Biochemistry I Lecture 17 Oct 8, 2007 1 Lecture 17: Enzyme Kinetics Assigned reading in Lehninger: 6.3, pg 202-4 Key Terms: • Steady State Kinetics • Michaelis-Menton Equation: Vo = Vmax[S]/(Km+[S]) • Vmax = kcat[Etotal] • Km = (k-1+kcat)/k1 1. Utility of enzyme kinetic analyses • Elucidating enzyme mechanism and comparisons between enzymes • Predicting enzyme activity under various conditions • Development of enzyme inhibitors (or activators) as therapeutic agents 2. Definitions of terms k1 ! E + S " # $ ESkcat$ # $ E + P k-1 • k1 is the forward rate constant for substrate binding. • k-1 is the reverse rate constant for substrate binding. • ES is the enzyme-substrate complex. • kcat is the catalytic rate constant (containing terms related to stabilization of the transition state). • Km (k-1+k2/k1) is approximates the affinity of enzyme for substrate when kcat << k-1 3. Derivation of the Michaelis-Menton equation based on the Steady State Assumption: ! Vo=VMAX[S]Km+ [S] This is the Michaelis-Menton equation. 4. Important constants that describe the enzymatic reaction: • Km or Michaelis constant: ! Km=k"1+ kcatk1 This is almost the same as the KD (=k-1/k1) dissociation constant, except for the presence of the kcat term. Therefore it is related to the affinity or strength of binding of a substrate to the enzyme. Km is equal to the substrate concentration that gives 1/2 of the maximal velocity, in a similar manner that KD is given by the [L] that gives Y=0.5.Biochemistry I Lecture 17 Oct 8, 2007 2 • Vmax = kcat[ET] This is the highest rate of product production possible. It is obtained at high substrate levels ([S]>>Km). Under these conditions all of the enzyme is in the [ES] form (i.e. [ES]=[ET]). • Turnover number or catalytic efficiency: kcat = Vmax/[ET] This is the number of reactions performed by a single enzyme molecule in a certain period of time (e.g. moles of product formed in a defined period by a mole of enzyme) when the enzyme is fully saturated with substrate. 5. Throughput, or efficiency, of enzyme systems ! Vo=VMAX[S]Km+ [S] • Low Substrate: At low substrate concentrations ([S]<<Km) the efficiency of an enzyme will depend both on how efficiently it can bind substrate as well as how well it can perform the chemical step (kcat). In other words, the intrinsic efficiency of an enzyme at low substrate levels is given by kcat/Km. The overall rate depends on the total amount of enzyme, [ET], and [S]: ! Vo=kcatkm[ET][S] • High Substrate: At high substrate concentrations ([S]>>Km) the rate becomes independent of [S] since all of the enzyme molecules are saturated with [S]. The overall rate depends only on kcat and the total amount of enzyme, [ET]. ! Vo= VMAX= kcat[ET]. Therefore the intrinsic efficiency of an enzyme at high substrate concentrations is given by
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