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(A) According to the Hardy-Weinberg principle,the sum of allele frequencies is $p + q = 1$,and the genotype frequencies are $p^2 + 2pq + q^2 = 1$.Here

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$0.49$

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(A) According to the Hardy-Weinberg principle,the sum of allele frequencies is $p + q = 1$,and the genotype frequencies are $p^2 + 2pq + q^2 = 1$.Here,$p^2$ represents the frequency of homozygous dominant individuals,and $2pq$ represents the frequency of heterozygous individuals.The sum of these two,$p^2 + 2pq$,represents the frequency of individuals possessing at least one dominant allele.Given that $p^2 + 2pq = 51\% = 0.51$.Since the total frequency of all genotypes is $p^2 + 2pq + q^2 = 1$,the frequency of recessive individuals $(q^2)$ is calculated as:$q^2 = 1 - (p^2 + 2pq)$$q^2 = 1 - 0.51 = 0.49$.Therefore,the frequency of recessive individuals in the population is $0.49$.

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

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