🚀 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.  

Assuming that mutations at the different positions are independent (which may
or may not be a realistic assumption) the binomial distribution is correct:
the probability of exactly k mutations is binom(n,k) p^k q^(n-k) with
n=180, p=10^(-5) and q=1-p.  Since 180 is fairly large and 10^(-5) is small, 
the Poisson distribution with lambda = np = .0018 would be a good approximation: 
P(k) = lambda^k exp(-lambda)/k!.  Thus for k=0 (no mutations) the binomial
distribution gives .998201610 while the Poisson gives .998201619.  On the 
other hand, with good computational tools available these days, there's no
real need to use the Poisson approximation.  The Gaussian (normal) approximation
would not be appropriate in your case because np is too small.



-- 
Robert Israel                            israel@math.ubc.ca
Department of Mathematics             
University of British Columbia
Vancouver, BC, Canada V6T 1Y4