Genetics From Genes To Genomes 6th Edition by Leland Hartwell – Test Bank

 

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Sample Test

Genetics, 6e (Hartwell)

Chapter 3   Extensions to Mendel’s Laws

 

1) Two alleles of gene C control hair color in horses: C¹ and C². Horses homozygous for allele C¹ are red, heterozygotes are yellow, and C² homozygotes are cream. What type of allele interaction is described?

1.   A) pleiotropy

2.   B) complete dominance

3.   C) incomplete dominance

4.   D) recessive lethality

5.   E) codominance

 

Answer:  C

Section:  03.01

Topic:  Extensions to Mendel for Single-Gene Inheritance

Learning Objective:  03.01.01 Categorize allele interactions as completely dominant, incompletely dominant, or codominant.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

2) If a phenotype is controlled by the genotypes at two different loci, a possible interaction between alleles of these genes can be called

1.   A) epistasis.

2.   B) epigenetics.

3.   C) dominance.

4.   D) codominance.

5.   E) incomplete dominance.

 

Answer:  A

Section:  03.02

Topic:  Extensions to Mendel for Multifactorial Inheritance

Learning Objective:  03.02.02 Infer from the results of crosses the existence of interactions between alleles of different genes including: additivity, epistasis, redundancy, and complementation.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

 

3) A pure-breeding fruit fly with curled wings mates with a pure-breeding fruit fly with normal (straight) wings. The F1 mate with each other to produce an F2 generation that consists of 160 flies with curled wings and 80 with straight wings. What can you infer from this observation?

1.   A) Curled wings is a recessive trait.

2.   B) The dominant curled wing allele is also a recessive lethal.

3.   C) Wing shape is controlled by two codominant alleles.

4.   D) Two interacting genes determine wing shape.

5.   E) All of the hybrid F1flies had straight wings.

 

Answer:  B

Section:  03.01

Topic:  Extensions to Mendel for Single-Gene Inheritance

Learning Objective:  03.01.02 Recognize progeny ratios that imply the existence of recessive lethal alleles.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

4) What can explain the phenomenon where the same genotype might give rise to different phenotypes?

1.   A) pleiotropy

2.   B) codominance

3.   C) incomplete dominance

4.   D) complete dominance

5.   E) penetrance and expressivity

 

Answer:  E

Section:  03.03

Topic:  Extensions to Mendel for Multifactorial Inheritance

Learning Objective:  03.03.01 Discuss that factors that can cause different individuals with the same genotype to be phenotypically dissimilar.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

 

5) Two alleles of gene C control hair color in horses: C¹ and C². Horses homozygous for allele C¹ are red, heterozygotes are yellow, and C² homozygotes are cream. In the offspring of matings between heterozygotes, what phenotypic ratio is expected?

1.   A) 2 yellow: 1 red

2.   B) 3 red: 1 cream

3.   C) 1 red: 2 yellow: 1 cream

4.   D) all red

5.   E) 9 red: 3 yellow: 4 cream

 

Answer:  C

Section:  03.01

Topic:  Extensions to Mendel for Single-Gene Inheritance

Learning Objective:  03.01.01 Categorize allele interactions as completely dominant, incompletely dominant, or codominant.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

6) If a trait is controlled by two codominant alleles of one gene, what phenotypic ratio is expected in the offspring of a mating of two heterozygotes?

1.   A) 2:1

2.   B) 3:1

3.   C) 1:2:1

4.   D) 1:1

5.   E) 4:1

 

Answer:  C

Section:  03.01

Topic:  Extensions to Mendel for Single-Gene Inheritance

Learning Objective:  03.01.01 Categorize allele interactions as completely dominant, incompletely dominant, or codominant.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

7) Which genotypic ratio indicates a recessive lethal allele when two heterozygotes are mated?

1.   A) 2:1

2.   B) 3:1

3.   C) 1:2:1

4.   D) 1:1

5.   E) 4:1

 

Answer:  A

Section:  03.01

Topic:  Extensions to Mendel for Single-Gene Inheritance

Learning Objective:  03.01.02 Recognize progeny ratios that imply the existence of recessive lethal alleles.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

8) ABO blood type demonstrates which of the following inheritance patterns? (Select all that apply.)

1.   A) complete dominance

2.   B) incomplete dominance

3.   C) codominance

4.   D) pleiotropy

5.   E) three genes are involved

 

Answer:  A, C

Section:  03.01

Topic:  Extensions to Mendel for Single-Gene Inheritance

Learning Objective:  03.01.01 Categorize allele interactions as completely dominant, incompletely dominant, or codominant.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

9) A particular flower can be purple, blue, red, or white. A pure-breeding red plant is crossed with a pure-breeding white one, and the F1 are then crossed to produce an F2 generation. Which of the following phenotypic ratios in the F2 indicate that flower color in these plants is controlled by two genes?

1.   A) 2:1

2.   B) 3:1

3.   C) 1:2:1

4.   D) 9:3:4

 

Answer:  D

Section:  03.02

Topic:  Extensions to Mendel for Multifactorial Inheritance

Learning Objective:  03.02.01 Conclude from the results of crosses whether a single gene or two genes control a trait.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

 

 

10) If a gene for a trait is monomorphic in a population and two random individuals mate, what would be the most likely phenotypic ratio for that trait in the offspring?

1.   A) 2:1

2.   B) 1

3.   C) 3:1

4.   D) More information is needed

 

Answer:  B

Section:  03.01

Topic:  Extensions to Mendel for Single-Gene Inheritance

Learning Objective:  03.01.03 Predict from the results of crosses whether a gene is polymorphic or monomorphic in a population.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

11) What genetic phenomenon might a 2:1 phenotypic ratio indicate?

1.   A) additivity

2.   B) codominance

3.   C) complete dominance

4.   D) recessive epistasis

5.   E) recessive lethal

 

Answer:  E

Section:  03.01

Topic:  Extensions to Mendel for Single-Gene Inheritance

Learning Objective:  03.01.02 Recognize progeny ratios that imply the existence of recessive lethal alleles.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

 

12) A particular flower can be purple, blue, red, or white. Two different pure-breeding white plants are crossed and the F1are then crossed to produce an F2 generation. What might a 9:7 phenotypic ratio in the F2 indicate?

1.   A) reciprocal recessive epistasis

2.   B) codominance

3.   C) dominant epistasis

4.   D) recessive lethality

5.   E) additivity

 

Answer:  A

Section:  03.02

Topic:  Extensions to Mendel for Two-Gene Inheritance

Learning Objective:  03.02.02 Infer from the results of crosses the existence of interactions between alleles of different genes including: additivity, epistasis, redundancy, and complementation.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

13) A particular flower can be purple, blue, red, or white. A pure-breeding purple plant is crossed with a pure-breeding white plant and the F1 are then crossed to produce an F2 generation. What might a 9:3:4 phenotype ratio in the F2 indicate?

1.   A) reciprocal recessive epistasis

2.   B) codominance

3.   C) dominant epistasis

4.   D) recessive epistasis

5.   E) additivity

 

Answer:  D

Section:  03.02

Topic:  Extensions to Mendel for Two-Gene Inheritance

Learning Objective:  03.02.02 Infer from the results of crosses the existence of interactions between alleles of different genes including: additivity, epistasis, redundancy, and complementation.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

 

 

14) Which ratio would indicate that a phenotype is controlled by multiple genes?

1.   A) 3:1

2.   B) 2:1

3.   C) 1:2:1

4.   D) 9:3:3:1.

 

Answer:  D

Section:  03.02

Topic:  Extensions to Mendel for Multifactorial Inheritance

Learning Objective:  03.02.01 Conclude from the results of crosses whether a single gene or two genes control a trait.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

In rats, the P gene allele for pigmentation (P) is dominant to the allele for albinism (p). The B gene allele for black pigmentation (B) is dominant to the allele for cream pigmentation (b). The pp homozygous recessive genotype is epistatic to any allele combination at gene B.

 

15) Predict the genotypes and phenotypes of the F1 progeny of a cross between a pure-breeding black rat and an albino that is also homozygous for cream.

1.   A) PP BB, black

2.   B) Pp Bb, black

3.   C) Pp Bb, albino

4.   D) pp Bb, albino

 

Answer:  B

Section:  03.02

Topic:  Extensions to Mendel for Two-Gene Inheritance

Learning Objective:  03.02.02 Infer from the results of crosses the existence of interactions between alleles of different genes including: additivity, epistasis, redundancy, and complementation.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

 

 

16) Predict the phenotypic ratio of the F2 progeny of a parental cross between a pure-breeding black rat and an albino that is also homozygous for cream.

1.   A) 1 black: 2 cream: 1 albino

2.   B) 9 black: 3 cream: 4 albino

3.   C) 9 black: 7 albino

4.   D) 12 black: 3 cream: 1 albino

5.   E) 15 black: 1 albino

 

Answer:  B

Section:  03.02

Topic:  Extensions to Mendel for Two-Gene Inheritance

Learning Objective:  03.02.02 Infer from the results of crosses the existence of interactions between alleles of different genes including: additivity, epistasis, redundancy, and complementation.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

In the common daisy, genes A and B control flower color. Both genes have a dominant allele (A or B) and a recessive allele (a or b). At least one copy of each dominant allele is required for flowers to be colorful instead of white.

17) Predict the genotypes and phenotypes of the F1 progeny of a cross between two white-flowered plants, one homozygous for A and the other homozygous forB.

1.   A) AA bb, white

2.   B) aa BB, white

3.   C) Aa Bb, colorful

4.   D) Aa Bb, white

5.   E) aa bb, colorful

 

Answer:  C

Section:  03.02

Topic:  Extensions to Mendel for Two-Gene Inheritance

Learning Objective:  03.02.02 Infer from the results of crosses the existence of interactions between alleles of different genes including: additivity, epistasis, redundancy, and complementation.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

 

 

18) Predict the phenotypic ratio of the F2 progeny of a cross between two white-flowered plants, one homozygous for A and the other homozygous for B.

1.   A) 3 colorful : 1 white

2.   B) 9 colorful : 7 white

3.   C) 9 white : 7 colorful

4.   D) 15 white : 1 colorful

5.   E) 15 colorful : 1 white

 

Answer:  B

Section:  03.02

Topic:  Extensions to Mendel for Two-Gene Inheritance

Learning Objective:  03.02.02 Infer from the results of crosses the existence of interactions between alleles of different genes including: additivity, epistasis, redundancy, and complementation.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

19) The inheritance pattern of daisy flower color provides an example of what type of gene interaction?

1.   A) additivity

2.   B) recessive epistasis

3.   C) reciprocal recessive epistasis

4.   D) dominant epistasis

5.   E) redundancy

 

Answer:  C

Section:  03.02

Topic:  Extensions to Mendel for Two-Gene Inheritance

Learning Objective:  03.02.02 Infer from the results of crosses the existence of interactions between alleles of different genes including: additivity, epistasis, redundancy, and complementation.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

 

 

20) A disease is caused by homozygosity for the g allele (G is the corresponding wild-type allele). However, the penetrance of the disease is 75%. Two individuals known to be heterozygotes have a child. What is the probability that the child exhibits the disease?

1.   A) 1/4

2.   B) 3/4

3.   C) 1/8

4.   D) 3/16

5.   E) 9/16

 

Answer:  D

Explanation:  The disease is seen only in gg individuals. Because the disease is 75% penetrant, 75% of gg individuals will show disease symptoms. The chance that the child of two heterozygotes (Gg × Gg) is gg is 1/4. If the child is gg, there is a 75% (or 3/4) chance the child will show disease symptoms. Because both of those events must happen for the child to have the disease, the probabilities are multiplied (1/4 × 3/4) to equal 3/16.

Section:  03.03

Topic:  Extensions to Mendel for Multifactorial Inheritance

Learning Objective:  03.03.01 Discuss that factors that can cause different individuals with the same genotype to be phenotypically dissimilar.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

21) Some multifactorial traits are determined by many interacting genes, each of which may have several alleles. If such a trait is measured in a population, what phenotypic pattern is expected?

1.   A) All individuals will have the same phenotype.

2.   B) Two types of individuals will exist, and most will have the dominant phenotype.

3.   C) Different individuals will each have one of a few discrete phenotypes.

4.   D) The population will have continuous variation in phenotypic expression.

 

Answer:  D

Section:  03.03

Topic:  Extensions to Mendel for Multifactorial Inheritance

Learning Objective:  03.03.02 Explain how Mendelian genetics is compatible with the fact that many traits, such as human height and skin colors, exhibit continuous variation.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

 

22) In some flowers, a purple pigment is synthesized from a red precursor pigment. In the absence of all pigment, flowers are white. A pure-breeding plant with red flowers was crossed to a pure-breeding plant with white flowers. All of the F1 plants had white flowers. The F1 plants were crossed to each other, and the F2 consisted of 165 plants: 123 with white flowers, 32 with purple flowers, and 11 with red flowers. How do these results suggest that flower color is determined?

1.   A) One gene with two alleles exists and heterozygotes have a different phenotype than either homozygote.

2.   B) The dominant allele of one gene masks the effect of a second gene.

3.   C) The recessive allele of one gene masks the effect of a second gene.

4.   D) A dominant allele of two genes is necessary for purple flowers.

5.   E) A dominant allele of either of two genes is sufficient for purple flowers.

 

Answer:  B

Section:  03.02

Topic:  Extensions to Mendel for Two-Gene Inheritance

Learning Objective:  03.02.01 Conclude from the results of crosses whether a single gene or two genes control a trait.; 03.02.02 Infer from the results of crosses the existence of interactions between alleles of different genes including: additivity, epistasis, redundancy, and complementation.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

Achondroplasia is a form of dwarfism in humans. It is caused by a mutant allele of the fibroblast growth factor receptor gene (FGFR) that produces an overactive protein. Having one copy of the mutant allele results in dwarfism. Two copies of the mutant allele results in death before birth.

 

23) Based in this information, what can you infer about the inheritance of the FGFR alleles?

1.   A) The mutant FGFR allele is pleiotropic.

2.   B) Some achondroplastic dwarfs are heterozygous for the mutant allele and some are homozygous.

3.   C) The wild-type and mutant FGFR alleles are codominant.

4.   D) The mutant FGFR allele shows incomplete penetrance.

 

Answer:  A

Section:  03.01

Topic:  Extensions to Mendel for Single-Gene Inheritance

Learning Objective:  03.01.02 Recognize progeny ratios that imply the existence of recessive lethal alleles.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

 

 

24) If two people with achondroplasia have a child together, what is the probability that their child will also have achondroplasia?

1.   A) 0

2.   B) 1/2

3.   C) 2/3

4.   D) 3/4

5.   E) 1

 

Answer:  C

Section:  03.01

Topic:  Extensions to Mendel for Single-Gene Inheritance

Learning Objective:  03.01.02 Recognize progeny ratios that imply the existence of recessive lethal alleles.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

25) A particular flower can be purple, blue, red, or white. A pure-breeding purple plant is crossed with a pure-breeding white plant and the F1 are then crossed to produce an F2 generation. Which phenotypic ratio in the F2 may indicate that flower color in these plants is controlled by two genes that interact additively?

1.   A) 3:1

2.   B) 1:2:1

3.   C) 9:3:3:1

4.   D) 9:3:4

5.   E) 15:1

 

Answer:  C

Section:  03.02

Topic:  Extensions to Mendel for Two-Gene Inheritance

Learning Objective:  03.02.02 Infer from the results of crosses the existence of interactions between alleles of different genes including: additivity, epistasis, redundancy, and complementation.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

 

 

In primroses, the dominant allele of gene K is necessary to synthesize blue flower pigment. Blue pigment synthesis is inhibited by a dominant allele of gene D. In other words, plants with the genotype K– D– will not produce pigment (and their flowers will be white) because of the presence of the D allele.

 

26) If two dihybrid plants (Kk Dd) are crossed, what is the ratio of blue to white offspring in the progeny?

1.   A) 3 blue: 1 white

2.   B) 7 blue: 9 white

3.   C) 4 blue: 12 white

4.   D) 3 blue: 13 white

5.   E) 1 blue: 15 white

 

Answer:  D

Section:  03.02

Topic:  Extensions to Mendel for Two-Gene Inheritance

Learning Objective:  03.02.02 Infer from the results of crosses the existence of interactions between alleles of different genes including: additivity, epistasis, redundancy, and complementation.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

27) What type of gene interaction determines flower color in primroses?

1.   A) one gene with two completely dominant alleles

2.   B) two-genes with recessive epistasis

3.   C) two redundant genes

4.   D) two genes with dominant epistasis

 

Answer:  D

Section:  03.02

Topic:  Extensions to Mendel for Two-Gene Inheritance

Learning Objective:  03.02.01 Conclude from the results of crosses whether a single gene or two genes control a trait.; 03.02.02 Infer from the results of crosses the existence of interactions between alleles of different genes including: additivity, epistasis, redundancy, and complementation.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

Two pure-breeding phlox plants were crossed, one with dark-blue flowers and the other with pink flowers. The F1 generation all had dark-blue flowers. When the F1 were crossed with each other, the F2 generation consisted of 176 plants: 101 with dark-blue flowers, 33 with light-blue flowers, 30 with red flowers, and 10 with pink flowers.

 

 

 

28) How is flower color controlled in phlox?

1.   A) one gene with four alleles

2.   B) one gene with two codominant alleles

3.   C) two genes whose alleles interact additively

4.   D) two genes with dominant epistasis

 

Answer:  C

Section:  03.02

Topic:  Extensions to Mendel for Two-Gene Inheritance

Learning Objective:  03.02.01 Conclude from the results of crosses whether a single gene or two genes control a trait.; 03.02.02 Infer from the results of crosses the existence of interactions between alleles of different genes including: additivity, epistasis, redundancy, and complementation.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

29) What progeny types would result from crossing a homozygous plant with light-blue flowers to a homozygous plant with red flowers?

1.   A) all would be dark blue

2.   B) all would be red

3.   C) 1 light blue : 1 red

4.   D) 9 dark blue : 3 light blue : 3 red : 1 pink

 

Answer:  A

Section:  03.02

Topic:  Extensions to Mendel for Two-Gene Inheritance

Learning Objective:  03.02.02 Infer from the results of crosses the existence of interactions between alleles of different genes including: additivity, epistasis, redundancy, and complementation.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

Genetics, 6e (Hartwell)

Chapter 5   Linkage, Recombination, and the Mapping of Genes on Chromosomes

 

In Drosophila, the genes y, f, and v are all X-linked. Females who are homozygous for recessive alleles of all three genes (y f v / y f v) are crossed to wild-type males. The resulting trihybrid F1 females are testcrossed. The F2 are distributed as follows:

 

y  f  v	3210
y  f  v+	72
y  f+  v	1024
y  f+  v+	678
y+  f  v	690
y+  f   v+	1044
y+  f+  v	60
y+  f+  v+	3222
	10,000

 

1) Which of the following linkage maps correctly shows the order and distance between the y, f, and v genes?

1.   A) f——35 m.u.——y——15 m.u.——v

2.   B) f——22 m.u.——y——15 m.u.——v

3.   C) y——35 m.u.——f——22 m.u.——v

4.   D) y——22 m.u.——v——15 m.u.——f

5.   E) y——15 m.u.——v——22 m.u.——f

 

Answer:  E

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

 

 

2) What is the coefficient of coincidence in this region?

1.   A) 0

2.   B) 0.2

3.   C) 0.4

4.   D) 0.6

5.   E) 0.8

 

Answer:  C

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

Females heterozygous for the recessive second chromosome mutations px, sp, and cn are mated to a male homozygous for all three mutations. The offspring are as follows:

 

px  sp  cn	1461
px  sp  cn+	3497
px  sp+  cn	1
px  sp+  cn+	11
px+  sp  cn	9
px+  sp  cn+	0
px+  sp+  cn	3482
px+  sp+  cn+	1539
	10,000

 

3) What is the genotype of the females that gave rise to these progeny?

1.   A) px+sp cn / px sp+cn+

2.   B) px+sp cn+ / px sp+cn

3.   C) px+sp+ cn+ / px spcn

4.   D) pxsp cn+ / px+ sp+cn

5.   E) insufficient data

 

Answer:  D

Section:  05.01; 05.03

Topic:  Gene Linkage and Recombination; Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.; 05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

4) Which of the three genes is in the middle?

1.   A) px

2.   B) sp

3.   C) cn

4.   D) insufficient data

 

Answer:  A

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

5) Which of the following linkage maps correctly shows the order and distance between the px, sp, and cn genes?

1.   A) sp——0.21 m.u.——px——30.01 m.u.——cn

2.   B) sp——30.01 m.u.——px——0.21 m.u.——cn

3.   C) sp——0.2 m.u.——px——30 m.u.——cn

4.   D) px——0.2 m.u.——sp——30.2 m.u.——cn

5.   E) px——30.2 m.u.——sp——0.2 m.u.——cn

 

Answer:  A

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

6) What is the coefficient of coincidence in this region?

1.   A) 0

2.   B) 0.16

3.   C) 0.33

4.   D) 0.5

5.   E) 0.66

 

Answer:  B

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

 

In peas, tall (T) is dominant to short (t), red flowers (R) is dominant to white flowers (r), and wide leaves (W) is dominant to narrow leaves (w). A tall plant that has red flowers and wide leaves is crossed to a short plant that has white flowers and narrow leaves. The resulting progeny are shown in the table.

 

tall, red, wide	381
tall, white, wide	122
short, red, wide	118
short, white, wide	379
	1000

 

7) What is the genotype of the tall plant that has red flowers and wide leaves?

1.   A) T R W / t r w

2.   B) T R W / t r W​

3.   C) T R W / T R W​

4.   D) T R W / T r w​

5.   E) T r W / t R W​

 

Answer:  B

Section:  05.01

Topic:  Gene Linkage and Recombination

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

8) This cross is not useful to determine if one of the genes is linked to the others. Which gene?

1.   A) gene T

2.   B) gene R

3.   C) gene W

4.   D) This cross shows that all three genes are linked.

 

Answer:  C

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.01 Establish relative gene positions using two-point cross data.; 05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

9) If two or more of the genes are linked, what map distance separates them?

1.   A) 4 m.u.

2.   B) 12 m.u.

3.   C) 24 m.u.

4.   D) 50 m.u.

5.   E) None of the genes are linked to each other.

 

Answer:  C

Section:  05.01; 05.03

Topic:  Gene Linkage and Recombination; Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.01.04 Explain how a testcross can provide evidence for or against linkage.; 05.03.01 Establish relative gene positions using two-point cross data.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

 

 

A dihybrid testcross is made to determine if genes C and D are linked. The results are shown in the table.

 

Parent genotypes:	Cc Dd × cc dd
	Cc Dd	222
	Cc dd	280
Progeny genotypes:	cc Dd	280
	cc dd	218
		1000

 

 

10) The chi-square value is the sum for all progeny classes of (observed-expected)2/expected. Using the chi-square test for goodness of fit, calculate the chi-square value to test the null hypothesis that genes C and Dare unlinked. What is the chi-square value?

1.   A) 0

2.   B) 0.0576

3.   C) 10.8

4.   D) 14.4

5.   E) cannot be determined

 

Answer:  D

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.02 Discuss the concept of the null hypothesis and its use in data analysis.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

 

 

11) Using the chi-square test for goodness of fit, how many degrees of freedom are in this data set?

1.   A) 1

2.   B) 2

3.   C) 3

4.   D) 4

 

Answer:  C

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.03 Evaluate the significance of experimental data based on the chi-square test.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

12) Given this data, use Table 5.2 to find the most accurate range within which the p value falls.

1.   A) 0.001 <p< 0.01

2.   B) 0.01 <p< 0.05

3.   C) 0.05 <p< 0.10

4.   D) 0.10 <p< 0.50

 

Answer:  A

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.03 Evaluate the significance of experimental data based on the chi-square test.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

13) What is a reasonable conclusion based on the chi-square analysis?

1.   A) There is a high probability that the deviation from the expected results is due chance.

2.   B) One can say with a high degree of confidence that genes C and D are linked.

3.   C) The analysis supports the null hypothesis.

4.   D) Genes C and D are most likely unlinked.

 

Answer:  B

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.03 Evaluate the significance of experimental data based on the chi-square test.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

 

14) If only 100 progeny had been counted and the same proportions of progeny genotypes observed, how would the p value and the conclusion drawn about linkage change?

1.   A) The p value would increase, and the likelihood of linkage decreases.

2.   B) The p value would decrease, and the likelihood of linkage increases.

3.   C) Neither the p value nor the likelihood of linkage would change.

4.   D) The p value would decrease, and the likelihood of linkage decreases.

 

Answer:  A

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.03 Evaluate the significance of experimental data based on the chi-square test.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

15) The R and S genes are linked and 10 map units apart. In the cross R s / r S × r s / r s what fraction of the progeny will be R S / r s?

1.   A) 5%

2.   B) 10%

3.   C) 25%

4.   D) 40%

5.   E) 45%

 

Answer:  A

Section:  05.01; 05.02

Topic:  Gene Linkage and Recombination; Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.; 05.02.03 Discuss the relationship between the recombination frequency and the map distance separating two loci on a chromosome.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

16) If the map distance between genes A and B is 20 map units and the map distance between genes B and C is 35 map units, what is the map distance between genes A and C?

1.   A) 15 map units

2.   B) 55 map units

3.   C) More information is needed to distinguish between 15 and 55 map units.

4.   D) Gene C must be located on a different nonhomologous chromosome.

 

Answer:  C

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.01 Establish relative gene positions using two-point cross data.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

17) In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild-type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct+ is crossed to a sn+ct male. The F1 flies are interbred. The F2 males are distributed as follows:

 

sn ct	13
sn ct+	36
sn+ct	39
sn+ct+	12

 

What is the map distance between sn and ct?

1.   A) 12 m.u.

2.   B) 13 m.u.

3.   C) 25 m.u.

4.   D) 50 m.u.

5.   E) 75 m.u.

 

Answer:  C

Section:  05.01; 05.03

Topic:  Gene Linkage and Recombination; Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.01.04 Explain how a testcross can provide evidence for or against linkage.; 05.03.01 Establish relative gene positions using two-point cross data.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

 

18) In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct+ is crossed to a sn+ct male. The F1 flies are interbred. The F2 males are distributed as follows:

 

sn ct	13
sn ct+	36
sn+ ct	39
sn+ ct+	12

 

Of these 4 genotypic classes of offspring, which arose from a parental gamete produced by the F1 females? (Select all that apply.)

1.   A) sn+ ct+

2.   B) sn ct

3.   C) sn+ ct

4.   D) sn ct+

 

Answer:  C, D

Section:  05.01

Topic:  Gene Linkage and Recombination

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

19) Suppose the L and M genes are on the same chromosome but separated by 100 map units. What fraction of the progeny from the cross  L M / l m  ×  l m / l m would be L m / l m?

1.   A) 10%

2.   B) 25%

3.   C) 50%

4.   D) 75%

5.   E) 100%

 

Answer:  B

Section:  05.02

Topic:  Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.02.04 Explain why the value of the recombination frequency between any two genes is limited to 50%.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

 

20) The pairwise map distances for four linked genes are as follows: A-B = 22 m.u., B-C = 7 m.u., C-D = 9 m.u., B-D = 2 m.u., A-D = 20 m.u., A-C = 29 m.u. What is the order of these four genes?

1.   A) ABCD

2.   B) ADBC

3.   C) ABDC

4.   D) BADC

5.   E) CADB

 

Answer:  B

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.01 Establish relative gene positions using two-point cross data.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

21) The zipper-like connection between paired homologs in early prophase is known as a

1.   A) spindle fiber.

2.   B) synaptic junction.

3.   C) synaptonemal complex.

4.   D) chiasma.

5.   E) None of the choices is correct.

 

Answer:  C

Section:  05.02

Topic:  Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.02.02 Describe the role of chiasmata in chromosome segregation during meiosis.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

22) The measured distance between genes D and E in a two-point testcross is 50 map units. Where are genes D and E in relation to each other? (Select all that apply.)

1.   A) D and E are on different homologous chromosomes.

2.   B) D and E are on the same chromosome, at least 50 map units apart.

3.   C) D and E are on the same chromosome, exactly 50 map units apart.

4.   D) D and E are on the same chromosome, less than 50 map units apart.

 

Answer:  A, B

Section:  05.02

Topic:  Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.02.03 Discuss the relationship between the recombination frequency and the map distance separating two loci on a chromosome.; 05.02.04 Explain why the value of the recombination frequency between any two genes is limited to 50%.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

23) The R and S genes are linked and 10 map units apart. In the cross R s / r S × r s / r s what percentage of the progeny will be R s / r s?

1.   A) 5%

2.   B) 10%

3.   C) 25%

4.   D) 40%

5.   E) 45%

 

Answer:  E

Section:  05.01; 05.02

Topic:  Gene Linkage and Recombination; Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.; 05.02.03 Discuss the relationship between the recombination frequency and the map distance separating two loci on a chromosome.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

24) If the recombination frequency between two genes is close to 50%, what could be true about the location of the two genes? (Select all that apply.)

1.   A) They are on nonhomologous chromosomes.

2.   B) They are far apart on the same chromosome.

3.   C) They are very close together on the same chromosome.

4.   D) None of the choices could be true.

 

Answer:  A, B

Section:  05.02

Topic:  Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.02.04 Explain why the value of the recombination frequency between any two genes is limited to 50%.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

 

25) A dihybrid testcross is made between genes H and I. Four categories of offspring are produced: H I, H i, h I, and h i. How many degrees of freedom would there be in a chi-square test for goodness of fit of the null hypothesis that the H and I genes are unlinked?

1.   A) 1

2.   B) 2

3.   C) 3

4.   D) 4

5.   E) 0

 

Answer:  C

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.03 Evaluate the significance of experimental data based on the chi-square test.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

26) Which process(es) can generate recombinant gametes? (Select all that apply.)

1.   A) crossing-over between two linked heterozygous loci

2.   B) independent assortment of two unlinked heterozygous loci

3.   C) segregation of alleles in a homozygote

4.   D) crossing-over between two linked homozygous loci

 

Answer:  A, B

Section:  05.01

Topic:  Gene Linkage and Recombination

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

27) Crossing-over takes place in bivalents (tetrads) consisting of ________ chromatids, and one crossover involves ________ chromatids.

1.   A) 2; 2

2.   B) 2; 4

3.   C) 4; 2

4.   D) 4; 4

5.   E) 8; 4

 

Answer:  C

Section:  05.02

Topic:  Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.02.01 Explain the physical process by which recombination takes place.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

 

 

28) In Drosophila, the genes y (yellow body) and car (carnation eyes) are located at opposite ends of the X chromosome. In doubly heterozygous females (y+car+ / ycar), a single chiasma is observed somewhere along the X chromosome in 90% of the examined oocytes. No X chromosomes with multiple chiasmata are observed. What percentage of the male progeny from such a female would be recombinant for y and car?

1.   A) 5%

2.   B) 10%

3.   C) 45%

4.   D) 55%

5.   E) 90%

 

Answer:  C

Section:  05.02; 05.03

Topic:  Recombination – A Result of Crossing-Over During Meiosis; Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.02.04 Explain why the value of the recombination frequency between any two genes is limited to 50%.; 05.03.01 Establish relative gene positions using two-point cross data.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

29) The Q gene locus is 10 map units from the R gene locus which is 40 map units from the S gene locus:

 

Q——10 m.u.——R——40 m.u.——S

 

Which interval would likely show the higher ratio of double to single chiasmata?

1.   A) Q-R

2.   B) R-S

3.   C) The ratios would be the same in the two intervals.

4.   D) Two chiasmata never occur in the same interval.

 

Answer:  B

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

 

30) The map of a chromosome interval is:

 

A——10 m.u.——B——40 m.u.——C

 

From the cross A b c / a B C × a b c / a b c, how many double crossovers would be expected out of 1000 progeny?

1.   A) 5

2.   B) 10

3.   C) 20

4.   D) 40

5.   E) 80

 

Answer:  D

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

31) The cross L p q / l P Q × l p q / l p q is carried out. If the L gene is in the middle, between genes P and Q, what would be the genotypes of the double crossover gametes in this cross?

1.   A) L P Q and l p q

2.   B) L p Q and l P q

3.   C) l p Q and L P q

4.   D) L p q and l P Q

5.   E) cannot be determined

 

Answer:  A

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

 

 

32) Suppose a three-point testcross was conducted involving genes X, Y, and Z. If the most abundant classes of progeny are X Y z and x y Z and the rarest classes are x Y Z and X y z, which gene is in the middle?

1.   A) X

2.   B) Y

3.   C) Z

4.   D) cannot be determined

 

Answer:  B

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

33) In Drosophila, the genes b, c, and sp are linked and arranged as shown below:

 

b——30 m.u.——c——20 m.u.——sp

 

This region exhibits 90% interference. How many double crossovers would be recovered in a three-point cross involving b, c, and sp out of 1000 progeny?

1.   A) 3

2.   B) 6

3.   C) 54

4.   D) 60

5.   E) 600

 

Answer:  B

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

 

34) In tetrad analysis, which result would indicate that two genes are linked?

1.   A) NPD = T.

2.   B) PD = T.

3.   C) PD = NPD.

4.   D) PD > NPD.

5.   E) PD > T.

 

Answer:  D

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.03 Describe how the relative numbers of PDs and NPDs can be used to establish linkage.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

35) In tetrad analysis, NPD asci can result from

(Select all that apply.)

1.   A) independent assortment of unlinked genes.

2.   B) double crossovers between linked genes.

3.   C) single crossovers between linked genes.

4.   D) single crossovers between a gene and a centromere.

 

Answer:  A, B

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.02 Differentiate between parental ditype (PD), nonparental ditype (NPD), and tetratype (T).

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

36) When analyzing octads (ordered tetrads), second-division (MII) segregations result from

1.   A) single crossovers between linked genes.

2.   B) double crossovers between linked genes.

3.   C) single crossovers between a gene and a centromere.

4.   D) independent assortment of unlinked genes.

5.   E) nondisjunction of homologs.

 

Answer:  C

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.01 Explain the meaning of the term tetrad as applied to the asci produced by certain fungi.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

37) Tetrad analysis shows that crossing-over occurs at the four-strand stage (i.e., after replication) because, when two genes are linked,

1.   A) NPD > T.

2.   B) T > NPD.

3.   C) T > PD.

4.   D) PD > NPD.

5.   E) PD > T.

 

Answer:  B

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.02 Differentiate between parental ditype (PD), nonparental ditype (NPD), and tetratype (T).

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

38) Sturtevant’s detailed mapping studies of the X chromosome of Drosophila supported what genetic principle?

1.   A) That genes are arranged in a linear order on the chromosomes.

2.   B) That genes are carried on chromosomes.

3.   C) That sex determination is controlled by the X and Y chromosomes.

4.   D) That segregation of an allelic gene pair is accompanied by disjunction of homologous chromosomes.

5.   E) That different pairs of chromosomes assort independently.

 

Answer:  A

Section:  05.02

Topic:  Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.02.03 Discuss the relationship between the recombination frequency and the map distance separating two loci on a chromosome.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

39) Suppose an individual is heterozygous for alternate alleles of gene A (Aa). Under what conditions would a crossover in a somatic cell of this individual lead to a clone of cells that are homozygous for a? (Pick the most precise answer.)

1.   A) The crossover would have to occur between the A locus and the centromere and involve two homologous (nonsister) chromatids.

2.   B) The crossover would have to occur between the A locus and the end of the chromosome and involve two homologous (nonsister) chromatids.

3.   C) The crossover would have to occur between the A locus and the end of the chromosome and involve two nonhomologous chromosomes.

4.   D) The crossover would have to occur between the A locus and the centromere and involve two sister chromatids.

 

Answer:  A

Section:  05.06

Topic:  Mitotic Recombination and Genetic Mosaics

Learning Objective:  05.06.01 Explain how mitotic recombination leads to the mosaic condition termed twin spots.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

40) If an individual is heterozygous at two loci (A b / a B) that are located on the same chromosome arm with A closer to the centromere than B, under what conditions would a crossover in a somatic cell generate a twin spot?

1.   A) The crossover would have to occur between the A locus and the centromere and involve two homologous, nonsister chromatids.

2.   B) The crossover would have to occur between the A locus and the B locus and involve two homologous, nonsister chromatids.

3.   C) The crossover would have to occur between the B locus and the end of the chromosome and involve two homologous, nonsister chromatids.

4.   D) A double crossover would have to occur, with one crossover between the A locus and the centromere and a second crossover between the A and B loci, and both crossovers would have to involve two homologous, nonsister chromatids.

5.   E) No crossover in a somatic cell could generate a twin spot.

 

Answer:  A

Section:  05.06

Topic:  Mitotic Recombination and Genetic Mosaics

Learning Objective:  05.06.01 Explain how mitotic recombination leads to the mosaic condition termed twin spots.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

41) Individuals heterozygous for the RB+ and RB− alleles can develop tumors as a result of (Select all that apply.)

1.   A) a mitotic crossover that leads to homozygosity for RB+in some cells and RB−in other cells.

2.   B) a somatic mutation in the RB+allele that leads to homozygosity for RB−.

3.   C) a somatic mutation in the RB−allele that leads to homozygosity for RB+.

4.   D) the fact that RB−is dominant to RB+.

 

Answer:  A, B

Section:  05.06

Topic:  Mitotic Recombination and Genetic Mosaics

Learning Objective:  05.06.01 Explain how mitotic recombination leads to the mosaic condition termed twin spots.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

42) Which type of tetrad contains two recombinant and two parental spores?

1.   A) PD

2.   B) NPD

3.   C) T

4.   D) ordered tetrads

5.   E) None of these types contain two recombinant and two parental spores.

 

Answer:  C

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.02 Differentiate between parental ditype (PD), nonparental ditype (NPD), and tetratype (T).

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

 

 

In Drosophila, singed bristles (sn) and carnation eyes (car) are both caused by recessive X-linked alleles. The wild-type alleles (sn+ and car+) are responsible for straight bristles and red eyes, respectively. A sn car female is mated to a sn+car+ male and the F1 progeny are interbred. The F2 are distributed as follows:

 

sn car	55
sn car+	45
sn+ car	45
sn+ car+	55
	200

 

 

43) If you want to analyze this data for evidence of linkage between sn and car, what is the null hypothesis?

1.   A) Genes sn and car are linked.

2.   B) Genes sn and car are not linked.

3.   C) Genes sn and car are located close together on the same chromosome.

4.   D) Crossing-over sometimes occurs between sn and car.

 

Answer:  B

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.02 Discuss the concept of the null hypothesis and its use in data analysis.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

 

44) What is the χ2 value for a chi-square test for goodness of fit of the null hypothesis?

1.   A) 0.5

2.   B) 1.0

3.   C) 2.0

4.   D) 0.4

5.   E) 20

 

Answer:  C

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.03 Evaluate the significance of experimental data based on the chi-square test.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

45) What is the p value from this test? (Pick the most accurate choice.)

1.   A) p> 0.5

2.   B) 0.1 <p< 0.5

3.   C) p< 0.1

4.   D) p< 0.05

5.   E) p< 0.01

 

Answer:  A

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.03 Evaluate the significance of experimental data based on the chi-square test.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

46) What does the data analysis allow you to conclude about linkage between sn and car?

(Select all that apply.)

1.   A) There is a high probability that the deviations from the expected number of F2in each genotype class are due to chance.

2.   B) The data do not allow rejection of the null hypothesis.

3.   C) The p value is high meaning that the data is significant.

4.   D) There is good evidence thatcn and car are linked.

 

Answer:  A, B

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.03 Evaluate the significance of experimental data based on the chi-square test.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

In humans, the genes for red-green color blindness (R = normal, r = color blind) and hemophilia A (H = normal, h = hemophilia) are both X-linked and only 3 map units apart.

 

47) Suppose a woman has four sons, and two are color blind but have normal blood clotting and two have hemophilia but normal color vision. What is the probable genotype of the woman?

1.   A) H R / h r

2.   B) H r / h r

3.   C) h r / h R

4.   D) H r / h R

5.   E) H R / H r

 

Answer:  D

Section:  05.01; 05.02

Topic:  Gene Linkage and Recombination; Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.; 05.02.03 Discuss the relationship between the recombination frequency and the map distance separating two loci on a chromosome.

Bloom’s:  3. Apply

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48) A woman whose mother is color blind and whose father has hemophilia A is pregnant with a boy. If the alleles for color blindness and hemophilia A are rare in the population, what is the probability that the baby will have normal vision and normal blood clotting?

1.   A) 0

2.   B) 0.03

3.   C) 0.485

4.   D) 0.47

5.   E) 0.015

 

Answer:  E

Section:  05.01; 05.03

Topic:  Gene Linkage and Recombination; Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.; 05.03.01 Establish relative gene positions using two-point cross data.

Bloom’s:  3. Apply

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In Drosophila, the autosomal recessive pr and cn mutations cause brown and bright-red eyes, respectively (wild-type flies have brick-red eyes). Flies who are homozygous recessive at both pr and cn have orange eyes. A female who has wild-type eyes is crossed to an orange-eyed male. Their progeny have the following distribution of eye colors:

 

wild-type	8
brown	241
bright-red	239
orange	12
	500

 

49) Which phenotypes are parental?

1.   A) wild-type and orange

2.   B) brown and bright-red

3.   C) wild-type and brown

4.   D) bright-red and orange

5.   E) There is no way to determine this.

 

Answer:  B

Section:  05.01

Topic:  Gene Linkage and Recombination

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.; 05.01.04 Explain how a testcross can provide evidence for or against linkage.

Bloom’s:  4. Analyze

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50) What is the genotype of the wild-type mother of these progeny?

1.   A) pr cn / pr+cn+

2.   B) pr+cn / pr+cn

3.   C) pr+cn / pr cn+

4.   D) pr cn+ / pr cn+

5.   E) pr cn / pr cn

 

Answer:  C

Section:  05.01

Topic:  Gene Linkage and Recombination

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.; 05.01.04 Explain how a testcross can provide evidence for or against linkage.

Bloom’s:  4. Analyze

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51) The mother of these progeny resulted from a cross between two flies from true-breeding lines. What are the genotypes of these two lines?

1.   A) pr cn+/ pr cn+and pr+cn / pr+cn

2.   B) pr+cn+/ pr+cn+and pr cn / pr cn

3.   C) pr+cn+/ pr cn and pr cn / pr cn

4.   D) pr+cn / pr cn and pr cn+/ pr cn

5.   E) More than one of these could be true.

 

Answer:  A

Section:  05.01

Topic:  Gene Linkage and Recombination

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.; 05.01.04 Explain how a testcross can provide evidence for or against linkage.

Bloom’s:  4. Analyze

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