phillip smith wrote:
> I wonder if any one can help me? I need to check a calculation for a paper.
> 
>>From Kimura the time for a neutral mutation to be fixed is the mutation
> rate.
> If I have ten genes each of which can have a neutral mutation. What is the
> mean time taken  for there to be no individuals which are free of mutation.
> That is the whole population carries at least one of the ten mutations
> 
> I think its is  t =(1-u)^n
> 
> Where n is the number of genes that can have a mutation in this case 10 and
> u is the mutation rate t = replications
> 
And all mutations rates are equal!  :-)

> I think I have made a mistake some where so any one who can put me right
> would be appreciated
> 
Yes you have!  This would be the time taken if there was some rule that 
mutation 1 had to go extinct first, and when it went extinct, only then 
could mutation 2 could drift to extinction.

Did Kimura show what the distribution of times to extinction was?  Under 
some conditions, it'll probably an exponential distribution, in which 
case the time for the extinction of all 10 mutants will follow a gamma 
distribution (which is also a Chi-squared distribution).  Check the 
derivations of a gamma, and you should see where it's coming from.

So, the answer is that under the assumtion that the time to extinction 
of one mutant is exponentially distributed with mean u, the mean time to 
extinction of n muatants is un.

Bob

-- 
Bob O'Hara

Dept. of Mathematics and Statistics
P.O. Box 68 (Gustaf H„llstr”min katu 2b)
FIN-00014 University of Helsinki
Finland

Telephone: +358-9-191 51479
Mobile: +358 50 599 0540
Fax:  +358-9-191 51400
WWW:  http://www.RNI.Helsinki.FI/~boh/
Journal of Negative Results - EEB: http://www.jnr-eeb.org
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