FBICH 410 1st Edition Lecture 19Outline of Last Lecture - EnzymesOutline of Current Lecture - Kinetics- the study of rates of reactionso Rate is proportional to [S] o K is proportionality constant and has unit time^-1o Uncatalyzed reactions v=k[S] k=slope first order reactiono WITH ENZYME no longer first order rather dependent on enzyme E+S<->ES<->E+Po 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 Derivationo 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 substrateo Initial velocity assumption- enzyme accelerates forward and backwards reaction- measurement of rate of product formation/break down for only first 10% of reaction=Voo 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 breakdowno Catalytic efficiency- Velocityo VELOCITYmax=k2[ES]o Velocity= (K2ET x [S])/(Km +[S]) km=rate constants with units of Molarityo S is large compared to P so doesn’t go backwardso Under max velocity conditions, rate of reaction-k2 times total amount of enzyme equals the velocity--- not depended on S but enzymeo V0/Vmax= [S]/(km+[S]) --- percentage of max velocityo Substrate concentration vs Vel hyperbolic curveo Km and Kd are equivalent therefore Km=S and Vo=VmaxS/2S= Vmax/2 When S=Km we are at ½ Vmax If [S] is very small (less than km) then Vo=Vmax[S]/Km- 1st 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/vmaxThese 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.- 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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