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H2 = — to give an average observed heterozygosity of

H, = —--1--= 0.30. If p is the frequency of one allele

(open circles) and q the frequency of the alternate allele (filled circles), then p1 = 13/20 = 0.65 and q1 = 1 - p1 = 0.35 while p2 = 7/20 = 0.35 and q2 = 1 - p2 = 0.65. The average expected heterozygosity in the two subpopulations is then HS = 1/2[2(0.65)(0.35) + (2(0.35)(0.65)] = 0.455. In the total population the average allele frequencies are p = 1/2(0.65 + 0.35) = 0.50 and q = 1/2(0.35 + 0.65) = 0.50 giving an expected heterozygosity in the total population of Ht = 2pq = 2(0.5)(0.5) = 0.5.

frequencies of H, = — and H = — • Together 1 10 2 10

these yield an average observed heterozygosity of

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