Question

The reaction of forming 3-phosphoglyceric acid in glycolysis is
a. Cleavage
b. Oxidative phosphorylation
c. Dephosphorylation
d. Oxidative decarboxylation

Answer 1

Glycolysis is a metabolic process. It is found in both aerobic and anaerobic respiration. In this process, glucose is converted to pyruvate. 3-phosphoglyceric acid helps in yielding ATP molecules by the irreversible reaction.

Complete answer:

Option A:
Cleavage is the breaking of covalent bonds in a molecule which leads to smaller molecules. In glycolysis, the glucose molecule is broken down into 2 NADH, 2 pyruvate, and 2 ATP molecules. In stage 2, cleavage of fructose results in carbon fragments. Option A is incorrect.

Option B:
In Oxidative Phosphorylation ATP is formed. This result is due to the transfer of electrons from FADH2 or NADH to oxygen by a series of electron carriers. This process is the major source of ATP in aerobic organisms and takes place in mitochondria. The chemiosmosis and electron transport chain together make for oxidative phosphorylation. Option B is incorrect.

Option C:
In Dephosphorylation, a phosphate group is added to form ATP and 3-phosphoglycerate. The enzyme phosphoglycerate kinase and the cofactor magnesium convert 1,3-bisphosphoglycerate to 3-phosphoglycerate. This yields two ATP molecules. This ATP formation is called substrate phosphorylation or direct phosphorylation. The reactions of phosphorylation and dephosphorylation are irreversible.

Option D:
Oxidative decarboxylation takes place after the process of glycolysis and before the process of the Krebs Cycle. In this, NADH is produced in mitochondria. In Krebs Cycle, Oxidative decarboxylation converts pyruvate into acetyl-CoA. Option D is incorrect.

Hence, the correct answer is option (C).

Note: The reactions of phosphorylation and dephosphorylation are irreversible. In the process of glycolysis 4ATP and $2NADH_2$ are formed and the net gain is 2ATP and $2NADH_2$. Glycolysis takes place in the cytoplasm and is also called an EMP pathway.