Enzyme Kinetics PDF

Kinetics: Rate of enzymatic reactions Dr. Nadine Karaki  1. A. Chemical reactions 1 3 2 K1 k2 In case of multistep reactions :A  I P (made of 3 elementary reactions) I symbolize inte...

Kinetics: Rate of enzymatic reactions Dr. Nadine Karaki  1. A. Chemical reactions 1 3 2 K1 k2 In case of multistep reactions :A  I P (made of 3 elementary reactions) I symbolize intermediates The rate of overall reaction is determined primarily by the slowest step called the rate limiting step.  Reaction Rates A P A  P  Change in [reactant] or [product] per unit time=> v  -  t t A P  Rate law= rate equation aA + bB  P P Rate= K [A]m[B]n K= rate constant= proportionality constant The rate constant k and the reaction orders m and n must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed.  Reaction Order aA + bB  P => Rate= K [A]m[B]n Order= Molecularity Molecularity: the nb of molecules that must simultaneously collide in the elementary Rx. Sum of the exponents in the rate equation m+n rate = k[NO]2[H2]=>overall order?  Reaction order A P V= K [A]0 zero order reaction K= rate constant V= K [A]0 => v= K (M.s-1) UNIT OF K IS M/S or M.s-1 A reaction is zero order if the change in [reactant] produces no effect.  Reaction order Unimolecular (M. s-1 ) s-1 M A reaction is 1st order if doubling the [reactant] causes the rate to double.  Reaction order Bimolecular (M. s-1 ) (M-1. s-1 ) M2 Bimolecular (M. s-1 ) (M-1. s-1 ) A reaction is 2nd order if doubling the [reactant] causes a quadruple increase in rate.  Half Time (t1/2) How long it takes to consume 50% of reactants.  Slope= -K t1  A 0 2 2k Concentration-time equation Half of initial reacted, [A]= [A]0/2 First order lnA t  - kt  lnA 0 ln[A] T0.5 and k have same unit( s or min) Half time is cte and independent of [reactant]i  Second order : 1 reactant 2A → P 1 1  kt  slope = k At A0 1 [A] Half-life is Inversely 1 [A]o proportional to [A]0 1 t1  kA 0 time 2 K is in M-1. s-1 From the unit of K we can determine the order of the reaction  Model of kinetics Transition State Theory Also known as the absolute rate theory How reactants convert to products Molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation  Transition state theory Minimum amount of energy required for reaction: the activation energy, Ea. Molecules must possess sufficient energy to get over the activation energy barrier. Ea Transition State diagram= Rx coordinate diagram Activated complex (transition state) Ea (Min free energy pathway of a reaction) Helps to visualize the energy changes throughout a process  If reactants ≠ products The transition state diagram is no longer symmetrical (Ea) because there is a free energy difference between reactants and products.  d. Catalysis Reduces G‡ ratecatalyzed > rateuncatalyzed General mechanism of an enzyme: It actsGibbs free energy by reducing theofactivation activation,energy ∆G‡,  2. ENZYME Enzymes are biological catalyst that bind the substrate. They form a temporary bound intermediate complexes with their substrates. E + S  ES complex (stabilized by molecular interactions)  2.1 Enzyme Kinetics Is the study of the chemical reactions that are catalyzed by enzymes. It a measure of the reaction rate It investigates the effects of varying the conditions of the reaction.  General observations Enzymes are able to exert their influence at very low [E] The initial rate (velocity ) is linear with [E] The initial velocity increases with the [S] at low [S] The initial velocity approaches a maximum at high [S] “saturation”  Michaelis–Menten They proposed a model to explain how an enzyme can cause kinetic rate enhancement of a reaction and explains how reaction rates depends on the [S]. The simplest mechanism is to consider that: one Substrate, one intermediate and one product rate limiting step The catalytic step is the rate- limiting step Michaelis–Menten plot  Derivation of Michaelis–Menten Equation [ES] remains ~ cte until S is exhausted. [ES] maintains a steady state d[ES]/dt =0 V appearance of ES = V disappearance of ES (Assumption of equilibrium) => k1[E][S]= k-1[ES] + k2[ES] (1)  Derivation of Michaelis-Menten equation [ES] remains ~ cte until S is exhausted. [ES] maintains a steady state d[ES]/dt =0 V appearance of ES = V disappearance of ES (Assumption of equilibrium)=> k1[E][S]= k-1[ES] + k2[ES] (1) steady state assumption: d[ES]/dt =0  k1[E][S] - k-1[ES] - k2[ES] =0 (2) [E]T = [E] + [ES] Replace [E] by [E]T-[ES] in equation (2)   k1[E][S] - k-1[ES] - k2[ES] =0  k1( [E]T - [ES] ) [S] - (k-1 + k2) [ES]=0  k1( [E]T - [ES] ) [S] = (k-1 + k2) [ES]  K1[E]T [S] – K1 [ES][S]= (k-1 + k2) [ES] K1 [E]T[S] = (k-1 +K2+ K1 [S])[ES] k1 [E]T [S] [ES]= (÷ by K1) (k−1 + k2+ K1 [S]) [E]T [S] 𝒌−𝟏+𝒌𝟐 [ES]= with kM= kM+ [S] 𝒌𝟏 V= k2 [ES] rate limiting step k [E] [S] => V= 2 T kM+ [S]  V= k2 [ES] (rate limiting step) k2[E]T [S] Hyperbolic curve => V= kM+ [S] Max velocity (Vmax) occurs at high [S] when the E is entirely in [ES]: Vmax= K2.[E]T Michaelis- Menten equation KM is the [S] at which Vo= Vmax/2 Low KM => E has high affinity for the substrate (S strongly binds to the E)  KM varies widely with: 1. Identity of the enzyme 2. Nature of the substrate 3. Temperature and pH KM is therefore also a measure of the affinity of the enzyme for its substrate providing k2/k1 is small compared With Ks, that is, k2 < k–1  Analysis of Kinetic Data Several methods exist for determining the kinetic constants in the Michaelis-Menten equation. (1) Vo vs [S] plots require extrapolation of Vmax (asymptote—not accurate Vmax nor KM)  (2) Lineweaver-Burk (double reciprocal) plot: 1/v o vs 1/[S] *Most experimental measurements involve relatively high [S]=> crowded onto the left side of the graph. * Small errors in vo lead to large errors in 1/vo => large errors in KM and Vmax.  (3) Eadie-Hofstee (not required) (double reciprocal × V max ) Improved version of Lineweaver-Burk that is less sensitive to small errors  kcat/KM Is a Measure of Catalytic Efficiency Kcat= catalytic constant= turnover number= Definition Vmax Kcat= If the absolute [E] is unknown, Kcat cannot be [E]T determined experimentally. (high K cat=>fast E) Catalytic efficiency (kcat/KM): how often enzyme and substrate encounter one another in solution. b. Some Enzymes Have Attained Catalytic Perfection This ratio is maximal when k2 >> k–1, (formation of P from the ES, is fast compared to its decomposition back to S and E). Then kcat/KM = k1. Catalase, superoxide dismutase, fumarase, acetylcholinesterase, and possibly carbonic anhydrase.  Since the active site of an E is small compared to its total surface area, how can any enzyme catalyze a Rx every time it encounters a S ? The charged groups on the enzyme’s surface electrostatically guide the charged S to the E active site (superoxide dismutase (SOD) and acetylcholinesterase)