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Enzymes II

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Enzymes IIWhat we’ll discussKinetics, continuedMore complex casesBimolecular reactionForward and backwardMulti-step reactionsMichaelis-Menten kineticsMichaelis-Menten ratesEquating the ratesDerivation, continuedMaximum velocityUsing Vmax in M-M kineticsGraphical interpretationPhysical meaning of KmkcatPhysical meaning of kcatSpecificity constant, kcat/KmDimensionsTypical units for kinetic parametersKinetic MechanismsHistorical thoughtSequential, ordered reactionsSequential, random reactionsPing-pong mechanismInduced fitKinase reactionsiClicker quiz, question 1iClicker quiz question 2Hexokinase conformational changesMeasurements and calculationsLineweaver-BurkHow to use thisDemonstration that the X-intercept is at -1/KmGraphical form of L-BAre those values to the left of 1/[S] = 0 physical?Advantages and disadvantages of L-B plotsDon’t fall into the trap!Sanity checksiClicker quiz: question 3iClicker quiz #4iClicker question 5Another physical significance of KmExample: hexokinase isozymesUsing kinetics to determine mechanismsL-B plots for ordered sequential reactionsL-B plots for ping-pong reactionsUsing exchange reactions to discern mechanismsSucrose & maltose phosphorylaseWhy study inhibition?The concept of inhibitionInhibitors and acceleratorsWhy more inhibitors than accelerators?Distinctions we can makeTypes of inhibitors10/21/2010Biochem: Enzymes IIEnzymes IIAndy HowardIntroductory Biochemistry21 October 201010/21/2010Biochem: Enzymes IIP. 2 of 55What we’ll discussEnzymesEnzyme kineticsMichaelis-Menten kinetics: overviewKinetic ConstantsKinetic MechanismsInduced FitEnzymes (concluded)Bisubstrate reactionsMeasurements and calculational toolsInhibitionWhy study it?The conceptTypes of inhibitors10/21/2010Biochem: Enzymes IIP. 3 of 55Kinetics, continuedIn most situations more product will be produced per unit time if A0 is large than if it is small, and in fact the rate will be linear with the concentration at any given time:d[B]/dt = v = k[A]where v is the velocity of the reaction and k is a constant known as the forward rate constant.Here, since [A] has dimensions of concentration and d[B]/dt has dimensions of concentration / time, the dimensions of k will be those of inverse time, e.g. sec-1.[B]t10/21/2010Biochem: Enzymes IIP. 4 of 55More complex casesMore complicated than this if >1 reactant involved or if a catalyst whose concentration influences the production of species B is present.If >1 reactant required for making B, then usually the reaction will be linear in the concentration of the scarcest reactant and nearly independent of the concentration of the more plentiful reactants.In fact, many enzymes operate by converting a second-order reaction into a pair of first-order reactions!10/21/2010Biochem: Enzymes IIP. 5 of 55Bimolecular reactionIf in the reactionA + D  Bthe initial concentrations of [A] and [D] are comparable, then the reaction rate will be linear in both [A] and [D]:d[B]/dt = v = k[A][D] = k[A]1[D]1i.e. the reaction is first-order in both A and D, and it’s second-order overall10/21/2010Biochem: Enzymes IIP. 6 of 55Forward and backwardRate of reverse reaction may not be the same as the rate at which the forward reaction occurs.If the forward reaction rate of reaction 1 is designated as k1,the backward rate typically designated as k-1.10/21/2010Biochem: Enzymes IIP. 7 of 55Multi-step reactionsIn complex reactions, we may need to keep track of rates in the forward and reverse directions of multiple reactions.Thus in the conversion A  B  Cwe can write rate constantsk1, k-1, k2, and k-2as the rate constants associated with converting A to B, converting B to A, converting B to C, and converting C to B.10/21/2010Biochem: Enzymes IIP. 8 of 55Michaelis-Menten kineticsA very common situation is one in which for some portion of the time in which a reaction is being monitored, the concentration of the enzyme-substrate complex is nearly constant. Thus in the general reactionE + S  ES  E + Pwhere E is the enzyme, S is the substrate, ES is the enzyme-substrate complex (or "enzyme-intermediate complex"), and P is the productWe find that [ES] is nearly constant for a considerable stretch of time. [ES]t10/21/2010Biochem: Enzymes IIP. 9 of 55Michaelis-Menten ratesRate at which new ES molecules are being produced in the first forward reaction is equal to the rate at which ES molecules are being converted to (E and P) and (E and S).Formation of ES is first-order in both S and available [E]Therefore: rate of formation of ES from left =vf = k1([E]tot - [ES])[S]because the enzyme that is already substrate-bound is unavailable!10/21/2010Biochem: Enzymes IIP. 10 of 55Equating the ratesWe started with the statement that the rate of formation of ES and the rate of destruction of it are equalRate of disappearance of ES on right and left isvd = k-1[ES] + k2[ES] = (k-1+ k2)[ES]This rate of disappearance should be equal to the rate of appearanceUnder these conditions vf = vd.10/21/2010Biochem: Enzymes IIP. 11 of 55Derivation, continuedThus since vf = vd by assumption, k1([E]tot - [ES])[S] = (k-1+ k2)[ES] Km  (k-1+ k2)/k1 = ([E]tot - [ES])[S] / [ES][ES] = [E]tot [S] / (Km + [S])But the rate-limiting reaction is the formation of product: v0 = k2[ES]Thus v0 = k2[E]tot [S] / (Km + [S])10/21/2010Biochem: Enzymes IIP. 12 of 55Maximum velocityWhat conditions would produce the maximum velocity?Answer: very high substrate concentration ([S] >> [E]tot),for which all the enzyme would be bound up with substrate. Thus under those conditions we getVmax = v0 = k2[ES] = k2[E]tot10/21/2010Biochem: Enzymes IIP. 13 of 55Using Vmax inM-M kineticsThus sinceVmax = k2[E]tot,v0 = Vmax [S] / (Km+[S])That’s the famous Michaelis-Mentenequation10/21/2010Biochem: Enzymes IIP. 14 of 55Graphical interpretationMichaelis-Menten kinetics00.0010.0020.0030.0040.0050.0060.0070.0080.0090.010 0.1 0.2 0.3 0.4 0.5 0.6 0.7Substrate conc, MInitial velocity v0, Ms-1Vmax = 0.01 Ms-1Km = 0.03M[E]tot = 10-7Mkcat = 105 s-110/21/2010Biochem: Enzymes IIP. 15 of 55Physical meaning of KmAs we can see from the plot, the velocity is half-maximal when [S] = KmTrivially derivable: if [S] = Km, thenv0 = Vmax[S] / ([S]+[S]) = Vmax /2We can turn that around and say that the Km is defined as the concentration resulting in half-maximal velocityKm is a property


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