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(C) According to the Hardy-Weinberg principle,the equation is $p^2 + 2pq + q^2 = 1$,where $p$ is the frequency of the dominant allele and $q$ is the f

Vedclass · Accepted Answer

$0.48$

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(C) According to the Hardy-Weinberg principle,the equation is $p^2 + 2pq + q^2 = 1$,where $p$ is the frequency of the dominant allele and $q$ is the frequency of the recessive allele.Given that the frequency of recessive individuals $(q^2)$ is $0.16$.Therefore,$q = \sqrt{0.16} = 0.4$.Since $p + q = 1$,we have $p = 1 - 0.4 = 0.6$.The frequency of heterozygous individuals is represented by $2pq$.$2pq = 2 	imes 0.6 	imes 0.4 = 0.48$.Thus,the frequency of heterozygous individuals is $0.48$.

In a population of $800$ rabbits showing Hardy-Weinberg equilibrium,the frequency of recessive individuals was $0.16$. What is the frequency of heterozygous individuals?

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