Lecture 27 Transition States and Enzyme Catalysis Reading for Today: Chapter 15 sections B and C Chapter 16 – next two lectures MCB65 4/8/16 1 Pop Question 9 Binding data for your thesis protein (YTP), binding to its ligand, Hoopes. You are hoping that these data provide some clue as to its possible oligomeric state, especially because you have already determined that there is one Hoopes molecule bound per molecule of YTP. What quantitative value(s) can you extract about Hoopes binding to YTP? 0 < nH <1 (nH ~ 0.2) – thus negative cooperativity Slope = 1 -8.5 KD for the high-affinity state ~10 -5.5 Slope = 1 KD for the low-affinity state ~10 Slope = n <1 H What can you determine about the YTP oligomeric state? Because there is cooperativity, YTP is an oligomer – there has to be more than one binding site per -8.5 -5.5 KD(high) ~10 KD(low) ~10 functional unit to observe allostery. However, the Hill coefficient value does not provide direct information about the number of protomers in an oligomer. Note: If it is not a simple binding equilibrium (i.e. if the trend is not linear with a slope of 1), MCB65 then the measure of [L] at f=1/2 is not a true KD. 4/8/16 2 Today’s Goals • We’re continuing to explore reaction kinetics: • Reactions at steady state – where the concentrations of reactants and products are constant • Equilibrium reactions – additional constraints on the rate constants • Kinetics of ligand interactions • Transition state theory – what determines the rate constant • Role of enzymes as catalysts • Introducing Michaelis-Menten kinetics of enzymes MCB65 4/8/16 3 Reactions at steady state • Most biological reactions are not at equilibrium, but are at steady state MCB65 4/8/16 4 Reactions at steady state • Most biological reactions are not at equilibrium, but are at steady state • Formation and utilization of metabolites are equal • Concentrations of the relevant reactants and products do not change with time (or only slowly) • Case of: A k1 B k2 C • Where [A] = [A]0 at all times d[B] k [A] k [B] dt 1 0 2 MCB65 4/8/16 5 Reaction at steady state • Can be integrated to: k1 < k2 k1 k2t [B] [A]0 (1 e ) k2 A k1 B k2 C • As t becomes large: k1 [B]ss [A]0 k2 k1 > k2 d[B] • k1[A]0 k2[B] [B]ss is set by the dtratio of rate constants and the concentration of reactant MCB65 4/8/16 6 Conditions of a true steady state • Concentrations are constant • There must be a flow through the system • Formation and utilization of intermediates is equal • Requirement for reaching steady state: the rate at each step beyond the 1st step must depend on [reactant] • Necessary so that if the [reactant] increases, the rate increases accordingly • Total flux is determined by the slowest step • Concentrations of intermediates before the slowest step are higher than those after the slowest step MCB65 4/8/16 7 If the change in free energy is small, the reaction is reversible • When the DG is not much larger than RT, then the reverse reaction needs to be considered kf A B kr d[A] • Applicable rates: k [A] k [B] dt f r d[B] k [A] k [B] dt f r • To integrate the equations, we also know that: • The sum of [A] and [B] is constant • At equilibrium, kf[A]eq = kr[B]eq MCB65 4/8/16 8 Reversible reactions (k f kr )t [A] ([A]0 [A]eq )e [A]eq • When t is large, [A]~[A]eq • Similar equation can be developed for [B]: (k f kr )t [B] [B]eq ([B]0 [B]eq )e • Equations correspond to an exponential approach to final concentrations • The effective rate constant is the sum of the forward and reverse rate constants MCB65 Figure from The Molecules of Life (© Garland Science 2008) 4/8/16 9 Rate constants and equilibrium constants kf A B kr d[A] d[B] • At equilibrium, 0 and 0 dt dt d[A] k [A] k [B] 0 • Therefore dt f eq r eq • Forward and backward rates are equal: k f [A]eq kr [B]eq k f [B]eq • And Keq kr [A]eq • Although thermodynamics cannot be used to predict the kinetics, it does constrain the ratio of the rate constants MCB65 4/8/16 10 Binding kinetics of imatinib (Gleevec) • Binding of kinase inhibitor to kinase leads to a decrease in fluorescence of Trp residues in the kinase kon(i.e. kf) P + L PL koff (i.e. kr) MCB65 Figure from The Molecules of Life (© Garland Science 2013) 4/8/16 11 Kinetics of a ligand binding equilibrium k on Under conditions where P + L PL [L]>>[P] kon ' kon[L] koff d[P] kon[P][L] koff [PL] kon '[P] koff [PL] • Rates: dt d[PL] k [P][L] k [PL] k '[P] k [PL] dt on off on off • Integrates to: (kon 'koff )t [P] ([P]0 [P]eq )e [P]eq kobs kon 'koff MCB65 4/8/16 12 Binding kinetics of imatinib (Gleevec) • From kobs kon 'koff and kon ' kon[L] • Linear dependence on [L]: kobs kon[L] koff • Binding of kinase inhibitor to kinase leads to a decrease in fluorescence of Trp residues in the kinase MCB65 Figure from The Molecules of Life (© Garland Science 2008) 4/8/16 13 Ligand binding kinetics kobs kon[L] koff • Plot kobs vs [L] • Slope is kon • Note: [L] >> [P] for all experiments • Extrapolate to [L] = 0 to get koff at the y-intercept MCB65 Figure from The Molecules of Life (© Garland Science 2008) 4/8/16 14 Ligand binding kinetics • Measured rate constants for Abl kinase binding to its inhibitor imatinib (Gleevec), vs. Src, a non-specific target 6 -1 -1 kon (Abl) 0.14610 M s 6 -1 -1 kon (Src) 0.00310 M s -1 koff (Abl) 0.00219 s -1 koff (Src) 0.126 s kon P + L PL koff [Imatinib] k -1 koff 0.00219 s 6 KD (Abl) 0.01510 M D k 6 -1 -1 kon 0.14610 M s -1 Ratio of 2800 koff K (Src) off 0.126 s 6 KD (Src) 6 -1 -1 42 10 M kon kon 0.00310 M s MCB65 Figure from The Molecules of Life (© Garland Science 2008) 4/8/16 15 What determines the rate constant? d[A] A B C rate k[A][B] dt • What determines the value of k? • Rate of collisions 10 -1 -1 • In gases, ~6x10 M s = kcollision • In liquids, it depends on diffusion: kcollision 8RT / 3h • h is the viscosity of the solution 10 -1 -1 • kcollision ~ 10 M s for a typical aqueous solution MCB65 4/8/16 16 Diffusion-limited reactions vs. orientation- dependent reactions 2I I2 • The delocalized electrons in the outermost orbitals come together to form the bond, and this can happen during ~any collision • In this case, the collision has to happen in a specific orientation – only a fraction of the collisions are effective • Even 106 M-1 s-1 are considered fast rate constants • ~1 in 104 collisions are productive • Define factor A = kcollision x fp, where fp is the fraction of collisions at the correct angle MCB65 Figure from The Molecules of Life (© Garland Science 2008) 4/8/16 17 Activation energy • A second factor is the energy required to reach the transition state – activation energy, Ea • Here the methyl hydrogens have to be pushed away from their low energy tetrahedral geometry to a planar arrangement MCB65 Figure from The Molecules of Life (© Garland Science 2008) 4/8/16 18 Source of the activation energy • The kinetic energy of molecules is converted to potential energy as the reactants approach in the appropriate orientation • The fraction of molecules that have sufficient energy is given by e-Ea/RT, related to the Boltzmann distribution • The rate constant is a product of the collision factor, A, and the energetic factor, e-Ea/RT: k Ae Ea / RT • A is called the “pre-exponential factor” = kcollision x f MCB65 4/8/16 19 Proline peptide bond isomerization • The activation energy of a trans-to-cis isomerization of proline is 54 kJ/mol MCB65 Figure from The Molecules of Life (© Garland Science 2008) 4/8/16 20 Arrhenius plot provides the activation energy value Ea 1 k Ae Ea / RT ln k ln A R T • The slope of the Arrhenius plot gives Ea/R • 54 kJ mol-1 for proline isomerization MCB65 Figure from The Molecules of Life (© Garland Science 2008) 4/8/16 21 Deviations from linearity E 1 ln k a ln A • What kind of reactions might show a plotR likeT this one? MCB65 Figure from The Molecules of Life (© Garland Science 2008) 4/8/16 22 Transition state theory • The process of getting to the transition state is an equilibrium: ‡ Keq k‡ A + B A·B‡ C + D [AB‡ ] d[C] K k ‡[AB‡ ] k ‡ K ‡ [A][B] eq [A][B] dt eq DG‡ DH ‡ DS ‡ ‡ RT RT R Keq e e e k T Reflects the relevant vibration frequency k ‡ ~ B h leading the reaction forward Where h is Planck’s constant MCB65 4/8/16 23 Another equation for the rate constant d[C] k‡K‡ [A][B] dt eq DS‡ DH ‡ d[C] k T B e R e RT [A][B] dt h Analogous to the angle Analogous to the activation restriction for the collisions energy term • Provides clear connections to thermodynamic concepts • Useful to predict effects on reaction rates MCB65 4/8/16 24 Catalysts alter the rate constant rate k (concentration factors) A kcollision f p • A catalyst can modify both the pre-exponential and the exponential factors • Favorable interaction with the transition state • Increase the rate of collisions k Ae Ea / RT • Favor the productive orientation MCB65 Figure from The Molecules of Life (© Garland Science 2008) 4/8/16 25 Electrostatic potential to accelerate reactant encounters • The substrate of acetylcholine esterase, acetylcholine (a neurotransmitter), is positively charged • The negatively charged active site provides an electrostatic force “pulling” the substrate into the site MCB65 Figure from The Molecules of Life (© Garland Science 2008) 4/8/16 26 Catalysts alter mechanism but not thermodynamics of the reaction equilibrium • Single step: Two steps: A B C A cat E E B C cat [C] [E] [cat][C] K K K eq [A][B] 1 [A][cat] 2 [E][B] DGeq DG1 DG2 cat A BK1 E B K 2 Keq K1K2 C cat MCB65 4/8/16 27 Catalysts alter mechanism but not thermodynamics of the reaction equilibrium • Single step: Two steps: A B C A cat E E B C cat [C] [E] [cat][C] K K K eq [A][B] 1 [A][cat] 2 [E][B] [E] [C][cat] Catalyst is not consumed K K K during the reaction, but affects eq 1 2 the mechanism and therefore [A][cat] [E][B] the kinetics.