11-1 MCB 450 Lecture 11 Effect of [substrate] and [enzyme] on reaction rate Km and Vmax kcat kcat / Km Bi-substrate reactions Enzyme KineticsAmount of E present pH Temperature Big picture: in an enzyme-catalyzed reaction, amount of 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) 11-2 Presence of inhibitors Amount 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 time on a single E molecule when E is saturated with S total E present maximal velocity• [Substrate] is key factor affecting rate of enzyme-catalyzed reactions • Study of rates of enzyme-catalyzed reactions & how they change in response to experimental parameters = enzyme kinetics Concepts 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. S Effect of [substrate] on reaction rate 11-3- 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 reaction under a given set of conditions (pH, temp…) - Where rate depends on [S] → 1st order reaction, and k has units of time-1 Example: 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 sec Rate constants and 1st order reactions 11-4- Where rate depends on [ ] of 2 different compounds, or if reaction is between 2 molecules of the same compound, reaction is 2nd order, and V = k[S1][S2] with k having units of M-1.time-1 Rate constants and 2nd order reactions 11-5 S1 + S2 P1 + P2 2S P V = k[S1][S2] V = k[S]2V0 = V measured early in a conversion, when [S] hasn't significantly been lowered by conversion to P LINEAR RANGE V0, initial velocity = Rate of P formation early in reaction, when rate of S→P conversion is linear Time Product concentration 11-6S""""""""""""P"V0 as a function of [S] in a 1st order reaction 11-7- 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 order 11-8 [S1]1%%%%%%%%%[S1]2%%%%%%%%%[S1]3%%%%%%%%%[S1]4%%%%%%%%%E 2E 3E 4E V0 is proportional to the amount of E at a fixed, saturating [S] 11-9V0, initial velocity, depends on [S] LINEAR RANGE !11-10Reflects formation of ES complex and filling of all catalytic sites with S All enzyme molecules are reacting as fast as they can Effect of [substrate] on V0 of enzyme-catalyzed reaction 11-11 Approaches “saturation” rectangular hyperbola11-1211-13 One more assumption…. ES has two possible fates: it can dissociate to E + S, or it can dissociate to E + P Michaelis and Menten (1913) came up with a model to account for these kinetic characteristics11-14 The Michaelis-Menten equation MICHAELIS CONSTANT Km = COMBINED RATE CONSTANT:Km on the V0 versus [S] plot 11-15At [S] >> Km, Km term in denominator of M-M equation becomes insignificant, and equation simplifies to: plateau at high [S] Michaelis-Menten eq. fits experimental observations V0 = Vmax [S] Km + [S] 11-16 At [S] << Km, [S] is an insignificant addition to Km, so M-M equation simplifies to: ….and V0 has linear dependence on [S]… …..as seen experimentally11-17Small 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 FOR EFFECTIVE CATALYSIS TO OCCUR 11-18V0 = Vmax [S] Km + [S] 11-19Experimental determination of Km and Vmax: Determine Vo for reaction at a range of [substrate] Plot 1 Vo [S] 1 versus From intercepts, calculate Vmax and Km Determination of Km and Vmax from double reciprocal plot 11-201![S]!!3.125!2.5!1.67!1.0!0.5! 1!V0!!10.3!8.8!6.7!5.0!3.75! [S] (mM) 0.32 0.40 0.60 1.00 2.00 V0 (nmoles min-1) 0.097 0.114 0.15 0.20 0.267 Example 11-21 “You assay the activity of enzyme X at a range of substrate concentrations, and obtain the data in the table below. Assume the enzyme obeys Michaelis-Menten kinetics. Plot these data using a Lineweaver-Burk plot, and determine Km and Vmax” 1.0! 2.0! 3.0! 4.0!0.0!10! 1!V0! 1![S] (mM)!-2.0!-1!Km!~ -0.95! 1!Vmax!From graph, 1/Vmax = 2.4, so Vmax = 0.42 nmoles min-1 -1/Km ~ -0.95, so Km = 1.05 mM MULTIPLY 1/x AXIS INTERCEPT by -111-22Km of an E for its S is often near the concentration of that S in the cell Generalization about Km...... 11-23 Km value is just above the steeply rising portion of the V0 versus [S] rate curve, so an E for which Km ≥ cellular [S] is most able to respond proportionally to changes in [S]. Vmax / 2 Km This can help prevent the accumulation of S in the cell if metabolic conditions change11-24 OF THE AFFINITYkcat : rate constant for the rate-limiting step in an enzyme-catalyzed reaction 11-25 Total amount of enzyme = Et For , rate constant for limiting step when enzyme is saturated with S is k2 Vmax 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 unit time 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 saturationkcat kcat Vmax [Et] kcat = AT SATURATION W. [S] 11-26 WHAT'S kcat?kcat : rate constant for the rate-limiting step in an enzyme-catalyzed reaction 11-2711-28 For a one-substrate, enzyme-catalyzed reaction, double-reciprocal plots were generated at three different enzyme concentrations. Which of these three sets of plots would you expect to 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 rapidly THE HIGHER Kcat /Km THE GREATER THE ENZYMEʼS
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