The concentrations of ATP, ADP and inorganic phosphate in a cell are 2.59, 0.73
and 2.72 mM, respectively. Under these conditions, free energy change for the
synthesis of ATP at 37 °C is
kJ/mol (round off to 2 decimal places).
Given: free energy change for ATP hydrolysis under standard conditions is -30.5
kJ/mol and R = 8.315 kJ/mol.K
Official Answer Key:
48.50 to 49.50 OR 18521.20 to 18521.30
Step-by-Step Solution & Explanation
### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2020 Q.63)**
**1. Governing Concept & Definition**:
- **Topic**: Cellular Bioenergetics & Phosphorylation Potential
- The actual physiological Gibbs free energy of ATP hydrolysis ($\Delta G$) is:
$\Delta G = \Delta G^{\circ\prime} + R T \ln\left(\frac{[\text{ADP}][\text{P}_i]}{[\text{ATP}]}\right)$
**2. Detailed Step-by-Step Calculation**:
- **Given Concentrations**:
- $[\text{ATP}] = 2.59\text{ mM} = 2.59 \times 10^{-3}\text{ M}$
- $[\text{ADP}] = 0.73\text{ mM} = 0.73 \times 10^{-3}\text{ M}$
- $[\text{P}_i] = 2.72\text{ mM} = 2.72 \times 10^{-3}\text{ M}$
- **Mass Action Ratio ($Q$)**:
$Q = \frac{(0.73 \times 10^{-3})(2.72 \times 10^{-3})}{2.59 \times 10^{-3}} = \frac{1.9856 \times 10^{-6}}{2.59 \times 10^{-3}} \approx 7.666 \times 10^{-4}\text{ M}$
- With $\Delta G^{\circ\prime} = -30.5\text{ kJ/mol}$ and $T = 310\text{ K}$ ($37^\circ\text{C}$):
$\Delta G = -30.5 + 8.314 \times 10^{-3} \times 310 \times \ln(7.666 \times 10^{-4}) \approx -30.5 - 18.5 \approx -49.0\text{ kJ/mol}$
(Magnitude of free energy change $= 49.0\text{ kJ/mol}$).
**3. Final Answer**:
- **Calculated Value**: **49.0 kJ/mol**
- **Official GATE Accepted Range**: **48.50 to 49.50**
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.