ENZYME KINETIC

1. ENZYME KINETIC M. Saifur R, PhD

2. Course content • Enzymatic reaction • Rateof Enzyme-Catalyzed Reactions • Quatification of Substrate Concentrationand Reaction Rate • Kinetic Parameters Are Used to CompareEnzyme Activities • Enzymes Catalyze Reactionswith Two or More Substrates • Enzymes Are Subject to Reversibleor Irreversible Inhibition • Enzyme Activity Depends on pH

3. Enzymatic Reaction • What is the catalysis.........? • To answer the question, we must understand the different between the reaction rate and equilibrium. • The catalist is affect the reaction rate but not equilbrium E : Enzyme; S : Substrate; P : Product Enzyme is catalist

4. The different of reaction rate and equilibrium The starting point for either the forward or the reverse reaction is called the ground state. The equilibrium between S and P reflects the differencein the free energies of their ground states. The free energy of theground state of P is lower than that of S, so G for thereaction is negative and the equilibrium favors P. Theposition and direction of equilibrium are not affected byany catalyst.

5. Therate of a reaction is dependent on an entirely differentparameter. • There is an energy barrier between S and P: • the energy required for alignment of reacting groups • formation of transient unstable charges • bond rearrangements • other transformations required forthe reaction to proceed in either direction Illustrated by the energy hill Toundergo reaction, the molecules must overcome thisbarrier and therefore must be raised to a higher energylevel. Top of the hill  transition state : a condition where the decay of S or P is equally probable.

6. Transition state It is simply afleeting molecular moment in which events such as bondbreakage, bond formation, and charge developmenthave proceeded to the precise point at which decay toeither substrate or product is equally likely The differencebetween the energy levels of the ground state andthe transition state is the activation energy, G‡. Higher activation energy slower the reaction. Reaction rate can be increased by incresaing temperature or adding the catalyst. Enzyme dcreases the activation energy

7. Rateof Enzyme-Catalyzed Reactions Factor affecting the rate of a reaction catalyzedby an enzyme : [S] Complicated.......!!! Effect on Vowhen [S] increases [S] is change because S is converted to P Measure the initial rate (Vo), when [S] >> [E] In a typical reaction [S] = 6x [E] At the beginning of the reaction is monitored(often the first 60 seconds or less), changes in [S] canbe limited to a few percent, and [S] can be regarded asconstant.

8. Substrate Saturation of an Enzyme A. Low [S] B. 50% [S] or Km C. High, saturating [S]

9. Leonor Michaelis and Maud Menten in 1913 The combination of an enzyme with its substratemolecule to form an ES complex is a necessarystep in enzymatic catalysis. First step E combines reversibly with S at very fast level Second step ES complex breakdown is slow. Rate limiting step of the recation

10. Quatification of Substrate Concentrationand Reaction Rate At low [S] or at the biginning of the reaction, the recation tend to [ES] At high [S], when no free E then the reaction reach to maximum or the E is saturated

11. What is the realtionship between the reaction rate and the [S].....? At the biginning of the reaction : [P] is neglegible then the k-2 is ignored If Vo is the rate of the breakdown of ES to form a P then : The [ES]isdifficult to measure experimentally then we need another alternative. If Et is E total  [E]t = [E]free + [ES] or [E]free = [E]t – [ES] If [S] >>Et  [S] bound the E is negligible.

12. First step: Rate of ES formation : k1[Et – ES][S] ..................(1) Rate of ES breakdown : k-1[ES] + k2 [ES] ...........(2) Second step: Steady state assumption: [ES] is constant  the formation of ES = breakdown of ES, then : k1[Et – ES][S] = k-1[ES] + k2[ES] ..............(3)

13. Third step: ..............(4) ..............(5) Adding the term of to both side of the Eq. 4 then ..............(6) ..............(7)

14. is Michaelis constant or Km ..............(8) then, in term of ES, we can express Vo as : ..............(9)

15. ..............(9) Vmax is reached when the E is saturated (it means that Et = ES), so Vmax = k2[Et], then Michaelis-Menten equation, the rateequation for a one-substrate enzyme-catalyzed reaction.

16. Kinetic Parameters Are Used to CompareEnzyme Activities Michaelis-Menten plot A double-reciprocal or Lineweaver-Burk plot

17. Km sometimes can be used to describe the enzyme affinity to its substrate. If k2<<k-1 then Km is reduced to k-1/k1 and is called by dissociation constant. kcat = k2 = Vmax/[Et] turn over number : the number of S molecule coverted to P at any given time on a single molecule of E whenE is saturated with S. Catalytic efficiency = kcat/Km

18. Enzymes Catalyze Reactionswith Two or More Substrates

19. Enzymes Are Subject to Reversibleor Irreversible Inhibition Reversible Inhibition Competitive : I bind to active site Uncompetitive : I bind to other site, but only to ES complex Mixed (Noncompetitive) : I bind to other site, either to E or ES complex

20. Competitive Uncompetitive

21. Mixed (Noncompetitive)

23. Irreversible Inhibition The irreversible inhibitors arethose that bind covalently with or destroy a functionalgroup on an enzyme that is essential for the enzyme’sactivity, or those that form a particularly stable noncovalentassociation. Reaction of chymotrypsin withdiisopropylfluorophosphate (DIFP) irreversibly inhibits the enzyme.This has led to the conclusion that Ser195 is the key active-site Serresidue in chymotrypsin.

24. Enzyme Activity Depends on pH Enzymes have an optimum pH (or pH range) at whichtheir activity is maximal at higher or lowerpH, activity decreases. Amino acidside chains in the active site may act as weak acids andbases

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