7. The Hardy-Weinberg equation allows us to make some calculations about the local population.

(A) What is the frequency of "aa" individuals?


Given 1/685 children have the disease and are aa, this means q² = .00146


(B) What is the frequency of the "a" allele in the gene pool?


q=.0382


(C) What is the frequency of the "A" allele in the gene pool?


Since p + q = 1 and q = .0382, then 1 - .0382 = p = 0.9618


(D) What is the frequency of heterozygous individuals?


2 pq = .073 or 1 person in 14 are Aa individuals.
The frequency of aa (homozygous recessive) q² = .00146
The frequency of AA (homozygous dominants) p² = 0.925


(E) Assuming random breeding, what are the chances that two carriers will marry?


.073 x .073 = .0053 or 1 marriage in 190 will involve two carriers