MS-Fit | MS-Tag | MS-Seq | MS-Pattern | MS-Homology | MS-Digest | MS-Bridge | MS-NonSpecific | MS-Product | MS-Comp | MS-Isotope | MS-Viewer
This document provides instructions for MS-Comp.
- Search Times
- General features of links from program output
- Link from the peptide sequence in program output to MS-Product
- Constant Modifications
- User Specified Amino Acid
- Mass (m/z)
- Mass type
- Charge (z)
- AA Composition Ions
- Absent Amino Acids
- Modified Amino Acids Possibly Present
- Instrument
Amino Acid
Lists the amino acid combinations consistent with the search conditions. Some of these will have identical elemental compositions.
Peptide ElementalLists the unique elemental compositions from the list of amino acid combinations reported by the Amino Acid option.
ElementalLists the elemental compositions consistent with the search conditions. The elemental compositions returned will obey the nitrogen rule and will have a double bond equivalent within the range expected for a peptide. The elemental compositions are, however, not guaranteed to have corresponding peptides. This option will work at much higher mass than the first two options.
You can select one or more possible ion types for your m/z value. The report will list all the possibilities for each ion type in turn.
The nitrogen rule states that for an organic compound with even number of nitrogens (including 0) the nominal mass of the molecular ion will be even. Note that this rule was first observed for EI spectra of small molecules, where the molecular ion is not protonated. Hence the rule for peptides is that the nominal mass for an MH+ equivalent must be odd.
The nitrogen rule stems from the fact that most of the common elements that have even nominal masses have even valence:
12C, valence = 4;
16O, valence = 2;
28Si, valence = 4;
32S, valence = 2.
On the other hand most of the elements with odd nominal masses have odd valence:
1H, valence = 1;
19F, valence = 1;
31P, valence = 3;
35Cl, valence = 1.
Nitrogen is an exception in that it has an even nominal mass but an odd valence:
14N, valence = 3.
The double bond equivalent (DBE) is the number of rings or double bonds that an ion contains. It can be calculated from the elemental formula as follows:
DBE = 1 - a/2 + c/2 + d
where:
a = number of atoms with a valence of 1 (H, F, Cl).
b = number of atoms with a valency of 2 (O, S).
c = number of atoms with a valency of 3 (N, P).
d = number of atoms with a valency of 4 (C, Si).
If the value calculated ends in 0.5 then this should be subtracted to get the true value.
Amino Acid | DBE | Elemental Formula | Calculation |
---|---|---|---|
A | 1.0 | C3 H5 N1 O1 | 3 - 5/2 + 1/2 |
C | 1.0 | C3 H5 N1 O1 S1 | 3 - 5/2 + 1/2 |
D | 2.0 | C4 H5 N1 O3 | 4 - 5/2 + 1/2 |
E | 2.0 | C5 H7 N1 O3 | 5 - 7/2 + 1/2 |
F | 5.0 | C9 H9 N1 O1 | 9 - 9/2 + 1/2 |
G | 1.0 | C2 H3 N1 O1 | 2 - 3/2 + 1/2 |
H | 4.0 | C6 H7 N3 O1 | 6 - 7/2 + 3/2 |
I | 1.0 | C6 H11 N1 O1 | 6 - 11/2 + 1/2 |
K | 2.0 | C6 H12 N2 O1 | 6 - 12/2 + 2/2 |
L | 3.0 | C6 H11 N1 O1 | 6 - 11/2 + 1/2 |
M | 1.0 | C5 H9 N1 O1 S1 | 5 - 9/2 + 1/2 |
N | 3.0 | C4 H6 N2 O2 | 4 - 6/2 + 2/2 |
P | 2.0 | C5 H7 N1 O1 | 5 - 7/2 + 1/2 |
Q | 3.0 | C5 H8 N2 O2 | 5 - 8/2 + 2/2 |
R | 2.0 | C6 H12 N4 O1 | 6 - 12/2 + 4/2 |
S | 1.0 | C3 H5 N1 O2 | 3 - 5/2 + 1/2 |
T | 1.0 | C4 H7 N1 O2 | 4 - 7/2 + 1/2 |
V | 1.0 | C5 H9 N1 O1 | 5 - 9/2 + 1/2 |
W | 8.0 | C11 H10 N2 O1 | 11 - 10/2 + 2/2 |
Y | 5.0 | C9 H9 N1 O2 | 9 - 9/2 + 1/2 |
The terminal groups and cation then contribute H3 O to the overall elemental formula reducing the DBE by 1.5. Also there is one to add on from the original formula.