Thread: Mass Spec: how do I calculate neutral molecular masses?

1. How do I calculate the neutral molecular masses of proteins from their m/z value?

for example, 810.5 m/z

2.

3. Originally Posted by TAMB0
How do I calculate the neutral molecular masses of proteins from their m/z value?

for example, 810.5 m/z
You need to search in the literature for a fragmentation library that matches the peak location aka 810.5m/z and use it to derive the number of fragments. When you have the number of fragments you have the number of charges and you have everything. In other words, your life is ****ed.

4. maybe this will help.. im very lost.

Deconvolution of protein ESI mass spectra
The accompanying figure shows the positive ion electrospray mass spectrum of aserum protein acquired with a quadrupole instrument (all masses are averagevalues). This protein was isolated by immunoprecipitation followed by reversedphase HPLC. The spectrum shows a distribution of charge states typical for aprotein molecule. There are a group of peaks for each charge state indicating that amixture of variants was immunoprecipitated. The inset shows the expanded massrange for one of the charge states. Please calculate the neutral molecular masses ofthe five protein variants shown in the inset (
m/z 810.5, 812.3, 815.1, 817.4, 819.4).This will entail determining the charge state of the masses in the expanded regionand then calculating the neutral protein masses accordingly. Please note, you are
calculating the average neutral mass of the protein.

untitled.JPG

5. The equation is

m/z = [(M+n(1.00794)]/n

I am having a hell of a time figuring out what "n" is.

HELP!

6. Hi,

I think I'm in your class. So what I did is to solve for n by using the highest peak of the protein at each charge state.

If m/z = [M+n(1.00794)]/n

Plug a peak (819.4) into this equation as m/z

You then get 819.4 = (m+1.00794n)/n

Solve for m in terms of n, m=818.392n

Plug this value into another equation using another peak. I chose the one to the left of it having a value of 773.8. I also assumed that this peak corresponded to something with a charge of one greater than the previous peak because its m/z is the next less on the chromatogram.

So using the same equation, you have: 773.8 = [818.392n + (n+1)H]/(n+1)
Solve for n
n~=17 (16.95)

So then I used this charge value to calculate the neutral charge of the protein by just plugging in 17 for n with the m/z values: 810.5, 812.3, 815.1, 817.4, 819.4

I ended up getting a range looking like:
 13761.5 13792.1 13839.7 13878.8 13912.8

I'm not sure if this is right, but its the best I have now!

Cheers,

Michael (the guy with the bike)

7. Yeah, I think i saw u on linkedin too. Thanks for the help. Mass spec always gave me trouble.

8. howd u do on 1b? i got peak 812.3 m/z

 Bookmarks
Bookmarks
 Posting Permissions
 You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement