6
$\begingroup$

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!

$\endgroup$
  • 5
    $\begingroup$ Man, your questions always scare me :). $\endgroup$ – terdon Feb 26 '14 at 18:44
6
$\begingroup$

Well, I think I found the very simple mistake I made…

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 $\frac{h_N^2}{2} = slope$

Here was my mistake!

$\endgroup$
  • $\begingroup$ I thought the range of heritability is between 0 and 1. But values of parent-offspring regression times 2 calculated this way can certainly be higher than 1. Am I wrong about heritability? $\endgroup$ – sterid May 9 '17 at 6:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.