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UIUC MCB 450 - Lecture 11 for posting

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PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 40Slide 41Slide 42Slide 43Slide 44Slide 45Slide 4611-1MCB 450Lecture 11Effect of [substrate] and [enzyme] on reaction rateKm and Vmaxkcatkcat / KmBi-substrate reactionsEnzyme KineticsAmount of E presentpHTemperatureBig picture: in an enzyme-catalyzed reaction, amountof S to P conversion depends on...Metal ions (often)Allosteric regulators= low mol. wt. compounds that bind to another site on E(requires multisubunit proteins)11-2Presence of inhibitorsAmount of substrate present...Km = [S] for efficient catalysis to occur~ measure of the affinity of E for its substrate(s)Vmax: depends on kcat and Et= “intrinsic property” of E= # S molecules converted to P in a given unit timeon a single E molecule when E is saturated with Stotal E presentmaximalvelocity• [Substrate] is key factor affecting rate of enzyme-catalyzed reactions• Study of rates of enzyme-catalyzed reactions & how they changein response to experimental parameters = enzyme kineticsConcepts and terms:• Reaction velocity V• Initial velocity V0 (V measured early in a conversion)• Substrate concentration [S]• Enzyme-Substrate complex ES• Saturation all catalytic sites on E filled w. SEffect of [substrate] on reaction rate11-3- Reaction rate is governed by [reactant(s)] and a rate constant k- For S  P, rate or velocity of the reaction, V, (= amount of S that reacts per unit time) is expressed by 1st order rate equation V = k[S] (V = k[A] in book)- k is a proportionality constant that reflects the probability of reactionunder a given set of conditions (pH, temp…)- Where rate depends on [S]  1st order reaction, and k has units of time-1Example: if a 1st order reaction has a rate constant k of 0.03 s-1,then ~3% of S will be converted into P in 1 secRate constants and 1st order reactions11-4- Where rate depends on [ ] of 2 different compounds, or if reaction is between2 molecules of the same compound, reaction is 2nd order,and V = k[S1][S2] with k having units of M-1.time-1Rate constants and 2nd order reactions11-5S1 + S2 P1 + P22S PV = k[S1][S2]V = k[S]2V0 = V measured earlyin a conversion,when [S] hasn'tsignificantly beenlowered by conversion to PLINEARRANGEV0, initial velocity= Rate of P formation early in reaction, when rate of SP conversion is linearTimeProductconcentration11-6S PV0 as a function of [S] in a 1st order reaction11-7- If [S2] is ≥≥ [S1], then reaction will be 1st order with respect to S1,and not appear to be dependent on [S2]Bimolecular reactions can become pseudo 1st order11-8[S1]1 [S1]2 [S1]3 [S1]4E 2E 3E 4E V0 is proportional to the amount of Eat a fixed, saturating [S]11-9V0, initial velocity, depends on [S]LINEARRANGE 11-10Reflects formation ofES complex and fillingof all catalytic sites with SAll enzyme moleculesare reacting as fast asthey canEffect of [substrate] on V0 of enzyme-catalyzed reaction 11-11Approaches “saturation”rectangularhyperbola11-1211-13 One more assumption….ES has two possible fates:it can dissociate to E + S, orit can dissociate to E + PMichaelis and Menten (1913) came up with a model toaccount for these kinetic characteristics11-14 The Michaelis-Menten equationMICHAELIS CONSTANT Km =COMBINED RATE CONSTANT:Km on the V0 versus [S] plot11-15At [S] >> Km, Km term in denominatorof M-M equation becomesinsignificant, and equationsimplifies to:plateau athigh [S]Michaelis-Menten eq. fits experimental observationsV0 =Vmax [S]Km + [S]11-16At [S] << Km, [S] is an insignificantaddition to Km, so M-Mequation simplifies to:….and V0 has lineardependence on [S]……..as seenexperimentally11-17Small Km means 1/2 maximal V achieved at relatively low [S]Large Km means 1/2 maximal V achieved at relatively high [S]IS A MEASURE OF THE [S] REQUIRED FOREFFECTIVE CATALYSIS TO OCCUR11-18V0 =Vmax [S]Km + [S]11-19Experimental determinationof Km and Vmax:Determine Vo for reactionat a range of [substrate]Plot 1 Vo [S]1 versus From intercepts,calculate Vmaxand KmDetermination of Km and Vmax from double reciprocal plot 11-201[S]3.1252.51.671.00.5 1V010.38.86.75.03.75 [S](mM)0.320.400.601.002.00V0(nmoles min-1)0.0970.1140.150.200.267Example11-21“You assay the activity of enzyme X at a range of substrate concentrations, and obtain the data in the table below. Assumethe enzyme obeys Michaelis-Menten kinetics. Plot these data using a Lineweaver-Burk plot, and determine Km and Vmax”1.0 2.0 3.0 4.00.010 1V0 1[S] (mM)-2.0-1Km~ -0.95 1VmaxFrom graph,1/Vmax = 2.4, so Vmax = 0.42 nmoles min-1-1/Km ~ -0.95, so Km = 1.05 mMMULTIPLY 1/x AXIS INTERCEPT by -111-22Km of an E for its S is often near theconcentration of that S in the cellGeneralization about Km......11-23Km value is just above the steeply rising portionof the V0 versus [S] rate curve, so an E for whichKm ≥ cellular [S] is most able to respondproportionally to changes in [S].Vmax / 2KmThis can help prevent theaccumulation of S in the cellif metabolic conditions change11-24OF THE AFFINITYkcat : rate constant for the rate-limiting stepin an enzyme-catalyzed reaction11-25Total amount of enzyme = EtFor , rate constant for limiting step when enzyme is saturated with S is k2Vmax is dependent on [Et]Knowing Vmax and [Et], we have the TURNOVER NUMBER of the enzyme,= # of substrate molecules an enzyme can convert into product per unittime when the enzyme is saturated with substrate.So, Vmax = k2 [Et] Rearranging, k2 = Vmax / [Et] = the TURNOVER NUMBER= kcat = Limiting rate of any enzyme-catalyzed reaction at saturationkcatkcatVmax[Et]kcat =AT SATURATION W. [S]11-26WHAT'S kcat?kcat : rate constant for the rate-limiting stepin an enzyme-catalyzed reaction11-2711-28For a one-substrate, enzyme-catalyzed reaction, double-reciprocal plots were generatedat three different enzyme concentrations. Which of these three sets of plots would you expectto be obtained? Why?Problem Hints: What do the x and y intercepts signify?What do you expect to happen when [E] is raised?Which intercepts will be affected when [E] is raised?11-2911-30At high kcat/Km, E binds S more tightly (low Km value),and also catalyzes the reaction relatively rapidlyTHE HIGHER Kcat /KmTHE GREATER THE ENZYME’S CATALYTIC


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