A population is in Hardy-Weinberg equilibrium | vedclass

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(B) For a gene with two alleles,$A$ and $a$,if the frequency of $A$ is $p$ and the frequency of $a$ is $q$,then the frequencies of the three possible

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$0.42$

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(B) For a gene with two alleles,$A$ and $a$,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 $aa$) are expressed by the Hardy-Weinberg equation: $p^2 + 2pq + q^2 = 1$.Here,$p$ is the frequency of allele $A$ and $q$ is the frequency of allele $a$.Given that $p = 0.7$,we know that $p + q = 1$,so $q = 1 - 0.7 = 0.3$.The frequency of the heterozygous genotype $Aa$ is represented by $2pq$.Substituting the values: $2pq = 2 	imes 0.7 	imes 0.3 = 0.42$.

$A$ population is in Hardy-Weinberg equilibrium for a gene with only two alleles. If the gene frequency of an allele $A$ is $0.7$,the genotype frequency of $Aa$ is

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