Genetic drift lets a neutral gene take over a finite population

A population that copies itself at random

Take $N$ individuals, each carrying one copy of a gene in one of two versions, called alleles: the mutant $A$ or the original $a$. Let $i$ be the number of $A$ individuals and $p=i/N$ their frequency. Time is counted in generations, and parents die when their children are born.

To build the next generation, each of its $N$ members picks a parent at random from the current one, with every parent equally likely, and copies its allele. The picks are independent. The number of $A$ children is then a binomial count: $N$ trials, each a success with probability $p$. This is the Wright–Fisher model.

Nothing in the rule favours either allele. Every change in frequency comes from the luck of the draw, and population geneticists call it genetic drift.

Source: K. S. Korolev · N. Berestycki

The average stays put while the spread grows

A binomial count with $N$ trials has mean $Np$ and variance $Np(1-p)$. Dividing by $N$ gives the frequency in the next generation:

The mean does not move. Each generation adds a random step whose typical size is $\sqrt{p(1-p)/N}$, and the steps add up like a random walk. The walk ends when the frequency reaches $0$ or $1$, because a population with no copy of an allele can never draw one again. The allele is then lost or fixed.

Now average over many populations that start at the same $p_0$. The average frequency is the same in every generation, so at the end it still equals $p_0$. At the end each population sits at $0$ or $1$, and the average is the fraction that reached $1$. A neutral allele therefore fixes with probability $u(p_0)=p_0$.

Source: N. Berestycki · K. S. Korolev

Heterozygosity: the chance that two individuals differ

To follow diversity, draw two individuals at random, with replacement. They carry different alleles with probability $H=2p(1-p)$, called the heterozygosity. It is largest at $p=1/2$ and zero once one allele has fixed.

Average $H$ over the next draw. The mean square of the new frequency is the square of its mean plus its variance, $\mathrm{E}[p_{t+1}^2]=p_t^2+p_t(1-p_t)/N$. Substituting gives a one-line recursion:

Each generation removes a fraction $1/N$ of the expected diversity. A population of $50$ keeps about $e^{-1}\approx0.37$ of it after $50$ generations. A population ten times larger needs ten times as long.

Source: K. S. Korolev

Worked example: one neutral mutant among fifty

Problem. A population of $N=50$ starts with one neutral mutant. How likely is the mutant to fix, how long does fixation take when it happens, and how much diversity is left after $50$ generations?

Fixation. The starting frequency is $p_0=1/50=0.02$, so $u=0.02$. In $49$ populations out of $50$ the mutant disappears.

Time. For large $N$ the frequency behaves like a diffusion, and one unit of diffusion time equals $N$ Wright–Fisher generations. The mean time to fixation, counted only in populations where the mutant fixes, is then given by the formula below. At $p=0.02$ it gives $99.0$ generations. Solving the finite Wright–Fisher chain exactly gives $96.97$ generations; the diffusion formula is a few per cent high because it treats $N$ as large.

Diversity. $H_0=2\times0.02\times0.98=0.0392$, and after $50$ generations $H_{50}=0.0392\times0.98^{50}=0.0143$.

Check. The drift Lab runs this case as $12000$ seeded replicate populations. The mutant fixed in a fraction $0.0212\pm0.0013$ of them, within one standard error of $0.02$, and the populations where it fixed took $96.2\pm3.4$ generations.

Source: N. Berestycki · K. S. Korolev