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Ronald Fisher discovered what he, with humility, called the Fundamental Theorem of Natural Selection. This theorem says (in its modern terminology):

The rate of increase in the mean fitness of any organism at any time ascribable to natural selection acting through changes in gene frequencies is exactly equal to its genetic variance in fitness at that time.

As I understand it, it sounds alike the standard equation that we learn in the first class of Introduction to evolutionary biology

$$R = S \cdot \frac{V_G}{V_p}$$

In words: the response to selection equals the selection differential times the genetic variance of the trait under consideration divided by the total phenotypic variance of the trait under consideration

But how can we prove/demonstrate that Fisher's fundamental theorem of natural selection holds true?

I don't ask for empirical evidences that support this claim but for a theoretical/mathematical proof/demonstration of this claim.

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3 Answers 3

I don't think that you need to look further than the Price equation, which is basically a proof of a generalized version of Fisher's fundamental theorem. Price had a series of papers in the 70's that derived and applied the Price equation (e.g. Price, 1970; Price, 1972a), but most relevant for your question in probably Price (1972b).

A nice summary of Price's legacy can be found in Frank (1995).

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There's a Frank paper from JEB (2007?) on it too which is good - we have an ongoing discussion group of the price equation here at the moment. (Any good suggestions of papers with quantitative genetic applications of price equation?) –  GriffinEvo Feb 11 at 20:51
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Good to know. I've seen this one as well, but haven't read it: Frank. 2012. Natural selection. IV. The Price equation –  fileunderwater Feb 11 at 21:36
    
That's the one, why did I think it was 2007?? I only read it last week! –  GriffinEvo Feb 12 at 8:18

Here's a paper with an historical bibliography of mathematical analyses in the introduction.

As you can see when you demand a mathematical treatment of something as poorly understood as the genetic inheritance of traits, the word proof is to be used in a qualified way. In this case a multilocus fitness with variation in the population with no linkage between the alleles. Its probably a case that shows up frequently enough, but how many alleles are we talking about? With large numbers of alleles which are scrambled randomly at each generation (zero linkage) I would imagine that the number of individuals that contain a lot of diversity.

Most GWAS studies show lots of less probable variants may contribute to many genetic traits (like height, diabetes, etc). So this is quite a reasonable number of the cases.

The fundamental theorem is true for any number of variants involved though, so hopefully this helps.

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up vote 1 down vote accepted

The following answer is not complete and only give some intuitive grasp on Fisher's fundamental theorem of Natural Selection. A better devlopment can be found in Ewen's book

Let's first define what is the Additive Genetic Variance

Consider a quantitative character that is determined entirely by a locus $A$ which two alleles are $A_1$ and $A_2$. the measurement $m$ of this quantitative character for an individual is given by their genotypes so that the genotypes $A_1A_1$, $A_1A_2$ and $A_2A_2$ have measurements $m_{11}$, $m_{12}$ and $m_{22}$ respectively. Suppose that random mating obtains with respect to this character and that the frequencies of $A_1A_1$, $A_1A_2$ and $A_2A_2$ are $x^2$, $2x(1-x)$ and $(1-x)^2$, respectively. Then, the mean value $\bar m$ of this measurement is

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

and the variance in the measurement is

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

The covariance between the fathers and the sons (assuming no change in allele frequency) is

$$x(1-x)\left(xm_{11} + (1-2x)m_{12} - (1-x)m_{22}\right)^2$$

The correlation between the fathers and the sons is found by dividing the covariance by the variance (since the variance of fathers equal the variance of sons) is

$$\frac{x(1-x)\left(xm_{11} + (1-2x)m_{12} - (1-x)m_{22}\right)^2}{\sigma ^2}$$

which can be decomposed into dominance and additive variance

$$\sigma _A ^2 = 2x(1-x) (xm_{11} + (1-2x)m_{12} - (1-x)m_{22})^2$$ $$\sigma _D ^2 = x^2(1-x)^2 (2xm_{12} - m_{11} - m_{22})^2$$

Replacing the measurement by the fitness, you get the additive genetic variance in fitness.

$$\sigma _A ^2 = 2x(1-x) (xw_{11} + (1-2x)w_{12} - (1-x)w_{22})^2$$ $$\sigma _D ^2 = x^2(1-x)^2 (2xw_{12} - w_{11} - w_{22})^2$$


Now, the mean fitness in the population is given by (as already said)

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

Using Wright-Fisher equation, the change in mean fitness is

$$\Delta \bar w = 2x(1-x) (xw_{11} + (1-2x)w_{12} - (1-x)w_{22})^2 \cdot (w_{11}x^2 + (w_{12} + \frac{w_{11} + w_{22}}{2})x(1-x) + w_{22}(1-x)^2)\bar w ^{-2}$$

which can be approximated

$$\Delta \bar w ≈ 2x(1-x) (xw_{11} + (1-2x)w_{12} - (1-x)w_{22})^2 = \sigma _A ^2$$


Reference

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