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(B) According to the Hardy-Weinberg principle,for a population in genetic equilibrium,the genotype frequencies are given by the expansion of $(p + q)^

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$0.48$

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(B) According to the Hardy-Weinberg principle,for a population in genetic equilibrium,the genotype frequencies are given by the expansion of $(p + q)^2 = p^2 + 2pq + q^2 = 1$.Here,$p$ is the frequency of the dominant allele $(A)$ and $q$ is the frequency of the recessive allele $(a)$.Given: $p = 0.6$ and $q = 0.4$.The frequency of heterozygotes $(Aa)$ is represented by $2pq$.Substituting the values: $2 	imes 0.6 	imes 0.4 = 2 	imes 0.24 = 0.48$.Therefore,the frequency of heterozygotes in the population is $0.48$.

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

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