# Fixation index (FST)

Subpopulation 1 Subpopulation 2 Subpopulation 3 Total
Genotype AA 125 50 100
Aa 250 30 500
aa 125 20 400
Number of individual 500 100 1000 1600
Number of alleles 1000 200 2000 3200
Step 1. Calculate the gene   (allele) frequencies
Observed allele frequency A (p) (125*2+250)/1000=0.5 (50*2+30)/200=0.65 (2*100+500)/2000=0.35
a (q) 0.5 0.35 0.65
Step 2. Calculate the expected   genotypic counts under Hardy-Weinberg Equilibrium, and then calculate the   excess or deficiency of homozygotes in each subpopulation.
Summary of homozygote deficiency or excess relative to HWE:
Pop. 1. Observed = Expected: perfect fit
Pop. 2. Excess of 15.5 homozygotes: some inbreeding
Pop. 3. Deficiency of 45 homozygotes: outbred or   experiencing a Wahlund effect (isolate breaking).
Expected allele frequency AA 500*0.5^2 = 125 (= observed) 100*0.65^2 = 42.25 (observed has excess of 7.75) 1,000*0.35^2 = 122.5   (observed has deficiency of 22.5)
Aa 500*2*0.5*0.5 = 250 (=   observed) 100*2*0.65*0.35 = 45.5 (observed has deficit of 15.5) 1,000*2*0.65*0.35 = 455   (observed has excess of 45)
aa 500*0.5^2 = 125 (= observed) 100*0.35^2 = 12.25 (observed has excess of 7.75) 1,000*0.35^2 = 422.5   (observed has deficiency of 22.5)
Step 3. Calculate the local   observed heterozygosities of each subpopulation (we will call them Hobs s,   where the s subscript refers to the sth of n populations — 3 in this   example).
Local observed   heterozygosities 250/500 = 0.5 (Hobs 1) 30/100 = 0.3 (Hobs 2) 500/1000 = 0.5(Hobs 3)
Step 4. Calculate the local   expected heterozygosity, or gene diversity, of each subpopulation
Hexp = 2pq
Local expected   heterozygosity 2*0.5*0.5=0.5 (Hexp 1) 2*0.65*3.5=0.455 (Hexp 2) 2*0.35*0.65=0.455 (Hexp 2)
Step 5. Calculate the local   inbreeding coefficient of each subpopulation
F = (Hexps -Hobs)/Hexp
[positive F means fewer heterozygotes than expected indicates   inbreeding]
[negative F means more heterozygotes   than expected means excess outbreeding]
F1=(0.5—0.5)/0.5=0 F2=(0.455—0.3)/0.455=0.341 F3=(0.455—0.5)/0.455=-0.099
Step 6. and 7. Calculate p-bar   (p-bar, the frequency of allele A) over the total population.
Calculate q-bar (q-bar, the frequency of allele a) over the total   population.
Check: p-bar + q-bar = 1.0
the frequency of allele over the total population p-bar (0.5*1000+0.65*200+0.35*2000)/3200=0.4156
q-bar (0.5*1000+0.35*200+0.65*2000)/3200=0.5844
Step 8. Calculate the global   heterozygosity indices (over Individuals, Subpopulations and Total   population)
HI based on observed heterozygosities in individuals in subpopulations
HS based on expected heterozygosities in subpopulations
HT based on expected heterozygosities for overall total population
HI (observed) (0.5*500+0.3*100+0.5*1000)/1600=0.4875
HS (expected) (0.5*500+0.455*100+0.455*1000)/1600=0.4691
HT (in overall total population) 2*p-bar *q-bar    = 2 * 0.4156 * 0.5844 = 0.4858
Step 9. Calculate the global   F-statistics
Compare and contrast the global FISbelow with the “local inbreeding   coefficient” Fs of Step 5.
Here we are using a weighted average of the individual heterozygosities   over all the subpopulations.
Both FIS and Fs are, however, based   on the observed heterozygosities,
whereas FST and FIT are based   on expected heterozygosities.
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
Step 10 conclusions 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), and
Pop. 3 may have disassortative mating or be experiencing a Wahlund effect   (more heterozygotes than expected).
3) Conclusion concerning overall   degree of genetic differentiation (FST)
Subdivision of populations, possibly due to genetic drift,
accounts for approx. 3.4% of the total genetic variation
(result of Eqn FST.8 FST   calculation in Step 9),
4) No excess or deficiency of   heterozygotes over the total population (FIT    is nearly zero).