Allele Frequency Calculation

Question

Mourant et al. (1976) cite data on 400 Basques from Spain, of which 250 were Rh+ and 150 were Rh- (in H-W equilibrium)

Calculate the allele frequencies of D and d (DD individuals have the Rh+ phenotype. dd individuals have the Rh- phenotype. The phenotype of Dd is Rh+). How many of the Rh+ individuals would be expected to be heterozygous?

Answer 1

Hi Lauren, thanks for the question, I'll try to explain it as thoroughly as I can.

-First, we need to recall the Hardy-Weinberg equation: 1=p²+2pq+q² this equation can be used to determine the frequency of each allele in the population.

-In this case, it would be 1=D²+2Dd+d²

-We also need to remember that p+q=1 (or, in this case, D+d=1)

-We know that DD and Dd result in the Rh+ phenotype and that dd results in the Rh- phenotype. Therefore 150 out of 400 Basques are known to have a dd genotype and 250 out of 400 have either DD or Dd.

-Based off what we know, we can do the math:

d²=150/400

√d²=√(150/400)

d=0.612

Since we know d+D=1, we can solve for D:

1-d=D

1-0.612=0.388=D

We know the allele frequencies now, and we can calculate the expected frequencies of any other the three possible genotypes by applying the Hardy-Weinberg equation.

To calculate the expected frequency of heterozygotes, we simply need the middle term of the Hardy-Weinberg equation: 2pq.

2(0.612)(0.388)=0.475

We expect 47.5% of Basques to be heterozygous, or 190 out of 400.