20 Heritability from a linear regression

20.1 Background

Heritability measures how much of the variation in a trait within a population is due to additive genetic differences between individuals. It is a proportion, between 0 and 1, and it tells us how closely offspring resemble their parents. Heritability matters in plant and animal breeding, in medicine and in conservation, because it determines how fast a trait can respond to selection.

Narrow-sense heritability (Neal writes it \(h^2_N\)) is a ratio of two variances: the additive genetic variance divided by the total phenotypic variance.

\(h^2 = \frac{V_A}{V_P}\)

Only the additive part of the genetic variance counts, because parents pass on alleles, not whole genotypes. Effects that depend on particular combinations of alleles (dominance and interactions between genes) are broken up every generation, so they do not make offspring resemble their parents in a predictable way.

The rest of the phenotypic variance is mainly environmental. If we ignore non-additive genetic effects, as we do in this practical, \(V_P = V_A + V_E\), where \(V_E\) is the environmental variance.

A heritability close to 1 means that most of the variation among individuals is additive genetic, so offspring closely resemble their parents. A heritability close to 0 means that parents’ trait values tell us little about their offspring’s: the variation is mostly due to the environment, to chance, or to non-additive genetic effects.

Heritability is not a fixed property of a trait. It belongs to a particular population in a particular environment, because it changes whenever either variance changes:

  • More environmental variation (a patchier habitat, more variable weather or food) increases \(V_P\) and so lowers \(h^2\), even if \(V_A\) is unchanged.
  • Selection removes unfavourable alleles and so reduces \(V_A\). A trait under strong, persistent selection can end up with low heritability. (Traits closely related to fitness also tend to have low heritability because their environmental variance is large; see the lecture.)
  • Genetic architecture (how many genes affect the trait, and how large their effects are) differs between traits and species.

We can estimate heritability from the resemblance between parents and offspring. If we plot the offspring value in each family (one offspring, or the mean of several) against the mid-parent value (the mean of the two parents), the slope of the regression line estimates \(h^2\). If you only know one parent, the slope estimates \(h^2/2\), because each parent passes on only half of its alleles. Using the mean of several offspring per family gives a more precise estimate, but one offspring per family (as in this practical) also works. This is the method used in the lecture for East’s tobacco plants and for Darwin’s finches.

Reading: Neal, chapter Polygenic Inheritance and Quantitative Genetics.

Reference

Wray, N. & Visscher, P. (2008) Estimating trait heritability. Nature Education 1(1):29

Learning Outcomes:

  • Estimate narrow-sense heritability, with its uncertainty, from a mid-parent–offspring regression in R.
  • Use heritability and phenotypic variance to estimate additive genetic variance, and explain why heritability can differ between populations.
  • Evaluate the assumptions behind regression estimates of heritability.

20.2 Before you start: predict

Imagine plotting offspring wing length against mid-parent wing length.

  • What slope would you expect if offspring were, on average, exactly like their parents?
  • What slope would you expect if offspring were no more like their own parents than like any other bird in the population?
  • What does a slope between these two tell you?

Write down your answers, then check them against your results at the end.

20.3 Your Task

20.3.1 Estimating heritability (15 minutes)

You are provided with two datasets for wing length (mm) in a small songbird. Each row is one family: Midparent_value is the mean wing length of the two parents, and Offspring_value is the wing length of one of their offspring, measured as an adult. Each dataset contains 100 families.

Data set 1

Data set 2

  1. Download the first data set and import it into R (RStudio). The starter code assumes you have saved it in a folder called data inside your project; change the path if you saved it somewhere else.
  2. Plot the data, with mid-parent value on the x-axis and offspring value on the y-axis.
  3. Fit a linear regression model of the form y ~ x.
  4. Look at the slope of the relationship: this is your estimate of the heritability. Use confint() to find its 95% confidence interval.

Starter code in R:

pop_1 <- read.csv("data/heritability_pop1.csv")
plot(pop_1$Midparent_value, pop_1$Offspring_value,
     xlab = "Mid-parent wing length (mm)", ylab = "Offspring wing length (mm)")
mod1 <- lm(Offspring_value ~ Midparent_value, data = pop_1)
abline(mod1, col = "red")
summary(mod1)
coef(mod1)[2]     # slope = heritability estimate
confint(mod1)[2, ] # 95% confidence interval for the slope

Reading a confidence interval: the 95% confidence interval is the range of values for \(h^2\) that are compatible with your data. The narrower it is, the more precise your estimate. If the interval includes zero, the data cannot rule out \(h^2 = 0\), but that does not show that \(h^2\) is zero: the upper limit tells you how large it could plausibly be.

Questions

  • What is your estimate of \(h^2\), and how precise is it?
  • What does this heritability tell us about the variation in wing length in this population? What does it not tell us about any one bird?
  • What else might explain the remaining variation?
  • Calculate \(V_P\) as the variance of the offspring values (var(pop_1$Offspring_value)). Now calculate var(pop_1$Midparent_value) and compare the two. Why is the variance of the mid-parent values only about half of \(V_P\), and so why should we not use it as \(V_P\)? (Hint: each mid-parent value is the mean of two birds.)
  • Calculate the additive genetic variance, \(V_A = h^2 \times V_P\).
  • Imagine a breeder selects parents whose mean wing length is 2 mm above the population mean. Use the breeder’s equation, \(R = h^2 S\), to predict the change in mean wing length in the next generation.

20.3.2 Comparative Analysis (10 minutes)

Now analyse the second data set. It is for the same trait in the same species, but from a different population living in a different environment. Repeat steps 1 to 4, and calculate \(V_P\) and \(V_A\) for this population too. Copy this table into your notes and fill it in for both populations.

Population 1 Population 2
\(h^2\)
95% confidence interval
\(V_P\) (mm²)
\(V_A\) (mm²)
\(V_P - V_A\) (mm²), an estimate of \(V_E\)

Questions

  • How does the heritability compare with population 1? Do the two confidence intervals overlap?
  • Use your table to compare \(V_A\) and \(V_P - V_A\) between the two populations. Which differs more, and what does that suggest about the cause of the difference in heritability?
  • Can you suggest environmental differences between the two populations that could explain this?
  • For the same selection differential (\(S\) = 2 mm), which population would respond faster to selection?
  • Can you think of any real-world applications where understanding heritability would be important?

20.3.3 Assumptions (10 minutes)

Estimating heritability from a mid-parent–offspring regression assumes the following.

  • Additive genetic effects: the slope estimates the additive genetic variance only. Dominance and interactions between genes do not make offspring resemble parents, and the method does not account for genotype-by-environment interactions.
  • No shared environment: parents and offspring do not resemble each other for non-genetic reasons. If they share a territory, food supply or parental care, the heritability will be overestimated. Cross-fostering experiments (swapping eggs or chicks between nests) break this link.
  • Correct parentage: the measured parents are the genetic parents. In many birds, some chicks are sired by a male other than the social father (extra-pair paternity), which makes the heritability look lower than it is.
  • Linearity: the relationship between parent and offspring trait values is a straight line.
  • Measurement accuracy: measurement error in the parents’ values makes the slope shallower, so heritability is underestimated; error in the offspring values makes the estimate less precise.
  • Random mating: mates do not choose each other on the basis of the trait (no assortative mating).
  • Representative sample: the families measured are a random sample of the population. If the offspring measured are only those that survived a harsh winter, for example, selection on wing length can bias the estimate.
  • Independence: each family contributes one data point. Using the same parents in several families, or several offspring from one family as separate data points, violates this.

Questions

  • For shared environment, correct parentage and measurement accuracy: give a specific example of how each assumption could be violated in a wild bird population, and explain why that would make the heritability estimate too high or too low.
  • Optional extension: do the same for the other five assumptions. Some have no clear direction, or cause a different problem (for example, a confidence interval that is too narrow), and one may surprise you.

20.4 Takeaways

  • The slope of offspring on mid-parent estimates narrow-sense heritability, \(h^2 = V_A/V_P\); report it with its uncertainty.
  • Heritability belongs to a population in an environment: two populations with the same additive genetic variance can have very different heritabilities if their environments differ.
  • High heritability does not mean a trait is fixed or unaffected by the environment.