neutral theory question

Joe Felsenstein joe at evolution.genetics.washington.edu
Mon Oct 30 19:16:51 EST 1995

In article <470eft$iuk at news.bu.edu>, Chris Colby <colby at bio.bu.edu> wrote:
>Given a constant population size, what is the average time 
>until equilibrium heterozygosity (H = 4Nv/[4Nv + 1]) is reached
>starting from a monomorphic locus? 

The equation for homozygosity (= 1- heterozygosity) which we call F is

   F(t+1)  =  (1-v)^2 [1/(2N)  + (1 - 1/(2N)) F(t) ]

If we write this as  F(t+1) = a + b F(t)  where

   a  =  1/(2N)  

   b  =  (1-v)^2 (1 - 1/(2N))

then the equilibrium value of F is   a/(1-b)   (which is approximated
by one minus the usual expression you have given).   If we call the
eqiulibrium value F(e), then we can also show rather easily that

    F(t+1) - F(e)  =  b  (F(t) - F(e))

So in effect you go a fraction b of the way to equilibrium each
generation.    Thus the average time to reach equilibrium is ...
infinity.  But a more informative number is the half-life of the return
there, which will be the solution of  b^t = 0.5 which is

    t(half) =  ln (0.5) / ln (b)

where ln (0.5) is 0.693147.   ln(b) will be approximately

   2v + 1/(2N).

So the half life of restoring heterozygosity will be approximately

           0.7 / (2v + 1/(2N))    generations.

Interestingly enough, this is also, using the approximation for the
equilibrium heterozygosity H that you cited,  0.7 H / v.

Thus if the heterozygosity at equilibrium is, say, 0.15, and v = 10^(-7),
one would need   (0.7)(0.15)/0.0000001 =  1,050,000 generations to get
halfway back.

Joe Felsenstein         joe at genetics.washington.edu     (IP No.
 Dept. of Genetics, Univ. of Washington, Box 357360, Seattle, WA 98195-7360 USA

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