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Scaffold S11 — FST and migration

Five rounds using Wright's two-population approximation: FST ≈ 1 / (1 + 4 Ne m). Three rounds go forward — you are given Ne and m and predict FST. Two rounds go inverse — you are given a measured FST from the Atlantic-cod example and predict Ne·m, the number of migrants per generation.

Locked — answer the pretest above first.

Running tally — FST vs. Ne·m curve

What you just did has a name

Look at the tally curve above. Rounds 1–3 were forward — you guessed FST given Ne and m. Rounds 2 and 3 had identical FST (~0.71) despite wildly different Ne and m. That is because the product Ne·m was the same (≈ 0.1).

The one quantity that matters is Ne·m — how many migrants actually arrive per generation. Above ~10 migrants per generation (FST below 0.025), populations behave as one. Below ~1 migrant per generation (FST above 0.2), they drift apart. The Atlantic-cod rounds (4 and 5) — same species, same coastline, one inner fjord — span both sides of that boundary. Topography alone controls the drift–migration balance.

A small FST does not mean small populations or weak drift. It means the product Ne·m is large. Trade off Ne for m, or vice versa, and FST is unchanged. This is why the Risør and Søndeled cod differ 20-fold in FST — not because they are different species, but because the fjord geometry reduces incoming migrants.