Ask Question
28 December, 11:01

Assume that a population is in Hardy-Weinberg equilibrium for a particular gene with two alleles, A and a. The frequency of A is p, and the frequency of a is q. Because these are the only two alleles for this gene, p + q = 1.0. If the frequency of homozygous recessive individuals (aa) is 0.04, what is the value of q? What is the value of p

+1
Answers (1)
  1. 28 December, 12:00
    0
    q = 0.2, p = 0.8

    Explanation:

    According to Hardy-Weinberg equilibrium,

    p + q = 1

    p² + 2pq + q² = 1 where,

    p = frequency of dominant allele

    q = frequency of recessive allele

    p² = frequency of homozygous dominant genotype

    2pq = frequency of heterozygous genotype

    q² = frequency of homozygous recessive genotype

    Here,

    q² = aa = 0.04

    q = √0.04 = 0.2

    p = 1 - q

    = 1 - 0.2

    = 0.8
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Assume that a population is in Hardy-Weinberg equilibrium for a particular gene with two alleles, A and a. The frequency of A is p, and the ...” in 📙 Biology if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers