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