i am building a mutation library in Mycobacterium, and after i collected the colonies of the library, the total CFU will be more than 10^9 CFU. there are 4000 genes in the genome, so i can say there are 4000 kinds of mutation in the library.
i have to take 100 CFU every time to infect the fish, but the issue is i do not know how many times i have to take to have a 95% chance to cover all the mutation kinds.
i got this formula from a friend but i do not know whether it is right, can you guys help me double check it?
the X represents the times that 100CFU are taken from the whole library.
P represents the chance.
So, you want to know the minimum value of x for P to be greater than or equal to 0.95, right?
According to the formula, if you sampled the population just once (i.e. x were 1), P would be (3999/4000)^(100*1) = 0.975, which is obviously wrong, as there are only 100 cfu in the sample. Or am I misunderstanding something?
Assuming that the probability of having a mutation is the same for all genes, that there is only one type of mutation for each gene, and that each individual of the population carries exactly one mutation in one random gene, I think the number of cfu to screen in order to cover the entire mutational 'universe' with 95% probability can be calculated with the same formula used to determine the number of primary clones to screen in a genomic library: N=[ln(1-P)]/[ln(1-(f))] (where P is the coverage probability and f is the frequency of each event, or in that case, average insert length divided by total genome size).
In your case, it would be N=[ln(1-0.95)]/[ln(1-(1/4000))] which is 11981 cfu, that is, 120 samples of 100 cfu each.
The derivation of that formula appears in an old paper by Brian Seed (PMID 6293929)
Hope it helps!
- Badges
- Science topic
- Similar topics
- Biological Science
- Genetics
- Genetic Analysis