How fragmentation reads a sequence
In a tandem mass spectrometry experiment the peptide is isolated and broken, most commonly at the amide bonds along the backbone. The pieces that retain the N-terminus are b ions and the pieces that retain the C-terminus are y ions, and each series steps by one residue at a time.
The mass difference between consecutive members of a series is the mass of the residue between them. Reading those differences along the series reads the sequence, which is what makes fragmentation the definitive identity test rather than a supporting one.
Why b and y pair up
Each cleavage produces one b ion and one y ion, and together they account for the whole molecule. So b1 pairs with y(n-1), b2 with y(n-2), and each pair sums to the intact neutral mass.
This is a useful check on any fragment table: if a b and its partner y do not sum to the precursor, something in the table is wrong.
What the series can and cannot resolve
Leucine and isoleucine are isomers with identical masses, so no mass-based method distinguishes them. Glutamine and lysine differ by 0.036 daltons, which needs high resolution. Everything else in the standard twenty is separated by at least a dalton.
A gap in the series, where a consecutive pair is missing, leaves the order of two residues undetermined. Proline is a frequent cause: cleavage on its N-terminal side is disfavoured, so the corresponding ion is often weak or absent.
Modifications and where they show up
A modification shifts every ion in a series that contains the modified residue and leaves the rest unchanged. The point at which the shift appears identifies the position, which is the only way to locate a modification rather than merely detect it.
Methionine oxidation is the routine case. Selecting it in this tool adds 15.995 daltons to every methionine, which propagates through the series in exactly the way an observed spectrum would.
How the fragment masses are calculated
Cumulative sums of monoisotopic residue masses from each end, with water added to the C-terminal series and nothing added to the N-terminal one. Charge is applied last, so the neutral masses can be read directly.
b(i) = SUM of residues 1..i (no water) y(j) = SUM of the last j residues + 18.010565 (with water) precursor = SUM of all residues + 18.010565 m/z = (neutral + z x 1.007276) / z check: b(i) + y(n-i) = precursor, for every i
- Use monoisotopic residue masses. Fragmentation is analysed on high-resolution instruments that resolve isotope patterns, so monoisotopic is the correct basis rather than average mass.
- Accumulate b ions from the N-terminus. b1 is the first residue, b2 the first two, and so on. No water is added: the b fragment keeps the N-terminus and the acylium carbon, not the terminal hydroxyl.
- Accumulate y ions from the C-terminus. y1 is the last residue plus water, y2 the last two plus water. The water is the C-terminal hydroxyl and the hydrogen transferred during cleavage.
- Apply modifications per residue. Methionine oxidation adds 15.995 to each methionine before the sums are taken, so the shift propagates through exactly the ions containing that residue.
- Convert to m/z at the chosen charge. Neutral mass plus one proton per charge, divided by the charge. The table shows the neutral mass alongside, since that is the figure to compare against a deconvoluted spectrum.
What this method cannot tell you
- •It generates b and y ions only. Real spectra also contain a ions, c and z ions from electron-transfer methods, and neutral losses of water and ammonia.
- •It does not predict intensities. Which fragments are actually observed depends on the sequence, the charge state and the fragmentation method.
- •It cannot distinguish leucine from isoleucine, and separates glutamine from lysine only at high resolution.
- •It assumes free termini. An amidated C-terminus or an acetylated N-terminus shifts a whole series.
Fragment mass calculator: frequently asked questions
The two fragment series produced when a peptide breaks at an amide bond. b ions keep the N-terminal end; y ions keep the C-terminal end.
Each cleavage produces one of each, so b and y are complementary and together account for the whole molecule.
Because of where the bond breaks and where the hydrogen goes. The y fragment retains the C-terminal hydroxyl and picks up a hydrogen from the cleavage; the b fragment retains neither.
The difference is 18.011 daltons, and it is why b1 is simply the first residue mass while y1 is the last residue mass plus water.
Take the difference between consecutive ions in a series. That difference is the mass of the residue between them.
A step of 113.084 is a leucine or isoleucine; 128.095 is a lysine; 57.021 is a glycine. Walking the differences along the series reads the order.
They are structural isomers with the same atoms, so their masses are identical and no mass measurement separates them.
Distinguishing them requires fragmentation chemistry energetic enough to break the side chain, or independent sequence information.
That the table is internally consistent. Each pair, b(i) and y(n-i), comes from the same cleavage and together they contain the whole molecule.
It is a quick check on any fragment calculation. A pair that does not sum to the precursor indicates an error.
Because fragmentation is not uniform. Some bonds break more readily than others, and intensity varies with sequence, charge state and method.
Cleavage on the N-terminal side of proline is particularly disfavoured, which is why proline-containing peptides often show a gap in the series at that position.
Singly charged for small peptides, doubly for most tryptic-length ones. Fragment ions carry fewer charges than the precursor they came from.
The table shows the neutral mass beside each m/z so it can be compared against a deconvoluted spectrum regardless of the charge state chosen.
It adds 15.995 daltons to every ion containing that methionine, and leaves every ion that does not contain it unchanged.
The position at which the series shifts is what localises the modification, which the intact mass alone cannot do.
b ions that have lost carbon monoxide, so 27.995 daltons lighter. They appear in most spectra at lower intensity than the corresponding b ions.
This tool does not generate them. An a ion is readily identified as a peak exactly 28 daltons below a b ion.
The fragment series produced by electron-transfer and electron-capture dissociation, which cleave the N to alpha carbon bond rather than the amide bond.
They are valuable for modified peptides, because these methods preserve labile modifications such as phosphorylation that collision-based fragmentation tends to knock off.
Monoisotopic, always. Fragmentation is performed on instruments that resolve isotope patterns, and the monoisotopic peak is what they report.
A fragment that has lost a small neutral molecule, most often water at 18.011 or ammonia at 17.027 daltons.
Water losses are common from serine and threonine, ammonia losses from asparagine, glutamine, lysine and arginine. They appear as satellite peaks below the main fragment.
The y series would all be 0.98 daltons lighter than shown, because the amide replaces the free acid. The b series is unaffected.
The tool assumes free termini, so subtract 0.98 from every y ion for an amidated peptide.
Enough consecutive ions to cover the sequence. Complete coverage of both series is unusual; a continuous stretch of one series covering most of the peptide is a strong result.
Search engines score exactly this: how many of the predicted ions are present and how well their intensities are accounted for.
No. It gives the expected masses. Predicting intensity requires a model of the fragmentation chemistry and is a considerably harder problem.
1.007276 daltons, which is a hydrogen atom less one electron. Using the hydrogen atom mass of 1.00794 instead introduces an error of about 0.0007 daltons per charge.
That is negligible on most instruments and detectable on a high-resolution one at low m/z.
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