Lecture 11 Enzymes: Kinetics

Lecture 11 Enzymes: Kinetics Reading: Berg, Tymoczko & Stryer, 6th ed., Chapter 8, pp. 216-225 Key Concepts Kinetics is the study of reaction rates (velocities). Study of enzyme kinetics is useful for measuring concentration of an enzyme in a mixture (by its catalytic activity), its purity (specific activity), its catalytic efficiency and/or specificity for different substrates comparison of different forms of the same enzyme in different tissues or organisms, effects of inhibitors (which can give information about catalytic mechanism, structure of active site, potential therapeutic agents...) Dependence of velocity on [substrate] is described for many enzymes by the Michaelis-Menten equation: kinetic parameters: K m (the Michaelis constant) k cat (the turnover number, which relates V max, the maximum velocity, to [E t ], the total active site concentration) k cat /K m (the catalytic efficiency of the enzyme) can't be greater than limit imposed by diffusion control, ~10 8-10 9 M 1 sec 1 Kinetic parameters can be determined graphically by measuring velocity of enzyme-catalyzed reaction at different concentrations of substrate (V o vs. [substrate]). LEC 11, Enzymes - Kinetics 1

Learning Objectives Terminology: active site, enzyme-substrate complex, induced fit, initial velocity, steady state, V max, K m, k cat, turnover number, K ES, enzyme efficiency. Write out a simple Michaelis-Menten kinetic mechanism for an enzymecatalyzed reaction. Recognize the Michaelis-Menten equation, and sketch a graph of V o vs. [S] for an enzyme-catalyzed reaction that illustrates V max and K m. Define K m in terms of the rate constants in the Michaelis-Menten kinetic mechanism; give the operational definition of K m that holds no matter what the actual kinetic mechanism is for a particular enzyme. Explain the relationship of k cat to V max, and the relationship of K m to K ES. State the units of K m, k cat, and V max. Express the ratio of occupied active sites to total enzyme active sites ([ES]/[E T ]) in terms of V o and V max. What is the maximum possible value of that ratio? Given a plot of V o /V max vs. [S], find the value of K m from the plot. What two things is the parameter k cat / K m used to indicate? What sets the upper limit for the value of k cat /K m for an enzyme? What is the approximate range of numerical values for that upper limit of k cat /K m, with units? REVIEW: How do enzymes reduce activation energy (ΔG )? 1. by lowering the free energy of the transition state ( ), e.g., by binding the transition state tightly 2. by changing the reaction pathway by which reactants react to form products; e.g., taking a 1-step uncatalyzed reaction and accomplishing the same result by a different route, with several intermediate reactions en route. Each reaction step has its own transition state with its own activation energy (ΔG ). If all of the individual steps' ΔG s are lower than the activation energy of the uncatalyzed reaction, the overall rate of product formation will be greater in the presence of the catalyst. The overall rate of the catalyzed reaction is dictated by the slowest step in a multistep reaction. Given a free energy diagram like the one in Nelson & Cox, Lehninger Principles of Biochemistry, 4th ed. (2004) Fig. 6-3 (previous lecture notes), how do you identify the rate-limiting (slowest) step on the reaction coordinate? LEC 11, Enzymes - Kinetics 2

BINDING = the essence of enzyme action! binding of SUBSTRATE to form an ES COMPLEX binding of TRANSITION STATE more tightly than the substrate Binding occurs at ACTIVE SITE of enzyme. Subsequent chemical events can then occur. Active site: relatively small part of whole enzyme structure 3-dimensional cleft with participating components from different parts of primary structure water often excluded so substrates and intermediates are in non-aqueous environment (unless H 2 O is a reactant) Binding uses multiple weak interactions: 1. hydrogen bonds 2. salt links 3. van der Waals interactions 4. hydrophobic effect Lysozyme (residues from different parts of AA sequence come together in active site) Berg et al., Fig. 8-7 LEC 11, Enzymes - Kinetics 3

Specificity of binding depends on active site crevice being sterically and chemically complementary to groups it is binding (best complementarity may be present in ES complex but NOT in free enzyme -- induced fit) Enzymes flexible -- conformational changes can occur when substrate binds during the reaction, to get maximal complementarity to the transition state. induced fit: conformational changes giving tighter binding in a new conformation For many (probably most) enzymes, the active site assumes shape complementary to S only when S is bound. Berg et al., Fig. 8-10 Why study enzyme kinetics (reaction rates)? measurement of velocity = reaction rate compare enzymes under different conditions, or from different tissues or organisms understand how differences relate to physiology/function of organism e.g., physiological reason for different K m values for hexokinase vs. glucokinase (discussed later in course) compare activity of same enzyme with different substrates (understand specificity) measure amount or concentration of one enzyme in a mixture by its activity measure enzyme purity (specific activity = amount of activity/amount of protein) study/distinguish different types of inhibitors info about enzyme active sites and reaction mechanism development of specific drugs (enzyme inhibitors) LEC 11, Enzymes - Kinetics 4

Simple Enzyme-Catalyzed Reaction: Measurement of velocity: V = rate of appearance of product = change in [P] per unit time units of velocity (V)? Plot of [P] vs. time Experimentally, forward velocity V F = slope of plot of [P] vs. time because V F = k F [S] = Δ[P]/Δt Berg et al., Fig. 8.11 (slightly modified) Determining initial velocity, V o Why we measure initial velocity, V o, the slope of [P] vs. [time] at very early time after mixing enzyme with substrate Problem: As S is converted to P, concentration of S decreases, so forward velocity gets slower and slower. Furthermore, as [P] increases, rate of reverse reaction (P --> S) becomes significant, and eventually, when V F = V R, reaction is at equilibrium: d[p]/dt = 0. Solution: Measure V at very early times in reaction, before [S] decreases signficantly (so [P] = ~0). Velocity measured immediately after mixing E + S, at beginning of reaction (initial velocity), is called V o. LEC 11, Enzymes - Kinetics 5

V o vs. [S] plots Simple uncatalyzed S P reaction shows linear dependence of V o on [S]: What s the slope of plot of V F vs. [S]? Enzyme-catalyzed reactions show a hyperbolic dependence of V o on [S]: 1. Rate is saturable: there's a maximum rate (velocity = V max ) for any single concentration of enzyme. 2. Rate (velocity) is proportional to [E total ]: at any single concentration of [S], V o = c[e] (double [E] double V o ). 3. Half-maximal velocity occurs at a specific substrate concentration, independent of [E]. K m = substrate concentration that gives V o = 1/2 V max. Dependence of V o on [S] in enzyme-catalyzed reaction: Plot of V o vs. [S]) for an enzyme that follows Michaelis-Menten kinetics plots as a rectangular hyperbola approaches V max (= maximum velocity) at high [S] 3 parts of [S] concentration range 1. At very low [S]: V o is proportional to [S]; doubling [S] double V o. 2. In mid-range of [S], V o is increasing less as [S] increases (where V o is around 1/2 V max ). K m = [S] that gives V o = 1/2 V max. 3. At very high [S], V o is independent of [S]: V o = V max. 1 2 3 Berg et al., Fig. 8.12 LEC 11, Enzymes - Kinetics 6

Michaelis-Menten model to explain hyperbolic dependence of V o on [S] in enzyme catalyzed reactions Before the chemistry occurs, enzyme binds substrate to make a noncovalent ES complex. Turnover number (def.): number of substrate molecules converted into product by one molecule of enzyme active site per unit time, when enzyme is fully saturated with substrate. = k2 in M-M scheme above = k cat (general term) k cat = turnover number (general term used, independent of specific kinetic mechanism) NOTE: k cat (the turnover number) = k 2 in the specific Michaelis-Menten kinetic mechanism above. The Michaelis-Menten Equation describes a rectangular hyperbola. The Michaelis-Menten Equation where V max = k 2 [E T ] (so V max is indeed proportional to [E T ]) K m : an "aggregate" constant (sum of rate constants for breakdown of ES divided by rate constant for formation of ES): Michaelis-Menten equation explains hyperbolic V o vs. [S] curve: 1. At very low [S] ([S] << K m ), V o approaches (V max /K m )[S]. V max and K m are constants, so linear relationship between V o and [S] at low [S]. 2. When [S] = K m, V o = 1/2 V max 3. At very high [S], ([S] >> K m ), V o approaches V max (velocity independent of [S]) LEC 11, Enzymes - Kinetics 7

Meaning of K m By definition, K m = substrate concentration at which velocity (V) is exactly 1/2 of V max (operational definition that holds for ANY kinetic mechanism). K m is a SUBSTRATE CONCENTRATION. What is the equilibrium dissociation constant for ES complex (K ES )? Compare: K m = K ES (dissoc. constant for ES complex) only if k 2 << k 1. Kinetic parameters used to characterize enzyme activity Kinetic parameters useful for many comparisons, e.g., effects of enzyme inhibitors on K m and V max = basis of diagnosis of type of inhibition. effects immediately apparent on a double reciprocal plot (see Enzyme Inhibition notes) 1. K m (units: concentration units, e.g., M or mm or µm) IF k 2 << k 1 (not always true), then K m = k -1 /k 1 = K ES and K m can be taken as a measure of the dissociation constant for the ES complex (an INVERSE measure of the binding affinity of enzyme for that substrate). 2. k cat (directly related to V max ) (units: inverse time, e.g., s 1 ) calculate k cat from V max and concentration of enzyme active sites [E T ]: V max = k 2 [E T ] = k cat [E T ] so 3. k cat / K m (the criterion of catalytic efficiency and "kinetic perfection") (units: inverse conc. inverse time, e.g., M 1 s 1 ) LEC 11, Enzymes - Kinetics 8

Wide range of catalytic rate constants, k cat (turnover numbers) among enzymes: Lysozyme: k cat = 0.5 s 1 Catalase: k cat = 4 x 10 7 s 1 In vivo, most enzymes operate below saturation (V < V max ) (important for regulation) Activity of some metabolic enzymes is regulated: V can be "tuned" up or down depending on cell's metabolic needs. Remember: V o = k cat [ES] = rate of appearance of P, and V max = k cat [E T ] so (divide 1st equation by the second): = fractional saturation! Ratio of actual velocity at a given substrate concentration (V o ) to V max indicates ratio of occupied active sites to total active sites at that [S]. Michaelis-Menten Equation: Divide both sides by V max : Can calculate fractional saturation {ratio of (occupied sites / total active sites)} as What is the maximum possible value of [ES]/[E T ]? What would be the shape of a plot of V o /V max vs. [S], and how would you find K m on the plot? LEC 11, Enzymes - Kinetics 9

k cat /K m k cat /K m is the criterion of substrate specificity, catalytic efficiency and "kinetic perfection. units of k cat /K m = conc 1 time 1. k 1 sets an upper limit on value of k cat /K m. k cat /K m can never be greater than k 1 (rate constant for association of E + S) for a given enzyme-substrate system. (You're not responsible for the explanation of this.) How fast 2 molecules can react (or bind in the case of E + S) is limited by how fast the molecules can diffuse to "bump into each other" in solution so they can react. For molecules the size of an enzyme and a typical substrate, the maximum (diffusion-limited) k 1 = ~ 10 8 10 9 M 1 sec 1. Thus max. possible k cat /K m for an enzyme = ~ 10 8 10 9 M 1 sec 1. Any enzyme with k cat /K m value in that range approaches the limit of diffusion control, and thus has achieved something very close to "kinetic perfection": the rate at which that enzyme's active site can convert substrate into product is limited only by the rate at which it encounters the substrate in solution. Uses of k cat /K m k cat /K m used as a measure of 2 things: 1. enzyme's substrate preference 2. enzyme's catalytic efficiency 1. enzyme's preference for different substrates (substrate specificity) The higher the k cat /K m, the better the enzyme works on that substrate. e.g., chymotrypsin: protease that clearly "prefers" to cleave after bulky hydrophobic and aromatic side chains. chymotrypsin specificity: active site best accommodates substrates with a bulky hydrophobic or aromatic residue contributing carbonyl group to peptide bond to be hydrolyzed. Berg et al., 5th ed., Fig. 9.1 LEC 11, Enzymes - Kinetics 10

Substrate preferences of chymotrypsin Chymotrypsin also catalyzes hydrolysis of ESTER bonds whose carboxylic acid component has a bulky, hydrophobic and/or aromatic R group. Amino acids listed in Berg et al. Table 8-6 (below) are the groups that contribute the carbonyl group to the ester bond being hydrolyzed by the enzyme. Which amino acid is most preferred by chymotrypsin for contribution of carbonyl group in ester bond to be cleaved, among substrates above? Catalytic efficiency of enzymes the higher the k cat /K m, the more "efficient" the enzyme maximum possible k cat /K m dictated by the diffusion limit If K M = K ES, which enzyme above binds its substrate the most tightly? Which enzyme has the most rapid catalytic turnover when the enzyme is saturated with substrate? Which enzymes have the highest catalytic efficiency? Are they near the limit of diffusion control? NOTE: k cat /K M for an enzyme can have a value close to the limit of diffusion control either because its k cat is very high, or because its K m is very low, or some combination. LEC 11, Enzymes - Kinetics 11

Appendix: How was the Michaelis-Menten Equation Derived? To derive Michaelis-Menten Equation (describe hyperbolic V o vs. [S] plot): Start with the following 3 equations/assumptions: 1. Enzyme active site concentration: [E Total ] = [E free ] + [ES] [total active sites] = [empty sites] + [occupied sites] 2. Total concentration of [S] is high enough that [S free ] = [S total ] i.e., an insignificant fraction of total [S] is bound even when all the enzyme active sites are occupied. 3. In the STEADY STATE, concentration of [ES] is constant (the steady state assumption): rate of formation of ES = rate of breakdown of ES (breakdown rate = sum of rates of breakdown forward + backward) (See last 2 slides, Appendix, if you re interested in where to start the derivation, but you re not responsible for it.) M-M Equation Derivation, Steady State Assumption Changes in conc. of enzyme-catalyzed reaction participants with time: "Steady state" is the part of the reaction period that concentration of ES is not changing: rate of formation of ES = rate of breakdown of ES k 1 [E][S] = k 1 [ES] + k 2 [ES] Berg et al., 5th ed. Fig. 8-13a LEC 11, Enzymes - Kinetics 12