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 '18 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 '18 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 '18 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 '18 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 '18 at 19:45

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)!

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.