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

Frequency of individuals with AA genotype is
`0.81`

Step-by-step 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`