The pioneering work of John Fenn (2002 Nobel Prize) on the use of electrospray ionization (ESI) for — Physical Chemistry — Kinetics Chemistry Question
Electro-spray Ionization Mass-spectrometry of Peptides
The pioneering work of John Fenn (2002 Nobel Prize) on the use of electrospray ionization (ESI) for mass spectrometry opened new possibilities for analyzing biologically important non-volatile molecules. ESI has since been used in numerous biological applications, resulting in emergence of proteomics that aims at large-scale characterization of proteins in organisms.
A bio-analytical chemist considered the use of ESI mass spectrometry to measure the relative abundance of myoglobin in two protein mixtures. Realizing the challenges of whole protein analysis, this chemist decided to reduce the problem to the peptide level. The relative concentrations of a peptide in two samples can be measured by isotope tagging. Consider the analysis scheme described below.
First, the proteins in two samples were digested using trypsin, and the digested samples were lyophilized (the solvent was evaporated, leaving behind the peptides). For isotope tagging of the peptides two methanolic solutions were prepared by dropwise addition of 160 µdm³ of acetyl chloride to methanol cooled in an ice bath using 1 cm³ of CH₃OH in one case and 1 cm³ CD₃OH in the second case.
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The CH₃OH solution was added to the digested lyophilized peptide sample 1. The CD₃OH solution was combined with digested lyophilized peptide sample 2. After 2 hours both methanolic solutions were evaporated to dryness. 10 µdm³ of 0.1% acetic acid in water was used to dissolve each of the residues and the resulting solutions were mixed. The mixture was then injected into a high-performance liquid chromatography-ESI mass spectrometer where the tagged peptides were separated and detected by a mass spectrometer.
The summary of the workflow is shown below:
[VISUAL]
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Peptides undergo multiple protonation during ionization to form cations with the overall charge of +1, +2, +3 etc. As a result, a peptide with monoisotopic mass M (molecular mass based on most abundant isotopes of elements) can produce in its ESI mass spectrum signals of [M+H]⁺, [M+2H]²⁺, and [M+3H]³⁺ ions. The ion charge (“charge state”) corresponding to a given peak in a mass spectrum can be determined from the mass-to-charge (m/z) spacing between the isotopic peaks.
A series of peaks corresponding to a tagged peptide in the mass spectrum of the mixture of two samples (Mix 1 and Mix 2) was found at m/z values of 703.9 (100), 704.4 (81), 704.9 (36), 705.4 (61), 705.9 (44), and 706.4 (19). The numbers in parentheses show the relative areas under the peaks.
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Analysis of the peptide mass and fragmentation patterns led the chemist to the conclusion that this series of peaks belongs to a tagged peptide originating from myoglobin.
Write equation(s) for the chemical reaction(s) involved in the preparation of methanolic solutions of acetyl chloride.
Model Answer
2 CH3OH + CH3COCl → CH3COOCH3 + CH3OH2+ + Cl–
2 CH3OD + CH3COCl → CH3COOCH3 + CH3OD2+ + Cl–
or
CH3OH + CH3COCl → CH3COOCH3 + HCl
CH3OD + CH3COCl → CH3COOCH3 + DCl
What chemical modification of peptides occurs in isotopic tagging reactions, resulting in Tag 1 and Tag 2? What is the role of acetyl chloride?
Model Answer
Methyl esterification at the carboxylic end of the peptide occurs as a result of subjecting the peptides to methanolic HCl. Acetyl chloride in methanol releases HCl which catalyzes the esterification reaction.
What is the charge state of the tagged peptide in this series of peaks?
Model Answer
The isotopic peaks are 0.5 Dalton apart, meaning that the detected ions are doubly charged with the general formula of [M+2H]2+.
Identify the monoisotopic peak corresponding to the light isotope tagged peptide and calculate the monoisotopic mass of the tagged peptide based on this peak.
Model Answer
The lowest mass peak in the series corresponds to the monoisotopic peak of light isotope tagged peptide. M = 2 × 703.9 – 2 × 1 = 1406
Which m/z values have contributions from the heavy isotope tagged peptide?
Model Answer
The isotopic peaks show a descending order for the first three peaks in the series. A sudden jump in the intensity for the fourth peak indicates contribution from an isotopically labeled peptide. Accordingly, the isotopic peaks from the heavy isotope tagged peptide should appear at m/z 705.4, 705.9, and 706.4.
Calculate the monoisotopic mass of the untagged peptide.
Model Answer
The monoisotopic mass of heavy isotope tagged peptide is 2 × 705.4 – 2 = 1409 which is 3 amu larger than the monoisotopic mass of light isotope tagged peptide. This indicates that only one d3-methanol is incorporated into the peptide structure by esterification. Accordingly, the monoisotopic mass of unmodified peptide can be calculated by subtracting mass of CH2 from the monoisotopic mass of light isotope tagged peptide. M untagged = 1406 – 14 = 1392.
Assuming that ionization efficiency is not affected by isotopes, calculate the relative abundance of myoglobin in the two protein samples using the relative areas of peaks in the series.
Model Answer
The ratio of myoglobin in two samples is reflected in the ratio of isotopic peaks. The heaviest peak in the series can be used as the representative of heavy isotope tagged peptide because the contribution to this m/z from the light isotope tagged peptide is minimum and negligible. This m/z corresponds to the third isotopic peak of heavy isotope tagged peptide and should be compared to the third isotopic peak of light isotope tagged peptide at m/z 704.9. Therefore:
Myoglobin sample2 / Myoglobin sample1 = I706.4 / I704.9 = 19/36 = 0.53.
What would the relative peak intensities be if our chemist used 13CH3OH rather than CD3OH? The isotopic distribution patterns can be assumed to be the same for 12CH3OH and 13CH3OH tagged peptides within the experimental errors of mass spectrometric measurements.
Model Answer
First we calculate the isotopic distribution of light isotope tagged peptide by subtracting the contributions from heavy isotope tagged peptide. We then add the contributions from 13CH3OH tagging as follows:
m/z of isotopic peaks: 703.9, 704.4, 704.9, 705.4, 705.9, 706.4
- Peak areas from mixture of CH3OH and CD3OH tagged peptides: 100, 81, 36, 61, 44, 19
- Contributions from CH3OH tagged peptide: 100, 81, 36, (61 - 100 * 0.53) = 8, (44 - 81 * 0.53) = 1, (19 - 36 * 0.53) = 0
- Calculated peak areas for mixture of CH3OH and 13CH3OH tagged peptides with relative abundance of 0.53 for the heavy isotope tagged peptide:
- 703.9: 100
- 704.4: 81 + 100 * 0.53 = 134
- 704.9: 36 + 81 * 0.53 = 79
- 705.4: 8 + 36 * 0.53 = 27
- 705.9: 1 + 8 * 0.53 = 5
- 706.4: 0 + 1 * 0.53 = 0.5 (or approx. 0)
Which of the reagents is a better choice for relative quantification of samples: 13CH3OH or CD3OH?
Model Answer
The isotopic peaks from the light isotope tagged peptide create a background for the peaks corresponding to heavy isotope tagged peptide. This interference becomes less important as the difference between the masses of tagged peptides becomes larger. Therefore, tagging with CD3OH will result in wider dynamic range for quantification of relative concentrations compared to tagging with 13CH3OH.