详细的解释Fst,Fis和Fit

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1、WorkedexampleofcalculatingF-statisticsfromgenotypicdata:ReturntoMainIndexpage                                   GotoLecture35               GotoLecture36   Genotype  AAAaaaSubpopulation1125250125Subpopulation2503020Subpopulation3100500400N(numberofindividualsgenotyped.Thesumofeachof

2、therowsinthetableabove):Population1:  500Population2:  100Population3:  1,000RememberthatthenumberofallelesisTWICEthenumberofgenotypes.Step1. Calculatethegene(allele)frequencies:Eachhomozygotewillhavetwoalleles,eachheterozygotewillhaveoneallele. NotethatthedenominatorwillbetwiceNi(t

3、wiceasmanyallelesasindividuals).             EqnsFST.1Step2.CalculatetheexpectedgenotypiccountsunderHardy-WeinbergEquilibrium,andthencalculatetheexcessordeficiencyofhomozygotesineachsubpopulation.Pop.1       ExpectedAA=500*0.52                   =125    (=observed)                  

4、     ExpectedAa= 500*2*0.5*0.5          =250    (=observed)                       Expectedaa= 500*0.52                     =125    (=observed)Pop.2       ExpectedAA=100*0.652                  =42.25 (observedhasexcessof7.75)                       ExpectedAa= 100*2*0.65*0.35       = 

5、45.5  (observedhasdeficitof15.5)                       Expectedaa= 100*0.352                     =12.25 (observedhasexcessof7.75)                    Notethatsumoftwotypesofhomozygoteexcess=amountofheterozygotedeficiency.                                   Thesequantitieshavetobalance

6、(it'samathematicalnecessity,giventhatp+q=1. Pop.3       ExpectedAA=1,000*0.352                  =122.5 (observedhasdeficiencyof22.5)                       ExpectedAa= 1,000*2*0.65*0.35       = 455  (observedhasexcessof45)                       Expectedaa= 1,000*0.352                

7、     =422.5 (observedhasdeficiencyof22.5)SummaryofhomozygotedeficiencyorexcessrelativetoHWE:   Pop.1. Observed=Expected:perfectfit   Pop.2. Excessof15.5homozygotes:someinbreeding   Pop.3. Deficiencyof45homozygotes:outbredorexperiencingaWahlundeffect(isolatebreaking).Step3.Calculatet

8、helocalobservedhete