Lesson 7 — Tracing how much of a parent ends up in their child

The same line-fitting you have done since Lesson 3, now with a parent on one axis and their child on the other.

A — Two independent traits, no relationship

Simulate a population where two traits are generated independently. Read three summary numbers; reseed and read them again.

Locked — confirm your name above to begin.

Scenario

Two simulated traits, drawn independently from Normal(0, 1). n individuals. Read the three numbers that each measure how two things move together, and watch them jump around as you reseed.

Scatter of two independent traits

n: 200  |  cov(x, y): —  |  link: —  |  Spearman ρ: —

Prediction (required before sliders unlock)

  1. Q1. You simulate two traits that are truly independent (no real connection) and measure 200 individuals. What do you predict the measured link will be?
  2. Q2. You keep simulating the same two independent traits but increase n from 200 → 2,000 → 20,000. What happens to the typical sample r?
Try at least 5 seed/n combinations to unlock Stage B. 0/5 combos

Controls

200
42

R code — independent traits, three summaries

set.seed(42)n <- 200x <- rnorm(n); y <- rnorm(n)cov(x, y)cor(x, y)                # Pearsoncor(x, y, method = "spearman")   # Spearman rank correlation

B — Let the child inherit part of the parent

Now each child's trait is built partly from its parents'. One slider sets how much carries over. Move it.

Complete Stage A (submit prediction, try 5 combos) to unlock this section.

Scenario

Simulated pairs: each child and the average of its two parents. Move the slider — how much of a parent's departure from the crowd's average is handed to the child — and watch the cloud tilt.

At one end of the slider, the child is built entirely from the parents. At the other, the parents tell you nothing.

Parents' average vs. child

carry-over you set: 0.50  |  move-together: —  |  tightness: —

Prediction (required before sliders unlock)

  1. Q1. As h² moves from 0 to 1, how much the parents' average and the child move together will:
  2. Q2. At h² = 1, the cloud of (midparent, offspring) points looks like:
Try at least 6 carry-over settings to unlock Stage C. 0/6 settings

Controls

0.50
300
42

R code — couple two traits via h²

set.seed(42)n <- 300h2 <- 0.50mu <- 68; sigma <- 3father <- rnorm(n, mu, sigma); mother <- rnorm(n, mu, sigma)midparent <- (father + mother) / 2offspring <- mu + h2 * (midparent - mu) +             rnorm(n, 0, sigma * sqrt(1 - h2))cov(midparent, offspring); cor(midparent, offspring)

C — Fit offspring on midparent

Same simulator. Fit lm(offspring ~ midparent). Compare β̂ to the true h² slider.

Complete Stage B (submit prediction, try 6 h² values) to unlock this section.

Scenario

Same simulator as Stage B. Now we explicitly fit lm(offspring ~ midparent) and display the slope β̂ next to the true h² slider.

Drag h². Watch β̂.

The predictor is the midparent average of both parents, so this slope reads h² directly. A regression on one parent alone would give ½h² (you'd double it to recover h²) — keep track of which slope you're reading.

Midparent × offspring with fitted line

true h²: 0.50  |  fitted slope β̂: —  |  |β̂ − h²|: —
R²: —  |  SE(β̂): —

Prediction (required before slider unlocks)

  1. Q1. β̂ is the slope of the regression of offspring on midparent. Pick the description that captures what β̂ does for you:
  2. Q2. You set the true h² yourself with the slider. Whenever h² is below 1, a child whose parents are well above average is predicted to land:
  3. Q3. (Reflection — not scored.) Suppose β̂ came out as 1.0 in some other population. In one sentence, what would that say about the heritability of height in that population? Write your answer below.
Drag h² through at least 8 distinct values to unlock Stage D. 0/8 values

Controls

0.50
500
42

R code — read h² off the regression slope

set.seed(42)n <- 500h2 <- 0.50mu <- 68; sigma <- 3f <- rnorm(n, mu, sigma); m <- rnorm(n, mu, sigma)midparent <- (f + m) / 2offspring <- mu + h2 * (midparent - mu) + rnorm(n, 0, sigma * sqrt(1 - h2))fit <- lm(offspring ~ midparent)coef(fit)[2]               # the slope IS h-hat-squaredsummary(fit)$coefficients[2, 2]   # SE on h²-hat

D — Galton's 1885 data: 934 children, 197 families

Real father–mother–child records. Same regression you just built. Read the slope, then bootstrap it.

Complete Stage C (submit prediction, try 8 h² values) to unlock this section.

Scenario

934 child-records, 197 Victorian-era families. Midparent height = (father + 1.08·mother) / 2 (Galton's sex correction).

Fit lm(childHeight ~ midparentHeight). Bootstrap it. The slope is what you came here for.

Galton family data: midparent height vs child height

N children: —  |  families: —  |  slope (= ĥ²): —
intercept: —  |  R²: —  |  95% bootstrap on ĥ²: —

Prediction (required before bootstrap unlocks)

  1. Q1. The slope of childHeight on midparentHeight in Galton's data will be:
  2. Q2. Split the data by child sex. The two sex-specific slopes will:
Run the bootstrap and split-by-sex toggle to wrap up. 0/2 actions

Controls

200
42

R code — Galton in 5 lines

g <- read.csv("data/clean/galton_families.csv")g$midparent <- (g$father + 1.08 * g$mother) / 2fit <- lm(childHeight ~ midparent, data = g)coef(fit)[2]   # the slope IS h-hat-squaredB <- 200replicate(B, {  k <- sample(nrow(g), nrow(g), replace = TRUE)  coef(lm(childHeight ~ midparent, data = g[k, ]))[2]}) |> quantile(c(.025, .975))

Stretch challenge (optional)

Galton's mother coefficient of 1.08 is rough. Refit using an un-corrected mother coefficient of 1.0 and report how the slope changes. Then run a per-sex fit: male offspring on midparent (no correction), then female offspring on midparent. Which approach gives the cleanest ĥ²? Hit "I tried it" after you have an answer.

Not yet attempted.