<blockquote>
<p>With the data attached above, would my N50 and N90 length in BASE PAIRS so excluding the kb, be 7464 and 14314, respectively?</p>
</blockquote>
<p>Looks like this is inverse so your N50 is 1032bp and your N90 is 106bp. N90 cannot be longer than N50 by definition. You L50 and L90 are 7464 and 14314</p>
<blockquote>
<p>Would the number of contigs produced by this assembly be 7, since there are 7 different contigs produced or would it be 25,534?</p>
</blockquote>
<p>25534</p>
<p>Now for the real answer. <strong>Do not calculate or report N50 on transcript assemblies. It is more than useless.</strong> Transcriptomes are - by nature - dynamic in length. Longer is not better in this regard. Are your short transcripts lower quality assemblies, probably not, but N50 treats them as such. Please read the <a href="https://github.com/trinityrnaseq/trinityrnaseq/wiki/Transcriptome-Assembly-Quality-Assessment" rel="nofollow noreferrer">Trinity wiki</a> for other, better ways to QC your transcript assembly.</p>

CC BY-SA 4.0
2020-11-12T23:22:53.760Z
Hi @BastianSchiffthaler, thank you for your help! Ok so basically, one is the count and one is the length. Therefore, would my N50 length be 1032bp or is that my count? Then would 7464 be my length?
AlphaQueUp
CC BY-SA 4.0
2020-11-12T23:50:18.333Z
Also, say if my N50 length is base pairs was 1032 and I wanted to find out "how many contigs are as long or longer than then the length of 1032; would this simply just be 25,534 - 1032 = 24,502? Thanks! @Bastian Schiffthaler
AlphaQueUp
CC BY-SA 4.0
2020-11-13T00:03:31.750Z
Let's say you order your contigs from longest to shortest. After that, you calculate the sum of all contig lengths, i.e. the total length of your assembly. Now you go from the longest contig to the shortest until the cumulative sum reaches >=50% of the total length. The length of the contig at that position is the N50, the number of contigs you had to include to reach 50% is the L50. N50 = length = 1032bp. L50 = count = 7464. 25534 − 7464 = 18070 contigs are _shorter_ than your N50.
Bastian Schiffthaler
CC BY-SA 4.0
2020-11-13T03:02:34.437Z
So then that would simply just mean that 7464 contigs are as long or longer than my N50 @BastianSchiffthaler and 14314 for N90? I apologize for the repetitive questions. I am just starting to learn bioinformatics, so this is all a new language to me. I appreciate all your help!
AlphaQueUp
CC BY-SA 4.0
2020-11-13T15:41:55.140Z
Yes. That's correct
Bastian Schiffthaler
CC BY-SA 4.0
2020-11-14T03:07:17.473Z
Thank you very much! @BastianSchiffthaler
AlphaQueUp
CC BY-SA 4.0
2020-11-14T22:01:06.373Z
Hi @BastianSchiffthaler, I was wondering if I could get your advice for one of my commands? Thanks!
AlphaQueUp
CC BY-SA 4.0
2020-11-14T22:01:11.187Z
Hi @Bastian Schiffthaler, I was wondering if I could get your advice for one of my commands? Thanks!
AlphaQueUp
CC BY-SA 4.0
2020-11-15T00:59:40.417Z
Please open a separate question.
Bastian Schiffthaler
CC BY-SA 4.0
2020-11-12T20:09:30.733Z
2020-11-12T20:09:30.733Z
Bastian Schiffthaler
<p>I am having trouble with some underlying questions about my project.</p>
<p>I have ran a Trinity tool to create a trinity.fasta file.</p>
<p>To determine some underlying questions, I used the utility asm_stats to gather some basic statistics.</p>
<p>I am just having some issues of calculating data.</p>
<p>DATA:</p>
<p>Main genome scaffold N/L50: 7464/1.032 KB<br />
Main genome contig N/L50: 7464/1.032 KB<br />
Main genome scaffold N/L90: 14314/106<br />
Main genome contig N/L90: 14314/106</p>
<p>Number of contigs (scaffold length):</p>
<p>25,534 (ALL)<br />
25,534 (100)<br />
21,609 (250)<br />
10,818 (500)<br />
907 (1 KB)<br />
38 (2.5KB)<br />
5 (5 KB)<br />
2 (10KB)</p>
<p>With the data attached above, would my N50 and N90 length in <strong>BASE PAIRS</strong> so excluding the kb, be 7464 and 14314, respectively?</p>
<p><em>Would the number of contigs produced by this assembly</em> be 7, since there are 7 different contigs produced or would it be 25,534?</p>
<p>Thanks!</p>

<rna-seq><fasta><fastq><file-formats>
How to find the number of contigs produced? N50 length in base pairs? N90 length in base pairs?
# How to find the number of contigs produced? N50 length in base pairs? N90 length in base pairs?

(Asked by: username_0 on 2020-11-12)

I am having trouble with some underlying questions about my project.

I have ran a Trinity tool to create a trinity.fasta file.

To determine some underlying questions, I used the utility asm\_stats to gather some basic statistics.

I am just having some issues of calculating data.

DATA:

Main genome scaffold N/L50: 7464/1.032 KB  
Main genome contig N/L50: 7464/1.032 KB  
Main genome scaffold N/L90: 14314/106  
Main genome contig N/L90: 14314/106

Number of contigs (scaffold length):

25,534 (ALL)  
25,534 (100)  
21,609 (250)  
10,818 (500)  
907 (1 KB)  
38 (2.5KB)  
5 (5 KB)  
2 (10KB)

With the data attached above, would my N50 and N90 length in **BASE PAIRS** so excluding the kb, be 7464 and 14314, respectively?

*Would the number of contigs produced by this assembly* be 7, since there are 7 different contigs produced or would it be 25,534?

Thanks!

**Tags:** rna-seq, fasta, fastq, file-formats

---

## 1 answers

### Answer by username_1 on 2020-11-12. Score: 1 | Accepted answer

> With the data attached above, would my N50 and N90 length in BASE PAIRS so excluding the kb, be 7464 and 14314, respectively?

Looks like this is inverse so your N50 is 1032bp and your N90 is 106bp. N90 cannot be longer than N50 by definition. You L50 and L90 are 7464 and 14314

> Would the number of contigs produced by this assembly be 7, since there are 7 different contigs produced or would it be 25,534?

25534

Now for the real answer. **Do not calculate or report N50 on transcript assemblies. It is more than useless.** Transcriptomes are - by nature - dynamic in length. Longer is not better in this regard. Are your short transcripts lower quality assemblies, probably not, but N50 treats them as such. Please read the [Trinity wiki](https://github.com/trinityrnaseq/trinityrnaseq/wiki/Transcriptome-Assembly-Quality-Assessment) for other, better ways to QC your transcript assembly.

> **Comments:**
> **username_0**: Hi @BastianSchiffthaler, thank you for your help! Ok so basically, one is the count and one is the length. Therefore, would my N50 length be 1032bp or is that my count? Then would 7464 be my length?  
> **username_0**: Also, say if my N50 length is base pairs was 1032 and I wanted to find out "how many contigs are as long or longer than then the length of 1032; would this simply just be 25,534 - 1032 = 24,502? Thanks! @Bastian Schiffthaler  
> **username_1**: Let's say you order your contigs from longest to shortest. After that, you calculate the sum of all contig lengths, i.e. the total length of your assembly. Now you go from the longest contig to the shortest until the cumulative sum reaches >=50% of the total length. The length of the contig at that position is the N50, the number of contigs you had to include to reach 50% is the L50. N50 = length = 1032bp. L50 = count = 7464. 25534 − 7464 = 18070 contigs are _shorter_ than your N50.  
> **username_0**: So then that would simply just mean that 7464 contigs are as long or longer than my N50 @BastianSchiffthaler and 14314 for N90? I apologize for the repetitive questions. I am just starting to learn bioinformatics, so this is all a new language to me. I appreciate all your help!  
> **username_1**: Yes. That's correct  
> **username_0**: Thank you very much! @BastianSchiffthaler  
> **username_0**: Hi @BastianSchiffthaler, I was wondering if I could get your advice for one of my commands? Thanks!  
> **username_0**: Hi @Bastian Schiffthaler, I was wondering if I could get your advice for one of my commands? Thanks!  
> **username_1**: Please open a separate question.  

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