The conversion of NADPH to NADP+ results in a decrease in absorbance at 340 nm. This reaction can also be monitored by a decrease in fluorescence, with 340 nm excitation and 460 nm emission. The reaction can be monitored continuously so the rate of change can be measured from a single reaction. If conversion of the other substrate to product also results in a significant absorbance change at 340 nm or a fluorescence change at the stated wavelengths, this could complicate the interpretation, however.

I think MTT forms an insoluble product. If you want to measure the concentration of NADPH remaining at the end of the reaction, it might be preferable to use XTT and diaphorase. XTT is reduced by diaphorase in the presence of NADPH to form a soluble reddish product.

Dear Adam, appreciate your inputs. I have tested in the lab to measure the fluorescence of NADPH and indeed it has maximum emission at 460 nm. However, I found out that my control, where there was supposed to be no reaction happened, has its emission at 460 nm decreasing overtime. I am using 50 mM Tris-HCl buffer pH ~6. I am aware that NADPH is not stable in presence of acetate and phosphate ions, and acidic pH may cause degradation, but my reductase enzyme was reported to be optimal at pH around 6. Is this continuous decrease in emission expected in my control reaction?

Regarding the MTT (or XTT) assay, my initial thought was to introduce it right from the start of the reaction and monitor the change in absorbance at around 585 nm (corresponding to formazan formed as the reaction proceeds). From what I understood, the MTT would 'steal' the electrons and hydride ions from NADPH-reductase complex, and hence my original reaction product would not be formed. This thought coming from my understanding that MTT (or XTT), would not react with NADPH without the reductase enzyme.

Appreciate if anyone could share their understanding.

If you think the spontaneous rate of 460 nm fluorescence decrease is too high, consider increasing the pH to 7. The enzyme activity may be a little lower, but the loss may be tolerable. If the spontaneous rate of fluorescence decrease is not too severe, then just measure it and subtract it from the enzyme catalyzed rate.

You probably can't add the MTT or XTT at the start of the reductase reaction because the reductase might catalyze reduction of the dye in the presence of NADPH. In other words, the dye might act as an alternate substrate, competing with the intended substrate. If you have a means of stopping the reductase reaction by adding a specific inhibitor or filtering out the enzyme (using an ultrafiltration membrane), then you can quantify the remaining NADPH using the XTT+diaphorase method or any other convenient analytical method. Since diaphorase is an enzyme, you can't use a harsh quenching treatment to stop the reductase reaction, like adding acid.

Dear Adam,

Thank you for your input. That what exactly I thought. Without the original substrate, the MTT would not be reduced in presence of NADPH, even with the enzyme present. On the other hand, in presence of the original substrate, the MTT would be reduced in presence of the enzyme and NADPH, but the original substrate is not reduced. And I always thought that in presence of the enzyme and NADPH, the rate of MTT reduction is proportional, if not equivalent, to the rate of original substrate reduction. Please correct me if I misunderstand anything. Thank you!