Lesson 13 — Where deleterious alleles get held in place

A harmful variant keeps getting weeded out, yet never quite disappears. Where does it settle, and why there?

The frequency of an allele in the population goes back to what frequencies always are: how often you get it yourself, plus how often you inherit it. The two ways to get it. The frequency of a bad mutation works out to μ/(H·S) — the mutation rate divided by how bad it is and how heritable that badness is.— 202_lec17_02

A — Mutation alone — q drifts up at rate μ

Without selection, a new mutation arises at rate μ per generation. In a finite population, it usually goes extinct from drift; rarely it sweeps to fixation.

Locked — confirm your name above to begin.

Scenario

Start with q = 0. Each generation, each non-mutant allele can mutate to "a" with rate μ. Plot q over time. Without selection, this is a Wright-Fisher process plus an injection rate of μ per allele per generation.

q over time (mutation only)

μ: 1e-4  |  N: 1000  |  mean q at t=2000: —

Prediction

  1. Q1. With mutation rate μ = 10⁻⁴ and no selection in N = 1000, after 2000 generations the equilibrium mean q across replicates will be:
Try at least 4 (μ, N) combos. 0/4 combos

Controls

1e-4
1000
42

R code — mutation only

set.seed(42)mu <- 1e-4N <- 1000q <- 0; traj <- qfor (g in 1:2000) {  q_mut <- (1-q)*mu + q*(1-mu)   # symmetric for simplicity  q <- rbinom(1,2*N,q_mut)/(2*N)  traj <- c(traj, q)}

B — Mutation + selection — equilibrium at μ/(hs)

Add selection: aa has fitness 1 − s; Aa has fitness 1 − hs (h is dominance). The deleterious allele settles at q ≈ √(μ/s) for recessive lethals, q ≈ μ/(hs) for dominant.

Complete Stage A.

Scenario

q stabilizes when mutation influx (μ × frequency of A) equals selection efflux (≈ hs × 2pq for partially dominant, ≈ s × q² for recessive lethal). For partially dominant (h > 0): q ≈ μ/(hs). For recessive (h = 0): q ≈ √(μ/s).

q trajectory and theoretical equilibrium

μ: 1e-5  |  s: 0.1  |  h: 0.5  |  theory q̂: —  |  observed q (gen 5000): —

Prediction

  1. Q1. For h = 0.5, μ = 10⁻⁵, s = 0.1, the equilibrium frequency q̂ will be approximately:
Try at least 4 (μ, s, h) combos. 0/4 combos

Controls

1e-5
0.100
0.50

R code — the standing balance level

mu <- 1e-5; s <- 0.1; h <- 0.5q <- 0; gens <- 5000; traj <- numeric(gens+1)for (g in 1:gens) {  p <- 1-q; w_AA <- 1; w_Aa <- 1-h*s; w_aa <- 1-s  wbar <- p^2*w_AA + 2*p*q*w_Aa + q^2*w_aa  q <- (p*q*w_Aa + q^2*w_aa) / wbar  q <- q*(1-mu) + (1-q)*mu  traj[g+1] <- q}tail(traj, 1)

C — Estimate s from observed q

If you can measure q (population frequency of the allele), μ (per-generation rate), and h (dominance), you can solve for s: s ≈ μ/(h·q). The pink-katydid back-calculation.

Complete Stage B.

Scenario

Pink katydid example. Population frequency q ≈ 5 × 10⁻⁴. Mutation rate μ ≈ 5 × 10⁻⁵. Dominance h ≈ 1 (visible heterozygote). Solve: s ≈ μ/(h·q) ≈ 0.1. The selection coefficient is 10% — a huge effect.

Inferred s from (μ, h, q)

μ: 5e-5  |  h: 1.0  |  observed q: 5e-4  |  inferred s: —

Prediction

  1. Q1. The pink katydid sits at q ≈ 5 × 10⁻⁴ with μ ≈ 5 × 10⁻⁵. The inferred selection coefficient is closest to:
Try at least 3 (q, μ) combos. 0/3 combos

Controls

5e-5
1.00
5e-4

R code — back-calculate s

mu <- 5e-5; h <- 1; q <- 5e-4# Mutation-selection balance: q = mu/(h*s) for h > 0s_inferred <- mu / (h * q)s_inferred

D — Cystic fibrosis — when the formula breaks

CF allele frequency in Northern European populations is ~0.022. If h = 0 and s = 1 (recessive lethal), μ/(hs) doesn't apply; the recessive formula gives q ≈ √(μ/s). With μ ≈ 10⁻⁶ and s ≈ 1, q ≈ 10⁻³. Observed is 22× higher. Something else.

Complete Stage C.

Scenario

Test three models against q_observed = 0.022: (1) recessive-lethal standing balance q = √(μ/s); (2) partially dominant model q = μ/(hs); (3) heterozygote advantage at AA fitness 1−sₐ, Aa fitness 1, aa fitness 1−s_aa. Only the third reaches the observed q.

Predicted q under three models

observed q: 0.022  |  recessive q (√(μ/s)): —  |  partial dominance q: —  |  het advantage q: —

Prediction

The basic making-vs-removing balance model assumes two causal arrows: μ → q and s_aa → q. It predicts q ≈ 0.001. The observed q is 22× higher. Which additional arrows in the causal model could be doing the work? Check every plausible one.

Q1. What's keeping CF at q ≈ 0.022 above the simple making-vs-removing balance? (Check any that you think contribute.)
Adjust the het-advantage slider and find a configuration that matches 0.022. 0/2 attempts

Controls

1e-6
1.00
0.020

R code — three CF models

# Recessive lethal balancemu <- 1e-6; s_aa <- 1; q_rec <- sqrt(mu/s_aa)# Partial dominance balanceh <- 0.1; q_pd <- mu/(h*s_aa)# Heterozygote advantage equilibrium: q_eq = s_AA / (s_AA + s_aa)s_AA <- 0.02; q_ha <- s_AA/(s_AA + s_aa)c(rec = q_rec, partial = q_pd, hetadv = q_ha, observed = 0.022)