Evolution 1st edition bergstrom test bank

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Chapter 9: Evolution at Multiple Loci MULTIPLE CHOICE 1. What two major concepts in biology were unified in the middle of the twentieth century? a. Natural selection and drift b. Discontinuous and continuous variation c. Mutation and migration d. Natural selection and Mendelian genetics e. Population genetics and quantitative genetics ANS: D MSC: Factual

DIF: Easy

REF: 9.1

TOP: I

2. The rediscovery of Mendelian inheritance caused increased scrutiny of Darwin’s theory of natural selection because of an apparent conflict in the two concepts. This conflict was based on the observation that Darwin focused on continuous traits, whereas Mendelian genetics deals with a. discrete characters. b. alleles. c. flowers. d. genotypes. e. All of the above ANS: A MSC: Applied

DIF: Easy

REF: 9.1

TOP: I.A

3. The observation of an offspring phenotype outside of the range of parental phenotypes is most likely due to a. additive genetic effects. b. mutation. c. selection. d. latent variation. e. C and D ANS: E MSC: Factual

DIF: Easy

REF: 9.1

TOP: I.C

4. Selection can generate novel phenotypes by a. drift causing the creation of a new phenotype. b. increasing the probability of novel allele combinations. c. linkage disequilibrium and selective sweeps. d. B and C e. None of the above ANS: B MSC: Factual

DIF: Easy

REF: 9.1

5. A haplotype is a. all of the alleles at a single locus. b. a product of epistasis. c. a set of alleles at different loci along a chromosome. d. all of the alleles in a population. e. None of the above

TOP: I.D

ANS: C MSC: Factual

DIF: Easy

REF: 9.2

TOP: II.A

6. Consider a population with two loci (A, B), each with two alleles (A, a; B, b). What are the haplotype frequencies in this population if the allele frequencies are A = 0.5, a = 0.5, B = 0.7, b = 0.3? a. AB = 0.35, Ab = 0.15, aB = 0.35, ab = 0.15 b. AB = 0.45, Ab = 0.05, aB = 0.25, ab = 0.25 c. AB = 0.4, Ab = 0.1, aB = 0.3, ab = 0.2 d. AB = 0.125, Ab = 0.25, aB = 0.25, ab = 0.125 e. Cannot be determined based on the information provided ANS: E MSC: Applied

DIF: Medium

REF: 9.2

TOP: II.A

7. Consider two loci on a single chromosome. These loci a. have alleles that will always segregate together. b. are physically linked. c. are in linkage disequilibrium. d. interact with each other. e. will be broken up during recombination.

ANS: B MSC: Applied

DIF: Medium

REF: 9.2

TOP: II.A

8.

The figure above demonstrates that Hardy–Weinberg frequencies are a. uniquely determined by allele frequencies. b. determined by haplotype frequencies. c. depend on the recombination rate. d. All of the above e. None of the above ANS: B MSC: Factual

DIF: Easy

REF: 9.2

TOP: II.B

9. The statistical association of alleles at different loci is referred to as a. physical linkage. b. haplotypes. c. epistasis. d. linkage equilibrium. e. linkage disequilibrium.

ANS: E MSC: Factual

DIF: Easy

REF: 9.2

TOP: II.C

10. What is the largest possible value that linkage disequilibrium can reach if all allele frequencies are 0.5? a. 0.25 b. 0.5 c. 0.75 d. 1 e. None of the above ANS: A MSC: Applied

DIF: Easy

REF: 9.2

TOP: II.D

11. If the value of linkage disequilibrium (D) is equal to –0.25, one can infer that the alleles are a. in repulsion. b. in coupling. c. in linkage disequilibrium. d. A and C e. None of the above ANS: D MSC: Applied

DIF: Medium

REF: 9.2

TOP: II.D

12. Which of the following can generate linkage disequilibrium? a. Drift b. Selection c. Mutation d. Migration e. All of the above ANS: E MSC: Factual

DIF: Easy

REF: 9.2

TOP: II.E

13. Which of the following processes that can create linkage disequilibrium is illustrated in the figure below?

a. b. c. d. e.

Drift Selection Mutation Migration All of the above

ANS: D MSC: Factual

DIF: Easy

REF: 9.2

TOP: II.E

14. In the figure below, the grey boxes represent the offspring of parents that are heterozygous at locus 1 (Aa) and locus 2 (Bb). Which of the grey boxes represent repulsion double heterozygote offspring?

a. b. c. d. e.

Light grey Medium grey Dark grey B and C All of the above

ANS: C MSC: Factual

DIF: Easy

REF: 9.2

TOP: II.F

15. Assuming the value of linkage disequilibrium (D) is positive, recombination will generate a. more repulsion pairs.

b. c. d. e.

more coupling pairs. no change, since recombination does not change expected haplotype frequencies. an increase in D. None of the above

ANS: A MSC: Applied

DIF: Medium

REF: 9.2

TOP: II.F

16.

In the figure above, a new beneficial allele, B, arises on a chromosome containing the ordered loci A, B, C, D, and E, depicted in the top panel. After reproduction, those chromosomes possessing the B allele increase in frequency. Over time, recombination breaks up the association of the B allele with the A, C, D, and E alleles. Assuming that the five loci are spaced evenly along the chromosome, the first pair of allele combinations to be broken up will be a. B and A. b. B and C. c. B and D. d. B and E. e. All alleles will break up at an equal rate ANS: D MSC: Applied

DIF: Easy

REF: 9.2

TOP: II.G

17.

In the figure above, each new line represents the frequency of a new beneficial mutation. The figure describes the process of a. selective sweeps. b. periodic selection. c. genetic hitchhiking. d. background selection. e. A and B ANS: E MSC: Factual

DIF: Medium

REF: 9.2

TOP: II.G

18. Understanding phenotype–genotype relationships is difficult because of a. pleiotropy. b. dominance. c. epistasis. d. A and C e. All of the above ANS: E MSC: Factual

DIF: Easy

REF: 9.3

TOP: III.A

19.

In the figure above, the population indicated by the dots a. has a high fitness. b. exhibits two distinct morphs c. is experiencing background selection. d. is experiencing drift. e. is in equilibrium.

ANS: A

DIF: Easy

REF: 9.3

TOP: III.A

MSC: Applied 20. The following figure demonstrates the genotypic adaptive landscape for bacteria growing in media without antibiotics. The height of the bars associated with a genotype indicates the fitness of that genotype. The R allele confers resistance to antibiotics, and the C allele compensates for the loss of fitness that the R allele confers in the absence of antibiotics. The rc haplotype lacks both compensation and resistance. Based on this landscape, why are resistance and compensation unlikely to be lost even in the absence of antibiotics?

a. b. c. d. e.

A lack of mutation will prevent the haplotypes from changing.

Epistasis will maintain haplotype frequencies.

Haplotype frequencies do not change in Hardy–Weinberg populations.

RC populations are stuck on a local maximum.

Bacteria evolve slowly.

ANS: D MSC: Applied

DIF: Medium

REF: 9.3

TOP: III.B

21. Genotypic adaptive landscapes may better represent true adaptive landscapes than phenotypic adaptive landscapes when a. single mutations can have a large effect on the phenotype. b. phenotypic landscapes are hard to mathematically model. c. phenotypic landscapes do not account for drift. d. genotypic landscapes are easier to visualize. e. genotypic landscapes are smooth.

ANS: A MSC: Factual

DIF: Easy

REF: 9.3

TOP: III.B

22. In quantitative genetic terms, a phenotypic value can be expressed as a. dominance and recessiveness.

b. c. d. e.

epistasis and pleiotropy. genotypic value and environmental deviation.

A and B All of the above

ANS: C MSC: Factual

DIF: Easy

REF: 9.4

TOP: IV.A

23. Narrow-sense heritability is defined as the total a. genetic variation divided by the total phenotypic variation. b. additive genetic variation divided by the total phenotypic variation. c. dominance genetic variation divided by the total phenotypic variation. d. additive genetic variation divided by the total dominance variation. e. epistatic genetic variation divided by the total dominance variation.

ANS: B MSC: Factual

DIF: Easy

REF: 9.4

TOP: IV.B

24. In the figure below, what does the straight line indicate?

a. b. c. d. e.

The broad-sense heritability The narrow-sense heritability The norm of reaction of nocturnal restlessness None of the above All of the above

ANS: B MSC: Applied

DIF: Medium

REF: 9.4

TOP: IV.B

25. A researcher tracks the escape behavior of a population of squirrels on campus across two generations.

When she plots the escape behavior of the offspring generation against the parental behavior, she notes that the offspring’s behavior exactly matches the parental behavior. Nonetheless, the narrow-sense heritability equals zero. How can the heritability equal zero in this example? a. The additive genetic variance equals zero. b. All the escape behavior alleles are fixed in the population. c. There is no phenotypic variation. d. A and B e. All of the above ANS: D MSC: Applied

DIF: Medium

REF: 9.4

TOP: IV.B

26. The figure below describes the percent oil content in corn from different generations. What is the expected slope of a parent–offspring regression for corn oil content?

a. b. c. d. e.

0.69 1.6 2.1 2.5 Cannot be determined based on the information provided

ANS: A MSC: Applied

DIF: Difficult

REF: 9.4

TOP: IV.C

27. Texas longhorn cattle have been bred to increase the tip to tip width of their horns, and the heritability of this trait has been well studied by Professor David Hillis at the University of Texas, Austin. In the current generation, the mean horn length is 52 inches. If only a subset of this population, with a mean horn length of 61 inches, are allowed to reproduce, and the narrow-sense heritability of this trait = 0.2, what is the expected horn length in the offspring generation? a. ~1.8 inches b. ~9 inches c. ~54 inches d. ~61 inches e. Cannot be determined based on the information provided ANS: C MSC: Applied

DIF: Difficult

REF: 9.4

TOP: IV.D

SHORT ANSWER 1. A species of lotus produces a nearly continuous spectrum of flower colors, ranging from white to deep red. The color of the flower is determined by 10 loci, each with two codominant alleles. What concept does this example illustrate?

ANS: Polygenic phenotypic variation DIF: Medium

REF: 9.1

TOP: I.B

MSC: Applied

2. In a population of dragonflies, wing structure is determined by two loci (A, B), each with two alleles (Aa, Bb). Abnormal wings are produced in aa individuals unless they possess at least one B allele. This is an example of ANS: epistasis.

DIF: Medium

REF: 9.1

TOP: I.C

MSC: Applied

3. Consider two microsatellite loci (A, B) each with two alleles determined by the number of repeats of the microsatellite motif (A = 15 or 17; B = 27 or 31). You genotype an individual in this population and determine that its haplotype is 15/27 and 15/27. Are these alleles in coupling and/or repulsion?

Explain your answer.

ANS: These alleles could be in either coupling or repulsion depending on how the researcher decides to refer to the four alleles.

DIF: Medium

REF: 9.2

TOP: II.D

MSC: Conceptual

4. Consider two loci (A, B) each with two alleles (Aa, Bb). The frequency (f) of A is known to be 0.3, the frequency (f) of b is known to be 0.25, and the coefficient of linkage disequilibrium (D) between loci A and B is known to be 0.15. What is the expected haplotype frequency of haB? Please show your work.

ANS: Equation: D = fafB – haB Rearrange to: haB = fafB – D fa = 1 – 0.3 = 0.7

fB = 1 – 0.25 = 0.75 D = 0.15 Then substitute and solve: haB = (0.7)(0.75) – 0.15 haB = 0.525 – 0.15 haB = 0.375 DIF: Medium

REF: 9.2

TOP: II.D

MSC: Conceptual

5. In Lake Malawi cichlid fish, one can observe two color variants, Black Barred (BB) and Orange Blotch (OB). Sex in these fish is not determined by a sex chromosome but by a gene on an autosome, which produces either males or females. Both of these phenotypes are controlled by one biallelic locus each.

OB females have a selective advantage in terms of their survivability: The OB phenotype allows females to blend into their background. OB males have a selective disadvantage because their reproductive coloration is disrupted and they have little mating success. How would this process influence the association of alleles at the color locus and sex determining locus?

ANS: Natural selection would generate linkage disequilibrium between the two loci such that the female allele would become associated with the OB allele and the male allele would become associated with the BB allele.

DIF: Difficult

REF: 9.2

TOP: II.E

MSC: Conceptual

6. If recombination between locus A and locus B is 0.15, and the initial value of linkage disequilibrium is 0.20, what will the value of linkage disequilibrium be after one generation of recombination? Show your work.

ANS: D = –rD D = –0.15  0.20 D = –0.03 D(t + 1) = 0.20 – 0.03 = 0.17 DIF: Medium

REF: 9.2

TOP: II.F

MSC: Applied

7. Multiple drug-resistant Staphylococcus aureus (MRSA) is a bacterium that has developed resistance to a wide range of antibiotics. Its resistance to penicillin, methicillin, and oxacillin is a result of the serial use of these antibiotics, which created a selective environment that favored resistance to the drug being used at the time. This resistance provides the bacteria with the ability to thrive in an environment with standard antibiotics, thereby making it a dangerous human health hazard. Why would one expect MRSA to retain penicillin resistance even if this antibiotic is rarely used?

ANS: The MRSA strain has experienced several rounds of periodic selection, which has generated a single nonrecombining chromosome that conveys resistance to multiple drugs. In the absence of sexual reproduction and recombination, these alleles are unlikely to be disrupted.

DIF: Difficult

REF: 9.2

TOP: II.G

8. Explain why a population may never reach its most fit phenotype.

ANS:

MSC: Conceptual

If the adaptive landscapes have multiple peaks, the population will tend to climb the nearest peak, even if it is not the highest peak. In the absence of drift, natural selection will keep the population on this local maximum, thereby preventing it from reaching the global maximum.

DIF: Difficult

REF: 9.3

TOP: III.A

MSC: Conceptual

9. Why would we expect genotypic adaptive landscapes to have rugged contours?

ANS: Single mutations can have large phenotypic effects.

DIF: Difficult

REF: 9.3

TOP: III.B

MSC: Conceptual

10. A study of susceptibility to major depressive disorder in monozygotic twins separated at birth revealed that the broad-sense heritability of this trait is 0.10, and its total phenotypic variance is 0.25. What is the variance of the genotypic value of this disorder? Show your work.

ANS: H2 = VG/(VG+VE) and VP = VG+VE Therefore: H2 = VG/VP VP  H2 = VG 0.25  0.1 = 0.025 VG = 0.025 DIF: Difficult

REF: 9.4

TOP: IV.A

MSC: Applied

11. Without using mathematical symbols, explain the difference between broad- and narrow-sense heritability, including why one is more relevant to evolutionary studies.

ANS: Broad-sense heritability describes the proportion of the total phenotypic variance described by the genetic variance. Narrow-sense heritability describes the proportion of the total phenotypic variance described by the additive genetic variance. The narrow-sense heritability is more valuable to evolutionary studies as it focuses strictly on the additive genetic variation, and it is the additive genetic variance on which natural selection operates. In contrast, the broad-sense heritability includes many variance components whose impact on the phenotype are context dependent. Therefore, these variance components may not be exposed to the action of natural selection.

DIF: Difficult

REF: 9.4

TOP: IV.B

MSC: Conceptual

12. Explain why one cannot compare the estimated heritability of intelligence tests from different populations from the United States.

ANS: All heritability estimates are within-population measures not between-population measures. This is due partly to the different environmental variances found in different populations.

DIF: Difficult

REF: 9.4

TOP: IV.C

MSC: Conceptual