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

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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: 


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