The univariate breeders equation is defined as,

$\ R = h^2 s$

where $\ R $ is the response, $\ h^2 $ is the heritability (additive genetic variation), $\ s $ is the selection differential. The multivariate equation is similar to this in principle but includes a multiple trait variance-covariance matrix and multiple selection differentials.

Essentially the breeders equation tells us how strong the response to selection will be as a result of the additive genetic variance within a trait (and between traits in the multivariate) and the selection applied to that variation ($\ s = cov (w,x) $ = the covariance between fitness ($w$) and the trait $x$).

For example, in trait $x$, $s$ = 0.8 and $h^2$ = 0.5 the value of the response for $x$ is $R$ = 0.4. Compared to another trait, $y$, where $s$ = 0.8 and $h^2$ = 0.1 and thus $R$ = 0.08.

My question is, if we look at a trait in a population, measure it's population mean and define $h^2$ and $s$ values, can we then predict the change in the population? IE. can we predict (with the values of $s, h^2$ and the population mean) what the mean population mean trait will be in the next generation directly from the result of the breeders equation? Does the value of $R$ give us anything of use?

Following from the example above, I examine a trait (wing length in chickens) and find the mean to be 24 cm. I select birds to start my next generation in a way that causes $s$ = 0.8. The additive genetic variance for the wing length is quite high, $h^2$ = 0.5. The predicted response is $R$ = 0.8 x 0.5 = 0.4. What does this tell me about the next generation?


2 Answers 2


The breeder's equation as you wrote it:

$$R = h^2S$$

The heritability that is the ratio of additive genetic variance over the total phenotypic variance is called the heritability in the narrow sense and is noted $h_N^2 = \frac{V_a}{V_P}$, where $V_A$ and $V_P$ are the additive genetic and phenotypic variance respectively. In contrast the heritability that consider the total genetic variance is called the heritability in the broad sense and is noted $h_B^2 = \frac{V_G}{V_P} = \frac{V_A + V_D}{V_P}$, where $V_G$ is the total genetic variance and is equal the sum of the additive genetic variance $V_A$ and the genetic dominance variance $V_D$. We can rewrite the breeder's equation using the standard notation for the heritability in the narrow sense:

$$R = h_N^2S$$

The so-called response to selection $R$ is the mean deviation of the trait in the next generation. In other words, $R$ is the difference between the mean trait of the parents and the mean trait of the offspring, So to answer your question, yes, the breeder's equation gives the value of the mean trait in the following generation. It does not give any information about the frequency and number of alleles though.

$S$ is the the mean trait of the individuals which reproduce weighted by their reproductive success minus the mean trait of the parents. In other words, it is

$$S=\frac{1}{\bar w}\sum_{i=1}^{\text{nb inds}}m_iw_i - \bar m$$

, where $m_i$ is the trait measurement and $w_i$ is the fitness of this individual, $\bar m$ is the mean trait of the parents and $\bar w$ is the mean fitness.

The mean trait of the parent generation does not appear in the breeder's equation but only their mean trait weighted by their reproductive success. In the standard case of truncated selection (common in artificial selection programs) all $w_i$ are equal and $S$ becomes the mean trait of the individuals that we allow to reproduce in the parent population. And it is exactly the same thing for the multivariate case except that the variance-covariance matrix should be taken into account.

  • $\begingroup$ @GriffinEvo Let me know if I answered your question or if I missed your point. $\endgroup$
    – Remi.b
    Commented Apr 23, 2014 at 16:58

Following on from @Remi.b and @Ell :

z score = (x - μ)/ σ from https://ncalculators.com/statistics/z-score-calculator.htm

Z score (predicted response in standard deviations) = X - adult population mean / standard deviation of normal distribution

0.4 = (X) - 24 cm/ (1) solve for X X= 24.4 cm, which would be the expected value of the mean trait in the following generation.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .