Question

If you have a 90%/10% mix of two sequences in a PCR product how many colonies will you need to sequence to have a >95% chance of sequencing at least one copy of each?

Answer 1

The success chance when using single colony is 0.1 and the failure chance is 0.9.

The probability should be lower than 0.05

So, Log10(0.9) < log10(0.05)

log10(0.9) is negative So the inequality sign could be reversed.e. n>log10(0.05)/log10(0.9)

n >28.4

0.9^n = 0.05

(n*log(0.9) = log(0.05) <-> n = log(0.05)/log(0.9) = 28.4).

At least 29 colonies have to be sequenced to obtain to have a >95% chance of sequencing at least one copy of each.

log(0.05) = -1.30102999566

log(0.9 = -0.04575749056

log(0.05)/log(0.9) = -1.30102999566/-0.04575749056 =28.4331588061