详细的解释Fst,Fis和Fit.docx

Worked example of calculating F-statistics from genotypic data:Return to Main Index page?????????? ???????????????? ????????? Go to Lecture 35??????????????? Go to Lecture 36? ??Genotype??AAAaaaSubpopulation 1125250125Subpopulation 2503020Subpopulation 3100500400N (number of individuals genotyped. The sum of each of the rows in the table above):Population 1:?? 500Population 2:?? 100Population 3:?? 1,000Remember that the number of alleles is TWICE the number of genotypes.Step 1.? Calculate the gene (allele) frequencies:Each homozygote will have two alleles, each heterozygote will have one allele.? Note that the denominator will be twice Ni (twice as many alleles as individuals).????????????? Eqns FST.1Step 2. Calculate the expected genotypic counts under Hardy-Weinberg Equilibrium, and then calculate the excess or deficiency of homozygotes in each subpopulation.Pop. 1??????? Expected AA = 500*0.52??????????????????? = 125???? (= observed)??????????????????????? Expected Aa =? 500*2*0.5*0.5?????????? = 250???? (= observed)??????????????????????? Expected aa =? 500*0.52????????????????????? = 125???? (= observed)Pop. 2??????? Expected AA = 100*0.652?????????????????? = 42.25? (observed has excess of 7.75)??????????????????????? Expected Aa =? 100*2*0.65*0.35??????? =? 45.5?? (observed has deficit of 15.5)??????????????????????? Expected aa =? 100*0.352????????????????????? = 12.25? (observed has excess of 7.75)???????????????????? Note that sum of two types of homozygote excess = amount of heterozygote deficiency.??????????????????????????????????? These quantities have to balance (its a mathematical necessity, given that p + q =1.? Pop. 3??????? Expected AA = 1,000*0.352?????????????????? = 122.5? (observed has deficiency of 22.5)??????????????????????? Expected Aa =? 1,000*2*0.65*0.35??????? =? 455?? (observed has excess of 45)??????????????????????? Expected aa =? 1,000*0.352????????????????????? = 422.5? (observed has deficiency of 22.5)Summary of homozygote defici