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(C) According to the Hardy-Weinberg principle,the sum of allele frequencies is $p + q = 1$.Here,let $p$ be the frequency of allele '$A$' and $q$ be th

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$0.6$ and $0.48$

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(C) According to the Hardy-Weinberg principle,the sum of allele frequencies is $p + q = 1$.Here,let $p$ be the frequency of allele '$A$' and $q$ be the frequency of allele '$B$'.Given $p = 0.4$,the frequency of allele '$B$' $(q)$ is $1 - 0.4 = 0.6$.The frequency of the heterozygous genotype is represented by $2pq$.Substituting the values: $2 	imes 0.4 	imes 0.6 = 0.48$.Therefore,the frequency of the '$B$' allele is $0.6$ and the frequency of the heterozygous genotype is $0.48$.

If the frequency of the '$A$' allele is $0.4$,find the frequency of the '$B$' allele and the heterozygous genotype in a random mating population at equilibrium.

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