Answer:
The correct answer is 48 percent.
Explanation:
Let the dominant allele be signified by letter T and the recessive allele be signified by letter t.
On the basis of Hardy-Weinberg equation, the frequency of T is denoted by p and the frequency of t by q. Therefore, p+q = 1.
p^2 signifies the frequency of TT, 2pq the frequency of Tt and q^2 the frequency of tt, which is denoted by the equation:
p^2 + 2pq + q^2 = 1
Based on the given information, 64 percent of the population are tasters, thus, 36 percent would be the non-tasters, and would exhibit the homozygous recessive genotype, that is, tt.
Thus, the frequency of tt will be 0.36, that is, q^2 will be 0.36 or q = 0.6.
p+q = 1, p = 1-q and so p = 1-0.6, p = 0.4
The frequency of Tt is 2pq = 2*0.6*0.4 = 0.48.
Hence, 48 percent of the population would be the heterozygous for this trait.
