In a classroom taste test for PTC the genotype frequency was determined to be P = .35 and Q = .65. After applying the Hardy-Weinberg equation, the percentage of genotypes was found to be as follows:

Question

Quizplus · Accepted Answer

When applying the Hardy-Weinberg equation, the percentage of genotypes should be P2 + 2pq + q2 = 1, where P and Q represent the frequencies of the two alleles and p and q represent the frequencies of the two genotypes. Using the given genotype frequencies, we can calculate P and Q as follows:

P = F1F1 = (0.35 * 2 + 0.65) / 2 = 0.675
Q = F1S1 = (0.35 + 0.65 * 2) / 2 = 0.325

Then, we can use the equation to calculate the genotype frequencies:

F1F1F1S1F1S1S1S1 = p2 = (0.675)2 = 0.456
F1F1F1S1F1S1S1U1 = 2pq = 2 * 0.675 * 0.325 = 0.438
F1S1F1S1F1S1S1S1 = q2 = (0.325)2 = 0.105

Therefore, the correct answer is B.

In a Classroom Taste Test for PTC the Genotype Frequency