The forward rate constant for an uncatalyzed reaction is 10-4 s-1; an enzyme binds to the transition state of this reaction with the free energy ΔG = -50 kJ/mol; temperature = 300K.
a) Find the kcat.
b) Assuming that the enzyme can be represented by a sphere with r = 5 nm and the substrate can be
represented by a sphere with radius r = 2 nm, determine the rate of this reaction in sucrose solution with
viscosity of 3 cP. The substrate concentration is 100 mM, the enzyme concentration is 1 nM. Quarter of all
collisions result in a complex formation.
Additional assumptions and points to consider:
The enzyme-catalyzed reaction can be described as E S ES E P a cat The uncatalyzed reaction is S P uncat k
Viscosity has no effect on the enzyme motions involved in catalysis and that due to steric restrictions only 25% of collisions between the enzyme and substrate result in productive collision (i.e. in the formation of the ES complex).
From an orthogonal experiment you know that once bound, the substrate undergoes chemical transformation (i.e. dissociation rate constant of the substrate from the ES complex (k-1) is negligibly small when compared to kcat)
To determine the overall reaction rate you will need to take into account both catalyzed and uncatalyzed reactions
Debye-Smoluchowski equation for the second order association rate: