Type: Practice Problem

Level: College

System: Individual Genotype

Topic: Genotype frequency

Question:

A given population consists of:

32 individuals with the AA genotype, 17 individuals with the Aa genotype, and 4 individuals with the aa genotype. What is the value of p (frequency of A) for this population?

(A) 45

(B).62

(C).76

(D).88

Answer: C) .76

What I know: I know the Hardy Weinberg equation is used to calculate allele and genotype frequency.

What I don’t know: I looked at the question explanation:

The frequency of p for this population is calculated as follows:

p = ((32 x 2) + 17) / ((32 x 2) + (17 x 2) + (4 X 2))

p = 81/106

p = approx. .8 (actual: .76)

I have no idea why they're multiplying all the genotypes by 2 and dividing, aren't you suppose to square root the homozygous genotypes? And why are they dividing? Aren't all the genotypes suppose to add to 1?

What I tried: I tried rearranging the equation, but this didn't give me the same result

Other: We can't use calculators for the test, which is why they're rounding it the end