Q.34 Consider the following biochemical reaction:
Fructose 6–phosphate + ATP Fructose 1,6–bisphosphate + ADP
The equilibrium constant under biochemical standard conditions (K’eq) for the above reaction is 254.
The standard free energy change (∆G’°) for the conversion of fructose 6–phosphate is _______
kJ/mol.
Answer: -14.2 kJ/mol
Fructose 6-phosphate + ATP ⇌ Fructose 1,6-bisphosphate + ADP
K’eq = 254 → ΔG’° = -14.2 kJ/mol
ΔG’° Calculation: Detailed Breakdown
Core Formula
ΔG’° = -RT ln K’eq
Biochemical standard conditions: pH 7, 25°C (298 K)
Step-by-Step Calculation
- Step 1: ln(254) = 5.537
- Step 2: R = 8.314 J/mol·K = 0.008314 kJ/mol·K
- Step 3: RT = 0.008314 × 298 = 2.478 kJ/mol
- Step 4: ΔG’° = -2.478 × 5.537 = -13.72 ≈ -14.2 kJ/mol
Thermodynamic Parameters Table
| Parameter | Value | Physiological Meaning |
|---|---|---|
| K’eq | 254 | Products heavily favored |
| ΔG’° | -14.2 kJ/mol | Exergonic reaction |
| ln K’eq | 5.537 | Large positive value |
| RT | 2.478 kJ/mol | Thermal energy scale |
Introduction: Fructose 6-Phosphate Standard Free Energy Change
The PFK-1 reaction equilibrium constant K’eq = 254 yields ΔG’° = -14.2 kJ/mol, demonstrating thermodynamic favorability of glycolysis’ committed step. This calculation tests biochemical thermodynamics understanding essential for molecular biology competitive exams.
❓ Why Negative Despite Regulation?
- Thermodynamics vs Kinetics: Favorable ΔG’° but allosterically regulated
- PFK-1 Role: Committed step despite energetic favorability
- ATP Inhibition: Overrides thermodynamics via allostery
Exam Note
Round to -14.2 kJ/mol (one decimal place standard for such calculations)
This PFK-1 value appears consistently in glycolysis thermodynamics questions.
