Answer:
The correct order is:
1.- pyruvate and thiamine pyrophosphate react form hydroxyethyl-TPP, CO2 is released
2.- an acetyl group is added to a coenzyme bound to PDH, forming acetyl-dihydrolipoamide
3.- the acetyl group is transferred to CoA, forming acetyl-CoA
4.- the reduced coenzyme bound to PDH is reoxidized using electrons from FAD
5.- electrons from FADH2 are transferred to NAD+ forming NADH and H+
Explanation:
Pyruvate reacts with thiamine pyrophosphate (TPP) on the complex enzyme pyruvate dehydrogenase (E1). This process cleaves CO2 and generates acetaldehyde bound to TPP or hydroxyethylamine. E2 is Dihydrolipoyl-transacetylase and has Lipoamide attached. In the transfer from E1 to E2, the hydroxyethyl is oxidized to acetyl, at the same time that the disulfide bridge of the lipoamide is reduced to dihydro-lipoamide, although as this is acetylated, the product is acetyl-dihydrolipoamide. Acetyl group of acetyl-lipoamide (thio-acyl bond) is transferred to CoA-SH, to form Acetyl-S-CoA (thio-acyl bond).Acetyl-CoA is released into the mitonchondrial matrix. Lipoamide is reduced as Dihydro-lipoamide. The last reaction is catalyzed by dihydrolipoyl dehydrogenase, where it is reestablished in the disulfide bond of lipoamide to make it reactive again. For this, FAD is used as an intermediate, which is reduced to FADH2 and oxidizes the sulfides (which are then linked again). Finally, FADH2 reacts with NAD, which is reduced to NADH and H+ released, restoring FAD for a new reaction.
