Muller's ratchet

Following (Gordo & Charlesworth, 2000).

This Wright-Fisher model starts with a haploid asexual population at a mutation-selection balance. The population size is $N$, the mutation rate is $u$ and the selection coefficient is $s$.

Denote the frequency of the best class by $x$ and its initial value $x_0 = e^{-u/s}$. The variance of $x$ due to binomial sampling of $N$ individuals from the previous generations is $$ b(x) = \frac{x(1-x)}{N} \approx \frac{x}{N}, $$

assuming $x \ll 1$. This follows from the variance of a binomial distribution: $Var(Bin(n,p)) = np(1-p)$.

The expected change in $x$ due to mutation and selection is, assuming $\bar{\omega}$ is the current population mean fitness and $\Delta \bar{\omega}$ is the difference between the mutation selection balance (MSB) and the current population mean fitness: $$ a(x) = \frac{x(e^{-u} - \bar{\omega})}{\bar{\omega}} = x \frac{\Delta \bar{\omega}}{\bar{\omega}}, $$ where the MSB mean fitness is $e^{-u}$.

This follows from the standard difference equation of the frequency of type $z$ after one generation: $$ f'(z) = f(z) Pr(no \; mutation) \frac{\omega(z)}{\bar{\omega}} $$

assuming the number of mutations is Poisson distributed with mean $u$ and that the best class has fitness 1.

Next, assume that throughout the process the mean fitness is close enough to the MSB value so that $\Delta \bar{\omega}$ can be modeled as a perturbation. Furthermore, the perturbations are assumed to be a result of fluctuations in the frequency of the best class, $x$.

The mean fitness close to the MSB as a function of $x/x_0$ is expressed as a Taylor expansion around 1: $$ \bar{\omega} (\frac{x}{x0}) \approx \bar{\omega}{eq} + [\frac{\partial \bar{\omega}}{\partial \frac{x}{x0}}]{eq} (\frac{x}{x_0}-1) + O((\frac{x}{x_0}-1)^2). $$

This gives a linear approximation for $\Delta \bar{\omega}$ when it is small, that is, when $x\approx x_0$: $$ \Delta \bar{\omega} \approx K (1-\frac{x}{x_0}), \; K = x0 [\frac{\partial \bar{\omega}}{\partial x}]{eq}. $$

Set $K=0.6 s e^{-u}$ and use this for $a(x)$: $$ a(x) \approx 0.6 s (1-\frac{x}{x_0}) x. $$

Using $a(x)$ and $b(x)$ as the drift and diffusion coefficients in a diffusion equation, the time spent in the frequency interval $[0,x0]$ is: $$ T{0,x_0} = \int_0^{x_0}{\frac{2N}{x G(x)} {\int_0^x{G(x')d x'}} dx}, $$

and the time spent in the interval $[x0,1]$ is: $$ T{x0,1} = \int{x_0}^1{\frac{2N}{x G(x)} \int_0^{x_0}{G(x') dx'} dx }, $$ where $$ G(\xi) = exp[-2 \int_0^{\xi} {\frac{a(z)}{b(z)}dz}] = \\ exp[\frac{2N 0.6s}{x_0} \xi (\frac{\xi}{2}-x_0)]. $$

The time to the loss of the best class is then $$ T(N,u,s) = T_{0,x0} + T{x_0,1}. $$