Question

Consider a locus with two alleles - A and a. These alleles are codominant, meaning that the fitness of the heterozygote is halfway between either homozygote. Consider further a population of randomly mating green frogs where the genotype counts are AA = 500, Aa = 250, and aa = 250. In this population the relative fitnesses of each genotype are AA = 1.00, Aa = 0.80, and aa = 0.60. What is the mean realtive fitness within this population? Please give your answer to two decimal places.

Consider a locus with two alleles - A and a. These alleles are codominant, meaning that the fitness of the heterozygote is halfway between either homozygote. Consider further a population of randomly mating green frogs where the genotype counts are AA = 500, Aa = 250, and aa = 250. In this population the relative fitnesses of each genotype are AA = 1.00, Aa = 0.80, and aa = 0.60. What is the expected allele frequency change for A after one generation with selection? Please give your answer to two decimal places.

Answer 1

The mean relative fitness of the population is 0.85. The expected allele frequency change for allele A after one generation with selection is a decrease to approximately 0.62 due to the relative fitness differences among the genotypes.

Step-by-step explanation:

The question at hand requires an understanding of genetic structure, allele frequency, and the Hardy-Weinberg principle to calculate both the mean relative fitness within a population and the expected allele frequency change for allele A after one generation with selection.

Mean Relative Fitness Calculation:

To calculate the mean relative fitness (w), we multiply the relative fitness of each genotype by its proportional representation in the population and sum the results:

• w(AA) = 1.00 × (500/1000) = 0.50,

• w(Aa) = 0.80 × (250/1000) = 0.20,

• w(aa) = 0.60 × (250/1000) = 0.15.

The sum is the mean relative fitness: wμ = 0.50 + 0.20 + 0.15 = 0.85.

Expected Allele Frequency Change:

The next step is to calculate the expected change in allele frequency using the Hardy-Weinberg equation. First, determine the frequency of the alleles (p for A and q for a). The current frequency of A (p) = (2×500 + 250) / (2×1000) = 0.625. The frequency of a (q) = 1 - p = 0.375. To calculate the expected frequency of allele A after one generation of selection, we must account for the relative fitness of each genotype:

• p' = (p² × w(AA) + pq × w(Aa)) / wμ,

• p' = (0.625² × 1.00 + 0.625 × 0.375 × 0.80) / 0.85 ≈ 0.62.

Hence, the expected frequency of A after one generation is approximately 0.62, indicating a slight decrease due to selection.

Answer 2

Question 1.

The mean relative fitness is 0.85

Question 2.

The expected allele frequency change of A, p = 0.63: AA, p2 =0.63 x 0.63 = 0.39.

Step-by-step explanation:

Mean relative fitness is given as

p2W(AA) + 2pqW(Aa) + q2W(aa).

Where W(AA)= 1.00, W(Aa) = 0.80, W(aa) = 0.60

Firstly, find p2, 2pq, q2 which are the frequencies of AA, Aa and aa respectively.

AA= 500,Aa=250 ,aa=250. Total is 1000

Frequency of A, p=

250/2)+500= 625.

625/1000= 0.625.

Therefore A, p is 0.625

p2 = 0.625x0.625= 0.39

For q, since p+q =1

q= 1 - 0.625 = 0.375

q2= 0.375x0.375=0.14.

2pq= 2x0.625x0.375= 0.47.

Mean relative fitness is

p2W(AA) + 2pqW(Aa) + q2W(aa).

0.39x1+ (0.47x0.80) + (0.14x0.60)=0.85.

Question 2.

From Hardy-Weinberg equilibrium

F = p2 + 2pq + q2.

Frequency of A, p= 0.625 and AA, p2 is 0.625 x 0. 0625= 0.39.