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Mass spectrometry: identity, not purity

HPLC tells you how much of what it can see sits in the main peak. It does not tell you what the main peak is. That question belongs to mass spectrometry, or MS, which measures the mass of molecules. For a synthetic peptide the central check is simple: does the measured mass match the mass calculated from the intended sequence?

How a mass spectrometer works

Every mass spectrometer does three things: it gives molecules an electric charge, sorts the charged molecules (ions) by mass, and counts them.

Schematic

Schematic of a mass spectrometer: an ion source (electrospray) gives molecules a charge; a mass analyzer sorts the ions by mass-to-charge ratio; a detector counts them; the output is a spectrum of intensity against m/z.Four-stage flow diagram, not to scale.Ion sourceelectrospray:charges moleculesMass analyzersorts ionsby m/zDetectorcounts ionsSpectrum: intensity against m/z
Schematic. The three stages inside a mass spectrometer, and the spectrum they produce.
Text version of this figure
  1. Ion source: electrospray gives the molecules an electric charge and moves them into the gas phase.
  2. Mass analyzer: sorts the ions by their mass-to-charge ratio (m/z).
  3. Detector: counts the ions arriving at each m/z.
  4. Spectrum: a plot of ion intensity against m/z.

Why one peptide gives several peaks

In electrospray, a peptide usually picks up more than one proton. Each proton adds one positive charge and 1.00728 Da of mass. An ion carrying z extra protons therefore appears at

m/z = (M + z × 1.00728) ÷ z

where M is the neutral mass of the molecule. Take an invented peptide, Peptide A, whose neutral monoisotopic mass is 1,800.00 Da:

So one molecule shows up as a family of peaks, called the charge-state envelope, and none of them sits at 1,800. This is why a raw electrospray spectrum can look puzzling at first glance.

Example

Two invented spectra. Top, intensity against m/z from 300 to 1,900: sticks at m/z 451.01 labeled 4+, 601.01 labeled 3+ (the tallest) and 901.02 labeled 2+. Bottom, the deconvoluted spectrum against mass from 1,780 to 1,830 Da: one tall stick at 1,800.02 Da and a small stick at 1,816.01 Da labeled oxidized form.Invented mass spectrum and its deconvoluted mass.Measured spectrum (intensity against m/z)4008001,2001,6004+3+2+m/zDeconvoluted (intensity against mass in Da)1,7801,7901,8001,8101,8201,8301,800.02 Daoxidized form, 1,816.01 DaMass (Da)
Example. An invented electrospray mass spectrum of Peptide A (neutral monoisotopic mass 1,800.00 Da by construction). Top: the measured ions appear at m/z 451.01, 601.01 and 901.02, carrying four, three and two extra protons. An oxidized form, 16 Da heavier, has its own small ions just above each of these, too close to see at this scale. Bottom: after deconvolution, one main mass of 1,800.02 Da and the small oxidized form at 1,816.01 Da.
Text version of this figure
Example peaks (invented)
Measured m/zCharge (z)Neutral mass from this peak (Da)
901.01822 × (901.018 − 1.00728) = 1,800.02
601.01333 × (601.013 − 1.00728) = 1,800.02
451.01244 × (451.012 − 1.00728) = 1,800.02

Small companion peaks at m/z 455.01, 606.34 and 909.01 correspond to a molecule 16 Da heavier (1,816.01 Da), which usually indicates an oxidized form. The deconvoluted spectrum shows 1,800.02 Da as the main mass.

Deconvolution: from several peaks back to one mass

Two neighboring peaks in the envelope are enough to work out both the charge and the mass. If a peak at m/z m1 carries z charges, and its neighbor at the lower value m2 carries z + 1, then

z = (m2 − 1.00728) ÷ (m1 − m2), and M = z × (m1 − 1.00728)

In the invented example the lab measured peaks at m/z 901.018 and 601.013. The charge is (601.013 − 1.00728) ÷ (901.018 − 601.013) = 600.006 ÷ 300.005 = 2.00, and the mass is 2 × (901.018 − 1.00728) = 1,800.02 Da. Software does the same for every peak in the envelope and reports a single, deconvoluted mass. The lab then compares it with the expected mass calculated from the sequence: 1,800.02 Da found against 1,800.00 Da expected, a difference of 0.02 Da.

There is a second way to read the charge. Zoom into any one peak and it turns out to be a small cluster, because about 1.07% of the carbon found in nature is the heavier isotope carbon-13, which adds 1.00335 Da. Inside a cluster the peaks are 1.00335 ÷ z apart: about 0.50 for a 2+ ion and about 0.33 for a 3+ ion.

Schematic

Two zoomed isotope clusters for invented Peptide A. Left, the 2+ ion: four sticks at m/z 901.01, 901.51, 902.01 and 902.51, spaced 0.50 apart. Right, the 3+ ion: four sticks at m/z 601.01, 601.34, 601.68 and 602.01, spaced 0.33 apart. Stick heights fall from left to right after the first two.Two isotope clusters, schematic heights.2+ ion of Peptide A901.01901.51902.01902.51spacing 0.503+ ion of Peptide A601.01601.34601.68602.01spacing 0.33
Schematic, calculated positions. Each peak is really a cluster of isotope peaks, mainly because about 1.07% of the carbon found in nature is carbon-13; other elements add smaller contributions. The spacing inside a cluster is 1.00335 divided by the charge, so it reads the charge directly: about 0.50 apart for 2+, about 0.33 apart for 3+.
Text version of this figure
Isotope spacing by charge (schematic)
ChargeSpacing between isotope peaks (m/z)Example peaks (m/z)
2+1.00335 / 2 = 0.50901.01, 901.51, 902.01, 902.51
3+1.00335 / 3 = 0.33601.01, 601.34, 601.68, 602.01
4+1.00335 / 4 = 0.25451.01, 451.26, 451.51, 451.76

That cluster also explains why a certificate should say which mass it reports. The monoisotopic mass counts every atom as its most abundant isotope. The average mass uses each element’s average atomic mass, which reflects the proportions of its isotopes found in nature. For a peptide of around 1,800 Da the two differ by about one dalton, so a comparison is only fair if the expected and found values are the same kind of mass.

What a mass match proves, and what it does not

A measured mass that agrees with the calculated one is strong evidence that the main component is the intended peptide. That is what identity means on a certificate. It is not evidence of purity, for three reasons.

A spectrum can still flag some impurities. A form 16 Da heavier (15.995 Da) usually means one added oxygen atom, an oxidized form. A mass one residue lighter than expected points to a deletion sequence, a chain missing one amino acid (see How peptides are made). A mass about 22 Da above the expected one (22 divided by the charge, in m/z) is usually a sodium adduct: the same molecule carrying a sodium ion picked up during ionization, rather than a different molecule in the sample. Tandem mass spectrometry (MS/MS) goes further: it breaks the ion into fragments and reads the sequence piece by piece, which is stronger identity evidence than an intact mass alone.

Reading the MS line in a lab report

Then read it next to the HPLC line. Identity by MS and purity by HPLC answer different questions, and a certificate needs both; see How to read a certificate of analysis.