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## Chapter 13 Enzyme Kinetics

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**Outline**• What characteristic features define enzymes ? • Can the rate of an enzyme-catalyzed reaction be defined in a mathematical way ? • What equations define the kinetics of enzyme-catalyzed reactions ? • What can be learned from the inhibition of enzyme activity ? • What is the kinetic behavior of enzymes catalyzing bimolecular reactions ? • How can enzymes be so specific ? • Are all enzymes proteins ? • Is it possible to design an enzyme to catalyze any desired reaction ?**Virtually All Reactions in Cells Are Mediated by Enzymes**• Living systems use enzymes to accelerate and control the rates of vitally important biochemical reactions (see Figure 13.1). • Enzymes provide cells with the ability to exert kinetic control over thermodynamic potentiality. (glucose pyruvate) • Enzymes are the agents of metabolic function. • Most enzymes are proteins. • Some enzymes require cofactors or coenzymes.**Virtually All Reactions in Cells Are Mediated by Enzymes**Figure 13.1 Reaction profile showing the large free energy of activation for glucose oxidation. Enzymes lower ΔG‡, thereby accelerating rate.**13.1 What Characteristic Features Define Enzymes ?**• Catalytic power is defined as the ratio of the enzyme-catalyzed rate of a reaction to the uncatalyzed rate. • Specificity is the term used to define the selectivity of enzymes for their substrates. • Regulation of enzyme activity ensures that the rate of metabolic reactions is appropriate to cellular requirements. • Coenzymes and cofactors are nonprotein components essential to enzyme activity.**13.1 What Characteristic Features Define Enzymes ?**• Enzymes can accelerate reactions as much as 1016 over uncatalyzed rates. • Urease is a good example: • Catalyzed rate Uncatalyzed rate Ratio 3x104/sec 3x10-10/sec 1x1014 • Enzymes selectively recognize proper substrates over other molecules. • Enzymes produce products in very high yields - often much greater than 95%. • Specificity is controlled by structure - the unique fit of substrate with enzyme controls the selectivity for substrate and the product yield.**Enzymes are the Agents of Metabolic Function**Figure 13.2 The breakdown of glucose by glycolysis provides a prime example of a metabolic pathway with many sequential steps. All pathways are regulated but not all enzymes are regulated . Regulation of glycolysis occurs at several points, the major control point is through the enzyme phosphofructokinase I.**90% yield in each step = 35% over 10 steps**Figure 13.3 Yields in biological reactions must be substantially greater than 90%.**Enzymes Nomenclature**As in organic, there is a system of nomenclature devised for enzymes as well as a lot of commonly used names. There are six main classes used for naming enzymes. Know these six classes. There is also a numerical assignment for each enzyme developed by the Enzyme Commission. E.g.: EC 1.1.1.1 denotes alcohol dehydrogenase Its systematic name is: alcohol:NAD+ oxidoreductase donor:acceptor main class**The Six Classes of Enzymes**Class# Class and reaction catalyzed 1.Oxidoreductases (dehydrogenases) oxidation-reduction reactions 2.Transferases group transfer reactions 3. Hydrolases hydrolysis reactions 4. Lyases lysis, forming a double bond 5. Isomerases isomerization reactions 6.Ligases (synthetases) joining of two substrates, uses ATP**Enzyme Nomenclature Provides a Systematic Way of Naming**Metabolic Reactions Know the six main classes.**Coenzymes and Cofactors Are Nonprotein Components Essential**to Enzyme Activity Coenzymes are organic cofactors.**13.2 A Mathematical statement of the Rate of an**Enzyme-Catalyzed Reaction • Kinetics is the branch of science concerned with the rates of reactions. • Enzyme kinetics seeks to determine the maximum reaction velocity that enzymes can attain and binding affinities for substrates and inhibitors. • Analysis of enzyme rates yields insights into enzyme mechanisms and metabolic pathways. • This information can be exploited to control and manipulate the course of metabolic events.**Several kinetics terms to understand**• rate or velocity of reaction • rate constant • rate law • order of a reaction • molecularity of a reaction**Chemical Kinetics Provides a Foundation for Exploring Enzyme**Kinetics • Consider a reaction of overall stoichiometry as shown: • The rate is proportional to the concentration of A**Chemical Kinetics Provides a Foundation for Exploring Enzyme**Kinetics • The simple elementary reaction of A→P is a first-order reaction, v = k[A] and exponent on [A] is 1. • Figure 13.4 shows the course of a first-order reaction as a function of time. • This is a unimolecular reaction. • For a second order reaction, the rate law is: • v = k[A][B]. This reaction may be bimolecular or unimolecular depending on the mechanism. • Kinetics cannot prove a reaction mechanism. • Kinetics can only rule out various alternative hypotheses because they don’t fit the data.**The Time-Course of a First-Order Reaction**Figure 13.4 Plot of the course of a first-order reaction. The half-time, t1/2 is the time for one-half of the starting amount of A to disappear.**Catalysts Lower the Free Energy of Activation for a Reaction**• A typical enzyme-catalyzed reaction must pass through a transition state. • The transition state sits at the apex of the energy profile in the energy diagram. • The reaction rate is proportional to the concentration of reactant molecules with the transition-state energy. • This energy barrier is known as the free energy of activation, ΔG‡. • Decreasing ΔG‡ increases the reaction rate. • The activation energy is related to the rate constant by:**Catalysts Lower the Free Energy of Activation for a Reaction**Figure 13.5 Energy diagram for a chemical reaction (A→P) and the effects of (a) raising the temperature from T1 to T2, or (b) adding a catalyst.**The Transition State**Understand the difference between G and G‡ • The overall free energy change for a reaction is related to the equilibrium constant. • The free energy of activation for a reaction is related to the rate constant. • It is extremely important to appreciate this distinction.**13.3 What Equations Define the Kinetics of Enzyme-Catalyzed**Reactions ? • Simple first-order reactions display a plot of the reaction rate as a function of reactant. concentration that is a straight line (Figure 13.6) • Enzyme-catalyzed reactions are more complicated. • At low concentrations of the enzyme substrate, the rate is proportional to S, as in a first-order reaction. • At higher concentrations of substrate, the enzyme reaction approaches zero-order kinetics. • This behavior is a saturation effect.**13.3 What Equations Define the Kinetics of Enzyme-Catalyzed**Reactions ? Figure 13.6 A plot of v versus [A] for the unimolecular chemical reaction, A→P, yields a straight line having a slope equal to k.**As [S] increases, kinetic behavior changes from 1st order to**zero-order kinetics Figure 13.7 Substrate saturation curve for an enzyme-catalyzed reaction.**The Michaelis-Menten Equation is the Fundamental Equation of**Enzyme Kinetics • Louis Michaelis and Maud Menten's theory. • It assumes the formation of an enzyme-substrate complex. • It assumes that the ES complex is in rapid equilibrium with free enzyme. • Breakdown of ES to form products is assumed to be slower than 1) formation of ES and 2) breakdown of ES to re-form E and S. • Briggs and Haldane later introduced the steady state assumtion.**The Michaelis-Menten Equation is the Fundamental Equation of**Enzyme Kinetics k1 k2 E + S ES E + P k-1 E = enzyme concentration. S = Substrate concentration. ES = Enzyme-substrate complex concentration (noncovalent). P = product concentration. k1 = rate constant for formation of ES from E + S. k-1 = rate constant for decomposition of ES to E + S. k2 = rate constant for decomposition of ES to E + P.**k1**k2 E + S ES E + P k-1 Development of the Michaelis-Menton Equation 1. The overall rate of product formation: v = k2 [ES] 2. Rate of formation of [ES]: vf = k1[E][S] 3. Rate of decomposition of [ES]: vd = k-1[ES] + k2 [ES] 4. The steady state assumption requires that: Rate of ES formation = Rate of ES decomposition 5. So: k1[E][S] = k-1[ES] + k2 [ES]**Michaelis-Menton Derivation**6. In solving for [ES], use the enzyme balance to eliminate [E]. ET = [E] + [ES] 7. k1 (ET - [ES])[S] = k-1[ES] + k2 [ES] k1 ET[S] - k1[ES][S] = k-1[ES] + k2 [ES] 8. Rearrange and combine [ES] terms: k1 ET[S] = (k-1 + k2 + k1 [S])[ES] k1 ET[S] 9. Solve for [ES] = ----------------------- (k-1 + k2 + k1 [S])**Michaelis-Menton Derivation**ET[S] 10. Divide through by k1:[ES] = ----------------------- (k-1 + k2)/k1 + [S] 11. Defined Michaelis constant: KM = (k-1 + k2) / k1 12. Substitute KM into the equation in step 10. 13. Then substitute [ES] into v = k2 [ES] from step1 and replace Vmax with k2 ET to give: Vmax[S] vo = ----------- KM + [S]**[ES] Remains Constant Through Much of the Enzyme Reaction**Time Course Figure 13.8 Time course for a typical enzyme-catalyzed reaction obeying the Michaelis-Menten, Briggs-Haldane models for enzyme kinetics. The early state of the time course is shown in greater magnification in the bottom graph.**The dual nature of the Michaelis-Menten equation**Combination of 0-order and 1st-order kinetics • When S is low, the equation for rate is 1st order in S. • When S is high, the equation for rate is 0-order in S. • The Michaelis-Menten equation describes a rectangular hyperbolic dependence of v on S. • The relation of the “rectangular hyperbola” to the enzyme kinetics profile is described in references at the end of the chapter.**Understanding Vmax**The theoretical maximal velocity • Vmax is a constant. • Vmax is the theoretical maximal rate of the reaction - but it is NEVER achieved in reality. • To reach Vmax would require that ALL enzyme molecules are tightly bound with substrate. • Vmax is asymptotically approached as substrate is increased.**Understanding Km**The "kinetic activator constant" • Km is a constant. • Km is a constant derived from rate constants. • Km is, under true Michaelis-Menten conditions, an estimate of the dissociation constant of E from S. • A measure of ES binding. • Small Km means tight binding; large Km means weak binding. Where k2 is small then Km ≈ Kd.**Table 13.3 gives the Km values for some enzymes and their**substrates**Table 13.3 gives the Km values for some enzymes and their**substrates**The Turnover Number Defines the Activity of One Enzyme**Molecule A measure of catalytic activity • kcat, the turnover number, is the number of substrate molecules converted to product per enzyme molecule per unit of time, when E is saturated with substrate. A measure of rate of enzyme activity. • If the M-M model fits, k2 = kcat = Vmax/Et. • Values of kcat range from less than 1/sec to many millions per sec.**The Turnover Number Defines the Activity of One Enzyme**Molecule**The Ratio kcat/Km Defines the Catalytic Efficiency of an**Enzyme The catalytic efficiency: kcat/KmAn estimate of "how perfect" the enzyme is • kcat/Km is an apparent second-order rate constant. • It measures how the enzyme performs when S is low. • The upper limit for kcat/Km is the diffusion limit - the rate at which E and S diffuse together. • The maximum rate of diffusion for small molecules is 109 M-1-sec-1.**The Ratio kcat/Km Defines the Catalytic Efficiency of an**Enzyme Superoxide dismutase O2¯· (radical) 1 x 106 5 x 10-4 2 x 109 α-Chymotrypsin Acetyl-Phe-amide 1.4 x 10-1 1.5 x 10-2 9.3**Units of Enzyme Activity**Terms in discussing enzyme activity Units of enzyme activity: :mol S/min mol S/sec = katal Specific activity: (to follow purification, see p 99) :mol S/min/mg E mol S/sec/kgm E Molecular activity: (turn-over number, TON = kcat ) :mol S/min/:mol E mol S/sec/mol E**Turnover Number**Example calculation An enzyme (1.84 :gm, MW 36800), in presence of excess substrate catalyzes at a rate of 4.2 :mol substrate/min. Calculate the TON. (:mol S/min/:mol E) 1.84:gm :mol E: = --------------------- = 5 x 10-5:mol E 36800 :gm/:mol Vmax 4.2 :mol S/min TON = ------ = ----------------------- = 84000 min-1 Et 5 x 10-5:mol E**Michaleis-Mention Equation**Example calculation The rate of an enzyme catalyzed reaction is 35:mol/min at [S] = 10-4 M. KM for this substrate is = 2 x 10-5 M. Calculate the rate where [S] = 2 x 10-6 :mol/min. VM [S] VM (10-4) v = ------------- so 35 = --------------------- KM + [S] (2 x 10-5) + (10-4) And VM = 1.2(35) = 42 :mol/min (42)(2 x 10-6) (84 x 10-6) v = -------------------------- = ------------- = 3.8 :mol/min-1 (2 x 10-5) + (2 x 10-6) (22 x 10-6)**Specific Activity**• Assume that an assay of 0.8 ml of the crude extract gives 0.518 activity units. • (3800/0.8)(0.518) = 2460 units in the 3.8 l. • 2460 units/22800 mg protein = 0.108 units/mg • As purification proceeds, the specific activity increases due to loss of extraneous protein. • By purifying to a constant specific activity, one has reached a limit in purification. Table 5-1, p. 99**Graphical Determination of KM and VM**The Michaleis-Menton plot only permits an estimate of Vmax, so KM is also an estimate at VM/2. There are several graphical methods which provide a better determination of VM and KM. • We will focus on the Lineweaver-Burk equation which is obtained by taking the reciprocal of the Michaleis-Menton equation (See the next slide). • A Lineweaver-Burke plot is frequently referred to as a double reciprocal plot since one plots 1/v vs 1/[S]. • The plot gives a straight line which has a slope of KM/VM, a y-intercept of 1/VM and an x-intercept of -1/KM.**Linear Plots Can Be Derived from the Michaelis-Menten**Equation Be able to develop this equation Lineweaver-Burk: Begin with v = Vmax[S]/(Km + [S]) and take the reciprocal of both sides. Rearrange to obtain the Lineweaver-Burk equation: A plot of 1/v versus 1/[S] should yield a straight line.**Linear Plots Can Be Derived from the Michaelis-Menten**Equation The Lineweaver-Burk equation. Figure 13.9 The Lineweaver-Burk double-reciprocal plot.**Enzymatic Activity is Strongly Influenced by pH**• Enzyme-substrate recognition and catalysis are greatly dependent on pH. • Enzymes have a variety of ionizable side chains that determine its secondary and tertiary structure and also affect events in the active site. • Substrate may also have ionizable groups. • Enzymes are usually active only over a limited range of pH. • The effects of pH may be due to effects on Km or Vmax or both.**Enzymatic Activity is Strongly Influenced by pH**Figure 13.11 The pH activity profiles of four different enzymes.**The Response of Enzymatic Activity to Temperature is Complex**• Rates of enzyme-catalyzed reactions generally increase with increasing temperature. • However, at temperatures above 50° to 60° C, enzymes typically show a decline in activity. • Two effects here: • Enzyme rate typically doubles in rate for ever 10°C as long as the enzyme is stable and active. • At higher temperatures, the protein becomes unstable and denaturation occurs.**The Response of Enzymatic Activity to Temperature is Complex**Figure 13.12 The effect of temperature on enzyme activity.**13.4 Inhibition of Enzyme Activity**• Enzymes may be inhibited reversibly or irreversibly. • Reversible inhibitors may bind at the active site or at some other site. • Reversible inhibitors typically change VM, KM or both. • There are three common types of reversible inhibition: • Competitive • Non-competitive • Uncompetitive