Question

The binding free energy of a ligand to its receptor protein is \( -11.5 \, \text{kJ mol}^{-1} \) at 300 K. What is the value of the equilibrium binding constant?

• A. 0.01
• B. 1.0
• C. 4.6
• D. 100.5

Hint:

The equilibrium binding constant \( K \) can be calculated from the binding free energy using the equation \( \Delta G = -RT \ln K \).

Answer 1

The Correct Option is D

To find the equilibrium binding constant \( K \), we use the relationship between the binding free energy \( \Delta G \) and the equilibrium constant \( K \), which is given by the following equation: \[ \Delta G = -RT \ln K \] Where:
- \( \Delta G = -11.5 \, \text{kJ mol}^{-1} = -11.5 \times 10^3 \, \text{J mol}^{-1} \) (since \( 1 \, \text{kJ} = 1000 \, \text{J} \)),
- \( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \) is the gas constant,
- \( T = 300 \, \text{K} \) is the temperature in Kelvin,
- \( K \) is the equilibrium binding constant (which we need to calculate).
First, we rearrange the equation to solve for \( K \): \[ K = \exp\left(\frac{-\Delta G}{RT}\right) \] Substitute the given values into the equation: \[ K = \exp\left(\frac{-(-11.5 \times 10^3)}{8.314 \times 300}\right) \] \[ K = \exp\left(\frac{11.5 \times 10^3}{2494.2}\right) \] \[ K = \exp(4.62) \] \[ K \approx 100.5 \] Thus, the value of the equilibrium binding constant is \( 100.5 \), so the correct answer is (D).