In Wright-Fisher model of population size N and initial mutation frequency of 1/N, how does the fixation probability vary over generations. So, mathematically, what is the function that maps the generation to its fixation probability?

  • $\begingroup$ Is your question "What is the probability P(t) that a neutral mutations at frequency 1/2N fixes after t generation given that it will reach fixation?"? $\endgroup$
    – Remi.b
    Feb 16, 2018 at 19:19
  • $\begingroup$ @Remi.b Yes. I know that fixation probability is 1/N eventually. In a sense that, if we let the population evolve for enough generations, fixation probability becomes 1/N. But how does it actually vary over generations? Thank you. $\endgroup$
    – happystick
    Feb 16, 2018 at 19:24
  • 1
    $\begingroup$ The answer is probably in Kimura and Ohta 1968. From there, the expected time to fixation of an allele at frequency $p$ is $\bar t(p_0)=-4N\left(\frac{1-p_0}{p_0}\right)\ln(1-p_0)$ $\endgroup$
    – Remi.b
    Feb 16, 2018 at 19:24
  • $\begingroup$ But how can we describe the variation of fixation probability over time using that equation? Thank you. $\endgroup$
    – happystick
    Feb 16, 2018 at 19:37
  • $\begingroup$ Are you thinking of a case where a population where a population starts without any genetic diversity and hence there is little fixation at early generations and then the rate of fixation plateaus or are you looking for a variance in the number of fixation events from generation to generation in a equilibrium population? $\endgroup$
    – Remi.b
    Feb 16, 2018 at 19:45

1 Answer 1


The function probably look like $r(t) = A + B e^{-C t}$, where $r(t)$ is the rate of fixation at time $t$ (in generation). The reason for this exponential is that the probability that a mutation happens after $t$ generation is given by the exponential distribution.

As the rate of fixation at equilibrium must be at $2N\mu \frac{1}{2N} = \mu$, $A=\mu$, that is $r(t) = \mu + B e^{-C t}$

As at $t=0$, the rate must be 0. As $e^{0} = 1$, you should have $B=-\mu$. Hence,

$$r(t) = \mu - \mu e^{-C t}$$

$C$ here represents the steepness of the curve. I don't know how you could calculate it a priori but it must depends upon your population size $N$. As the expected time to fixation starting at frequency $p_0$ is (from Kimura and Ohta 1968)

$$\bar t(p_0)=-4N\left(\frac{1-p_0}{p_0}\right)\ln(1-p_0)$$

For $p_0 = \frac{1}{2N}$

$$\bar t\left(\frac{1}{2N}\right)=-4N\left(2N-1\right)\ln\left(1-\frac{1}{2N}\right)$$


$$\bar t\left(\frac{1}{2N}\right)=\left(4N-8N^2\right)\ln\left(\frac{2N-1}{2N}\right)$$

if you prefer. Maybe $C = \bar t\left(\frac{1}{2N}\right)$ leading to

$$r(t) = \mu \left(1 - e^{-\left( \left(4N-8N^2\right)\ln\left(\frac{2N-1}{2N}\right) \right) t}\right)$$

Let me know if it matches (I'd pretty amazed)!


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