Q57 The free energy change of ATP hydrolysis at 25°C is −32 kJ mol−1. The free energy change for the formation of α-glycerophosphate from glycerol and ATP is −8.2 kJ mol−1 (at 25°C). Using the above information, the free energy change for hydrolysis of α-glycerophosphate is ________ kJ mol−1 (rounded off to the nearest integer).
Free Energy Change for Hydrolysis of α-Glycerophosphate: Detailed Calculation with ATP Data
The free energy change for the hydrolysis of α-glycerophosphate is -24 kJ mol-1. This value is derived by coupling the given reactions using Hess’s law for standard free energies. The calculation accounts for the phosphorylation of glycerol by ATP and the reverse hydrolysis process.
Reaction Breakdown
ATP hydrolysis releases energy: ATP + H2O → ADP + Pi, ΔG = -32 kJ mol-1.
The formation reaction given is Glycerol + ATP → α-glycerophosphate + ADP, ΔG = -8.2 kJ mol-1, which implies coupling ATP hydrolysis to drive glycerol phosphorylation.
The target hydrolysis is α-glycerophosphate + H2O → Glycerol + Pi; reversing the formation gives ΔG = +8.2 kJ mol-1, but full coupling yields the direct hydrolysis value.
Step-by-Step Calculation
Write the formation as the sum: (ATP + H2O → ADP + Pi, -32 kJ mol-1) + (Glycerol + Pi → α-glycerophosphate + H2O, +8.2 kJ mol-1, reverse of hydrolysis).
Net: Glycerol + ATP → α-glycerophosphate + ADP, ΔG = -32 + 8.2 = -23.8 kJ mol-1.
Reverse for hydrolysis: ΔG = +23.8 kJ mol-1, rounded to 24 kJ mol-1.
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Core Concept: Coupled Reactions
In biochemistry exams like IIT JAM, calculating the free energy change for hydrolysis of α-glycerophosphate is a common coupled reaction problem. Given ATP hydrolysis ΔG = -32 kJ mol-1 at 25°C and the free energy change for the formation of α-glycerophosphate from glycerol and ATP = -8.2 kJ mol-1, students must find the hydrolysis ΔG for α-glycerophosphate.
Free energy changes add in coupled reactions per Hess’s law. ATP hydrolysis (ATP + H2O → ADP + Pi) provides energy (-32 kJ mol-1) to phosphorylate glycerol (Glycerol + Pi → α-glycerophosphate + H2O), which is endergonic.
Net formation ΔG = -32 + ΔG_phosphorylation = -8.2 kJ mol-1.
Thus, ΔG_phosphorylation (hydrolysis reverse) = -8.2 – (-32) = +23.8 kJ mol-1.
No Options to Explain
This numerical question has no multiple-choice options; it’s a fill-in-the-blank (integer answer). Common errors include forgetting to reverse signs (-8.2 directly) or misadding values (+40 or -40). Always verify units (kJ mol-1) and temperature (25°C standard).
Exam Tips
- Round -23.8 to -24 (nearest integer).
- Practice similar: phosphoenolpyruvate or creatine phosphate couplings.
- Answer: -24 kJ mol-1
