- asmanian devils once inhabited most of present day Australia, but only an isolated population on the island of Tasmania has survived to present day. Which of the following processes has likely affected Tasmanian devils as a result of this history:
- a higher mutation rate
- stronger natural selection
- a genetic bottleneck
- gene flow
- The effectiveness of selection on an allele depends in part on:
- the frequency of the allele
- the magnitude of average excess fitness
- the average fitness of the population
- all of the above
- none of the above
- Inbreeding:
- increases heterozygosity in populations
- creates deleterious recessive alleles
- increases homozygosity in populations
- all are correct
- b and c are correct
- Genetic drift:
- will always lead to higher fitness of individuals in the population
- reduces genetic variation within a population
- can lead to divergence between populations
- b and c are correct
- a, b, and c are correct
- In a population of infinite size, which statement accurately describes the eventual fate of a new beneficial allele?
(a) If it is dominant, it will reach fixation; if it is recessive it will rise to high frequency but not reach fixation
(b) If it is dominant it will rise to high frequency but will not reach fixation; if it is recessive it will reach fixation
(c) Since it is advantageous, it will reach fixation regardless of whether it is dominant or recessive
(d) Regardless of whether it is dominant or recessive, it will rise to high frequency but not reach fixation
- Lively and Dybdahl studied parasite infection rates in a population of asexual clonal snails. The graph above shows relative infection rates for the four most common clone genotypes and for several rare genotypes (all lumped together). Based on these data they hypothesized that parasites adapted to infecting the most common clone genotypes in the population, and thus these genotypes had lower fitness. This is consistent with ________operating in the population. Further evidence would be provided if:
- genetic drift; heterozygosity declined over time in the population
- genetic drift; rare clones were lost from the population frequency-dependent selection;
- negative frequency-dependent selection; rare clones became common in the next generation, but then declined in frequency in the following generation
- negative frequency-dependent selection; rare clones became more common until they completely replaced the clones that were originally common
- Assuming that a deleterious allele is maintained in a population by mutation-selection balance, which scenario below describes the case where you would expect the equilibrium frequency of the allele to be highest?
- the mutation rate is low; the allele is highly deleterious
- the mutation rate is low; the allele is slightly deleterious
- the mutation rate is high; the allele is highly deleterious
- the mutation rate is high; the allele is slightly deleterious
- Inbreeding results in a higher frequency of ________ in a population. Inbreeding depression occurs because _______.
- deleterious alleles; individuals with deleterious alleles have high mortality
- heterozygosity; heterozygotes have lower fitness
- homozygosity; deleterious recessive alleles are expressed more often
- heterozygosity; deleterious dominant alleles are expressed more often
- Below you see graphs that depict the change in frequency of a neutral allele in four populations that differ in size. Which population would you predict is the smallest? Answer: D
- The frequency of a slightly deleterious allele maintained at an equilibrium frequency by mutation-selection balance would be higher:
- if the mutation rate is high
- if the mutation rate is low
- if the selection coefficient is high
- if the population size is small
- Genetic drift:
- reduces genetic variation within a population
- increases divergence (variation) between populations
- affects only neutral alleles
- a and b are correct
- a, b, and c are correct
- Many plant species are hermaphroditic and run the risk of self-mating. Some species carry self-incompatibility alleles that can prevent this from occurring. If a pollen grain with self-incompatibility allele S1 lands on a stigma that also carries the S1 allele the pollen will not germinate and fertilization does not occur. Thus, this mechanism not only prevents selfing, but also has the unfortunate effect of preventing mating with any other plant that carries the same allele. However, if the pollen lands on a stigma of a plant with a different allele, fertilization occurs. Imagine a population of plants in which the allele frequency of S1=0.9 and the allele frequency of S2=0.1. All other things being equal, individuals with the ____ allele will have higher fitness on average. This is an example of______.
- S1; positive selection
- S2; positive selection
- S1; negative frequency-dependent selection
- S2; negative frequency-dependent selection
- You collect the following data on genotypes for a sunflower population: AA: 40, AB: 20, BB: 40. Based on Hardy-Weinberg predictions you expected the following numbers: AA: 25, AB: 50, BB: 25. Which of the following is a plausible explanation for the deviation?
- Balancing selection
- Negative frequency-dependent selection
- Inbreeding
- Genetic drift
- Some Drosophila melanogaster larvae use a “sitting” foraging strategy in which they feed more or less in the same location, while “rovers” wander around the substrate looking for more food sources. In the graph above, the red line corresponds to sitters and the purple line corresponds to rovers. This is an example of _______; over time we expect________:
- Negative frequency-dependent selection; both strategies to persist in the population
- pleiotropy; both strategies to persist in the population
- frequency-dependent selection; the rover strategy to replace the sitter strategy because it has the highest fitness
- pleiotropy; the rover strategy to replace the sitter strategy because it has the highest fitness
- Mutations in the GDF9 gene in sheep have been linked to changes in female fecundity. The following are the relative fitnesses of different genotypes in the population:
Relative fitness
+/ – 1
+/ + 0.7
-/ – 0.1
- this is an example heterozygote advantage; genetic variation will be maintained over time
- this is an example of negative frequency-dependent selection; genetic variation will be maintained over time
- this is an example of heterozygote advantage; genetic variation will be lost over time
- this is an example of negative frequency-dependent selection; genetic variation will be lost over time
- The graph above shows the change in allele frequency for a beneficial allele over time (x-axis shows generations). Based on the shape of the curve this allele is most likely:
(a) homozygous
(b) dominant
(c) recessive
(d) heterozygous
(e) additive
- Considering the principles of mutation, natural selection, and genetic drift do you expect adaptive evolution to occur more rapidly in small or large populations? What about non adaptive evolution? For each answer, please explain your reasoning.
Adaptive evolution will occur more rapidly in a large population. Adaptation occurs through the process of natural selection, which is more effective in large populations because there are more mutations (some may be beneficial) and the effects of random genetic drift are relatively weak compared to a small population. Non-adaptive evolution would be more likely in a small population because genetic drift is stronger. Genetic drift is the result of random sampling error, without regard to the fitness effects of alleles. Thus, in a small population, slightly deleterious alleles may be fixed by drift.
- You are studying a population of 100 flowers that has two alleles at a locus for flower color, blue (B) and green (G). There are 15 individuals with the BB genotype, 70 individuals with the BG genotype, and 15 individuals with the GG genotype.
(a) (6 pts) What are the allele frequencies of B and G in the starting population? Show your calculations.
Freq B= 15 x 2 +70 = 100/200=0.5
Freq G= 15 x 2 +70 = 100/200=0.5
(b) (5 pts) Is this population in Hardy-Weinberg equilibrium? Show your calculations.
Expected genotype frequencies from HW equation:
BB = 0.52 = 0.25
GG = 0.52 = 0.25
BG = 2 x 0.5 x 0.5 = 0.5
Actual genotype frequencies:
BB= 15/100 = 0.15
GG = 15/100 = 0.15
BG = 70/100 = 0.7
No, the population is not in HW equilibrium because expectations are not met.
(c) (3 pts) Given the results of part b and the distribution of genotypes, offer a hypothesis that could explain the results—explain your reasoning.
There is an excess of heterozygotes, which could be explained by heterozygote advantage.
- Discuss the effectiveness of genetic drift and natural selection in small vs. large populations. Please be sure to explain why each process is stronger/weaker depending on the population size.
Natural selection is stronger in a large population because there are more mutations, some of which may be beneficial. Moreover, because drift is weak in large populations selection is more likely to overcome the random loss of new beneficial mutations. Natural selection is weaker in small populations because there are fewer mutations, and the random effects of drift are likely to overpower selection in determining the fate of an allele.
Drift is stronger is small populations because it results from random sampling error, which is more prominent in a small population. Sampling error is reduced in a large population, making drift relatively weaker.
- Please describe two factors that would increase the likelihood of fixing a beneficial allele in a population of finite size.
- A higher starting frequency means an allele is less likely to be lost by drift and more likely to be fixed by selection.
- A larger excess fitness means selection for the allele will be stronger and it will be more likely to fix.
- The larger the population size the more likely the allele will fix by selection. Drift will be weaker in a large population meaning that the beneficial allele is less likely to be lost by drift.
- (6 pts) A researcher performs an experiment on fruit flies to monitor the change in allele frequency of an allele called “A.” She starts with 24 populations, each with an initial starting frequency for A of 0.5. Flies are maintained for 10 generations by transferring the offspring from each generation to a new vial, where they produce the next generation. For half of the populations she randomly selects 20 flies to transfer, while for the other half she randomly selects 200 flies to transfer. After 10 generations she collects the following allele frequency data:
Treatment 1: 0.55, 0.6, 0.2, 0.9, 0.45, 0.35, 0.1, 0.65, 0.65. 0.55, 0.75, 0.35, Treatment 2: 0.85, 0.8, 0.75, 0.8, 0.75, 1.0, 0.8, 0.85, 0.9, 0.8, 0.85, 0.8
What is a plausible explanation for the differences between the treatments? Please make sure to explain your logic.
Considering the fact that the allele increased in frequency in all replicates of treatment two, we can assume this allele is beneficial. In treatment one, even though the allele is beneficial it increases in some replicates but not in others. This discrepancy can be explained by the fact that treatment one had a smaller population size transferred each generation. As a result, drift was strong and, since it is a random process, the allele frequency fluctuated more or less randomly.
- The graph above depicts the change in frequency for an advantageous allele in two different populations, both of infinite size. The strength of selection is the same in both populations.
(a) What type of allele is this? Explain how you know.
This is a recessive allele. You can tell by the shape of curve. Beneficial recessive alleles at low frequency are rarely present in the homozygous recessive genotype, which is the only genotype that has the advantageous phenotype. As a result, low frequency beneficial recessives rise very slowly in frequency until they become more common. This is clearly the case in both simulations above.
(b) Why does the frequency of the allele in one of the populations rise faster than the frequency in the other population?
The starting frequency is higher in the one that rises faster.
(c) If given enough time, will the allele become fixed in each of these populations?
Yes, in both cases the allele will be fixed by natural selection. In this case, the dominant allele is deleterious and can be eliminated by selection because the heterozygote expresses the dominant phenotype.
- (a) Contrast evolution by natural selection with evolution by genetic drift?
Evolution by natural selection is non-random and adaptive, while evolution by genetic drift is random and non-adaptive.
- The earth’s biotic and abiotic environments are changing rapidly due, in part, to human activities. For example, the introduction of non-native invasive species into new habitats and climate change highlight two ways in which humans are altering the environment experienced by other species. Some species will probably adapt to these changes while others may not. Considering the processes of mutation, natural selection, and genetic drift comment on the likelihood of adaptation to environmental change for species that have small population sizes vs. species with large population sizes. At a minimum, a fully correct answer will incorporate all three of these processes into the answer.
Large populations are more likely to adapt than small populations. In large populations, there will be more mutations, increasing the chances for a beneficial mutation to occur. Natural selection (which results in adaptation) will be more effective not only because there are more beneficial mutations, but also because the random effects of drift are weak in a large population. This means that new beneficial mutations are less likely to be lost by drift—even slightly beneficial mutations can be fixed by selection. The converse is true in small populations. There will be a smaller number of new mutations, and genetic drift, which is random and non-adaptive, will overpower natural selection.
- (a) The graphs above show the results of simulations of the effect or selection on deleterious alleles. Population size is infinite in both simulations and the starting frequency and the strength of selection are the same. Based on the shape of the curves, why do the results of the simulations differ? Explain your answer.
In the top simulation the deleterious allele is dominant. You can tell because the frequency drops rapidly, and the allele is completely eliminated by the end. A deleterious dominant can be eliminated because it will be expressed in the heterozygous genotype, meaning that the last copy of the allele in the population is visible to selection.
In the bottom simulation the deleterious allele is recessive. You can tell because the rate of frequency decline decelerates as the frequency drops and the allele is not fully eliminated by the end. This happens because when deleterious recessive alleles are rare, they are most often found in heterozygotes, where the allele is not visible to selection because heterozygotes display the dominant phenotype.
(b) The allele in the bottom simulation is not eliminated entirely from the population. Would this change if the population was finite in size? Why or why not?
Yes, it would most likely be eliminated. Selection would bring the allele down to low frequency. Since the population is finite drift would also have an effect, and would most likely eliminate the allele (low frequency alleles are likely to be lost by drift).
- The graph above shows results of two simulations, both depicting the rise in frequency of beneficial allele in a population of infinite size. The selection coefficient and the starting frequency are the same, but in one simulation the beneficial allele is dominant and in the other it is recessive. Neither allele is fixed by 500 generations.
(a) Which simulation shows results for a dominant and which shows results for a recessive allele? How can you tell?
The allele in simulation one is dominant and the allele in simulation two is recessive. You can tell because dominant alleles will rise in frequency more rapidly because the advantageous allele will be expressed in homozygotes and heterozygotes. Recessive alleles will rise slowly in frequency at first because the allele is only expressed in homozygotes. When the allele is rare, it will most likely be present in heterozygotes.
(b) Neither of the alleles reaches fixation by 500 generations. If given enough time, will both of these alleles reach fixation in the population? Why or why not?
The dominant allele will not reach fixation because the deleterious recessive allele cannot be eliminated by selection alone because it can “hide” from selection in the heterozygous genotype. The recessive allele will reach fixation because the deleterious dominant can be eliminated by selection because the allele is visible to selection in heterozygotes.
- The graph above depicts the rise in resistance to warfarin in a rat population. Notice that after reaching a peak of 100% resistance, resistance in the population declined. Please provide a plausible evolutionary explanation for this.
Genes conferring resistance to warfarin appear to have pleiotropic effects, resulting in a trade-off. When selection pressure from poisoning is present resistant individuals have the highest fitness. When the poisoning program ceased, this selection pressure was removed and non-resistant individuals had the highest fitness.
Test bank, Chapter 7
- The residues (“tailings”) of mines often contain such high concentrations of toxic metals (e.g., copper, lead) that most plants are unable to grow on them. However, some hardy species (e.g. certain grasses) are able to spread from the surrounding uncontaminated soil into such areas. These plants evolve resistance to the toxic metals while their ability to grow on uncontaminated soil decreases. Because grasses are wind pollinated, breeding between the resistant and nonresistant populations continues, but offspring of crosses between the two populations are intermediate and have low fitness. This is an example of:
- stabilizing selection
- directional selection
- frequency-dependent selection
- disruptive selection
(e) underdominance
- If the heritability of a trait in a population is low:
- few individuals have the trait
- most of the phenotypic variation for the trait in the population is due to environmental differences experienced by individuals
- most of the phenotypic variation for the trait in the population is due to genetic differences among individuals
- there are no genes that code for the trait
- b and d are correct
- The selection differential for a quantitative trait is ____________. Assuming heritability is the same in all cases, a larger selection differential ___________.
- the amount that the mean of the trait changes in the next generation; means there was more change in the trait between generations
- the difference between the mean for the trait in the original population and the mean of those individuals that reproduced; means there will be more change in the trait in the next generation.
- the difference between the mean for the trait in the original population and the mean of those individuals that reproduced; means there will be less change in the trait in the next generation.
- the amount that the mean of the trait changes in the next generation; means there was less change in the trait between generations
- The greyhound is a dog breed known for its amazing speed. This breed was produced by taking the fastest animals and breeding them together generation after generation. Though artificial, this is an example of __________.
- stabilizing selection
- disruptive selection
- overdominance
- directional selection
- underdominance
- The heritability of a trait describes:
(a) the proportion of phenotypic variation in the population that is due to genetic differences between individuals
(b) whether selection on a trait will result in evolutionary change
(c) the proportion of an individual’s phenotype that is due to its genotype
(d) a and b are correct
(e) a, b, and c are all correct
- Given the graph above, and assuming that corolla flare is heritable, what type of selection should occur?
- Directional
- Stabilizing
- Disruptive
- Overdominance
- Frequency-dependent
- A trait with heritability = 0:
- must have no genetic basis
- means that most of the phenotypic variation in the population is explained by differences in the alleles that individuals have
- means that there is no phenotypic variation
- means that the phenotypic variation in the population is explained by differences in the environments experienced by individuals
- a and d are correct
- The graph above plots survival vs. birth weight for human babies. What type of selection is operating?
- stabilizing
- disruptive
- directional
- negative frequency-dependent
- Broad sense heritability differs from narrow sense heritability in that:
- for broad sense heritability, VG includes only additive effects of alleles
- for broad sense heritability, VG includes additive, dominance, and epistatic effects
- broad sense heritability includes only components of variance that cause offspring to resemble their parents
- all of the above
- none of the above
- The genetic component of variance for narrow sense heritability includes:
- additive effects
- dominance effects
- G X E interactions
- A and b are correct
- A, b, and c are correct
- Below are some different scenarios for the heritability of a trait and the selection differential in a population. Which scenario will produce the largest response to selection?
(a) h2=0.9; S=0.6
- h2=0.5; S=0.9
- h2=0.2; S=0.1
- h2=0.5; S=0.5
- Which of the following is true of QTL association studies:
- they pinpoint the specific gene(s) responsible for phenotypic variation
- F1 offspring are examined to look for genetic markers linked to phenotypic variation
- they rely on recombination to create a wide range of phenotypic variation
- all of the above
(e) none of the above
- If VGXE > 0:
- there is genetic variation for phenotypic plasticity
- phenotypic plasticity is at least partially heritable
- genotypes would need to be grown in different environments to achieve a true estimate of narrow sense heritability
- all of the above
- none of the above
- (a) Consider selection on a quantitative trait such as height. Please graph a fitness function (fitness on the Y; height on the X) for each of the following types of selection: directional, stabilizing, disruptive.
See textbook for examples.
- What effect does each type of selection have on the mean and variance of the trait in the next generation?
Directional: Mean changes in direction of selection; variance reduced.
Stabilizing: Mean stays the same; variance reduced
Disruptive: Mean stays the same; variance increased
- You are studying directional selection on neck length in two giraffe populations. Below you can see data on neck length of the whole population and breeding parents for each of the two populations.
| Neck length of individuals in population 1 | Neck length of breeders in population 1 | Neck length of individuals in population 2 | Neck length of breeders in population 2 |
| 50 | 90 | 55 | 90 |
| 80 | 80 | 75 | 60 |
| 90 | 75 | 45 | 70 |
| 45 | 95 | 60 | 75 |
| 60 | 60 | 40 | 80 |
| 35 | | 90 | |
| 95 | | 45 | |
| 40 | | 70 | |
| 75 | | 80 | |
(a) Calculate the selection differential for each population. Show your work.
S = mean of population – mean of breeders
Pop. 1: 63 – 80 = -17
Pop. 2: 62 – 75 = – 13
You also have some data you on the length of necks for parents and their offspring (offspring corresponding to each midparent listed in adjacent cell).
| Midparent pop 1 | Midoffspring pop 1 | Midparent pop 2 | Midoffspring pop 2 |
| 85 | 84 | 80 | 65 |
| 90 | 88 | 75 | 90 |
| 75 | 75 | 85 | 55 |
| 65 | 62 | 65 | 75 |
| 70 | 69 | 70 | 85 |
| 80 | 82 | 95 | 75 |
| 95 | 94 | 60 | 80 |
(b) Describe how you would use this data to calculate narrow sense heritability.
Plot midparent on the X-axis vs. midoffspring on the Y-axis and run a regression line through the data points. The slope of this line is equal to narrow sense heritability.
(c) Based on parts ‘a’ and ‘b,’ in which population would you expect to see the largest evolutionary change in neck length in the next generation? Explain your answer.
R = h2S
Population one has the larger selection differential and offspring also resemble their parents more, so population one should experience the most evolutionary change.
- You are studying the effect of directional selection on body height in three populations (a, b, and c).
(a) What is the selection differential? Show your calculation.
S = 65 – 70 = -5
(b) Which population has the highest narrow sense heritability for height? Explain your answer.
Population C because narrow sense heritability is estimated by the slope of the regression line. The slope is steeper in population C indicating higher heritability.
(c) If you examined the offspring in the next generation in each population, which population would have the highest mean height? Why?
R = h2S; The selection differential is the same in all populations, but population C has the highest heritability. Therefore, this population will experience the most change.
- As a result of warming climate, most butterfly species in California mature 24 days sooner than they did 30 years ago. (a) Please describe a scenario that would account for this change through phenotypic plasticity. (b) Please describe a scenario that would account for this change through evolution by natural selection.
(a) Individual development time is influenced by environmental variables such as temperature. For example, if weather is warmer, individuals develop more rapidly, resulting in less time to maturity. Thus, warming weather could account for the change over the last 30 years.
(b) There is genetic variation for development time in butterfly populations. Over the last 30 years, individuals with alleles that lead to faster maturity have had higher reproductive success. Thus, these alleles have increased in frequency in populations.
- Please describe a scenario where selection occurs on a trait, but there is no evolutionary change.
Individuals may vary in phenotype, with certain traits leading to a reproductive advantage. Evolution will not result unless these phenotypic differences are heritable. Thus, if the differences are attributed to environmental causes, the population will not evolve even if selection occurs.
- Stalk-eyed flies are so named because their eyes are situated at the end of long stalks extending away from their head. Stalk length is a quantitative trait that is variable in populations. Describe the steps you would need to take in order to find quantitative trait loci linked to differences in eye stalk length.
- Create highly inbred lines for long and short stalk length by selectively breeding long with long and short with short for several generations.
- Create a genome map using SNPs or some other variable genetic marker.
- Cross-breed the two lines to create an F2 generation that varies in stalk length due to recombination.
- Score the phenotype of the F2 flies and score individuals at the SNPs or other genetic markers to determine if any are non-randomly associated with body size differences.
| Stalk length | Locus 1 | Locus 2 | Locus 3 | Locus 4 |
| 100 | B | B | B | A |
| 58 | A | A | B | B |
| 102 | A | B | A | A |
| 110 | A | B | B | B |
| 40 | B | A | C | B |
| 45 | B | A | A | A |
| 99 | A | B | A | B |
| 70 | A | A | B | A |
| 105 | B | B | A | B |
- You are conducting a QTL analysis to find genetic regions associated with differences in eye stalk length in stalk-eyed flies. The table above gives information on stalk length and genotype at four variable loci used to create a genetic map for F2 flies. What would you conclude and why?
Locus 2 appears to be linked to differences in stalk length. All flies with long stalk length carry an “B” at locus two, while those with short stalk length carry an “A.” This suggests that locus two is linked to alleles responsible for differences in stalk length.
- Each graph above shows the phenotype produced by three different genotypes in different environmental conditions (each line is a genotype). For each graph write “plasticity” or “no plasticity” depending on whether the graph depicts phenotypic plasticity. Also, write “G X E” or “no G X E” depending on whether the graph depicts a genotype by environment interaction.
- No plasticity; No G X E
- Plasticity; No G X E
- Plasticity; G X E
- Plasticity; G X E
- Each graph above depicts reaction norms for three different genotypes reared in different environments. In which case(s) is there a capacity for phenotypic plasticity to further evolve (you can assume that graphs for parents and their offspring would be the same)? Explain your answer.
In order for phenotypic plasticity to further evolve there needs to be phenotypic plasticity in the first place. Graphs b, c, and d demonstrate phenotypic plasticity. Furthermore, there needs to be a G X E interaction in order for plasticity to further evolve. Only graphs C and D show such an interaction. Thus, only C and D show the capacity for further evolution of phenotypic plasticity (assuming offspring show similar responses to their parents).
- Given the warming climate, you are interested in how seed production of a certain plant species if affected by temperature. In particular, you are interested in whether seed production shows a plastic response to temperature, and, if so, whether there is the possibility for adaptive evolution of phenotypic plasticity in response to climate change. Please describe the steps that would be needed to demonstrate there is the capacity for adaptive evolution in the population you are studying.
- Grow several plant genotypes over a range of temperature to generate reaction norms for each genotype. If a single genotype produces different phenotypes at different temperatures and the slopes of the reaction norms are not parallel then there is a G X E interaction.
- To estimate VGXE, grow parents and their offspring over the same range of temperatures and produce an offspring-parent regression. If the slope of the regression line is greater than zero, this indicates heritable genetic variation for the response to different temperatures and the possibility of future evolution of the plastic response. For example, genotypes that maintain higher seed production at high temperature may be favored as the climate warms.
- Agree or disagree with the following statement and explain your answer: If a trait such as height has high narrow sense heritability (say 0.7), this means that genetics is more important than environment in determining the height of any given individual.
Disagree. Heritability is a measure that is only relevant at the population level, as it explains what accounts for phenotypic differences among individuals in a population. Just because heritability is high, does not mean that genetics is more important than environment for every individual in a population.
