At a particular locus, frequency of allele ' A ' is 0.6 and that of...

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 ' $\mathrm{a}$ ' is $q$, then the frequencies of the three possible genotypes (AA, Aa and aa) can be expressed by the Hardy-Weinberg equation: $p^2+2 p q+q^2=1$ where $p^2=$ Frequency of AA (homozygous dominant) individuals $q^2=$ Frequency of aa (homozygous recessive) individuals $2 p q=$ Frequency of Aa (heterozygous) individuals. So, $p=0.6 and q=0.4$ (given) $$\begin{aligned} \therefore 2 p q(\text{ frequency of heterozygote }) & =2 \times 0.6 \times 0.4 \\ & =0.48 .\end{aligned}$$