GATE BT 2022 | Question: 5

Question

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

Use $\text{R = 8.314 J mol}^{-1} \text{K}^{-1}.$

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

Answer 1

Ans: 1.0

delta G= -RTlnKeq

Given, deltaG= -11.5KJ/mol; R=8.314 J/mol/K; T=300K

Subtitute in formula

-11.5= -8.314* 300* ln Keq

-11.5/-8.314*300= ln Keq

ln Keq= 0.0046

Keq= e^0.0046

Keq= 1.0046

Keq = 1.0