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 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

Answer 1

q = 0.2, p = 0.8

Step-by-step 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