In a panmictic population, the probability of fixation of an allele at a neutral locus is equal to its frequency at that time. I will refer to this probability of fixation as calculated at time $t$ as $P_{fix,t}$.

If $p_t$ is the frequency of the allele $A_1$ at time $t$, then the probability of the allele $A_1$ to reach fixation (rather than disappearing) is $P_{fix,t}=p_t$. Typically, the generation when the mutation occurred, the probability of the new allele to get fixed is $P_{fix,0}=p_0=\frac{1}{2N}$, where $N$ is the population size.


This simple and classic result makes very good intuitive sense to me. However, I would fail to provide a mathematical proof.

Can you please demonstrate that $P_{fix,t}=p_t$?


1 Answer 1


Several proofs are given here (p. 9). My favorite comes from the genealogical argument:

Consider the situation where there are $2N$ alleles: $A_1$, $A_2$, $A_3$ ... $A_{2N}$.

By the genealogical argument, we may state that at $t = \infty$, all alleles at this locus will be direct descendants of one particular allele present at $t = 0$.

Allelic variants at this locus are selectively neutral, so $Pr(A_{1fix})$ $=$ $Pr(A_{2fix})$ $=$ $Pr(A_{3fix})$ $= ... =$ $Pr(A_{2Nfix})$. For any given allele present at $t = 0$, the probability of fixation is therefore $\frac{1}{2N}$.

Now define allelic variants $A$ and a as complementary, non-overlapping groups of the initial alleles, such that $n_A$ + $n_a$ $=$ $2N$. From the above, the probability of fixation of an allele within group $A$ is $n_A * \frac{1}{2N} = \frac{n_A}{2N} = p_0$.

  • $\begingroup$ Sorry for digging this topic after 2 years. I am studying evolution these days and was reading basics from the book by Hartl and Clarke. There, this proof is given using the Markov chain approach. However, the way it is proved doesn't make sense to me (also shown in the slides that you have linked). In the Markov chain approach the probability is a vector with each element denoting the probability of having i number of alleles in the population. Can you, if possible, please show how to prove this result using the Markov chain approach? $\endgroup$
    Commented May 14, 2018 at 12:16

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