21 q 079 ht 2pq 033 locus 2 p 059 q 41 ht

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Unformatted text preview: le chi- square: [(19- 21.25)^2]/ 21.25 = 0.238 total chi- square = 0.079 + 0.238 = 0.318 1 degree of freedom P = 0.573 observed phenotypic ra,o is consistent with expecta,on of single Mendelian locus with complete dominance a two locus example LOCUS 1 subpopulation pi H 1 0.10 0.18 2 0.43 0.53 3 0.55 0.50 4 0.20 0.45 5 0.20 0.32 6 0.19 0.40 7 0.15 0.26 8 0.10 0.40 9 0.11 0.20 10 0.03 0.10 LOCUS 2 pi H 0.70 0.50 0.70 0.60 0.60 0.40 0.40 0.50 0.40 0.50 0.40 0.50 0.40 0.70 0.32 0.55 1.00 0.00 1.00 0.00 HI = observed proportion of heterozygotes over all subpopulations locus 1: HI = 0.33 locus 2: HI = 0.43 HS = expected proportion of heterozygotes within subpopulations assuming random mating locus 1: HS = 0.28 locus 2: HS = 0.37 over both loci, average HS = (0.28+0.37)/2 = 0.32 HT = expected proportion of heterozygotes over entire population locus 1: p = 0.21; q = 0.79; HT = 2pq = 0.33 locus 2: p = 0.59; q = 41; HT = 2pq = 0.48 => average HT over loci = (.33+.48)/2 = 0.41 inbreeding locus 1 FIS =(HS-HI)/HS = (0.32-0.33)/0.32 = -0.03 locus 2 FIS =(HS-HI)/HS = (0.32-0.43)/0.32 = -0.34 remember these are like F; inbreeding coefficient =substantial outcrossing in these populations overall FST (using average values of HT and HS across loci) FST = (HT- HS)/HT = (0.41- 0.32)/0.41 = 0.22 ⇒ the two subpopula,ons show substan,al isola,on ⇒ could be by drij or sele...
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This document was uploaded on 02/27/2014.

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