CHEM3013 NMR spectroscopy and mass spectrometry of small and large molecules 2020
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Semester 1 – End of Semester, 2020
CHEM3013 NMR spectroscopy and mass spectrometry of small and large molecules
Question 1 (1S). What structural information do NOESY spectra provide? (1 mark)
A NOESY cross-peak between two 1H-NMR resonances indicates that the two protons are closer in space than 5 Å .
Question 2 (1S). What is a cross-peak as opposed to a diagonal peak? (2 marks)
A cross-peak has different frequencies in different spectral dimensions, whereas a diagonal peak has the same frequency (in at least two dimensions).
Question 3. Explain how the jump-return sequence works with vectors representing the magnetisation of (a) water and (b) a protein resonance. Why will one half of the spectrum appear
positive and the other half negative? (4 marks)
See NMR_Otting_part_B.pdf, page 14 and Problem 3.
Question 4. The figure below shows the pulse sequence of a 3D NOESY-15N-HSQC experiment. Explain how this pulse sequence leads to a 3D NMR spectrum. Which nuclear frequencies are observed in each of the three dimensions? Compared to a 2D NOESY spectrum, does the 3D NOESY-15N-HSQC spectrum contain more or fewer peaks? Explain your
reasoning. (5 marks)
The FID is recorded during t3 . t2 is systematically incremented from FID to FID to obtain a 2D 15N-HSQC spectrum. t1 is systematically incremented from 2D spectrum to 2D spectrum to obtain the 3rd dimension. The 3D NMR spectrum is obtained by Fourier transform of the data in all three dimensions. F1 : 1H; F2 : 15N; F3 : 1H. The spectrum contains fewer peaks than a 2D NOESY spectrum, because the 15N-HSQC acts as a filter that lets 1H magnetisation pass only for protons bonded to 15N.
Question 5. The pulse sequence 90ox – t – 180ox – t is applied to a spectrum containing two doublets, each showing the coupling constant J. The signal produced by the spin-echo sequence with increasing echo delay t follows the equation exp(-2t/T2) cos(pJ2t). Explain why T2 rather
than T2 * describes the signal decay, i.e. why magnetic field inhomogeneity does not need to be
taken into account. (2 marks)
A spin-echo refocuses chemical shifts, including the chemical shift differences caused by an inhomogeneous magnetic field.
Question 6. A protein was produced, where all alanine residues are 100 % enriched with 15N, while all other amino acids are at natural isotopic abundance. A 15N-HSQC spectrum was recorded in the absence and presence of a ligand molecule. The protein contains 5 alanine residues.
(a) (1S) Why are no cross-peaks observable for other amino acid residues?
(b) (1S) Give a reason why the cross-peak of Ala87 (circled) is shifted in the presence of the ligand, whereas the other cross-peaks stay at the same locations.
(3 marks)
(a) The natural isotopic abundance of 15N is very low (0.3 %), so that only cross-peaks of 15N-labelled amino acids are intense in a 15N-HSQC spectrum.
(b) If the ligand binds in the vicinity of Ala87, it changes its chemical environment and therefore its chemical shift.
Question 7. Sketch the multiplet fine-structure of the COSY cross-peak between spins M and X in a linear 3-spin system A-M-X, where JAM > JMX > 0 and JAX = 0; plot the chemical shift of M in the F1 dimension and the chemical shift of X in the F2 dimension. Identify the
coupling constants in the multiplet fine-structure of the cross-peak. (4 marks)
Question 8. A static sample of urea powder is placed into an NMR spectrometer with 600 MHz 1H NMR frequency.
The static 13C spectrum at 150 MHz NMR frequency is shown below.
The same spectrum is acquired again, this time with strong 1H decoupling applied during acquisition. Some key features have been highlighted.
(a) (1S) What spin interaction has been removed upon applying 1H decoupling?
The scalar and dipolar coupling.
(b) (1S) What spin interaction remains present in the spectrum?
The chemical shift anisotropy.
The same spectrum is acquired again, this time with magic angle spinning (MAS). Some key features have been highlighted on the spectrum below:
(c) What MAS rate was used in the acquisition of this spectrum? (Note that 1 ppm corresponds to 150 Hz. Show your working.)
193.6- 153.9 = 39.7 ppm, corresponding to 39.7 * 150 = 5.955 kHz.
(d) At what MAS rate would a single peak be observed in the spectrum (explain your reasoning) and what would its chemical shift be? (4 marks)
To cover the entire width of the peak, MAS spinning at a rate of (244. 1 – 89.35)*150 = 23.2 kHz would be required.
Question 9. Give one advantage of performing a top-down mass spectrometry analysis on a protein, and one advantage for using bottom-up mass spectrometry analysis. (2 marks)
Question 10. Shown below are the mass spectra of the FK506-binding protein. The top spectrum was recorded of the protein with natural isotope abundance. The bottom spectrum was recorded of a protein sample produced from amino acids depleted of 13C and 15N isotopes. Explain the differences in the mass spectra, and give one reason why it is beneficial to isotope-deplete large proteins for mass spectrometric analyses.
(3 marks)
Question 11. Determine the average molecular weight and the monoisotopic molecular
weight of the FK506-binding protein from the spectra in question 10. (2 marks)
Question 12. Determine the molecular weight of the molecule that was analysed by
negative ion ESI-MS, producing the mass spectrum shown below. (2 marks)
Question 13. Briefly explain what “Multiple Reaction Monitoring” (MRM) in mass spectrometry is, what type of mass spectrometer is (typically) used and why, and give a
possible application where it could be applied. (3.5 marks)
2022-05-25