GATE BT 2025 • Q65 NAT • 2 Marks

Section 2: General Biotechnology

Bioenergetics and Metabolism

Subtopic: Thermodynamics & ATP Bioenergetics

Question Statement:

For the coupled reactions given below Glucose 6-phosphate + H2O → Glucose + Pi (Reaction 1) ATP + Glucose → ADP + Glucose 6-phosphate (Reaction 2) the standard free energy change of ATP hydrolysis at 25 °C is ____ kJ/mol. The equilibrium constants for Reaction 1 and Reaction 2 are 360 and 800, respectively; Gas constant R = 8.314 J mol-1 K-1. (Round off to two decimal places)

Official Answer Key:

-32.00 to -30.00

Step-by-Step Solution & Explanation

### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2025 Q.65)** **1. Governing Concept & Definition**: - **Topic**: Biochemistry & Bioenergetics of Coupled Reactions - **Target Reaction (ATP Hydrolysis)**: $\text{ATP} + \text{H}_2\text{O} \rightleftharpoons \text{ADP} + \text{P}_i$ - Coupling Reaction 1 and Reaction 2: - Reaction 1: $\text{G6P} + \text{H}_2\text{O} \rightleftharpoons \text{Glucose} + \text{P}_i \quad (K_1 = 360)$ - Reaction 2: $\text{ATP} + \text{Glucose} \rightleftharpoons \text{ADP} + \text{G6P} \quad (K_2 = 800)$ - Adding Reaction 1 + Reaction 2 yields overall ATP hydrolysis: $K_{\text{overall}} = K_1 \times K_2$ **2. Detailed Step-by-Step Calculation**: - **Step 1: Calculate overall equilibrium constant**: $K_{\text{overall}} = 360 \times 800 = 288,000$ - **Step 2: Calculate standard Gibbs free energy change ($\Delta G^\circ$)**: $\Delta G^\circ = -R T \ln(K_{\text{overall}})$ - **Given**: - $T = 25^\circ\text{C} = 298.15\text{ K}$ - $R = 8.314\text{ J mol}^{-1}\text{K}^{-1}$ - $\ln(288,000) \approx 12.5707$ - **Step 3: Substitution**: $\Delta G^\circ = -8.314 \times 298.15 \times 12.5707 \approx -31,160\text{ J/mol} = -31.16\text{ kJ/mol}$ **3. Final Answer**: - **Calculated Value**: **-31.16 kJ/mol** - **Official GATE Accepted Range**: **-32.00 to -30.00**

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