Question

In which of the following reactions of glycolysis, a molecule of water is removed from the substrate?
A. Glucose → Glucose-6-phosphate
B. Fructose-6-phosphate → Fructose-1,6 bisphosphate
C. 2-phosphoglycerate → Phosphoenolpyruvate
D. Phosphoenolpyruvate → Pyruvate

Answer 1

Glycolysis is the mechanism by which one glucose molecule is converted into two pyruvate molecules, two hydrogen ions, and two water molecules. This method synthesizes the 'high energy' molecules $ATP$ and $NADH$. The pyruvate molecules then lead to a link reaction in which $acetyl-coA$ is formed.

Complete answer:
The process of glycolysis occurs in 10 stages which are explained below:
Step 1: A phosphate group is added to glucose through the action of the enzyme hexokinase in the cell cytoplasm. In this, from $ATP$ a phosphate group is transferred to form $glucose$ $6-phosphate$,
Step 2: $Glucose-6-phosphate$ is isomerized by the enzyme $phosphoglucomutase$ into $fructose\text{ }6-phosphate$.
Step 3: The other $ATP$ molecule passes the phosphate group to $fructose\text{ }6-phosphate$ and, through the action of the enzyme phosphofructokinase, transforms it into $fructose\text{ }1,6-bisphosphate$.
Step 4: $fructose\text{ }1,6-bisphosphate$is converted by the enzyme aldolase into $glyceraldehyde\text{ }3-phosphate$ and $dihydroxyacetone\text{ }phosphate$, which are isomers of each other.
Step 5: Dihydroxyacetone phosphate is converted into $glyceraldehyde\text{ }3-phosphate$ by triose-phosphate isomerase, which is the substrate in the glycolysis step.
Phase 6: There are two reactions to this step:
The $glyceraldehyde\text{ }3-phosphate\text{ }dehydrogenase$ enzyme transfers 1 molecule of hydrogen to form $NAD{{H}^{+}}and\text{ }{{H}^{+}}$from glyceraldehyde phosphate to nicotinamide adenine dinucleotide.
To form $1,3-bisphosphoglycerate,\text{ }glyceraldehyde\text{ }3-phosphate\text{ }dehydrogenase$ adds phosphate to the oxidized glyceraldehyde phosphate.
Step 7: With the aid of phosphoglycerokinase, phosphate is transferred from $1,3-bisphosphoglycerate$ to ADP to form $ATP$. Two $phosphoglycerate$ and $ATP$molecules are thus obtained at the end of this reaction.
Step 8: From the third to the second carbon, the phosphate of both $phosphoglycerate$ molecules is transferred to create two 2- $phosphoglycerate$ molecules by the phosphoglyceromutase enzyme.
Step 9: To form phosphoenolpyruvate, the enzyme enolase extracts a water molecule from 2- $phosphoglycerate$.
Step 10: By the action of pyruvate kinase, a phosphate from phosphoenolpyruvate is transferred to ADP to form pyruvate and $ATP$. Two pyruvate ions and $ATP$ are the end products.
Options A and B reflect the addition, respectively, of the phosphate group to glucose and fructose 6 phosphates.
A water molecule is eliminated during the formation of Phosphoenolpyruvate (PEP) from 2-phosphoglycerate. The enzyme enolase catalyzes this reaction. This is the glycolysis stage. This reaction in the cytoplasm occurs. Reversible removal of a molecule of water from $2-phosphoglycerate$ requires dehydration of $2-phosphoglycerate$ to $phosphoenolpyruvate$.
Option D illustrates phosphoenolpyruvate dephosphorylation into pyruvate.

So, choice C-2-phosphoglycerate → Phosphoenolpyruvate is the correct answer.

Note: Humans use $NADH$ supplements as medicines. $NADH$ is used to enhance mental clarity, alertness, focus, and memory, as well as to treat Alzheimer's disease and dementia. Due to its role in energy production, $NADH$ is also used to improve athletic performance and treat chronic fatigue syndrome (CFS).