# Fixation index (FST)

## 文章目录

# Fixation index (FST)

整理来源 http://www.uwyo.edu/dbmcd/popecol/maylects/fst.html，把这个课程的步骤用表格来表示

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 | 50020.5*0.5 = 250 (= observed) |
10020.65*0.35 = 45.5 (observed has deficit of 15.5) |
1,00020.65*0.35 = 455 (observed has excess of 45) |
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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 | 20.50.5=0.5 (Hexp 1) |
20.653.5=0.455 (Hexp 2) |
20.350.65=0.455 (Hexp 2) |
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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 | (0.51000+0.65200+0.35*2000)/3200=0.4156 |
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q-bar | (0.51000+0.35200+0.65*2000)/3200=0.5844 |
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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) | (0.5500+0.3100+0.5*1000)/1600=0.4875 |
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HS (expected) | (0.5500+0.455100+0.455*1000)/1600=0.4691 |
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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

- 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).

- 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),

- No excess or deficiency of heterozygotes over the total population (FIT is nearly zero).

文章作者 zzx

上次更新 2020-11-02