Question

Which of the following is involved in the catalysis of link reaction during aerobic respiration?
(a) Vitamin A
(b) Vitamin $B_{1}$
(c) Vitamin $B_{6}$
(d) Vitamin K

Answer 1

The link reaction is the one that links the products of glycolysis in the cytoplasm to the reactions that take place in the Krebs cycle in the mitochondrial matrix. This reaction requires a vitamin cofactor to the enzymes involved.

Complete answer:
The link reaction which connects glycolysis to Krebs cycle is pyruvate decarboxylation of pyruvate oxidation. It is the conversion of pyruvate into acetyl-CoA catalyzed by the enzyme complex pyruvate dehydrogenase complex.
- The pyruvate dehydrogenase complex contains a coenzyme - thiamine pyrophosphate, which contains the vitamin thiamine, or Vitamin $B_{1}$.
- The conversion of pyruvate, the last product of glycolysis, to acetyl-CoA, the reactant at the beginning of the Krebs cycle require the cofactors $Mg^{+}$ ions, thiamine pyrophosphate, $NAD^{+}$, Co-A, and lipoic acid.

Additional Information:
Many vitamins in our body are required as coenzymes for helping enzymes in reactions.
- The H atoms produced from the oxidation of substrates in glycolysis, the link reaction, and
Kreb’s cycle can enter the ETC via NAD or FAD.
- Vitamin A is a lipid-soluble vitamin that is necessary for the development of retina by aiding in the formation of rhodopsin.
- Vitamin $B_{6}$ or Pyridoxine is required as a coenzyme for reactions in amino acid, glucose, and lipid metabolism.
- Vitamin K or Menaquinone is necessary for synthesis of prothrombin, blood clotting, and reactions in the electron transport chain.

So, the correct answer is ‘(b) Vitamin $B_{1}$’.

Note:
The formation of acetyl-CoA is as follows - the pyruvate molecules enter the mitochondrion, they get decarboxylated and oxidized to form acetyl (2C) molecules. Each 2C molecule combines with Coenzyme A to form acetyl-CoA.
- The $CO_{2}$ is released as a gas during decarboxylation and the H released during oxidation reduces $NAD^{+}$ to NADH.