Question

seventy percent of all white north america can taste the chemical phenylthiocarbmide; 30% cannot taste it. The ability to taste is determined by the dominant allele T, and inability to taste is determined by the recessive allele t. a) if the North Americans are assumed to be at Weinberg equilibrium, then what are the frequencies of T and t? b) what is the frequency of homozygous tasters ? c) what is the frequency of heterozygous tasters.

Answer 1

A) Because 70 percent are tasters (T/T), 30 percent must be nontasters (t/t). This homozygous recessive frequency is equal to q ²; so, to obtain q, we will take the value of square root of 0.30:

q = √0.30 = 0.55

As we know, p + q =1

So, p = 1-q =1-0.55 = 0.45

Therefore, frequencies of T = 0.45

And frequencies of t = 0.55

B) The frequency of homozygous tasters:

Frequency of T/T = P² = 0.45² = 0.20 (T/T)

Frequency of t/t = q² = 0.55² = 0.30 (t/t)

C) The frequency of heterozygous tasters:

2pq = 2 X 0.45 X 0.55 = 0.495 (T/t)