CH908, Problem Set 9. Electron Capture Dissociation (ECD) of Peptides

CH908, Problem set 9. Electron Capture Dissociation (ECD) of peptides

1. Calculate the ECD fragments that would be expected from the following peptide sequence: RGHTTPSDELWIVK. What can you say about expected relative cleavage intensity? Create a table of the expected fragments and mark them with "strong, medium, weak" for expected intensities.

Relative fragmentation efficiency will be pretty even, with the exception of a lack of cleavage at proline - because it's ECD - and low abundance at glycine.

Because the n-terminus has arginine, and the c-terminus has lysine, the neutralized proton will be primarily the one on the c-terminal lysine, so n-terminal fragments will dominate. Thus you'll see mostly c-ions, with only a few z-ions.

Also, because of the many hydrogen bonds in this system (many -OH groups and acids to bond with the charge sites), the fragment intensity for a 2+ ion will be low unless a bit of heating (activated ion ECD) is used to break the hydrogen bonds - and even then, formation of c• and z (even electron) fragment ions will be readily observable.

There's a reported, slight tendency for higher cleavage at tryptophan, but it's barely noticeable compared to the proline effect in CAD.

2. Calculate the ECD fragments that would be expected from the following peptide sequence: RGHTpTPpSDELpYIVK. Note: pS, pT, and pY refer to phosphoserine, phosphothreonine, and phosphotyrosine, respectively. What can you say about expected relative cleavage intensity? Create a table of the expected fragments and mark them with "strong, medium, weak" for expected intensities.

Same fragments as above. However, the phosphates will primarily remain intact on the peptide rather than being cleaved as they would in CAD.

There will be a strong charge state dependence of fragmentation in this case because of tendency of phosphate groups to be negatively charged. It will also often be difficult to observe significant positive ion signal for this species for the same reason.

3. ECD can be used to distinguish leucine and isoleucine. How? Draw a probable radical rearrangement that could distinguish these two. Under what conditions?

Isoleucine:

Leucine:

This will occur under slightly more energetic conditions than normal ECD, with electrons in the 10 eV range, rather than the usual <1 eV range. This experiment is known as "hot" ECD or HECD.

4. In the following paper, a novel hemoglobin variant was discovered using electron capture dissociation. Which one? How?

J. P. Williams, A. J. Creese, D. R. Roper, B. N. Green, H. J. Cooper, Hot Electron Capture Dissociation Distinguishes Leucine from Isoleucine in a Novel Hemoglobin Variant, Hb Askew, beta 54(D5)Val -> Ile. J. Am. Soc. Mass Spectrom. 2009, 20. 1707-1713, DOI: 10.1016/j.jasms.2009.05.002.

Doh! The answer is in the title. Valine54 to Isoleucine or V54I.

It was detected using ECD because it's the only current method that can distinguish leucine and isoleucine.
5. In the following paper, variants of Amyloid Beta (Aβ) were generated that contained a mixture of Aspartic acid and IsoAspartic acid at particular residues. How were they generated? Which residues? How was this tested? Were the results conclusive?

N. P. Sargaeva, C. Lin, P. B. O'Connor, Identification of Aspartic and Isoaspartic Acid Residues in Amyloid beta Peptides, Including A beta 1-42, Using Electron-Ion Reactions. Anal. Chem. 2009, 81. 9778-9786, DOI: 10.1021/ac901677t.

Note: these papers is also posted on the CH908 website on the recommended texts section.

Doh! Another mistake, the variants were aspartic acid and isoaspartic acid, not leucine and isoleucine.

Ms. Sargaeva used a couple of commercially available variants, D7N and D23N which are known to correlate with increased risk of Alzheimer's disease. Aging these samples converted the asparagines back to aspartic acid, but as a mixture of two isoforms, normal aspartic acid (Asp), and beta-linked aspartic acid (called isoaspartic acid, isoAsp).

ECD clearly distinguishes Asp and isoAsp because of cleavage at the cα-cβ bond, which results in two diagnostic fragments, c+57 and z-57. They are definitely conclusive, although one was a bit less convincing from the intensities.