How many molecules are formed from the cleavage of glucose during the first step of glycolysis?

The correct answer is- 2 molecules with 3 carbons

Explanation:

Glycolysis is the first step in both aerobic and anaerobic respiration. In glycolysis, glucose is broken down in the absence of oxygen into two molecules of pyruvate which consists of three carbons.

So in glycolysis, glucose is partially oxidized into two molecules of pyruvate. Then in aerobic respiration, this glucose is fully oxidized into CO2 in kreb's cycle. Glucose is a six-carbon compound so it gives 2 molecule of 3 carbon compounds during the first step of glycolysis.

So during this partial oxidation of glucose 2 ATP and 2 NADH is produced with 2 molecule of pyruvate which is a 3 carbon compound.