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I would like to know where the advantages of the HNCO spectrum stem from. I read that it is the most sensitive multi-D NMR spectrum, but all i know is, that it produces an absorptive spectrum, which is an obvious advantage.

Can someone explain to me why it is the most sensitive 3D-NMR spectrum and why it is popular?

asked Apr 17 '13 at 02:07

nmrguy's gravatar image


updated Apr 17 '13 at 02:28

3 Answers:
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To compliment Matts answer, I would like to draw your attention to the classical review for multidimensional NMR experiments: M. Sattler et al., Progress in Nuclear Magnetic Resonance Spectroscopy 1999, 34, p93-158.

Especially table 1, where you would get an impression of where the other experiments stand in comparison to 3D HNCO experiment. Hope this helps.


answered Apr 19 '13 at 07:59

Bharathwaj's gravatar image


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The simplest answer is that the sensitivity for these experiments is dependent on the coupling constant between the nuclei, the bigger the coupling the more sensitive the experiment. The C(i-1) to N(i) coupling constant is 15 Hz while the N(i)-Ca(i) is ~11Hz and the Ca(i-1) -N(i) is much smaller ( ~ 7Hz). Therefore the HNCO is a good first spectrum to test your protein, if you don't get good signal from your HNCO then it is probably not worth running the other triple res experiments , also it does give you the carbonyl chemical shifts that can be useful in secondary structure prediction. It is not particularly useful for assignments unless you also run the much less sensitive HN(CA)CO ( 1Hz coupling)


answered Apr 19 '13 at 05:54

Matthew%20Revington's gravatar image

Matthew Revington

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To compliment the earlier answers --

The overall sensitivity of a 3D experiment depends on the product of the efficiency of coherence transfer of each of the steps. For HNCO, we need to consider the efficiency of coherence transfer pathways from 1H to 15N and from 15N to 13C of carbonyl (and back). The efficiency of a coherence transfer process depends on the ratio of the magnitude of the J-coupling to the line-width; in addition other available nonproductive/undesirable (branching) coherence transfer paths would lower the sensitivity of the experiment.

Hence, in addition to the magnitude of the coupling constant, (discussed in earlier posts), we also need to consider the linewidths. The linewidths of the carbonyl carbons are substantially lower than that of the linewidths of Ca atoms. This is yet another reason for the much higher sensitivity of HNCO compared to HNCA. The larger linewidth of the Ca is due to relaxation induced by attached hydrogen (Ha). The carbonyl carbon in the peptides/proteins is not covalently bonded to any hydrogen and hence a major relaxation mechanism is absent, resulting in a narrow linewidth.
A long range HNCO was used to observe 0.2 Hz 3J-NC' coupling in hydrogen bonded GG base pairs - this was possible due to the favorable relaxation properties of the system. (Grzesiek, JBNMR, 2000, 16, 279; Chap 9, BioNMR in Drug research, Wiley-VCH)

Yet another reason for the higher sensitivity of HNCO compared to HNCA is due the presence of multiple coherence transfer pathways originating from 15N in the HNCA for proteins. Coherence transfer occurs from 15N to Ca of the same residue as well as to the Ca of the previous residue, due to the relatively small difference in the coupling constants from 15N to the Ca of same and previous residues - this results in a reduction in sensitivity. Furthermore, the unresolved coupling of Ca to Cb of side-chain may become effective if the resolution (and hence acquisition time) in the carbon dimension increases. Since the carbonyl carbon does not have significant J-coupling to any other carbon atom with similar chemical shift, such alternative coherence transfer pathways are minimized in the HNCO experiments on proteins.


answered May 25 '13 at 14:13

sekhar%20Talluri's gravatar image

sekhar Talluri

updated May 25 '13 at 14:20

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