I have a question about how to determine the probability that a mutation is lost or fixed after 1 or 2 generations in population genetics.

Let's say we have a randomly mixing population, with N diploid individuals (N=5), and there is one mutation that appears. I know it should follow a binomial distribution, but I also heard it may follow a Poisson distribution... and all I feel like saying is that the probability to get fixed is the same as the frequency at time 0, which is 1/2N, or here 1/10.

Furthermore, how do you calculate the probability that it exists in 2 copies?

  • $\begingroup$ your question assumes that a mutation is something that is "broken" - might want to reword a bit $\endgroup$ Commented Nov 16, 2016 at 1:12
  • $\begingroup$ Thanks a lot for your reply ! It helps a lot. However, shouldn't we use 10 among 2, instead of 5 among 2 in the binomial calculation, as we're looking at alleles in a diploid population ? $\endgroup$
    – user27813
    Commented Nov 16, 2016 at 8:07
  • $\begingroup$ @VanceLAlbaugh 'fixed' is a technical term in population genetics. It has nothing to do with a mutation being 'broken'. Fixation is when a mutation reproduces through a population until essentially 100% of the population has the mutation. Loss is the opposite process, where a mutation disappears from a population. Fixation and loss can be entirely chance processes, or they can be driven by a selective advantage or disadvantage of the mutation. $\endgroup$ Commented Nov 16, 2016 at 16:40
  • $\begingroup$ @CharlesE.Grant excellent - thanks for clarifying... might want to edit question to clarify that - as users unfamiliar with that (like myself) clearly interpreted that differently $\endgroup$ Commented Nov 16, 2016 at 16:43

1 Answer 1


How to determine the probability that a mutation is lost / fixed?

The probability that a neutral mutation gets fixed after an infinite amount of time is equal to its frequency $p$ as you said. Therefore the probability of being lost is $1-p$. This post offers an explanation but there many ways to make the demonstration. You might want to have a look at any good book in population genetics for this demo. Here are book recommendations.

how do you calculate the probability that it exists in 2 copies?

A probability always depends on a priori. What are your a priori? Let's assume that we know that the allele frequency was $\frac{4}{10}$ in the previous time step.

Under the Wright-Fisher model the probability of having 2 copies in the next generation is given by the binomial distribution. Let $N=5$ and therefore $2N=10$ and let the frequency of the allele of interest be $\frac{4}{10}$, the probability of having two alleles in the next generation is ${10 \choose 2} \left(\frac{4}{10}\right)^2 \left(\frac{6}{10}\right)^8 ≈ 0.12$.

Under the Moran model this probability is zero. Moran's model is a birth-death model (Markov model) and therefore transition between time steps can only add or subtract (or make no change) a single allele. You will note that the time step does not mean the same thing for the two models. Loss of heterozygosity is twice as fast under the Wright-Fisher model but this discussion is definitely not what you were asking for!

  • $\begingroup$ @user27813 Thanks for your comment. You are right, you said diploid population with $N=5$ and I considered a haploid population in my answer. It is fixed now. $\endgroup$
    – Remi.b
    Commented Nov 17, 2016 at 6:26
  • $\begingroup$ If you know the probability of an individual surviving, both without & with the change, you can calculate the probability of the gene surviving. This isn't actually knowable, so the details of the model used to estimate seem inescapably relevant to any answer to me. Am I missing something, here? $\endgroup$
    – The Nate
    Commented Nov 17, 2016 at 16:28
  • $\begingroup$ @TheNate I don't really understand your comment. Typically, what do you mean by "the change". These models are models of genetic drift, they assume all individuals have the same fitness. The probability of an individual to produce k offspring is given by a poisson distribution with rate \lambda=1. $\endgroup$
    – Remi.b
    Commented Nov 17, 2016 at 16:32
  • $\begingroup$ "The change" I meant was the mutation of interest. I understand/understood what the models represent; you answered the question. That set of (apparently customary?) simplifying assumptions is exactly what I was missing. $\endgroup$
    – The Nate
    Commented Nov 17, 2016 at 16:46

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