Answer:
p = 0.7, q = 0.3 and p² = 0.49 2pq = 0.42 and q²= 0.09
Explanation:
The alleleic frequency of T = p is 70% which is 70/100= 0.7
The alleleic frequency of t= q is 30% which is 30/100= 0.3.
check: p+q = 1= 0.7 + 0.3 = 1.
The tasters could exist in genotypes as TT (p) or as heterozygous Tt (2pq) while the none tasters exist generally as tt (q).
thus the genotypic frequency of TT (p²) = 0.7² = 0.49
the genotypic frequency of Tt (2pq) = 2 x 0.7 x 0.3 = 0.42
the genotypic frequency of tt (q²) = 0.32 = 0.09
check: p²+ 2pq + q² = 1