This is the only information provided, and I am unsure how to solve for FST. Allelic Frequencies for each Subpopulation are listed in the graph.
N (number of individuals genotyped. The sum of each of the rows in the table)
population 1: 1
population 2: 1
population 3: 1
Remember that the number of alleles is TWICE the number of genotypes
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 Ni (twice as many alleles as individuals)
p1 = 2 X 0.1 + 0.3 / 2 = 0.25
q1 = 1 - P1 = 1 - 0.25 = 0.75
p2 = 2 X 0.2 + 0.3 / 2 = 0.35
q2 = 1 - P2 = 1 - 0.35 = 0. 65
p3 = 2 X 0.3 + 0.3/ 2 = 0.45
q3 = 1 - P3 = 1 - 0.45 = 0.55
2. Calculate the local observed heterozygosity of each subpopulation. Here we count genotypes:
Hobs 1 = 0.3 / 1 = 0.3
Hobs 2 = 0.3 / 1 = 0.3
Hobs 3 = 0.3/ 1 = 0.3
3. Calculate the local expected heterozygosity, or gene diversity, of each subpopulation
Hexp1 = 1 - (p12 + q12) = 1 - (0.0625 + 0.5625) = 0.375
Hexp2 = 1 - (p22 + q22) = 1 - (0.1225 + 0.4225) = 0.455
Hexp3 = 1 - (p32 + q32) = 1 - (0.2025 + 0.3025) = 0.495
Fs = Hexp - Hobs / Hexp
F1 = 0.375 - 0.3 / 0.375 = 0.2
F2 = 0.455 - 0.3 / 0.455 = 0.34
F3 = 0.495 - 0.3 / 0.495 = 0.39
4. Calculate P- (p-bar, the frequency of allele 1) over the total population
2 X 0.1 + 0.3 + 2 X 0.2 + 0.3 + 2 X 0.3 + 0.3 / 2 + 2 + 2 = 0.35
5. Calculate q- (q-bar, the frequency of allele 3) over the total population
2 x 0.6 + 0.3 + 2 x 0.5 + 0.3 + 2 x 0.4 + 0.3 / 2 + 2 + 2 = 0.65
6. HS based on expected heterozygosity in subpopulations
Hexp1 x N1 + Hexp2 X N2 + Hexp3 X N3 / Ntotal = 0.375 x 1 + 0.455 x 1 + 0.495 x 1 / 3 = 0.411
7. HT based on expected heterozygosity for overall total population
1 - (p-2 + q-2) = 1 - (0.1225 + 0.4225) = 0.455
8. FST = HT - HS / HT = 0.455 - 0.411 / 0.455 = 0.0967
This is the only information provided, and I am unsure how to solve for FST. Allelic...
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