Fragments and probability law correction

This post is a correction of my previous post. Before I describe calculations that my supervisor and I made to calculate the probability of a simple uptake motif (lets say a dimer AA) be in a 30bp degenerate fragment.

The calculation was:

Given 4 bases = A, C, G, T probability of having an A is 0.25 (1/4)

The probability of having two AA is 0.25*0.25 = 0.0625, or of the 16 dimer combinations, we have 1 success (1/16).

0.0625*29 base positions = 1.8125

The calculated value 1.8125 is the mean number of AA present in a 30bp random fragment (when only one strand is considered).

Before I stated that this calculation was wrong; however this calculation is in fact correct.

I verified this by calculating the mean number of AA’s based on the frequency distribution that I calculated before:

30bp freq dist one strand

I did this by: sum(AA’s per fragment * frequency per 30bp fragments).

Mean number of AA’s was 1.8125 equal to what was previously calculated.

My confusion was originated by non clearly stated what I was calculating.


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