<p>I actually disagree with that.. I guess I write this down as a discussion.</p>
<p>The expected value for an average gene is 1200. For this gene at 1/1200 expression, you expect 1 read.</p>
<p>However, because of sampling, sometimes you get 0, 1 or 2. If we assume the sampling follows a poisson distribution, we can calculate the probability of getting 0, 1 , 2 ... reads:</p>
<pre><code>plot(dpois(0:10,lambda=1),ylab=&quot;Probability&quot;,xlab=&quot;No of reads&quot;)
</code></pre>
<p><a href="https://i.sstatic.net/R3q8V.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/R3q8V.png" alt="enter image description here" /></a></p>
<p>The probabilities are like:</p>
<pre><code>data.frame(n=0:5,p=dpois(0:5,lambda=1))

  n           p
1 0 0.367879441
2 1 0.367879441
3 2 0.183939721
4 3 0.061313240
5 4 0.015328310
6 5 0.003065662
</code></pre>
<p>The probability of getting exactly 0 or 1 read should be ~0.37. The probability of getting at least 1 read would be 1 - P(n=0) = 1- 0.367879441 = 0.6321206</p>

CC BY-SA 4.0
2020-11-19T14:20:05.340Z
I guess the author just assumes uniform distribution instead of poisson.
haci
CC BY-SA 4.0
2020-11-19T14:20:38.690Z
possible.. but why 50:50 ?
StupidWolf
CC BY-SA 4.0
2020-11-19T15:47:46.373Z
I was curious about this question but didn't know what the distribution would be, so I left a very clumsy monte-carlo simulation running yesterday.  After about 8,500 iterations I get exactly these values.
Jesse
CC BY-SA 4.0
2020-11-19T15:57:52.867Z
One other thing: Could the author have taken the probability determined for the lambda=1200 value of getting >=1200 (about 50%) and incorrectly assumed it'd stay the same for lambda=1?
Jesse
CC BY-SA 4.0
2020-11-19T19:29:41.343Z
There are a lot of back-of-the-envelope calculations in the original statement. I don't think 50:50 is supposed to be an exact number. The point is that you cannot be very confident in detecting that gene.
burger
CC BY-SA 4.0
2020-11-19T14:06:02.783Z
2020-11-19T14:06:02.783Z
StupidWolf
<p>I am reading a book about RNAseq analysis and it says</p>
<blockquote>
<p>&quot;To calculate the probability that a read will map to a specific gene, we can assume an average gene size of 4000 nt (100 M nt divided by 25,000 genes). At 30 M reads equivalent to 30× coverage, at single read 100 nt (or paired-end read 50 nt) length, we can expect a single read to map to the average expressed and length gene 4000 nt× 30 coverage/100 nt 1200 times. Thus, if the gene is expressed at a level of 1/1200 compared to the average gene, then we have a 50:50 probability to have a read map to it.&quot;</p>
</blockquote>
<p>I understand that a single read would map to the average expressed 1200 times but I don't understand how the 50:50 probability is calculated. Please could you explain how this probability is determined.</p>

<rna-seq><reads><coverage>
If a gene is expressed at a level of 1/1200 compared to the average gene, how is probability 50:50 that we have a read mapped to it?
# If a gene is expressed at a level of 1/1200 compared to the average gene, how is probability 50:50 that we have a read mapped to it?

(Asked by: username_0 on 2020-11-18)

I am reading a book about RNAseq analysis and it says

> "To calculate the probability that a read will map to a specific gene, we can assume an average gene size of 4000 nt (100 M nt divided by 25,000 genes). At 30 M reads equivalent to 30× coverage, at single read 100 nt (or paired-end read 50 nt) length, we can expect a single read to map to the average expressed and length gene 4000 nt× 30 coverage/100 nt 1200 times. Thus, if the gene is expressed at a level of 1/1200 compared to the average gene, then we have a 50:50 probability to have a read map to it."

I understand that a single read would map to the average expressed 1200 times but I don't understand how the 50:50 probability is calculated. Please could you explain how this probability is determined.

**Tags:** rna-seq, reads, coverage

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## 1 answers

### Answer by username_1 on 2020-11-19. Score: 3

I actually disagree with that.. I guess I write this down as a discussion.
The expected value for an average gene is 1200. For this gene at 1/1200 expression, you expect 1 read.
However, because of sampling, sometimes you get 0, 1 or 2. If we assume the sampling follows a poisson distribution, we can calculate the probability of getting 0, 1 , 2 ... reads:
plot(dpois(0:10,lambda=1),ylab="Probability",xlab="No of reads")


The probabilities are like:
data.frame(n=0:5,p=dpois(0:5,lambda=1))

  n           p
1 0 0.367879441
2 1 0.367879441
3 2 0.183939721
4 3 0.061313240
5 4 0.015328310
6 5 0.003065662

The probability of getting exactly 0 or 1 read should be ~0.37. The probability of getting at least 1 read would be 1 - P(n=0) = 1- 0.367879441 = 0.6321206

> **Comments:**
> **username_2**: I guess the author just assumes uniform distribution instead of poisson.  
> **username_1**: possible.. but why 50:50 ?  
> **username_3**: I was curious about this question but didn't know what the distribution would be, so I left a very clumsy monte-carlo simulation running yesterday.  After about 8,500 iterations I get exactly these values.  
> **username_3**: One other thing: Could the author have taken the probability determined for the lambda=1200 value of getting >=1200 (about 50%) and incorrectly assumed it'd stay the same for lambda=1?  
> **username_4**: There are a lot of back-of-the-envelope calculations in the original statement. I don't think 50:50 is supposed to be an exact number. The point is that you cannot be very confident in detecting that gene.  

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