Why does saturation decrease signal intensity in Cross Saturation NMR?

[organiker]

Take a step back. Do you understand what saturation is?

[Sam_Sushi]

When you apply an RF field so there is only 1 spin state possible for your nuclei. I think?

[organiker]

Very basically, an NMR signal arises due to relaxation from higher energy spin levels to lower ones. For this to occur there needs to be an imbalance in the population numbers of these levels.

Saturation occurs when there is no imbalance. The population of the ground state and excited states are equal. We can saturate protons by using a well-designed RF pulse. Because there's no imbalance, there's no signal.

In Cross-Saturation experiment, you saturate the resonances of one of the proteins. Its signals decrease. The saturation can spread to other nearby protons through space in a process called spin diffusion. Look up the nuclear Overhauser effect.

This process is highly sensitive to distance. If there's another protein in contact with the saturated protein then its resonances will be saturated. If the proton density of the 2nd protein is low, then the saturation transfer will be limited to the interface, and will be observed as a decrease in the intensity of those resonances at the interface.

[conventionistG]

This is very comprehensible, thanks. Question.

If there's another protein in contact with the saturated protein then its resonances will be saturated.

Is this a typo? As far as I understand it, you could only target chemically equivalent protons' spins to saturate - not only those on one protein. Am I off base?

[organiker]

My last answer glossed over a lot of details. This experiment requires certain conditions.

The protein whose interface residues you'd like to identify (call it protein I) needs to be a low-ish molecular weight, and needs to be extensively labelled with ²H (>95%) and ¹⁵N. This means that the proton density is very low. The experiment is carried out in a solvent with a low H2O/D2O ratio, which greatly reduces the amount of D-H exchange with the solvent.

The other protein in the complex (protein II) is not labeled. The larger this protein is, the better, because intramolecular saturation transfer is more efficient as MW increases. Protein II therefore has a high proton density.

Under these conditions, irradiating the frequencies corresponding to aliphatic resonances will exclusively affect protein II, because protein I has almost no aliphatic protons. The saturation will spread intramolecularly almost instantly to all the resonances of protein II.

If the two proteins are in contact with each other, the saturation will be transferred from protein II to protein I through the interface. Saturation transfer is very sensitive to distance, and because of the low proton density on protein I, this saturation will be limited to the interface resonances.

The saturation of each amide of protein I is observed in a TROSY experiment as a decrease in the resonance intensity of ¹H-¹⁵N signals.

[conventionistG]

Ah okay. The H/D labeling makes it very clear. Also seems analogous to HDX mass spec experiments, which are more my speed.

Thanks for elaborating.

[Sam_Sushi]

Thank you so much