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(C) According to the Hardy-Weinberg principle,the sum of allele frequencies is $p + q = 1$,where $p$ is the frequency of the dominant allele and $q$ i

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$0.91$

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(C) According to the Hardy-Weinberg principle,the sum of allele frequencies is $p + q = 1$,where $p$ is the frequency of the dominant allele and $q$ is the frequency of the recessive allele.Individuals with at least one recessive allele include both homozygous recessive $(q^2)$ and heterozygous $(2pq)$ genotypes.The frequency of individuals with at least one recessive allele is given by $q^2 + 2pq = 0.51$.Since $p^2 + 2pq + q^2 = 1$,we can write $p^2 = 1 - (q^2 + 2pq)$.Substituting the given value: $p^2 = 1 - 0.51 = 0.49$.The frequency of the dominant phenotype is represented by $p^2$ (homozygous dominant) plus $2pq$ (heterozygous),which is $1 - q^2$. However,the question asks for the frequency of the dominant phenotype,which is $p^2 + 2pq = 1 - q^2$. Wait,the dominant phenotype is $p^2 + 2pq$. Since $p^2 + 2pq + q^2 = 1$,the dominant phenotype frequency is $1 - q^2$. Actually,the frequency of the recessive phenotype is $q^2$. Given $q^2 + 2pq = 0.51$,we know $p^2 = 1 - (q^2 + 2pq) = 1 - 0.51 = 0.49$. Since $p^2 = 0.49$,$p = 0.7$. Then $q = 1 - 0.7 = 0.3$.The dominant phenotype frequency is $p^2 + 2pq = (0.7)^2 + 2(0.7)(0.3) = 0.49 + 0.42 = 0.91$.

If $51\%$ of the members in a population possess at least one recessive allele,find the frequency of the dominant phenotype.

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