# CINTICA ENZIMTICA Chemical Kinetics A A I 1

- Slides: 62

CINÉTICA ENZIMÁTICA

Chemical Kinetics A A I 1 P I 2 P I 1 and I 2: Intermediates in the reaction Reaction Order At constant temperature, the rate of an elementary reaction is proportional to the frequency with which the reacting molecules come together First order reaction A P k : rate constant The instantaneous rate of appearance of products or disappearance of reactant is called the velocity (v) of the reaction k has units of (s-1)

Reaction kinetics: 1 st order reactions • Decay reactions, like radio-activity; SN 1 reactions [A] Rate: - • • • Rewriting: - t Integration gives: So: ln[A]t – ln[A]0 = -kt or: -kt • • •

The time for half of the reactant initially present to decompose, its half-time or half-life, t 1/2 , is a constant and hence independent of the initial concentration of reactant. By substituting the relationship [A] = [A 0] / 2 when t = t 1/2 into ln [A]=ln [A]0 - kt and rearranging: t 1/2 = ln 2/k = 0. 693/k

Rate Equations Substances that are inherently unstable, such as radioactive nuclei, decompose through first order reactions Radionuclide Half-life Type of Radiationa

Second-order reaction 2 A The half-time for a second order reaction is expressed P t 1/2 = 1/k [A]0 and therefore, in contrast to a first order reaction depends on the initial reactant concentration. A+B P Here, the reaction is said to be first order in A and first order in B. Unimolecular and bimolecular reactions are common. Termolecular reactions are unusual because the simultaneous collision of three molecules is a rare event. Fourth and higher order reactions are unknown.

Reaction kinetics: 2 nd order reactions So: - =- When [A] [B], this equation is mathematically rather complicated. A simplification reads as follows: take [P] = x, then [A] = [A]0 – x and [B] = [B]0 – x The rate then becomes: n = k 2([A]0 -x)([B]0 -x) so Integration gives: Plotting against t gives a straight line with slope k 2([B]0 -[A]0)

Special cases: • [A]0>>[B]0 (pseudo-first order kinetics) Example: - in which k'=k 2[H 2 O] This is a pseudo-first order reaction, since [H 2 O] is constant. The second-order rate constant k 2 can be calculated from k' and [H 2 O]. In a dilute aqueous solution, [H 2 O]=55 M.

Reversible reactions Take the simplest possibility: On t = 0: [A] = [A]0 [B] = 0 t = t: [A] = [A]0 -x [B] = x [A]0 [A]t xe [B]t t = k 1[A] – k-1[B] = k 1([A]0 – x) – k-1 x = k 1[A]0 – (k 1 + k-1)x Integration gives: At equilibrium, the net reaction rate = 0, so [B]t is constant (=[B]e = xe), so: k 1[A]e = k-1[B]e = k-1 xe (1)

There is an equilibrium constant: so: (2) Combining eq (1) with (2) gives: This is the rate equation for a first order process! Determination of (k 1 + k-1) by plotting Eq (2) gives against t 2 equations, 2 unknowns Individual values of k 1 and k-1 can be determined

Preequilibria Very complicated kinetics, unless you assume that [A·B] is constant during a large part of the reaction (steady state approach) k 1[A][B] = k-1[A·B] + k 2[A·B] = (k-1 + k 2)[A·B] = k 1[A][B] / k-1 + k 2 So the rate equation now becomes: n = k 2[A·B] =

n = k 2[A·B] = A 0 Two possibilities: [C] - rapid breakdown of A·B, k 2>>k-1, so n = k 1[A][B] [A·B] t - slow breakdown of the complex: k 2<<k 1, k-1, so: n = k 2[A·B] = = A 0 xe [A·B] [C] k 2 K[A][B] t

Enzyme Kinetics ß-fructofuranosidase: Sucrose + H 2 O glucose + fructose When [S] » [E] : the rate is zero order with respect to sucrose. Initial rate no longer increases at S higher than S 4

The Michaelis-Menten Equation ET = E + ES v= Vmax = k 2 ET v/Vmax = k 2 ES / k 2 ET = ES/ ET v/Vmax = ES/ E + ES ¿How to know ES? This equation cannot be explicitly integrated, however, without simplifying assumptions, two possibilities are: 1. Assumption of equilibrium. Leonor Michaelis and Maud Menten, building on the work of Victor Henri, assumed that k-1 » k 2, so that the first step of the reaction reaches equilibrium. ES = E * S / Ks Ks is the dissociation constant of the first step in the enzymatic reaction

The Michaelis-Menten Equation 1. Assumption of steady-state. Figure illustrates the progress curves of the various participants in reaction under the physiologically common conditions that substrate is in great excess over Enzyme ([S] » [E]). ES maintains a steady state and [ES] can be treated as having a constant value: The so called steady state assumption, a more general condition that of equilibrium, was first proposed in 1925 by G. E. Briggs and B. S. Haldane

The Michaelis-Menten Equation Solving for [ES]: ES = E * S /(k-1+k 2)/k 1 The Michaelis constant, KM , is defined as Therefore: ES = E * S / KM

The Michaelis-Menten Equation The expression of the initial velocity (v 0) of the reaction, the velocity at t=0, thereby becomes v/Vmax = ES/ (E + ES) v/Vmax = (E*S)/Km/ (E + (E*S)/Km ) v/Vmax = S/ Km / (1 + S/Km) v/Vmax = S / Km + S This expression, the Michaelis-Menten equation, is the basic equation of enzyme kinetic. The maximal velocity of a reaction, Vmax occurs at high substrate concentrations when the enzyme is saturated, that is, S>> Km, and ET is entirely in the ES form v= Vmax when

Significance of the Michaelis Constant The Michaelis constant, KM, has a simple operational definition. At the substrate concentration at which [S] = KM, this equation yields v 0 = Vmax/2 so that KM is the substrate concentration at which the reaction velocity is half maximal

Significance of the Michaelis Constant The magnitude of KM varies widely with the identity of the enzyme and the nature of the substrate. It is also a function of temperature and p. H. The Michaelis constant can be expressed as Since Ks is the dissociation constant of the Michaelis complex, as Ks decreases, the enzyme’s affinity for substrate increases. KM in therefore also a measure of the affinity of the enzyme for its substrate, provided k 2/k 1 is small compared to Ks, that is k 2 ‹ k-1 so that the ES P reaction proceeds more slowly than ES reverts to E + S kcat/KM Is a Measure of Catalytic Efficiency We can define the catalytic constant, kcat, of an enzyme as This quantity is also known as the turnover number of an enzyme because it is the number of reaction processes (turnovers) that each active site catalyzes per unit time.

Turn Over Numbers of Enzymes Substrate Catalase H 2 O 2 Carbonic anhydrase HCO 3 - 400, 000 Acetylcholinesterase Acetylcholine 140, 000 b-Lactamase Benzylpenicillin Fumarase Fumarate Rec. A protein (ATPase) ATP kcat (s-1) 40, 000 2, 000 800 0. 4 The number of product transformed from substrate by one enzyme molecule in one second Adapted from Nelson & Cox (2000) Lehninger Principles of Biochemistry (3 e) p. 263

kcat/KM Is a Measure of Catalytic Efficiency When [S] « KM, very little ES is formed. Consequently, [E] ≈ [E]T, so reduces to a second-order rate equation: The quantity kcat/KM is a measure of an enzyme’s catalytic efficiency. There is an upper limit to the value of kcat/KM : It can be not greater than k 1; that is, the decomposition of ES to E + P can occur no more frequently than E and S come together to form ES. The most efficient enzymes have kcat/KM values near to the diffusion-controlled limit of 108 to 109 M-1. s-1

Chymotrypsin Has Distinct kcat / Km to Different Substrates = – – – = O R O H 3 C–C–N–C–C–O–CH 3 H H kcat / Km R= Glycine –H 1. 3 ╳ 10 -1 Norvaline –CH 2–CH 3 3. 6 ╳ 102 Norleucine –CH 2–CH 3 3. 0 ╳ 103 Phenylalanine –CH 2– 1. 0 ╳ 105 (M-1 s-1) Adapted from Mathews et al (2000) Biochemistry (3 e) p. 379

Lineweaver-Burk or double-reciprocal plot Analysis of Kinetic Data

S >> Km vi=Vmax= k 2 Et

Al iniciar: t = 0, S = So A cualquier tiempo: T=t S=S X = (So-S)/So - d. S/dt = vi = So d. X/dt

Temperature Dependence of Enzymes • • As is the case with most reactions, an increase in temperature will result in an increase in kcat for an enzymatic reaction. From general principles, it can be determined that the rate of any reaction will typically double for every 10°C increase in temperature. Many enzymes display maximum temperatures around 40°C, which is relatively close to body temperature. There are enzymes that are isolated from thermophilic organisms that display maxima around 100°C, and some that are isolated from psychrophilic organisms that display maxima around 10°C.

Kinetics and Transition State Theory Consider a bimolecular reaction that proceeds along the following pathway X‡ is the transition state k is the ordinary rate constant of the elementary reaction and k’ is the rate constant for the decomposition of X‡ to products. Although X‡ is unstable, it is assumed to be in rapid equilibrium with the reactants; that is: K‡ is an equilibrium constant T is the absolute temperature R is the gas constant (8. 3145 J. K-1 mol -1) Combining the equations yields three preceding This equation indicates that the rate of a reaction not only depends on the concentration of its reactants, but also decreases exponentially with ∆G‡

Kinetics and Transition State Theory The larger the difference between the free energy of the transition state and that of the reactants, that is, the less stable the transition state, the slower the reaction proceeds. k’ is the rate at which X‡ decomposes v is the vibrational frequency X‡ of the bond that breaks as X‡ decomposes to products k, the transmission coefficient, is the probability that the breakdown of X‡ will be in the direction of products formation rather than back to reactants. For most spontaneous reactions, k is assumed to be 1. 0 (although this number, which must be between 0 and 1, can rarely be calculated with confidence) Planck’s law states that is the average energy of the vibration that leads to the decomposition of X‡ h is Planck’s constant (6. 6261 x 10 -34 J. s) At temperature T, the classical energy of an oscillator is k. B is the Boltzmann constant (1. 3807 x 10 -23 J. K-1) As the temperature rises, so that there is increased thermal energy available to drive the reacting complex over the activation barrier, the reaction speeds up.

Interpretation of rate constants: the Arrhenius equation X‡ (TS) Every reaction has to overcome an energy barrier: the transition state (TS, X‡). At higher temperature, more particles are able to overcome the energy barrier. Eact E S P Reaction coordinate Arrhenius equation: Ea can be determined by measuring kobs at two different temperatures:

Arrhenius: Idem, from statistical mechanics (collision theory) P = probability factor (not every collision is effective) Z = collision number (number of collisions per second)

Idem, from transition state theory: X‡ (TS) Eact E or [X‡] = K‡[A][B] S P Reaction coordinate n = k‡[X‡] = k‡K‡[A][B] = k[A][B], so k = k‡K‡ Statistical mechanics gives us the following relation: so k. B = Boltzmann’s constant; h = Planck’s constant

For all equilibria we can write: DG 0 = - RT ln K, so for our case we get: DG‡ = - RT ln K‡ Expressing K‡ in terms of DG‡ and RT gives the following equation for k: (1) Since DG‡ = DH‡ - TDS‡, we can also write: (2) Eq (1) and (2) are called the Eyring equations

The Eyring and Arrhenius equations resemble each other: ln A Arrhenius: slope = so: ln k 1/T Eyring: so Ea = DH‡ + RT In order to determine DH‡ and DS‡ it is easier to differentiate ln (k/T) to 1/T: slope = 1/T

So, the procedure to determine activation parameters is: - determine k at different temperatures - plotting ln(k/T) against 1/T gives DH‡ - then gives DS‡ and when you have DH‡ and DS‡, you also have DG‡ since DG = DH-TDS

Interpretation of activation parameters • DG‡, the Gibbs free energy of activation, determines at which rate a certain reaction will run at a given temperature • DH‡ is a measure for the amount of binding energy that is lost in the transition state relative to the ground state (including solvent effects) • DS‡ is a measure for the difference in (dis)order between the transition state and the ground state – for monomolecular reactions: DS‡ 0 J/mol. K – for a bimolecular reaction: DS‡ << 0 J/mol. K (two particles have to come together in the transition state to form one particle, demanding a much greater order)

Example: D G‡ = 62. 8 k. J/mol (very fast rx) DH‡ = 33. 0 k. J/mol (rel. low, compensation of C-H bond cleavage by hydration TS) DS‡ = -100 J/mol. K (bimolecular rx)

Another example: D H‡ = 85 k. J/mol (relatively high: no new bonds to be formed, no compensation for the partial cleavage of the C-C bond in the transition state; acetonitrile is aprotic, compensation of DH‡ by solvation will be less than in water DS‡ = 0 J/mol. K (monomolecular reaction)

Application of activation parameters for the elucidation of reaction mechanisms: A DS‡ of +12 J/mol. K was found monomolecular process A DS‡ of -117 J/mol. K was found bimolecular process; rate determining step in this case is the attack of water on the carbonyl group. Look in your course book for the exact reaction mechanisms!

Solvation (solvent effects) Influence of solvation on the reaction rate: k(H 2 O) = 10 -7 l. mol-1. s-1; k(DMF) = 10 -1 l. mol-1. s-1 so DDG‡ ~ 30 k. J/mol DG‡DMF DG‡H 2 O E DMF H 2 O DG‡DMF < DG‡H 2 O reaction progress

What is the background of this strong solvent effect? In H 2 O there is more solvation than in DMF, due to hydrogen bonds. Note the changes in entropy: loss of DS‡ because of orientation of the substrates, gain of DS‡ because of the liberation of water (less solvated transition state). The balance is not easy to predict! In general, in case of ions, the ground state is more solvated than the transition state: TS (‡) is hardly solvated due to the spreading of charge. Again a strong solvent effect here: k(H 2 O)= 7. 4 x 10 -6 s-1; k(DMF) = 37 s-1

Solvation effects in (bio)polymers Polymers or enzymes may have apolar pockets, which leads to: - less solvation and therefore higher reaction rates; - changes in p. Ka’s of acidic/basic groups: R = CH 3: p. Ka = 9. 7 R = polymer: p. Ka = 7. 7 [Py. N][H+] Ka = [Py. NH+] E. g. lysine, R-NH 2 + H+ R-NH 3+ p. Ka (H 2 O) = 10. 4, in some enzymes p. Ka = 7 !

1/v = 1/ Vmax + Km/Vmax (1/S) + (1 / Ki Vmax) S

S pequeñas 1/v = 1/ Vmax + Km/Vmax (1/S) + (1 / Ki Vmax) S S grandes 1/v = 1/ Vmax + Km/Vmax (1/S) + (1 / Ki Vmax) S

Enzyme Inhibition Many substances alter the activity of an enzyme by reversibly combining with it in a way what influence the binding of substrate and/or its turnover number. Substances that reduce an enzyme’s activity in this way are known as inhibitors Competitive Inhibition A substance that competes directly with a normal substrate for an enzyme’s substratebinding site is known as a competitive inhibitor. Here it is assumed that I, the inhibitor, bind reversibly to the enzyme and is in a rapid equilibrium with it so that And EI, the enzyme-inhibitor complex, is catalytically inactive. A competitive inhibitor therefore reduces the concentration of free enzyme available for substrate binding.

Enzyme Inhibition Competitive Inhibition This is the Michaelis-Menten equation that has been modified by a factor, , which is defined as Is a function of the inhibitor’s concentration and its affinity for the enzyme. It cannot be less than 1.

Enzyme Inhibition Competitive Inhibition Recasting in the double-reciprocal form yields A plot of this equation is linear and has a slope of KM/Vmax, a 1/[S] intercept of -1/ KM, and a 1/v 0 intercept of 1/ Vmax

Enzyme Inhibition Uncompetitive Inhibition In uncompetitive inhibition, the inhibitor binds directly to the enzyme-substrate complex but not to the free enzyme In this case, the inhibitor binding step has the dissociation constant The uncompetitive inhibitor, which need not resemble the substrate, presumably distorts the active site, thereby rendering the enzyme catalytically inactive.

Enzyme Inhibition Uncompetitive Inhibition The double-reciprocal plot consists of a family of parallel lines with slope KM/Vmax, 1/v 0 intercepts of ’/Vmax and 1/[S] intercept of - ’/KM

Enzyme Inhibition Mixed Inhibition (noncompetitive inhibition) A mixed inhibitor binds to enzyme sites that participate in both substrate binding and catalysis. The two dissociation constants for inhibitor binding Double-reciprocal plots consist of lines that have the slope KM/Vmax, with a 1/v 0 intercept of ’/Vmax and 1/[S] intercept of - ’/ KM

Steady State Kinetics Cannot Unambiguously Establish a Reaction Mechanism Measurements of a multistep reaction can be likened to a “black box” containing a system of water pipes with one inlet and one drain The steady state kinetics analysis of a reaction cannot unambiguously establish its mechanism

Bisubstrate Reactions Almost all of these so called bisubstrate reactions are either transferase reactions in which enzyme catalyzed the transfer of a specific functional group, X, from one of the substrates to the other: or oxidation-reduction reactions in which reducing equivalents are transferred between two substrates. Sequential Reactions in which all substrates must combine with the enzyme before a reaction can occur and products be released are known as Sequential reactions

Bisubstrate Reactions Sequential Reactions Ordered bisubstrate reaction A and B : substrates in order that they add to the enzyme P and Q : products in order that they leave the enzyme Random bisubstrate reaction Ping Pong Reactions Group-transfer reactions in which one or more products are released before all substrates have been added are known as Ping Pong reactions

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