The data below demonstrate the frequency of tasters and nontasters of a certain compound in four isolated populations that are in Hardy-Weinberg equilibrium. The allele for nontasters is recessive. In which population is the frequency of the recessive allele highest?

$$\begin{array}{|c|c|c|c|} \hline \text{Population} & \text{Tasters} & \text{Nontasters} & \text{Size of Population} \\ \hline \text{A} & 110 & 32 & 142 \\ \hline \text{B} & 8{,}235 & 4{,}328 & 12{,}563 \\ \hline \text{C} & 215 & 500 & 715 \\ \hline \text{D} & 11{,}489 & 2{,}596 & 14{,}085 \\ \hline \end{array}$$

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The frequency of the recessive allele is given by $q$, where,

$$p^2+2pq+\underbrace{q^2}_{\text{homozygous recessive}}=1$$

We can calculate this for each population. Since the allele for nontasters is recessive, the frequency of nontasters corresponds to $q^2$. Take the square root of this value to obtain $q$.

$$\begin{array}{|c|c|c|c|c|c|} \hline \text{Pop.} & \text{Tasters} & \text{Nontasters} & \text{Size of Population} & q^2 & q \\ \hline \text{A} & 110 & 32 & 142 & 32/142 & 0.47\\ \hline \text{B} & 8{,}235 & 4{,}328 & 12{,}563 & 4{,}328/12{,}563 & 0.59 \\ \hline \text{C} & 215 & 500 & 715 & 500/715 & 0.84 \\ \hline \text{D} & 11{,}489 & 2{,}596 & 14{,}085 & 2{,}596/14{,}085 & 0.43 \\ \hline \end{array}$$