Lecture 2: Michaelis-Menten Kinetics and Enzyme Assay Considerations PDF

Lecture #2: Michaelis-Menten Kinetics and considerations for setting up an enzyme assay 1. Review of the basics of reaction kinetics 2. Derivation of the Michaelis-Menten model 3. How to use the Michaelis-Menten model to set up an enzyme assay...

Lecture #2: Michaelis-Menten Kinetics and considerations for setting up an enzyme assay 1. Review of the basics of reaction kinetics 2. Derivation of the Michaelis-Menten model 3. How to use the Michaelis-Menten model to set up an enzyme assay 1 Review Review: From first- year chemistry we know that the rate of reaction at time t is proportional to the concentration of the reactants at time t. For the reaction shown below k1 A B k− 1 the rate equation can be readily writt en as: − d[A] d[B] V (t) = = = k1[A]t = k− 1[B]t dt dt this reaction is said to be first order because it depends on the c oncentration of one reactant (i.e. A ). Another example: 2  dt dt Review this reaction is said to be first order because it depends on the c oncentration of one reactant (i.e. A ). Another example: k1 A+ B C k− 1 − d[A] − d[B] d[C] V (t) = = = = k1[A]t [B]t = k− 1[C]t dt dt dt this reaction is said to be second order overall because it depends on the c oncentration of two reactant (i.e. A and B ). However, it is first order with respect to each of A and B. Another example: k1 2A B k− 1 − d[A] d[B] V(t) = = = k1[A]2t = k− 1[B]t dt dt this reaction is said to be second order overall and second order with respect to A. 3 Let's go back to our lovely little reaction: k1 A B k− 1 The rate that we measure at the onset of the reaction, i.e. at t = 0, is called the init ial velocit y and is denot ed as V0: V0 = k1[A]0 It is easy to show that the rate of this reaction is the fastest at the very begining. After all, [A]0 only drops over time. [A]0 = A3 Derivation of the Michaelis- Menten model: Reaction rate (=[B]) A. Introduction: A3 > A2 > A1 We mix our enzyme E with a substrate S, they proceed to bind [A]0 = Aother to each 2 in 1:1 ratio and form a complex E S. The enzyme E acts upon the substrates, generates product P and the enzyme is regenerated according to [A]0 = A1 k1 k2 E+ S ES E+ P (1) k−Time 1 k− 2 4 Pitfalls of measuring enzyme reaction rates at longer time points Specifically for an enzyme reaction, initial velocities are especially suitable. Rates at longer reactions time might be reduced due to: 1- denaturation of enzyme. 2- product inhibition. 3- substrate exhaustion (decrease of enzyme saturation). 4- inactivation of coenzyme. 5- increase of reverse reaction rate. 5 Always measure initial velocities! By choosing an arbitrary incubation time, the incubation time itself now becomes a variable Can be avoided by measuring initial velocities Core Topics in Biochem. Stenesh (1993) 6 Derivation of ofthe Derivation Michaelis-Menten the Michaelis- Menten model: (MM) model The purpose of this script is to derive Michaelis- Menten model. A. Introduction: We mix our enzyme E with a substrate S, they proceed to bind to each other in 1:1 ratio and form a complex E S. The enzyme E acts upon the substrates, generates product P and the enzyme is regenerated according to k1 k2 + E+ S ES E+ P (1) k− 1 k− 2 The only information we have about our is system is the total starting concentration of our enzyme and substrate denoted as [E]k01 and [S] k2 0, respectively. E+ S ES E+ P (1) k− 1 k− 2 Note that whenever we set up any enzyme assay, the only inf ormation we have about our system is the total starting concentration of our enzyme and substrate denoted as [E]0 and [S]0, respectively. Therefore it makes sense for us to focus on the init ial velocit y where we know exactly what the [E]0 and [S]0 values are. 7 B. Assumptions: B. Assumptions: 1. The E- S complex is in a steady- state: The rate at which the complex is being formed from E and S is equal to the rate at which it is consumed by being broken down to E + P or E + S d[E S] = 0 (2) dt 2. The formation of P ∝ [ES] : implies that the breakdown of ES to E + P is rate- limiting step RLS of the reaction, therfeore the reverse reaction k2 E+ P EP ES k− 2 can be ignored. Mathematically it means 8  d[E S] = 0 (2) dt 2. The formation of P ∝ [ES] : implies that the breakdown of ES to E + P is rate- limiting step RLS of the reaction, therfeore the reverse reaction k− 2 E+ P EP ES k2 can be ignored. Mathematically it means k− 2 = 0 (3) and V0 = k2[ES] (4) 3. [S]0 is much larger than [E]0: The change in [S] as a result of the f ormation of the ES complex is negligible. Therefore [S] can be taken to be a constant. 9 C. Derivation: can be taken to be a constant. k1 k2 E+ S ES E+ P (1) C. Derivation: k− 1 k− 2 Note that Rate of whenever formation of we ESset= up any enzyme k1[E][S] + k− 2assay, [E][P]the only inf ormation we have about our system is the total starting concentration of our enzyme and substrate denoted as [E]0 and Rate, of [S] consumption respectively. of ES =it makes Therefore k− 1[ES] + kfor sense 2[Eus S]to focus on the init ial velocit y where we 0 know exactly what the [E]0 and [S]0 values are. From assumption 1 we can write: B. Assumptions: d[ES] [dt ] = 0 1. The E- S complex is in a steady- state: Therefore: The rate at which the complex is being formed from E and S is equal to the rate at which it the is rate of formation consumed by beingofbroken ES = Rate downofto consumption E + P or Eof+ES S k1[E ][S] + k− 2[Ed[E ][P] S]= k− 1[E S] + k2[ES] (5) = 0 (2) dt from assumption 2 we know that k− 2 = 0 therefore 2. The formation of P ∝ [ES] := k− 1[ES] + k2[E S] k1[E][S] (6) implies The lawthat the breakdown of conservation of ES of mass to E + that dictates P is rate- limiting step RLS of the reaction, therfeore the reverse reaction [E]0 = [E] + [ES] (7) 10 k2  [E]0 = [E] + [ES] (7) which can be readily rearranged to yield: [E] = [E]0 − [ES] (8) Substitution of (8) into (6) yields: k1[S]([E ]0 − [E S]) = k− 1[ES] + k2[E S] (9) which can be readily rearranged to yield: k1[S]([E ]0 − [E S]) = (k− 1 + k2)[E S] (10) k− 1 + k2 [S]([E ]0 − [E S]) = [E S] (11) k1 let's define K M the Michaelis constant as: k− 1 + k2 KM = (12) k1 therefore: [S]([E]0 − [ES]) = K M [ES] (13) 11 solve for [E S]:  [S]([E]00 − [ES]) = K M [S]([E] [E S] M [ES] (13) (13) solve for [E S]:: solve for [E [S] − [E S][S] = K M [E]]00[S] [E S] M [ES] (14) (14) K MM [E K [E S] + [E S][S] = [E ]00[S] (15) (15) [E S](K M [ES](K M + [S]) = [E [E]]00[S] (16) (16) [E]00[S] [E S] = (17) KM + [S] M + From assumption 2 we can write: V0 = k2[E S] (18) therefore: k2[E]0[S] V0 = (19) K M + [S] Let's define Vmax = k2[E ]0 (20) 12 Substitution of (20) into (19) yields the Michaelis- Menten equat ion: Vmax [S] V0 = (21) KM = 20 µM K M + [S] Vmax = 55 µM s-1 Let's show what the Michaelis- Menten equation yields if we set: [S] = K M (22 Substitution of (22) into (21) yields: This is a rectangular hyperbolic shape Vmax K M V V0 = = max (21) KM + KM 2 considerations for setting up an enzyme assay 13  Vmax [S] V0 = (21) K M + [S] Let's show what the Michaelis- Menten equation yields if we set: [S] = K M (22 Substitution of (22) into (21) yields: Vmax K M V V0 = = max (21) KM + KM 2 considerations for setting up an enzyme assay KM is [S] at which one half of Vmax of a reaction is observed 14 The influence of KM on reaction rates KM = 5 µM KM = 20 µM KM = 80 µM Vmax = 55 µM s-1 15 The influence of Vmax on reaction rates KM = 20 µM Vmax = 55.0 µM s-1 Vmax = 27.5 µM s-1 Vmax = 13.8 µM s-1 16 ion of mass dictates that Important [E] = [E] + [ES]facts 0 (7) about KM rearranged to yield: 1) KM has units of concentration [E] = [E]0 − [ES] (8) 2) KM is a constant only under rigorously defined conditions of pH, T, ionic strength, etc to (6) yields: -if reaction has 2 or more substrates, the true KM for a given substrate is when all other [S] are at = k1[S]([E ]0 − [E S]) saturation k− 1[ES] + k2[E S] (9) -Vmax is that observed when all substrates are at saturation concentration rearranged to yield: 3)] Values for KM vary greatly from(10)enzyme to enzyme k1[S]([E 0 − [E S]) = (k − 1 + k2)[E S] -generally range 10-8 to 1.0 M -largekK −M1 :+ means k2 that one half of Vmax is achieved at a relatively large [S] [S]([E ]0 − [E S]) = KM: means -small [E S]that(11) 1/2 of Vmax occurs at relatively low [S] k1 4) KM constant he Michaelis is not an equilibrium as: constant, but rather a complex constant, composed of various rate constants k− 1 + k2 KM = (12) k1 5) KM is not a dissociation/association constant (KD) of the ES complex. More on this later! [S]([E]0 − [ES]) = K M [ES] (13) 17 considerations forupsetting considerations for setting an enzyme assay up an enzyme assay to K Knowing Knowing the K the M value permits M value permits adjustment [E ] adjustment of [S] [S] of assay assay conditions, [S] conditions, one one cc an an vary vary according according to the the needs needs of of assay, assay, can can then then measure or [S] 0 measure [E] 0 or 0 0 If: If: [S] [S] < > K M the Michaelis- Menten equation 19  Vmax V0 = [S] (22) considerations for setting up an enzyme Vmax KM assay Given that K is a constant we can write: M V0 = k′ [S] (23) In other words, the reaction is first order with respect to [S]. If: [S] > > K M the Michaelis- Menten equation Vmax [S] V0 = (21) K M + [S] simplifies to V0 = Vmax = k2[E ]0 (24) In other words, the reaction is zero order with respect to [S] but is sensitive to [E ]0. 20  Given that K is a constant we can write: M V0 = k′ [S] (23) Quantitative enzyme assays In other words, the reaction is first order with respect to [S]. Measure the amount of EIf:in the sample E is present at much lower concentration than S [S] > > K M E is present in limiting amounts and S is present in excess the Michaelis- Menten equation assay conditions correspond to levelled off portions of vo versus [S] curves thus Vo is proportional to [E] V = Vmax [S] (21) 0 K M + [S] Zero order reaction simplifies to V0 = Vmax = k2[E ]0[S]0(24) In other words, the reaction is zero order with respect to [S] but is sensitive to [E ] 21