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. 128.95.12.41)
Dept. of Genetics, Univ. of Washington, Box 357360, Seattle, WA 98195-7360 USA
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