If a population is in Hardy-Weinberg equilibr | vedclass

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(A) According to the Hardy-Weinberg principle,the frequency of the recessive phenotype is represented by $q^2$.Given that $25\%$ of the population exh

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$0.5$

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(A) According to the Hardy-Weinberg principle,the frequency of the recessive phenotype is represented by $q^2$.Given that $25\%$ of the population exhibits the recessive trait,we have $q^2 = 25\% = 0.25$.To find the frequency of the recessive allele $(q)$,we take the square root of $q^2$:$q = \sqrt{0.25} = 0.5$.Therefore,the frequency of the recessive allele in the gene pool is $0.5$.

If a population is in Hardy-Weinberg equilibrium and $25\%$ of the individuals exhibit the recessive trait,find the frequency of the recessive allele in the gene pool.

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