Enzyme Kinetics: Reaction Rates & Graphical Analysis in Chemical Reactions, Exams of Biochemistry

Download Enzyme Kinetics: Reaction Rates & Graphical Analysis in Chemical Reactions and more Exams Biochemistry in PDF only on Docsity! 1 Enzyme Kinetics • Chemical Kinetics – Rates, – Reaction Order – Rate equations • Enzyme Kinetics – Simplifying assumptions • Pre-equilibrium • Steady state – Michaelis-Menten Kinetics – Graphical Analysis – Bisubstrate reactions • Inhibition – Mechanism – Graphical analysis • Regulation – Allosteric enzymes – ATCase Chemical Kinetics • The conversion of a substrate to a product can occur directly or through a series of steps. • The rate of conversion through each elementary reaction step depends on a rate constant and the concentration(s) of the reacting species First Order Rate Equations • For a 1st order reaction S –>P • The rate of change in S and P depends only on the concentration of S v = d[P] dt = − d[S] dt = k[S] − d[S] dt = k[S] −d[S] = k[S]dt 1 [S] d[S] = −kdt 2 Integrated rate equations • Regrouping variables gives one concentration integral and one time integral • The integral of 1/S is ln(S) • Limits of integration from time=0 (when [S] = [S]0 to time t 1 [S] d[S] [S]o [S] ∫ = −kdt 0 t ∫ 1 [S] d[S] [S]o [S] ∫ = −k dt 0 t ∫ ln[S] − ln[S]o = −k(t − 0) First order rate equations lead to exponential functions • Both integrals give differences between final and initial states • ln[S] vs t gives a line with slope -k ln[S] − ln[S]o = −k(t − 0) ln[S] = ln[S]o − kt exp(ln[S]) = exp(ln[S]o − kt) [S] = [S]o exp(−kt) First order reactions have defined half lives • No matter what the starting concentration of S, the time required to reduce it by half is constant ln[S] = ln[S]o − kt ln [S] [S]o = −kt when [S] [S]o = 1 / 2 − ln(2) = −kt t = ln(2) k = 0.693 k ≡ t1/ 2 5 Michaelis-Menten Kinetics • Usually v is measured as initial velocity – Before substrate is significantly depleted – At multiple substrate concentrations • When [S]=KM v=Vmax/2 • KM is the concentration of substrate necessary for half saturation [S][E]T KM + [S] = [ES] d[P] dt = k2[ES] in the steady state v = d[P] dt = k2 [S][E]T KM + [S] At maximal velocity S >> KM Vmax = k2[E]T v = Vmax[S] KM + [S] Michaelis-Menten Kinetics • kcat is the catalytic constant • The turnover number – number of products produced per enzyme at saturating substrate concentrations • kcat/KM is the catalytic efficiency, the limiting second order rate constant at low [S] v = Vmax[S] KM + [S] kcat = Vmax / [E]T when [S] << KM and [E] ≈ [E]T V = kcat KM [E][S] Linear double reciprocal plots provide estimates of Vmax and KM • Inverting the Michaelis-menten equation provides a linear relationship between 1/v and 1/[S] – Slope depends on Km and Vmax – Intercept depends only on Vmax v = Vmax[S] KM + [S] 1 / v = KM + [S] Vmax[S] 1 / v = KM Vmax[S] + [S] Vmax[S] 1 / v = KM Vmax ( 1 [S] ) + 1 Vmax 6 Graphical Analysis • Initial velocities are measured as a function of substrate concentration • Vo vs S is a hyperbolic curve • 1/Vo vs 1/S Slope = Km/Vmax Intcpt = 1/Vmax Initial Velocities 0 0.2 0.4 0.6 0.8 1 1.2 0 50 100 150 time (Seconds) 1 2 5 10 20 Vo vs [S] 0 2 4 6 8 10 0 5 10 15 20 25 [S] (mM) Double Reciprocal Plot 0 0.1 0.2 0.3 0.4 0.5 0 0.5 1 1.5 1/[S] Bisubstrate reactions • Most enzymatic reactions have multiple substrates and products • In sequential (single displacement) reactions all products are bound before substrates are released – Ordered mechanisms - A binds first then B; P released first then Q (by convention) – Random mechanisms - either substrate binds first and product release can occur in either order • In Ping Pong (double displacement) reactions substrate binding alternates with product release Cleland diagrams E EA EAB EPQ EQ E A B P Q Sequential Ordered Sequential Random E EAB EPQ E EA EQ EPEB A B B A P Q Q P A P B Q Ping Pong E EA E*P E* E*B EQ E