Question

In a population, the frequency of the dominant allele A is 0.6 and that of the recessive allele a is 0.4. What would be the frequency of the heterozygous genotype in a random mating population at equilibrium?

• A. 0.36
• B. 0.16
• C. 0.24
• D. 0.48

Answer 1

The correct option is D

0.48

Hardy-Weinberg equation is used calculate the frequency of occurrence of alleles in a population.
It has been given that the frequency of dominent allele 'p' = 0.6 and recessive allele 'q' = 0.4.
Applying the Hardy-Weinberg Equation, $p^2+2pq+q^2=1$
The frequency of the heterozygous genotype (Aa) = 2pq
Substituting for p and q, we get
2pq = 2 x 0.4 x 0.6 = 0.48