Question

In a population at Hardy Weinberg equilibrium, allele frequency of 'A' is 0.3, the expected frequency of 'Aa' individuals is

• A. 0.21
• B. 0.42
• C. 0.63
• D. 0.18

Answer 1

The correct option is B 0.42

To estimate the frequency of alleles in a population, we can use the Hardy-Weinberg equation.
According to this equation:-
p = the frequency of the dominant allele (represented here by A)
q = the frequency of the recessive allele (represented here by a)
For a population in genetic equilibrium:
p + q = 1.0 (The sum of the frequencies of both alleles is 100%.)
(p + q)${}^2$ = 1 so p${}^2$ + 2pq + q${}^2$ = 1
The three terms of this binomial expansion indicate the frequencies of the three genotypes:
p${}^2$ = frequency of AA (homozygous dominant)
2pq = frequency of Aa (heterozygous)
q${}^2$ = frequency of aa (homozygous recessive).
p+q = 1
0.3 + q = 1
q = 1 - 0.3 = 0.7
Aa = 2pq = 2* 0.3 * 0.7 = 0.42.
Thus, the correct answer is option B.