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How do I compute the +3 charge state m/z for a peptide of 3751.0 Da?

Asked 22 Jul 2024Modified 20 months agoViewed 14k times
8

Setup, so nobody has to ask: +3 · 3751.0 Da.

The units are where I keep going wrong, so please be explicit about them.

I have sanity-checked the order of magnitude and it seems right, which is not the same as being right.

How many significant figures are actually justified here?

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askedines_delacruz16k1622 Jul 2024

5 Answers

Accepted answer first, then by votes
119

Accepted answer

m/z = 1251.34 at 3+. Electrospray charges a peptide by adding protons, so the observed ion is the neutral mass plus 3 protons, all divided by the charge: (3751 + 3 × 1.00728) ÷ 3 = 3754.022 ÷ 3 = 1251.34. The proton term is the one people drop, and because it is z protons over z charges it shifts m/z by 1.007 at every charge state — small, and far larger than the mass accuracy of the instrument. The neighbouring charge state sits at 938.76, and seeing the two of them where they belong is better identity evidence than either one alone. Use the average mass against an average-mass calculation and the monoisotopic mass against a monoisotopic one; mixing them costs you a couple of daltons on a peptide this size.

Start from what electrospray ionisation does: it ionises the peptide without fragmenting it, creating singly or multiply charged species that the mass analyser then separates by their mass-to-charge ratio.

A mass shift of minus eighteen usually means dehydration or a succinimide intermediate, which is pH-dependent and can be reversible.

Concretely, high-resolution mass spectrometry can distinguish a Lys-containing peptide from an Arg-containing peptide of similar mass because of the isotope difference.

Peptide mapping — enzymatic digestion followed by tandem mass spectrometry — can confirm the primary sequence and is the method of choice when identity is ambiguous.

One qualification: high-resolution mass spectrometry gives high mass accuracy but low speed, and the reverse is true for low-resolution instruments.

Always run a blank between samples and check for carry-over.

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answered · acceptedorla_ferriter89k1489 Oct 2024
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46

It helps to be literal here: two ions with the same nominal mass but different molecular formulae have different exact masses, and only high-resolution mass spectrometry can distinguish them.

A monoisotopic mass includes only the lightest isotope of each element, while the average mass weights by natural isotope abundance, and small peptides use monoisotopic mass.

More usefully, a mass shift of plus one usually means deamidation at asparagine or glutamine, which creates a secondary amine instead of an amide and changes the mass by exactly one.

Electrospray ionisation soft-ionisation behaviour is well-characterised and standards exist for m/z calibration and mass accuracy assessment.

The practical summary: use mass spectrometry for identity, not for purity.

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answeredmarta_okonkwo190k25821 Oct 2024
8Which wavelength was the purity integrated at? It changes the number more than people think. – Dr_Aoife_Brennan 2 months ago
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37

Source contamination and carry-over between samples are the failure modes most specific to mass spectrometry, and they are invisible without a blank injection between samples.

Electrospray ionisation creates multiple charge states of the same peptide — a 4 kDa peptide might appear at +2, +3 and +4 — and all of them must be accounted for in the spectrum.

The m/z accuracy achievable depends on the mass analyser type — quadrupole gives low accuracy, time-of-flight gives moderate accuracy, and Orbitrap gives high accuracy.

False positives from contamination are common in mass spectrometry work, and running a blank between every sample and a solvent background are standard practice.

A correct mass is necessary for identity but not sufficient — you also need the chromatography to confirm it.

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M4
answeredmz_4113101k3581 Nov 2024
4The system-suitability data is the part that tells you whether to believe the rest. – Dr_Nadia_Farsi 6 days ago
3I would gently push back on the second point — inter-laboratory spread is wider than stated. – rhian_prydderch 8 months ago
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29

A mass shift of exactly zero with a shifted retention time points to an isomer — a scrambled disulfide or a racemised residue — which mass spectrometry alone cannot identify.

The charge state distribution depends on the solution pH, the structure of the peptide and the source conditions, so the same peptide can look different under different conditions.

The caveat is that a correct mass does not mean the peak is correct — isomers and co-eluting species can have the same m/z.

If you only pay for one test, pay for quantified content. Purity is the number everyone quotes and content is the number that changes what you do.

edited 20 Nov 2024 by tandem_gradient — corrected a unit error in the worked example

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TG
answeredtandem_gradient61k24812 Nov 2024
26

Coupling HPLC to a mass spectrometer adds identity information to the chromatographic separation, but the mass spectrometer's ionisation conditions can distort the HPLC peak shape.

For a large peptide with multiple peaks in the mass spectrum, comparing the observed isotope pattern to the calculated pattern is a quick check that the formula matches.

The limitation is that mass spectrometry tells you the mass and almost nothing else, so it needs to be paired with chromatography or other identity information.

In practice: ask for the chromatogram, check the method section, check the lot number against the vial, and set your accept threshold before you see the result rather than after.

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LM
answeredlucia_marchetti19k2726 Aug 2024
8Small correction: the limit of quantitation, not the limit of detection, is the relevant one there. – laminar_bench 5 months ago
7Worth adding that the method section is where the answer usually is. – Dr_Wren_Halliday 3 months ago
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