A classical result of population genetic is that the rate of fixation of netreual alleles is the mutation rate $\mu$. The reason is that each generation $PN_e\mu$ mutations enter the population, where $P$ is the ploidy number (e.g. 2 for diploids) and $N_e$ is the effective population size. The probability of each neutral mutation to reach fixation is simply its frequency $p$. When the mutation occur $p=\frac{1}{PN_e}$ and therefore the rate of fixation is:

$$\lambda = PN_e\mu \frac{1}{PN_e} = \mu$$

This result s typically very much used in phylogenetic, as assuming a constant mutation rate only, one can estimate the divergence time between two extant lineages.

The above is also explained on wiki>fixation_rate


How robust is the result $\lambda = \mu$?

I understand that the result $\lambda=\mu$ is independent of the effective population size but is it also independent of..

  • changes in population size?
  • background selection?
  • selective sweep?
  • population structure?
  • selfing rate?
  • etc..
  • $\begingroup$ It is special case of Wright-Fisher game right ? In that case how it is independent of population size? $\endgroup$ – Dexter Sep 23 '15 at 19:20
  • $\begingroup$ What do you mean by Wright-Fisher game? It is independent of $N_e$ (at least for constant population size) as I showed in the background. It is also explained in this wiki article $\endgroup$ – Remi.b Sep 23 '15 at 20:15
  • $\begingroup$ See Write-Fisher model of genetic drift. And see introduction of these notes. $\endgroup$ – Dexter Sep 24 '15 at 4:28
  • $\begingroup$ I know the Wright-Fisher model of genetic drift. We also talk about Wright-Fisher population but I've never heard of Wright-Fisher game. $\endgroup$ – Remi.b Sep 24 '15 at 13:23

The answer to your headline question is that no, fixation rate is not always equal to the mutation rate for neutral alleles. For instance:

Fixation rates for neutral alleles are affected by changes in population size, given a constant mutation rate. In general, fixation rates are lower in growing populations (Waxman 2012).

This makes instinctive sense if you consider the single-generation fixation probability of a neutral allele which only exists at all but one copy in the population, i.e., the common allele (A) exists at 2N(t)-1 copies, whereas the less common allele (a) exists at 1 copy. To simplify, let's assume complete random mating. Under negative population growth (say N(t+1) = 0.9N(t)), there will be 0.9*(2N) places that the rare allele could possibly exist at t+1, and it has a 1/(2*N(t)) chance of filling each of them. Under positive population growth (say, N(t+1) = 1.1*N(t)), the rare allele could potentially occupy 1.1*(2N) places at t+1, but still has 1/(2*N(t)) chance of filling each of them. It is clear in this case that A has a reduced generation-wise likelihood of fixation under positive population growth than under negative population growth, simply because the rare allele has an increased chance of doggedly persisting. The finding applies to any allele near fixation.

If I have understood Kim and Stephan (2000) correctly, then overall fixation rates are also reduced by background selection. The main mechanism for this is that background selection purges linked neutral alleles, decreasing overall heterozygosity and thereby reducing the chance that a new, rare allele will become established. The same surely applies to selective sweeps, given linked neutral alleles.

  • $\begingroup$ Thank you for your answer +1. I suppose that your answer also suggest that population structure and selfing aren't modifying the result $\lambda=\mu$, right? Is there anything else that you can think about that might affect this result? Thank you! $\endgroup$ – Remi.b Sep 24 '15 at 13:55
  • 1
    $\begingroup$ I didn't mean to imply that population structure and selfing wouldn't modify the λ=μ relationship - I didn't find literature that tested those hypotheses. My instinct is that both structuring and selfing would reduce the fixation rate by reducing population-wide heterozygosity, but I would want to do some more reading (or run some simulations) before I included that in an answer. $\endgroup$ – bshane Sep 24 '15 at 14:34

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