STUDY HINTS
A biologist might define evolution as a change in allele frequencies over time. These changes in allele frequencies could be the result of selection, drift, mutation, migration, or combinations of these population variables. These Study Hints are primarily concerned with selection and the effects of relative differences in reproductive success.
The problem-related questions in this section might best be introduced by first considering a general selection model:
|
Genotype |
A1A1 |
A1A2 |
A2A2 |
Total |
|
Frequencies before selection |
p2 |
2pq |
q2 |
1 |
|
Fitness |
W1 |
W2 |
W3 |
|
|
Relative contribution to next generation |
p2W1 |
2pqW2 |
q2W3 |
|
|
Frequencies after selection |
p2W1 |
2pqW2 |
q2W3 |
1 |
|
|
|
|
In this model, p and q are the allele frequencies of A1 and A2, respectively, and the W’s refer to the relative fitness of each genotype. The relative fitness is a measure of the relative contribution that a genotype makes to the next generation, and it can be measured in terms of the intensity of selection (s), where W = 1 − s and s can differ for each genotype (0 
s 1).
In the case of selection against the homozygous recessive, for example, W1 = W2 = 1 and W3 would be less than 1 (e.g., .2, where the intensity of selection is .8). Substitution into the table for the case where the allele frequencies are identical before selection (p = q = .5) yields the following results:
|
Genotype |
A1A1 |
A1A2 |
A2A2 |
Total |
|
Frequencies before selection |
.25 |
.50 |
.25 |
1.00 |
|
Fitness |
1.00 |
1.00 |
.20 |
|
|
Relative contribution to next generation |
.25 |
.50 |
.05 |
.80 |
|
Frequencies after selection |
.25/.80 |
.50/.80 |
.05/.80 |
|
|
= .31 |
= .62 |
= .07 |
Allele frequencies calculated from these genotype frequencies are p = .62 and q = .38. Selection against the A2A2 homozygote has reduced the frequency of the A2 allele in the sample.
For selection against the dominant, the relative fitnesses of the genotypes are W1 = 1 − s for A1A1 individuals, W2 = 1 − s for heterozygotes, and W3 = 1 for A2A2 homozygotes. The proportion of zygotes after selection is, therefore, p2W1 = p2(1 − s), 2pqW2 = 2pq(1 − s), and q2W3 = q2. The gene frequency of p after selection is
The change in p is
This formula tells us that the rate of change in allele frequency is faster than in selection against the recessive, because the dominant gene is not hidden in heterozygotes.
For selection against both homozygotes (sometimes called overdominance, or heterosis), the relative fitnesses of the three genotypes are W1 = 1 − s, W2 = 1, and W3 = 1 − t, where s and t can be different intensities of selection. The proportion of zygotes after selection is therefore p2(1 − s), 2pq, and q2(1 − t). To determine the change in allele frequency of the recessive, q, the gene frequency after selection (q1) is
The change in q is
Finally, for selection against the heterozygote, the relative fitnesses of the three genotypes are W1 = 1, W2 = 1 − s, and W3 = 1. The proportion of zygotes after selection is p2, 2pq(1 − s), and q2. The new gene frequency of q is
If q < 1/2, then Δq is negative, and q decreases under selection. If q > 1/2, then Δq is positive, and q increases. There is consequently an unstable equilibrium at q = 1/2. Note also that only in the case of selection against both homozygotes is a stable polymorphism possible.
When we turn our attention from selection on a single locus to selection on the polygenic determinants of a phenotype, we find that it is convenient to divide selection into three basic categories. These and their consequences for the phenotypic mean and variance are diagramed on the preceding page.
Directional selection shifts the mean toward one extreme. Although it may temporarily increase phenotypic variance, the primary effect is on the phenotypic mean. Stabilizing selection, on the other hand, reduces phenotypic variance without changing the mean, whereas disruptive selection favoring two or more phenotypes in the same population increases phenotypic variance.
Much of the concern of population genetics centers on attempts to measure parameters such as selection intensity, relative fitness, and the interactions among genotypes and environment that influence these measures. Although this is a complex application of genetic principles, it is clear that there are many types of selection, that heritability is important in predicting the rate of response to selection, and that the population structure and environmental variables are important to any explanation of evolution. The difficulties and successes of some of these studies are discussed in your text. There can be no doubt that it is a challenging arena of genetic investigation.
IMPORTANT TERMS
Allopatric population
Cyclical selection
Directional selection
Disruptive selection
Genetic homeostasis
Genetic load
Hybrid zone
Incipient species
Isolating mechanisms
Segregational load
Selection pressure (intensity)
Stabilizing selection
Sympatric population
PROBLEM SET 22
1. A plant-breeding institute establishes four new strains of winter grain (G1, G2, G3, and G4). The genetic variation in each of these for resistance to an infectious fungus (rust) is measured by comparing the F1 and F2 phenotypic variances after crosses of each to a standard grain strain (G). The following data are collected. Which one of these four strains would be expected to respond most readily to artificial selection for rust resistance?
|
Strain |
F1 variance |
F2 variance |
|
G1 |
0.41 |
0.57 |
|
G2 |
0.40 |
0.62 |
|
G3 |
0.43 |
0.55 |
|
G4 |
0.46 |
0.66 |
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2. A rancher has just bought a flock of sheep in which a recessive allele for black wool is segregating. At the time he bought the flock, the allele was at a frequency of .5 (p = q = .5). Theoretically, how many generations of complete selection against the black sheep would be required to reduce the allelic frequency to .1?
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3. According to Ernst Mayr, sympatric isolating mechanisms can be placed into two broad categories: premating and postmating. Give some examples of each of these. Which of the two is more adaptive, in terms of the metabolic energy investment of the population?
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4. What evidence is there to support the concept of sympatric speciation?
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5. Would the heritability for a trait that is important for organism survival and reproduction (a “fitness trait”) likely be larger than, smaller than, or the same as the heritability for a trait not directly affecting survival or fitness? Please explain your answer and discuss the expected relative effectiveness for directional selection on a fitness trait.
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6. Please test whether the following data are in Hardy–Weinberg equilibrium by using a χ2 test to compare the observed and the expected genotype frequencies. If the data are not in equilibrium, what explanations can you propose to account for any deviation? Phenotypes have been omitted in order to enable you to have full freedom in forming your explanations.
|
Genotypes |
RR |
Rr |
rr |
Total |
|
Numbers of individuals |
602 |
290 |
108 |
1,000 |
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7. The initial frequency of a recessive eye-color mutant in a Drosophila population is .4. What is the maximum change in frequency that can occur in 10 generations of complete selection against the recessive, if no other factors, such as mutation, are acting in the population?
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8. Suppose you have a large herd of horses in which three phenotypes are segregating: chestnut, carmello, and palomino. Both carmello and palomino are the result of a modifier gene (D′) influencing the expression of the chestnut-coat color gene. The alternative allele (D) has no effect. D′D′ plus chestnut gives carmello, DD′ plus chestnut gives palomino, and DD plus chestnut gives chestnut-colored horses. Suppose you would like to have a herd made up primarily of palomino horses, but you would also like to have more chestnuts than carmellos. Therefore you put selection pressure on this herd by selling all carmello horses and 50 percent of the chestnut horses before they mature. The initial allele frequencies of D and D′ were equal (p = q = .5), and the population was in Hardy–Weinberg equilibrium when you started selection.
(a) What is the relative fitness of each genotype?
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(b) What would the gene frequencies be after one generation of selection?
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(c) What would the zygotic frequencies be in the second generation?
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(d) What is the equilibrium frequency of q?
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9. Beginning with the same original population of horses as in problem 8, suppose you decided to remove all the chestnuts as well as all carmello horses from the first generation of progeny. What would the gene frequencies and the zygotic frequencies be in the next generation?
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10. Instead of selecting for the palominos, you decide that you want more chestnut and carmello horses than palominos. Starting at the same gene frequencies (p = q = .5) and selecting against the palominos (s= 50 percent),
(a) what is the relative fitness of each genotype?
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(b) what will the frequency of the D′ allele determining carmello be after one generation of selection?
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(c) What would the frequency of carmellos be after one generation of selection, if the starting frequency of D′ were .3?
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ANSWERS TO PROBLEM SET 22
1. The heritabilities in each strain can be measured as shown in Chapter 9.
The heritability of each strain is obtained by substituting into this formula, with the following results:
G1 = 0.28
G2 = 0.35
G3 = 0.22
G4 = 0.30
Strain G2 has the largest heritability for rust resistance and thus, by definition, has the largest proportion of genetic (heritable) variation. It would respond most successfully to artificial selection for this trait, assuming that all other characteristics of the strains are equivalent.
2. There are two ways to solve this problem. One is to solve for q after each generation of selection. If p2 + 2pq + q2 = 1, then after one generation of selection against q2,
|
AA |
Aa |
aa |
Frequency of a |
|
|
Before |
p2 |
2pq |
q2 |
q |
|
After |
|
|
0 |
|
Repeating this calculation, however, is time-consuming. The more general form is
where qo is the original frequency of the recessive, qn is the frequency at generation n, and n is the number of generations of selection. This can be solved for n:
Substituting from the problem, n = (0.5 − 0.1)/(0.5)(0.1) = 8 generations to make the desired reduction in allele frequency.
3. Premating barriers include seasonal or habitat isolation, behavioral isolation, and mechanical isolation. Postmating barriers include gametic mortality, zygotic mortality, hybrid inviability, and hybrid sterility. Premating isolation mechanisms are more energetically economical, because gametes are not wasted and doomed developmental activities are not initiated.
4. One of the most generally accepted theories of speciation is allopatric speciation. Many workers believe that populations can accumulate significant genetic differences only when they are separated to prevent gene exchange. Others, including K. Mather and J. M. Thoday, have proposed that selection for adaptation to different niches within a single locality could lead to polymorphism and that, if the disadvantages of the hybrids were severe enough, speciation might occur. Thoday and J. B. Gibson, for example (see reference at end of this answer), selected for high and low bristle number in a single culture of Drosophila melanogaster, and they obtained significant divergence between these extremes, in spite of the fact that extremes were given the opportunity to interbreed freely. Clearly, geographic separation is not required for divergence to occur. Many mechanisms could encourage sympatric divergence, such as differences in flowering time in two groups of plants in the same field or localized heterogeneity in the environment (e.g., high levels of toxic waste at a mining site). These and other possibilities are discussed more thoroughly in a review by J. M. Thoday (Proceedings of the Royal Society of London (B) 182 [1972]: 109–143).
5. Heritability measures the proportion of phenotypic variation that can be traced to genetic segregation. For heritability to be high, a large number of polygenic heterozygotes must be present; that is, a large number of “increasing” and “decreasing” polygenic alleles must be maintained in the population. This is not likely to be true for a critical fitness trait, since there would seldom be a selective advantage to maintaining anything less than the greatest possible expression of such characters. Heritability would therefore tend to be higher for traits not directly affecting survival or fitness.
6. The 1,000 individuals in this population make up a gene pool of 2,000 copies of each gene. To test the fit to Hardy–Weinberg, we must first calculate the allele frequencies. Since the homozygote has two alleles of one type, and the heterozygote has one (RR and Rr, respectively),
At Hardy–Weinberg equilibrium,
The fit to Hardy–Weinberg can be calculated using a χ2 analysis.
The number of degrees of freedom is 1; 3 (the number of classes) minus 1 (for using the sample size to calculate the expectations) minus 1 (for estimating one parameter for the data, the allele frequency p, which automatically establishes q and the genotype frequencies) = 1. The probability value for this magnitude of χ2 is < .001, indicating a significant discrepancy between the observed and expected frequencies. There appears to be a relative deficiency of heterozygotes, suggesting selection favoring the homozygous genotypes. Another way of getting such a pattern is to have this sample of 1,000 individuals really represent a mixture of individuals from two different local populations, each of which had been selected for an alternative allele or in which drift had led to the increase of alternative alleles.
|
RR |
Rr |
rr |
Total |
|
|
Observed (O) |
602 |
290 |
108 |
1,000 |
|
Expected (E) |
558 |
378 |
64 |
1,000 |
|
O − E |
+44 |
−88 |
+44 |
0 |
|
(O − E)2/E |
1,936/558 |
7,744/378 |
1,936/64 |
7. From problem 2,
Substituting qo = .4 and n = 10,
8. From the problem, the frequency of D = p = .50, and the frequency of D = q = .50. Since p2 + 2pq + q2 = 1, the phenotypic frequencies in the herd are p2 = .25 for carmellos, 2pq = . 50 for palominos, and q2 = .25 for chestnuts. The selection pressure against carmellos is s = 1 and against chestnuts is t = .5.
(a) The fitness of
The contribution of each genotype to the next generation is
(b) After one generation of selection,
Substituting p = q = .5, s = 1, and t = .5, we find that q1 = .6, and p1 = .4.
(c) The zygotic frequencies can be calculated by using p = .4 and q = .6. For carmellos, p2 = .16; for palominos, 2pq = .48; and for chestnuts, q2 = .36.
(d) At equilibrium, ps = qt. Substituting 1 − q for p, (1 − q)s = qt
Correspondingly,
In this problem, at equilibrium q = .667 and p = .333.
9. The selection intensities would be s = 1 and t = 1.
This makes sense if you recognize that all remaining horses will be heterozygotes, so p must equal q. From this point on, the population will be in genotypic (and therefore phenotypic) equilibrium.
10.
(a) W1 = 1, W2 = 1 – s, and W3 = 1. Since s = 5, W2 must be .5.
(b) After one generation of selection,
The gene frequencies did not change.
(c) Letting q be the frequency of D’ and substituting q = .3 into the formula just given in (b),
Since q1 < .5, the frequency of the D′ allele will decrease.
CROSSWORD PUZZLE 22
Selection and Evolution
Across
2. ___ evolution: change that splits a single ancestral lineage into two or more branches
8. Evolution in one or more species in response to changes in another associated species
9. Measure of the effects of selection (loss of fitness) of one genotype relative to a best fit genotype that is given a fitness value of 1
10. Small colonizing population that has only a sample of the genetic variation found in the population from which it came
12. Selection against variations from the “normal”
15. Genetic mechanisms that prevent the production of hybrid zygotes
17. Basic taxonomic category, which describes lineages evolving separately; one definition involves reproductive isolation
20. Selection against one extreme and the intermediate, while another extreme phenotype has an advantage
21. Reduction in genetic variation due to the effect of genetic drift on a population that was reduced in size (crashed) and then at a later date increases in numbers
22. Selection for the resemblance of one species to another species (model) that gives the impersonator some form of protection or advantage
Down
1. Presence of similar adaptations in widely diverse species having structures with similar function but different embryonic origins
3. - hybridization; incorporation of genetic information from one gene pool into another gene pool as a result of hybrid fertility
4. Selection that is often used to explain altruistic behavior
5. Causes differential reproductive survival of genetically variable populations
6. Divergence from the normal gamete ratios produced by heterozygotes (melotic drive)
7. Reproductive mechanisms that prevent hybrid zygotes from producing fertile offspring
9. Related populations that are not isolated geographically from each other
11. Modification of specific population traits such that the population diverges into different habitats and a production of a large variety of species or groups results
13. Describes the average loss of genetic fitness in individuals of a population because they carry harmful alleles
14. Selection that tends to favor the survival of the extreme phenotypes and eliminate the intermediate phenotypes
16. Mechanisms that act as barriers to genetic exchange between populations
18. Closely related species that are geographically isolated from each other
19. Describes the relative reproductive success of one genotype over others; opposite of selection
