E3 catalyzes rxns 4 and 5

Question 1

E3 has what ?

Answer 1

E3 has covalently bound FAD which is a prosthetic group.
Also binds NAD+ but NONcovalently (will leave as NADH)

Question 2

Oxidized E3

Answer 2

has a redox active disulfide bond…
Is between cys43 and cys48. (low and high number cys).
E3 will accept an e- pair to form a dithiol. This is a redox rxn, involves the disulfide interchange which includes:
- proton transfer
- e- transfer

Will be transferring one as a proton and one as a hydride.

Question 3

What is a disulfide interchange

Answer 3

when you have one disulfide bond and swap to the other one???

Question 4

Where does substrate binding occur

Answer 4

Occurs at subunit interface, near the redox active R-S-S-R (disulfide bond)

Question 5

What type of mechanism is this?

Answer 5

ping-pong mechanism. Doing a rxn where something leaves midway and leaves something behind attached to the enzyme.

Question 6

stage one

Answer 6

LH2 is converted to L
(dihydrolipoamide converted to lipoamide)

Question 7

stage two

Answer 7

NAD+ is going to NADH

Question 8

If NAD+ is not present

Answer 8

Tyrosine is going to lay over the binding pocket like a gate (solution shield)
reason: prevents non-specific e- transfer (doing the wrong redox rxn, which could be with Oxygen because this is taking pkace in the mitochondria)

Question 9

If NAD+ is present

Answer 9

If NAD+ is present, tyrosine must move out of the way, reason: so the FAD and NAD+ rings can be parallel and in Va der waals contact.

vdW: touching each other at their outermost e- , in order to trasnfer e- ….

Where do these transfer from and to??
NAD+ C4 and N5 of FADH-

Question 10

Stereospecificity of Hydride transfer

Answer 10

H:- is added to the si face of NAD+ at C4 from N5 of FADH- to give PRO-S (means youve already lost one of the hydrogens from FADH2, lsot as a proton so will now lose the hydride to go to FAD)

Question 11

Active site is where?

Answer 11

at the subunit interface…

Question 12

What happens in the reation?

Answer 12

1. Dihydrolipoamide has its proton removed so it can do a nucleophilic displacement as part of the disulfide interchange that will happen.
2. creates a mixed disulfide between lipoamide and lower numbered cysteine.
3. high number Cys attacks C4a (ring junction C between rightmost rings)
4. forms a charge transfer complex (CTC), which transfers charge from disulfide system to FAD system via temporary covalent bond (higher numbered cys and FAD ring sys at C4a)
5. complete disulfide interchange by having cysteines reform disulf bond… will break CTC bond between high number Cys as C4a which completes charge transfer to the FAD, which gets protonated.
6. FADH2 is oxidized by NAD+ being reduced as well.