Metabolism — glycolysis (step 9)
Enolase: dehydration to phosphoenolpyruvate
How the Enolase works, in 3 steps. Each step shows the structures and the curved arrows that move the electrons.
In short. A 3-step mechanism. Typical reagents: 2-Phosphoglycerate, 2 Mg²⁺, enolase (Lys345 base, Glu211 acid). Conditions: Cytosol, pH ≈ 7, 37 °C. Bonds formed: C=C (enol phosphate); H₂O; bonds broken: C2–H; C3–OH. Stereochemistry: Anti elimination of H and OH; reversible. Rate and selectivity: ΔG°′ ≈ +1.8 kJ/mol; PEP has a very high phosphoryl-transfer potential.
At a glance
Step 1: Mg²⁺ binding acidifies C2–H
2-Phosphoglycerate binds with two Mg²⁺ ions that coordinate its carboxylate and phosphate, which makes the α-proton (the C2–H) acidic enough to be pulled off by an active-site lysine.
Step 2: Lysine removes the C2 proton
Removing that proton forms an enolate stabilized by the Mg²⁺ ions. This is an E1cB-type elimination — the same pattern as aldol dehydration — and the C3 hydroxyl is protonated by a general acid so that it can leave.
Step 3: Hydroxide leaves as water
Loss of water forms the C=C bond of phosphoenolpyruvate (PEP). Dehydration alone converts a low-energy phosphate ester into one of the highest-energy phosphate compounds in the cell, because the enol phosphate cannot tautomerize while it stays phosphorylated. That is what allows pyruvate kinase to make ATP in the final step.
Abbreviations and terms
- H2
- Hydrogen gas. Reductant for catalytic hydrogenation.
- E1cB
- Elimination, unimolecular conjugate base. Deprotonation first to a carbanion, then loss of the leaving group.
- ATP
- Adenosine triphosphate. The cell’s energy currency; transfers phosphate groups.
How it compares with related reactions
| Enolase | GAPDH | Hexokinase | |
|---|---|---|---|
| Reagents | 2-Phosphoglycerate, 2 Mg²⁺, enolase (Lys345 base, Glu211 acid) | Glyceraldehyde-3-phosphate (G3P), NAD⁺, inorganic phosphate (Pi), GAPDH (Cys149, His176) | Glucose, Mg·ATP, hexokinase |
| Conditions | Cytosol, pH ≈ 7, 37 °C | Aqueous, pH ≈ 7.4, 37 °C (cytosol) | Aqueous, pH ~7.4, 37 °C |
| Stereochemistry | Anti elimination of H and OH; reversible | Hydride delivered to one face of NAD⁺ (pro-R) | In-line SN2-like, inversion at P |
| Rate / selectivity | ΔG°′ ≈ +1.8 kJ/mol; PEP has a very high phosphoryl-transfer potential | Fast and reversible; couples oxidation to phosphate capture | Irreversible (ΔG°′ ≈ −17 kJ/mol); glucose induces domain closure |
Test yourself
Try to answer each question before opening it.
Which reagents and conditions are typical for the Enolase?
2-Phosphoglycerate, 2 Mg²⁺, enolase (Lys345 base, Glu211 acid). Conditions: Cytosol, pH ≈ 7, 37 °C.
Which bonds are formed and which are broken?
Formed: C=C (enol phosphate); H₂O. Broken: C2–H; C3–OH.
What is the stereochemical outcome?
Anti elimination of H and OH; reversible
What controls the rate or the selectivity?
ΔG°′ ≈ +1.8 kJ/mol; PEP has a very high phosphoryl-transfer potential
Study guides that help with this reaction
- How to learn reaction mechanisms (without memorising everything)
- Curved-arrow notation: rules and the mistakes students make