, platypuses possess 10 sex chromosomes. Females have five pairs of X chromosomes (X1X1X2X2X3X3X4X4X5X5) and males have five pairs of X and Y chromosomes (X1Y1X2Y2X3Y3X4Y4X5Y5). Do each of the XY chromosome pairs in males assort independently in meiosis? Why or why not?

Section 4.1

Question 4.15

What is the sexual phenotype of fruit flies having the following chromosomes?

Sex chromosomes Autosomal chromosomes
a. XX       all normal          
b. XY       all normal          
c. XO       all normal          
d. XXY    all normal          
e. XYY    all normal          
f. XXYY all normal          
g. XXX    all normal          
h. XX       four haploid sets
i. XXX    four haploid sets
j. XXX    three haploid sets
k. X         three haploid sets
l. XY       three haploid sets
m. XX       three haploid sets

Question 4.16

If nondisjunction of the sex chromosomes takes place in meiosis I in the male in Figure 4.5, what sexual phenotypes and proportions of offspring will be produced?

Question 4.17

For each of the following chromosome complements, what is the phenotypic sex of a person who has

  • a. XY with the SRY gene deleted?
  • b. XX with a copy of the SRY gene on an autosomal chromosome?
  • c. XO with a copy of the SRY gene on an autosomal chromosome?
  • d. XXY with the SRY gene deleted?
  • e. XXYY with one copy of the SRY gene deleted?

Question 4.18

A normal female Drosophila produces abnormal eggs that contain all (a complete diploid set) of her chromosomes. She mates with a normal male Drosophila that produces normal sperm. What will the sex of the progeny from this cross be?

Question 4.19

In certain salamanders, the sex of a genetic female can be altered, changing her into a functional male; these salamanders are called sex-reversed males. When a sex-reversed male is mated with a normal female, approximately of the offspring are female and are male. How is sex determined in these salamanders? Explain the results of this cross.

Question 4.20

In some mites, males pass genes to their grandsons, but they never pass genes to their sons. Explain.

Question 4.21

In organisms with the ZZ-ZW sex-determining system, from which of the following possibilities can a female inherit her Z chromosome?

Yes No
Her mother’s mother __________ __________
Her mother’s father __________ __________
Her father’s mother __________ __________
Her father’s father __________ __________

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Section 4.2

Question 4.22

When Bridges crossed white-eyed females with red-eyed males, he obtained a few red-eyed males and white-eyed females (see Figure 4.13). What types of offspring would be produced if these red-eyed males and white-eyed females were crossed with each other?

Question 4.23

Joe has classic hemophilia, an X-linked recessive disease. Could Joe have inherited the gene for this disease from the following persons?

Yes No
a. His mother’s mother __________ __________
b. His mother’s father __________ __________
c. His father’s mother __________ __________
d. His father’s father __________ __________

Question 4.24

In Drosophila, yellow body is due to an X-linked gene that is recessive to the gene for gray body.

[Courtesy Masa-Toshi Yamamoto, Drosophila Genetic Resource Center, Kyoto Institute of Technology.]
  • a. A homozygous gray female is crossed with a yellow male. The F1 are intercrossed to produce F2. Give the genotypes and phenotypes, along with the expected proportions, of the F1 and F2 progeny.
  • b. A yellow female is crossed with a gray male. The F1 are intercrossed to produce the F2. Give the genotypes and phenotypes, along with the expected proportions, of the F1 and F2 progeny.
  • c. A yellow female is crossed with a gray male. The F1 females are backcrossed with gray males. Give the genotypes and phenotypes, along with the expected proportions, of the F2 progeny.
  • d. If the F2 flies in part b mate randomly, what are the expected phenotypic proportions of flies in the F3?

Question 4.25

Coat color in cats is determined by genes at several different loci. At one locus on the X chromosome, one allele (X+) encodes black fur; another allele (Xo) encodes orange fur. Females can be black (X+X+), orange (Xo Xo), or a mixture of orange and black called tortoiseshell (X+Xo). Males are either black (X+Y) or orange (XoY). Bill has a female tortoiseshell cat named Patches. One night Patches escapes from Bill’s house, spends the night out, and mates with a stray male. Patches later gives birth to the following kittens: one orange male, one black male, two tortoiseshell females, and one orange female. Give the genotypes of Patches, her kittens, and the stray male with which Patches mated.

Question 4.26

Red–green color blindness in humans is due to an X-linked recessive gene. Both John and Cathy have normal color vision. After 10 years of marriage to John, Cathy gave birth to a color-blind daughter. John filed for divorce, claiming that he is not the father of the child. Is John justified in his claim of nonpaternity? Explain why. If Cathy had given birth to a color-blind son, would John be justified in claiming nonpaternity?

Question 4.27

Red–green color blindness in humans is due to an X-linked recessive gene. A woman whose father is color blind possesses one eye with normal color vision and one eye with color blindness.

  • a. Propose an explanation for this woman’s vision pattern. Assume that no new mutations have spontaneously arisen.
  • b. Would it be possible for a man to have one eye with normal color vision and one eye with color blindness?

Question 4.28

Bob has XXY chromosomes (Klinefelter syndrome) and is color blind. His mother and father have normal color vision, but his maternal grandfather is color blind. Assume that Bob’s chromosome abnormality arose from nondisjunction in meiosis. In which parent and in which meiotic division did nondisjunction take place? Assume no crossing over has taken place. Explain your answer.

Question 4.29

Xg is an antigen found on red blood cells. This antigen is caused by an X-linked allele (Xa) that is dominant over an allele for the absence of the antigen (X). The inheritance of these X-linked alleles was studied in children with chromosome abnormalities to determine where nondisjunction of the sex chromosomes took place. For each type of mating in parts a through d, indicate whether nondisjunction took place in the mother or in the father and, if possible, whether it took place in meiosis I or meiosis II (assume no crossing over).

  • a. Xa Y × X X → Xa (Turner syndrome)
  • b. Xa Y × Xa X → X (Turner syndrome)
  • c. Xa Y × X X → Xa X Y (Klinefelter syndrome)
  • d. Xa Y × Xa X → X X Y (Klinefelter syndrome)

Question 4.30

The Talmud, an ancient book of Jewish civil and religious laws, states that, if a woman bears two sons who die of bleeding after circumcision (removal of the foreskin from the penis), any additional sons that she has should not be circumcised. (The bleeding is most likely due to the X-linked disorder hemophilia.) Furthermore, the Talmud states that the sons of her sisters must not be circumcised, whereas the sons of her brothers should be. Is this religious law consistent with sound genetic principles? Explain your answer.

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Question 4.31

Craniofrontonasal syndrome (CFNS) is a birth defect in which premature fusion of the cranial sutures leads to abnormal head shape, widely spaced eyes, nasal clefts, and various other skeletal abnormalities. George Feldman and his colleagues, looked at several families in which offspring had CFNS and recorded the results shown in the following table (G. J. Feldman. 1997. Human Molecular Genetics 6:1937–1941).

  • a. On the basis of these results, what is the most likely mode of inheritance for CFNS?
  • b. Give the most likely genotypes of the parents in family 1 and in family 10a.

Question 4.32

Miniature wings (Xm) in Drosophila result from an X-linked allele that is recessive to the allele for long wings (X+). Give the genotypes of the parents in each of the following crosses.

Male parent Female parent Male offspring Female offspring
a. long long 231 long, 560 long
250 miniature
b. miniature long 610 long 632 long
c. miniature long 410 long, 412 long,
417 miniature 415 miniature
d. long miniature 753 miniature 761 long
e. long long 625 long 630 long

Question 4.33

In chickens, congenital baldness is due to a Z-linked recessive gene. A bald rooster is mated with a normal hen. The F1 from this cross are interbred to produce the F2. Give the genotypes and phenotypes, along with their expected proportions, among the F1 and F2 progeny.

Question 4.34

If the blue F1 females in Figure 4.15b are backcrossed to the blue males in the P generation, what types and proportions of offspring will be produced?

Question 4.35

Red–green color blindness is an X-linked recessive trait in humans. Polydactyly (extra fingers and toes) is an autosomal dominant trait. Martha has normal fingers and toes and normal color vision. Her mother is normal in all respects, but her father is color blind and polydactylous. Bill is color blind and polydactylous. His mother has normal color vision and normal fingers and toes. If Bill and Martha marry, what types and proportions of children can they produce?

Question 4.36

A Drosophila mutation called singed (s) causes the bristles to be bent and misshapen. A mutation called purple (p) causes the fly’s eyes to be purple in color instead of the normal red. Flies homozygous for singed and purple were crossed with flies that were homozygous for normal bristles and red eyes. The F1 were intercrossed to produce the F2, and the following results were obtained.

Cross 1

P male, singed bristles, purple eyes × female, normal bristles, red eyes
F1 420 female, normal bristles, red eyes
426 male, normal bristles, red eyes
F2 337 female, normal bristles, red eyes
113 female, normal bristles, purple eyes
168 male, normal bristles, red eyes
170 male, singed bristles, red eyes
56 male, normal bristles, purple eyes
58 male, singed bristles, purple eyes

Cross 2

P female, singed bristles, purple eyes × male, normal bristles, red eyes
F1 504 female, normal bristles, red eyes
498 male, singed bristles, red eyes
F2 227 female, normal bristles, red eyes
223 female, singed bristles, red eyes
225 male, normal bristles, red eyes
225 male, singed bristles, red eyes
78 female, normal bristles, purple eyes
76 female, singed bristles, purple eyes
74 male, normal bristles, purple eyes
72 male, singed bristles, purple eyes
  • a. What are the modes of inheritance of singed and purple? Explain your reasoning.
  • b. Give genotypes for the parents and offspring in the P, F1, and F2 generations of Cross 1 and Cross 2.

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Question 4.37

The following two genotypes are crossed: Aa Bb Cc X+ Xr × Aa BB cc X+Y, where a, b, and c represent alleles of autosomal genes and X+ and Xr represent X-linked alleles in an organism with XX-XY sex determination. What is the probability of obtaining genotype aa Bb Cc X+ X+ in the progeny?

Question 4.38

Miniature wings in Drosophila are due to an X-linked gene (Xm) that is recessive to an allele for long wings (X+). Sepia eyes are produced by an autosomal gene (s) that is recessive to an allele for red eyes (s+).

  • a. A female fly that has miniature wings and sepia eyes is crossed with a male that has normal wings and is homozygous for red eyes. The F1 flies are intercrossed to produce the F2. Give the phenotypes, as well as their expected proportions, of the F1 and F2 flies.
  • b. A female fly that is homozygous for normal wings and has sepia eyes is crossed with a male that has miniature wings and is homozygous for red eyes. The F1 flies are intercrossed to produce the F2. Give the phenotypes, as well as their expected proportions, of the F1 and F2 flies.

Question 4.39

Suppose that a recessive gene that produces a short tail in mice is located in the pseudoautosomal region. A short-tailed male mouse is mated with a female mouse that is homozygous for a normal tail. The F1 mice from this cross are intercrossed to produce the F2. Give the phenotypes, as well as their proportions, of the F1 and F2 mice?

Question 4.40

A color-blind woman and a man with normal vision have three sons and six daughters. All the sons are color blind. Five of the daughters have normal vision, but one of them is color blind. The color-blind daughter is 16 years old, is short for her age, and has not undergone puberty. Explain how this girl inherited her color blindness.


Section 4.3

Question 4.41

How many Barr bodies would you expect to see in a human cell containing the following chromosomes?

  • a. XX
  • b. XY
  • c. XO
  • d. XXY
  • e. XXYY
  • f. XXXY
  • g. XYY
  • h. XXX
  • i. XXXX

Question 4.42

A woman with normal chromosomes mates with a man who also has normal chromosomes.

  • a. Suppose that, in the course of oogenesis, the woman’s sex chromosomes undergo nondisjunction in meiosis I; the man’s chromosomes separate normally. Give all possible combinations of sex chromosomes that this couple’s children might inherit and the number of Barr bodies that you would expect to see in each of the cells of each child.
  • b. What chromosome combinations and numbers of Barr bodies would you expect to see if the chromosomes separate normally in oogenesis, but nondisjunction of the sex chromosomes takes place in meiosis I of spermatogenesis?

Question 4.43

What is the most likely sex and genotype of the cat shown in Figure 4.18?

Question 4.44

Anhidrotic ectodermal dysplasia is an X-linked recessive disorder in humans characterized by small teeth, no sweat glands, and sparse body hair. This trait is usually seen in men, but women who are heterozygous carriers of the trait often have irregular patches of skin with few or no sweat glands (see the illustration below).

[After A. P. Mance and J. Mance, Genetics: Human Aspects (Sinauer, 1990), p. 133.]
  • a. Explain why women who are heterozygous carriers of a recessive gene for anhidrotic ectodermal dysplasia have irregular patches of skin lacking sweat glands.
  • b. Why does the distribution of the patches of skin lacking sweat glands differ among the females depicted in the illustration, even between the identical twins?

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