The EGFR protein kinase domain has inactive (P′) and active conformation (P). Only the active conformation (P) binds ATP (L) with a KD = 1.200 mM. T=300K. Use 3 decimal places for protein and ATP concentrations in mM and 1 decimal places for inhibitor concentrations in nM for the problems below.
A Solution has 1.000 Mm protein kinases only ([P] + [P']) and it was found at equilibrium that 80% of the protein is in active from (P). Use 3 decimal places for concentaruins of protein and ATP in mM when you solve the problems beliw. What is the free energy difference between the two conformations ∆G = G(P) - G(P') ub kJ/mol? round ti two decimal places places (10 points)