11 1 MCB 450 Lecture 11 Enzyme Kinetics Effect of substrate and enzyme on reaction rate Km and Vmax kcat kcat Km Bi substrate reactions 11 2 Big picture in an enzyme catalyzed reaction amount of S to P conversion depends on Amount of E present Temperature pH Metal ions often Amount of substrate present Km Vmax S for efficient catalysis to occur measure of the affinity of E for its substrate s depends on kcat and Et maximal velocity total E present 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 Presence of inhibitors Allosteric regulators low mol wt compounds that bind to another site on E requires multisubunit proteins 11 3 Effect of substrate on reaction rate 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 11 4 Rate constants and 1st order reactions 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 11 5 Rate constants and 2nd order reactions 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 S1 S2 P1 P2 V k S1 S2 2S P V k S 2 11 6 V0 initial velocity Rate of P formation early in reaction when rate of S P conversion is linear Product concentration V0 V measured early in a conversion when S hasn t significantly been lowered by conversion to P LINEAR RANGE Time 11 7 V0 as a function of S in a 1st order reaction S P 11 8 Bimolecular reactions can become pseudo 1st order If S2 is S1 then reaction will be 1st order with respect to S1 and not appear to be dependent on S2 S1 4 S1 3 S1 2 S1 1 11 9 V0 is proportional to the amount of E at a fixed saturating S 4E 3E 2E E 11 10 V0 initial velocity depends on S LINEAR RANGE 11 11 Effect of substrate on V0 of enzyme catalyzed reaction rectangular hyperbola Approaches saturation Reflects formation of ES complex and filling of all catalytic sites with S All enzyme molecules are reacting as fast as they can 11 12 11 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 characteristics 11 14 The Michaelis Menten equation MICHAELIS CONSTANT Km COMBINED RATE CONSTANT 11 15 Km on the V0 versus S plot 11 16 Michaelis Menten eq fits experimental observations At S Km S is an insignificant addition to Km so M M equation simplifies to V0 Vmax S Km S At S Km Km term in denominator of M M equation becomes insignificant and equation simplifies to and V0 has linear dependence on S as seen experimentally plateau at high S 11 17 11 18 Small 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 19 V0 Vmax S Km S 11 20 Determination of Km and Vmax from double reciprocal plot Experimental determination of Km and Vmax Determine Vo for reaction at a range of substrate Plot 1 1 versus Vo S From intercepts calculate Vmax and Km 11 21 Example 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 S mM 1 S V0 nmoles min 1 0 32 0 40 0 60 1 00 2 00 1 V0 3 125 2 5 1 67 1 0 0 5 0 097 0 114 0 15 0 20 0 267 10 3 8 8 6 7 5 0 3 75 1 V0 10 MULTIPLY 1 x AXIS INTERCEPT by 1 1 Vmax From graph 1 Km 0 95 so Km 1 05 mM 2 0 1 Vmax 2 4 so Vmax 0 42 nmoles min 1 0 0 1 Km 0 95 1 0 2 0 1 S mM 3 0 4 0 11 22 11 23 Generalization about Km Km of an E for its S is often near the concentration of that S in the cell 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 change 11 24 OF THE AFFINITY 11 25 kcat rate constant for the rate limiting step in an enzyme catalyzed reaction Total amount of enzyme Et For rate constant for limiting step when enzyme is saturated with S is k2 So Vmax k2 Et 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 Rearranging k2 Vmax Et the TURNOVER NUMBER kcat Limiting rate of any enzyme catalyzed reaction at saturation 11 26 AT SATURATION W S kcat kcat Vmax Et kcat WHAT S kcat 11 27 kcat rate constant for the rate limiting step in an enzyme catalyzed reaction 11 28 Problem 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 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 29 11 30 11 31 Small Km What kcat Km tell us about an enzyme means 1 2 maximal V achieved at relatively low S Large Km means 1 2 maximal V achieved at relatively high S THE HIGHER Kcat Km THE GREATER THE ENZYME S CATALYTIC EFFICIENCY TOWARDS A PARTICULAR …
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