# The convergence of mean fitness towards the mutation-selection balance

## Overview

In an eariler post I described how the mean fitness of a population at the mutation-selection balance can be analysed. I assumed that the population is asexual, that only deleterious mutations occur, that there is no drift or recombination, and that selection is constant.

In this post I would like to continue with these assumpions. I will show how to find a simple formula for the mean fitness after an arbitrary number of generations. The formulation and derivation will follow one of my favorite papers - "Nonequilibrium model for estimating parameters of deleterious mutations" by Gordo & Dionisio (2005).

## Motivation

But first, why are we interested in the mean fitness of a population that is not at the mutation-selection balance?

The general answer is that most populations are probably not at the mutation-selection balance. The mutation-selection balance is a nice idea, but it serves as a reference, as a null hypothesis, not as a general rule. The assumption that selection is constant is not relevant to many populations - selection is probably fluctuating, both in amplitude and direction. This means that at least some of the natural populations we encounter in the wild are not evolving around a mutation-selection balance, but rather evolving towards a mutation-selection balance, never actually reaching one.

The more specific answer is that even in laboratory conditions which induce a constant selection regime it may take a long time for populations to reach the mutation-selection balance, and this time might be longer than what we intuitively expect (or desire).

This issue is presented by Gordo & Dionisio (2005). They wanted to use traditional models for calculating the mutation rate of bacterial populations in the lab. These models assume that the populations are at a mutation-selection balance, but Gordo and Dionisio showed that in the timeframe they planned for their study their populations will not reach an equilibrium, especially if selection is weak (see Figure 1 in (Gordo & Dionisio, 2005)).

This is interesting, because we saw in the eariler post that the mean fitness at the mutation-selection balance is e^-U^, where U is the mutation rate and is independent of s, the selection coefficient. But Figure 1 in (Gordo & Dionisio (2005) shows that the mean fitness before the mutation-selection balance is dependent on s, and that s determines the rate of convergence towards the balance.

## Model

This follows the model from Gordo & Dionisio (2005) but I elaborated and included all the rigouros steps.

### Definitions

Denote the pmf (probability mass function) of a Poisson distribution with parameter $\lambda$: $\varphi{\lambda} (x)$ Consider an asexual infinite population evolving in a constant environment. The number of new mutations per individual per generation is Poisson distributed with parameter U, the mutation rate. All mutations are deleterious with a multiplicative effect s, so that an individual with $i$ mutant allele has a fitness $\omega{i}:=(1-s)^{i}$ with selection coefficient $0<s<1$. The frequency of individuals with i mutant at generation g is denoted $f{i}(g)$ and $\sum{i}{f{i}(g)}=1$. From the former two statements it follows that the population mean fitness at generation g is $\bar{\omega }{g}=\sum{i}{f{i}(g)\omega _{i}}$. Assume that the selection occurs before mutation.

### The distribution of mutant alleles

The change in the frequency of individuals with i mutant alleles from generation g to generation g+1 is written as:

$$f{i}(g+1)=\sum{k=0}^{i}{\frac{ \omega {k} }{ \bar{\omega}{g} } f{k}(g) \varphi{U} (i-k) }$$

The model starts with a mutation-free population, so the initial condition is $f{0}(0)=1, f{i>0}(0)=0$.

Let's calculate the frequency of individuals with i mutant alleles after one generation:

$$f{i}(1)=\sum{k=0}^{i}{f{k}(0)\frac{\omega {k}}{\bar{\omega}{1}}\cdot \frac{U^{i-k}}{(i-k)!}e^{-U}}$$ and because $$f{k}(0)=\bigg{ \begin{gathered} 1, k=0 \ 0, otherwise \ \end{gathered}$$ we get $$f{i}(1) = f{0}(0)\frac{\omega {0}}{\bar{\omega }{1}} \cdot \frac{U^{i}}{i!}e^{-U} = \frac{U^{i}}{i!}e^{-U}$$ so the frequency of individuals with i mutant alleles after one generation is Poisson distributed with parameter U.

To verify this - the expected number of mutations after one generation is $$\sum{i\ge 0}{i\cdot f{i}(1)} = \sum{i\ge 1}{i\cdot \frac{U^{i}}{i!}e^{-U}} =\ e^{-U}U\sum{i\ge 1}{\frac{U^{i-1}}{(i-1)!}} = \ e^{-U}U\sum_{i\ge 0}{\frac{U^{i}}{i!}} = \ Ue^{-U}e^{U} = U$$ as expected.

To go on the next step, we will find the frequency of individuals with i mutant alleles at the second generation, after selection but before mutations - this is marked by $f{i}^{s}(2)$: $$f{i}^{s}(2) = \frac{\omega {i}}{\bar{\omega}{1}}f{i}(1) = \ \frac{(1-s)^{i}}{\sum{k=0}^{\infty }{f{k}(1)\omega{k}}}\frac{U^{i}}{i!}e^{-U} = \ \frac{(1-s)^{i}}{\sum{k=0}^{\infty}{\frac{U^{k}}{k!}e^{-U}(1-s)^{k}}}\frac{U^{i}}{i!}e^{-U} = \ \frac{(U(1-s))^{i}}{i!\sum{k=0}^{\infty}{\frac{(U(1-s))^{k}}{k!}}} = \ \frac{(U(1-s))^{i}}{i!}e^{-U(1-s)}$$

So, $f{i}^{s}(2)$ is Poisson distributed with parameter $U(1-s)$, and the frequency after mutation will be: $$f{i}(2) = \sum{k=0}^{i}{f{k}^{s}(2)\frac{U^{i-k}}{(i-k)!}e^{-U}} = \ \sum{k=0}^{i}{\frac{(U(1-s))^{k}}{k!}e^{-U(1-s)}\frac{U^{i-k}}{(i-k)!}e^{-U}} = \ e^{-U(1-s)-U}\sum{k=0}^{i}{\frac{U^{i-k}(U(1-s))^{k}}{k! \cdot (i-k)!}} = \ \frac{U^{i}}{e^{U(1-s) + U}} \sum{k=0}^{i}{\frac{(U(1-s))^{k}}{U^{k} \cdot k! \cdot (i-k)!}} = \ \frac{U^{i}}{e^{U(1-s) + U}} \sum{k=0}^{i}{\frac{(1-s)^{k}}{k! \cdot (i-k)!}}$$

Note that $\sum{k=0}^{i}{\frac{q^{k}}{k!\cdot (i-k)! }} = \frac{(q+1)^{i}}{i!}$ And $\sum{k=1}^{i}{\frac{q^{k}}{k!\cdot (i-k)! }}=\frac{(q+1)^{i}-1}{i!}$, and therefore:

$$f_{i}(2) = \frac{U^{i}}{e^{U(1-s)+U}}\cdot \frac{(2-s)^{i}}{i!} = \ \frac{(U(1-s)+U)^{i}}{i!}e^{-(U(1-s)+U)} = \ \phi(i | U(1-s)+U)$$

Similar to the former expansion and because the Poisson process is memoryless, the frequency of individuals with i mutant alleles after mutation at generation 2 will be Poisson distributed with parameter $U(1-s)+U$.

The same argument makes it clear that the distribution of mutant alleles at any generation g is Poisson distributed.

## The expected number of mutant alleles

After seeing how the distribution of mutant alleles changes at the first couple of generations, and verifying that this is a Poisson distribution, we can now formulate the recurrence relation for the expected number of mutant alleles - or the parameter of the Poisson distribution, $\lambda$:

$$\lambda (g+1) = \sum{k=0}^{\infty }{k\cdot f{k}(g+1)} = \ \sum{k=0}^{\infty}{\sum{i=0}^{k}{k\frac{(1-s)^{i}}{\bar{\omega}{g}}f{i}(g)\frac{U^{k-i}}{(k-i)!}e^{-U}}} = \ \sum{i=0}^{\infty}{\sum{k=i}^{\infty }{k\frac{(1-s)^{i}}{\bar{\omega}{g}}f{i}(g)\frac{U^{k-i}}{(k-i)!}e^{-U}}} = \ \sum{i=0}^{\infty}{e^{-U}\frac{(1-s)^{i}}{\bar{\omega}{g}}U^{-i}f{i}(g)\sum{k=i}^{\infty}{k\frac{U^{k}}{(k-i)!}}} = \ \sum{i=0}^{\infty}{e^{-U}\frac{(1-s)^{i}}{\bar{\omega}{g}}U^{-i}f{i}(g)e^{U}U^{i}(i+U)} = \ \frac{1}{\bar{\omega}{g}}\sum{i=0}^{\infty }{f{i}(g)(1-s)^{i}(i+U)} = \ \frac{U}{\bar{\omega}{g}}\sum{i=0}^{\infty }{f{i}(g)(1-s)^{i}}+\frac{1}{\bar{\omega}{g}}\sum{i=1}^{\infty }{i\cdot f{i}(g)(1-s)^{i}} = \ U + \frac{\sum{i=1}^{\infty }{i\cdot f{i}(g)(1-s)^{i}}}{\sum{i=0}^{\infty}{f{i}(g)(1-s)^{i}}} = \ U + \frac{\sum{i=1}^{\infty }{i\cdot \frac{e^{\lambda (g)}(\lambda (g)(1-s))^{i}}{i!}}}{\sum{i=0}^{\infty}{\frac{e^{\lambda(g)}(\lambda (g)(1-s))^{i}}{i!}}} = \ U + \frac{\lambda(g)(1-s)\cdot e^{\lambda (g)(1-s)}}{e^{\lambda(g)(1-s)}} = \ \lambda (g)(1-s) + U$$

So we got a nice recurrence formula - $\lambda (g+1)=\lambda (g)(1-s)+U$ with an initial condition $\lambda (1)=U$. The recurrence means that the expected number of mutatnt alleles per individual in the population is reduced every generation by selection by multiplying it by (1-s) and increased by mutation by adding U.

The solution to this recurrence relation is (why? this will require another post): $$\lambda(g)=\frac{U}{s}(1-(1-s)^{g})\xrightarrow{g\to \infty }\frac{U}{s}$$

For a sinlge-locus deterministic model, the expected frequency of the wild-type allele at the mutation-selection balance (assuming that $s>>U$) is $1 - \frac{U}{s}$, which agrees well with the frequency of mutation-free individuals we can now calculate: $\phi(0 | \frac{U}{s}) = \frac{\frac{U}{s}^0}{0!} e^{-\frac{U}{s}} = e^{-\frac{U}{s}} = 1-\frac{U}{s} + O(\frac{U}{s}^{2})$. The last approximation uses the Taylor expansion of the exponential function around 0.

## The population mean fitness

After g generations the population mean fitness is:

$$\bar{\omega}{g} = \sum{i=0}^{\infty }{\frac{\lambda(g)^{i}}{i!}e^{-\lambda(g)}(1-s)^{i}} = \ e^{-\lambda(g)}\sum{i=0}^{\infty}{\frac{ (\lambda(g) (1-s))^{i}}{i!}} = \ e^{-\lambda(g)}e^{\lambda(g)(1-s)} \Rightarrow \ \bar{\omega}{g} = e^{-\lambda(g) s}$$

Which gives us yet [another way][earlier post] to calculate the mean fitness at the mutation-selection balance: $$\bar{\omega }^* = lim_{g\to \infty} e^{-\lambda(g) s} = e^{-\frac{U}{s}s} = e^{-U}$$

The second moment of the population fitness is given by:

$$E[\omega{g}^2] = \sum{i=0}^{\infty }{\frac{\lambda(g)^{i}}{i!}e^{-\lambda(g)}(1-s)^{2i}} = \ e^{-\lambda(g)} \sum_{i=0}^{\infty }{\frac{ (\lambda(g)(1-s)^2)^{i} }{i!}} = \ e^{-\lambda(g)} e^{\lambda(g)(1-s)^2} = \ e^{\lambda(g)(1-2s+s^2)-\lambda(g)} = \ e^{\lambda(g)(s^2-2s)}$$

## Estimation of mutation rates

In mutation-accumulation (MA) experiments, a population is undergoing a sequence of bottlenecks that cause the accumulation of deleterious mutations due to the random sampling effect of genetic drift. In these experiments the investigators usually measure the mean fitness of the experimental population thorugh time and use these measurements to estimate s and U.

Denote the expected mean fitness at bottleneck B by $\bar{\omega{B}}$. We start with a mutation-free population, and $\bar{\omega{0}} = 1$. At the first bottleneck, the population is assumed to reach a mutation-selection balance, so the expected mean fitness is now $\bar{\omega{1}} = e^{-U}$. At the next bottleneck, the expected mean fitness is reduced again by the same factor, so it is now $\bar{\omega{1}} = e^{-U}e^{-U} = e^{-2U}$. After B bottlenecks, the mean fitness is $\bar{\omega{B}} = e^{-UB}$. So one can take the log of the measured mean fitness after B bottlenecks, $-BU$, and use a linear regression model to estimate U. By using a similar linear regression on the ratio of the second moment of fitness and the square of the expected mean fitness, one can estimate s (see below).

However, what happens if the population doesn't reach a mutation-selection balance? In a population of bacteria, for example, the number of generations between bottlenecks may be insufficient for the population to reach a mutation-selection balance (see Figure 1 in [@Gordo2005]). In this case one must use a the non-equilibrium value of the mean fitness which was derived above - $\bar{\omega}_{g} = e^{-\frac{U}{s} (1-(1-s)^{g})}$. This concept is the subject of the paper by @Gordo2005, titled: Nonequilibrium model for estimating parameters of deleterious mutations. The paper explores a statistical model based on the above calculations. The statistical model is tested with simulations. In a later paper, (Trindade et al. 2010), the statistical model was used on results of a mutation-accumulation experiment with E. coli.

In the above development, we saw that the expected mean fitness after g generations is $\bar{\omega}{g} = e^{-\lambda(g) s}$. If the number of generations between bottlenecks is constant, say g=24 (for a population in which a generation is estimated at 30 minutes, resulting in a bottleneck period of 12 hours), then we can denote $\lambda {:=} \lambda(g)$ and use $\bar{\omega{B}} = e^{-\lambda s B}$. Then we can use a linear regression model to estimate $-\lambda s$.

But how do we proceed? The estimator we found using linear regression, $\lambda s= \lambda(g) s = U(1-(1-s)^{g})$, doesn't directly yield either U or s. Well, this is why we calculated the second moment. We will not use it directly, but instead define an F-statistic (not to be confused with the F-statistic used to describe heterozigosity or population structuring): $F_{B} = \frac{\bar{\omega^2}}{\bar{\omega}^2}$. This is a little confusing so I'll write it in a probablistic presentation:

$$F_{B} = \frac{E[\omega^2]}{E^2[\omega]}$$

This is the ratio of the second moment to the square of the first moment. What is this ratio? Let's have a look:

$$F_{B} = \frac{E[\omega^2]}{E^2[\omega]} = \ \frac{e^{\lambda (s^2-2s) B}}{e^{-2 \lambda s B}} =\ e^{( \lambda s^2-2\lambda s+2 \lambda s)B} = \ e^{( \lambda s^2)B}$$

So the log of the F-statistic is $\lambda s^2 B$, and linear regression can give us $\lambda s^2$. Now we can estimate both s and U using the estimation for $-\lambda s$, $\lambda s^2$, and the relation $\lambda = \frac{U}{s}(1-(1-s)^g)$.