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Genotype  




Population 1 



Population 2 



Population 3 



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.
p_{2} = (2*50 + 30)/200 = 0.65
q_{2} = 0.35
p_{3} = (2*100 + 500)/2,000 =
0.35
q_{3} = 0.65
Observed heterozygosity (here we are counting heterozygous genotypes):
H_{obs1} = 250/500 = 0.5
H_{obs2} = 30/100 = 0.3
H_{obs3} = 500/1000 = 0.5
Expected heterozygosity (2pq):
H_{exp1} = 2*0.5*0.5 = 0.5 [Observed = Expected]
H_{exp2} = 2*0.65*0.35 = 0.46 [Observed < Expected]
H_{exp3} =2*0.35*0.65 = 0.46 [Observed > Expected]
Population inbreeding coefficient [F = (H_{exp}  H_{obs}) /H_{exp}]
F_{1} = (0.5 � 0.5)/0.5 = 0
F_{2} = (0.46 � 0.3)/0.46 = 0.341 [fewer heterozygotes than expected indicates inbreeding]
F_{3} = (0.46 � 0.5)/0.46 = 0.099 [more heterozygotes than expected means excess outbreeding]
pbar (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
qbar (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: pbar + qbar = 1.0 (correct!)
Heterozygosity indices (over individuals, subpopulations and total population)
H_{I }based on observed heterozygosities in populations
[= (H_{obs1}*N_{1} + H_{obs2}*N_{2} + H_{obs3}*N_{3})/N_{TOTAL}] = (0.5*500 + 0.3*100 + 0.5*1000) / 1,600 = 0.4875
H_{S} based on expected heterozygosities in populations
[= (H_{exp1}*N_{1} + H_{exp2}*N_{2} + H_{exp3}*N_{3})/N_{TOTAL}] = (0.5*500 + 0.46*100 + 0.46*1000) / 1,600 = 0.4691
H_{T }based on expected heterozygosities overall:
[= 2*pbar *qbar] = 2 * 0.4156 * 0.5844 = 0.0.4858
FINALLY, THE FSTATISTICS:
F_{IS} [= (H_{S}  H_{I})/H_{S}] = (0.4691  0.4875) / 0.4691 = 0.0393
F_{ST} [= (H_{T}  H_{S})/H_{T}] = (0.4858  0.4691) / 0.4858 = 0.0344
F_{IT} [= (H_{T}  H_{I})/H_{T}] = (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 (F_{IT} is nearly zero).