In a population at Hardy-Weinberg equilibrium | vedclass

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Question

(C) According to the Hardy-Weinberg principle,the genetic equilibrium in a population is represented by the equation: $(p + q)^2 = p^2 + 2pq + q^2 = 1

Vedclass · Accepted Answer

$0.48$

Vedclass · Answer

(C) According to the Hardy-Weinberg principle,the genetic equilibrium in a population is represented by the equation: $(p + q)^2 = p^2 + 2pq + q^2 = 1$.Here,$p$ represents the frequency of the dominant allele '$A$',and $q$ represents the frequency of the recessive allele '$a$'.Given: $p = 0.6$ and $q = 0.4$.The frequency of heterozygotes $(Aa)$ is represented by the term $2pq$.Substituting the values: $2pq = 2 	imes 0.6 	imes 0.4 = 0.48$.Therefore,the frequency of heterozygotes in the population is $0.48$.

In a population at Hardy-Weinberg equilibrium,the frequency of allele '$A$' is $0.6$ and the frequency of allele '$a$' is $0.4$. What will be the frequency of heterozygotes in the population?

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