Lecture 11: Enzymes: Kinetics [PDF] Reading: Berg, Tymoczko & Stryer, Chapter 8, pp. 216-225 Updated on: 2/4/07 at 9:00 pm Key Concepts Kinetics is the study of reaction rates. Study of enzyme kinetics is useful for measuring concentration of an enzyme in a mixture (by its catalytic activity), its purity (specific activity), measurement of the catalytic efficiency and/or the specificity of an enzyme 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. 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 enzyme-catalyzed 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. Explain the definition of 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? Page 1 of 11
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, and what is the approximate range of numerical values for that upper limit of k cat /K M, with units? Enzyme Kinetics REVIEW: How do enzymes reduce the 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? 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: hydrogen bonds salt links van der Waals interactions hydrophobic effect (We'll come back to binding and reduction of ΔG when we discuss enzyme mechanisms, how (chemically) enzymes catalyze reactions.) Berg, Tymoczko & Stryer, 6th ed., Fig. 8.5: Structure of an ES complex (a cytochrome P450 bound to camphor) Page 2 of 11
Berg, Tymoczko & Stryer, 6th ed. Fig. 8.7: Lysozyme, with color coded active site residues that come from different parts of AA sequence SPECIFICITY of binding depends on active site crevice being sterically and chemically precisely complementary to the groups it is binding (best complementarity may be present in E S 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 Berg, Tymoczko & Stryer, 6th ed. Fig. 8.9: old "Lock and key" model for E-S interaction Berg, Tymoczko & Stryer, 6th ed. Fig. 8.10: Induced fit For many (probably MOST) enzymes, the active site Page 3 of 11
Active site of free enzyme complementary to shape of S even without substrate bound. assumes shape complementary to S only when S is bound. Why study enzyme kinetics (reaction rates)? 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 and 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 measurement of VELOCITY = reaction rate measure enzyme purity (specific activity = amount of activity/amount of protein) study/distinguish different types of INHIBITORS *info about enzyme active sites & reaction mechanism *development of specific DRUGS (enzyme inhibitors) SIMPLE ENZYME-CATALYZED REACTION: Measurement of velocity V = rate of appearance of product = change in [P] per unit time (V has units of conc time 1 ) Experimentally, forward velocity V F = SLOPE of plot of [P] vs. time (Fig. 8.11 below), because V F = k F [S] = d[p]/dt 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, you're 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. Berg, Tymoczko & Stryer, 6th ed. 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: Page 4 of 11
As starting [S] increases (S 1, S 2, etc. concentrations on plot), velocity increases in direct proportion to [S], as expected if V F = k F [S]. Later in time, as [P] increases and [S] decreases in concentration, velocity decreases, both because V F = k F [S] and also because of reverse reaction (V R = k R [P]) becoming significant and taking P back to S. Simple uncatalyzed S P reaction gives straight line plot of V o vs. [S]: If you double [S], you double velocity, throughout the [S] concentration range: V F = k F [S] Simple uncatalyzed S--> P does NOT give a hyperbolic plot of V o vs. [S] like the one on the plots below. Plotting v o vs. [S] for an enzyme-catalyzed reaction, what do we see? Enzyme-catalyzed reactions show a HYPERBOLIC dependence of V o on [S]: 1. Rate is saturable; i.e., there's a maximum rate (velocity = V max ) for any single concentration of enzyme. 2. Rate (velocity) is linearly dependent on enzyme concentration, i.e., at any single concentration of [S], V o = c[e] 1. (Double [E] double V o.) 3. Half-maximal velocity occurs at a specific substrate concentration, independent of enzyme concentration. K M is the substrate concentration that gives V o = 1/2 V max. Page 5 of 11
Berg, Tymoczko & Stryer, 6th ed. Fig. 8.12: Dependence of velocity on substrate concentration in enzymecatalyzed reaction (plot of V o vs. [S]) for an enzyme that follows Michaelis- Menten kinetics. plots as a rectangular hyperbola "maxes out" as [S] gets high V max = maximum velocity, at high [S] Note 3 parts of [S] concentration range on abscissa (x axis): 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 ). 3. At very high [S], V o is independent of [S]: V o = V max. (another figure showing same thing, but also showing how V o vs. [S] plot would look if V o WERE directly proportional to [S] even at high [S] concentrations): Nelson & Cox, Lehninger Principles of Biochemistry, 4th ed. Fig. 6-12: Velocity on V o vs. [S] plot asymptotically approaches a maximum velocity, V max. Early 20th century: Leonor Michaelis & Maud Menten's simple model was proposed, explaining the odd kinetics of enzyme-catalyzed reactions: Before the chemistry occurs, enzyme binds substrate to make a noncovalent ES complex: Page 6 of 11
MICHAELIS-MENTEN MODEL Turnover number (definition): the number of substrate molecules converted into product by one enzyme active site per unit time, when enzyme is fully saturated with substrate. 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 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 [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 (forward + backward) Berg, Tymoczko & Stryer, 5th ed., Fig. 8.13: Changes in concentration of enzyme-catalyzed reaction participants with time "Steady state" is the part of the reaction period that concentration of ES is not changing. The Michaelis-Menten Equation (derived algebraically starting with the steady state assumption, above): where V max = k 2 [E T ] (so V max is indeed proportional to [E T ]) Page 7 of 11
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. When [S] is very low ([S] << K M ), V o becomes (V max /K M ) [S]. V max & K M are constants, so linear relationship between V o and [S] at low [S]. See Fig. 6-12 from Nelson & Cox textbook, above. At HIGH [S], ([S] >> K M ), V o becomes = V max. (velocity independent of [S]) MEANING OF K M : By definition, K M = SUBSTRATE CONCENTRATION at which velocity (V) is exactly 1/2 of maximum velocity (V max ). This operational definition holds for ANY kinetic mechanism. NOTE: K M is a substrate CONCENTRATION. What is the equilibrium dissociation constant for ES complex (K ES )? Compare:. 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). PARAMETERS USED TO CHARACTERIZE ENZYME ACTIVITY: K M Page 8 of 11
k cat (which can be obtained from V max and the concentration of enzyme active sites [E T ], since V max = k 2 [E T ] = k cat [E T ] so k cat /K M (the criterion of catalytic efficiency &"kinetic perfection") -- see below Effects of 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). UNITS of catalytic parameters K M : always has units of CONCENTRATION (M or mm or µm, etc.) (definition of K M = substrate concentration at which velocity is half the maximal velocity for that enzyme concentration) V max obviously has velocity units: concentration/time (conc time -1 ); e.g., mm/min or µm sec -1 k cat has units of inverse time (e.g. sec -1 or min -1 ) WIDE RANGE OF CATALYTIC RATE CONSTANTS (TURNOVER NUMBERS): 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 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 Ratio of actual velocity at a given substrate concentration to V max indicates ratio of occupied active sites to total active sites at that substrate concentration. Reminder: Michaelis-Menten Equation: Divide both sides by V max : Can calculate ratio of (occupied sites / total active sites) as Page 9 of 11
What is the maximum possible value of [ES]/[E T ]? k cat /K M (the criterion of catalytic efficiency and "kinetic perfection"): UNITS of k cat /K M = conc -1 time -1. k cat /K M can never be greater than k 1 (rate constant for association of E + S) for a given enzymesubstrate system (You're not responsible for the explanation of this.) k 1 sets an upper limit on value of k cat /K M. 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. 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. k cat /K M used as a measure of 2 things: enzyme's substrate preference, and enzyme's catalytic efficiency 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, clearly "prefers" to cleave after bulky hydrophobic and aromatic side chains. Berg, Tymoczko & Stryer, 5th ed., Fig. 9.1: Specificity of chymotrypsin a protease whose active site best accommodates substrates with a bulky hydrophobic and aromatic residue contributing the carbonyl group to the peptide bond to be hydrolyzed. Peptide bonds expected to be hydrolyzed are in RED, on carboxyl side of the (shaded) residues with the "specificity groups", hydrophobic and/or aromatic R groups. Chymotrypsin also catalyzes hydrolysis of ESTER bonds whose carboxylic acid component has a bulky, Page 10 of 11
hydrophobic and/or aromatic R group catalytic efficiency The higher the k cat /K M, the more "efficient" the enzyme maximum possible k cat /K M dictated by the diffusion limit Enzyme Substrate k cat (sec -1 ) K m (M) k cat /K M (M -1 sec -1 ) Catalase H 2 O 2 4.0x10 7 1.1 4.0x10 7 Carbonic anhydrase CO 2 1.0x10 4 1.2x10-2 8.3x10 7 Acetylcholine esterase acetylcholine 1.4x10 4 9.0x10-5 1.6x10 8 Fumarase fumarate 8.0x10 2 5.0x10-6 1.6x10 8 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 thereof. 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? zieglerm@u.arizona.edu Department of Biochemistry & Molecular Biophysics The University of Arizona Copyright ( ) 2007 All rights reserved. Page 11 of 11