1. ENZYME KINETIC M. Saifur R, PhD

2. Course content  Enzymatic reaction  Rate of Enzyme-Catalyzed Reactions  Quatification of Substrate Concentration and Reaction Rate  Kinetic Parameters Are Used to Compare Enzyme Activities  Enzymes Catalyze Reactions with Two or More Substrates  Enzymes Are Subject to Reversible or 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 difference in the free energies of their ground states. The free energy of the ground state of P is lower than that of S, so G for the reaction is negative and the equilibrium favors P. The position and direction of equilibrium are not affected by any catalyst.

5.  The rate of a reaction is dependent on an entirely different parameter.  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 for the reaction to proceed in either direction Illustrated by the energy hill To undergo reaction, the molecules must overcome this barrier and therefore must be raised to a higher energy level. Top of the hill  transition state : a condition where the decay of S or P is equally probable.

6. Transition state It is simply a fleeting molecular moment in which events such as bond breakage, bond formation, and charge development have proceeded to the precise point at which decay to either substrate or product is equally likely The difference between the energy levels of the ground state and the 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. Rate of Enzyme-Catalyzed Reactions Factor affecting the rate of a reaction catalyzed by an enzyme : [S] Complicated.......!!! [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] can be limited to a few percent, and [S] can be regarded as constant. Effect on V o when [S] increases

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

9. Leonor Michaelis and Maud Menten in 1913 The combination of an enzyme with its substrate molecule to form an ES complex is a necessary step 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 Concentration and 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 V o 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 E t is E total  [E] t = [E] free + [ES] or [E] free = [E] t – [ES] If [S] >>Et  [S] bound the E is negligible.

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

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

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

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

16. Kinetic Parameters Are Used to Compare Enzyme Activities A double-reciprocal or Lineweaver-Burk plot Michaelis-Menten 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 Reactions with Two or More Substrates

19. Enzymes Are Subject to Reversible or 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 are those that bind covalently with or destroy a functional group on an enzyme that is essential for the enzyme’s activity, or those that form a particularly stable noncovalent association. Reaction of chymotrypsin with diisopropylfluorophosphate (DIFP) irreversibly inhibits the enzyme. This has led to the conclusion that Ser195 is the key active-site Ser residue in chymotrypsin.

24. Enzyme Activity Depends on pH Enzymes have an optimum pH (or pH range) at which their activity is maximal at higher or lower pH, activity decreases. Amino acid side chains in the active site may act as weak acids and bases

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