Biochemistry I Lecture 17 Oct 10, 20051Lecture 17: Enzyme KineticsAssigned reading in Campbell: Chapter 6.6Key Terms:• Steady State Kinetics• Michaelis-Menton Equation: vo = Vmax [S] / (KM+[S])• Vmax = kcat [Etotal]• KM = (k-1+k2) / k1Enzyme kinetics: the study of the rates of enzyme-catalyzed reactions:Provides indirect information regarding the specificities and catalytic mechanisms of enzymes.17.1 Chemical Kinetics:Consider a simple bimolecular reaction:† E + S ´ PThe rate is proportional to the product of the reactants:rate † µ [E]1[S]1 (2nd order because sum of exponents is two)A 2nd order reaction has the following properties:• The forward rate constant (k1) has units of M-1sec-1.• The reverse rate constant (k-1) has units of sec-1.The rate, or velocity, of any reaction can be determined bymeasuring the increase in the amount of product formedD[P] during a given amount of time Dt:† v =D[P]Dt=d[P]dt17.2 Steady State Enzyme Kinetics:For many enzymes, the rate of catalysis, v, varies with substrateconcentration [S]. At fixed enzyme concentrations, v is almostlinearly proportional to [S]. However, at high [S], v is nearlyindependent of [S].A model to account for these observations was proposed byMichaelis & Menton. The critical feature of this model is that EScomplex formation is a necessary intermediate in catalysis.This model accounts for the kinetic properties of many enzymes. k1† E + S ¨ Æ æ ESkcatæ Æ æ E + P k-1• k1 is the forward rate constant for substrate binding.• k-1 is the reverse rate constant for substrate binding.• ES is the enzyme-substrate complex.• kcat is the catalytic rate constant (containing termsrelated to stabilization of the transition state).Biochemistry I Lecture 17 Oct 10, 20052The goal is to use this model to generate an expressionthat relates the rate of catalysis to the concentrationsof enzyme, substrate, and the individual rate constants.The starting point is:† v =d[P]dt= kcat[ES]Now we need to express [ES] in terms of readily measurableexperimental parameters, such as:1. The total amount of enzyme ET = [E] + [ES]2. The concentration of substrate: [S]3. The rate of the reaction v = kcat [ES][ES] is determined by the rates of its formation and breakdown:Rate of [ES] formation =† k1[E][S]Rate of [ES] breakdown =† k-1[ES] + kcat[ES]To simplify a complicated kinetic equation, we makea key assumption. The steady state assumptionpostulates that there is a period of time (called thesteady state) (coincides with the initial velocity period)when [ES] does not change:d[ES]/dt = 0Then k1[E][S] = k-1[ES] + kcat [ES]If we express the rate of product formation as† v = kcat[ES]and express [ES] in terms of the kinetic rate constants,[S], and the total amount of enzyme present [ET] = [E]+[ES],† v = kcat[ET][S]k-1+ kcatk1+ [S]Now if we abbreviate† k-1+ kcatk1 as KMAnd substitute VMAX for kcat [ET], we get the followingsimple equation:† v =VMAX[S]KM+ [S] This is the Michaelis-Menton equation.Biochemistry I Lecture 17 Oct 10, 2005317.3 Important constants that describe theenzymatic reaction:1) The KM or Michaelis constant:• † KM=k-1+ kcatk1• This is almost the same as the KD (=k-1/k1) dissociation constant, except for the presenceof the kcat term. Therefore it is related to the affinity or strength of binding of a substrateto the enzyme.• KM is equal to the substrate concentration that gives 1/2 of the maximal velocity, in asimilar manner that KD is given by the [L] that gives Y=0.5.† v =VMAX[S]KM+ [S]2) Vmax = kcat [ET]:This is the highest rate of product production possible. It is obtained at high substratelevels ([S]>>KM).† v =VMAX[S]KM+ [S]Under these conditions all of the enzyme is in the [ES] form (i.e. [ES]=[ET]). Thus VMAXis a reflection of the ability of an enzyme to perform the catalytic step.3) Turnover number: kcat = Vmax / [ET]This is the number of reactions performed by a single enzyme molecule in a certainperiod of time (e.g. moles of product formed in a defined period by a mole of enzyme)when the enzyme is fully saturated with substrate.VMaxvo0[S]Biochemistry I Lecture 17 Oct 10, 2005417.4 Throughput, or efficiency, of enzyme systems† v =VMAX[S]KM+ [S]VMAX = kcat [ET]Low Substrate: At low substrate concentrations ([S]<<KM) the overall rate of product formationdepends on both the total amount of enzyme [ET] and substrate [S]. The rate constant iscomprised of both KM and kcat (kcat/KM). Thus under these conditions, the efficiency of anenzyme will depend both on how efficiently it can bind substrate (KM), as well as how well it canperform the chemical step (kcat). In other words, the intrinsic efficiency of an enzyme at lowsubstrate levels is given by kcat / KM.† v =kcatKM[ET][S]High Substrate: At high substrate concentrations ([S]>>Km) the rate becomes independent of[S] since all of the enzyme molecules are saturated with [S]. Therefore the intrinsic efficiency ofan enzyme is given simply by kcat. The overall rate depends only on kcat and the total amount ofenzyme, [ET]. v = kcat [ET] or v =