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Thank you for looking at my question. I am trying to solve this homework question.

Consider the problem of sequencing genome by random reads. If G is the length of the entire sequence, L is the length of the read and n is the number of reads, then coverage is defined as nL/G. Now, if we want 50% of the original long sequence to be covered by at least one fragment, how much coverage do we need?

I read Lander-Waterman http://www.genetics.wustl.edu/bio5488/lecture_notes_2005/Lander.htm model to understand the concept. But didn't quite get how to solve this problem. I thought to consider the given 50% as probability and y as 1 (the one from Poisson distribution) and calculate lambda (that is the coverage). But I don't think I am on right track. I thought of considering y as 1 because the question says 50% of the original long sequence to be covered by atleast one fragment, which means that those bases are sequenced atleast once.

I may be wrong.

Experts can you guide me please.

Thank you.

Andreas
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smandape
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2 Answers2

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If you look at it as a continuous problem (as in, n is large and L is much smaller than G), and assume the placement of each read is completely random, the likelihood that each additional read is wasted is proportional to the existing coverage. This leads to the following math:

d(coverage)/d(n) = (L/G) * (1 - coverage)

As this is homework, I will leave the solution as an exercise for the reader. (I would like to point out, though, that this is a rare case of actually using calculus to solve a CS problem 8^)

Solving the above equation, starting with coverage=0 at n=0, yields:

   ln(1 - coverage) = - (L/G) * n
-> coverage = 1 - exp(- (L/G) * n)

As a reality check, note that this is what you should expect to see: if your sampling is truly random, the uncovered part of G should decay exponentially, like a radioactive element.

comingstorm
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  • Maybe my doubt is a bit silly, but are you taking differentiation of coverage with respect to n? – smandape Dec 08 '11 at 02:14
  • Its alright. Can you please elaborate it a bit more, to be honest I tried getting it but I am unable to do so. – smandape Dec 08 '11 at 02:27
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I think you're not off so far. As I read it y=1 means that the bases are read "exactly once" and not "at least once". You would need to calculate the coverage under the probability of 0.5 for P(y=1) + P(y=2) + ... which is equal to ...

Andreas
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