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

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INSTANT DOWNLOAD COMPLETE TEST BANK WITH ANSWERS

 

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

 

Sample  Questions

 

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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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.
  4. cytoplasmic inheritance.

 

Answer: a

Section 5.1

Comprehension

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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), br (red), bg (green), and by (yellow). B is dominant to all the other alleles, and by is recessive to all the other alleles. The bg allele is dominant to by but recessive to br. You cross a pure-breeding brown spider with a pure-breeding green spider. Predict the genotype of the progeny.

 

  1. B/bg
  2. Br/bg
  3. br/by
  4. by/bg
  5. B/by

 

Answer: a

Section 5.1

Application

 

  1. 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), br (red), bg (green), and by (yellow). B is dominant to all the other alleles, and by is recessive to all the other alleles. The bg allele is dominant to by but recessive to br. 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

 

  1. In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla; gray color), ch (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, cch is dominant to ch and c, ch is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > cch > ch > c. Indicate the phenotypic ratios expected if rabbits with the cross Ccch ´ Cch.

 

  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

 

  1. In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla; gray color), ch (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, cch is dominant to ch and c, ch is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > cch > ch > c. Indicate the phenotypic ratios expected if rabbits with the cross Cch ´ chc.

 

  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

 

  1. In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla; gray color), ch (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, cch is dominant to ch and c, ch is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > cch > ch > c. Indicate the phenotypic ratios expected if rabbits with the cross Cch ´ 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

 

  1. In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla; gray color), ch (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, cch is dominant to ch and c, ch is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > cch > ch > c. Indicate the phenotypic ratios expected if rabbits with the cross cchch ´ chc.

 

  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

 

  1. In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla; gray color), ch (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, cch is dominant to ch and c, ch is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > cch > ch > c. Indicate the phenotypic ratios expected if rabbits with the cross Cc ´ chc.

 

  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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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 +.

 

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

 

 

 

1 XM/X+ (melanotic female zygote)

1 XM/y (inviable male zygote)

1 X+/X+ (normal female zygote)

1 X+/y (normal male zygotes)

 

Section 5.1

Application

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

Test Bank for

Chapter 7: Linkage, Recombination, and Eukaryotic Gene Mapping

 

Multiple Choice Questions

 

  1. Linked genes always exhibit

 

  1. phenotypes that are similar.
  2. recombination frequencies of less than 50%.
  3. homozygosity when involved in a testcross.
  4. a greater number of recombinant offspring than parental offspring when involved in a testcross.
  5. a lack of recombinant offspring when a heterozygous parent is testcrossed.

 

Answer: b

Section 7.1

Comprehension

 

  1. Linked genes

 

  1. assort randomly.
  2. can’t crossover and recombine.
  3. are allelic.
  4. co-segregate.
  5. will segregate independently.

 

Answer: d

Section 7.1

Comprehension

 

  1. Recombination occurs through

 

  1. crossing over and chromosome interference.
  2. chromosome interference and independent assortment.
  3. somatic-cell hybridization and chromosome interference.
  4. complete linkage and chromosome interference.
  5. crossing over and independent assortment.

 

Answer: e

Section 7.1

Comprehension

 

  1. A genetic map shows which of the following?

 

  1. The distance in numbers of nucleotides between two genes
  2. The number of genes on each of the chromosomes of a species
  3. The linear order of genes on a chromosome
  4. The location of chromosomes in the nucleus when they line up at metaphase during mitosis
  5. The location of double crossovers that occur between two genes

 

Answer: c

Section 7.2

Comprehension

 

  1. A testcross includes

 

  1. one parent who is homozygous recessive for one gene pair and a second parent who is homozygous recessive for a second gene pair.
  2. one parent who is homozygous dominant for one or more genes and a second parent who is homozygous recessive for these same genes.
  3. two parents who are both heterozygous for two or more genes.
  4. one parent who shows the dominant phenotype for one or more genes and a second parent who is homozygous recessive for these genes.
  5. one parent who shows the recessive phenotype for one or more genes and a second parent who is homozygous dominant for these genes.

 

Answer: d

Section 7.2

Comprehension

 

  1. Recombination frequencies can be calculated by

 

  1. counting the number of recombinant and parental offspring when an individual who is heterozygous for two genes is testcrossed.
  2. performing a chi-square test on the F2 progeny when an individual who is homozygous for two genes is initially crossed to an individual who is homozygous recessive for these two genes.
  3. counting the number of offspring who are expressing the dominant phenotype when a heterozygous individual for two genes is testcrossed.
  4. performing a chi-square test of the progeny of a cross between parents who are both heterozygous for the same two genes.
  5. counting the number of offspring that are found in the cross of an individual who is heterozygous for two genes with another parent who is homozygous dominant for one of these genes and homozygous recessive for the other gene.

 

Answer: a

Section 7.2

Comprehension

 

  1. Assume that an individual of AB/ab genotype is testcrossed and four classes of testcross progeny are found in equal frequencies. Which of the following statements is TRUE?

 

  1. The genes A and B are on the same chromosome and closely linked.
  2. The genes A and B are on the same chromosome and very far apart.
  3. The genes A and B are probably between 10 and 20 map units apart on the same chromosome.
  4. The genes A and B are likely located on different chromosomes.
  5. Either b or d could be correct.

 

Answer: e

Section 7.2

Comprehension

 

  1. Is it possible for two different genes located on the same chromosome to assort independently?

 

  1. No, if two genes are on the same chromosome, they will be linked and the recombination frequency will be less than 50%.
  2. Yes, if the two genes are close enough to each other, there are a limited number of crossover events between them.
  3. No, there will be a very high crossover interference such that the recombination frequency will be reduced significantly.
  4. Yes, if the genes are far enough apart on the same chromosome, a crossover occurs between them in just about every meiotic event.
  5. Yes, but only if the two genes are both homozygous.

 

Answer: d

Section 7.2

Comprehension

 

  1. Genetic distances within a given linkage group

 

  1. cannot exceed 100 cM.
  2. are dependent on crossover frequencies between paired, non-sister chromatids.
  3. are measured in centiMorgans.
  4. cannot be determined.
  5. Both b and c are correct.

 

Answer: e

Section 7.2

Comprehension

 

  1. Crossing over occurs during

 

  1. late anaphase.
  2. early anaphase

 

Answer: b

Section 7.2

Comprehension

 

  1. What major contribution did Barbara McClintock and Harriet Creighton make to the study of recombination?

 

  1. Genetic recombination of alleles is associated with physical exchange between chromosomes.
  2. Genes were locate on chromosomes and the map distance between them could often be measured by the number of nucleotides in the DNA.
  3. Determining map distances in humans could be done by using pedigrees and calculating lod scores.
  4. Association studies allow genes that have no obvious phenotype to be accurately mapped.
  5. Crossing over does not occur in male Drosophila, so there is no genetic recombination.

 

Answer: a

Section 7.2

Comprehension

 

  1. You are doing lab work with a new species of beetle. You have isolated lines that breed true for either blue shells and long antenna, or green shells and short antenna. Crossing these lines yields F1 progeny with blue shells and long antenna. Crossing F1 progeny with beetles that have green shells and short antenna yield the following progeny:

 

Blue shell, long antenna                      82

Green shell, short antenna                   78

Blue shell, short antenna                     37

Green shell, long antenna                    43

Total                                                                240

 

A chi-square test is done to test for independent assortment. What is the resulting chi-square value and how many degree(s) of freedom should be used in its interpretation?

 

  1. 1 and one degree of freedom
  2. 9 and three degrees of freedom
  3. 9 and two degrees of freedom
  4. 1 and three degrees of freedom
  5. 42 and two degrees of freedom

 

Answer: d

Section 7.2

Application

 

  1. You are doing lab work with a new species of beetle. You have isolated lines that breed true for either blue shells and long antenna, or green shells and short antenna. Crossing these lines yields F1 progeny with blue shells and long antenna. Crossing F1 progeny with beetles that have green shells and short antenna yield the following progeny:

 

Blue shell, long antenna                      82

Green shell, short antenna                   78

Blue shell, short antenna                     37

Green shell, long antenna                    43

Total                                                                 240

 

Assuming that the genes are linked, what is the map distance between them in cM?

 

  1. 3 cM
  2. 0 cM
  3. 5 cM
  4. 0 cM
  5. The genes are actually assorting independently.

 

Answer: a

Section 7.2

Application

 

  1. Disney has been raising exotic fruit flies for decades. Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb. He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw. He is able to establish flies that are homozygous for both mutant traits. He mates these two strains with each other. Dr. Disney then takes phenotypically wild-type females from this cross and mates them with double recessive males. In the resulting testcross progeny, he observes 500 flies that are of the following make-up:

 

41      with baby blue eyes and pink wings

207    with baby blue eyes only

210    with pink wings only

42            with wild-type phenotype

 

Assuming the wild-type alleles for these two genes are b+ and pw+, what is the correct testcross of the F1 flies?

 

  1. b+ pw+/b pw ´ b pw/b pw
  2. b+ pw+/b pw ´ b pw+/b+ pw
  3. b+ pw/b pw+ ´ b pw/b pw
  4. b+ pw/b pw+ ´ b+ pw+/b pw
  5. b+ pw+/b pw ´ b+ pw/b pw+

 

Answer: c

Section 7.2

Application

 

  1. Disney has been raising exotic fruit flies for decades. Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb. He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw. He is able to establish flies that are homozygous for both mutant traits. He mates these two strains with each other. Dr. Disney then takes phenotypically wild-type females from this cross and mates them with double recessive males. In the resulting testcross progeny he observes 500 flies that are of the following make-up:

 

41      with baby blue eyes and pink wings

207    with baby blue eyes only

210    with pink wings only

42            with wild-type phenotype

 

What is the relationship with respect to location between the two genes?

 

  1. They are far apart on the same chromosome and assorting independently.
  2. They are linked and the map distance between them is 41.5 cM.
  3. They are on different chromosomes and assorting independently.
  4. They are linked and 16.6 cM apart.
  5. They are linked and 50.0 cM apart.

 

Answer: d

Section 7.2

Application

 

  1. Two linked genes, (A) and (B), are separated by 18 cM. A man with genotype Aa Bb marries a woman who is aa bb.  The man’s father was AA BB. What is the probability that their first child will be Aa bb?

 

  1. 18
  2. 41
  3. 09
  4. 25
  5. 50

 

Answer: c

Section 7.2

Application

 

  1. Two linked genes, (A) and (B), are separated by 18 cM. A man with genotype Aa Bb marries a woman who is aa bb. The man’s father was AA BB. What is the probability that their first two children will both be ab/ab?

 

  1. 168
  2. 0081
  3. 032
  4. 062
  5. 13

 

Answer: a

Section 7.2

Application

 

  1. You are studying two linked genes in lizards. You have two females and you know that both are the same genotype, heterozygous for both genes (A/a and B/b). You test-cross each female to a male that is fully homozygous recessive for both genes (a/a and b/b) and get the following progeny with the following phenotypes:

 

Female 1                      Female 2

AB – 37                      AB – 5

ab – 33                        ab – 4

Ab – 4                         Ab – 35

aB – 6                          aB – 36

 

How can you explain the drastic difference between these two crosses?

 

  1. The two genes are assorting independently in female 1 and are linked in female 2.
  2. The two genes are linked in female 1 and are assorting independently in female 2.
  3. The two alleles are in the coupling configuration in female 1 but in the repulsion configuration in female 2.
  4. The two alleles are in the repulsion configuration in female 1 but in the coupling configuration in female 2.
  5. The two genes are likely located on a sex chromosome in female 1 and are likely located on an autosome in female 2.

 

Answer: c

Section 7.2

Application

 

  1. Assume that A and B are two linked genes on an autosome in Drosophila. A testcross is made where AB/ab flies are crossed to ab/ab flies and the progeny are counted and shown below. However, it is known that the Aa/bb genotype is lethal before the flies hatch and does not appear among the testcross progeny counted. What is the most precise map distance that can be calculated from these data?

 

Aa Bb = 235

aa bb = 225

aa Bb = 20

  1. 2 cM
  2. 0 cM
  3. 4 cM
  4. 0 cM
  5. 50 cM

 

Answer: d

Section 7.2

Application

 

  1. If the recombination frequency between genes (A) and (B) is 5.3%, what is the distance between the genes in map units on the linkage map?

 

  1. 53 cM
  2. 3 CM
  3. 53 cM
  4. 6 cM
  5. 25 cM

 

Answer: b

Section 7.2

Application

 

  1. You are examining the following human pedigree and want to determine if the rare dominant disease allele (D) is linked to a specific DNA sequence location you are using as a molecular marker. Parental and progeny genotypes and phenotypes are indicated. Note that the father is a dihybrid at both loci, but the mother is homozygous at both loci. There is complete penetrance of the trait and a linkage phase of D-R1/d-R2 in the father. Assuming that the marker and the gene are linked, what is the best estimate of the map distance between the two loci?

 

 

D/d, R1/R2                             d/d, R1/R1

 

 

 

 

 

R1 marker detected:                           yes      yes                    yes                   yes                     yes       yes       yes         yes

R2 marker detected:                           no       yes         yes       no         yes        no        no         yes

 

  1. 12 cM
  2. 50 cM
  3. 16 cM
  4. 5 cM
  5. 25 cM

Answer: e

Section 7.2

Application

 

  1. In soreflies (a hypothetical insect), the dominant allele, L, is responsible for resistance to a common insecticide called Loritol. Another dominant allele, M, is responsible for the ability of soreflies to sing like birds. A true-breeding mute, Loritol-resistant sorefly was mated to a true-breeding singing, Loritol-sensitive sorefly, and the singing, Loritol-resistant female progeny were test-crossed with true-breeding wild type (i.e., mute, Loritol-sensitive) males. Of the 400 total progeny produced, 117 were mute and Loritol-resistant, 114 could sing and were Loritol-sensitive, 83 could sing and were Loritol-resistant, and 86 were mute and Loritol-sensitive. A chi-square test is done to determine if there is equal segregation of alleles at the L locus. What will be the chi-square value obtained and how many degrees of freedom would be used to interpret this value?

 

  1. 09 and one degree of freedom
  2. 56 and two degrees of freedom
  3. 0 and one degree of freedom
  4. 72 and four degrees of freedom
  5. A chi-square test is not the appropriate statistical test to answer this question.

 

Answer: c

Section 7.2

Application

 

  1. In soreflies (a hypothetical insect), the dominant allele, L, is responsible for resistance to a common insecticide called Loritol. Another dominant allele, M, is responsible for the ability of soreflies to sing like birds. A true-breeding mute, Loritol-resistant sorefly was mated to a true-breeding singing, Loritol-sensitive sorefly, and the singing, Loritol-resistant female progeny were test-crossed with true- breeding wild-type (i.e., mute, Loritol-sensitive) males. Of the 400 total progeny produced, 117 were mute and Loritol-resistant, 114 could sing and were Loritol-sensitive, 83 could sing and were Loritol-resistant, and 86 were mute and Loritol-sensitive. What will be the results of a chi-square test for independent assortment?

 

  1. 70 with three degrees of freedom
  2. 63 with three degrees of freedom
  3. 48 with four degrees of freedom
  4. 54 with one degree of freedom
  5. Because there are four classes of offspring, the genes must be assorting independently.

 

Answer: a

Section 7.2

Application

 

  1. Why are the progeny of a testcross generally used to map loci? Why not the F2 progeny from an F1 × F1 cross?

 

  1. Only recombinant offspring would be found in the progeny of an F1 ´ F1 cross.
  2. The progeny of an F1 ´ F1 cross would be found in a 9:3:3:1 ratio when two genes are involved, whereas the progeny of a testcross would result in a 1:1:1:1 ratio.
  3. It is easier to classify recombinant and parental offspring of a testcross than with the progeny of an F1 ´ F1 cross.
  4. In a testcross more of the progeny would be expected to display the dominant phenotype than in the progeny of an F1 ´ F1 cross.
  5. A testcross is more useful for mapping genes that are located near each other but when genes are quite far apart on the same chromosome, an F1 ´ F1 cross actually is more useful.

 

Answer: c

Section 7.2

Application

 

  1. In corn, small pollen (sp) is recessive to normal pollen (sp+) and banded necrotic tissue, called zebra necrotic (zn), is recessive to normal tissue (zn+). The genes that produce these phenotypes are closely linked on chromosome 10.  If no crossing over occurs between these two loci, give the types of progeny expected from the following cross:

 

×

 

  1. sp+ zn+/sp zn; sp zn/sp zn
  2. sp+ zn/sp zn; sp zn+/sp zn
  3. sp+ zn+/sp+ zn+; sp+ zn+/sp zn; sp zn/sp zn
  4. sp+ zn/sp+ zn;sp+ zn/sp zn+;sp zn+/sp+ zn; sp zn+/sp zn+
  5. sp+ zn+/sp zn; sp+ zn/sp zn

 

Answer: c

Section 7.2

Application

 

  1. In flower beetles, pygmy (py) is recessive to normal size (py+), and red color (r) is recessive to brown (r+). A beetle heterozygous for these characteristics was test crossed to a beetle homozygous for pygmy and red. The following are progeny phenotypes from this testcross.

 

py+      r+         180

py+      r           22

py                    r+         19

py                    r           191

Total                412

 

 

Carry out a series of chi-square tests to determine if there is equal segregation of alleles at the py locus.What is the correct chi-square value and how many degree(s) of freedom should be used in its interpretation?

 

  1. 16 with one degree of freedom
  2. 16 with three degrees of freedom
  3. 48 with one degree of freedom
  4. 48 with two degrees of freedom
  5. 56 with one degree of freedom

 

Answer: a

Section 7.2

Application

 

  1. In flower beetles, pygmy (py) is recessive to normal size (py+), and red color (r) is recessive to brown (r+). A beetle heterozygous for these characteristics was test crossed to a beetle homozygous for pygmy and red. The following are progeny phenotypes from this testcross.

 

py+      r+         180

py+      r           22

py                    r+         19

py                    r           191

Total                412

 

Carry out a series of chi-square tests to determine if the two loci are assorting independently. What is the correct chi-square value and how many degree(s) of freedom should be used in its interpretation?

 

  1. 112 with one degree of freedom
  2. 265 with three degrees of freedom
  3. 367 with four degree of freedom
  4. 5 with three degrees of freedom
  5. 367 with three degrees of freedom

 

Answer: b

Section 7.2

Application

 

  1. A series of two-point crosses among fruit flies is carried out between genes for brown eyes (bw), arc wings (a), vestigial wings (vg), ebony body color (e), and curved wings (cv). The following number of nonrecombinant and recombinant progeny were obtained from each cross:

 

Genes in cross Progeny (NR) Progeny (R)
a, bw 2224 117
a, cv 2609 823
a, e 3200 3200
a, vg 5172 2379
bw, cv 4614 1706
bw, e 4150 4150
bw, vg 2796 1434
cv, e 3116 3116
cv, vg 2102 305
vg, e 4559 4559

 

Using these data from two-point crosses, what it the best genetic map (in cM) that can be developed?

 

  1. cv 5 bw 13  a 34 vg with e assorting independently
  2. bw 5 cv 24 vg 32 a with e assorting independently
  3. a 5 bw 13 vg 24 e with vg assorting independently
  4. cv 13 bw 5 a 27 vg with e assorting independently
  5. bw 5 a 24 cv 13 vg with e assorting independently

 

Answer: e

Section 7.2

Application

 

  1. An individual has the following genotype. Gene loci (A) and (B) are 15 cM apart. What are the correct frequencies of some of the gametes that can be made by this individual?

 

A                   b

 

a                    B

 

  1. Ab = 7.5%; AB = 42.5%
  2. ab = 25%; aB = 50%
  3. AB = 7.5%; aB = 42.5%
  4. aB = 15%; Ab = 70%
  5. aB = 70%; Ab = 15%

 

Answer: c

Section 7.2

Application

 

  1. Two genes, A and B, are located 30 map units apart. The dihybrid shown below is mated to a tester aa bb. What proportion of the offspring is expected to be dominant for both traits?

 

 

 

 

 

 

 

  1. 0%
  2. 15%
  3. 30%
  4. 35%
  5. 70%

 

Answer: d

Section 7.2

Application

 

  1. You just bought two black guinea pigs, one male and one female, of the same genotype from the pet store that are known to be heterozygous (Bb). You also know that black fur (BB) is dominant over white fur (bb), and that a lethal recessive allele is located only one cM away from the recessive b allele and your animals are both heterozygous for this gene also. You decide to start raising your own guinea pigs, but after mating these animals several times, you discover they produce only black offspring among the first 12 progeny. How would you best explain this result?

 

  1. The B locus is on the X chromosome, so it can never produce a white phenotype.
  2. The B allele is actually codominant with the b allele, so a white phenotype cannot be produced.
  3. The recessive l allele is in tight coupling linkage with the b allele, so almost all of the bb offspring will also be ll and thus die before they can be observed.
  4. The dominant L allele is in tight repulsion linkage with the B allele, so it will be impossible to produce the Bb genotype that would express the white phenotype.
  5. Normally, it would be expected that 25% of the offspring would be white, but in this case, random deviations from the expected resulted in no white offspring.

 

Answer: c

Section 7.2

Application

 

  1. You just bought two black guinea pigs, one male and one female, of the same genotype from the pet store that are known to be heterozygous (Bb). You also know that black fur (BB) is dominant over white fur (bb) and that a lethal recessive allele is located only one cM away from the recessive b allele, and your animals are both heterozygous for this gene also. What is the probability of finding a white individual among the progeny if you cross these two animals?

 

  1. 25
  2. 002475
  3. 000025
  4. 004975
  5. 495

 

Answer: d

Section 7.2

Challenge

 

  1. What does lod stand for?

 

  1. Linkage over DNA
  2. Linkage of dihybrids
  3. Long overall distances (with respect to map distances)
  4. Linker of DNA
  5. Logarithm of odds

 

Answer: e

Section 7.3

Comprehension

 

  1. Lod scores measure

 

  1. the relatedness of two individuals.
  2. the number of crossover events that occur along an entire chromosome.
  3. how often double crossovers occur.
  4. the length of a linkage group.
  5. the likelihood of linkage between genes.

 

Answer: e

Section 7.3

Comprehension

 

  1. Three-factor testcrosses are only informative in gene mapping when which of the following occurs?

 

  1. One parent is homozygous recessive for the three genes, and the other parent is homozygous dominant.
  2. All three genes are located on separate chromosomes, and one parent is homozygous dominant for at least two of these genes.
  3. Both parents are homozygous for the three genes.
  4. One parent is heterozygous for the three genes, and the other parent is homozygous recessive.
  5. One of the genes must be located on a sex chromosome and be heterozygous, and the other two genes must be located on an autosome and be homozygous.

 

Answer: d 

Section 7.3

Comprehension

 

  1. A low coefficient of coincidence indicates that

 

  1. far fewer double-crossover recombinant progeny were recovered from a testcross than would be expected from the map distances of the genes involved.
  2. crossing over has been enhanced for genes that are located near the centromere of chromosomes because there is less interference of one crossover on the occurrence of a second crossover event.
  3. single-crossover recombinant classes in the progeny have been increased because the genes involved produce lethal phenotypes when in parental gene combinations.
  4. there is a large map distance between one of the outside genes in the heterozygous parent and the middle gene, while there is a short map distance between the middle gene and the other outside gene.
  5. the physical distance between two genes is very short compared with the genetic map distance between these two genes.

 

Answer: a 

Section 7.3

Comprehension

 

  1. The map distances for genes that are close to each other are more accurate than map distances for genes that are quite far apart because

 

  1. with genes that are far apart, double crossovers and other multiple-crossover events often lead to lethal recombinants that reduce the number of recombinant progeny.
  2. with genes that are far apart, double crossovers and other multiple-crossover events often lead to nonrecombinant or parental offspring and thus reduce the true map distance.
  3. crossover interference will cause more double crossovers and other multiple crossover events to occur than would be expected and thus result in a higher number of recombinant progeny than expected to occur with genes that are far apart.
  4. double crossovers and other multiple-crossover events occur more often when genes are close to each other and can be readily detected, so these map distances are more accurate than those for genes that are far apart.
  5. when genes are far apart, single-crossover recombinant classes are more difficult to detect than when genes are close together.

 

Answer: b

Section 7.3

Comprehension

 

  1. Interference occurs when

 

  1. two genes are assorting independently.
  2. two genes are far apart on a genetic map.
  3. one crossover inhibits another.
  4. the number recombinant progeny classes in the testcross of a heterozygote exceeds the number of parental progeny.
  5. a crossover causes the termination of the meiosis event in which the crossover is occurring.

 

Answer: c

Section 7.3

Comprehension

 

  1. A situation where the coefficient of coincidence greater than 1.0 would indicate that

 

  1. the interference is high and one crossover suppresses the occurrence of a second one.
  2. no double crossovers were found in the progeny of a testcross, even though some were expected based on probability.
  3. double crossovers were found in the progeny of a testcross, but there were fewer of them than would be expected based uon probability.
  4. there were more double crossovers in the progeny than would be expected based on probability.
  5. the genes involved were actually assorting independently.

 

Answer: d

Section 7.3

Comprehension

 

  1. In addition to determining genotypes, two- and three-factor testcrosses can be used to

 

  1. map gene loci.
  2. screen recessive mutants.
  3. measure heritability.
  4. determine parental origin.
  5. determine the physical location of genes.

 

Answer: a

Section 7.3

Comprehension

 

  1. A physical map often measures _____, whereas a genetic map measures ____.

 

  1. distances between chromosomes; distances between genes map units between genes;
  2. physical distances along the chromosome centiMorgans; base pairs distances in base
  3. pairs along the chromosome; centiMorgans
  4. between genes map units between genes; centiMorgans

 

Answer: d

Section 7.3

Comprehension

 

  1. What is a major difference in using lod-score analysis compared to using association studies in determining gene locations in humans?

 

  1. Lod-score analysis relies on family or pedigree data, while association studies use population data.
  2. Lod-score analysis requires that the loci being mapped must be on different chromosome arms, while association studies can map genes on different chromosomes.
  3. Association studies compare genotypes between parents and their children, while lod-score analysis compares genotypes between siblings of the same family.
  4. Lod-score analysis requires isolated human populations, while association studies require very large family pedigrees.
  5. Lod-score analysis requires a large number of genes with multiple alleles, while association studies can use genes that have only two alleles.

 

Answer: a

Section 7.3

Comprehension

 

  1. The results of linkage analysis for DNA marker A and the p53 gene are shown below. What is the best estimate for the approximate genetic distance between marker A and the p53 gene in humans?

 

Recombination values (cM)       1              5           10           20        30        40

lod score                          2.13         2.54       3.14       4.10    4.96     3.22

 

  1. 1 cM
  2. 5 cM
  3. 10 cM
  4. 20 cM
  5. 30 cM

 

Answer: e

Section 7.3

Application

 

  1. In maize (corn), assume that the genes A and B are linked and 30 map units apart. If a plant of Ab/aB is selfed, what proportion of the progeny would be expected to be of ab/ab genotype?

 

  1. 25%
  2. 15%
  3. 9%
  4. 30%
  5. 5%

 

Answer: a

Section 7.3

Application

 

  1. A testcross is performed on an individual to examine three linked genes. The most frequent phenotypes of the progeny were Abc and aBC, and the least frequent phenotypes were abc and ABC. What was the genotype of the heterozygous individual that is testcrossed with the correct order of the three genes?

 

  1. Abc aBC
  2. BAC/bac
  3. bcA/BCa
  4. aBc/AbC
  5. bAc/BaC

 

Answer: e

Section 7.3

Application

 

  1. In Drosophila melanogaster, cut wings (ct) is recessive to normal wings (ct+), sable body (s) is recessive to gray body (s+), and vermilion eyes (v) is recessive to red eyes (v+). All three recessive mutations are X-linked. A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes. The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross. The following are the progeny resulting from the testcross.

 

v          ct         s           510

v+        ct         s           1

v+        ct+       s           14

v+        ct+       s+        500

v+        ct         s+        73

v          ct         s+        20

v          ct+       s           81

v          ct+       s+        1

Total    à                    1200

 

What is the correct genetic map with respect to gene order and distances (in cM) for these three genes?

 

  1. s 13              ct      3        v
  2. s 3       v      13                ct
  3. v 13              ct      3        s
  4. s 26               v      3        ct
  5. ct 13              s      3        v

 

Answer: b

Section 7.3

Application

 

  1. In Drosophila melanogaster, cut wings (ct) is recessive to normal wings (ct+), sable body (s) is recessive to gray body (s+), and vermilion eyes (v) is recessive to red eyes (v+). All three recessive mutations are X-linked. A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes. The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross. The following are the progeny resulting from the testcross.

 

v          ct         s           510

v+        ct         s           1

v+        ct+       s           14

v+        ct+       s+        500

v+        ct         s+        73

v          ct         s+        20

v          ct+       s           81

v          ct+       s+        1

Total    à                    1200

 

What the interference value shown by this cross?

 

  1. 42
  2. 25
  3. 58
  4. −0.42
  5. 13

 

Answer: c

Section 7.3

Application

 

  1. You are studying three genes X, Y, and Z that are linked (in that order) in the Imperial Scorpion Pandinus imperator. The distance between X and Y is 10 cM, the distance between Y and Z is 8 cM. You conduct a testcross by crossing a heterozygous female with a homozygous recessive male and obtain 1500 testcross progeny. When the progeny are analyzed, you find 5 double-crossover offspring. What is the interference value shown by this cross?

 

  1. 008
  2. 42
  3. 12
  4. 58
  5. 22

 

Answer: d

Section 7.3

Application

 

  1. Consider the following three-point (trihybrid) testcross:

 

+                      +                               c

(x)  abc/abc

a                       b                               +

 

10.6 cM        13.4 cM

 

Calculate the number of double crossovers you would expect to observe if 1000 progeny result from this testcross assuming a coefficient of coincidence of 0.25.

 

  1. About 14
  2. About 26
  3. About 10
  4. About 4
  5. None, because there is crossover interference

 

Answer: d

Section 7.3

Challenge

 

  1. Consider the following three-factor (trihybrid) testcross:

 

+                      +                               c

(x)  abc/abc

a                       b                               +

 

10.6 cM        13.4 cM

 

Calculate the number of individuals of a+a bb c+c genotype if 1000 progeny result from this testcross.

 

  1. About 102
  2. About 46
  3. About 130
  4. About 65
  5. About 250

 

Answer: b

Section 7.3

Challenge

 

  1. Linked genes are

 

  1. on different chromosomes.
  2. on the same chromosome.
  3. recessive lethal.

 

Answer: e

Section 7.4

Comprehension

 

  1. A cell possessing two nuclei derived from different cells through cell fusion is called

 

  1. a heterokaryon.
  2. a haplotype.
  3. None of the above is correct.

 

Answer: a

Section 7.4

Comprehension

 

  1. Compared to a physical map, a genetic map

 

  1. is more accurate.
  2. is less accurate.
  3. is equally accurate.
  4. measures different things.
  5. cannot be made for humans.

 

Answer: d

Section 7.4

Comprehension

 

  1. In using somatic-cell hybridization experiments, a human gene was found to be located on chromosome 6. However, when lod-score analysis was done to detect linkage between this gene and a DNA marker locus also known to be on chromosome 6, no linkage could be found between the marker locus and the gene. What is the most likely explanation for this result?

 

  1. Somatic-cell hybridization experiments are not very accurate, and the gene may be on chromosome 5 or chromosome 7 instead of chromosome 6.
  2. Too few recombinants could be found to indicate linkage in the lod-score analysis.
  3. A lod-score analysis cannot be used when a DNA marker locus needs to mapped with respect to a gene locus.
  4. The gene and the DNA marker locus are so far apart on chromosome 6 that they assort independently.
  5. There were probably too few double-crossover events occurring between the two loci, so the lod score could not be determined accurately.

 

Answer: d

Section 7.4

Comprehension

 

  1. You and a colleague are working on a rare Peruvian llama that appears to be susceptible to diabetes, a disease related to insulin function. Your colleague has established a somatic-cell hybrid panel, and you would like to figure out to which llama chromosome the gene that encodes the llama insulin receptor maps. You also have an assay that allows you to detect which of the somatic-cell lines can produce the insulin receptor. You assay the colleague’s somatic-cell hybrid panel and get the following results. Which of the 74 llama chromosomes is the gene on?

 

 

Line # Llama chromosomes Assay
1 7, 10, 15, 29, 41, 55, 68  –
2 4, 8, 16, 31, 44, 56, 70  –
3 9, 12, 18, 52, 54, 61, 73  +
4 1, 8, 13, 22, 33, 45, 59  –
5 2, 11, 14, 18, 23, 49, 62  +
6 5, 12, 15, 24, 35, 47, 63  –
7 3, 17, 21, 32, 43, 58, 72  –
8 7, 21, 25, 36, 53, 64  –
9 1, 3, 9, 28, 37, 48, 65  –
10 13, 22, 27, 38, 50, 66, 19  –
11 4, 6, 12, 18, 20, 40, 67  +
12 2, 16, 26, 34, 51, 69  –
13 10, 14, 17, 39, 46, 57, 74  –
14 5, 11, 19, 30, 42, 60, 71  –

 

  1. 55
  2. 7
  3. 41
  4. 68
  5. 18

 

Answer: e

Section 7.4

Application

 

  1. A panel of cell lines was created from mouse-human somatic cell fusions. The following table indicates which human chromosomes are found in five cell lines (A, B, C, D, and E). A (+) or (–) in the following table indicates whether the enzyme glutathione S-transferase is present or absent in each cell line. Assume that a student is given the information provided in the table, along with information about the presence or absence of the enzyme in each cell line. On what human chromosome is the gene for glutathione S-transferase located?

 

Cell line    Enzyme             1        2        3       4      5        6      7       8      20     21      22   X

A         (+)       –          +          –          –          –          +          –          –          –          –          –   +

B         (+)       +          –          –          +          –          +          –          –          –          –          –   +

C         (–)        +          –          +          –          –          –          +          –          –          –          –   –

D         (–)        –          –          –          +          –          –          +          –          –          –          –   +

E          (+)       –          +          +          +          –          +          –          –          –          –          –   –

 

  1. X
  2. 2
  3. 8
  4. 6
  5. The gene must be located on a human chromosome not present in any of the cell lines above.

 

Answer: d

Section 7.4

Application

 

  1. You are working with five disease-resistance genes (A, B, C, D, and E) that are on chromosome 4 of Arabidopsis. You cross a line that is fully homozygous recessive for these five genes to each of five lines (recessive lethal lines) that are heterozygous for a different deletion. You observe the following results (wt = wild type; m = 50% of progeny displayed mutant phenotype):

 

Recessive lethal lines

1          2          3          4          5

A   wt        m         m         m         wt

Disease      B         m         m         wt        m         wt

Resistance C         m         wt        wt        m         wt

Phenotype D         wt        wt        m         wt        wt

E    m         wt        wt        wt        wt

 

What is the correct gene order for your five mutants?

 

  1. C A     B     D     E     (Either orientation is fine.)
  2. A D     B     C     E     (Either orientation is fine.)
  3. A D     E     C     B     (Either orientation is fine.)
  4. D A     B     C     E     (Either orientation is fine.)
  5. E B     A     C     D     (Either orientation is fine.)

Answer: d                                    

Section 7.4

Application

 

  1. A study is done on a group of families in Sweden that are segregating a genetic disorder. Lod-score analysis indicates that the gene involved in the disorder shows a strong likelihood of linkage with a particular DNA marker locus. However, a second study done in Italy with other families segregating the same genetic disorder results in lod-score values that strongly indicate the lack of linkage between the gene and the same DNA marker locus. Assuming that both studies were performed appropriately, what is the most likely explanation for the different outcomes?

 

  1. No recombinants were found in the families studied in Sweden.
  2. The allele that caused the disorder was in coupling linkage with one of the DNA marker alleles in the Swedish families but was in repulsion linkage in the Italian families.
  3. This disorder is caused by mutations in either of two different genes; one of these genes is linked to the DNA marker locus and the other gene is not.
  4. In the Italian families, the gene involved with the disorder is near a lethal allele at another locus and most of the parental or nonrecombinant genotypes contain the lethal allele, and this reduces the number of nonrecombinants observed.
  5. Linkage should have been observed in the Italian families, but there were only two alleles at the DNA marker locus that prevented recombinant offspring from appearing.

 

Answer: c

Section 7.3

Application

 

Short-Answer Questions

 

  1. What is the chi-square test used for, and what does it tell you?

 

Answer: The chi-square test is a statistical tool for analyzing data generated from genetic experiments to determine if observed results are consistent with a hypothesis proposed to explain them. The calculated chi-square statistic may be used to determine whether or not to accept or reject the proposed hypothesis within pre-determined confidence limits.

Section 7.2

Comprehension

 

  1. Discuss the differences, and at least one similarity, between recombination and independent assortment.

 

Answer: Genetic recombination (i.e., recombination due to crossing over) involves precise cleavage, physical exchange, and splicing together of chromatin between paired homologous chromosomes. By contrast, independent assortment (i.e., recombination in the case of physically unlinked genes) involves random alignment of homologous chromosome pairs along the metaphase plate, which results in a random assortment of haploid sets of individual homologs in the gametes. Both cases involve random rearrangement of alleles and contribute to the immense genetic variation present in the gametes of sexually reproducing species.

Section 7.2

Comprehension

 

  1. Geneticists often assume that map distances less than 7 to 8 map units or cM are quite accurate. Map distances that exceed this threshold significantly are assumed to be less accurate and the level of accuracy decline increases as map distances increase. Briefly explain this observation.

 

Answer: The further apart two genes are the more likely it is that multiple crossovers will occur. Double crossovers (or an even number of crossovers) between two genes may lead to parental genotypes in the offspring that will not be counted as recombinants, even though crossing over has taken place. Since genetic maps are created by counting observable recombinants in the offspring, double crossovers will lead to an underestimation of the true map distance.

Section 7.2

Application

 

  1. In a two-point linkage analysis, genes a and b have been found to be 26 cM apart on the same chromosome. A third gene, c, has just been discovered and found to be located between a and b. A three-point linkage analysis with a, b, and c indicates that a and b are actually 33 cM apart, rather than 26 cM. Why does the three-point analysis give a different map distance for a and b than does the two-point linkage analysis, and which is more accurate?

 

Answer: Multiple crossovers, particularly double crossovers, are likely to occur between a and b. In the two-point linkage analysis, some of these double crossovers are not detected since they result in parental or nonrecombinant genotypes in the offspring. This leads to an underestimation of the true map distance between a and b since they will not be counted as recombinants even though crossing over has taken place. With three-point linkage analysis, many of these double crossovers that went undetected in        the two-point analysis are now detected as recombinants because of the middle gene, c. This results in a longer and more accurate map distance between a and b. Because of this, three-point linkage analysis is more accurate and is an important reason why three-point linkage analysis is preferred when it can be done.

Section 7.3

Application

 

  1. A geneticist finds that a human gene and a particular DNA marker locus are located on chromosome 8 on the basis of somatic-cell hybridization studies. However, when lod-score analysis is done with these two loci using family pedigrees, no evidence for linkage between the two loci can be found. Assuming that both types of studies were done correctly and the results are valid, how would you explain the different outcomes?

 

Answer: Somatic-cell hybridization studies allow for the physical location of genetic loci to be assigned, which normally means revealing on which chromosome they are located. The two loci in this case have been both assigned to chromosome 8, but it is not known how close to each other they are. The lack of linkage revealed by the lod-score analysis indicates that the two loci are quite far apart on chromosome 8 and are not        linked. It is not uncommon for two loci to be on the same chromosome but not be linked.

Section 7.4

Application

 

  1. Assume that you discover a new human gene that you believe is located on the Y chromosome although not in the region (pseudoautosomal) of the Y that is homologous with part of the X chromosome. How would you map this gene with respect to the other genes on the Y chromosome?

 

Answer: A gene located on the Y should be strictly transmitted from fathers to sons. Mapping such genes is difficult because there is normally only one copy of the Y and therefore one copy of a gene on the Y. This means that no normal recombination is possible and analysis of recombination is the main way that genetic mapping studies are done. One possibility is to find rare XYY men where two copies of genes on the   Y will be present and recombination may be observed in the sons of such men who are fertile.

Section 7.4

Application

Test Bank for     
Chapter 9: Bacterial and Viral Genetic Systems

 

Multiple Choice Questions

 

  1. Bacterial mutants that require supplemental nutrients in their growth media are called

 

 

Answer: e

Section: 9.1

Comprehension Question

 

  1. Which of the following statements about nutritional requirement and growth of bacteria is NOT true?

 

  1. Culture media developed for bacteria must contain carbon source and essential elements for the survival of the bacteria.
  2. Auxotrophic mutants can grow on medium that lack carbon source because they can synthesize their own nutrients.
  3. Each bacterium has specific nutritional needs and conditions for successful cultivation.
  4. Prototrophic bacterial strains can grow on minimal media.
  5. The growth rate of bacteria on specific media can be assessed by the number and size of bacterial colonies.

 

Answer: b

Section: 9.1

Comprehension Question

 

  1. Bacterial strains that can produce all the necessary compounds and therefore grow on minimal media are called

 

 

Answer: c

Section: 9.1

Comprehension Question

 

4.      Which of the following facts would NOT be considered as an advantage for using bacteria and viruses for genetic studies?

 

a.       Rapid reproduction and high progeny number

  1. Haploid genome for expressing mutations
  2. Complete absence of recombination, which maintains the integrity of genome
  3. Low cost to maintain and little storage space required
  4. Genomes being small and readily subjected to genetic manipulation

 

Answer: c

Section: 9.1

Comprehension Question

 

  1. Which of the following statements about bacterial genome is NOT true?

 

  1. All bacteria contain a single circular double stranded DNA as their genome.
  2. Some bacteria may have linear chromosomes instead of circular one.
  3. In addition to chromosome, many bacteria possess small extrachromosomal DNA called plasmid.
  4. Each plasmid contains an origin of replication that allows independent replication for its maintenance.
  5. The F factor, which is important for bacterial conjugation is found as a circular episome of E.coli.

 

Answer: a

Section: 9.1

Comprehension Question

 

  1. What is the result of conjugation between F’ and F cells?

 

  1. One F+ cells
  2. Two F’ cells
  3. Two F+ cells
  4. One Hfr and one F– cells
  5. Two Hfr cells

 

Answer: b

Section: 9.1

Application Question

 

  1. Bacterial cells containing an F plasmid that has acquired bacterial chromosomal genes are called

 

  1. F+.
  2. F′.
  3. F.

 

Answer: b

Section: 9.2

Comprehension Question

 

  1. A bacterial cell transfers chromosomal genes to F cells, but it rarely causes them to become F+. The bacterial cell is

 

  1. F+

 

Answer: a

Section: 9.2

Comprehension Question

 

  1. Which of the following statements about genetic exchange in bacteria is NOT true?

 

  1. In some viruses, the DNA that encodes one gene product can overlap with DNA that encodes a different gene product.
  2. Plasmids do not have to integrate into the host cell chromosome in order to be replicated.
  3. Interrupted conjugation results in the production of Hfr strains.
  4. The order of gene transfer is not the same for different Hfr strains.
  5. Antibiotic resistance can be transferred from one bacterial cell to another by conjugation.

 

Answer: c

Section: 9.2

Comprehension Question

 

  1. Which of the following will have the least influence on the efficiency of transformation in coli bacteria?

 

  1. Calcium chloride treatment
  2. Heat shock
  3. Electrical field
  4. Chilling on the ice
  5. The amount of foreign DNA

 

Answer: d

Section: 9.2

Comprehension Question

 

  1. Which of the following horizontal gene transfer mechanisms would specifically use time as a basic unit of mapping?

 

  1. Transformation
  2. Crossing-over
  3. Transduction
  4. Conjugation
  5. Recombination

 

Answer: d

Section: 9.2

Comprehension Question

 

  1. The transfer of DNA from a donor cell to a recipient cell through a cytoplasmic connection is called

 

  1. lysogenic cycle.
  2. lytic cycle.

 

Answer: e

Section: 9.2

Comprehension Question

 

  1. When the F integrates into the coli chromosome, the result is an _______ strain.

 

  1. Hfr
  2. F
  3. F+
  4. F’
  5. F+/–

 

Answer: b

Section: 9.2

Comprehension Question

 

  1. Cotransformation between two genes is more likely if they are

 

  1. close to one another.
  2. far apart from one another.
  3. both next to the F factor.
  4. both oriented in the same direction.
  5. not located on the same chromosome.

 

Answer: a

Section: 9.2

Comprehension Question

 

15.  Which of the following statements about antibiotic resistance in bacteria is NOT true?

 

  1. Antibiotic resistance cannot be conferred by conjugation as conjugation only affects the fertility of bacteria.
  2. The antibiotic resistance gene can be transmitted to bacteria via transformation or transduction.
  3. Environments where antibiotics are frequently used such as hospitals are under the higher risk of developing antibiotic resistance.
  4. Antibiotic resistance often originates from the microbes that produce antibiotics for their own survival.
  5. The plasmid containing the antibiotic resistance gene can pass the genes to genetically unrelated bacteria.

 

Answer: a

Section: 9.2

Comprehension Question

 

  1. leu bacteria are mixed in a flask with leu+ bacteria, and soon all bacteria are leu+. However, if the leu cells are on one side of a U-tube and the leu+ cells are on the other, the leu cells do not become prototrophic. Which process is likely to produce this observed result?

 

  1. Conjugation
  2. Transduction
  3. Transformation
  4. Reciprocal translocation
  5. Transfection

 

Answer: a

Section: 9.2

Application Question

 

  1. How are Hfr strains of bacteria different from F+ strains?

 

  1. Cells of Hfr strains are able to transfer chromosomal genes, whereas cells of F+ strains cannot.
  2. Cells of Hfr strains cannot initiate conjugation with F
  3. The F factor is integrated into the bacterial chromosome in all or most cells of an Hfr strain but in only a few cells in an F+
  4. Cells of Hfr strains carry F’ plasmids, whereas F+ cells do not.
  5. Cells of Hfr strains can initiate conjugation with F+ cells or other Hfr cells.

 

Answer: c

Section: 9.2

Application Question

 

  1. You perform interrupted-mating experiments on three Hfr strains (A, B, and C). Genes are transferred (from last to first) in the following order from each strain: strain A, thi-his-gal-lac-pro; strain B, azi-leu-thr-thi-his; strain C, lac-gal-his-thi-thr. How are the F factors in these strains oriented?

 

  1. A and B are oriented in the same direction.
  2. B and C are oriented in the same direction.
  3. A and C are oriented in the same direction.
  4. All of them are oriented in the same direction.
  5. It cannot be determined from the information given.

 

Answer: a

Section: 9.2

Application Question

 

  1. A bacterium of genotype a+b+c+d+ is the donor in a cotransformation mapping. The recipient is abcd. Data from the transformed cells are shown below. What is the order of the genes?

 

a+ and b+ 2

a+ and c+ 0

a+ and d+ 5

b+ and c+ 5

b+ and d+ 0

c+ and d+ 0

 

  1. a c b d
  2. a d c b
  3. c b a d
  4. c a d b
  5. b c d a

 

Answer: c

Section: 9.2

Application Question

 

  1. The figure below shows a partial chromosome map of an coli Hfr strain. Each mark equals 10 minutes. If transfer of genes begins at “*” and goes in the direction of the arrow, which of the predicted results from this map is highly likely observed?

 

 

 

 

 

 

 

 

 

 

 

  1. gal will be the first and ton will be the last gene to be transferred.
  2. lac and azi will rarely be transferred together.
  3. Ten minutes after transfer of ton, azi will be transferred.
  4. It would take 30 minutes to transfer all of the genes that are shown.
  5. All the chromosomal genes will be transferred by the end.

 

Answer: b

Section: 9.2

Application Question

 

  1. The figure below shows the results of interrupted-mating experiments with three different Hfr strains. What is the order of the genes, starting with C?

 

Hfr strain

Order of transfer

1 A, B, E, D, F
2 D, F, C, G, A
3 D, E, B, A, G

 

  1. C, G, A, D, F, B, E
  2. C, F, D, B, A, E, G
  3. C, B, E, D, F, G, A
  4. C, G, A, B, E, D, F
  5. C, D, F, G, A, B, E

 

Answer: d

Section: 9.2

Application Question

 

  1. The process of transferring DNA from one bacterium to another through a bacteriophage is

 

 

Answer: d

Section: 9.3

Comprehension Question

 

  1. Integrated, inactive phage DNA is called a

 

 

Answer: b

Section: 9.3

Comprehension Question

 

  1. HIV belongs to a group of viruses called

 

  1. dsDNA viruses.
  2. ssDNA viruses.
  3. ssRNA-RT viruses.
  4. dsDNA-RT viruses.
  5. ssRNA viruses.

 

Answer: c

Section: 9.3

Comprehension Question

 

  1. The life cycle of virulent phages that always kill their host cell and never become inactive prophages would be the

 

  1. lethal cycle.
  2. lytic cycle.
  3. temperate cycle.
  4. strict cycle.
  5. lysogenic cycle.

 

Answer: b

Section: 9.3

Comprehension Question

 

  1. Which type of transduction is used to map distances between phage genes?

 

  1. Generalized transduction
  2. Specialized transduction
  3. Targeted transduction
  4. Random transduction
  5. Discontinuous transduction

 

Answer: a

Section: 9.3

Comprehension Question

 

  1. What does the enzyme reverse transcriptase do?

 

  1. Using the amino acid sequence of a protein as a template, it makes an RNA molecule.
  2. Using RNA as a template, it makes a DNA molecule.
  3. Using RNA as a template, it makes an RNA molecule.
  4. Using DNA as a template, it makes an RNA molecule.
  5. Using DNA as a template, it makes DNA molecule.

 

Answer: b

Section: 9.3

Comprehension Question

 

  1. Which of the following statements about retroviruses is false?

 

  1. All retroviruses contain oncogenes, which can induce the formation of tumors.
  2. All retroviruses contain gag genes whose product forms the viral protein coat.
  3. All retroviruses require pol genes, which are critical for retrotranscription.
  4. All retroviral genomes have gag, pol, and env
  5. Not all RNA viruses are retroviruses.

 

Answer: a

Section: 9.3

Comprehension Question

 

  1. Two different strains of a mutant phage infect a single bacterium. One phage strain is d and the other is e. Some of the progeny phages are genotype d+e+, and some are de. What genetic phenomenon does this demonstrate?

 

  1. Complementation
  2. Specialized transduction
  3. Generalized transduction
  4. Recombination

 

Answer: d

Section: 9.3

Application Question

 

  1. Two different strains of a mutant phage infected a single bacterium. One phage strain is de+ and the other is d+e. The coinfected phages produced the wild-type phenotype in the bacterium: One phage supplies the wild-type gene product from a d+ allele, and the other supplies the wild-type gene product from the e+ What genetic phenomenon does this demonstrate?

 

  1. Complementation via transduction event
  2. Co-transformation
  3. Recombination via conjugation
  4. Random mutation
  5. Interrupted transduction

 

Answer: a

Section: 9.3

Application Question

  

Short-Answer Questions

 

  1. What are plasmids and what purposes do they serve?

 

Answer: Plasmids are small, circular, extra-chromosomal DNA molecules found naturally in bacteria. They carry extra genes and can transfer these genes from one bacterial cell to another. They are also used extensively in genetic engineering.

Section: 9.1

Comprehension Question

 

 

  1. In order to better understand arginine biosynthesis in bacteria, a microbial geneticist might first isolate mutant bacterial strains.

 

  1. What characteristics must these mutant bacteria have?
  2. Outline a strategy for isolating such mutants.
  3. List three possible methods for mapping the genetic location of the mutations in these strains.

 

Answer:

  1. Each mutant strain must be auxotrophic for the amino acid arginine. This characteristic of mutant strains can be used to determine the number of genes, the locations of genes, and the functions of genes that encode proteins used for arginine biosynthesis.
  2. Bacteria would first be plated on complete media (i.e., media containing arginine and other essential nutrients). Next, bacteria would be replica-plated both to Petri plates containing media that lacked arginine and to duplicate plates containing media that contained arginine. Analysis of the growth of the strains on these two types of media would reveal auxotrophs: Those that grow on media containing arginine but do not grow on media lacking arginine.
  3. Three methods are (1) interrupted conjugation, (2) cotransformation, and (3) generalized transduction.

Section: 9.1

Application Question

 

  1. What causes an F cell to be converted to F+?

 

Answer: The F factor plasmid in an F+ cell is replicated, and one copy is transferred to the F cell through conjugation.

Section: 9.2

Comprehension Question

 

  1. What causes an F cell to be converted to Hfr in the presence of F+ cells?

 

Answer: The F cell must first receive an F factor plasmid by conjugation with an F+ cell. Once inside the recipient cell, the F plasmid can integrate into the bacterial chromosome, converting the cell to Hfr.

Section: 9.2

Comprehension Question

 

  1. Outline the steps involved in mapping a bacterial chromosome by conjugation.

 

Answer:

  • Select a donor Hfr strain and a recipient F The two strains must be different in genotype for several marker genes—for example, leu and leu+, azir and azs.
  • The two strains are mixed together in a nutrient medium to allow conjugation to begin. After a few minutes, the culture is diluted to prevent new conjugations.
  • At regular intervals, conjugation is interrupted by shearing the mating bacteria away from each other.
  • For each time point, cells that mated are selected—for instance, by requiring that a prototrophy or antibiotic resistance from each parent be present in one cell.
  • Cells that mated are checked for transfer of other genes from the Hfr to the F
  • Order and distance of transferred genes are determined by comparing the time required for them to transfer from the Hfr to the F

Section: 9.2

Comprehension Question

 

  1. Outline the steps involved in mapping a bacterial chromosome by cotransformation.

 

Answer:

  • A donor strain is chosen that has a number of prototrophies or antibiotic resistances.
  • The recipient strain’s genotype is different from that of the donor.
  • DNA from the donor strain is fragmented and is used to transform the recipient strain.
  • Transformants are selected—for instance, for a prototrophy or antibiotic resistance.
  • Transformants are tested for other genes they acquired in the transformation.
  • Order and distance of transferred genes are determined by comparing cotransformation frequencies.

Section: 9.2

Comprehension Question

 

  1. You perform interrupted conjugation using an a+b+c+d+l+m+n+o+ Hfr strain and an F strain that is abcdlmno. You observe the following genes transferred together in order from last to first:

 

n+a+c+m+

o+m+c+a+n+

o+b+d+l+n+

 

What is the map order of the genes?

 

Answer: The map is circular: –mcanldbo-.

Section: 9.2

Application Question

 

  1. (a) Explain how chromosomal genes are transferred from donors to recipients when cells of an F+ strain are mixed with F (b) Explain why transfer of chromosomal genes occurs at a higher frequency when cells of an Hfr strain are mixed with F cells.

 

Answer:

  • Within a few cells of an F+ strain the F factor will integrate into one of several positions of the bacterial chromosome and become Hfr. When the integrated F factor initiates conjugation, genes of the bacterial chromosome will be transferred behind the leading part of the F factor. Once inside the recipient, the transferred       chromosomal genes can be recombined into the recipient’s chromosome.  Transfer of chromosomal genes occurs at only a low frequency because relatively few cells in an F+ strain are Hfr.
  • Almost all of the cells of an Hfr strain contain an F factor that is integrated into the bacterial chromosome. All of these cells have the ability to transfer chromosomal genes to F− cells, so the frequency of transfer is high.

Section: 9.2

Application Question

 

  1. (a) What is an F’ plasmid and how is it formed? (b) Explain how an F’ can be used to construct a bacterial strain that is partially diploid. (c) Explain how partial diploid strains can be used to assess interactions between different alleles (e.g., lac + and lac).

 

Answer:

  • An F’ is an F factor that carries one or two genes from the bacterial chromosome. An F’ forms when an F factor excises imprecisely from the chromosome, carrying a small part of the chromosome with it.
  • When a cell carrying an F’ contacts an F cell, the F’ is copied and transferred to the F The recipient cell will then possess two copies of the gene or genes that are carried on the F’.
  • It is more difficult to assess interactions between alleles in bacteria because bacteria normally are haploid and the alleles normally do not occur together in a cell or strain. However, F’ strains can be used to construct partial diploid strains that allow for these interactions to be tested. For example, a strain can be established that is heterozygous for the lac gene, allowing one to determine the dominance relationship between lac+ and lac.

Section: 9.2

Application Question

 

  1. Explain the significance of horizontal gene transfer to bacterial evolution and to our ability to discern relationships between different groups of bacteria.

 

Answer: Horizontal gene transfer is the transfer of genes from one type of bacteria to             another through conjugation, transformation, or (less likely) transduction.  In some cases, genes can be transferred horizontally between bacteria that are not closely related. Horizontal gene transfer allows bacteria of different types to exchange genetic variation and to acquire new genetic traits that can be tested by natural selection. Some evidence suggests that horizontal gene transfer is extensive in nature. For example, about 17% of the E. coli genome is thought to have come from other types of bacteria. As a result, some bacteriologists question whether concepts of species even apply in bacteria, given that a particular bacterium can contain parts of a variety of genomes from different groups.

 

Horizontal gene transfer complicates molecular analysis of relationships between different types of bacteria because such analyses assume that the genes being compared have been acquired by each group through vertical gene transfer (inheritance). To the extent that assumption is violated by horizontal gene transfers, molecular analysis of relationships is made much more difficult.  In reality, such analyses are valid only for that part of the genome that has been transferred vertically by inheritance. In other words, the history of the group is not reflected perfectly in the history of its genome.

Section: 9.2

Application Question

 

  1. Outline the steps involved in mapping bacterial genes by generalized transduction.

 

Answer:

  • Use two strains of bacteria whose genotypes differ for each gene to be mapped.
  • Phages that have infected one bacterial strain (the donor) are collected and used to infect the other strain.
  • Recipient bacteria with recombinant genotypes are selected.
  • Order and distance of genes are determined by comparing frequencies of cotransduction.

Section: 9.3

Comprehension Question

 

  1. How does a virulent phage differ from a temperate phage?

 

Answer: Virulent phages reproduce using only the lytic cycle, whereas temperate        phages can use the lytic or lysogenic cycle.

Section: 9.3

Comprehension Question

 

  1. Both retroviruses and lysogenic bacteriophages employ a mechanism that allows them to be replicated and passed from cell to cell without producing viruses. What is the common mechanism that these two very different viruses use?

 

Answer: Both integrate their viral genome into a host genome, so that when the cell DNA is replicated, the viral genetic information is replicated as well.

Section: 9.3

Comprehension Question

 

  1. A retrovirus has an RNA genome but integrates into the DNA chromosome of a host cell. Explain how it does this.

 

Answer: The retrovirus genome encodes a reverse transcriptase enzyme. RTase makes a DNA copy of the viral RNA genome. The DNA copy integrates into the host cell’s DNA.

Section: 9.3

Comprehension Question

 

  1. List and describe three different ways that DNA from one bacterium can be transferred into bacterial cells.

 

Answer:

  • Transformation: competent cells take up DNA from the environment.
  • Transduction: DNA is carried into the cell in a virus.
  • Conjugation: DNA from one cell is transferred in a pilus to another cell.

Section: 9.3

Comprehension Question

 

  1. What is the difference between specialized and generalized transduction?

 

Answer: Generalized transduction transfers fragments from anywhere in the chromosome; specialized transduction transfers only DNA that is adjacent to the prophage insertion site.

Section: 9.3

Comprehension Question

 

  1. You are using phages to map three toxin-production genes (R, Y, and G) in a new bacterium. You grow phages in a strain of the bacteria that produces all three toxins (R+, Y+, G+), isolate the phage, and then infect a second bacterial strain that cannot produce any of the toxins (R, Y, G). The recipient bacteria are then grown on colorimetric media (media that change color in response to toxin presence) to see which toxin genes are transferred together by the phage. The data are as follows:

 

R         Y         G

R         —        76%     0%

Y                     —        3%

G                                 —

 

  1. What kind of mapping is this called?
  2. Which gene is in the middle?
  3. Which of the outside genes is closer to the middle gene?

 

Answer:

  1. Transduction
  2. Y
  3. R

Section: 9.3

Application Question

 

  1. HIV has a high mutation rate. What causes this, and how might this be advantageous to the virus?

 

Answer: The reverse transcriptase of HIV has an unusually high error rate, so that many mutations occur as the HIV genome is converted from RNA to DNA. A high   mutation rate, though potentially creating nonfunctional viral proteins, also leads to rapid evolution of the virus, allowing it continually to adapt to new threatening conditions even within one host.

Section: 9.3

Application Question

 

  1. Two phage phenotypes are controlled by the genes a and b. In a mapping experiment, a culture of bacteria is infected simultaneously with an ab+ strain and an a+b When plaques are analyzed, five out of 1000 have the a+b+ or ab phenotype. Based on the information, how far apart are genes a and b?

 

Answer: They are 0.5 map units apart.

Section: 9.3

Application Question

 

  1. A virulent bacteriophage is used to infect a prototrophic bacterial culture. Phages are collected from the culture and are used to infect a new bacterial strain that has several auxotrophies. After infection, rare prototrophs are found:

 

met+ leu+          0

met+ pro+         1

met+ his+          0

pro+ leu+          2

pro+ his+          0

leu+ his+           1

 

How are the auxotrophy genes organized on the bacterial chromosome?

 

Answer:

  • met and pro are close to each other.
  • pro and leu are close to each other.
  • leu and his are close to each other.
  • The gene order is met pro leu his.

Section: 9.3

Application Question

 

  1. Using the rII mutant bacteriophage and coli strains B and K, Seymour Benzer studied gene structure by both recombination and complementation. Wild-type phages form plaques on strains B or K; rII forms plaques on B only.

In one set of experiments, Benzer infected E. coli strain B simultaneously with two different rII mutants. Progeny phages were collected and used to infect strain K, and, occasionally, plaques formed.

 

In a second set of experiments, Benzer infected strain K simultaneously with high amounts of two different rII mutant phages. Occasionally, plaques were observed from this infection.

 

Which set of experiments shows recombination, and which demonstrates complementation? Explain the difference.

 

Answer:

(1)  The first set of experiments shows recombination, the second complementation.

(2)  Recombination requires exchange between genomes. In an experiment like this, in which each initial genome contains one mutation, recombination produces completely wild-type DNA (or genomes or chromosomes), along with DNA (or genomes or chromosomes) with the two mutations.

(3)  In complementation, each mutant genome (or chromosome or DNA) produces a gene product for which the other mutant genome is defective.

Section: 9.3

Application Question

 

  1. Using the rII mutant bacteriophage and coli strains B and K, Seymour Benzer studied gene structure by both recombination and complementation. Wild-type phages form plaques on strains B or K; rII forms plaques on B only.

 

In one set of experiments, Benzer infected E. coli strain B simultaneously with two different rII mutants. Progeny phages were collected and used to infect strain K, and, occasionally, plaques formed.

 

In a second set of experiments, Benzer infected strain K simultaneously with high amounts of two different rII mutant phages. Occasionally, plaques were observed from this infection.

 

If progeny were collected from the plaques on strain K in the first experiment and then were used in low numbers to infect strain K, what proportion of the infection events (low or high) would produce plaques? Why?

 

Answer: A high proportion of the infections would lead to a plaque. These plaques on strain K are formed by a wild-type recombinant phage. Their genomes (or             chromosomes) contain no mutations. Therefore, virtually every infection would lead to a plaque, unless a phage had acquired a rare, new mutation.

Section: 9.3

Application Question

 

  1. Using the rII mutant bacteriophage and E. coli strains B and K, Seymour Benzer studied gene structure by both recombination and complementation. Wild-type phages form plaques on strains B or K; rII forms plaques on B only.

 

In one set of experiments, Benzer infected E. coli strain B simultaneously with two different rII mutants. Progeny phages were collected and used to infect strain K, and, occasionally, plaques formed.

 

In a second set of experiments, Benzer infected strain K simultaneously with high amounts of two different rII mutant phages. Occasionally, plaques were observed from this infection.

 

If progeny were collected from the plaques on strain K in the second experiment and then were used in low numbers to infect strain K, what proportion of the infection events (low or high) would produce plaques? Why?

 

Answer: A low proportion of the infections would lead to a plaque. These plaques on strain K formed from the complementation of gene products produced within the infected bacterial cell. Thus, each infected cell contained both phage types, but each phage genome still contains the original mutation. If packaged into a phage, the phages would be mutant and unable to infect strain K on their own. Therefore, virtually no infection would lead to a plaque, unless recombinant phages had formed        in the initial infection.

Section: 9.3

Application Question

 

  1. You are studying a new phage that infects pylori. You have isolated two mutant strains of the phage, each producing a different plaque phenotype due to a specific mutation: rough (r) and big (b). You co-infect H. pylori with both strains by adding a mixture of phages to a culture of cells. You collect the cell lysate containing progeny phages; plate diluted phages on a lawn of H. pylori cells; and observe 970 rough plaques, 890 big plaques, 0 rough and big plaques, and 500 normal, wild-type plaques.

 

  1. What is the recombination frequency between the r locus and the b locus?
  2. Can r and b be different alleles of the same locus?
  3. How can you explain the results?

 

Answer:

  1. No recombinant plaques were observed, so the recombination frequency is 0.
  2. No, the 500 wild-type plaques are a result of complementation between the two phage strains in doubly infected cells.
  3. The genes could be overlapping so that no recombination occurs between them.

Section: 9.3

Challenge Question

 

  1. (a) Explain the mechanism that leads to rapid evolution of the virus that causes influenza. (b) Distinguish between antigenic drift and antigenic shift, and explain the significance of each to influenza evolution and the occurrence of influenza in humans.

 

Answer:

  • The enzyme that copies the RNA genome of the influenza virus is prone to replication errors, leading to a high rate of mutation and rapid production of new genetic variants upon which natural selection can act.
  • Antigenic drift is the continual change in influenza proteins that result from mutations due to replication errors. Antigenic drift involves fairly small changes in the virus from year to year. Although the changes are not dramatic, they are significant in that most humans are not immune to the new variants and a new flu vaccine must be created each year. Antigenic shift is the production of a fundamentally new type of influenza virus through reassortment that occurs when two or more different strains of virus infect a single cell. Some of the progeny will carry fundamentally new combinations of genetic traits. Typically, very few people will be immune to the new strain, giving rise to the potential for widespread infection or pandemic. Significantly, reassortment is unpredictable, so that vaccines typically are not available until after the new strain has spread throughout the world.

Section: 9.3

Challenge Question