Question 5.12

Match each of the following terms with its correct definition (parts a through i).

________________phenocopy

________________pleiotrophy

________________polygenic trait

________________penetrance

________________sex-limited trait

________________genetic maternal effect

________________genomic imprinting

________________sex-influenced trait

________________anticipation

• a. the percentage of individuals with a particular genotype that express the expected phenotype
• b. a trait determined by an autosomal gene that is more easily expressed in one sex
• c. a trait determined by an autosomal gene that is expressed in only one sex
• d. a trait that is determined by an environmental effect and has the same phenotype as a genetically determined trait
• e. a trait determined by genes at many loci
• f. the expression of a trait is affected by the sex of the parent that transmits the gene to the offspring
• g. the trait appears earlier or more severely in succeeding generations
• h. a gene affects more than one phenotype
• i. the genotype of the maternal parent influences the phenotype of the offspring

Question 5.13

Palomino horses have a golden yellow coat, chestnut horses have a brown coat, and cremello horses have a coat that is almost white. A series of crosses between the three different types of horses produced the following offspring:

• a. Explain the inheritance of the palomino, chestnut, and cremello phenotypes in horses.
• b. Assign symbols for the alleles that determine these phenotypes, and list the genotypes of all parents and offspring given in the preceding table.

Coat color: (a) palomino; (b) chestnut; (c) cremello.

[Part a: Keith J. Smith/Alamy. Part b: jiri jura/iStockphoto.com. Part c: Olga_i/Shutterstock.]

Question 5.14

The L^M and L^N alleles at the MN blood-group locus exhibit codominance. Give the expected genotypes and phenotypes and their ratios in progeny resulting from the following crosses.

• a. L^ML^M × L^ML^N
• b. L^NL^N × L^NL^N
• c. L^ML^N × L^ML^N
• d. L^ML^N × L^NL^N
• e. L^ML^M × L^NL^N

Question 5.15

Assume that long ear lobes in humans are an autosomal dominant trait that exhibits 30% penetrance. A person who is heterozygous for long ear lobes mates with a person who is homozygous for normal ear lobes. What is the probability that their first child will have long ear lobes?

Question 5.16

Club foot is one of the most common congenital skeletal abnormalities, with a worldwide incidence of about 1 in 1000 births. Both genetic and nongenetic factors are thought to be responsible for club foot. C. A. Gurnett et al. (2008. American Journal of Human Genetics 83:616–622) identified a family in which club foot was inherited as an autosomal dominant trait with reduced penetrance. They discovered a mutation in the PITXI gene that caused club foot in this family. Through DNA testing, they determined that 11 people in the family carried the PITXI mutation, but only 8 of these people had club foot. What is the penetrance of the PITXI mutation in this family?

Question 5.17

When a Chinese hamster with white spots is crossed with another hamster that has no spots, approximately of the offspring have white spots and have no spots. When two hamsters with white spots are crossed, of the offspring possess white spots and have no spots.

• a. What is the genetic basis of white spotting in Chinese hamsters?
• b. How might you go about producing Chinese hamsters that breed true for white spotting?

Question 5.18

In the early 1900s, Lucien Cuénot studied the genetic basis of yellow coat color in mice. He carried out a number of crosses between two yellow mice and obtained what he thought was a 3 : 1 ratio of yellow to gray mice in the progeny. The following table gives Cuénot’s actual results, along with the results of a much larger series of crosses carried out by Castle and Little (W E. Castle and C. C. Little. 1910. Science 32:868–870).

Progeny Resulting from Crosses of Yellow × Yellow Mice

• a. Using a chi-square test, determine whether Cuénot’s results are significantly different from the 3 : 1 ratio that he thought he observed. Are they different from a 2 : 1 ratio?
• b. Determine whether Castle and Little’s results are significantly different from a 3 : 1 ratio. Are they different from a 2 : 1 ratio?
• c. Combine the results of Castle and Cuénot and determine whether they are significantly different from a 3 : 1 ratio and a 2 : 1 ratio.
• d. Offer an explanation for the different ratios that Cuénot and Castle obtained.

Question 5.19

In the pearl-millet plant, color is determined by three alleles at a single locus: Rp¹ (red), Rp² (purple), and rp (green). Red is dominant over purple and green, and purple is dominant over green (Rp¹ > Rp² > rp). Give the expected phenotypes and ratios of offspring produced by the following crosses.

• a. Rp¹/Rp² × Rp¹/rp
• b. Rp¹/rp × Rp²/rp
• c. Rp¹/Rp² × Rp¹/Rp²
• d. Rp²/rp × rp/rp
• e. rp/rp × Rp¹/Rp²

Question 5.20

If there are five alleles at a locus, how many genotypes may there be at this locus? How many different kinds of homozygotes will there be? How many genotypes and homozygotes may there be with eight alleles at a locus?

Question 5.21

Turkeys have black, bronze, or black-bronze plumage. Examine the results of the following crosses:

Parents	Offspring
Cross 1: black and bronze	all black
Cross 2: black and black	black, bronze
Cross 3: black-bronze and	all black-bronze
black-bronze
Cross 4: black and bronze	black, bronze,
	black-bronze
Cross 5: bronze and	bronze, black-bronze
black-bronze
Cross 6: bronze and bronze	bronze, black-bronze

Do you think these differences in plumage arise from incomplete dominance between two alleles at a single locus? If yes, support your conclusion by assigning symbols to each allele and providing genotypes for all turkeys in the crosses. If your answer is no, provide an alternative explanation and assign genotypes to all turkeys in the crosses.

Question 5.22

In rabbits, an allelic series helps to determine coat color: C (full color), c^ch (chinchilla, gray color), c^h (Himalayan, white with black extremities), and c (albino, all-white). The C allele is dominant over all others, c^ch is dominant over c^h and c, c^h is dominant over c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > c^ch > c^h > c. The rabbits in the following list are crossed and produce the progeny shown. Give the genotypes of the parents for each cross:

Phenotypes of parents	Phenotypes of offspring
a. full color × albino	full color, albino
b. Himalayan × albino	Himalayan, albino
c. full color × albino	full color, chinchilla
d. full color × Himalayan	full color, Himalayan, albino
e. full color × full color	full color, albino

Question 5.23

In this chapter, we considered Joan Barry’s paternity suit against Charlie Chaplin and how, on the basis of blood types, Chaplin could not have been the father of her child.

• a. What blood types are possible for the father of Barry’s child?
• b. If Chaplin had possessed one of these blood types, would that prove that he fathered Barry’s child?

Question 5.24

A woman has blood-type AM. She has a child with blood-type AB MN. Which of the following blood types could not be that of the child’s father? Explain your reasoning.

Question 5.25

In chickens, comb shape is determined by alleles at two loci (R, r and P, p). A walnut comb is produced when at least one dominant allele R is present at one locus and at least one dominant allele P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant allele is present at the first locus and two recessive alleles are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive alleles are present at the first locus and at least one dominant allele is present at the second (genotype rr P_). If two recessive alleles are present at the first and at the second locus (rr pp), a single comb is produced. Progeny with what types of combs and in what proportions will result from the following crosses?

• a. RR PP × rr pp
• b. Rr Pp × rr pp
• c. Rr Pp × Rr Pp
• d. Rr pp × Rr pp
• e. Rr pp × rr Pp
• f. Rr pp × rr pp

Comb shape: (a) walnut; (b) rose; (c) pea; (d) single.

[Parts a and d: Robert Dowling/Corbis. Part b: Robert Maier/Animals Animals. Part c: Dapne Godfrey Trust/Animals Animals.]

Question 5.26

Tatuo Aida investigated the genetic basis of color variation in the Medaka (Aplocheilus latipes), a small fish found in Japan (T. Aida. 1921. Genetics 6:554–573). Aida found that genes at two loci (B, b and R, r) determine the color of the fish: fish with a dominant allele at both loci (B_R_) are brown, fish with a dominant allele at the B locus only (B_rr) are blue, fish with a dominant allele at the R locus only (bb R_) are red, and fish with recessive alleles at both loci (bb rr) are white. Aida crossed a homozygous brown fish with a homozygous white fish. He then backcrossed the F₁ with the homozygous white parent and obtained 228 brown fish, 230 blue fish, 237 red fish, and 222 white fish.

• a. Give the genotypes of the backcross progeny.
• b. Use a chi-square test to compare the observed numbers of backcross progeny with the number expected. What conclusion can you make from your chi-square results?
• c. What results would you expect for a cross between a homozygous red fish and a white fish?
• d. What results would you expect if you crossed a homozygous red fish with a homozygous blue fish and then backcrossed the F₁ with a homozygous red parental fish?

Question 5.27

A variety of opium poppy (Papaver somniferum L.) with lacerate leaves was crossed with a variety that has normal leaves. All the F₁ had lacerate leaves. Two F₁ plants were interbred to produce the F₂. Of the F₂, 249 had lacerate leaves and 16 had normal leaves. Give genotypes for all the plants in the P, F₁, and F₂ generations. Explain how lacerate leaves are determined in the opium poppy.

Question 5.28

E. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings, 853 virescent-white seedlings, and 260 yellow seedlings (E. W. Lindstrom. 1921. Genetics 6:91–110).

• a. Give the genotypes for the green, virescent-white, and yellow progeny.
• b. Explain how color is determined in these seedlings.
• c. Is there epistasis among the genes that determine color in the corn seedlings? If so, which gene is epistatic and which is hypostatic?

Question 5.29

A dog breeder liked yellow and brown Labrador retrievers. In an attempt to produce yellow and brown puppies, he mated a yellow Labrador male and a brown Labrador female. Unfortunately, all the puppies produced in this cross were black.

• a. Explain this result.
• b. How might the breeder go about producing yellow and brown Labradors?

Coat color in Labrador retrievers: (a) black; (b) brown; (c) yellow.

[Parts a and b: Juniors Bildarchiv/Alamy. Part c: C. Byatt-Norman/Shutterstock.]

Question 5.30

When a yellow female Labrador retriever was mated with a brown male, half of the puppies were brown and half were yellow. The same female, when mated with a different brown male, produced only brown offspring. Explain these results.

Question 5.31

A summer-squash plant that produces disc-shaped fruit is crossed with a summer-squash plant that produces long fruit. All the F₁ have disc-shaped fruit. When the F₁ are intercrossed, F₂ progeny are produced in the following ratio: disc-shaped fruit : spherical fruit : long fruit. Give the genotypes of the F₂ progeny.

Question 5.32

Some sweet-pea plants have purple flowers and other plants have white flowers. A homozygous variety of pea that has purple flowers is crossed with a homozygous variety that has white flowers. All the F₁ have purple flowers. When these F₁ are self-fertilized, the F₂ appear in a ratio of purple to white.

• a. Give genotypes for the purple and white flowers in these crosses.
• b. Draw a hypothetical biochemical pathway to explain the production of purple and white flowers in sweet peas.

Question 5.33

• a. Why are Irish setters reddish in color?
• b. Can a poodle crossed with any other breed produce spotted puppies? Why or why not?
• c. If a St. Bernard is crossed with a Doberman, what will be the coat color of the offspring: solid, yellow, saddle, or bicolor?
• d. If a Rottweiler is crossed with a Labrador retriever, what will be the coat color of the offspring: solid, yellow, saddle, or bicolor?

Question 5.34

Male-limited precocious puberty results from a rare, sex-limited autosomal allele (P) that is dominant over the allele for normal puberty (p) and is expressed only in males. Bill undergoes precocious puberty, but his brother Jack and his sister Beth underwent puberty at the usual time, between the ages of 10 and 14. Although Bill’s mother and father underwent normal puberty, two of his maternal uncles (his mother’s brothers) underwent precocious puberty. All of Bill’s grandparents underwent normal puberty. Give the most likely genotypes for all the relatives mentioned in this family.

Question 5.35

In some goats, the presence of horns is produced by an autosomal gene that is dominant in males and recessive in females. A horned female is crossed with a hornless male. The F₁ offspring are intercrossed to produce the F₂. What proportion of the F₂ females will have horns?

Question 5.36

In goats, a beard is produced by an autosomal allele that is dominant in males and recessive in females. We’ll use the symbol B^b for the beard allele and B⁺ for the beardless allele. Another independently assorting autosomal allele that produces a black coat (W) is dominant over the allele for white coat (w). Give the phenotypes and their expected proportions for the following crosses.

• a. B⁺B^b Ww male × B⁺B^b Ww female
• b. B⁺B^b Ww male × B⁺B^b ww female
• c. B⁺B⁺ Ww male × B^bB^b Ww female
• d. B⁺B^b Ww male × B^bB^b ww female

Question 5.37

Cockfeathering in chickens is an autosomal recessive trait that is sex-limited to males. List all possible genotypes for the chicken shown in:

• a. Figure 5.13a
• b. Figure 5.13b
• c. Figure 5.13c

Question 5.38

J. K. Breitenbecher (1921. Genetics 6:65–86) investigated the genetic basis of color variation in the four-spotted cowpea weevil (Bruchus quadrimaculatus). The weevils were red, black, white, or tan. Breitenbecher found that four alleles (R, R^b, R^w, and r) at a single locus determine color. The alleles exhibit a dominance hierarchy, with red (R) dominant over all other alleles, black (R^b) dominant over white (R^w) and tan (r), white dominant over tan, and tan recessive to all others (R > R^b > R^w > r). The following genotypes encode each of the colors:

Color variation in this species is sex-limited to females: males carry color genes but are always tan regardless of their genotype. For each of the following crosses carried out by Breitenbecher, give all possible genotypes of the parents.

Question 5.39

Shell coiling of the snail Lymnaea peregra (discussed in the introduction to the chapter) results from a genetic maternal effect. An autosomal allele for a right-handed, or dextral, shell (s⁺) is dominant over the allele for a left-handed, or sinistral, shell (s). A pet snail called Martha is sinistral and reproduces only as a female (the snails are hermaphroditic). Indicate which of the following statements are true and which are false. Explain your reasoning in each case.

• a. Martha’s genotype must be ss.
• b. Martha’s genotype cannot be s⁺s⁺.
• c. All the offspring produced by Martha must be sinistral.
• d. At least some of the offspring produced by Martha must be sinistral.
• e. Martha’s mother must have been sinistral.
• f. All of Martha’s brothers must be sinistral.

Question 5.40

If the F₂ dextral snails with genotype s⁺s in Figure 5.17 undergo self-fertilization, what phenotypes and proportions are expected to occur in the progeny?

Question 5.41

Hypospadias, a birth defect in male humans in which the urethra opens on the shaft instead of at the tip of the penis, results from an autosomal dominant gene in some families. Females who carry the gene show no effects. Is this birth defect an example of (a) an X-linked trait, (b) a Y-linked trait, (c) a sex-limited trait, (d) a sex-influenced trait, or (e) genetic maternal effect? Explain your answer.

Question 5.42

In unicorns, two autosomal loci interact to determine the type of tail. One locus controls whether a tail is present at all; the allele for a tail (T) is dominant over the allele for tailless (t). If a unicorn has a tail, then alleles at a second locus determine whether the tail is curly or straight. Farmer Baldridge has two unicorns with curly tails: when he crosses them, of the progeny have curly tails, have straight tails, and do not have a tail. Give the genotypes of the parents and progeny in Farmer Baldridge’s cross. Explain how he obtained the 2 : 1 : 1 phenotypic ratio in his cross.

Question 5.43

In 1983, a sheep farmer in Oklahoma noticed in his flock a ram that possessed increased muscle mass in his hindquarters. Many of the offspring of this ram possessed the same trait, which became known as the callipyge mutant (callipyge is Greek for “beautiful buttocks”). The mutation that caused the callipyge phenotype was eventually mapped to a position on the sheep chromosome 18.

When the male callipyge offspring of the original mutant ram were crossed with normal females, they produced the following progeny: male callipyge, female callipyge, male normal, and female normal. When the female callipyge offspring of the original mutant ram were crossed with normal males, all of the offspring were normal. Analysis of the chromosomes of these offspring of callipyge females showed that half of them received a chromosome 18 with the allele encoding callipyge from their mother. Propose an explanation for the inheritance of the allele for callipyge. How might you test your explanation?

Question 5.44

Which of the following statements is an example of a phenocopy? Explain your reasoning.

• a. Phenylketonuria results from a recessive mutation that causes light skin as well as intellectual disability.
• b. Human height is influenced by genes at many different loci.
• c. Dwarf plants and mottled leaves in tomatoes are caused by separate genes that are linked.
• d. Vestigial wings in Drosophila are produced by a recessive mutation. This trait is also produced by high temperature during development.

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• e. Intelligence in humans is influenced by both genetic and environmental factors.

Question 5.45

Long ears in some dogs are an autosomal dominant trait. Two dogs mate and produce a litter in which 75% of the puppies have long ears. Of the dogs with long ears in this litter, are known to be phenocopies. What are the most likely genotypes of the two parents of this litter?

Question 5.46

The fly with vestigial wings shown in the lower-left hand corner of Figure 5.19 is crossed to the fly with normal wings shown in the upper-right hand corner of the figure. If the progeny are reared at 31^oC, what percentage will have vestigial wings?