I want to know how does genetic drift, and selection coefficient (s) together affect the fixation of an allele? For example, if there is a neutral allele (s=0), will only genetic drift cause the fixation or loss of an allele? Similarly, if an allele is beneficial which means that s > 0. In that case, how do genetic drift and selection coefficient together determine the fixation and loss of an allele? And what happens when s < 0 that is an allele is deleterious. Lastly, how do genetic drift and selection coefficient gets affected by the population size? I am reading about population genetics, and I am totally confused by this point. I would appreciate if someone can clear this to me.


It would take a lot of writing to fully answer your question. Below I give an answer without spending too much time on the underlying math assuming you are confortable with probability theory and Taylor series. If you really want to understand more of all of that you would just need to consider reading a book in population genetics (see end of the post for recommendation).

Probability of fixation under drift only

In absence of selection, mutation and migration, drift only may yield an allele to eventually reach fixation (or loss). The probability of an allele to get fixed is then just equal to its frequency $P_{fix}=p$. When the neutral mutation has just arisen, this probability is therefore $P_{fix}=p=\frac{1}{2N}$ for a diploid population of size $N$.

See this post for more information.

Drift and population size - Intuition

For some intuition about the effect of population size on drift you might want to have a look at Why is the strength of genetic drift inversely proportional to the population size?.

Drift and population size - Effective population size

To go a little further than the linked post, note that the the strength of genetic drift is defined by what we call the effective population size $Ne$. But before talking about that let's talk about the variance in possible allele frequencies in the next generation (which we'll call $var(p')$).

The number of alleles $A$ in the next generation follows a binomial distribution (if we follow Wright-Fisher model). The variance of this binomial distribution is $2 p N (1-p)$, where $p$ is the frequency of the allele $A$ in the previous generation. Dividing this variance by $(2N)^2$ (the square is caused by the fact that $var(k X) = k^2 var(X)$, where $X$ is a random variable), to get the variance in allele frequency in the next generation

$$var(p') = \frac{p(1-p)}{2N}$$

We define the strength of genetic drift as $var(p')$ and for intuition, we usually refers to it as the value of $N$ that corresponds and to avoid confusion with the actual population size we call this new measure $Ne$. Therefore,

$$var(p') = \frac{p(1-p)}{2Ne}$$

Solving for $Ne$ yields to

$$Ne = \frac{p(1-p)}{2var(p')}$$

and this effective population size $Ne$ is what we use to talk about strength of genetic drift.

Probability of not reproducing

As I will refer to this statistic later, I will introduce it here. What is the probability of a given individual to not reproduce? Let's call this quantity $P(k=0)$ as the probability that the number of offspring $k$ is $0$.

Imagine you sample $2N$ haplotypes (with replacement) to form the next generation and each of them has a probability $1-\frac{1}{2N}$ to be chosen, the the probability of not reproducing is $p(k=0) = \left(1-\frac{1}{2N}\right)^{2N}$. Taking the Taylor series of this expression for large values of $N$ yield to the approximation $p(k=0) = \left(1-\frac{1}{2N}\right)^{2N} ≈ e^{-1} ≈ 0.37$. So, any individual has a probability greater than a third to leave any offspring in the following generation.

Probability of fixation under both drift and selection

If an allele if beneficial, then its probability of fixation is greater than its frequency $p$ (and vice-versa if deleterious).


There are a number of ways to derive such probability and they all end up with some complicated form. An easy approximation (which I am not going to make any demonstration) for this probability is

$$P_{fix}≈\frac{1-e^{-4Ns} }{1-e^{-4Nsp}}$$

Just to give you some intuition about the form of the above approximation (but without doing a full demonstration that would be quite long), consider that the number of offspring an individual have follows a Poisson distribution with fitness as the rate. The fitness of the mutant individual is $1+s$, and therefore the probability of this individual of having $k$ offspring is

$$P(k) = \frac{(1+s)^k e^{-1-s}}{k!}$$

Think of the generation where the mutation just arised and only one individual has this mutation. The probability of this mutation to disappear immediately is therefore

$$P(0) = e^{-1-s}$$

, which for reasonable values of $s$ is not very different from the probability of not reproducing when $s=0$ and the probability become $e^{-1}$ (as already shown above via an other mean).

A less accurate approximation (but with a very easy form) for the probability of fixation is

$$P_{fix} = 2Ns$$

Some more calculations about the interplay of drift and selection can be found in this post.

Source of information

The post Book recommendation on population/evolutionary genetics? will give you great sources of information to learn more on the subject. It will also re-explain what I just said with a lower pace. I particularly recommend Population Genetics: A Concise Guide


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