3
$\begingroup$

Suppose my genome is 3 million bases and that my reads are 100 nucleotide long. I need to know how many reads I need to cover the entire genome.

I start from using the equation $C = \frac{N \cdot L}{G}$ where C is the coverage, N the number of reads, L the length of a read and G the length of the haploid genome. I also know that the average coverage follows a Poisson distribution, so I need this for C.

On my slides, in fact, I have the following:

$$ C = \frac{N \cdot L}{G} \approx ln \left( \frac{G}{L \cdot \epsilon}\right), $$

if we assume that this is the average coverage needed to cover the entire genome with probability $1-\epsilon$.

The problem is that I do not understand how to arrive at the expression on the right. I thought that if we want to be sure to have a coverage of at least 1 on average, we would have had to compute $1 - P(X=0) = 1 - e^{-\frac{N \cdot L}{G}} = 0.99$, if we take $\epsilon=0.01$. But I do not get the same result as with the equation above.

$\endgroup$

1 Answer 1

3
$\begingroup$

Let's see what your slides are claiming $\epsilon$ to be, using:

$$ C = \frac{NL}{G} \approx \ln \left( \frac{G}{L\epsilon} \right) $$

We can rearrange to get

$$ \epsilon \approx \frac{G}{L} \times e^{-C} $$

Now, what does it mean for an entire genome to be covered? Having a clear understanding of this definition is crucial, and your misunderstanding seems to partially stem from this.

I claim it's natural to consider a genome as consisting of $G/L$ bins. In this setting, we have $N$ balls that each randomly goes into a bin. Then, the entire genome is covered when all bins contain at least $1$ ball/read.

From the Poisson distribution, we know that P(a bin has 0 balls) = $e^{-C}$.

Then, we define $\epsilon$ to be the probability that at least one of the $G/L$ bins has 0 balls. Given the way we've modeled the problem, this is exactly equal to

$$ \epsilon = 1 - (1 - e^{-C})^{G/L}. $$

From above, your slides are claiming $ \epsilon \approx \frac{G}{L} \times e^{-C}$. This might be strange, as it appears to be equal to the expected number of bins with 0 balls, instead of the probability that there is at least 1 bin with 0 balls. Furthermore, it's strange because it contains only one exponential, compared to our exact derivation which has a double exponential.

The mismatch in the number of exponential is a hint that another mathematical approximation is at play. We can use a first-order approximation of the binomial series to obtain

$$ (1-e^{-C})^{G/L} \approx 1 - (G/L)e^{-C} $$ which holds particularly when $|e^{-C}| \times \frac{G}{L} << 1$.

In your case, $\frac{G}{L} = 30,000$, so the approximation only really starts getting accurate when $C > 15$ or so.

Following through with our approximation, we get

$$ \epsilon \approx 1 - (1 - (G/L) e^{-C}) $$

$$ \epsilon \approx \frac{G}{L} \times e^{-C}. $$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.