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Two subpopulations of a beetle species each contain 2000 individuals. In subpopulation X the frequency of allele a is 0.2, and in subpopulation Y it is 0.7. Calculate the fixation index F_ST using the formula F_ST = (H_T - H_S)/H_T, where H_T = 2p̄q̄ is the total expected heterozygosity (with p̄ the mean allele frequency across subpopulations) and H_S is the average heterozygosity within subpopulations (average of 2p_i q_i for each subpopulation).

hard 3 marks Class 12 · Biology · Evolution
Answer
0.25

The correct answer is 0.25 — Round to two decimal places: F_ST 0.25.

Step-by-step solution

  1. Find the mean allele frequency p̄ = (0.2 + 0.7)/2 = 0.45.
  2. Compute q̄ = 1 - p̄ = 0.55.
  3. Calculate total heterozygosity H_T = 2 0.45 0.55 = 0.495.
  4. For subpopulation X, heterozygosity H_X = 2 0.2 0.8 = 0.32.
  5. For subpopulation Y, heterozygosity H_Y = 2 0.7 0.3 = 0.42.
  6. Average within‑population heterozygosity H_S = (0.32 + 0.42)/2 = 0.37.
  7. Apply the formula: F_ST = (0.495 - 0.37) / 0.495 = 0.125 / 0.495 0.2525.
  8. Round to two decimal places: F_ST 0.25.
Common mistake: Using allele frequencies instead of heterozygosities in the numerator or forgetting to average H_S.

Concept tested: Genetic differentiation (F_ST)

Published · How this solution was prepared and checked

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