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(A) According to the Hardy-Weinberg principle,the frequency of the recessive phenotype $(q^2)$ is given by the ratio of individuals expressing the rec

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$42$

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(A) According to the Hardy-Weinberg principle,the frequency of the recessive phenotype $(q^2)$ is given by the ratio of individuals expressing the recessive trait to the total population.Total population = $200$.Number of individuals with recessive phenotype = $98$.Frequency of recessive phenotype $(q^2)$ = $98 / 200 = 0.49$.Therefore,the frequency of the recessive allele $(q)$ = $\sqrt{0.49} = 0.7$.Since $p + q = 1$,the frequency of the dominant allele $(p)$ = $1 - 0.7 = 0.3$.The frequency of heterozygotes is given by $2pq$.$2pq = 2 	imes 0.3 	imes 0.7 = 0.42$.To express this as a percentage,multiply by $100$: $0.42 	imes 100 = 42\%$.

If $98$ members out of $200$ in a population express the recessive phenotype,what is the percentage of heterozygotes in that population?

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