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Lecture 15

Lecture 15. Enzymology Today, Howard Salis (148 Baker) 3PM. Extra credit seminar-assignments due Monday. Enzyme-catalyzed reactions must involve formation of an enzyme - substrate complex, followed by one or more chemical steps. V max and K m are two key measurable properties of enzymes.

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Lecture 15

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  1. Lecture 15 • Enzymology • Today, Howard Salis (148 Baker) 3PM. Extra credit seminar-assignments due Monday.

  2. Enzyme-catalyzed reactions must involve formation of an enzyme-substrate complex, followed by one or more chemical steps • Vmax and Km are two key measurable properties of enzymes. Vmax: • kcat: the rate constant for the catalytic step carried out by the enzyme. • Vmax: the rate at which a given amount of enzyme catalyzes a reaction at saturating concentrations of substrate. kcat [Etotal] = Vmax • kcat is interpreted as a measure of the rate of chemical conversion of substrate to product.

  3. Enzyme-catalyzed reactions must involve formation of an enzyme-substrate complex, followed by one or more chemical steps • Vmax and Km are two key measurable properties of enzymes - Km Km: • Km is the ratio of the rate constants for the individual steps: Km = (k-1 + kcat)/k1. • If kcat is small compared to k-1, then Km = Ks, the dissociation constant for the enzyme-substrate complex. • Thus for many enzymes, Km can be interpreted as a measure of the affinity of the enzyme for its substrate. • Km also = [S] at which v = 1/2 Vmax

  4. Some Rules on Interpreting Michaelis-Menten Enzyme Kinetic Parameters (from A. Fersht text) • kcat is a 1st order rate constant that refers to the properties and reactions of the enzyme-substrate, enzyme-intermediate (‡) and enzyme product complexes. • Km is an apparent dissociation constant that may be treated as the overall dissociation constant of all enzyme bound species. • kcat/Km is an apparent second order rate constant that refers to properties of the free enzyme and the free substrate.

  5. k-2 k-1 v0 = Vmax[S] KM + [S] 1/v0 = KM +[S] Vmax[S] 1 1 KM + 1/v0 = Vmax [S] Vmax Michaelis-Menten Enzyme Kinetics k2 k1 E + P [ES] E + S Basic Michaelis-Menten Equation: If we take the inverse of MM eq:

  6. 1 1 KM + 1/V0 = Vmax [S] Vmax Lineweaver-Burk Inversion Use the equation for a straight line: y = ax + b slope y-intercept Plot 1/v (= y) versus 1/[S] (= x) slope = KM/Vmax 1/V x-intercept = -1/KM y-intercept = 1/Vmax 1/[S]

  7. Figure U2-2.3 Lineweaver-Burk or reciprocal kinetic plot of 1/v against 1/[S] 1. Original raw kinetic data 2. Transformed (inverse) data Note: this L-B method weights data incorrectly, puts emphasis on data near minimum rate (Vmax) : (e.g., 1/ 0.05 = 20, 1/0.1 = 10) You should perform non-linear fit of actual raw data to equation with computer software (KaleidaGraph, Sigmaplot, Orig, GraphPad Prism, kinetic fitting packages, etc.)

  8. V0 + - KM Vmax V0= [S] v0 = Vmax[S] KM +[S] Eadie-Hofstee Plot Another inversion method: multiply MM eq by V0Vmax Plot v (= y) versus v/[S] (= x) Vmax v slope = -KM Vmax/KM v/[S]

  9. 1 KM S/V0 = [S] + Vmax Vmax v0 = Vmax[S] KM +[S] Hanes-Wilkinson Plot Another inversion method: multiply MM eq by [S] slope y-intercept Plot 1/v (= y) versus 1/[S] (= x) [S]/V Slope =1/Vmax -KM KM/Vmax [S]

  10. v0 = Vmax[S] v0 = Vmax[S] v0 = Vmax[S] = Vmax 2 KM +[S] [S]+[S] 2[S] Enzyme Kinetics: General Principles • Remember, Vmax = maximal rate of an enzyme (Etotal = ES) • KM = (k-1 + k2)/k1 • KM is the [S] at which V = 1/2 (Vmax) Basic Michaelis-Menten Equation: If KM = [S]:

  11. k-2 k-2 k-1 k-1 Enzyme Kinetics: General Principles k2 k1 E + P ES E + S • If k-1 >> k2 k2+ k-1 k-1 for KM = = = KS k1 k1 Dissociation constant of the enzyme and substrate • If k2 >> k-1 k2 k1 E + P ES E + S Enzyme reacts every time it interacts with substrate. (see p. 480)

  12. Example of Michaelis-Menten Enzyme Kinetics Given this data, what is Vmax? What is KM? First, graph [S] vs. v to make sure it obeys MM kinetics Vmax is 60 by inspection v (µmol/min) [S]

  13. Example of Michaelis-Menten Enzyme Kinetics Given this data, what is Vmax? What is KM? Since Vmax = 60 we can solve for KM, plug this into MM eq. v0 = Vmax[S] KM = [S] Vmax -1 v0 KM +[S] If v = 48, [S]= 2 X 10-4, KM = 5.0 X 10-5 If v = 12, [S]= 1.3 X 10-5, KM = 5.2 X 10-5 These should agree with one another!

  14. 1 1 KM + 1/V0 = Vmax [S] Vmax Example of Michaelis-Menten Enzyme Kinetics Given this data, what is Vmax? What is KM? We can also check by Lineweaver-Burke plot 1/v 1/Vmax -1/KM Scale is important 1/[S]

  15. Graph of rate against total substrate concentration for a typical enzyme-catalyzed reaction • In this example, saturating concentrations of substrate B are converted to product more slowly than saturating concentrations of substrate A • (Vmax is lower for substrate B than for substrate A), but substrate B binds more tightly to the enzyme (reflected in its lower Km). • The value of kcat/Km is higher for substrate B than A, indicating that the enzyme is more specific for substrate B.

  16. Enzyme-catalyzed reactions can have multiple steps with several intermediates • Most enzyme-catalyzed reactions are more complex than we have presented so far. • Frequently multiple substrates are involved, and even when they are not, the reaction often proceeds with the formation of one or more semi-stable species. • These species, which do not dissociate from the enzyme and are on the pathway from substrate to final product, are called intermediates. • In such reactions, there will be one step that is slower than all the others. • In theory this can be any step in the reaction, including substrate binding (diffusion-controlled reactions) or product release, but often the slowest step is a chemical step involving the formation of one of the intermediates. • When the overall reaction rate is measured, what will be observed is the rate of this slowest step. Hence, it is referred to as the rate-determining step of the reaction (Figure U2-3.2).

  17. Figure U2-3.2 The rate-determining step • The rate-determining step of a reaction is the step with the largest energy barrier. • In this example, the height of the barrier between intermediates I1 and I2 is greater than between S and I1 or I2 and P, and the rate determining step is therefore I1 to I2.

  18. k-2 k-1 Michaelis-Menten Enzyme Kinetics What happens if we consider the back reaction of the products? k2 k1 E + P ES E + S VProduct = k2[ES] - k-2[P][E] If we assume the steady state d[ES]/dt =0, so Vproduct = Vreactant k1[S][E] + k-2 [P][E] = k-1[ES] + k2[ES] Solving for ES gives us: [ES] = (k1+ k-2[P])[E] 2 equations, 2 unknowns k-1+ k2 and we know: [ES] = [Etotal -E]

  19. k2 k1 E + P ES E + S k-2 k-1 Michaelis-Menten Enzyme Kinetics What happens if we consider the back reaction of the products? Solving [E] = (k-1+ k-2)[Etotal ] k1[S] + k-2[P] + k-1+ k2 [ES] = (k1 [S]+ k-2[P])[Etotal ] k1[S] + k-2[P] + k-1+ k2

  20. Michaelis-Menten Enzyme Kinetics Substitute back into the initial equation and simplify: VProduct = k2[ES] - k-2[P][E] [E] = (k-1+ k2)[Etotal ] [ES] = (k1 [S]+ k-2[P])[Etotal ] k1[S] + k-2[P] + k-1+ k2 k1[S] + k-2[P] + k-1+ k2 VProduct = k2((k1 [S]+ k-2 [P])[Etotal ]) - k-2[P]((k-1+ k2)[Etotal ]) k1[S] + k-2[P] + k-1+ k2 VProduct = ((k2k1[S] - k2k-2[P])- (k-1k-2[P] + k2k-2[P]))[Etotal ] k1[S] + k-2[P] + k-1+ k2

  21. k2 k1 E + P ES E + S k-1 k-2 k2+ k-1 k2+ k-1 for KM = for KP = k-2 k1 Michaelis-Menten Enzyme Kinetics Substitute back into the initial equation and simplify: VProduct = (k2k1[S] - k-2k-1[P])[Etotal ] k1[S] + k-2[P] + k-1+ k2 If we define and for Vmax k2[Etotal] k-1[Etotal] Vmax = Vmax = forward reverse

  22. k2+ k-1 k2+ k-1 for KM = for KP = k-2 k1 k2[Etotal] k-1[Etotal] Vmax = Vmax = forward reverse Michaelis-Menten Enzyme Kinetics Substitute and rearrange: VProduct = (k2k1[S] - k-2k-1[P])[Etotal ] k1[S] + k-2[P] + k-1+ k2 VProduct = k2k1[S][Etotal ] - k-2k-1[P ][Etotal ] k1[S] + k-2[P] + k-1+ k2

  23. k2[Etotal] k-1[Etotal] Vmax = Vmax = forward reverse Michaelis-Menten Enzyme Kinetics Substitute and rearrange: VProduct = k2k1[S][Etotal ] - k-2k-1[P ][Etotal ] k1[S] + k-2[P] + k-1+ k2 reverse forward VProduct = Vmaxk1[S]- k-2 Vmax[P ] k1[S] + k-2[P] + k-1+ k2

  24. k2+ k-1 k2+ k-1 for KM = for KP = k-2 k1 Michaelis-Menten Enzyme Kinetics Substitute and rearrange: reverse forward VProduct = Vmaxk1[S]- k-2 Vmax[P ] k1[S] + k-2[P] + k-1+ k2 Solve in terms of k1 and k-2: k2+ k-1 k2+ k-1 k-2 = k1 = KM KP forward reverse VProduct = Vmax((k2+k-1)/KM)[S]- ((k2+k-1)/KP)Vmax[P ] ((k2+k-1)/KM)[ S] + ((k2+k-1)/KP)[P] + k-1+ k2

  25. Michaelis-Menten Enzyme Kinetics Substitute and rearrange: reverse forward V = Vmax((k2+k-1)/KM)[S]- ((k2+k-1)/KP)Vmax[P ] ((k2+k-1)/KM)[ S] + ((k2+k-1)/KP)[P] + k-1+ k2 Divide by k2+k-1 forward reverse V = ((Vmax/KM)[S]- (Vmax /KP)[P ] ) [S] /KM + [P] /KP + 1 Multiply by KmKp forward reverse V = KPVmax[S]- KMVmax[P ] KMKP + KP[S] + KM [P]

  26. Michaelis-Menten Enzyme Kinetics Substitute and rearrange: forward reverse Vproduct = KPVmax[S]- KMVmax[P ] KMKP + KP[S] + KM [P] But, V(dP/dt) = Vforward - Vreverse, so we break it down into the V forward and V reverse reactions. forward Vforward = Vmax[S] KM + [S] + [P] (KM/ KP) reverse Vreverse = -Vmax[P] KP + [P] + [S] (KP/ KM) Note at P=0, we get the Michaelis-Menton Equation

  27. This derivation was for an enzyme that has two possible reactions • E + S • E + P • [P] inhibits the forward reaction by combining with E. • Note that as [S] increases, then the maximum rate is Vmax [P](KM/KP) is insignificant at high [S]. • Because the Product and Substrate are competing for the same site, this is the model for competitive inhibition.

  28. Enzyme reactions can be slowed by the presence of inhibitors • Other inhibitors bind noncovalently and reversibly to their target enzymes. • These are usually divided into three broad classes, competitive, noncompetitive, and uncompetitive, depending on their manner of binding. • Kinetic analysis can distinguish among these inhibitors if the reaction rate is measured against substrate concentration at different inhibitor concentrations.

  29. Figure U2-4.1 Competitive, noncompetitive and uncompetitive inhibition

  30. E Figure U2-4.1a Competitive inhibition ES (a) A competitive inhibitor (I) binds to the same site as does the substrate (S; top). The inhibitor changes the apparent Km for the reaction, but not the Vmax because enough substrate can keep any inhibitor from binding. A Lineweaver-Burk plot (bottom) for the reaction at various concentrations of the inhibitor reflects this behavior. EI

  31. v0 = Vmax[S] KM +[S] +[I] (KM/KI) KI = [E][I] [EI] Competitive Inhibition ES P S + E + I EI Vmax w/o I w/ I v (µmol/min) [S]

  32. v0 = Vmax[S] KM +[S] +[I] (KM/KI) KM 1/V0 = 1 1 (1+ ) + [I] Vmax Vmax [S] KI KM 1/V0 = 1 () + 1 Vmax [S] Vmax I w/o I 1/v0 -1/KM 1/Vmax -1 -1 1/[S] [I] KM KM (1+ ) KI Competitive Inhibition ES P S + E + I EI -1/KM’ or apparent KM

  33. Competitive inhibition •  = 1 +[I]/KI • As [I] increases, v decreases (1/v increases) • As [I] increases, KM decreases (1/KM increases) • Vmax is the same and the inhibition can be overcome by high [S]

  34. Figure U2-4.1b Noncompetitive inhibition E ES (b) A noncompetitive inhibitor (I) (green) does not bind to the substrate binding site and can bind to both the free enzyme or the ES complex. Usually a noncompetitive inhibitor resembles one substrate (S2) in a two-substrate reaction, as shown here, where both substrates are present on the enzyme at the same time. In the simplest cases, noncompetitive inhibitors don't change the Km for the first substrate (S), because they don't affect its binding. But providing the concentration of S2 is not high enough to out-compete all the inhibitor, the inhibitor does reduce the Vmax for the reaction. EI

  35. ES P S + E KI’ + I KI + I EIS EI + S KI = [E][I] KI’ = [ES][I] [EI] [ESI] Noncompetitive Inhibition (Mixed Inhibition) v0 = Vmax[S] KM +’[S]  = 1 +[I]/KI ’ = 1 +[I]/KI’ Vmax w/o I v (µmol/min) w/ I [S]

  36. ES P S + E KI’ + I KI + I EIS EI + S KM 1/V0 = ’ 1 () + Vmax [S] Vmax I w/o I 1/v0 -1/KM 1/Vmax 1/[S] Noncompetitive Inhibition (Mixed Inhibition) v0 = Vmax[S] KM +’[S]

  37. Noncompetitive inhibition •  = 1 +[I]/KI, ’ = 1 +[I]/KI’ • As [I] increases, v decreases (1/v increases) • As [I] increases, KM decreases (1/KM increases), but in the simplest case does not change (very slight). • Cannot be overcome by increasing [S] • May intersect above, blow or even on the line (x or y axis) • x-axis - no affect on KM (KM = KM’) • y-axis - no affect on Vmax

  38. Figure U2-4.1c Uncompetitive inhibition (c) An uncompetitive inhibitor (blue) binds only to the enzyme-substrate (ES) complex and slows down the reaction probably by inducing a conformational change in the enzyme. Both the apparent Km and Vmax are affected proportionally by such an inhibitor, leading to parallel Lineweaver-Burk plots for different inhibitor concentrations.

  39. KI’ = [ES][I] [ESI] KM 1/V0 = ’ 1 + Vmax [S] Vmax I Uncompetitive Inhibition ES P S + E v0 = Vmax[S] KI’ + I KM +’[S] EIS ’ = 1 +[I]/KI’ w/o I 1/v0 1/Vmax 1/[S]

  40. Uncompetitive inhibition • ’ = 1 +[I]/KI’ • Reacts only with ES complex • As [I] increases, v decreases (1/v increases) • As [I] increases, apparent KM increases (apparent 1/KM decreases), but there is no effect on binding of E to S. • Cannot be overcome by increasing [S] • Relatively rare in single substrate reactions but can be more common in complex cases.

  41. Enzyme reactions can be slowed by the presence of inhibitors • A key parameter that can be obtained from such an analysis is the affinity of the inhibitor for the enzyme, the inhibition constant Ki. • By convention, Ki is given as the dissociation constant for the enzyme-inhibitor equilibrium: Ki’ = [ES][I] Ki= [E][I] [ESI] [EI] • The lower the value of Ki the tighter the inhibitor binds. In pharmacology, the value of Ki is often used as a measure of the effectiveness of a drug. • A compound with a very low Ki, say 10-9 M (nanomolar) or less, can be given at very low doses and will still be able to bind its target.

  42. Enzyme Inhibitor Classification • Competitive inhibitors: • Most common class of reversible inhibitor consists of compounds that resemble the substrate. • Such molecules can fit into the substrate binding site, thereby blocking access from substrate molecules. These inhibitors compete with the substrate for the active site. • Example: Many HIV protease inhibitors that have proven to be effective in treatment of AIDS are competitive inhibitors that were designed to resemble the peptide substrate of the HIV protease.

  43. Enzyme Inhibitor Classification • Not all inhibitors compete with the substrate for the active site of the enzyme; other inhibitors bind to a separate site. • Noncompetitive inhibitors bind to both the free enzyme and the enzyme-substrate complex. • If an enzyme has two substrates that must bind simultaneously for the reaction to occur, an inhibitor might compete with one substrate but not the other.

  44. Enzyme Inhibitor Classification • Not all inhibitors compete with the substrate for the active site of the enzyme; other inhibitors bind to a separate site. • Uncompetitive inhibitors bind only to the enzyme-substrate complex. • In effect, the inhibitor reduces the amount of ES that can go on to form product. • Lineweaver-Burk plots characteristically show a series of parallel lines (Fig. U2-4.1c). • Uncompetitive inhibitors often stabilize an alternative conformation of the protein, and in this case they are called allosteric inhibitors. • There are also allosteric activators: molecules that activate enzymes by stabilizing a conformation of the enzyme that is more active than the conformation that exists in their absence.

  45. Figure 3-10 Ligand-induced conformational change activates aspartate transcarbamoylase Binding of the allosteric activator ATP to its intersubunit binding sites on the regulatory subunits (that between R1, outlined in purple, and R6 is arrowed) of the T state of ATCase (top) causes a massive conformational change of the enzyme to the R state (bottom). In this state the structure of the enzyme is opened up, making the active sites on the catalytic subunits (C) accessible to substrate. Al and Zn in the lower diagram indicate the allosteric regions and the zinc-binding region, respectively; cp and asp indicate the binding sites for the substrates carbamoyl phosphate and aspartate, respectively. The red and yellow regions are the intersubunit interfaces that are disrupted by this allosteric transition.

  46. Allostery • Does not follow Michaelis-Menton kinetics! • EIS goes to products at the same rate as ES but with lower affinity for [S] • Doesn’t affect Vmax. • Does affect KM • Can operate in both directions and involves a second binding site. • Positive-activator, cooperative • Negative-inhibitor, antagonistic • Enzyme controlled by binding at second site homotrophic (modifier is related to the substrate) heterotrophic (related to substrate).

  47. Enzyme-catalyzed reactions can have multiple steps with several intermediates • Multi-step reactions can have very complicated kinetics, e.g., “double-displacement” or “ping-pong” enzymes. • These are enzymes that use two or more substrates but catalyze reactions that are strictly ordered in the sequence in which substrates bind and products are released (Figure U2-3.3a). • Lineweaver-Burk plots of the velocity against one substrate concentration at a series of fixed concentrations of other substrate give a family of parallel lines (Fig. U2-3.3b). Insert: double reciprocal plot of observed initial velocities versus CO2 concentration for CA at different concentrations of TAPS buffer: 5 mM, 10 mM, 20 mM, 50 mM. Biochemistry 1999, 38, 13119-13128

  48. Enzyme-catalyzed reactions can have multiple steps with several intermediates Example of Ping-Pong Enzyme: Aspartate aminotransferase: Catalyzes the conversion of aspartate to glutamate with production of oxaloacetate and consumption of alpha-ketoglutarate. The reaction sequence starts with the binding of aspartate to the enzyme followed by its conversion to oxaloacetate, in the process of which the aspartate leaves behind its amino group bound to a cofactor in the active site. After oxaloacetate departs (the so-called ping step), alpha-ketoglutarate reacts with the amino group and is converted to glutamate (the so-called pong step), bringing the enzyme back to its original state. The two substrates, aspartate and alpha-ketoglutarate, never encounter each other on the enzyme. The kinetics are characteristically simple for this kind of reaction: Lineweaver-Burk plots of the velocity against aspartate concentration at a series of fixed concentrations of alpha-ketoglutarate, give a family of parallel lines (Fig. U2-3.3b).

  49. Figure U2-3.3 Ping-pong or double-displacement kinetic behavior • (a) In ping-pong reactions, two substrates bind sequentially to an enzyme. In this example, a chemical group (green) is transferred from substrate A (red) to substrate B (blue). • (b) A LineweaverBurk plot of 1/v against 1/[substrate A] at various fixed concentrations of substrate B shows a set of parallel lines which are diagnostic for the ping-pong reaction mechanism.

  50. Enzyme reactions can be slowed by the presence of inhibitors • The rate of an enzyme-catalyzed reaction can be affected by molecules that do not themselves participate in the chemical reaction. • Activators increase the reaction rate and inhibitors decrease the rate. • Many drugs, including aspirin, penicillin, statins and Viagra are enzyme inhibitors: • they achieve their pharmacological effects by reducing the rate of a key enzyme-catalyzed reaction. • In some cases the effect is achieved by forming a dead-end covalent complex between the inhibitor and enzyme. • Penicillin and aspirin work this way: they form stable chemical bonds with residues in the active sites of the enzymes they inhibit. • Such “suicide inhibitors” permanently inactivate their target enzyme molecules, and cells can only overcome their effects by synthesizing fresh enzyme.

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