In MS/MS, a peptide breaks mainly along its backbone amide bonds. Each break gives two pieces: the b ion keeps the N-terminus, the y ion keeps the C-terminus. The mass difference between neighbouring b ions, or between neighbouring y ions, is exactly one amino acid residue, so a ladder of b or y ions reads the sequence directly. Identity is confirmed when the precursor mass fits and an unbroken, or nearly unbroken, series of fragments matches the expected sequence within the mass accuracy of the instrument.
This article walks through the rules, a full worked example and the places where a matching spectrum still leaves the identity open.
How are b and y ions calculated?
For singly charged ions, using monoisotopic residue masses:
- b ion = sum of the residue masses from the N-terminus up to the break + 1.00728 (one proton)
- y ion = sum of the residue masses from the break to the C-terminus + 18.01056 (water) + 1.00728
The b ion carries no water because the C-terminal OH stays with the y fragment. Two consequences are useful for checking: b(i) + y(n − i) equals the singly protonated precursor plus one proton, and the y1 ion of a tryptic peptide is always 147.113 (lysine) or 175.119 (arginine).
A worked example: LVNELTEFAK
LVNELTEFAK is a well-known tryptic peptide of bovine serum albumin. Its neutral monoisotopic mass is 1162.6234 Da, so the doubly charged precursor appears at m/z 582.3190 and the singly charged one at 1163.6306.
| i | b ion | m/z | y ion | m/z |
|---|---|---|---|---|
| 1 | L | 114.0913 | K | 147.1128 |
| 2 | LV | 213.1597 | AK | 218.1499 |
| 3 | LVN | 327.2027 | FAK | 365.2183 |
| 4 | LVNE | 456.2453 | EFAK | 494.2609 |
| 5 | LVNEL | 569.3293 | TEFAK | 595.3086 |
| 6 | LVNELT | 670.3770 | LTEFAK | 708.3927 |
| 7 | LVNELTE | 799.4196 | ELTEFAK | 837.4352 |
| 8 | LVNELTEF | 946.4880 | NELTEFAK | 951.4782 |
| 9 | LVNELTEFA | 1017.5251 | VNELTEFAK | 1050.5466 |
Read the y series from the bottom up and the differences give the sequence from the C-terminus: 218.150 − 147.113 = 71.037 (A), 365.218 − 218.150 = 147.068 (F), 494.261 − 365.218 = 129.043 (E), and so on. The b series reads the same sequence from the other end. Check one complementary pair: b4 + y6 = 456.2453 + 708.3927 = 1164.6380, which is the [M+H]+ of 1163.6306 plus one proton.
In a real spectrum you rarely see all eighteen ions. b1 is usually absent, y ions dominate on many instruments for tryptic peptides because the C-terminal basic residue holds the charge, and doubly charged fragments appear for longer peptides. What matters is coverage: how many of the bonds are supported by at least one fragment.
What counts as a confirmed identity?
A defensible identity statement combines four things:
1. Precursor mass within the instrument tolerance, for example 5 ppm on an Orbitrap or Q-TOF. 2. Fragment coverage across the sequence, with consecutive ions in at least one series and both series where the sequence is ambiguous. 3. Fragment mass accuracy, each assigned fragment within a stated tolerance, for example 10 ppm or 0.02 Da. 4. No better explanation: the expected sequence explains the main peaks better than plausible alternatives, such as a sequence variant, a truncation or a modification.
For a synthetic peptide or a peptide drug substance, this is the evidence behind the identity test. For a biologic, the same logic applied to all peptides of a digest becomes a peptide map.
Where does a matching spectrum still leave the question open?
- Leucine and isoleucine have identical mass (113.084 Da). Standard CID or HCD cannot tell them apart. If the distinction matters, it needs w ions from ETD/EThcD fragmentation, amino acid analysis or knowledge of the synthesis.
- Glutamine and lysine differ by 0.036 Da. At high resolution they separate; at low resolution they do not.
- Deamidation (N to D, +0.984 Da) shifts every fragment that contains the residue. If the y ions after the asparagine shift and the ones before do not, the site is localised; if coverage around the site is missing, it is not.
- Oxidation of methionine (+15.995 Da) behaves the same way and is a frequent artefact of sample handling.
- Isobaric sequence swaps: two adjacent residues in reversed order give the same precursor and change only the fragments between them. Without a fragment at that bond, the order is not proven.
Practical checks before you report
- Confirm the charge state of the precursor from the isotope spacing, not from the software assignment alone.
- Annotate both series and mark which bonds are covered; state the coverage as a number of bonds out of n − 1.
- For modified peptides, show at least one fragment on each side of the modified residue.
- Keep the raw file, the theoretical fragment list and the matching tolerance together, so the assignment can be reproduced.
Frequently asked questions
What is the difference between b and y ions?
Both come from cleavage of a peptide bond. The b ion contains the N-terminal part and carries no extra water; the y ion contains the C-terminal part and carries the C-terminal water. For singly charged ions, b = residue sum + 1.00728 and y = residue sum + 18.01056 + 1.00728.
Why are y ions often stronger than b ions for tryptic peptides?
Trypsin leaves lysine or arginine at the C-terminus. Their basic side chains keep the proton, so C-terminal fragments are more likely to stay charged and be detected.
How many fragments do I need to confirm a peptide sequence?
There is no single number, but coverage of most backbone bonds with consecutive ions in at least one series, plus an accurate precursor mass, is the usual basis. Ambiguous positions, such as a modification site or two possible residue orders, need fragments that specifically resolve them.
Can MS/MS distinguish leucine from isoleucine?
Not with standard CID or HCD, because both have the same mass and give the same b and y ions. ETD or EThcD can produce side-chain specific w ions; otherwise the distinction comes from amino acid analysis or the known synthesis.
What does b + y equal?
For complementary singly charged ions from the same cleavage, b(i) + y(n − i) equals the singly protonated precursor mass plus one proton. In the example, b4 + y6 = 1164.6380, the [M+H]+ of 1163.6306 plus 1.0073.