MUSCLE PHYSIOLOGY

PHYSIOLOGY

MUSCLE ENERGETICS TCA CYCLE

Question [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
FAD is reduced in which of the reaction of the Kreb’s cycle?
A
Isocitrate to oxaloacetate
B
Succinyl CoA to Succinate
C
Fumarate to malate
D
Succinate to fumarate
Explanation: 

Detailed explanation-1: -During Krebs cycle, Flavin Adenine Dinucleotide or FAD gets reduced to FADH2 during oxidation of succinic acid to fumaric acid.

Detailed explanation-2: -In step 6 of the citric acid (Krebs) cycle, FAD is reduced by succinate to form FADH2 and fumurate. The FAD picks up to hydrogens from the central two carbons of succinate to form FADH2. The FADH2 is subsequently oxidized back to FAD by the molecule ubiquinone as it becomes ubiquinol in its reduced form.

Detailed explanation-3: -The succinate dehydrogenase complex (SDH), associated with the inner mitochondrial membrane, catalyzes the dehydrogenation of succinate to fumarate, reducing the FAD cofactor bound to the enzyme. This redox potential is then used in the electron transfer chain to drive a proton motive force to generate ATP.

Detailed explanation-4: -FAD can be reduced to FADH2 through the addition of 2 H+ and 2 e−. FADH2 can also be oxidized by the loss of 1 H+ and 1 e− to form FADH. The FAD form can be recreated through the further loss of 1 H+ and 1 e−. FAD formation can also occur through the reduction and dehydration of flavin-N(5)-oxide.

Detailed explanation-5: -In the sixth step, succinate is converted to fumarate; here, a dehydrogenation takes place because two protons are removed. FAD serves as coenzyme wich is bound covalently to the enzyme succinate dehydrogenase, so that usually the notation E-FAD is used.

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