Ask Question
30 October, 08:00

In a Hardy-Weinberg population with two alleles, A and a, that are in equilibrium, the frequency of allele a is 0.1. What is the frequency of individuals with AA genotype? A) 0.42B) 0.20C) 0.32D) 0.81

+5
Answers (1)
  1. 30 October, 11:00
    0
    The correct answer is D) 0,81

    When there is Hardy-Weinberg equilibrium like in this case of a single locus with two alleles denoted A and a with frequencies f (A) = p and f (a) = q, the expected genotype frequencies under random mating are f (AA) = p² for the AA homozygotes, f (aa) = q² for the aa homozygotes, and f (Aa) = 2pq for the heterozygotes. And we have:

    p²+2*p*q+q² = 1 p+q = 1 p=1-q

    q = 0.1 p = 1-0.1 = 0.9

    p²=0.9² = 0.81
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “In a Hardy-Weinberg population with two alleles, A and a, that are in equilibrium, the frequency of allele a is 0.1. What is the frequency ...” in 📘 Biology if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers