Question

Mutation rate using fluctuation testing and the Poisson distribution. [Genetics]

Answer 1

Okay so I like to determine the mutation rate for a specific gene and has determined to use fluctuation testing.

Terminology:
Mutation rate: the number of mutations per celldivision
Mutation frequency: the number of mutations per cells.

The Luria–Delbrück experiment showed that mutation frequency very much toke form as a Poisson distribution. This conclusion was among others on the sample mean and sample variance had a 1:1 ratio.

If we choose to assume that the number of celldivisions is equal to the total amount of cells (I consider it a fair approximation since the number is sky high), and that the number of mutations are given from the sample mean, we get to write:\[\Large \mu=\frac{ \bar{k} }{ N }\]Here is \(\mu\) the rate, \(k\) the sample mean and \(N\) the amount of cells (the amount of celldivisions).

From here I've set up two experimental procedures and only one is to be done:

The practical:
In this version, the sample mean need to be obtained directly from data. From 20 tubes we take 1 ml and terfere it to a petri dish with a medium selective over for the gene I wish to test. After some time (1 celldivision) we see how many mutations that have happened and calculate the sample mean\[\Large \bar{k}=\frac{ \sum_{i=1}^{n} k _{i} }{ n }\]


The partly theoretical:
The Poisson distribution has the equation.
\[\Large P _{x}=\frac{ \bar{k} ^{x} \times e ^{-\bar{k}} }{ x! }\]We can see up the nul frequency and determine the sample mean:\[\Large P_{0}=\frac{ \bar{k}^{0} \times e ^{- \bar{k}} }{ 0! }=\frac{ 1 \times e ^{- \bar{k}} }{ 1 }=e ^{- \bar{k}}\]The sample mean is then:\[\Large \bar{k}=-\ln(P _{0})\]
So all we gotta do is a count if there is 0 mutants or not on our petri dish, and from there calculate the sample mean.


Please help me evaluate the pros and cons.

@shrutipande9 @thomaster @blues @aaronq

Answer 2

Correction

And I like to add that like in the practical one we again do 20 dishes, but just evaluate, mutants or no mutants.