Question

At a particular locus, frequency of allele A is $0.6$ and that of allele a is $0.4$. What would be the frequency of heterozygotes in a random mating population at equilibrium?

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

Answer 1

Answer: (D)

$0.48$

In a stable population, for a gene with two alleles, 'A' (dominant) and 'a' (recessive), if the frequency of 'A' is p and the frequency of 'a' is q, then the frequencies of the three possible genotypes $(AA,Aa$ and $aa$) can be expressed by the Hardy-Weinberg equation:
$p^2+2pq+q^2=1$
where $p^2=$ Frequency of $A$ (homozygous dominant) individuals
$q^2=$ Frequency of $aa$ (homozygous recessive) individuals
$2pq=$ Frequency of $Aa$ (heterozygous) individuals.
so, $p=0.6$ and $q=0.4$ (given)
$\therefore 2pq$ (frequency of heterozygote) $=2\times 0.6\times 0.4$
$=0.48$.