SELECTION COEFFICIENTS AND FITNESS
CALCULATIONS
1.1 PROBLEM SET
1. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
2. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
3. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
4. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
5. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
6. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
7. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
8. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
9. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
10. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
11. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
12. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
13. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
14. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
15. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
16. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
17. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
18. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
19. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
20. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
21. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
22. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
23. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
24. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
25. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
26. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
27. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
28. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
29. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
30. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
31. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
32. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
33. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
34. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
35. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
36. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
37. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
38. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
39. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
40. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
41. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
42. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
43. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
44. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
45. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
46. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
47. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
48. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
49. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
50. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
51. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
52. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
53. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
54. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
55. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
56. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
57. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
58. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
59. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
60. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
61. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
62. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
63. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
64. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
65. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
66. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
67. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
68. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
69. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
70. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
71. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
72. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
73. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
74. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
75. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
76. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
77. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
78. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
79. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
80. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
81. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
82. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
83. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
84. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
85. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
86. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
87. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
88. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
89. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
90. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
91. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
92. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
93. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
94. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
95. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
96. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
97. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
98. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
99. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
100. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
101. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
102. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
103. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
104. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
105. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
106. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
107. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
108. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
109. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
110. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
111. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
112. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
113. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
114. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
115. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
116. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
117. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
118. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
119. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
120. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
121. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
122. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
123. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
124. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
125. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
126. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
127. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
128. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
129. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
130. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
131. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
132. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
133. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
134. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
135. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
136. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
137. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
138. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
139. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
140. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
141. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
142. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
143. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
144. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
145. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
146. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
147. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
148. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
149. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
150. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
151. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
152. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
153. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
154. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
155. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
156. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
157. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
158. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
159. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
160. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
161. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
162. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
163. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
164. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
165. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
166. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
167. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
168. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
169. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
170. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
171. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
172. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
173. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
174. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
175. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
176. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
177. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
178. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
179. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
180. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
181. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
182. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
183. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
184. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
185. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
186. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
187. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
188. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
189. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
190. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
191. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
192. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
193. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
194. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
195. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
196. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
197. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
198. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
199. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
200. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
201. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
202. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
203. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
204. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
205. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
206. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
207. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
208. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
209. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
210. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
211. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
212. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
213. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
214. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
215. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
216. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
217. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
218. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
219. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
220. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
221. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
222. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
223. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
224. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
225. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
226. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
227. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
228. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
229. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
230. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
231. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
232. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
233. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
234. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
235. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
236. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
237. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
238. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
239. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
240. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
241. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
242. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
243. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
244. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
245. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
246. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
247. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
248. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
249. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
250. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
251. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
252. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
253. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
254. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
255. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
256. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
257. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
258. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
259. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
260. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
261. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
262. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
263. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
264. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
265. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
266. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
267. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
268. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
269. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
270. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
271. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
272. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
273. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
274. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
275. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
276. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
277. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
278. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
279. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
280. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
281. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
282. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
283. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
284. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
285. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
286. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
287. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
288. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
289. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
290. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
291. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
292. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
293. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
294. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
295. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
296. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
297. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
298. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
299. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
300. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
301. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
302. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
303. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
304. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
305. In a population of moths, the wild-type allele (A) is dominant to the mutant allele (a).
The frequencies of genotypes AA, Aa, and aa are 0.49, 0.42, and 0.09, respectively. If
the relative fitness of aa is 0.85, calculate the selection coefficient against the aa
genotype.
Solution:
– The selection coefficient (s) is related to relative fitness (w) by the equation: w =
1 - s
– Given: waa = 0.85
– Therefore: 0.85 = 1 - s
– Solving for s: s = 1 - 0.85 = 0.15
The selection coefficient against the aa genotype is 0.15 or 15%.
306. A population has allele frequencies p = 0.7 for allele A and q = 0.3 for allele a. The
relative fitnesses are: wAA = 1, wAa = 0.9, and waa = 0.8. Calculate the mean fitness of
the population.
Solution:
– Mean fitness (𝑤‾ ) = p2wAA + 2pqwAa + q2waa
– 𝑤‾ = (0.72 × 1) + (2 × 0.7 × 0.3 × 0.9) + (0.32 × 0.8)
– 𝑤‾ = 0.49 + 0.378 + 0.072
– 𝑤‾ = 0.94
The mean fitness of the population is 0.94.
307. In a plant species, individuals with genotype AA have a fitness of 1, Aa have a fitness of
0.95, and aa have a fitness of 0.85. If the frequency of allele A is 0.6, what is the
selection coefficient against the a allele?
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.62 × 1) + (2 × 0.6 × 0.4 × 0.95) + (0.42 × 0.85)
– 𝑤‾ = 0.36 + 0.456 + 0.136 = 0.952
– The selection coefficient against the a allele (sa) is:
– sa = 1 - $\frac{\bar{w}\textsubscript{a}}{\bar{w}}$
– $\bar{w}\textsubscript{a}$ = (0.6 × 0.95) + (0.4 × 0.85) = 0.57 + 0.34 = 0.91
– sa = 1 - 0.91
0.952 = 0.044
The selection coefficient against the a allele is 0.044 or 4.4%.
308. In a population of butterflies, the frequency of allele A is 0.8 and allele a is 0.2. The
relative fitnesses are: wAA = 1, wAa = 0.98, and waa = 0.90. Calculate the change in
allele frequency of A in one generation.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.82 × 1) + (2 × 0.8 × 0.2 × 0.98) + (0.22 × 0.90)
– 𝑤‾ = 0.64 + 0.3136 + 0.036 = 0.9896
– The change in allele frequency (Δp) is given by:
– Δp = $\frac{pq[p(w\textsubscript{AA} - w\textsubscript{Aa}) +
q(w\textsubscript{Aa} - w\textsubscript{aa})]}{2\bar{w}}$
– Δp = 0.8×0.2[0.8(1−0.98)+0.2(0.98−0.90)]
2×0.9896
– Δp = 0.16[0.016+0.016]
1.9792 = 0.00512
1.9792 = 0.00259
The change in allele frequency of A in one generation is 0.00259 or 0.259%.
309. A population has allele frequencies p = 0.6 for allele B and q = 0.4 for allele b. The
relative fitnesses are: wBB = 1, wBb = 1.1, and wbb = 0.9. Calculate the equilibrium
frequency of allele B.
Solution:
– At equilibrium: p(wBB - wBb) = q(wBb - wbb)
– Let the equilibrium frequency of B be p*
– p*(1 - 1.1) = (1 - p*)(1.1 - 0.9)
– -0.1p* = 0.2 - 0.2p*
– 0.1p* = 0.2
– p* = 0.2 / 0.1 = 2
Since p* > 1, there is no stable equilibrium. The population will evolve towards fixation
of the B allele.
310. In a population of fish, the wild-type allele (C) is dominant to the mutant allele (c). The
initial frequency of C is 0.7. After selection, the frequency of C increases to 0.75. If the
relative fitness of cc is 0.8, calculate the relative fitness of Cc.
Solution:
– Let wCC = 1, wCc = 1 - hs, wcc = 1 - s
– Given: wcc = 0.8, so s = 0.2
– Initial p = 0.7, q = 0.3
– After selection: p’ = 0.75, q’ = 0.25
– Using the formula: p’ = 𝑝2+𝑝𝑞(1−ℎ𝑠)
𝑝2+2𝑝𝑞(1−ℎ𝑠)+𝑞2(1−𝑠)
– 0.75 = 0.49+0.21(1−ℎ×0.2)
0.49+0.42(1−ℎ×0.2)+0.09×0.8
– Solving this equation: 1 - hs ≈ 0.9524
– Therefore, wCc ≈ 0.9524
The relative fitness of Cc is approximately 0.9524.
311. A population has allele frequencies p = 0.4 for allele D and q = 0.6 for allele d. The
relative fitnesses are: wDD = 0.9, wDd = 1, and wdd = 0.8. Calculate the frequency of
allele D after one generation of selection.
Solution:
– First, calculate the mean fitness:
– 𝑤‾ = (0.42 × 0.9) + (2 × 0.4 × 0.6 × 1) + (0.62 × 0.8)
– 𝑤‾ = 0.144 + 0.48 + 0.288 = 0.912
– The frequency of D after selection (p’) is given by:
– p’ = $\frac{p(pw\textsubscript{DD} + qw\textsubscript{Dd})}{\bar{w}}$
– p’ = 0.4(0.4×0.9+0.6×1)
0.912
– p’ = 0.4(0.36+0.6)
0.912 = 0.384
0.912 = 0.4211
The frequency of allele D after one generation of selection is 0.4211 or 42.11%.
312. In a population of beetles, a recessive allele (e) causes a darker color. The frequency of
this allele is 0.3. If the relative fitness of the dark-colored beetles (ee) is 0.85, what will
be the frequency of the e allele after 10 generations of selection, assuming the
population is large and mating is random?
Solution:
– Initial q = 0.3, p = 0.7
– wee = 0.85, wEe = wEE = 1
– For each generation: q’ = $\frac{q^2w\textsubscript{ee}}{p^2 + 2pq +
q^2w\textsubscript{ee}}$
– q’ = 0.32×0.85
0.72+2×0.7×0.3+0.32×0.85 = 0.2845
– Repeating this calculation for 10 generations:
– Generation 1: q = 0.2845
– Generation 2: q = 0.2698
– ...
– Generation 10: q ≈ 0.2088
After 10 generations of selection, the frequency of the e allele will be approximately
0.2088 or 20.88%.
