Considering that all other dehydrogenases of glycolysis and the TCA cycle use NADH as the electron donor, why does succinate dehydrogenase, a component of the TCA cycle and the electron transfer chain, use FAD as the electron acceptor from succinate, rather than \(\mathrm{NAD}^{+}\) ? Note that there are two justifications for the choice of FAD here-one based on energetics and one based on the mechanism of electron transfer for FAD versus \(\mathrm{NAD}^{+}\).

Short Answer

Expert verified
Succinate dehydrogenase uses FAD as the electron acceptor from succinate rather than NAD+ due to differences in energetics and electron transfer mechanisms. The reaction of succinate to fumarate does not produce enough energy to reduce NAD+ to NADH, but is sufficient to reduce FAD to FADH2. Also, FAD has a greater tendency to accept electrons than NAD+ and is tightly bound to succinate dehydrogenase, resulting in more efficient electron transfer.

Step by step solution

01

Understand the role of FAD and NAD+

Flavin adenine dinucleotide (FAD) and nicotinamide adenine dinucleotide (NAD+) are both important coenzymes in biochemical reactions. They act as electron carriers, transporting electrons from one reaction to another within a cell.
02

Discuss the energetics

The energetics of the reaction of succinate to fumarate (catalyzed by succinate dehydrogenase) are not sufficient to reduce NAD+ to NADH. However, the energetics of the reaction are adequate for the reduction of FAD to FADH2. Therefore, energetically, it makes sense for succinate dehydrogenase to use FAD as the electron acceptor.
03

Discuss the mechanism of electron transfer

FAD is a stronger oxidizing agent than NAD+. Therefore, it has a greater tendency to accept electrons. Additionally, FAD is tightly bound to succinate dehydrogenase, unlike NAD+ which is freely diffuse in the cytosol. Tightly bound electron carriers in enzymes are more efficient in transferring electrons directly to the electron transport chain without 'leakage' of high-energy electrons, thereby improving the overall efficiency of energy production pathways of the cell.

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Most popular questions from this chapter

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