NADP+ / NADPH

[ahc336]

If NADP+ is positive, when it gains an H+ (in the photosystem, for example) why does it become NADPH?

Wouldn't the H+ cause it to become 2+, not cancel the positive charge?

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[sazerac]

If NADP+ is positive, when it gains an H+ (in the photosystem, for example) why does it become NADPH?

Wouldn't the H+ cause it to become 2+, not cancel the positive charge?

It gains an H- not an H+.

Or you can think of it as reacting with an H2 molecule, keeping an H+ and two electrons, and releasing the other H+.

 

[ahc336]

It gains an H- not and H+.

Or you can think of it as reacting with an H2 molecule, keeping an H+ and two electrons, and releasing the other H+.

You confused the **** out of me. Could you explain further?

 

[theonlytycrane]

As ^ said, a hydride nucleophile attacks the NADP+.

 

[NextStepTutor_3]

This is the same principle as the reduction of NAD+ to NADH, which is especially MCAT-relevant. The most common way this reaction is written out is:

NAD+ + 2e- + H+ ----> NADH
​

As you can see, NAD+ is gaining electrons to become its reduced form, NADH. But it can be confusing to think about where these electrons come from, since H+ itself has one proton and zero electrons. But remember, hydrogen isn't always in the form of H+. It can take the form of hydride (H-) and attack NAD+. Since hydrogen has one proton, a hydride anion must have one proton and two electrons. NAD+ gains these two electrons as it becomes NADH. Alternatively, you can think about NAD+ reacting with H2. H2 has two protons and two electrons in total; NAD+ keeps one of the protons and both the electrons, and releases the remaining H+. Again, this same principle works for the reduction of NADP+ as well as NAD+.

Going back to the original post here - remember, these are redox reactions. If NADP+ were to gain an H+ and become NADPH2+ (which, again, is not what happens), that would not be a redox reaction.

Good luck with your MCAT prep!

 

[ahc336]

NAD+ + 2e- + H+ ----> NADH​

Couldn't you also think of this as NAD+ gaining the H+ (Now at NAD2+) , then being neutralized from the negative on the 2e-?

I think that's the simplest way to explain it, even if it is technically attacked by a hybrid H-

 

[aldol16]

Couldn't you also think of this as NAD+ gaining the H+ (Now at NAD2+) , then being neutralized from the negative on the 2e-?

I think that's the simplest way to explain it, even if it is technically attacked by a hybrid H-

This is a very bad way of explaining it because in this method, the student doesn't make the connection between NADPH and reduction. For instance, NADPH is used to reduce carbonyls in certain instances. Just like in Orgo 2, you learned that reducing a carbonyl requires a hydride source (e.g. NaBH4, LiAlH4, etc.). In biological systems, "hydride" sources are NADH and NADPH. So one must make the connection that NADPH is a hydride source and thus a reducing agent.