Question

Assume that the allele frequencies of alleles A1 and A2 were 0.2 and 0.8 in one subpopulation and 0.4 and 0.6 in another subpopulation. Assuming that there are Hardy-Weinberg proportions within each subpopulation and that the subpopulations are of equal size, what would be the observed heterozygosity if these populations were lumped? How does this compare to what would be expected if there were one random-mating population?? Calculate the same values for the three heterozygotes when the subpopulations had frequencies of 0.2, 0.4, and 0.4 and 0.6, 0.2, and 0.2 for alleles A1, A2, and A3. How does the observed heterozygosity compare to the expected heterozygosity for each heterozygote and for all heterozygotes combined?

I need help solving it.

These are the answers:

3. H = , = ; P2 = = = 2PP = 024, 0.42pq 0.42 0.2, 2p P2 0.24, P, 0.2, . P23 = 0.2, 2ppg = 0.18, H = 0.6 versus 0.66 3

Answer 1

To calculate the heterozygosity, first of all calculate the H of both the populations, according to the Hardy-Weinberg eqution i.e. p²+2pq+q²=1

Heterozygosity of first population

p = 0.2, q = 0.8

2pq = (0.2)² + (0.8)² - 1 = 0.32

Heterozygosity of second population

2pq = (0.4)² + (0.6)² - 1 = 0.48

Now the observed heterozygosity of boththe population = (0.32+0.48) / 2 =0.4

So, H = 0.4

Average allelic frequency of

P12 = P 1+2 = 0.2^{2 +} 0.4² = .2

Average allelic frequency of

P13 = P 1+3 = 0.2^{2 +} 0.4² = .2

Heterozygosity for three allelic frequency =1 - (p² + q² + r²)

So, in this for first population heterozygosity = .64

n this for first population heterozygosity = .44

Average Heterozygosity = .64 +.44= .54