evolution - Fisher's Fundamental Theorem of Natural Selection

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.

Answer

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.

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

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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

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

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Reference