Question:
Which one of the following conversion in glycolysis shows formation of water molecules from the substrate?
Which one of the following conversion in glycolysis shows formation of water molecules from the substrate?
Show Hint
Enolase removes water from 2-phosphoglycerate.
Updated On: Nov 12, 2026
- Fructose 6 phosphate \(\rightarrow\) Fructose 1,6 diphosphate
- Phosphoenolpyruvate \(\rightarrow\) Pyruvic acid
- 2 Phosphoglycerate \(\rightarrow\) 2 Phosphoenolpyruvate
- Glyceraldehyde 3 phosphate \(\rightarrow\) 1,3 diphosphoglycerate
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
In glycolysis, the enzyme enolase acts on 2-phosphoglycerate and removes one water molecule (dehydration).
Step 2: Product
This gives phosphoenolpyruvate, which has a high-energy enol phosphate bond.
Step 3: Other steps
Fructose 6-P to fructose 1,6-bisP is phosphorylation. PEP to pyruvate forms ATP, and G3P to 1,3-bisPGA is oxidation with NADH formation. None of these releases water.
Final Answer:
Water is formed in the 2-phosphoglycerate to phosphoenolpyruvate step. \[ \boxed{\text{(C) }\text{2 Phosphoglycerate}\ \rightarrow\ \text{2 Phosphoenolpyruvate}} \]
In glycolysis, the enzyme enolase acts on 2-phosphoglycerate and removes one water molecule (dehydration).
Step 2: Product
This gives phosphoenolpyruvate, which has a high-energy enol phosphate bond.
Step 3: Other steps
Fructose 6-P to fructose 1,6-bisP is phosphorylation. PEP to pyruvate forms ATP, and G3P to 1,3-bisPGA is oxidation with NADH formation. None of these releases water.
Final Answer:
Water is formed in the 2-phosphoglycerate to phosphoenolpyruvate step. \[ \boxed{\text{(C) }\text{2 Phosphoglycerate}\ \rightarrow\ \text{2 Phosphoenolpyruvate}} \]
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