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Scaffold S10 — F and the heterozygote deficit

Five rounds. You are shown the observed genotype counts (AA, Aa, aa) for a single locus. Compute the allele frequency p, the observed heterozygote fraction Hobs, the HWE expectation Hexp, and finally F = 1 − Hobs/Hexp. The reveal shows each intermediate value and names the biological cause the observed F is consistent with.

Locked — answer the pretest above first.

Running tally — F across rounds

What you just did has a name

Five rounds of the same arithmetic: take counts, compute p, compute the HWE expectation for heterozygotes, compute the observed heterozygote fraction, and report F = 1 − Hobs/Hexp. The number is easy. The interpretation is not — F has the same magnitude for several very different biological scenarios.

Round 3 (sibling mating colony, F ≈ 0.25) was real inbreeding — shared ancestry reducing heterozygosity. Round 4 (Wahlund, F = 1) was not inbreeding at all — it was population substructure: two subpopulations each at fixation for opposite alleles, sampled together. Both produce hugely positive F. You cannot distinguish them from F alone; you need independent evidence (pedigree, subsampled counts, geographic information).

Round 5 showed F < 0 — excess heterozygotes. This is the signature of outbreeding, hybrid vigor, or balancing selection (heterozygote advantage). The general lesson: F detects deviation from panmixia; it does not identify the cause. Pattern ≠ process.