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How do I compute the +5 charge state m/z for a peptide of 4813.5 Da?

Asked 4 Oct 2024Modified 19 months agoViewed 51k times
27

The case in front of me: +5 · 4813.5 Da.

I want the working, not the result — I need to be able to redo it with different numbers.

I care about the precision as well as the value — I want to know how many figures are real.

How many significant figures are actually justified here?

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Mass spectrometry for identity confirmation: electrospray ionisation, multiple charge states, monoisotopic versus average mass, deconvolution, and…

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LF
askedleah_ferrers12k164 Oct 2024

5 Answers

Accepted answer first, then by votes
89

Accepted answer

m/z = 963.71 at 5+. Electrospray charges a peptide by adding protons, so the observed ion is the neutral mass plus 5 protons, all divided by the charge: (4813.5 + 5 × 1.00728) ÷ 5 = 4818.536 ÷ 5 = 963.71. 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 803.26, 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.

More usefully, the single most important fact about mass spectrometry for peptides is that it measures only the molecular weight and tells you almost nothing about whether the peak is actually your target.

The baseline noise on a mass spectrum sets the limit of detection, and a weak signal close to the noise is not reliable evidence for the presence of a species.

What each test answers

TestAnswersDoes NOT answer
RP-HPLC, area %What fraction of detected material is the targetHow much target is present
Quantified contentMilligrams of peptide per vialWhat the impurities are
ESI-MS identityWhether the molecular weight matchesPurity, or isomeric substitution
Peptide mappingSequence, localised to a fragmentQuantity
Karl FischerWater content of the solidSolvent content
LAL endotoxinPyrogen load in EU/mgSterility
Sterility testGrowth in defined media over 14 daysEndotoxin, or bioburden count

Mechanically, deconvolution of a mass spectrum with multiple charge states produces a reconstructed neutral mass, and errors in the deconvolution produce errors in the inferred mass.

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.

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

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TG
answered · acceptedtandem_gradient61k24825 Nov 2024
5Do you have the chromatogram for this, or just the summary figure? – Dr_Priya_Raghunathan 2 months ago
6Which wavelength was the purity integrated at? It changes the number more than people think. – thermal_mass 4 months ago
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34

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.

Stated carefully, 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.

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

Worth noting that source contamination is common and silent, so a result that looks too good to be true often is.

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

edited 30 Dec 2024 by mz_4113 — added the placebo-arm figures

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M4
answeredmz_4113101k3587 Dec 2024
25

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.

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

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

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.

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

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AL
answereda_lindgren58k2483 Nov 2024
4Small correction: the limit of quantitation, not the limit of detection, is the relevant one there. – Dr_Ravi_Selvarajah 7 months ago
5Adding for future readers: the certificate should carry the lot number, not just a batch code. – h_pergande 8 months ago
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20

The underlying point is that tandem mass spectrometry fragments the ions and measures the fragment masses, which provides sequence information and is the best tool for confirming identity.

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.

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

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.

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LW
answeredlinnea_wahlberg17k2712 Oct 2024
19

Stated carefully, identity confirmation from mass spectrometry means matching the observed m/z to the calculated m/z for your peptide at its known charge states.

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.

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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DR
answeredDr_Priya_Raghunathan49k13715 Nov 2024
2The impurity table is the part I now read first, and this explains why. – retest_please 3 months ago
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