The binding free energy of a ligand to its receptor protein is −11.5 kJ mol-1 at
300 K. What is the value of the equilibrium binding constant?
Use R = 8.314 J mol-1 K-1.
Options:
(A) Q.01
(B) 1.0
(C) 4.6
(D) 100.5
Official Answer Key:
D
Step-by-Step Solution & Explanation
### **Official Solution & Concept Analysis (GATE BT 2022 Q.15)**
**1. Governing Concept**:
- **Topic**: Biophysical Chemistry & Receptor-Ligand Thermodynamics
- Standard binding free energy is related to the association equilibrium constant $K_a$ (or dissociation constant $K_d = 1/K_a$) by:
$\Delta G^\circ = -RT \ln K_a = RT \ln K_d$
- Spontaneous binding occurs when $\Delta G^\circ < 0$, which corresponds to $K_a > 1$ (or $K_d < 1$).
- More negative $\Delta G^\circ$ corresponds to higher binding affinity (smaller $K_d$).
**2. Option Analysis**:
- **Option (A) is INCORRECT**: Sign inversion.
- **Option (B) is INCORRECT**: Missing temperature factor $RT$.
- **Option (C) is INCORRECT**: Inverts the logarithm dependency.
- **Option (D) is CORRECT**: Matches thermodynamic relation $\Delta G^\circ = -RT \ln K_a$.
**3. Final Answer**:
- **Correct Choice**: **D**
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