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# Class Note for BIOC 463A at UA

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This 13 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Arizona taught by a professor in Fall. Since its upload, it has received 16 views.

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Date Created: 02/06/15

BIOC 463A Enzyme SteadyState Kinetics and Competitive Inhibition SteadyState Kinetics Determination of Km Vmax and kcat Objectives We will emphasis the methods and techniques employed in steadystate kinetic analysis of AP We will not cover Transition State Theory in this class Rather we will try to fully understand how data is obtained and analyzed in the context of the MichaelisMenten equation Historical Perspective In a typical chemical reaction k A B gt products 600 500 m kcaItaIyszBl 400 V9ocity 300 o 200 o 39 m kuncatB WM 100 I WWWM 0 777quotquot m W 0 quotW K W 0 WIT I I I I I I o 10 20 30 40 50 60 70 A1 Velocity of reaction vs A B being held constant is linear If a catalyst is added the velocity increases but plot remains linear So rate of reaction v kobsA where k0s kB And kuncatalyzed ltlt kcatalyzed BIOC 463A Enzyme SteadyState Kinetics and Competitive Inhibition At beginning of 1900 s it was understood that enzymes were catalysts involved in biological reactions For enzymes a different plot of v vs S was observed Initial velocity V0 uMmin Substrate concentration S mM Michaelis and Menten suggested that the approach of v to a maximal level Vmax meant E and S were forming a complex of distinct lifetime Assumption 1 E and S react rapidly and are in equilibrium k1 E S ltgt ES K1 and Ks k1lk1 ESIES dissociation constant for ES BIOC 463A Enzyme SteadyState Kinetics and Competitive Inhibition Assumption 2 Product is formed in a 2nd DISTINCT first order step k2 ES 9 E P rate k2ES Assumption 3 free E created by this step can bind new S Taken together the MichaelisMenten mechanism k1 k2 E S ltgt ES 9 EP k1 Briggs and Haldane then derived the MM eqn we know today making the following assumptions 0 ES rapidly reaches a constant level that does not change during steadystate dES 0 dt 0 This means rate of ES formation rate of ES breakdown k1ES k1ES k2ES dissociation to product k1ES k1 k2ES o BampH defined Km BIOC 463A Enzyme SteadyState Kinetics and Competitive Inhibition Km k1 k2 ES look familiar k1 ES 0 Initial Velocity Assumption at t 0 P 0 this does two things Since many P are very effective competitive inhibitors eliminates math complications due to the back reaction E P 9 ES Taking ALL of these assumptions into consideration BampH derived the beloved MM eqn v0 Vma S where Vmax kcatEJ Km S There are three S conditions to consider S Vo ltlt Km LIME Kcatl 1 1 Km Km Km Mm 2 gtgt Km Vmax kcatEt BIOC 463A Enzyme SteadyState Kinetics and Competitive Inhibition V0 uMmin 5 mM Two Important Definitions of Km 1 Km k1 k2lk1 2 Km is the S that gives v0 VmaXIZ The units of Km are CONCENTRATION Km is a CONCENTRATIONlllllll Sound familiar Km is primarily a measure of S binding with a little catalysis thrown in if k2 is large relative to k1 VmX is a measure of catalysis only kcat and E Usually when people compare Km values it is implied they are comparing the binding of different substrates BIOC 463A Enzyme SteadyState Kinetics and Competitive Inhibition Km and Vmax are defined for an E and S under set conditions 0 pH 0 Ionic Strength 0 Temp pressure etc Significance of kcatKm 0 Under physiological conditions in vivo 001 lt SKm lt 1 Le S ltlt Km so Eltotal Elfree and v0 VmazlSl Km KLatIEM 1 Km 0 Thus kcatKm M391s391 is a secondorder rate constant for the reaction E S 9 E P Maximum value 109 M391s391 signifying diffusion control limits for molecules the size of proteins and substrates For many metabolic pathways the rate of flux is controlled by rate of the above reaction thus reflects kcatKm catalytic efficiency E Via gene expression and S BIOC 463A Enzyme SteadyState Kinetics and Competitive Inhibition How are SteadyState experiments performed Typically you have S and buffer in a cuvette you add some E and follow an absorbance change at some wavelength for either decrease in S or increase in P Find the slope of the line tangential to data as close to t 0 as possible This gives a v0 value for that S Then you repeat using a different S Initial velocity V0 wvrmin Substrate concentration S mM Data analysis problem here is same as seen in ligand binding 1 Extrapolate to infinite S 2 Determine Vmax 3 Determine Km BIOC 463A Enzyme SteadyState Kinetics and Competitive Inhibition Linear Transform LineweaverBurk Plot Starting with V0 Vmax S Km S Take inverse of both sides 1 5m L Ii Vo Vmax Vmax BIOC 463A Enzyme SteadyState Kinetics and Competitive Inhibition Some practical problems with doing steadystate studies 0 getting accurate vo values at low S if the reaction is fast sometimes a lot of data is lost in the dead mixing time In that case you miss the true v0 and the one you observe is lt the true value This influences the MichaelisMenten plot giving higher Km values Absorbance 10 sec deadtime v 10 sec 08 05 0 sec deadtime 0 4 vow sec 02 I I I I I I D 10 20 30 40 50 so Time sec 0 P inhibition again if the reaction is fast P will increase rapidly If P is a strong inhibitor of E KiltKm then the observed vo will be smaller How do you avoid these problems Simple try to mix things together as fast as possible It is important to monitor how much of the AAbsorbance you are missing in the dead time Ideally it should be lt 10 total AAtotap BIOC 463A Enzyme SteadyState Kinetics and Competitive Inhibition Competitive Inhibitors Determination of Ki Inhibitor binds to active site of E ES ESgtEP l Ki i i k3 El k3 k3 1 El 0 Effect can be overcome at high S by mass action so VmX is the same 0 Km however is APPARENTLY changed Kmapp gt Km Kmapp Km 1 JII K Note Km is not actually changed Km is a property of the binding of S to E and that is not altered by I III MM eqn for Competitive Inhibitor Vo Vmax S Km1 llKi S 10 BIOC 463A Enzyme SteadyState Kinetics and Competitive Inhibition LineweaverBurk form of eqn 1 KM1HWMH Lu Vo Vmax VmaX As I increases or Ki decreases how will slope change ik i4 V0 Vmax s Vmax A 5 Pg 12 1 TIE a1 39 No inhibitor m Slope max Vmax L 1 s m It is important to note that I Ki Vmax and Km are all CONSTANTS 1IS is the VARIABLE This is not only true for the plot it is also true for the way the experiment is done I is held constant at ALL S throughout experiment This is a LineweaverBurk METHOD of studying Inhibitors 11 BIOC 463A Enzyme SteadyState Kinetics and Competitive Inhibition A second method and way of plotting data is called the Dixon MethodPlot S is now held constant and I is varied this requires a rearrangement of the LineweaverBurk eqn 1 My 391 1KmS1 V VmaxKiIS Vmax 010 065 5110 uM 005 1V 004 1v11lvSZ 002 0 DD 10 4 D 10 20 30 1 uM In a Dixon expt you do a set of reactions at first S S1 then repeat at a second constant S S2 What is significance of point of intersection of two Hnes 1lv for S1 1lv for S2 12 BIOC 463A Enzyme SteadyState Kinetics and Competitive Inhibition Km 1KmS11 Km 1 Km8211 VmaxKiS1 Vmax VmaxKiSZ Vmax rearranging and eliminating terms Ki I Ki I 31 32 If S1 at 82 how can this equation be true Answerlt is true only if Ki I So the pt of intersection represents Ki Ki 4 uM In Summary Method Constant Variable LB l S Dixon S l 13

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