Question

You are studying a population of milkweeds and you find the following allele frequencies:

f(A₁A₁) = 0.36

f(A₁A₂) = 0.48

f(A₂A₂) = 0.16

If the fitness of the heterozygote (w₁₂) is 0.9 and the average fitness of the population is 0.916, what do we expect the new frequency of the heterozygote genotype to be in the next generation?

Answer 1

Answer:

$\rightarrow$ Calculation:

• New frequency of A1A1 homozygote genotype in the next generation will be:p²w₁₁ / W
• where p² = f(A1A1)=0.36 ;
• w_{11 =} fitness of homozygote =0.9 ;
• W = average fitness = 0.916

So, 0.36 * 0.9 /0 .916 = 0.35

So the answer is "0.35"

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