Question

The enzymes catalysing reactions 1, 2, and 3 respectively in the above steps are

• A. Phosphoglyceromutase, Enolase, Pyruvate kinase
• B. Enolase, Phosphofructoknase, Pyruvate kinase
• C. Phosphoglycerokinase, Enolase, Pyruvate kinase
• D. G-3-P dehydrogenase, Enolase, Phosphoglyceromutase

Answer 1

The correct option is A Phosphoglyceromutase, Enolase, Pyruvate kinase

Glycolysis is the metabolic process that involves the breakdown of one molecule glucose into two molecules of pyruvic acid to extract energy for cellular metabolism.

The steps of EMP pathway are as follows:

1) Glucose and fructose are phosphorylated to give rise to glucose-6-phosphate by the activity of the enzyme hexokinase and by utilizing ATP.

2) Conversion of glucose-6-phosphate to fructose-6-phosphate occurs due to the enzyme phosphoglucose isomerase.

3) Conversion of fructose 6-phosphate to fructose 1, 6-biphosphate occurs by the utilization of ATP molecule in the presence of phosphofructokinase.

4) Fructose 1, 6-biphosphate is split into dihydroxyacetone phosphate and glyceraldehyde-3-phosphate by aldolase enzyme.

5) When glyceraldehyde-3-phosphate gets converted into 1, 3-bisphosphoglyceric acid in the presence of glyceraldehyde-3-phosphate dehydrogenase, NAD+ reduces to NADH and H+.

6) The conversion of 1, 3-bisphosphoglyceric acid to 3-phosphoglyceric acid by phosphoglycerate kinase, is an energy-yielding process, i.e. ATP is released in this step.

7) 3-phosphoglyceric acid is converted into its isomer, 2-phosphoglyceric acid by phosphoglycerate mutase.

8) 2-phosphoglyceric acid loses a molecule of water, becoming phosphoenolpyruvate in the presence of enolase.

9) Phosphoenolpyruvate loses its phosphate group to get converted into pyruvic acid in the presence of pyruvate kinase with the release of ATP molecule.

So, the answer is A.