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 of individuals with AA genotype?

Answer 1

Answer:

Frequency of individuals with AA genotype is $0.81$

Explanation:

As per Hardy-Weinberg equation, frequency of allele "a" is $0.1$

Which means that $q=1$

where "q" denotes frequency of allele "a"

As per I equation of Hardy-Weinberg -

$p+ q =1$

where "p" denotes frequency of allele "A"

Substituting the value of "q" we get ,

$p+0.1=1\\p=1-0.1\\p=0.9$

Frequency of individuals with AA genotype is represented as $p^{2}$ which is equal to

$0.9^2\\= 0.81$

Frequency of individuals with AA genotype is $0.81$