BIO 202 — Lesson 12: Pushing the allele frequency with selection

Pine trees — do they want their seeds to be eaten? No. The trees that made pinecone seeds that were easy to eat — what happened to them? They died. The trees that survived are the ones that had pinecone seeds that weren't easy to eat. That's more or less all selection is doing — the ones that aren't good enough to make more of themselves don't make more of themselves.— 202_lec16_03

A — A steady thumb on the scale

Scenario

Allele A has fitness 1 + s relative to a; AA / Aa / aa get genotype fitnesses (1+s)², (1+s), 1 in the haploid approximation. The frequency p evolves: p' = p(1+s)/(p(1+s)+(1−p)). Plot trajectory from p₀.

Frequency trajectory under selection

s: 0.05 | p₀: 0.01 | time to p=0.5: — gen | time to p=0.99: — gen

Prediction

Q1. Halving s (from 0.05 to 0.025) changes the time to fixation by approximately:
Half Double 4× Unchanged — sweeps are mostly about p₀

Try at least 4 (s, p₀) combos. 0/4 combos

Controls

s0.050

p₀0.010

R code — deterministic selection

s <- 0.05p <- 0.01gens <- 500; traj <- numeric(gens+1); traj[1] <- pfor (g in 1:gens) { p <- p*(1+s)/(p*(1+s)+(1-p)); traj[g+1] <- p }

B — Selection + drift

Scenario

50 replicate populations of N = 100. Same s = 0.05. Same p₀ = 0.05. Watch the spread. Some fix the allele; some lose it. Probability of fixation for a single new beneficial allele is ≈ 2s (Haldane).

50 replicate trajectories

N: 100 | s: 0.05 | p₀: 0.05 | % replicates fixed: — | Haldane prediction (2s): 10%

Prediction

Q1. A new beneficial allele (1 copy, p₀ = 1/(2N)) with s = 0.05 in a population of N = 100 will fix about:
10% of the time 50% of the time 100% of the time (it's beneficial) Essentially 0% (it's just one copy)

Try at least 5 (N, s, p₀) combos. 0/5 combos

Controls

N100

s0.050

p₀0.050

seed42

R code — selection + drift

set.seed(42)N <- 100; s <- 0.05; p0 <- 0.05; reps <- 50trajs <- replicate(reps, {  p <- p0; traj <- p  for (g in 1:300) {    p_sel <- p*(1+s)/(p*(1+s)+(1-p))    p <- rbinom(1,2*N,p_sel)/(2*N); traj <- c(traj, p)  }; traj})

C — Work backwards from one observed change

Scenario

Simulate a sweep at a known push strength. Watch the frequency at the start and at time T, then work backward to the push. How tight your answer comes out depends on the population size, how long you watched, and how hard the push was.

Total miss across candidate push strengths

true s: 0.05 | best-fit s: — | 95% range: —

Prediction

Q1. With N = 200, T = 50 generations, and true s = 0.02 (weak), the fit will:
Pin s to within ±0.005 Come out as a wide range Fit s = 0 (drift indistinguishable from weak selection)

Try at least 3 (s, N) combos. 0/3 combos

Controls

true s0.050

N200

T (gens observed)50

seed42

R code — fit s

set.seed(42)N <- 200; s_true <- 0.05; T <- 50; p0 <- 0.05p <- p0; traj <- pfor (g in 1:T) { p_sel <- p*(1+s_true)/(p*(1+s_true)+(1-p)); p <- rbinom(1,2*N,p_sel)/(2*N); traj <- c(traj, p) }s_grid <- seq(-0.05, 0.2, 0.001)ssr <- sapply(s_grid, function(s) { p2 <- p0; pp <- p2; for (g in 1:T) { pp <- pp*(1+s)/(pp*(1+s)+(1-pp)); p2 <- c(p2, pp) }; sum((traj - p2)^2) })s_grid[which.min(ssr)]

D — Lenski's long-term E. coli experiment

Scenario

LTEE allele frequencies plotted over generations. Each rising trajectory is a beneficial mutation. Fit s per trajectory. Watch for clonal interference — alleles that rise then crash as a fitter mutation appears in the population.

LTEE allele frequencies

trajectories shown: — | median fit s: — | range: —

Prediction

Q1. Across LTEE beneficial mutations that successfully fix, the typical selection coefficient is around:
0.001 — very weak 0.01 — small but measurable 0.05 — moderate 0.5 — large

Inspect at least 2 trajectories. 0/2 inspections

Controls

subset seed42

R code — fit s per LTEE trajectory

ltee <- read.csv("data/clean/ltee_allele_freqs.csv")# For each mutation, fit s by least squares to logisticmuts <- split(ltee, ltee$mutation_id)fits <- sapply(muts, function(m) {  t <- m$generation - min(m$generation); pobs <- m$freq; p0 <- pobs[1]  s_grid <- seq(0, 0.3, 0.005)  ssr <- sapply(s_grid, function(s) sum((pobs - 1/(1 + (1/p0 - 1)*exp(-s*t)))^2))  s_grid[which.min(ssr)]})

E — Drift, selection, or both? Five trajectories, five labels.

Scenario

Five trajectories, generated under known (N, s, p₀) combinations. Some are drift-only (s = 0). Some are selection-only-feeling (s much larger than 1/N). Some are the boundary case (s ≈ 1/N) where you can't tell from a single trace.

For each trajectory, pick the best label. The next round won't unlock until you commit.

Round 1 of 5

progress: 0 / 5 | correct so far: 0

Classify this trajectory

Drift only (no selection — s ≈ 0) Selection dominates (s ≫ 1/N — the trajectory is mostly deterministic) Both — selection nudges, drift jitters (s comparable to 1/N)

Finish all 5 rounds. 0/5 rounds