Two subpopulations of a marine algae species each contain 1500 individuals. In subpopulation 1 the allele A frequency is 0.6, while in subpopulation 2 it is 0.2. Compute the fixation index $F_{S T}$ between these subpopulations using $F_{S T} = \frac{H _{T} - H _{S}}{H _{T}}$ , where $H_{T} = 2 \overset{p}{ˉ} \overset{q}{ˉ}$ and $H_{S} = \frac{( 2 p _{1} q _{1} + 2 p _{2} q _{2} )}{2}$ .

hard 4 marks Class 12 · Biology · Evolution

Answer

F_{S T} a pp r o x 0.36

The correct answer is $F_{S T} a pp r o x 0.36$ — 8. Final answer: $F_{S T} a pp r o x 0.17$ (rounded to two decimal places).

Step-by-step solution

1. Compute allele frequencies: $p_{1} = 0.6,; q_{1} = 0.4$ ; $p_{2} = 0.2,; q_{2} = 0.8$ .
2. Find mean allele frequency: $\overset{p}{ˉ} = \frac{p _{1} + p _{2}}{2} = \frac{0.6 + 0.2}{2} = 0.4$ ; thus $\overset{q}{ˉ} = 1 - \overset{p}{ˉ} = 0.6$ .
3. Calculate total heterozygosity: $H_{T} = 2 \overset{p}{ˉ} \overset{q}{ˉ} = 2 \times 0.4 \times 0.6 = 0.48$ .
4. Compute heterozygosity within each subpopulation: $H_{1} = 2 p_{1} q_{1} = 2 \times 0.6 \times 0.4 = 0.48$ ; $H_{2} = 2 p_{2} q_{2} = 2 \times 0.2 \times 0.8 = 0.32$ .
5. Find average heterozygosity within subpopulations: $H_{S} = \frac{H _{1} + H _{2}}{2} = \frac{0.48 + 0.32}{2} = 0.40$ .
6. Compute $F_{S T} = \frac{H _{T} - H _{S}}{H _{T}} = \frac{0.48 - 0.40}{0.48} \approx 0.1667$ .
7. Re‑check calculations: Realise the correct $H_{T}$ should use weighted mean if subpop sizes differ; here equal sizes, so $H_{T}$ is correct. Recalculate $H_{S}$ again: $(0.48 + 0.32) /2 = 0.40$ . Thus $F_{S T} a pp r o x 0.1667$ .
8. Final answer: $F_{S T} a pp r o x 0.17$ (rounded to two decimal places).

Common mistake: using unweighted mean of allele frequencies when subpopulations differ in size

Concept tested: fixation index calculation