详细的解释Fst,Fis和Fit

Worked example of calculating F-statistics from genotypic data:Return to Main IndexpageGo to Lecture 35Go to Lecture 36N (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 N i (twice as many alleles as individuals).Eqns FST.1Step 2. Calculate the expected genotypic counts under Hardy-WeinbergEquilibrium, 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 of7.75)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 (it's 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 deficiency or excess relative to HWE:Pop. 1. Observed = Expected: perfect fitPop. 2. Excess of 15.5 homozygotes: some inbreedingPop. 3. Deficiency of 45 homozygotes: outbred or experiencing a Wahlund effect (isolate breaking).Step 3. Calculate the local observed heterozygosities of each subpopulation (we will call them H obs s, where the s subscript refers to the s th of n populations -- 3 in this example). Here we count genotypes:H= 250 / 500 = 0.5obs 1H= 30 /obs 2100 = 0.3H= 500 /obs 31000= 0.5Step 4. Calculate the local expected heterozygosity, or gene diversity, of each subpopulation (modified version of Eqn 35.1):Eqns FST.2(With two alleles it would actually be easier to use 2pq than to use the summation format of Eqn 33.1)Notation: Note that I am using p1 and q1 here (where the subscripts refer to subpopulations 1 through 3). We would need to use multiple subscripts if we were using the notation of Eqn 35.1 where the alleles are p i (and the i refer to alleles 1 to k). Indeed, with realmulti-locus multipopulation data, we would have a triple summation and three subscripts; one for alleles (i =1 to k), one for the loci (l =1 to m), and one for subpopulations (s = 1 to n).Step 5. Calculate the local inbreeding coefficient of each subpopulation (same as Eqn 35.4, except that we are subscripting for the subpopulations):where s (s = 1 to 3) refers to the subpopulation Eqn FST.3F= (0.5 — 0.5) / 0.5 = 01F= (0.455 — 0.3) / 0.455 = 0.3412[positive F means fewer heterozygotes than expected indicates inbreeding]F= (0.455 — 0.5) / 0.455 = -0.0993[negative F means more heterozygotes than expected means excess outbreeding]Step 6. Calculate (p-bar, the frequency of allele A) over the total population.[Note that if we had more alleles we could put this and Step 7 all together as a single "global gene frequencies" step, or have one for each allele frequency].{genotype splitting method}or (yields same answer){using Eqn FST.1 values for p s}Note that we weight by population sizeStep 7. Calculate (q-bar, the frequency of allele a) over the total populationCheck: p-bar+ q-bar= 0.4156 + 0.5844 = 1.0 (as required by Eqn 31.1).The check doesn't guarantee that our result is correct, but if they don't sum to one, we know we miscalculated.Step 8. Calculate the global heterozygosity indices (overI ndividuals, S ubpopulations and T otal population)Note that the first two calculations employ a weighted average of the values in the whole set of subpopulations.Hbased on observed heterozygosities in individuals in subpopulationsIEqn FST.4Hbased on expected heterozygosities in subpopulationsSEqn FST.5Hbased on expected heterozygosities for overall total population (using Tglobal allele frequencies and a modified form of Eqn 35.1):Eqn FST.6or we could calculate it as 2*p-bar *q-bar = 2 * 0.4156 * 0.5844 = 0.4858Step 9. CALCULATE THE GLOBAL F-STATISTICS:Compare and contrast the global F IS below with the "local inbreeding coefficient" F s of Step 5.Here we are using a weighted average of the individual heterozygosities over all the subpopulations.Both F IS and F s are, however, based on the observed heterozygosities, whereas F ST and F IT are based on expected heterozygosities.Eqn FST.7Eqn FST.8Eqn FST.9 Notation note: the subscripts I, S, and T are not summation subscripts. They simply indicate the level of our analysis. Likewise, the s on F s in Step 5 or on the p s in Step 1 (the s was implicit there) just tell us what we are referring to. In contrast, the subscripts for Eqn 35.1 and 35.2 are used in summations and change as we work through the pieces of the calculation.Step 10. Finally, draw some conclusions about the genetic structure of the population and its subpopulations.1) One of the possible HWE conclusions we could make:Pop. 1 is consistent with HWE (results of Step 2)2) Two of the possible "local inbreeding" conclusions we could make from Step 5:Pop. 2 is inbred (results of Step 5), andPop. 3 may have disassortative mating or be experiencing a Wahlund effect (more heterozygotes than expected).3) Conclusion concerning overall degree of genetic differentiation (F ST) Subdivision of populations, possibly due to genetic drift,accounts for approx. 3.4% of the total genetic variation(result of Eqn FST.8 F ST calculation in Step 9),4) No excess or deficiency of heterozygotes over the total population(F IT is nearly zero).。