F BICH 410 1st Edition Lecture 19 Outline of Last Lecture Enzymes Outline of Current Lecture Kinetics the study of rates of reactions o Rate is proportional to S o K is proportionality constant and has unit time 1 o Uncatalyzed reactions v k S k slope first order reaction o WITH ENZYME no longer first order rather dependent on enzyme E S ES E P o Saturation effect no effect is seen in increase of S once saturation reached change only seen when more enzyme added When saturated at E in ES complex Michaelis Menten Derivation o V k2 ES o Conservation of enzyme oscillates between free form and enzyme substrate form o Relative concentrations of E and S enzyme is used at catalytic amount little bit of enzyme lots of substrate o Initial velocity assumption enzyme accelerates forward and backwards reactionmeasurement of rate of product formation break down for only first 10 of reaction Vo o Steady state assumption rate at which we form ES is rate at which ES is broken down E S ES E P rate of formation of ES rate of breakdown o Catalytic efficiency Velocity o VELOCITYmax k2 ES o Velocity K2ET x S Km S km rate constants with units of Molarity o S is large compared to P so doesn t go backwards o Under max velocity conditions rate of reaction k2 times total amount of enzyme equals the velocity not depended on S but enzyme o V0 Vmax S km S percentage of max velocity o Substrate concentration vs Vel hyperbolic curve o Km and Kd are equivalent therefore Km S and V o VmaxS 2S Vmax 2 When S Km we are at Vmax If S is very small less than km then Vo Vmax S Km 1 st order rxn If S is very large compared to km then Vo Vmax S S Vmax Vmax k2 E T only way to increase Vmax is to increase Enzyme When S 100km we are saturated 1 S vs 1 Vel slope km vmax the point where yintercepted 1 vmax These notes represent a detailed interpretation of the professor s lecture GradeBuddy is best used as a supplement to your own notes not as a substitute o The slope is Km Vmax known as Lineweaver burk double reciprocal plot Linear plot with y mx b X intercept is 1 km The smaller the Km the tighter the substrate binds Max velocity Kcat Et kcat turnover number in reciprocal time number of substrate molecules converted to product per enzyme active site per unit time When S is low rate is 1st order when S is high rate is 0 order To compare Vmax of different proteins must know Et of each Vmax allows measurement of enzyme conc Rate constant km under saturation known as catalytic constant Specific Activity number of substrates converted to total protein present Increasing enzyme will not increase SA under saturation because velocity is proportional to amount of E present
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