Worked example of calculating F-statistics from genotypic data:

 Genotype AA Aa aa Population 1 125 250 125 Population 2 50 30 20 Population 3 100 500 400

N (number of individuals genotyped):

Population 1:   500
Population 2:   100
Population 3:   1,000

Remember that the number of alleles is TWICE the number of genotypes (every genotype has two alleles).

Gene frequencies: Each homozygote will have two alleles, each heterozygote will have one allele.

p1 (frequency of allele A in Pop. 1) = (2*125 + 250)/1,000 = 0.50
q
1 = 0.5

p2 = (2*50 + 30)/200 = 0.65
q2 = 0.35

p3 = (2*100 + 500)/2,000 = 0.35
q
3 = 0.65

Observed heterozygosity (here we are counting heterozygous genotypes):

Hobs1 = 250/500                = 0.5

Hobs2 = 30/100                              = 0.3

Hobs3 = 500/1000                                        = 0.5

Expected heterozygosity (2pq):

Hexp1 = 2*0.5*0.5           = 0.5                                   [Observed = Expected]

Hexp2 = 2*0.65*0.35                   = 0.46                     [Observed < Expected]

Hexp3 =2*0.35*0.65                                  = 0.46       [Observed > Expected]

Population inbreeding coefficient [F = (Hexp - Hobs) /Hexp]

F1 = (0.5  0.5)/0.5         = 0

F2 = (0.46  0.3)/0.46     = 0.341              [fewer heterozygotes than expected indicates inbreeding]

F3 = (0.46  0.5)/0.46     = -0.099            [more heterozygotes than expected means excess outbreeding]

p-bar (frequency of allele A) over all populations

(2*125 + 250 + 2*50 + 30 + 2*100 + 500)/(1,000 + 200 + 2,000) = 1,330 / 3,200 =      0.4156

or

(0.5*1,000 + 0.65*200 + 0.35*2,000) / 3,200

q-bar (frequency of allele a) over all populations

(2*125 + 250 + 2*20 + 30 + 2*400 + 500) / (1,000 + 200 + 2,000) = 1,870 / 3,2000   = 0.5844

Check: p-bar + q-bar = 1.0 (correct!)

Heterozygosity indices (over individuals, subpopulations and total population)

HI based on observed heterozygosities in populations

[= (Hobs1*N1 + Hobs2*N2 + Hobs3*N3)/NTOTAL] = (0.5*500 + 0.3*100 + 0.5*1000) / 1,600 = 0.4875

HS based on expected heterozygosities in populations

[= (Hexp1*N1 + Hexp2*N2 + Hexp3*N3)/NTOTAL] = (0.5*500 + 0.46*100 + 0.46*1000) / 1,600 = 0.4691

HT based on expected heterozygosities overall:

[= 2*p-bar *q-bar] = 2 * 0.4156 * 0.5844 = 0.0.4858

FINALLY, THE F-STATISTICS:

FIS [= (HS - HI)/HS] = (0.4691 - 0.4875) / 0.4691         = -0.0393

FST [= (HT - HS)/HT] = (0.4858 - 0.4691) / 0.4858         = 0.0344

FIT [= (HT - HI)/HT] = (0.4858 - 0.4875)/0.4858           = -0.0036

All of which tells us that: Pop1 is consistent with HWE,
Pop2 is inbred, and
Pop3 may have disassortative mating or be experiencing a Wahlund effect (more heterozygotes than expected).

Subdivision of populations, possibly due to genetic drift, accounts for approx. 3.4% of the total genetic variation,

and the set of populations, as a whole, shows no signs of inbreeding (FIT  is nearly zero).