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(B) According to the Hardy-Weinberg principle,the genetic equilibrium is represented by the equation $(p + q)^2 = p^2 + 2pq + q^2 = 1$,where $p$ is th

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$32$

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(B) According to the Hardy-Weinberg principle,the genetic equilibrium is represented by the equation $(p + q)^2 = p^2 + 2pq + q^2 = 1$,where $p$ is the frequency of the dominant allele and $q$ is the frequency of the recessive allele.Given that the frequency of the recessive allele $(q)$ is $80\%$,we have $q = 0.8$.Consequently,the frequency of the dominant allele $(p)$ is $p = 1 - q = 1 - 0.8 = 0.2$.Carrier individuals are those who are heterozygous for the recessive trait,represented by the term $2pq$ in the Hardy-Weinberg equation.Substituting the values: $2pq = 2 	imes 0.2 	imes 0.8 = 0.32$.Converting this to a percentage: $0.32 	imes 100 = 32\%$.Therefore,the frequency of carrier individuals in the population is $32\%$.

In a population exhibiting random mating,the frequency of a disease-causing recessive allele is $80\%$. What will be the frequency of carrier individuals in the population (in $\%$)?

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