Frequency of 'A' allele is 0.6 and that of 'a' allele is 0.4. What would be frequency of heterozygotes in random mating population?

The correct option is B 0.48
According to the Hardy-Weinberg law, the allele frequencies in a population remain constant under absence of factors responsible for evolution (natural selection, migration, genetic drift). It states that the sum of all genotype frequencies can be represented as the binomial expansion of the square of the sum of p and q. This sum is equal to one. (p + q)2 = $p^2$ + 2pq + $q^2$ = 1. Here p is the frequency of the "A" allele and q is the frequency of the "a" allele in the population. Thus, $p^2$ represents the frequency of the homozygous genotype AA, $q^2$ is the frequency of the homozygous genotype aa, and 2pq represents the frequency of the heterozygous genotype Aa. Further, it also states that the sum of the allele frequencies for all the alleles at the locus should be 1, so p + q = 1. In the question, the frequency of recessive allele is (a) = (0.4) and frequency of dominant allele is (A) = (0.6). Since (p + q)2 = $p^2$ + 2pq + $q^2$ = 1; hence, frequency of heterozygotes= (2pq = 2 x 0.6 x 0.4 = 0.48). Frequency of homozygous genotype aa = ($q^2$ = 0.16). Frequency of homozygous dominant AA = ($p^2$ = 0.36). So, the correct option is D.