Lesson 9 — Building the population where nothing changes

Build a population with the ratios of Lesson 8 locked in, then switch off its rules one at a time and watch the counts slip out of place.

We're analogizing evolution to motion. Newton's first law: an object at rest tends to stay at rest unless acted on by an outside force. Hardy-Weinberg equilibrium is our object at rest. A gene frequency tends to stay at that frequency unless acted on by an outside force. The thing is, "outside force" here does not mean what you want it to mean. When you hear "force," you expect a thing that pushes — a separate thing. But the system not being infinite is itself enough. — 202_lec09_05

A — A population with nothing acting on it

Locked — confirm your name above to begin.

Scenario

One locus, two alleles. Allele A at frequency p. Allele a at frequency q = 1 − p. Under HWE: AA = p², Aa = 2pq, aa = q². Three numbers determined by one slider.

Hardy-Weinberg equilibrium is the fancy phrase we use for saying: this will be our baseline. Our expectation of no change. It is not real. It requires an infinite number of individuals alive at one time, and teleportation. Hardy-Weinberg never exists. But it's our baseline for comparison. — 202_lec08_05

Genotype frequencies vs p

p: 0.50  |  AA: 0.250  |  Aa: 0.500  |  aa: 0.250

Prediction

  1. Q1. At what allele frequency p is the heterozygote (Aa) frequency maximized?
Slide p to at least 4 different values to unlock Stage B. 0/4 values

Controls

0.50

R code — HWE genotype frequencies

p <- 0.50q <- 1 - pc(AA = p^2, Aa = 2*p*q, aa = q^2)

B — Turn the assumptions off

Complete Stage A (submit prediction, slide p ≥ 4 times) to unlock this section.

Scenario

Start at HWE with p = 0.5. Run for 100 generations. Toggle finite N, mutation, selection, non-random mating. Each one peels the population off the equilibrium curve in a characteristic way.

Genotype frequency trajectories

gen: 100  |  final p: —  |  final F: —

Prediction

  1. Q1. Switch each one on by itself and run it out. Three of them shift how common the two versions are. One leaves them exactly where they started — not close, exactly. Which one?
Toggle the four switches in different combinations at least 5 times. 0/5 combos

Controls

100
0.05
0.30
42

R code — HWE violations

set.seed(42)N <- 100; gens <- 100; p <- 0.5mu <- 1e-3; s <- 0.05; F <- 0.30for (g in 1:gens) {  q <- 1 - p  fAA <- p^2 + F*p*q;  fAa <- 2*p*q*(1-F);  faa <- q^2 + F*p*q  w <- c(1, 1, 1 - s)            # aa is selected against  fAA <- fAA*w[1]; fAa <- fAa*w[2]; faa <- faa*w[3]  tot <- fAA + fAa + faa  p <- (fAA + fAa/2) / tot  p <- p*(1-mu) + (1-p)*mu          # symmetric mutation  if (finite) p <- rbinom(1, 2*N, p) / (2*N)}

C — Is the miss bigger than chance?

Complete Stage B to unlock this section.

Scenario

One locus. Sample N individuals. Compute observed AA / Aa / aa counts. The HWE-expected counts come from the observed allele frequency p̂. a single mismatch score for how far the counts sit from what the tidy ratio expects.

Observed vs HWE-expected genotype counts

N: 200  |  p̂: —  |  F̂ (=1 − H_obs/H_exp): —  |  mismatch: —  |  P: —

Prediction

  1. Q1. Leave the shortfall dial fixed, so the group really is short of carriers by the same amount throughout. Now sample fewer and fewer individuals. Your ability to spot that shortfall will:
Try at least 3 (F, N) combinations to unlock Stage D. 0/3 combos

Controls

200
0.50
0.30
42

R code — how far observed counts sit from expectation

set.seed(42)N <- 200; p <- 0.5; F <- 0.30q <- 1 - pprobs <- c(p^2 + F*p*q, 2*p*q*(1-F), q^2 + F*p*q)obs <- rmultinom(1, N, probs)[,1]phat <- (2*obs[1] + obs[2]) / (2*N)exp_HWE <- N * c(phat^2, 2*phat*(1-phat), (1-phat)^2)sum((obs - exp_HWE)^2 / exp_HWE)   # χ² with 1 df

D — A wild locus — four candidate culprits

Complete Stage C to unlock this section.

Scenario

Genotype counts across loci in data/clean/italian_sparrow_loci.csv (or fallback synthetic). For each locus: estimate p̂, compute HWE-expected genotype counts, compute the mismatch. Rank loci by how far they sit from the tidy ratio.

Per-locus mismatch scores

loci: —  |  in HWE (P>0.05): —  |  out of HWE (P≤0.05): —

Prediction

  1. Q1. Across many loci in a wild population, the fraction that fails an HWE test will be:
  2. Q2. For a locus that does fail HWE in this population, which of the four candidate forces are plausibly operating? Check every one you'd want to investigate. (Italian sparrows are a hybrid lineage of house and Spanish sparrows.)
Click a locus to see its per-genotype breakdown at least 2 times. 0/2 inspections

Controls

42

R code — per-locus HWE test

loci <- read.csv("data/clean/italian_sparrow_loci.csv")apply(loci, 1, function(r) {  obs <- c(r["AA"], r["Aa"], r["aa"])  N <- sum(obs)  phat <- (2*obs[1] + obs[2]) / (2*N)  e <- N * c(phat^2, 2*phat*(1-phat), (1-phat)^2)  sum((obs - e)^2 / e)})