ENZYMATIC REACTIONS

4 ENZYMATIC REACTIONS Chapter The living cell is a cauldron in which thousands of chemical reactions proceed at the same time. Hardly any of these reactions would proceed at any noticeable rate if their starting materials, or substrates, were simply mixed in a test tube by an overoptimistic chemist. The chemist could possibly force the reactions by increasing the temperature or by using a nonselective catalyst, such as a strong acid or a strong base. But the human body is not in this lucky position because body temperature and pH must be kept within narrow limits. Therefore living things depend on highly selective catalysts called enzymes. By definition, a catalyst is a substance that accelerates a chemical reaction without being consumed in the process. Because it is regenerated at the end of each catalytic cycle (Fig. 4.1), a single molecule of the catalyst can convert many substrate molecules into product. Only a tiny amount of the catalyst is needed. The thermodynamic properties of a reaction are related to energy balance and equilibrium, whereas kinetic properties are related to the speed (velocity, or rate) of the reaction. Enzymes do not change the equilibrium of a reaction or its energy balance; they only make the reaction go faster. Enzymes change the kinetic but not the thermodynamic characteristics of the reaction. THE EQUILIBRIUM CONSTANT DESCRIBES THE EQUILIBRIUM OF THE REACTION In theory, all chemical reactions are reversible. The reaction equilibrium can be determined experimentally by mixing substrates (or products) with a suitable catalyst and allowing the reaction to proceed to completion. At this point, the concentrations of substrates and products can be measured to determine the equilibrium constant Kequ, which is defined as the ratio of product concentration to substrate concentration at equilibrium. For a simple reaction AGB the equilibrium constant is Kequ = [B] [A] [B] and [A] are the molar concentrations of product B and substrate A at equilibrium. When more than one substrate or product participate, their concentrations have to be multiplied. For the reaction A + BGC + D the equilibrium constant is Kequ = [C] × [D] [A] × [B] The alcohol dehydrogenase (ADH) reaction provides an example: ð1Þ H3C CH2OH + NAD+ Ethanol GH3C CHO + NADH + H+ Acetaldehyde NADþ (nicotinamide adenine dinucleotide) is a coenzyme that accepts hydrogen in this reaction (see Chapter 5). The equilibrium constant of the reaction is ð2Þ Kequ = [Acetaldehyde] × [NADH] x [H+] [Ethanol] × [NAD+] = 10 –11 M From Equation (2) we can calculate the relative concentrations of acetaldehyde and ethanol at equilibrium when [NADH] ¼ [NADþ] and pH ¼ 7.0: + ð3Þ [Acetaldehyde] = 10 –11 M × [NAD ] × 1 [Ethanol] [NADH] [H+] = 10 –11 M × 1 × 107 = 10 –4 There is 10,000 times more ethanol than acetaldehyde at equilibrium! Under aerobic conditions, however, NADþ is far more abundant than NADH in the cell. When [NADþ] is 1000 39 40 PRINCIPLES OF MOLECULAR STRUCTURE AND FUNCTION Product Enzyme Enzyme • Product Substrate Enzyme • Substrate Figure 4.1 The catalytic cycle. The substrate has to bind to the enzyme to form a noncovalent enzyme-substrate complex (Enzyme•Substrate). The actual reaction takes place while the substrate is bound to the enzyme. Note that the enzyme is regenerated at the end of the catalytic cycle. times higher than [NADH], Equation (3) assumes the numerical values of [Acetaldehyde] =10–11 × 1000 × 107 [Ethanol] =10–1 1 = 10 The pH also is important. At pH of 8.0 and [NADþ]/ [NADH] ratio of 1000, for example, Equation (3) yields [Acetaldehyde] = 10–11 × 1000 × 108 [Ethanol] = 100 =1 This example shows that a reaction can be driven toward product formation by raising the concentration of a substrate or lowering the concentration of a product. To adapt the equilibrium constant to physiological conditions, a “biological equilibrium constant,” K0 equ, is used. In the definition of K0 equ, a value of 1.0 is assigned to the water concentration if water participates in the reaction, and a value of 1.0 to a proton concentration of 10%7 mol/L (pH ¼ 7.0) if protons participate in the reaction. The K0 equ of the alcohol dehydrogenase reaction, for example, is not 10%11 mol/L but 10%4 mol/L. At a pH of 8.0 in the preceding example, the proton concentration would be given a numerical value of 10%1. THE FREE ENERGY CHANGE IS THE DRIVING FORCE FOR CHEMICAL REACTIONS During chemical reactions, energy is either released or absorbed. This is described as the enthalpy change DH: (4) DH ¼ DE þ P & DV DE is the heat that is released or absorbed. It is measured in either kilocalories per mole (kcal/mol) or kilojoules per mole (kJ/mol, where 1 kcal ¼ 4.184 kJ). By convention, a negative sign of DE means that heat is released; a positive sign indicates that heat is absorbed. P is the pressure, and DV is the volume change. P & DV is the work done by the system. It can be substantial in a car motor when the volume in the cylinder expands against the pressure of the piston, but volume changes are negligible in the human body. Therefore DH ' DE. The enthalpy change describes the difference in the total chemical bond energies between the substrates and products. Reactions are driven not only by DH but by the entropy change (DS) as well. Entropy is a measure of the randomness or disorderliness of the system. A cluttered desk is often cited as an example of a highentropy system. A positive DS means that the system becomes more disordered during the reaction. Entropy change and enthalpy change are combined in the free energy change DG: (5) DG ¼ DH % T & DS where T ¼ absolute temperature measured in Kelvin. DG is the driving force of the reaction. Like DE and DH in Equation (4), it is measured in kilocalories per mole (kcal/mol) or kilojoules per mole (kJ/mol). A negative sign of DG defines an exergonic reaction. It can proceed only in the forward direction. A positive sign of DG signifies an endergonic reaction. It can proceed only in the backward direction. At equilibrium, DG equals zero. Equation (5) shows that a reaction can be driven by either a decrease in the chemical bond energies of the reactants (negative DH) or an increase in their randomness (positive T & DS). Low energy content and high randomness are the preferred states. Like most students, Nature tends to slip from energized order into energy-depleted chaos. Entropy changes are small in most biochemical reactions, but diffusion is an entropy-driven process (Fig. 4.2, A). There is no making and breaking of chemical bonds during diffusion. Therefore the enthalpy change DH is zero. This leaves the T & DS part of Equation (5) as the only driving force. Thus diffusion can produce only a random distribution of the dissolved molecules. The human body is a very orderly system. To maintain this improbable and therefore thermodynamically disfavored state of affairs, biochemical reactions must antagonize the spontaneous increase in entropy. Equation (5) shows that a reaction can reduce entropy (negative T & DS) only when it consumes chemical bond energy (negative DH). In other words, the human body must consume chemical bond energy to maintain its low-entropy state. In the example of Figure 4.2, B and C, the cell maintains a sodium gradient across the membrane by pumping Enzymatic Reactions A Distilled 1 M glucose water solution Porous membrane Extracellular space 0.5 M glucose solution conditions. Standard conditions are defined by a concentration of 1 mol/L for all reactants (except protons and water) at a pH of 7.0. As in the definition of K0 equ, values of 1 are assigned both to the water concentration and to the proton concentration at pH 7. For the reaction AþB!CþD Intracellular [Na+]: 10 mM Cell Extracellular [Na+]: 150 mM ATP Na+ the standard free energy change DG00 is related to the real free energy change DG by Equation (6): ð6Þ ∆G = ∆G0′ + R × T × loge Sodium pump Extracellular space Cell [A] × [B] [C] × [D] = ∆G0′ + R × T × 2.303 × log [A] × [B] ADP + Pi B [C] × [D] where R ¼ gas constant, and T ¼ absolute temperature measured in Kelvin. The numerical value of R is 1.987 & 10%3 kcal & mol%1 & K%1. At a “standard temperature” of 25) C (298K), Equation (6) assumes the form of ð7Þ ∆G = ∆G0′ + 1.364 × log [C] × [D] [A] × [B] At equilibrium, DG ¼ 0, and Equation (7) therefore yields C Figure 4.2 Diffusion as an entropy-driven process. A, In a hypothetical two-compartment system, molecules diffuse until their concentrations are equal. This is the state of maximal entropy. B, The living cell maintains a gradient of sodium ions across its plasma membrane. The cell can maintain this gradient, which represents a low-entropy state, only by “pumping” sodium out of the cell. The pump is fueled by the chemical bond energy in adenosine triphosphate (ATP). ADP, Adenosine diphosphate; Pi, inorganic phosphate. C, The dead cell lacks ATP; therefore, it cannot maintain its sodium gradient. A high-entropy state develops spontaneously, with intracellular [Naþ] ¼ extracellular [Naþ]. sodium out of the cell (negative T & DS). Sodium pumping is driven by the hydrolysis of adenosine triphosphate (ATP) to adenosine diphosphate (ADP) and inorganic phosphate (Pi) (negative DH). Without ATP the gradient dissipates, the entropy of the system increases, and the cell dies. That is what death and dying are all about: a sharp rise in the entropy of the body. THE STANDARD FREE ENERGY CHANGE DETERMINES THE EQUILIBRIUM The free energy change DG is affected by the relative reactant concentrations. It is not a property of the reaction as such. To describe the energy balance of a reaction, the standard free energy change, DG00 , must be defined: DG00 is the free energy change under standard ð8Þ ∆G0′ = –1.364 × log [C] × [D] [A] × [B] The reactant concentrations under the logarithm are now the equilibrium concentrations. Their ratio defines the biological equilibrium constant K0 equ ð9Þ [C] × [D] [A] × [B] ′ = Kequ Substituting Equation (9) into Equation (8) yields (10) 0 DG0 ¼ %1:364 & log K0equ There is a negative logarithmic relationship between DG00 and the equilibrium constant K0 equ (Table 4.1). When DG00 is negative, product concentrations are higher than substrate concentrations at equilibrium; when it is positive, substrate concentrations are higher. ENZYMES ARE BOTH POWERFUL AND SELECTIVE There is no compelling reason why only proteins should catalyze reactions, and catalytic ribonucleic acids (RNAs) are known to exist. By and large, however, almost all enzymes are globular proteins. Enzymes can accelerate a chemical reaction enormously. Many reactions that proceed within minutes in the presence of an enzyme would require thousands of years to reach their equilibrium in the absence of a catalyst. The turnover number describes the catalytic power of the enzyme. It is defined as the maximal 41 PRINCIPLES OF MOLECULAR STRUCTURE AND FUNCTION Table 4.1 Relationship between the Equilibrium Constant K0 equ and the Standard Free Energy Change DG00 K0 equ DG00 (kcal/mol) 10%5 10%4 10%3 10%2 10%1 1 10 102 103 104 105 6.82 5.46 4.09 2.73 1.36 0 %1.36 %2.73 %4.09 %5.46 %6.82 Table 4.2 Enzymes Active site S 42 Enzyme Enzyme S Enzyme S A S Enzyme Approximate Turnover Numbers of Some Enzyme Carbonic anhydrase Catalase Acetylcholinesterase Triose phosphate isomerase a-Amylase Lactate dehydrogenase (muscle) Chymotrypsin Aldolase Lysozyme Fructose 2,6-bisphosphatase B %1 Turnover Number (s )* 600,000 80,000 25,000 4,400 300 200 100 11 0.5 0.1 *Turnover numbers are measured at saturating substrate concentrations. They depend on the assay conditions, including temperature and pH. number of substrate molecules converted to product by one enzyme molecule per second. Table 4.2 lists the turnover numbers of some enzymes. Another key property of enzymes is their substrate specificity. Typically, each reaction requires its own enzyme. For example, when an enzyme is inhibited by a drug or is deficient because of a genetic defect, only one reaction is blocked. THE SUBSTRATE MUST BIND TO ITS ENZYME BEFORE THE REACTION CAN PROCEED Enzymatic catalysis, like sex, requires intimate physical contact. It starts with the formation of an enzymesubstrate complex: E + SGE · S where E ¼ free enzyme, S ¼ free substrate, and E*S ¼ enzyme-substrate complex. In the enzyme-substrate complex, the substrate is bound noncovalently to the active site on the surface of the enzyme protein. The active site contains the functional groups for substrate binding and catalysis. If a prosthetic group participates in the reaction as a coenzyme, it is present in the active site. Figure 4.3 Two models of enzyme-substrate binding. A, Lock-and-key model. B, Induced-fit model. S, Substrate. According to the lock-and-key model, substrate and active site bind each other because their surfaces are complementary. In many cases, however, substrate binding induces a conformational change in the active site that leads to further enzyme-substrate interactions and brings catalytically active groups to the substrate. This is called induced fit (Fig. 4.3). The enzyme’s substrate specificity is determined by the geometry of enzyme-substrate binding. If the substrate is optically active, generally only one of the isomers is admitted. This is to be expected because the enzyme, being formed from optically active amino acids, is optically active itself. A three-point attachment (shown schematically in Fig. 4.4) is the minimal requirement for stereoselectivity. RATE CONSTANTS ARE USEFUL FOR DESCRIBING REACTION RATES The rate (velocity) of a chemical reaction can be described by a rate constant k: A k B In this one-substrate reaction, the reaction rate is defined by (11) V ¼ k & ½A, The rate constant has the dimension s%1 (per second), and the velocity V is the change in substrate concentration per second. For a reversible reaction, the forward and backward reactions must be considered separately: k1 ð12Þ AGB k –1 Enzymatic Reactions R CH3 CH C CH2 H3C CH3 Enzyme COO– + NH3 O (17) H Zero-order kinetics are observed only in catalyzed reactions when the substrate concentration is high, and the amount and turnover number of the catalyst, rather than the substrate availability, is the limiting factor. O C C –OOC CH2 CH3 O H Figure 4.4 Three-point attachment is the minimal requirement for stereoselectivity. In this hypothetical example, the enzyme-substrate complex is formed by a salt bond, a hydrogen bond, and a hydrophobic interaction. The substrate binds, whereas its enantiomer (bottom), is not able to form an enzyme-substrate complex. Vforward ¼ k1 & ½A, Vbackward ¼ k%1 & ½B, At equilibrium, Vforward ¼ Vbackward. Therefore the net reaction is zero: (14) ð15Þ k1 & ½A, ¼ k%1 & ½B, [B] [A] = k1 k –1 Reactions with a negative DG can occur. In reality, however, many of these reactions do not occur at a perceptible rate. The reason is that in both catalyzed and uncatalyzed reactions, the substrate must pass through a transition state before the product is formed. The structure of the transition state is intermediate between substrate and product, but its free energy content is higher. Therefore it is unstable and decomposes almost instantly to form either substrate or product. The formation of the transition state is the rate-limiting step in the overall reaction. The overall reaction shown in Figure 4.5 is exergonic because the product has lower free energy content than the substrate, but the formation of the transition state from the substrate is endergonic. The free energy difference between substrate and transition state is called the free energy of activation (DGact). It is an energy barrier that must be overcome by the kinetic = Kequ Transition state Equation (15) shows that the equilibrium constant Kequ, previously defined as [B]/[A] at equilibrium, is also the ratio of the two rate constants. The forward reaction in Equation (12) is a firstorder reaction. In a first-order reaction, the reaction rate is directly proportional to the substrate concentration. When the substrate concentration [A] is doubled, the reaction rate V is doubled as well. Uncatalyzed one-substrate reactions follow first-order kinetics. A classic example is the decay of a radioactive isotope. When two substrates participate, the reaction rate is likely to depend on the concentrations of both substrates. This is called a second-order reaction. For A+B k G ∆Gact uncatalyzed ∆Gact catalyzed Substrate ∆Greaction Product C+D the following is obtained: (16) V¼k ENZYMES DECREASE THE FREE ENERGY OF ACTIVATION R (13) A zero-order reaction is independent of the substrate concentration. No matter how many substrate molecules are present in the test tube, only a fixed number is converted to product per second: V ¼ k & ½A, & ½B, Doubling the concentration of one substrate doubles the reaction rate; doubling the concentrations of both raises it fourfold. Progress of the reaction Figure 4.5 Energy profile of a reaction. The enzyme facilitates the reaction by decreasing the free energy content of the transition state. , Uncatalyzed reaction; , catalyzed reaction. DGact, free energy of activation. 43 44 PRINCIPLES OF MOLECULAR STRUCTURE AND FUNCTION energy of the reacting molecules as they collide with each other. Most chemical systems are meta-stable, that is, they are thermodynamically unstable but kinetically stable. The human body is meta-stable in an oxygen-containing atmosphere. CO2 and H2O have lower free energy than do molecular oxygen and the organic molecules in the human body, but humans do not self-combust spontaneously because the free energy of activation is too high. Enzymes stabilize the transition state and decrease its free energy content. As a result, the enzyme increases the reaction rate by decreasing the free energy of activation. Forward and backward reactions are accelerated in proportion; therefore, the equilibrium of the reaction remains unchanged. where [ET] ¼ concentration of total enzyme, and kcat ¼ turnover number of the enzyme. The tightness of binding between enzyme and substrate in the enzyme-substrate complex is described by the “true” dissociation constant. This is the equilibrium constant for the reaction E · SGE + S: ð20Þ KD = 1. The reaction has only one substrate. 2. The substrate is present at much higher molar concentration than the enzyme. 3. Only the initial reaction rate is considered, at a time when product is virtually absent and the backward reaction negligible. 4. The course of the reaction is observed for only a very short time period; the changes in substrate and product concentrations that take place as the reaction proceeds are neglected. The conversion of enzyme-bound substrate to enzymebound product E · SGE · P requires the formation of the transition state. Therefore it is usually the ratelimiting step. At low product concentration, the backward reactions in steps 2 and 3 can be neglected. In addition, steps 2 and 3 can be lumped together to yield k1 E + SGE · S k–1 kcat E+P The velocity or rate (V) of product formation, which defines the rate of the overall reaction, is (18) V ¼ kcat & ½E * S, where kcat is the catalytic rate constant. The upper limit of V is approached when the substrate concentration is high and nearly all enzyme molecules are present as enzyme-substrate complex. Therefore the maximal reaction rate (Vmax) is (19) V max ¼ kcat & ½ET , k–1 k1 k1 E + SGE · S k–1 kcat E+P the following is obtained: k1 × [E] × [S] = ð21Þ Rate of formation of E · S k–1 × [E · S] + Rate of dissociation of E · S to E + S kcat × [E · S] Rate of product formation which yields (22) k1 & ½E, & ½S, ¼ ðk %1 þ k cat Þ & ½E * S, and ð23Þ Enzymatic reactions proceed in three steps: E + SGE · SGE · PGE + P [E · S] = Besides decomposing back to free enzyme and free substrate, the enzyme-substrate complex can undergo catalysis. Under steady-state conditions, the concentration of the enzyme-substrate complex is constant, and its rate of formation equals its rate of decomposition. For MANY ENZYMATIC REACTIONS CAN BE DESCRIBED BY MICHAELIS-MENTEN KINETICS In Michaelis-Menten kinetics, a few simple (or simplistic) assumptions are made about enzymatic catalysis: [E] × [S] [E] × [S] [E · S] = k–1 + kcat = Km k1 This is the definition of the Michaelis constant, Km. Because kcat usually is far smaller than k%1, Km is numerically similar to the true dissociation constant of the enzyme-substrate complex [Equation (20)]. The meaning of Km becomes clear when Equation (23) is remodeled to yield ð24Þ [E] [E · S] = Km [S] or ð25Þ [E · S] = [E] × [S] Km These equations show that when the substrate concentration [S] ¼ Km, the concentration of the enzymesubstrate complex E*S equals that of the free enzyme E: Km is the substrate concentration at which the enzyme is half-saturated with its substrate. Because the reaction rate is proportionate to the concentration of the enzyme-substrate complex [Equation (18)], Km is also the substrate concentration at which the reaction rate is half-maximal. Enzymatic Reactions The total enzyme (ET) is present as free enzyme and enzyme-substrate complex: (26) ½E, þ ½E * S, ¼ ½ET , V Vmax or (27) ½E, ¼ ½ET , þ ½E * S, To obtain the Michaelis-Menten equation, Equations (24) and (27) are first combined: ð28Þ [ET] – [E · S] = [E · S] 1 2 Vmax Km [S] [S] This becomes ð29Þ [ET] [E · S] Km –1 = Km Figure 4.6 Relationship between reaction rate (V) and substrate concentration ([S]) in a typical enzymatic reaction. Km, Michaelis constant; Vmax, maximal reaction rate. [S] and ð30Þ [ET] [E · S] = Km [S] +1= Km [S] + [S] [S] = Km + [S] [S] 1 V Combining Equations (18) and (19) yields ð31Þ Vmax V = [ET] × kcat [E · S] × kcat = [ET] [E · S] 1 Vmax Equations (30) and (31) now can be combined to obtain the Michaelis-Menten equation: ð32Þ Vmax V = Km Vmax Km + [S] [S] – or ð33Þ slope = V = Vmax × [S] Km + [S] Km AND Vmax CAN BE DETERMINED GRAPHICALLY The derivation of Km and Vmax is not merely a joyful intellectual exercise for the student. These kinetic properties can actually be used to predict reaction rates at varying substrate concentrations. Figure 4.6 shows what happens when a fixed amount of enzyme is incubated with varying concentrations of substrate. At substrate concentrations far below Km, the reaction rate is almost directly proportionate to the substrate concentration, and the reaction shows firstorder kinetics. Eventually, however, the reaction rate approaches Vmax. At substrate concentrations far higher than Km, the reaction becomes nearly independent of the substrate concentration and shows zero-order kinetics. Almost all enzyme molecules are present as enzymesubstrate complex, and the reaction is limited no longer by substrate availability but by the amount and turnover number of the enzyme. Km is the point on the x-axis that corresponds to ½Vmax on the y-axis. In a double-reciprocal plot, known as the LineweaverBurk plot (Fig. 4.7), the relationship between 1/V and 1 Km 1 [S] Figure 4.7 Lineweaver-Burk plot for a typical enzymatic reaction. It is derived from the equation 1/V ¼ 1/Vmax þ Km/ Vmax & 1/[S]. Km, Michaelis constant; V, reaction rate; Vmax, maximal reaction rate. 1/[S] becomes a straight line. It corresponds to the equation ð34Þ 1 1 Km 1 = + × V [S] V Vmax max which is obtained by turning the Michaelis-Menten equation [Equation (33)] upside down. The intersection of this line with the y-axis is 1/Vmax, and its intersection with the x-axis is %1/Km. The transition from first-order to zero-order kinetics with increasing substrate concentration can be compared to ticket sales in a bus terminal. When passengers are scarce, the number of tickets sold per minute depends directly on the number of passengers: Tickets are sold with first-order kinetics. However, during rush hour, when a line forms, the rate of ticket sales is no longer limited by passenger availability but by the turnover number of the ticket clerk. No matter how long the line, it progresses at a constant rate, Vmax. Tickets are now sold with zero-order kinetics. 45 46 PRINCIPLES OF MOLECULAR STRUCTURE AND FUNCTION SUBSTRATE HALF-LIFE CAN BE DETERMINED FOR FIRST-ORDER BUT NOT ZERO-ORDER REACTIONS Figure 4.8 shows how the substrate of an irreversible reaction gradually disappears by being converted to product. The slope of the curve is the reaction rate. The rate of a zero-order reaction remains constant over time; therefore, we get a straight line. The first-order reaction, in contrast, slows down as less and less substrate is left, and its rate approaches zero asymptotically. The half-life is the time period during which one half of the substrate is consumed in a first-order reaction. Zero-order reactions do not have a half-life. Many drugs are metabolized by enzymes in the liver. The drug concentration usually is so far below the Km of the metabolizing enzyme that it is metabolized with first-order kinetics. Consequently, the drug’s half-life can be determined by measuring its plasma concentrations at different points in time. Alcohol metabolism is very different. The alcohol level is so high after a few drinks that the metabolizing enzyme, alcohol dehydrogenase, is almost completely saturated. Therefore a constant amount of about 10 g/hr is metabolized no matter how drunk a person is (Fig. 4.9). Kcat/Km PREDICTS THE ENZYME ACTIVITY AT LOW SUBSTRATE CONCENTRATION Vmax depends directly on the enzyme concentration [Equation (19)]: Doubling the enzyme concentration doubles the reaction rate. Therefore the amount of an enzyme is most conveniently determined by measuring its activity at saturating substrate concentrations. This is done in the clinical laboratory when serum enzymes are determined for diagnostic purposes (see Chapter 15). Enzyme activities can be expressed as international units (IUs). One IU is defined as the amount of enzyme that converts one Blood alcohol concentration (g/L) Absorption from GI tract 1.5 1.0 0.5 First-order kinetics 2 4 6 8 10 Time after ingestion (hours) Figure 4.9 Blood alcohol concentration after the ingestion of 120 g of ethanol. The linear decrease of the alcohol level 2 to 10 hours after ingestion shows that a zero-order reaction limits the rate of alcohol metabolism. GI, Gastrointestinal. micromole (mmol) of substrate to product per minute. Because Vmax depends on temperature, pH, and other factors, the incubation conditions must be specified. However, most enzymes in the living cell work with substrate concentrations far below their Km. At these low substrate concentrations, the kcat/Km ratio is the best predictor of the actual reaction rate. This is apparent when Equations (18) and (25) are combined: ð35Þ V = kcat × [E] × [S] Km When the substrate concentration is far below Km, almost all the enzyme molecules are present as free enzyme rather than enzyme-substrate complex, and ½E, ' ½ET , Therefore Equation (35) yields Substrate concentration ð36Þ V ≈ kcat × [ET] × [S] Km It now is evident that at a very low substrate concentration, the reaction rate depends directly on enzyme concentration [ET], substrate concentration [S], and kcat/Km. Km is a measure of the affinity between enzyme and substrate, where a low Km signifies high affinity, and kcat is the turnover number of the enzyme. 100% 50% 25% 12.5% T12 T12 T12 Time Figure 4.8 Disappearance of substrate is traced for a zeroorder reaction ( ) and a first-order reaction ( ). The halflife (T½) is defined as the time period during which half of the substrate is converted to product in the first-order reaction. ALLOSTERIC ENZYMES DO NOT CONFORM TO MICHAELIS-MENTEN KINETICS Not all enzymes show simple Michaelis-Menten kinetics. For example, the sigmoidal relationship between substrate concentration and reaction rate (Fig. 4.10) is typical for an allosteric enzyme with more than one active site and positive cooperativity between the active Enzymatic Reactions V V Vmax B A C [S] Figure 4.10 Plot of velocity (V) against substrate concentration ([S]) for an allosteric enzyme with positive cooperativity. Line A, Enzyme alone; line B, with positive allosteric effector; line C, with negative allosteric effector; Vmax, maximal reaction rate. sites. This curve would not yield a straight line in the Lineweaver-Burk plot. More important are the responses of enzymes to allosteric effectors. Positive allosteric effectors activate the enzyme, and negative allosteric effectors inhibit it. These regulatory molecules bind to sites other than the substrate-binding site. Their binding is noncovalent and therefore reversible. Allosteric effectors can change both the enzyme’s affinity for substrate (Km) and its turnover number (kcat). Allosteric enzymes occupy strategic locations in metabolic pathways where they are regulated by substrates or products of the pathway. ENZYME ACTIVITY DEPENDS ON TEMPERATURE AND pH Chemical reactions are accelerated by increased temperature. The greater the activation energy DGact of the reaction, the greater is its temperature dependence. The Q10 value is the factor by which the reaction is accelerated when the temperature rises by 10) C. Most uncatalyzed reactions have Q10 values between 2 and 5. Enzymatic reactions have lower activation energies, so their Q10 values are most commonly between 1.7 and 2.5. At very high temperatures, however, enzymes become irreversibly denatured. This produces the relationship shown in Figure 4.11. The temperature dependence of enzymatic reactions contributes to the increased metabolic rate during fever. Presumably it also is responsible for the fact that humans cannot tolerate body temperatures higher than 42) C to 43) C. The most sensitive enzymes already start denaturing at temperatures above this limit. Protein denaturation is time dependent, and an enzyme that survives a temperature of 45) C for some minutes may well denature gradually during the course of several hours. Hypothermia is far better tolerated than hyperthermia, and the temperature of the toes can fall close to 0) C on a cold winter day. This temperature blocks nerve conduction and muscle activity, but it does not 10 20 30 40 50 Temperature (°C) Figure 4.11 Temperature dependence of a typical enzymatic reaction. V, Reaction rate. kill the cells. However, the metabolic rate is depressed at low temperatures. Therefore a slowdown in the vital functions of the brain and heart limits a person’s tolerance of hypothermia. Also, a vicious cycle develops when decreased metabolism reduces heat production during hypothermia. Hypothermia makes cells and tissues more resistant to hypoxia because it decreases their oxygen consumption. Organs used for transplantation can be preserved in the cold for many hours. Enzymes are also affected by pH (Fig. 4.12), mainly because the protonation state of catalytically active groups in the enzyme depends on pH. The pH values of tissues and body fluids are tightly regulated to satisfy the pH requirements of the enzymes. Deviations of more than 0.5 pH units from the normal blood pH of 7.4 are fatal. Inside the cells, typical pH values are 6.5 to 7.0 in the cytoplasm, 7.5 to 8.0 in the mitochondrial matrix, and 4.5 to 5.5 in the lysosomes. DIFFERENT TYPES OF REVERSIBLE ENZYME INHIBITION CAN BE DISTINGUISHED KINETICALLY Competitive inhibitors are structurally related to the normal substrate of the enzyme. They compete with the substrate by binding noncovalently to the active site of the V Alkaline phosphatase Pepsin Lysozyme 2 Figure 4.12 rate. 4 6 8 10 pH pH dependence of some enzymes. V, Reaction 47 48 PRINCIPLES OF MOLECULAR STRUCTURE AND FUNCTION enzyme. For example, the mitochondrial enzyme succinate dehydrogenase (SDH) catalyzes the following reaction: H C H –OOC H COO– + FAD C H Succinate SDH –OOC H C H C COO– + FADH2 Fumarate CLINICAL EXAMPLE 4.1: Methanol Poisoning Methanol is sometimes swallowed by people who read “methyl alcohol” on a label and mistake it for the real stuff. Methanol itself is not very obnoxious, but it is converted to the toxic metabolites formaldehyde and formic acid in the body (Fig. 4.13). Formaldehyde (otherwise used to preserve cadavers) is chemically reactive, and formic acid causes acidosis. Blindness and death can result from methanol poisoning. H C H H OH ADH (liver) O– H C H O H Formaldehyde Methanol C This reaction is competitively inhibited by malonate: H C H –OOC COO– Malonate Malonate binds to the enzyme by the same electrostatic interactions as the substrate succinate, but it cannot be converted to a product. In other cases, such as Clinical Example 4.1, the inhibitor is converted to a product: Two alternative substrates can compete for the enzyme. Competitive inhibitors do not change Vmax because inhibitor binding is reversible and can be overcome by high concentrations of the substrate. However, substrate binding to the enzyme is impaired, and the apparent binding affinity is decreased. Therefore competitive inhibitors increase Km (Fig. 4.14). V Vmax(A, B) A B 1 V (A, B) 2 max Vmax(C) C Vmax(D) O D Formic acid [S] H H C H H C H Ethanol OH ADH (liver) H H C H H C H O Acetaldehyde H C H O– A C Km Km (D) (A, C) Km (B) O Acetic acid Metabolic pathways CO2 + H2O 1 V D Figure 4.13 Role of alcohol dehydrogenase (ADH) in the metabolism of methanol and ethanol. The two substrates compete for the enzyme. Therefore ethanol inhibits the formation of toxic formaldehyde and formic acid from methanol. The methanol-metabolizing enzyme alcohol dehydrogenase can metabolize ethanol as well. It actually has a higher affinity (lower Km) for ethanol than for methanol. However, whereas methanol metabolites accumulate in the body, ethanol metabolites are channeled smoothly into the major metabolic pathways where they are rapidly oxidized to carbon dioxide and water. They do not accumulate to toxic levels. When ethanol is administered to a patient with methanol poisoning, the formation of the toxic methanol metabolites is delayed because ethanol competes with methanol for the enzyme. The patient remains drunk but alive. C B A B 1 [S] Figure 4.14 Effects of inhibitors on enzymatic reactions. Line A, Uninhibited enzyme; line B, competitive inhibitor; line C, noncompetitive inhibitor; line D, uncompetitive inhibitor. For noncompetitive inhibition, it is assumed that the inhibitor binds equally well to the free enzyme and the enzymesubstrate complex (see text for discussion). The kinetic effects of irreversible inhibitors resemble those shown for the noncompetitive inhibitor. A, Reaction rate (V) plotted against substrate concentration ([S]). B, In the Lineweaver-Burk plot, the effects of inhibitors on Vmax (maximal reaction rate) and Km (Michaelis constant) are reflected in changes of the intercepts with the y-axis and x-axis, respectively. Enzymatic Reactions Noncompetitive inhibitors are structurally unrelated to the substrate and bind to the enzyme protein outside of the substrate-binding site. They do not necessarily prevent substrate binding, but they block enzymatic catalysis. If the noncompetitive inhibitor binds equally well to the free enzyme and the enzyme-substrate complex, it reduces Vmax without changing Km. Uncompetitive inhibitors bind only to the enzymesubstrate complex but not to the free enzyme. They thereby reduce both Km and Vmax. Unlike competitive inhibitors, which are most effective at low substrate concentrations, uncompetitive inhibitors work best when the substrate concentration is high. COVALENT MODIFICATION CAN INHIBIT ENZYMES IRREVERSIBLY Competitive, noncompetitive, and uncompetitive inhibitors bind noncovalently to the enzyme. Therefore their actions are reversible. Enzyme activity is fully restored when the inhibitor is removed, for example, by extensive dialysis in the test tube, or by metabolic inactivation or renal excretion in the body. Irreversible inhibitors, however, form a covalent bond with the enzyme. The chemically modified enzyme is dead, and this type of inhibition can be overcome only by the synthesis of new enzyme. CLINICAL EXAMPLE 4.2: Organophosphate Poisoning Organophosphates are irreversible inhibitors of acetylcholinesterase, the enzyme that degrades the neurotransmitter acetylcholine at cholinergic synapses (see Chapter 16). The organophosphate inactivates acetylcholinesterase by forming a covalent bond with an essential serine residue in its active site: C HC O CH2 O OH + F P CH3 O CH CH3 CH3 HN Serine residue (in acetylcholinesterase) Sarin ENZYMES ARE CLASSIFIED ACCORDING TO THEIR REACTION TYPE Enzymes are most commonly named after their substrate and their reaction type, with the suffix -ase at the end. For example, monoamine oxidase is an enzyme that oxidizes monoamines, and catechol-O-methyltransferase transfers a methyl group to an oxygen in a catechol. According to their reaction type, enzymes are grouped into the following six classes. Oxidoreductases Oxidoreductases catalyze oxidation-reduction reactions: electron transfers, hydrogen transfers, and reactions involving molecular oxygen. Dehydrogenases transfer hydrogen between a substrate and a coenzyme, most commonly NAD (nicotinamide adenine dinucleotide), NADP (nicotinamide adenine dinucleotide phosphate), FAD (flavin adenine dinucleotide), or FMN (flavin mononucleotide). These enzymes are named after the substrate from which hydrogen is removed. For example, C HC HN O CH2 O O P CH3 CH3 O CH + HF CH3 Without the free hydroxyl group of the serine side chain, the enzyme is completely inactive. Acetylcholine is no longer degraded and accumulates at cholinergic synapses in skeletal muscles, the autonomic nervous system, and the brain. Overstimulation of acetylcholine receptors leads to paralysis, autonomic dysfunction, and delirium. Organophosphates have two uses. Those that are most potent on the acetylcholinesterase of insects are used as agricultural pesticides, and those that work best on the human enzyme are “nerve gases” that are of interest to terrorists and the military. 49 50 PRINCIPLES OF MOLECULAR STRUCTURE AND FUNCTION H C H H H C H Ethanol OH + Other examples of transferases include phosphorylases, which cleave bonds by the addition of inorganic phosphate; glycosyl transferases, which transfer a monosaccharide to an acceptor molecule; the transaminases (see Chapter 26); and the peptidyl transferase of the ribosome (see Chapter 6). NAD+ Alcohol dehydrogenase Hydrolases H H H C H + NADH + H+ C O Acetaldehyde Oxygenases use molecular oxygen as a substrate. Dioxygenases incorporate both oxygen atoms of O2 into their substrate; monooxygenases incorporate only one. Most hydroxylases are monooxygenases: H þ O2 þ NADPH þ Hþ # Substrate OH þ NADPþ þ H2 O Substrate In this reaction, the second oxygen atom reacts with the reduced coenzyme NADPH to form water. Peroxidases use hydrogen peroxide or an organic peroxide as one of their substrates. Catalase is technically a peroxidase. It degrades hydrogen peroxide to molecular oxygen and water: H2O2 Hydrolases cleave bonds by the addition of water. Digestive enzymes and lysosomal enzymes are hydrolases. Their names indicate the substrates or bonds on which they act. Examples include peptidases (proteases), esterases, lipases, and phosphatases. For example, acetylcholinesterase cleaves an ester bond in acetylcholine. If the substrate is polymeric, the cleavage specificity of the enzyme is indicated by the prefixes endo- (from Greek meaning “inside”) and exo- (from Greek meaning “outside”). For example, an exopeptidase cleaves amino acids from the end of a polypeptide, and an endopeptidase cleaves internal peptide bonds. Lyases Lyases remove a group nonhydrolytically, forming a double bond. Examples are the dehydratases: R1 H2O + 1 O2 2 HO OH + ATP CH H2C H2C OH R1 O– CH H2C O P O Glycerol C C H OH H H C C OH Glycerol kinase HO H R2 G Transferases Transferases transfer a group from one molecule to another. Kinases transfer phosphate from ATP to a second substrate. They are named according to the substrate to which the phosphate is transferred. For example, H2C H Glycerol 3-phosphate + ADP O– R2 + H2O Enzymatic Reactions and decarboxylases: carboxylases (see Chapter 21) also are classified as ligases. O R R C H+O C O ENZYMES STABILIZE THE TRANSITION STATE OH Many lyase reactions proceed in the opposite direction, creating a new bond and obliterating a double bond in one of the substrates. These enzymes are called synthases. Isomerases Isomerases interconvert positional, geometric, or optical isomers. Ligases Ligases couple the hydrolysis of a phosphoanhydride bond to the formation of a bond. These enzymes are often called synthetases. For example, glutamine synthetase couples ATP hydrolysis to the formation of the amide bond in glutamine: COO– (CH2)2 + NH3 + ATP + H3N COO– CH Glutamate Glutamine synthetase O C NH2 + ADP + Pi + CH 1. Entropy effect: The transition state can form only when the substrates of a two-substrate reaction collide in the correct geometric orientation and with sufficient energy to bring them to the transition state. The enzyme increases the likelihood of such an event by binding the two substrates to its active site in close proximity and in the correct geometric orientation. 2. Stabilization of the transition state: The enzyme forms favorable interactions with the transition state of the reaction, thereby reducing its free energy content and the free energy of activation. 3. General acid-base catalysis: Catalysis requires ionizable groups on the enzyme that accept or donate protons during the reaction. Enzymes can also provide electron pair donors and acceptors, which are known as Lewis bases and Lewis acids, respectively. Because the ionizable groups on the enzyme must be in the correct protonation state, general acid-base catalysis is the most important reason for the pH dependence of enzymatic reactions. For example, if the reaction requires a deprotonated glutamate side chain with a pK of 4.0 as a general base and a protonated histidine side chain with a pK of 6.0 as a general acid, only pH values between 4.0 and 6.0 will allow high reaction rates. 4. Covalent catalysis: The enzyme forms a transient covalent bond with the substrate. The serine proteases described in the following paragraph are the most prominent example. CHYMOTRYPSIN FORMS A TRANSIENT COVALENT BOND DURING CATALYSIS (CH2)2 H3N Enzymes can stabilize the transition state of the reaction by making its formation a more likely event, thereby increasing the entropy of the transition state, or by forming energetically favorable noncovalent interactions with the transition state, thereby reducing its enthalpy. Four mechanisms of enzymatic catalysis can be distinguished: COO– Glutamine DNA ligase and aminoacyl–tRNA synthetases (see Chapter 6) are other examples. The biotin-dependent The serine proteases cleave peptide bonds with the help of a serine residue in their active site. The pancreatic enzyme chymotrypsin is a typical example. When chymotrypsin binds its polypeptide substrate, it forms a hydrophobic interaction with an amino acid side chain in the substrate. The peptide bond that is formed by the carboxyl group of this hydrophobic amino acid is targeted for cleavage. This peptide bond is placed right 51 52 PRINCIPLES OF MOLECULAR STRUCTURE AND FUNCTION His57 N O Asp102 Ser195 C O– HN N H H O O C Enzyme-substrate complex H HN N CH R Hydrophobic pocket His57 O O– N Ser195 Asp102 C HN+ N H H O– O First tetrahedral intermediate H C HN N CH R His57 O Ser195 Asp102 C O– HN O O N C H H –NH2 + CH O Acyl-enzyme intermediate, first product released R His57 O O– N Ser195 Asp102 C HN+ N H H Second tetrahedral intermediate O– O H C HO N CH R His57 O Ser195 Asp102 C O– HN N H O O + C HO CH Second product released R Figure 4.15 Catalytic mechanism of chymotrypsin, a typical serine protease. on the hydroxyl group of the catalytic serine residue in the active site, Ser-195 (Fig. 4.15). In addition to Ser-195, catalysis requires a deprotonated histidine residue, His-57, and a deprotonated aspartate residue, Asp-102. The numbers indicate the positions of the amino acids in the polypeptide, counting from the N-terminus. Although widely separated in the amino acid sequence of the protein, these three amino acids are hydrogen bonded to each other in the active site of the enzyme. Chymotrypsin cleaves the bond in a sequence of two reactions. In the first reaction, the peptide bond in the substrate is cleaved, and one of the fragments binds covalently to the serine side chain to form an acylenzyme intermediate. In the second reaction, this intermediate is cleaved hydrolytically. The acyl-enzyme intermediate does not qualify as a transition state because it occupies a valley in the free energy graph, rather than a peak (Fig. 4.16). The transition states in both reactions are negatively charged Enzymatic Reactions cleaves bonds formed by glycine. Thrombin is highly selective for a small number of plasma proteins, including fibrinogen. Tetrahedral intermediates (transition states) G SUMMARY Acyl-enzyme intermediate Progress of reaction Figure 4.16 Energy profile for the reaction of a serine protease. G, Free energy. tetrahedral intermediates that are stabilized by hydrogen bonds with two N-H groups in the main chain of the enzyme. To form the negatively charged transition state, a proton must be transferred from the hydroxyl group of Ser-195 to His-57. Asp-102 remains negatively charged throughout the catalytic cycle. It forms a salt bond with the protonated but not the deprotonated form of His-57, thereby stabilizing the protonated form and increasing the proton affinity of the histidine. The serine proteases are a large family of enzymes that includes the digestive enzymes trypsin, chymotrypsin, and elastase, and blood clotting factors including thrombin. They all use the same catalytic mechanism but have different substrate specificities. Chymotrypsin cleaves peptide bonds on the carboxyl side of large hydrophobic amino acids. Trypsin contains a negatively charged aspartate residue in its substratebinding pocket and therefore prefers peptide bonds formed by positively charged amino acids. Elastase Chemical reactions proceed to an equilibrium state at which the rates of the forward and backward reactions are equal, driven by the free energy change that accompanies the reaction. Enzymes cannot change the equilibrium of the reaction. They can only increase the reaction rate. Enzymatic catalysis starts with the formation of a noncovalent enzyme-substrate complex. While bound to the enzyme, the substrate is converted to an unstable transition state that decomposes almost immediately to form either substrate or product. The enzyme accelerates the reaction by making the formation of the transition state more likely and by stabilizing the transition state energetically. The Michaelis constant Km is the substrate concentration at which the reaction rate is half-maximal. It is determined mainly by the binding affinity between enzyme and substrate. At substrate concentrations far lower than Km, the reaction rate increases almost linearly with increasing substrate concentration. The reaction shows first-order kinetics. However, at substrate concentrations far higher than Km, the reaction rate can hardly be increased by further increases of the substrate concentration because most of the enzyme is already present as enzyme-substrate complex. The reaction shows zero-order kinetics. The rate of enzymatic reactions increases with increasing temperature, typically with a doubling of the reaction rate for a temperature increase of about 10# C. The pH also is important because most enzymes use ionizable groups for catalysis. These groups must be in the proper protonation state. Many drugs and toxins act as specific enzyme inhibitors. Some bind to the enzyme reversibly, through noncovalent interactions. Others form a covalent bond with the enzyme, thereby destroying its catalytic activity permanently. QUESTIONS 1. During a drug screening program, you find a chemical that decreases the activity of the enzyme monoamine oxidase. A fixed dose of the chemical reduces the catalytic activity of the enzyme by the same percentage at all substrate concentrations, with a decrease in Vmax. Km is unaffected. This inhibitor is A. Definitely a competitive inhibitor B. Definitely a noncompetitive inhibitor C. Definitely an irreversible inhibitor D. Either a competitive or an irreversible inhibitor E. Either a noncompetitive or an irreversible inhibitor 2. The irreversible enzymatic reaction Oxaloacetate þ Acetyl coenzyme A ðCoAÞ SH ! Citrate þ CoA is inhibited by high concentrations of its own product citrate. This product inhibition can be overcome, and a normal Vmax can be restored, when the oxaloacetate 53 54 PRINCIPLES OF MOLECULAR STRUCTURE AND FUNCTION concentration is raised but not when the acetyl-CoA concentration is raised. This observation suggests that citrate is A. An irreversible inhibitor reacting with the oxaloacetate binding site of the enzyme B. A competitive inhibitor binding to the acetyl-CoA binding site of the enzyme C. A competitive inhibitor binding to the oxaloacetate binding site of the enzyme D. A noncompetitive inhibitor binding to the acetyl-CoA binding site of the enzyme E. A noncompetitive inhibitor binding to the oxaloacetate binding site of the enzyme 3. An enzymatic reaction works best at pH values between 6 and 8. This is compatible with the assumption that the reaction mechanism requires two ionizable amino acid side chains in the active site of the enzyme, possibly A. Protonated glutamate and deprotonated aspartate B. Protonated histidine and deprotonated lysine C. Protonated glutamate and deprotonated histidine D. Protonated arginine and deprotonated lysine E. Protonated cysteine and deprotonated histidine 4. A biotechnology company has cloned four different forms of the enzyme money synthetase, which catalyzes the reaction Garbage þ ATP ! Money þ ADP þ Phosphate þ Hþ The Km values of these enzymes for garbage and the Vmax values are as follows: Enzyme 1: Km ¼ 0.1 mmol/L, Vmax ¼ 5.0 mmol/min Enzyme 2: Km ¼ 0.3 mmol/L, Vmax ¼ 2.0 mmol/min Enzyme 3: Km ¼ 1.0 mmol/L, Vmax ¼ 5.0 mmol/min Enzyme 4: Km ¼ 3.0 mmol/L, Vmax ¼ 20 mmol/min Which of the four enzymes is fastest at a saturating ATP concentration and a garbage concentration of 0.01 mmol/L? A. Enzyme B. Enzyme C. Enzyme D. Enzyme 1 2 3 4 5. Which of the four forms of money synthetase is fastest at a saturating ATP concentration and a garbage concentration of 10 mmol/L? A. Enzyme B. Enzyme C. Enzyme D. Enzyme 1 2 3 4 6. If the money synthetase reaction is freely reversible, which of the following manipulations would be best to favor money formation over garbage formation and to increase the [Money]/ [Garbage] ratio at equilibrium? A. Decreasing the pH value B. Adding another enzyme that destroys ADP C. Using a very low concentration of ATP D. Increasing the temperature E. Adding a noncompetitive inhibitor