Genetics A Conceptual Approach 5th Edition by Benjamin A. Pierce – Test Bank

 

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Test Bank for

Chapter 5: Extensions and Modifications of Basic Principles

 

Multiple Choice Questions

 

1. The R locus determines flower color in a new plant species. Plants that are genotype RR have red flowers, and plants that are rr have white flowers. However, Rr plants have pink flowers. What type of inheritance does this demonstrate for flower color in these plants?

 

1. Complete dominance
2. Incomplete dominance
3. Codominance
4. Complementation
5. Lethal alleles

 

Answer: b

Section 5.1

Comprehension

 

2. Interactions among the human ABO blood group alleles involve _______ and ________.

 

1. co-dominance; complete dominance
2. codominance; incomplete dominance
3. complete dominance; incomplete dominance
4. epistasis; complementation
5. continuous variation; environmental variation

 

Answer: a

Section 5.1

Comprehension

 

3. In the endangered African watchamakallit, the offspring of a true-breeding black parent and a true-breeding white parent are all gray. When the gray offspring are crossed among themselves, their offspring occur in a ratio of 1 black:2 gray:1 white. Upon close examination of the coats, each hair of a gray animal is gray. What is the mode of inheritance?

 

1. One gene pair with black dominant to white
2. One gene pair with codominance
3. One gene pair with incomplete dominance
4. Two gene pairs with recessive epistasis
5. Two gene pairs with duplicate genes

 

Answer: c

Section 5.1

Comprehension

 

4. Suppose that extra fingers and toes are caused by a recessive trait, but it appears in only 60% of homozygous recessive individuals. Two heterozygotes conceive a child. What is the probability that this child will have extra fingers and toes?

 

1. 05
2. 10
3. 15
4. 25
5. 33

 

Answer: c

Section 5.1

Comprehension

 

5. Polydactyly is the condition of having extra fingers or toes. Some polydactylous persons possess extra fingers or toes that are fully functional, whereas others possess only a small tag of extra skin. This is an example of

 

1. variable expressivity.
2. complete dominance.
3. independent assortment.
5. cytoplasmic inheritance.

 

Answer: a

Section 5.1

Comprehension

 

6. Achondroplasia is a common cause of dwarfism in humans. All individuals with achondroplasia are thought to be heterozygous at the locus that controls this trait. When two individuals with achondroplasia mate, the offspring occur in a ratio of 2 achondroplasia:1 normal. What is the most likely explanation for these observations?

 

1. Achondroplasia is incompletely dominant to the normal condition.
2. Achondroplasia is codominant to the normal condition.
3. The allele that causes achondroplasia is a dominant lethal allele.
4. The allele that causes achondroplasia is a recessive lethal allele.
5. The allele that causes achondroplasia is a late-onset lethal allele.

 

Answer: d

Section 5.1

Comprehension

 

7. Crossing two yellow mice results in 2/3 yellow offspring and 1/3 nonyellow offspring. What percentage of offspring would you expect to be nonyellow if you crossed two nonyellow mice?

 

1. 25%
2. 33%
3. 66%
4. 75%
5. 100%

 

Answer: e

Section 5.1

Comprehension

 

8. In humans, blood types A and B are codominant to each other and each is dominant to O. What blood types are possible among the offspring of a couple of blood types AB and A?

 

1. A, B, AB, and O
2. A, B, and AB only
3. A and B only
4. A, B, and O only
5. A and AB only

 

Answer: b

Section 5.1

Comprehension

 

9. A mother of blood type A gives birth to a child with blood type O. Which of the following could NOT be the blood type of the father?

 

1. A
2. B
3. O
4. AB
5. Any of the above is a possible blood type of the father.

 

Answer: d

Section 5.1

Comprehension

 

10. You are studying body color in an African spider and have found that it is controlled by a single gene with four alleles: B (brown), b^r (red), b^g (green), and b^y (yellow). B is dominant to all the other alleles, and b^y is recessive to all the other alleles. The b^g allele is dominant to b^y but recessive to b^r. You cross a pure-breeding brown spider with a pure-breeding green spider. Predict the genotype of the progeny.

 

1. B/b^g
2. B^r/b^g
3. b^r/b^y
4. b^y/b^g
5. B/b^y

 

Answer: a

Section 5.1

Application

 

11. You are studying body color in an African spider and have found that it is controlled by a single gene with four alleles: B (brown), b^r (red), b^g (green), and b^y (yellow). B is dominant to all the other alleles, and b^y is recessive to all the other alleles. The b^g allele is dominant to b^y but recessive to b^r. You cross a pure-breeding brown spider with a pure-breeding green spider. Predict the phenotype of the progeny.

 

1. Half brown, half green
2. Three-fourths brown, one-fourth green
3. All brown
4. All green
5. All yellow

 

Answer: c

Section 5.1

Application

 

12. 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 to all others, c^ch is dominant to c^h and c, c^h is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > c^ch > c^h > c. Indicate the phenotypic ratios expected if rabbits with the cross Cc^ch ´ Cc^h.

 

1. 1 full color:1 chinchilla
2. 1 full color:1 Himalayan
3. 1 chinchilla:1 Himalayan
4. 3 full color:1 chinchilla
5. 2 full color:1 Himalayan:1 albino

 

Answer: d

Section 5.1

Application

 

13. 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 to all others, c^ch is dominant to c^h and c, c^h is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > c^ch > c^h > c. Indicate the phenotypic ratios expected if rabbits with the cross Cc^h ´ c^hc.

 

1. 1 full color:1 chinchilla
2. 1 full color:1 Himalayan
3. 1 chinchilla:1 Himalayan
4. 3 full color:1 chinchilla
5. 2 full color:1 Himalayan:1 albino

 

Answer: b

Section 5.1

Application

 

14. 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 to all others, c^ch is dominant to c^h and c, c^h is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > c^ch > c^h > c. Indicate the phenotypic ratios expected if rabbits with the cross Cc^h ´ cc.

 

1. 1 full color:1 chinchilla
2. 1 full color:1 Himalayan
3. 1 chinchilla:1 Himalayan
4. 3 full color:1 chinchilla
5. 2 full color:1 Himalayan:1 albino

 

Answer: b

Section 5.1

Application

 

15. 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 to all others, c^ch is dominant to c^h and c, c^h is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > c^ch > c^h > c. Indicate the phenotypic ratios expected if rabbits with the cross c^chc^h ´ c^hc.

 

1. 1 full color:1 chinchilla
2. 1 full color:1 Himalayan
3. 1 chinchilla:1 Himalayan
4. 3 full color:1 chinchilla
5. 2 full color:1 Himalayan:1 albino

 

Answer: c

Section 5.1

Application

 

16. 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 to all others, c^ch is dominant to c^h and c, c^h is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > c^ch > c^h > c. Indicate the phenotypic ratios expected if rabbits with the cross Cc ´ c^hc.

 

1. 1 full color:1 chinchilla
2. 1 full color:1 Himalayan
3. 1 chinchilla:1 Himalayan
4. 3 full color:1 chinchilla
5. 2 full color:1 Himalayan:1 albino

 

Answer: e

Section 5.1

Application

 

17. A mother with blood type A has a child with blood type A. Give all possible blood types for the father of this child.

 

1. O
2. B, AB
3. A, AB
4. A, B, O
5. A, B, AB, O

 

Answer: e

Section 5.1

Application

 

18. A mother with blood type B has a child with blood type O. Give all possible blood types for the father of this child.

 

1. O
2. B, AB
3. A, AB
4. A, B, O
5. A, B, AB, O

 

Answer: d

Section 5.1

Application

 

19. A mother with blood type A has a child with blood type AB. Give all possible blood types for the father of this child.

 

1. O
2. B, AB
3. A, AB
4. A, B, O
5. A, B, AB, O

 

Answer: b

Section 5.1

Application

 

20. A mother with blood type AB has a child with blood type B. Give all possible blood types for the father of this child.

 

1. O
2. B, AB
3. A, AB
4. A, B, O
5. A, B, AB, O

 

Answer: e

Section 5.1

Application

 

21. You are studying a gene that controls ossicone (horn) length in giraffes. The wild-type long-ossicone allele (L) is dominant to the mutant short-ossicone (l) allele. However the L allele is only 60% penetrant. You cross two heterozygous giraffes. What proportion of offspring would you expect to exhibit the long ossicone phenotype? Assume the penetrance of L applies equally to both homozygotes and heterozygotes.

 

1. 40
2. 45
3. 55
4. 60
5. 75

 

Answer: b

Section 5.1

Application

 

22. You are studying a gene that controls ossicone (horn) length in giraffes. The wild-type long-ossicone allele (L) is dominant to the mutant short-ossicone (l) allele. However, the L allele is only 60% penetrant. You cross two heterozygous giraffes. What proportion of offspring would you expect to exhibit the short ossicone phenotype? Assume the penetrance of L applies equally to both homozygotes and heterozygotes.

 

1. 25
2. 40
3. 45
4. 55
5. 60

 

Answer: d

Section 5.1

Application

 

23. Hair color is determined in Labrador retrievers by alleles at the B and E A dominant allele B encodes black pigment, whereas a recessive allele b encodes brown pigment. Alleles at a second locus affect the deposition of the pigment in the shaft of the hair; dominant allele E allows dark pigment (black or brown) to be deposited, whereas recessive allele e prevents the deposition of dark pigment, causing the hair to be yellow. What type of gene interaction does this represent?

 

1. Recessive epistasis
2. Dominant epistasis
3. Duplicate recessive epistasis
4. Duplicate dominant epistasis
5. Dominant and recessive epistasis

 

Answer: a

Section 5.2

Comprehension

 

24. Hair color is determined in Labrador retrievers by alleles at the B and E A dominant allele B encodes black pigment, whereas a recessive allele b encodes brown pigment. Alleles at a second locus affect the deposition of the pigment in the shaft of the hair; dominant allele E allows dark pigment (black or brown) to be deposited, whereas recessive allele e prevents the deposition of dark pigment, causing the hair to be yellow. A black female Labrador retriever was mated with a yellow male. Half of the puppies were black and half were yellow. If the genotype of the black female parent was Bb Ee, then what was the genotype of the other parent?

 

1. bb ee
2. bb EE
3. Bb ee
4. BB ee
5. BB EE

 

Answer: d

Section 5.2

Comprehension

 

25. Suppose that the “fabulous” phenotype is controlled by two genes, A and B, as shown in the diagram below. Allele A produces enough enzyme 1 to convert “plain” to “smashing.” Allele a produces no enzyme 1. Allele B produces enough enzyme 2 to convert “smashing” to “fabulous.” Allele b produces no enzyme 2. The A and B genes are both autosomal and assort independently.

 

 

 

What will be the phenotype(s) of the F₁ offspring of a true-breeding “fabulous” father and a true-breeding “plain” mother (aa bb)?

 

1. All “plain”
2. All “smashing”
3. All “fabulous”
4. Plain” females and “fabulous” males
5. “Fabulous” females and “smashing” males

 

Answer: c

Section 5.2

Comprehension

 

26. Suppose that the “fabulous” phenotype is controlled by two genes, A and B, as shown in the diagram below. Allele A produces enough enzyme 1 to convert “plain” to “smashing.” Allele a produces no enzyme 1. Allele B produces enough enzyme 2 to convert “smashing” to “fabulous.” Allele b produces no enzyme 2. The A and B genes are both autosomal and assort independently.

 

 

 

 

 

 

What will be the expected ratio of the F₂ offspring of the F₁ generation?

 

1. 9 “fabulous”:7 “plain”
2. 13 “fabulous”:3 “plain”
3. 9 “fabulous”:3 “smashing”:4 “plain”
4. 12 “plain”:3 “fabulous”:1 “smashing”
5. 15 “fabulous”:1 “smashing”

 

Answer: c

Section 5.2

Comprehension

 

27. Coat color is determined by two loci in large cats. Two pink panthers fall in love and produce a large litter of baby panthers with the following phenotypic ratios: 12/16 pink; 3/16 black; and 1/16 white. What is the genotype of the pink progeny?

 

1. A_ B_
2. A_ bb
3. aa B_
4. aa bb
5. A_ B_ and A_ bb

 

Answer: e

Section 5.2

Comprehension

 

28. Coat color is determined by two loci in large cats. Two pink panthers fall in love and produce a large litter of baby panthers with the following phenotypic ratios: 12/16 pink; 3/16 black; and 1/16 white. What is the genotype of the black progeny?

 

1. A_ B_
2. A_ bb
3. aa B_
4. aa bb
5. A_ B_ and A_ bb

 

Answer: c

Section 5.2

Comprehension

 

29. Coat color is determined by two loci in large cats. Two pink panthers fall in love and produce a large litter of baby panthers with the following phenotypic ratios: 12/16 pink; 3/16 black; and 1/16 white. What is the genotype of the white progeny?

 

1. A_ B_
2. A_ bb
3. aa B_
4. aa bb
5. A_ B_ and A_ bb

 

Answer: d

Section 5.2

Comprehension

 

30. Coat color is determined by two loci in large cats. Two pink panthers fall in love and produce a large litter of baby panthers with the following phenotypic ratios: 12/16 pink; 3/16 black; and 1/16 white. What kind of gene interaction is this?

 

1. Recessive epistasis
2. Dominant epistasis
3. Duplicate recessive epistasis
4. Duplicate dominant epistasis
5. Dominant and recessive epistasis

 

Answer: b

Section 5.2

Comprehension

 

31. Two loci control body color in beetles. In a cross between a black beetle and a white beetle you obtain a ratio of 9 black to 7 white beetles. What kind of gene interaction is this?

 

1. Recessive epistasis
2. Dominant epistasis
3. Duplicate recessive epistasis
4. Duplicate dominant epistasis
5. Dominant and recessive epistasis

 

Answer: c

Section 5.2

Comprehension

 

32. In order to determine if mutations from different organisms that exhibit the same phenotype are allelic, which test would you perform?

 

1. Test cross
2. Epistasis test
3. Complementation test
4. Allelic series test
5. Biochemical test

 

Answer: c

Section 5.2

Comprehension

 

33. In purple people eaters, purple is dominant to white. A true-breeding white mutant is mated with a different true-breeding white mutant. All of the F₁ are purple. When the purple F₁ offspring mate with each other, their offspring occur in the ratio of 9 purple:7 white. Which phenomenon explains the purple F₁ offspring?

 

1. Recessive epistasis
2. Dominant epistasis
3. Complementation
4. Mutation
5. Suppression

 

Answer: c

Section 5.2

Comprehension

 

34. The presence of a beard on some goats is determined by an autosomal gene that is dominant in males and recessive in females. Heterozygous males are bearded, while heterozygous females are beardless. What type of inheritance is exhibited by this trait?

 

1. Sex-linked
2. Sex-limited
3. Sex-influenced
4. Autosomal recessive
5. Autosomal dominant

 

Answer: c

Section 5.3

Comprehension

 

35. In the yawncat (a rare hypothetical animal), the dominant allele R causes solid tail color, and the recessive allele r results in white spots on a colored background. The black coat color allele B is dominant to the brown allele b, but these genes can only be expressed if the animal has an mm genotype at a third gene locus. Animals that are M_ are yellow regardless of which allele from the B locus is present. A mating between a solid yellow-tailed male yawncat and a solid brown-tailed female yawncat produces 16 offspring with the following tail phenotypes: six solid yellow, two spotted yellow, three solid black, one spotted black, three solid brown, and one spotted brown. What is the most likely genotype of the male parent?

 

1. MM BB RR
2. MM Bb RR
3. Mm Bb RR
4. Mm BB Rr
5. Mm Bb Rr

 

Answer: e

Section 5.2

Application

 

36. In the yawncat (a rare hypothetical animal), the dominant allele R causes solid tail color, and the recessive allele r results in white spots on a colored background. The black coat color allele B is dominant to the brown allele b, but these genes can only be expressed if the animal has an mm genotype at a third gene locus. Animals that are M_ are yellow regardless of which allele from the B locus is present. A mating between a solid yellow-tailed male yawncat and a solid brown-tailed female yawncat produces 16 offspring with the following tail phenotypes: six solid yellow, two spotted yellow, three solid black, one spotted black, three solid brown, and one spotted brown. What is the most likely genotype of the female parent?

 

1. mm bb rr
2. Mm bb rr
3. mm Bb rr
4. mm bb Rr
5. mm Bb Rr

 

Answer: d

Section 5.2

Application

 

37. In the yawncat (a rare hypothetical animal), the dominant allele R causes solid tail color, and the recessive allele r results in white spots on a colored background. The black coat color allele B is dominant to the brown allele b, but these genes can only be expressed if the animal has an mm genotype at a third gene locus. Animals that are M_ are yellow regardless of which allele from the B locus is present. A mating between a solid yellow-tailed male yawncat and a solid brown-tailed female yawncat produces 16 offspring with the following tail phenotypes: six solid yellow, two spotted yellow, three solid black, one spotted black, three solid brown, and one spotted brown. What is the probability of the next offspring from these same two parents having a spotted brown tail?

 

1. 1/2
2. 3/16
3. 1/4
4. 1/16
5. 9/16

 

Answer: d

Section 5.2

Application

 

38. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: Walnut crossed with single produces 1 walnut, 1 rose, 1 pea, and 1 single offspring.

 

1. RR PP × rr pp
2. RR Pp × rr pp
3. Rr PP × rr pp
4. Rr Pp × rr pp
5. Rr pp × rr pp

 

Answer: d

Section 5.2

Application

 

39. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: Rose crossed with pea produces 20 walnut offspring.

 

1. RR pp × rr PP
2. Rr pp × rr Pp
3. Rr pp × rr PP
4. RR pp × rr Pp
5. Rr pp × Rr Pp

 

Answer: a

Section 5.2

Application

 

40. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: Pea crossed with single produces 1 single offspring.

 

1. rr PP × rr pp
2. RR Pp × rr pp
3. Rr PP × rr pp
4. Rr Pp × rr pp
5. rr Pp × rr pp

 

Answer: e

Section 5.2

Application

 

41. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: Rose crossed with pea produces 2 walnut, 1 single, and 1 pea offspring.

 

1. RR pp × rr PP
2. Rr pp × rr Pp
3. Rr pp × rr PP
4. RR pp × rr Pp
5. Rr pp × Rr Pp

 

Answer: b

Section 5.2

Application

 

42. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: Rose crossed with single produces 31 rose offspring.

 

1. RR PP × rr pp
2. RR pp × rr pp
3. Rr PP × rr pp
4. Rr Pp × rr pp
5. Rr pp × rr pp

 

Answer: b

Section 5.2

Application

 

43. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: Rose crossed with single produces 10 rose and 11 single offspring.

 

1. RR PP × rr pp
2. RR Pp × rr pp
3. Rr PP × rr pp
4. Rr Pp × rr pp
5. Rr pp × rr pp

 

Answer: e

Section 5.2

Application

 

44. In a certain species of plant, flowers occur in three colors: blue, pink, and white. A pure-breeding pink plant is mated with a pure-breeding white plant. All of the F₁ are blue. When the blue F₁ plants are selfed, the F₂ occur in the ratio 9 blue:3 pink:4 white. How many gene pairs control the flower color phenotype?

 

1. 0
2. 1
3. 2
4. 3
5. 4

 

Answer: c

Section 5.2

Application

 

45. In a certain species of plant, flowers occur in three colors: blue, pink, and white. A pure-breeding pink plant is mated with a pure-breeding white plant. All of the F₁ are blue. When the blue F₁ plants are selfed, the F₂ occur in the ratio 9 blue:3 pink:4 white. What is the name for this type of interaction?

 

1. Recessive epistasis
2. Dominant epistasis
3. Duplicate recessive epistasis
4. Duplicate dominant epistasis
5. Dominant and recessive epistasis

 

Answer: a

Section 5.2

Application

 

46. In a certain species of plant, flowers occur in three colors: blue, pink, and white. A pure-breeding pink plant is mated with a pure-breeding white plant. All of the F₁ are blue. When the blue F₁ plants are selfed, the F₂ occur in the ratio 9 blue:3 pink:4 white. The genotype that produces white is ww. The presence of one W allele allows pink or blue color to occur. The alleles at the hypostatic locus are B (blue) and b (pink). What is the genotype of the pink parent?

 

1. bb WW
2. bb Ww
3. Bb Ww
4. Bb ww
5. BB ww

Answer: a

Section 5.2

Application

 

47. In a certain species of plant, flowers occur in three colors: blue, pink, and white. A pure-breeding pink plant is mated with a pure-breeding white plant. All of the F₁ are blue. When the blue F₁ plants are selfed, the F₂ occur in the ratio 9 blue:3 pink:4 white. The genotype that produces white is ww. The presence of one W allele allows pink or blue color to occur. The alleles at the hypostatic locus are B (blue) and b (pink). What is the genotype of the white parent?

 

1. bb WW
2. bb Ww
3. Bb Ww
4. Bb ww
5. BB ww

 

Answer: e

Section 5.2

Application

 

48. In a certain species of plant, flowers occur in three colors: blue, pink, and white. A pure-breeding pink plant is mated with a pure-breeding white plant. All of the F₁ are blue. When the blue F₁ plants are selfed, the F₂ occur in the ratio 9 blue:3 pink:4 white. The genotype that produces white is ww. The presence of one W allele allows pink or blue color to occur. The alleles at the hypostatic locus are B (blue) and b (pink). What is the genotype of the F₁ plants?

 

1. bb WW
2. bb Ww
3. Bb Ww
4. Bb ww
5. BB ww

 

Answer: c

Section 5.2

Application

 

49. In domestic chickens, some males display a plumage pattern called cock feathering. Other males and all females display a pattern called hen feathering. Cock feathering is an autosomal recessive trait that is exhibited in males only. What type of inheritance is exhibited by this trait?

 

1. Sex-linked
2. Sex-limited
3. Sex-influenced
4. Autosomal recessive
5. Autosomal dominant

 

Answer: b

Section 5.3

Comprehension

 

50. In domestic chickens, some males display a plumage pattern called cock feathering. Other males and all females display a pattern called hen feathering. Cock feathering is an autosomal recessive trait that is exhibited in males only. Two birds heterozygous for cock feathering are mated. What are the phenotypes of the parents?

 

1. Male with cock feathering, female with hen feathering
2. Male with hen feathering, female with cock feathering
3. Male with cock feathering, female with cock feathering
4. Male with hen feathering, female with hen feathering
5. Cannot be determined from the information given

 

Answer: d

Section 5.3

Comprehension

 

51. In domestic chickens, some males display a plumage pattern called cock feathering. Other males and all females display a pattern called hen feathering. Cock feathering is an autosomal recessive trait that is exhibited in males only. Two heterozygous birds are mated. What fraction of the total offspring is expected to exhibit cock feathering?

 

1. 0
2. 1/8
3. 1/4
4. 1/2
5. 3/4

 

Answer: b

Section 5.3

Comprehension

 

52. In domestic chickens, some males display a plumage pattern called cock feathering. Other males and all females display a pattern called hen feathering. Cock feathering is an autosomal recessive trait that is exhibited in males only. Two heterozygous birds are mated. What fraction of the male offspring is expected to exhibit cock feathering?

 

1. 0
2. 1/8
3. 1/4
4. 1/2
5. 3/4

 

Answer: c

Section 5.3

Comprehension

 

53. In domestic chickens, some males display a plumage pattern called cock feathering. Other males and all females display a pattern called hen feathering. Cock feathering is an autosomal recessive trait that is exhibited in males only. Two heterozygous birds are mated. What fraction of the female offspring is expected to exhibit cock feathering?

 

1. 0
2. 1/8
3. 1/4
4. 1/2
5. 3/4

 

Answer: a

Section 5.3

Comprehension

 

54. Male-limited precocious puberty results from a rare autosomal allele (P) that is dominant over the allele for normal puberty (p) and is expressed only in males. A male and female that both went through normal puberty have two sons. The first son undergoes precocious puberty but the second undergoes normal puberty. What is the genotype of the mother?

 

1. PP
2. Pp
3. pp
4. PP or Pp
5. Pp or pp

 

Answer: b

Section 5.3

Comprehension

 

55. Male-limited precocious puberty results from a rare autosomal allele (P) that is dominant over the allele for normal puberty (p) and is expressed only in males. A male and female that both went through normal puberty have two sons. The first son undergoes precocious puberty but the second undergoes normal puberty. What is the genotype of the father?

 

1. PP
2. Pp
3. pp
4. PP or Pp
5. Pp or pp

Answer: c

Section 5.3

Comprehension

 

56. Which organelle in an animal cell, in addition to the nucleus, contains genes?

 

1. Lysosome
2. Ribosome
3. Mitochondrion
4. Golgi body
5. Vesicle

 

Answer: c

Section 5.3

Comprehension

 

57. Which of the following is a characteristic exhibited by cytoplasmically inherited traits?

 

1. Present in both males and females
2. Usually inherited from one parent, typically the maternal parent
3. Reciprocal crosses give different results
4. Exhibit extensive phenotypic variation, even within a single family
5. All of the above

 

Answer: e

Section 5.3

Comprehension

 

58. Leber hereditary optic neuropathy (LHON) is a human disease that exhibits cytoplasmic inheritance. It is characterized by rapid loss of vision in both eyes, resulting from the death of cells in the optic nerve. A teenager loses vision in both eyes and is later diagnosed with LHON. How did this individual most likely inherit the mutant DNA responsible for this condition?

 

1. A nuclear gene from the father
2. A nuclear gene from the mother
3. A mitochondrial gene from the father
4. A mitochondrial gene from the mother
5. Any of the above are possible.

 

Answer: d

Section 5.3

Comprehension

 

59. Which statement correctly explains the difference between development of a queen bee and a worker bee?

 

1. Queens are females and workers are males.
2. Queens inherit a special chromosome that causes queen development.
3. Queens inherit a particular allele that causes queen development.
4. Queens are raised at a higher temperature, which alters gene expression.
5. Queens are fed a special diet, which alters gene expression.

 

Answer: e

Section 5.3

Comprehension

 

60. The bicoid mutation (bcd⁻) in fruit flies is inherited as a maternal effect recessive allele. What is the expected ratio of phenotypes in the offspring of a cross between a bcd⁺/bcd⁻ female and a bcd⁺/bcd⁻ male?

 

1. 1 normal:1 mutant
2. 3 normal:1 mutant
3. 3 mutant:1 normal
4. All normal
5. All mutant

 

Answer: d

Section 5.3

Comprehension

 

61. The phenomenon in which a gene’s expression is determined by its parental origin is called

 

1. sex-influenced.
2. sex-limited.
3. genomic imprinting.
4. maternal effect.
5. paternal effect.

 

Answer: c

Section 5.3

Comprehension

 

62. A deletion of a small region on the long arm of chromosome 15 causes a developmental disorder in children called Prader-Willi syndrome when the deletion is inherited from the father. However, the deletion of this same region of chromosome 15 can also be inherited from the mother, but this inheritance results in a completely different set of symptoms, called Angelman syndrome. What type of genetic phenomenon does this represent?

 

1. Sex-influenced
2. Genomic imprinting
3. Cytoplasmic inheritance
4. Maternal effect
5. Paternal effect

 

Answer: b

Section 5.3

Comprehension

 

63. What phenomenon describes a genetic trait that is expressed more strongly or earlier in development with each generation?

 

1. Epigenetics
2. Maternally determined progeny phenotypes
3. Epistasis
4. Anticipation
5. Genomic imprinting

 

Answer: d

Section 5.4

Comprehension

 

64. Huntington disease tends to strike earlier and lead to a more rapid progression of symptoms as it moves from generation to generation. Also, it tends to strike earlier and progress more rapidly when it is passed from the father to his children rather than from the mother. Which genetic phenomenon (or phenomena) is (are) likely operating here?

 

1. Incomplete penetrance
2. Sex-limited inheritance
3. Genetic anticipation
4. Parental imprinting
5. More than one of the above

 

Answer: e

Section 5.4

Comprehension

 

65. The Himalayan allele in rabbits produces dark fur at the extremities of the body—on the nose, ears, and feet. The dark pigment develops, however, only when a rabbit is reared at a temperature of 25°C or lower; if a Himalayan rabbit is reared at 30°C, no dark patches develop. What does this exemplify?

 

1. Dominance
2. Discontinuous characteristic
3. Genetic imprinting
4. Phenocopy
5. Temperature-sensitive allele

 

Answer: e

Section 5.5

Comprehension

 

66. The SRY gene is located on the Y chromosome. This single gene encodes a protein called a transcription factor that binds to DNA and stimulates the transcription of other genes that lead to the development of male sex characteristics, including physical, biochemical, and behavioral phenotypes. What concept in genetics best describes this example?

 

1. Dominance
2. Discontinuous characteristic
3. Polygenic characteristic
4. Phenocopy
5. Pleiotropy

 

Answer: e

Section 5.5

Comprehension

 

67. Multi-factorial traits are influenced by _______ and ________.

 

1. dominance; codominance
2. epistasis; pleiotropy
3. age; sex
4. genetic imprinting; reduced penetrance
5. polygenes; environment

 

Answer: e

Section 5.5

Comprehension

 

Short Answer Questions

 

68. A geneticist is examining a culture of fruit flies and discovers a single female with strange spots on her legs. The new mutation is named melanotic. When a female melanotic fly is crossed with a normal male, the following progeny are produced: 123 normal females, 125 melanotic females, and 124 normal males. In subsequent crosses with normal males, melanotic females are frequently obtained, but never any melanotic males. Provide a possible explanation for the inheritance of the melanotic mutation. (Hint: The cross produces twice as many female progeny as male progeny.)

 

Answer: These observations can be explained by a single X-linked locus with two segregating alleles. The skewed sex ratio (2 female:1 male) in the F₂ suggests a recessive lethal allele on the X chromosome that kills males that carry the lethal allele in one copy on their one X chromosome. The phenotype of the female parent also suggests that the allele is dominant for the melanotic trait. We will represent the mutant allele as M and the normal allele as +.

 

X^M/X⁺ (melanotic female parent) × X⁺/y (normal male parent)

 

 

 

1 X^M/X⁺ (melanotic female zygote)

1 X^M/y (inviable male zygote)

1 X⁺/X⁺ (normal female zygote)

1 X⁺/y (normal male zygotes)

 

Section 5.1

Application

 

69. How do incomplete and co-dominance differ?

 

Answer: Incompletely dominant traits show an intermediate phenotype in the heterozygote, while co-dominant traits show both phenotypes in the heterozygote (e.g., AB alleles of blood type).

Section 5.1

Application

 

70. Describe the differences between dominance, co-dominance, and incomplete dominance.

 

Answer:

• Dominance is the condition in which one allele of a gene pair completely masks or inhibits phenotypic expression of the other allele.
• Co-dominance is the condition in which the complete expression of both alleles of a given gene pair is observed in heterozygotes; that is, the expression of neither allele influences the expression of the other.
• Incomplete (or partial) dominance is the condition in which one allele only partially inhibits the expression of the other allele in the phenotype. Heterozygotes exhibit phenotypes that are intermediate between those of the two homozygotes.

Section 5.1

Application

 

71. How do incomplete penetrance and variable expressivity differ?

 

Answer:

• If some individuals in a population don’t express a trait, even though they have the corresponding genotype, the trait is said to exhibit incomplete penetrance in that population. When using the term penetrance, therefore, think of populations. Polydactyly (extra fingers and toes) exhibits incomplete penetrance in human populations.
• A trait exhibiting variable expressivity is not expressed at the same degree among all individuals expressing it. Male pattern baldness in humans is an example of a trait that exhibits variable expressivity.

Note: For incomplete penetrance, not everyone with the genotype will express the phenotype. For variable expressivity, everyone with the genotype expresses the phenotype to some degree. Of course, some traits may (and often do) exhibit both incomplete penetrance and variable expressivity.

Section 5.1

Application

 

72. How does epistasis differ from Mendel’s principle of dominance?

 

Answer: Phenotypic expression is often the result of products produced by multi-step metabolic pathways involving several different genes; each gene encodes an enzyme that regulates a specific biochemical step or event. Epistasis refers to the interaction among two or more genes that control a common pathway. For example, a mutation in any single gene contributing to a metabolic pathway can affect the expression of other genes in the pathway, and, of course, the final phenotype, depending on which biochemical step that gene controls.

 

Epistasis thus involves interaction among alleles located at different gene loci. This is in contrast to dominance, which involves interaction between alleles located at the same gene locus.

Section 5.2

Application

 

73. What is a dominant epistatic gene?

 

Answer: A dominant allele that, if present, determines the phenotype of a given trait regardless of which alleles at other loci are present.

Section 5.2

Application

 

74. A homozygous strain of corn that produces yellow kernels is crossed with another homozygous strain that produces purple kernels. When the F₁ are interbred, 280 of the F₂ are yellow and 70 are purple.

 

1. If kernel color is controlled by a single gene pair with yellow dominant to purple, what would by the expected ratio of yellow to purple in the F₂?
2. Do the observed data differ significantly from that expected in (a)? Explain your answer.
3. Provide an alternative explanation for the inheritance of kernel color and evaluate it by comparing the observed data to that expected from your alternative hypothesis.

 

Answer:

1. Because there are only two progeny classes, the simplest explanation is monohybrid inheritance with an expected ratio of 3 yellow:1 purple.
2. A chi-square test of the observed numbers using an expected 3:1 ratio suggests rejection of this hypothesis (Χ² = 4.7, 0.025 < p < 0.05).
3. So, the next hypothesis to test is dihybrid inheritance. However, the progeny clearly don’t segregate in a classic Mendelian dihybrid ratio (9:3:3:1)—there are only two phenotypic classes. Therefore, some kind of epistasis is likely. There are three epistatic ratios with two phenotypic classes to test: 9:7, 15:1, and 13:3. Dividing each phenotypic class by 16 suggests that the 13:3 ratio is the closest. The 13:3 ratio is standard for the kind of epistasis called “dominant and recessive” interaction. In this particular kind of epistasis, only two F₂ phenotypes are generated, because a dominant genotype (e.g., A_) present at one locus and the recessive genotype at the other locus (bb) produce identical phenotypes in a 13:3 ratio (e.g., A_ B_, A_ bb, and aa bb produce one phenotype, and aa B_ produces another phenotype). To further substantiate that this is the correct ratio, a chi-square test can be done. In fact, among the alternatives only the 13:3 ratio and accompanying genetic hypothesis should not be rejected (Χ² = 0.35 with 0.9 < p < 0.5).

Section 5.2

Application

 

75. A yeast geneticist isolates two different haploid mutant yeast strains, Strain A and Strain B, which cannot grow unless the amino acid leucine is added to the growth media. Wild-type yeast strains can make their own leucine and do not require that it be added to the growth media. The geneticist discovers that each mutant yeast strain contains a single recessive mutation that leads to the observed leucine-requiring phenotype. When she crosses the two mutant strains together, she observes that the resulting diploid can grow without leucine added to the growth media. Explain the allelic relationship between the mutations in these two strains.

 

Answer: The mutations in strains A and B are NOT allelic because complementation was observed. Strain A contains a mutation at gene A, which is recessive (a), and strain B contains a mutation at a separate genetic locus, gene B, which is also recessive. Strain A contains a wild-type B gene and strain B contains a wild-type A gene. These wild-type genes complement the corresponding mutant alleles in the diploid.

Section 5.2

Application

 

76. Discuss the difference between “cytoplasmic inheritance” and “genetic maternal effect.”

 

Answer:

• In cytoplasmic inheritance, the genes controlling a given trait are inherited exclusively from the mother (through cytoplasmic organelles such as mitochondria) and can be expressed in both male and female progeny.
• In the genetic maternal effect, each individual’s phenotype is determined by the genotype of the mother. Typically, the offspring’s phenotype is determined by mRNA or protein factors loaded into the oocyte and encoded by the mother’s genome. So while genes related to the trait are inherited from both parents (not so for the cytoplasmic inheritance), in a given generation, phenotype is determined exclusively by the mother’s, not the offspring’s, genotype.

Section 5.3

Application

 

77. Queen and worker bees inherit the same genetic information from their parents. Explain the mechanism by which queen bee development deviates from that of workers.

 

Answer: During development, queen bees are fed a special substance called royal jelly. Royal jelly somehow causes different genes to be active during development of queens compared to worker bees. Recent research has shown that royal jelly silences a gene called Dnmt3, whose product when active methylates DNA. In the absence of methylation of DNA by Dmnt3, DNA is less methylated in cells of developing queens, and many genes that are normally inactive in workers are activated, leading to queen development.

Section 5.3

Application

 

78. In some plant species, a single pair of alleles is involved in both flower color and stem color. For example, a plant with red flowers may also have red stems, whereas white-flowered varieties of the same species have green stems. How would you explain this observation?

 

Answer: This phenomenon, called pleiotropy, is the condition where a single gene affects multiple, apparently unrelated, phenotypic traits. In many other cases of pleiotropy, a single gene affects more than two phenotypic traits. For example, a mutant white-eye gene in Drosophila (fruit fly) also affects the structure and color of internal organs, causes reduced fertility, and decreases life expectancy. Another example involves sickle-cell anemia in humans (caused by a single nucleotide change in a hemoglobin gene), which has adverse effects on different organs and tissues.

Section 5.5

Comprehension

 

79. Two mice of the same species have different ear shapes. You find that one mouse, having normal shaped ears, was caught in a field in Kenya. The other mouse, with curled ears, was caught in the frozen tundra of Greenland. You have determined that both mice have identical genotypes at the gene loci controlling ear shape. How would you explain the differences in ear shape?

 

Answer: Because both mice have the same genotype at the relevant loci controlling ear shape, there is most likely an effect of environment on phenotype. Because Kenya and Greenland have quite different climates (but also different food sources, humidity, sunlight intensities, etc.), it is possible that the very different temperature ranges within each region resulted in differential expression of identical genotypes for ear shape in each mouse—for example, differential expression of temperature-sensitive allele(s) involved in ear development. However, the phenotypic differences could also be a result of differences in diet, light conditions, exposure to chemicals, nutrition, or a range of other non-genetic factors.

Section 5.5

Application

 

80. With regard to the ear shape phenotypes described in the previous question, how could you test the relative importance of environmental and genetic factors?

 

Answer: Develop true-breeding strains of mice for normal and curled ears in their original habitats. Then, rear and observe one pair of each (one control pair and one test pair) true-breeding strain in each habitat in Kenya and Greenland, mating them with only their fellow Greenland or Kenyan siblings and raising all offspring under the same conditions as originally present for the native species (e.g., the original location outside). If the phenotypes of the experimental and control offspring reared in Kenya are all the same (normal) and the phenotypes of both sets of mice in Greenland are all the same (curled), then the two-ear phenotypes are caused by environment. For example, they may result from temperature-sensitive alleles. Note that strict temperature sensitivity could be tested under laboratory conditions without the need to transport mice to different countries. If the phenotypes of the two sets of mice at any one geographic location are not all the same, then there is a genetic component involved in ear shape. Note that there may be both genetic and environmental components.

Section 5.5

Application

 

81. Explain the differences between incomplete dominance and continuous variation.

 

Answer:

• In incomplete dominance, there will be three distinct phenotypes because the phenotype of a heterozygote is intermediate in appearance between the phenotypes of the two homozygotes. Incomplete dominance involves a single gene locus.
• Continuous variation refers to phenotypic variation exhibited by quantitative traits that are overlapping and distributed from one extreme to another. The continuous variation of quantitative traits is usually controlled by several genes whose alleles have an additive effect on the phenotype.

Section 5.5

Application

 

82. You observe continuous variation in tail length in a wild population of rats. How would you determine whether this variation is an example of variable expressivity or polygenic inheritance?

 

Answer: Take male and female rats from each phenotypic extreme (shortest and longest tails). Interbreed short with short and long with long under controlled laboratory conditions for several generations. If this is polygenic inheritance, then you will be able to develop different homozygous lines for short and long tails. But, if after several generations each line continues to produce progeny classes exhibiting significant variance in tail length, you could assume variable expressivity is the primary basis for the variation because the genotypes for each extreme line are (theoretically) homozygous and isogenic. Therefore, variances in tail length observed within each line cannot be the result of variable polygenic genotypes.

Section 5.5

Application

 

83. You are studying a coat color gene (B, brown) in Mexican bats. You have isolated a recessive allele (b) that causes yellow coat color, but you suspect that the phenotype may be sensitive to environmental conditions. To test your hypothesis, you examine the segregation ratio of phenotypes in F₁ progeny from a cross between two heterozygotes. You do this once at normal laboratory temperatures (28°C) and once at temperatures closer to their native habitat (34°C) and record the following data:

 

Brown             Yellow

28°C    153                  47

34°C    170                  30

 

1. What ratio do you expect in each experiment if temperature does not affect the phenotype?
2. What test can you use to determine if the ratio you observed is significantly different from the expected ratio?
3. Using that statistical test, is either observed ratio more different from the expected ratio than one would expect from chance alone? If so, suggest a biological explanation.

 

Answer:

1. This is a simple monohybrid cross with brown dominant to yellow, so expect 3 brown:1 yellow.
2. The chi-square test.
3. The chi-square test for the treatment at 34°C yields a value of c² = 10.67, indicating a significant difference from the expected ratio of 3:1. This suggests that elevated temperatures reduce the penetrance of the yellow phenotype. The chi-square test for the treatment at 28°C yields a value of 0.08, indicating that this data fits the expected ratio of 3:1.

Section 5.5

Application

 

84. List at least four phenomena that can alter expected Mendelian phenotypic ratios in genetic crosses.

 

Answer:

• Linkage
• Epistasis
• X-linked genes
• Lethal recessive alleles
• Environmental effects
• Continuous traits
• Variable expressivity

Section 5.5

Application

 

85. Cloning is a procedure by which exact genetic duplicates are made. Using cloning techniques, you have produced 10 cloned cows. However, the fur color of each of the calves looks very different from one another. Explain why this might have occurred.

 

Answer: A given phenotype arises from a genotype that develops within a particular environment. How the phenotype develops is determined by the effects of genes and environmental factors, and the balance between these influences varies from character to character. Since we are told that the calves are genetically identical, there must be environmental variation that explains the phenotypic differences. Even within the “constant” environment of a cow’s womb, there is environmental variation!

Section 5.5

Application

 

86. Explain how a phenotype like height in a tree can be due to the influence of both genes and environment.

 

Answer: The height reached by a tree at maturity is a phenotype that is strongly influenced by environmental factors, such as the availability of water, sunlight, and nutrients. Nevertheless, the tree’s genotype still imposes some limits on its height: an oak tree will never grow to be 300 meters tall no matter how much sunlight, water, and fertilizer are provided.

Section 5.5

Application