Spectral Library and Data processing

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Spectral Library and Data processing damlaaygun09  2024-02-12 04:46
 

Hi,

I'm trying to use Skyline version 23.1.0.380 for my thermo.raw proteomics datas. I have 2 diseased and 1 healthy data, and each one has 3 fragmentation. I want to get group comparision result LFQ and also need shared and different proteins between subjects. I watched tutorials from YouTube (MQSS 2023 - Dmitry Alexeev). I solved many problem but some points are still unknown for spectral peptide library. I have downloaded from Peptide Atlas and NIST but software is not accept these file types. I'm not sure about am I wrong when I download the library? The other problem is when I can get result for group comparision I see the adjust p-value as 1.00 so I don't know what is the actually correct steps for my aim. Can you help me?Thank you!

Damla

 
 
Nick Shulman responded:  2024-02-12 07:02
I do not understand your question but if you send us your Skyline document we might be able to figure out what is going wrong.
In Skyline you can use the menu item:
File > Share
to create a .zip file containing your Skyline document and supporting files including extracted chromatograms.
If that .zip file is less than 50MB you can attach it to this support request.
You can upload larger files here:
https://skyline.ms/files.url

It sounds like you are also having trouble getting your NIST library to work.
The way to use a .msp file in Skyline is to go to:
Settings > Peptide Settings > Library
tab and then push the "Edit List" button and add the library.
After you have added a NIST library in that way, you can explorer it with the "View > Spectral Libraries" menu item.
-- Nick
 
damlaaygun09 responded:  2024-02-13 11:35
I attached documents about what I mean. Thank you!
 
Nick Shulman responded:  2024-02-13 12:45
Thank you for sending your Skyline document.
The reason that the p-values are all 1.0 and the confidence intervals are all "NaN" (not a number) is that Skyline thinks that you only have one sample in each cohort, so Skyline cannot make any assumptions about the observed variance.

In your Group Comparison settings, I see that the "Control Group Annotation" is set to "Subject ID" and the "Identity Annotation" is also set to "Subject ID". With the settings set that way, you will definitely get p-values of 1.

The "Identity Annotation" setting is used when you have technical replicates-- that is, when you have measured samples from the same individual more than once. It's also only useful if you also have other measurements that came from different individuals because otherwise, as you see, the p-value ends up being 1.0 if Skyline thinks you only have measurements from one individual.

The reason that you are supposed to use the "Identity Annotation" to tell Skyline about your technical replicates is that you would expect that the observed variance between your technical replicates will be smaller than the variance between replicates that came from different individuals. If Skyline did not know that some of the replicates came from the same individual, Skyline might erroneously conclude that there was a statistically different measurement between your two cohorts, when, really all that is going on is that there is variation between individuals in general, and it looks like a cohort difference because the same individual was measured multiple times within each cohort.

-- Nick
 
damlaaygun09 responded:  2024-02-13 23:13
Yes I have 3 replicates of 3 subjects. I tried every combination for Group Comparision settings that attached some of them but p-value is not less than 0,05.

I want to see both individually diffirences from healthy sample and which proteins are shared. So which steps I should do?
 
Nick Shulman responded:  2024-02-14 03:24
You should have the following settings:
Control group annotation: Condition
Control group value: Healthy
Value to compare against: Diseased
Identity annotation: SubjectId

If you do that then you will get some p-values, but none of your adjusted p-values will be less than 0.05 (the lowest adjusted p-value will be 0.23).
-- Nick
 
damlaaygun09 responded:  2024-02-14 04:07
Thank you for your answer. I did it like you said but if it's not less than 0.05, it's not significant. Am I wrong?