HSCI 324 Lecture Notes - Lecture 2: Linkage Disequilibrium, Effective Population Size, Zygosity
9 Dec 2015
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Therefore, the optimal frequency of males and females that would produce the highest or maximum effective human population is when nf=0. 5 and nm=0. 5, i. e. when they are equal size. Therefore, it is 1042 times more likely for cystic fibrosis to occur in children from first cousins versus unrelated individuals: for hbs allele: Ht = 2 = 2(0. 165)(0. 835) = 0. 27555 (population heterozygosity) 2pq = 2(0. 10)(0. 90) = 0. 1800 subpopulation a heterozygosity. 2pq = 2(0. 08)(0. 92) = 0. 1472 subpopulation b heterozygosity. 2pq = 2(0. 11)(0. 89) = 0. 1958 subpopulation c heterozygosity. 2pq = 2(0. 37)(0. 63) = 0. 4662 subpopulation d heterozygosity. Hs = avg 2pq = 0. 2473 (average subpopulation heterozygosity) Fst = (ht hs)/ht = (0. 27555 0. 2473) / 0. 27555 = 0. 102522. Ht = 2 = 2(0. 06)(0. 94) = 0. 1128. 2pq = 2(0. 07)(0. 93) = 0. 1302 subpopulation a heterozygosity. 2pq = 2(0. 06)(0. 94) = 0. 1128 subpopulation b heterozygosity. 2pq = 2(0. 04)(0. 96) = 0. 0768 subpopulation c heterozygosity. 2pq = 2(0. 07)(0. 93) = 0. 1302 subpopulation d heterozygosity.
