# Why does the slope of parent-offspring regression equals the heritability in the narrow sense?

Background

---- Notations and assumptions ----

let $W_{ij}$ be the fitness associated to the genotype $AiAj$. $x$ is the frequency of the allele $A1$ in the population. The frequency of the allele $A2$ is $1-x$ because we'll consider only the case of a bi-allelic locus. Also, we consider that only one locus influence the random variable $W$ (the fitness). We will assume that the environment does not influence the fitness. We also assume random mating in order to use hardy-Weinberg equilibrium.

---- calculations ----

The mean fitness is:

$$\bar W = x^2W_{11} + 2x(1-x)W_{12} + (1-x)^2W_{22}$$

The variance of fitness is:

$$\sigma^2 = x^2(W_{11}-\bar W)^2 + 2x(1-x)(W_{12}-\bar W)^2 + (1-x)^2(W_{22}-\bar W)^2$$

We then want to calculate the covariance between father and son with respect to fitness.

Suppose first that the father is $A_1A_1$ Then the son will be $A_1A_1$ if the mother transmits an $A_1$ gene to him, an event with probability x. Similarly the son will be $A_1A_2$ with probability $1-x$. The father himself will be $A_1A_1$ with probability $x^2$. Continuing this wa it is possible to draw up a table of the probabilities of the various father-son combinations in genotype and hence fitness.

Therefore, assuming no change in frequency of $A1$ between the two generations, the covariance father-son is:

$$cov = x^3W_{11}^2 + 2x^2(1-x)W_{11}W_{12} + x(1-x)W_{12}^2 + 2x(1-x)^2W_{12}W_{22} + (1-x)^3W_{22}^2-\bar W^2$$

which also equals

$$cov = x(1-x)(xW_{11} + (1-2x)W_{12} - (1-x)W_{22})^2$$

The correlation between the two fitnesses is found by dividing the co-variance by the variance (since the variance of sons is the same as that for fathers):

$$cor = \frac{x(1-x)(xW_{11} + (1-2x)W_{12} - (1-x)W_{22})^2}{\sigma ^2}$$

We then define dominance and additive variance:

$$\sigma_D^2 = x^2(1-x)^2(2*W_{12} - W_{11} - W_{22})^2$$

$$\sigma_A^2 = 2x(1-x)(xW_{11}+(1-2x)W_{12} - (1-x)W_{22})^2$$

such as

$$\sigma_D^2 + \sigma_A^2 = \sigma$$

My question

The slope of a parent-offspring regression equals the heritability in the narrow sense $h^2_N$.

Heritability in the narrow sense is defined as $h_N^2 = \frac{\sigma_A^2}{\sigma}$ (see Why is a heritability coefficient not an index of how “genetic” something is?).

We defined above that the correlation between parents and offspring is $cor = \frac{cov}{\sigma}$. So I'd expect the slope of the parent-offsprings regression to equal $cor = \frac{cov}{\sigma}$ and not $h_N^2 = \frac{\sigma_A^2}{\sigma}$

What am I missing? Is there a mistake in my background calculations? Is there a mistake in the way I defined some concepts? The answer might be much simpler than it looks like!

• Man, your questions always scare me :). – terdon Feb 26 '14 at 18:44

Looking again in my equations, I realize that (for some reason) $cor = 2 \cdot \frac{\sigma_A^2}{\sigma}$
And looking at this website, I see that the slope of the parent-offspring regression is $\frac{h_N^2}{2} = slope$