1 / 7

Efficiency of Enzyme Catalysis

Efficiency of Enzyme Catalysis.

marged
Download Presentation

Efficiency of Enzyme Catalysis

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Efficiency of Enzyme Catalysis For an example, in the reaction of decomposition of hydrogen peroxide, the activation energy Ea,o of the uncatalyzed reaction at 20oC is 18 kcal/mol, whereas that for chemically catalyzed (Pt) and enzymatically catalyzed (catalase) decomposition are 13 kcal/mol (Ea,c) and 7 kcal/mol (Ea, en), respectively. Compare the reaction rates at these three different conditions.

  2. Efficiency of Enzyme Catalysis The forward reaction rate r (moles/L-s), r = k*f(H2O2) k is the forward reaction rate constant, f(H2O2) is a function of substrate perioxide concentration. 2H2O2 2H2O + O2

  3. If k is represented by k0, kc, and ken , respectively, for the uncatalyzed, chemically catalyzed (Pt) and enzymatically catalyzed reactions, the corresponding reaction rates (r0, rc, ren) can be expressed as follows r0 = k0*f(H2O2) rc = kc*f(H2O2) ren = ken*f(H2O2)

  4. The ratio of the chemically catalyzed rate to the uncatalyzed rate can be calculated by as follows: rc /r0 = kc*f(H2O2)/k0*f(H2O2) Similarly, The ratio of the enzymatically catalyzed rate to the uncatalyzed rate can be calculated by as follows: ren /r0 = ken*f(H2O2)/k0*f(H2O2) At 20oC, with the same initial concentration of reactant hydrogen peroxide, f(H2O2) remains the same for the above three process at initial conditions, the initial reaction rates vary with the rate constants. rc /r0 = kc/k0 ren /r0 = ken/k0

  5. Using Arrhenius Equations yields, rc /r0 = kc/k0 = (A*exp(-Ea,c/R*T))/(A*exp(-Ea,o/R*T)) ren /r0 = ken/k0 = (A*exp(-Ea,en/R*T))/(A*exp(-Ea,o/R*T)) Where Ea,o, Ea,c and Ea,en are activation energy for uncatalyzed reaction, chemically catalyzed and enzymatically catalyzed reactions, respectively. A is a constant and remains same for the specific system, the rate ratios could be simplified to: rc /r0 = (exp(-Ea,c/R*T))/(exp(-Ea,o/R*T)) ren /r0 = (exp(-Ea,en/R*T))/(exp(-Ea,o/R*T))

  6. When T= 20oC = 273+20 (K) = 293 K, Ea,o = 18 kcal/mol, Ea,c = 13 kcal/mol (Ea,c) and Ea, en =7 kcal/mol, respectively. R = 8.314 J mol-1K-1 , 1 cal = 4.18 J. Then substituting the parameters in the rate ratio eqns. with these values, yields, rc /r0 = (exp(-Ea,c/R*T))/(exp(-Ea,o/R*T)) = (exp(-13000*4.18/8.314*293))/(exp(-18000*4.18/8.314*293)) =? ren /r0 = (exp(-Ea,en/R*T))/(exp(-Ea,o/R*T)) =(exp(-7000*4.18/8.314*293))/(exp(-18000*4.18/8.314*293) =?

  7. When T= 20oC = 273+20 (K) = 293 K, Ea,o = 18 kcal/mol, Ea,c = 13 kcal/mol (Ea,c) and Ea, en =7 kcal/mol, respectively. R = 8.314 J mol-1K-1 , 1 cal = 4.18 J. Then substituting the parameters in the rate ratio eqns. with these values, yields, rc /r0 = (exp(-Ea,c/R*T))/(exp(-Ea,o/R*T)) = (exp(-13000*4.18/8.314*293))/(exp(-18000*4.18/8.314*293)) =5.1X103 ren /r0 = (exp(-Ea,en/R*T))/(exp(-Ea,o/R*T)) =(exp(-7000*4.18/8.314*293))/(exp(-18000*4.18/8.314*293) =1.4X108

More Related