I have performed label-free phosphoproteomics and analysed the data with MaxQuant. Now I get phosphopeptides marked being doubly phsphorylated, but only one amino acid has a high probability of being phosphorylated (e.g. DSQAAEHS(0.999)PT(0.001)AAESWSIFNR). I do not understand how this can happen. Can anybody help?
You have to understand binomial distributions. If there is more than one phosphorylation site, then the average number of phopshorylations is determined by the combined independent probabilities of the n sites that could be phosphorylated. Simple probability theory. If you can't work through the math, you can simply set up a Monte Carlo simulation of several hundred protein chains and use a random number generator to determine the independent phosphorylation events at each possible phosphorylation site on each chain, then average over all the chains.
Thank you Luke! But in the above example, the combined probability is 1, right? Also, I have another for me strange example:
ETCTGIQFS(1)PCSQTNR, number of phosphosites 1; 2, multiplicity double. How can I understand this?
All assistance is very much appreciated!
Dear Chandra Ramakrishnan ,
When it comes to your original question I think the interpretation of the prediction/calculation as you think yourself is: the most likely site is the indicated Ser (with apparently according to their database a slight possibility for the other AA).
Your related second question has to do with the term multiplicity.
In mathematics this means according to Mathwords: How many times a particular number is a zero for a given polynomial. For example, in the polynomial function f(x) = (x – 3)4(x – 5)(x – 8)2, the zero 3 has multiplicity 4, 5 has multiplicity 1, and 8 has multiplicity 2.
In Maxquant it means (something else?) as you can find here:
http://www.coxdocs.org/doku.php?id=maxquant:manual:beginner
It states:
“Labels: The second tab is Group-specific parameters. You need to go there to specify your labels. The “Type” will usually be “Standard” (obviously), and the “Multiplicity” will be 1 for label-free quantification, 2 if you have light and heavy labels, and 3 if you have light, medium, and heavy. Any number of labels can be checked in the lists. In this example, the light sample is unlabelled and the heavy sample has been labelled with Arg10 and Lys8. (You can't see them both without scrolling.)”
PS. I am no expert of this program but I noticed an interesting discussion here on RG:
https://www.researchgate.net/post/How_To_Perform_Quantification_of_Phosphoproteomics
Further (background) information about the benefits and ‘problems’ of using these type of programs might be of interest as well:
Lee, D. C., Jones, A. R., & Hubbard, S. J. (2015). Computational phosphoproteomics: from identification to localization. Proteomics, 15(5-6), 950-963.
Good luck and best regards.
Chandra, Each site is acting independently, so the probability of any site being occupied is independent of the others. You could have 5 sites with p=0.4, which means the probability of all five sites being occupied is 0.4^5. The probability of any 4 sites being occupied is 0.4^4 and so on. The sum of all those probabilities for 0, 1, 2, 3, 4, and 5 sites must equal 1, not the sum of the site probabilities, which in my example would be 5*0.4 or 2. The average number of phosphorylations then is the number of phosphorylations times the product of the probabilities that are needed to get that number of phosphorylations. This calculation gets more difficult when the probabilities of phosphorylation at individual sites are not the same, then you have a permutation problem. Much easier to run a Monte Carlo simulation for that case than work out the numbers manually, at least for me. You can do that in a spreadsheet for a simple problem like this. I once brought a Cray I computer to it's knees (20 cpu min) doing a Monte Carlo simulation of a catalyzed polymerization reaction with catalyst exchange to get the MW distribution of the resulting polymer, but the same principles applied. The system admin was NOT happy, to make an understatement.
Dear Rob,
Thank you for your detailed answer. The problem in my case it that I have a label-free experiment, so I do not think we are looking at the same multiplicities. Or am I getting this wrong?
Thank you again!
Hi Luke,
Thank you for the answer. This makes sense. I appreciate your help!
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