🚀 go-pugleaf

In article <3noda2$evi@agate.berkeley.edu>,
Louis Hom <lhom@OCF.Berkeley.EDU> wrote:
>So, I used a binomial distribution to describe the populations of mutants in
>my system (I'll explain below) and during <gak> my practice prelim someone
>asked me why I used the binomial distribution rather than some other one, say,
>a Poisson, or why I didn't expect a Gaussian . . . and I didn't know.  So I'm
>hoping you folks can help me understand/figure it out.
>       The system is this:  I have a DNA sequence of 180 bases that I've put
>into a virus.  When this virus infects a cell, it mutates 1/10^5 bases.  So
>basically, it's a linear array of 180 positions, with a 10^-5 chance of 
>mutation at each spot (and 1-10^5 of it not mutating), and what I want to do
>is predict the distribution of infected cells that get no mutations, one
>mutation, two mutations . . . up to 180 mutations.  

Let p be the probability of mutation of an individual base.  Then, if the
mutation of each base is independent of the mutation of the other bases,
the probability of exactly k mutations in n bases is

            k      n-k
    C(n,k) p  (1-p)

In your case p = 10^-5 and n = 180.  This is just because there are C(n,k)
choices of the k bases to be mutated and for each choice of the mutated
bases, there is a p^k (1-p)^(n-k) probability of its occurrence.

Am I missing something?

Rob Johnson
Apple Computer, Inc.
rjohnson@apple.com