Finding Genome Coverage Using Random Reads
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12.8 years ago
Smandape ▴ 120

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. Maybe, I am not clear about the way to solve this.

Experts can you guide me please.

Thank you.

algorithm genome homework • 3.2k views
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I think you are having problems because the question is ill-posed. There is a probability involved. It could be the case that the reads stack by chance in one position. So I would rephrase the question like so: we want 50% of the original long sequence to be covered by at least one fragment with probability P, how much coverage do we need? or like so: what is the coverage required such that the expected value for the number of positions covered at least once is at 50% of the genome length?

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Yeah. I got it. Thank you.

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12.8 years ago

Hi,

In the assumption that the distribution of reads is Poissonian:

the probability that a base is covered by 0 fragments is exp(-lambda). Hence, the probability that it is covered by at least 1 fragment is 1-exp(-lambda)

You want this probability to be bigger or equal to 0.5, and 1-exp(-lambda)>=0.5 gives you lambda>=0.69

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Thank you. Well, I got the same answer but I was not sure if its correct.