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I am running a 15N HSQC on a protein sample which is in a 100% deuterium oxide solvent. The command rga in TopSpin gives a value of >10000 for the receiver gain. I understand this is happening because there is no water signal and hence the receiver gain is optimizing the S/N. However, should the experiment be run with that high value or should it be adjusted to a value of a size similar to the one used on water samples?

asked Nov 28 '13 at 05:34

norsouth's gravatar image


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At low receiver gain, the s:n will be reduced, because the digitizer noise is singificant with respect to thermal noise, increasing the rg will increase s:n, until the point where the thermal noise becomes significantly larger than the digitizer noise, after which you get no benefit. On older Bruker hardware an rg value of 2000 is almost certainly sufficient for this, so there's no advantage in using a higher gain, and more risk of clipping if for some reason you get more signal later on in the experiment (e.g. h2o suppression gets worse). So I would stick with rg=2000 rather than going higher, but for example if you typically only get 128 with 90% h2o samples, you should definitely go higher for samples in d2o.

On AVII/AVIII the rg value you need to reach if much lower - 128 is certainly more than sufficient for room temperature probes, and 203 more than sufficient for cryoprobes (thermal noise is lower in cryoprobes, so you need to multiply it up more before it becomes significantly larger than the digitizer noise).


answered Nov 29 '13 at 00:53

Pete%20Gierth's gravatar image

Pete Gierth

Thank you for your answer, it was very insightful. - norsouth (Dec 02 '13 at 11:16)

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Someone once told me that you are not going to see anything in 100 D2O N-H HSQC because all the NH are ND in your tube. Maybe that's why you automatic RG is so high: no signal -> increase RG -> no signal, …

Usually if I remember people typically use 90% H2O plus 10% D2O for the lock signal.

Hope could be useful, but I have to admit that I'm not an expert.


answered Nov 29 '13 at 03:01

PQdotL's gravatar image


I agree thast most if not all NH protons will eventually exchange with the D2O and dissapear. The time constant for some of the NHs might be long, however. - Kirk Marat (Nov 29 '13 at 07:24)

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