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)

$0.48$

Answer 2

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 $AA$ (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) $=2 \times 0.6 \times 0.4$
$=0.48$