In a population that is in Hardy-Weinberg equ | vedclass

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(D) According to the Hardy-Weinberg principle,$p + q = 1$ and $p^2 + 2pq + q^2 = 1$.Here,$q$ represents the frequency of the recessive allele,and $p$

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$96$

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(D) According to the Hardy-Weinberg principle,$p + q = 1$ and $p^2 + 2pq + q^2 = 1$.Here,$q$ represents the frequency of the recessive allele,and $p$ represents the frequency of the dominant allele.Given: $q = 0.20$.Therefore,$p = 1 - 0.20 = 0.80$.The individuals showing the dominant trait are those with genotypes $AA$ $(p^2)$ and $Aa$ $(2pq)$.The frequency of the dominant phenotype is $p^2 + 2pq = 1 - q^2$.Calculating $q^2$: $(0.20)^2 = 0.04$.Frequency of dominant phenotype = $1 - 0.04 = 0.96$.To express this as a percentage: $0.96 	imes 100 = 96\%$.

In a population that is in Hardy-Weinberg equilibrium,the frequency of the recessive allele for a certain genetic trait is $0.20$. What percentage of individuals in the next generation is expected to show the dominant trait (in $\%$)?

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