Snake venom is composed of a variety of polypeptides and other small molecules. Venom polypeptides h — Analytical Chemistry Chemistry Question
Mass spectrometry of a peptide
Snake venom is composed of a variety of polypeptides and other small molecules. Venom polypeptides have a range of biological effects including muscle necrosis and the disruption of neurotransmission. Characterisation of the components of snake venom is important in the development of lead-compounds for the pharmaceutical industry and also in the creation of antivenins.
Tandem mass spectrometry (MS-MS) provides a rapid approach for determining the sequence of polypeptides. This involves formation of a parent ion, which is then fragmented to form other smaller ions. In peptides fragmentation often occurs at the amide bond, giving rise to so-called ‘b ions’. The b ions formed from an alanine-valine-glycine polypeptide are shown below. Remember that by convention the first amino acid is that with the free –NH2 group.
[VISUAL]
Polypeptide X was isolated from the venom of the pit viper, B. insularis. The amino acid composition of polypeptide X may be found by acid hydrolysis of the peptide. Under the conditions used for the hydrolysis, Asp and Asn are indistinguishable and are termed Asx; similarly Glu and Gln are indistinguishable and termed Glx. The composition of polypeptide X was found to be: 1 × Asx, 2 × Glx, 1 × His, 1 × Ile, 4 × Pro and 1 × Trp.
How many unique decapeptide sequences can be formed from these aminoacids:
i) assuming Glx are both the same amino acid?
ii) assuming that one of the Glx amino acids is Glu, the other Gln?
Model Answer
i) If Glx are both the same amino acid then the number of unique sequences is given by 10!/(4! x 2!). This gives 75600 sequences.
ii) If Glx are two different amino acids then the number of unique sequences is given by 10!/4!. This gives 151200 sequences.
What are the possible masses for Polypeptide X?
Model Answer
There are six possible peptides that could be formed depending upon the identity of Asx and Glx:
Amino Acids | Peptide mass | Amino Acids | Peptide Mass
Asn, Gln, Gln | 1213 | Asp, Gln, Gln | 1214
Asn, Gln, Glu | 1214 | Asp, Gln, Glu | 1215
Asn, Glu, Glu | 1215 | Asp, Glu, Glu | 1216
In the mass spectrum of Polypeptide X the parent ion showed at peak at an m/z of 1196.8. It is known that although snake toxins are synthesised from the 20 common amino acids shown in the table some of these amino acids can be chemically modified after polypeptide synthesis. The mass spectrum of the parent ion suggests that one of the amino acids in Polypeptide X has been modified in a way that is not evident after acid hydrolysis.
Polypeptide X was sequenced using MS-MS. The masses of the b ions are shown in the table below:
ion | m/z | ion | m/z | ion | m/z
b1 | 112.2 | b4 | 509.7 | b7 | 872.0
b2 | 226.4 | b5 | 646.7 | b8 | 985.0
b3 | 412.5 | b6 | 743.8 | b9 | 1082.2
What is the sequence of Polypeptide X? You may use “Mod“ for the modified aminoacid.
Model Answer
The mass of ion b1 can be used to determine the identity of the first amino acid in the polypeptide:
Mr(amino acid 1) = mass(b1) + Mr(O) + Mr(H) = 129.2.
This does not correspond to the mass of any of the 20 amino acids typically found in proteins, therefore amino acid 1 must be Mod.
The identity of amino acids 2 to 9 can be determined using consecutive b-ions:
ion b2 (m/z 226.4) - b1 (m/z 112.0*) = mass diff 114.2 → amino acid 2 (Asn, mass 132.2)
ion b3 (m/z 412.5) - b2 (m/z 226.4) = mass diff 186.1 → amino acid 3 (Trp, mass 204.1)
ion b4 (m/z 509.7) - b3 (m/z 412.5) = mass diff 97.2 → amino acid 4 (Pro, mass 115.2)
ion b5 (m/z 646.7) - b4 (m/z 509.7) = mass diff 137.0 → amino acid 5 (His, mass 155.0)
ion b6 (m/z 743.8) - b5 (m/z 646.7) = mass diff 97.1 → amino acid 6 (Pro, mass 115.1)
ion b7 (m/z 872.0) - b6 (m/z 743.8) = mass diff 128.2 → amino acid 7 (Gln, mass 146.2)
ion b8 (m/z 985.0) - b7 (m/z 872.0) = mass diff 113.0 → amino acid 8 (Ile, mass 131.0)
ion b9 (m/z 1082.2) - b8 (m/z 985.0) = mass diff 97.2 → amino acid 9 (Pro, mass 115.2)
*Note: in calculation, b1 is taken as 112.0.
Finally the identity of amino acid 10 can be verified using the masses of the polypeptide X and ion b9:
Mr(amino acid 10) = Mr(X) – mass(b9) + Mr(H) = 115.6 (which corresponds to Pro).
The sequence is therefore:
Mod – Asn – Trp – Pro – His – Pro – Gln – Ile – Pro – Pro
What is the mass of the modified amino acid?
Model Answer
The mass of the modified amino acid is 129.2.
The 13C NMR spectra of “Mod“ in D2O is shown on the right. [VISUAL]
The 1H NMR spectra, taken in an organic solvent, and in D2O are shown below. [VISUAL]
Draw the structure of Mod and suggest which protons give rise to which signals in the 1H NMR spectrum. You need not explain the multiplicity of the signals.
Model Answer
It is known from the amino acid composition that Mod must be based on Gln or Glu. The mass and NMR spectra are consistent with the cyclic amino acid usually referred to as pyroglutamic acid (structure is drawn).
If the peaks in the 1H NMR spectrum of mod in organic solvent are numbered 1 to 6 from low to high chemical shift then the assignment is as follows:
- Peak 5: N-H proton
- Peak 4: C-H proton adjacent to carboxylic acid
- Peaks 1 & 3: aliphatic CH2 protons of the ring
- Peak 2: CH2 protons adjacent to carbonyl
- Peak 6: O-H proton of the carboxylic acid group
Note:
Students are not expected to be able to completely assign peaks 1 and 3.