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?
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$\begingroup$ 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.bCommented Feb 16, 2018 at 19:19
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$\begingroup$ @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. $\endgroup$– happystickCommented Feb 16, 2018 at 19:24
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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.bCommented Feb 16, 2018 at 19:24
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$\begingroup$ But how can we describe the variation of fixation probability over time using that equation? Thank you. $\endgroup$– happystickCommented Feb 16, 2018 at 19:37
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$\begingroup$ 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.bCommented Feb 16, 2018 at 19:45
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1 Answer
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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 $p_0 = \frac{1}{2N}$
$$\bar t\left(\frac{1}{2N}\right)=-4N\left(2N-1\right)\ln\left(1-\frac{1}{2N}\right)$$
or
$$\bar t\left(\frac{1}{2N}\right)=\left(4N-8N^2\right)\ln\left(\frac{2N-1}{2N}\right)$$
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)!
