Transcription

1Enzyme KineticsIn this exercise we will look at the catalytic behavior of enzymes. You will use Excel to answerthe questions in the exercise section. At the end of this session, you must hand in answers toall the questions, along with print outs of any plots you created.BackgroundEnzymes are the catalysts of biological systems and are extremely efficient and specific ascatalysts. In fact, typically, an enzyme accelerates the rate of a reaction by factors of at least amillion compared to the rate of the same reaction in the absence of the enzyme. Mostbiological reactions do not occur at perceptible rates in the absence of enzymes. One of thesimplest biological reactions catalyzed by an enzyme is the hydration of CO2. The catalyst inthis reaction is carbonic anhydrase. This reaction is part of the respiration cycle which expelsCO2 from the body. Carbonic anhydrase is a highly efficient enzyme – each enzyme moleculecan catalyze the hydration of 105 CO2 molecules per second.Enzymes are highly specific. Typically a particular enzyme catalyzes only a single chemicalreaction or a set of closely related chemical reactions. As is true of any catalyst, enzymes donot alter the equilibrium point of the reaction. This means that the enzyme accelerates theforward and reverse reaction by precisely the same factor. For example, consider theinterconversion of A and B.A B(1)Suppose that in the absence of the enzyme the forward rate constant (kf) is 10-4 s-1 and thereverse rate constant (kr) is 10-6 s-1. The equilibrium constant (Keq) is given by the ratio of thetwo rate constants.K eq[B] k f 10 4 100[A] k r 10 6(2)The equilibrium concentration of B is 100 times that of A whether or not an enzyme is present.However, in the absence of an enzyme the reaction could take more than an hour to approachthis equilibrium, whereas in the presence of an enzyme, equilibrium could be attained within a

2second. The enzyme lowers the height of the energy barrier to the reaction thereby increasingthe rate of the reaction, but since the rate of both the forward and reverse reaction are affectedby the same amount, the equilibrium constant is not affected by the presence of the enzyme.the same amount (see Figure 1)Figure 1where, EAf is the activation energy for the forward reaction (AÆB) in the absence of a catalystand E’Af is the activation energy for the forward reaction (AÆB) in the presence of a catalyst,and Go is the change in free energy for the reaction.The equilibrium constant is related to Go as follows:K eq e G o / RTSince Go is the same for the catalyzed and uncatalyzed reaction, Keq is the same for bothreactions.One reason for the efficiency and specificity of an enzyme is the way the enzyme interacts withthe reactant molecule, more commonly known as the substrate, in enzyme catalyzed reactions.The enzyme and substrate interact to form an enzyme-substrate complex. The interactionsbetween the substrate and active site are weak, noncovalent interactions (i.e. the substrate doesnot covalently bind to the active site but weakly interacts with it through interactions likehydrogen-bonding, van der Waals interactions, etc). The orientation in which the two interact

3is highly favorable for facilitating conversion of the substrate to product. In the enzymesubstrate complex, the substrate molecule binds to a very specific region of the enzymemolecule called the active site. These active sites are highly selective for a specific substratemolecule with which the enzyme binds. This is why enzymes are such highly specificcatalysts, catalyzing a single reaction, or a set of closely related reactions. There are twoproposed models to explain the specificity of the interaction between the substrate moleculeand the active site of an enzyme.(a) The “lock and key” model – in this model the substrate has a shape matching the enzyme’sactive site (see figure 2)Figure 2(b) The “induced fit model” – the active site has a shape complementary to that of the substrateafter the substrate is bound (see figure 3)

4Figure 3Enzyme kineticsThe mechanism of enzyme catalyzed reactions is often studied by making kineticmeasurements on enzyme-substrate reaction systems. These studies include measuring ratesof the enzyme-catalyzed reactions at different substrate and enzyme concentrations. Here wewill look at a simple model for the catalytic behavior of an enzyme and the kinetic model thatarises from this model.For many enzymes, if we were to plot the rate of catalysis, V (also known as the reactionvelocity), vs. the substrate concentration, [S] (at a fixed enzyme concentration) we would see aplot as shown in figure 4.Figure 4Looking at this plot, we see that V varies linearly with [S] for small [S]. As [S] increases, V“plateaus” indicating that V becomes independent of [S] at large values of [S]. The simplestmodel which accounts for this behavior is:k1k2E S ES E P(3)k 1where E is the enzyme, S the substrate, ES the enzyme-substrate complex, P the product of theenzyme-catalyzed reaction, k1 the rate constant of the forward reaction of E S, k-1 the rate ofthe reverse reaction where the enzyme-substrate complex, ES, falls apart to E S and k2 the rateconstant of the forward reaction of ES forming E P. In this model, it is assumed that none of

5the product reacts with the enzyme to form the enzyme-substrate complex, ES (this is trueduring the initial stages of the reaction when [P] is low, but towards the end of the reactionwhen [P] is high this may no longer be true).We need to derive an expression that relates the reaction velocity, V, to the concentrations ofthe substrate and enzyme and the rates of the individual steps. From equation (3) the reactionvelocity, V can be expressed as:V k2 [ES](4)Since ES is an intermediate and hence its concentration unknown, we have to express [ES] interms of known values. The rates at which [ES] is formed and falls apart are:Rate of formation of [ES] k1 [E] [S](5)Rate at which [ES] falls apart (k-1 k2) [ES](6)We can use the steady-state approximation to express V in terms of known quantities. Underthe steady-state approximation, the concentration of the intermediate [ES] stays a constant,while the concentrations of reactants and product change. The steady state occurs wheneqn 5 eqn 6 i.e.k1 [E] [S] (k-1 k2) [ES](7)[E][S]k1k 1 k 2(8)Rearranging,[ES] Define, KM, the Michaelis constant, asKM k 1 k 2k1(9)[E][S]KM(10)Substituting (9) into (8)[ES] Since in most situations the enzyme concentration is very small ([E] [S]), the concentrationof the uncombined S is almost equal to the total concentration of S. The concentration of

6uncombined E is equal to the total enzyme concentration [Eo] minus the concentration of thecomplex [ES][E] [Eo] – [ES](11)Substituting (11) into (10)[ES] ([E o ] [ES])[S]KM(12)Solving (12) for [ES],[ES] [E o ][S]/ K M1 S / KM(13)[ES] [E o ][S][S] K M(14)[S][S] K M(15)or,Substituting (14) into (4)V k 2 [E o ]The maximum reaction velocity, Vmax, is reached when all enzyme sites are saturated with thesubstrate. This will happen when [S] KM, so that [S]/([S] KM) approaches 1. In this limit,we can express Vmax (from (15)) as:Vmax k2 [Eo](16)Substituting (16) into (15),V Vmax[S][S] K M(17)If we were to plot V vs S the resulting plot will have a shape as shown in figure 4. Hence,equation (17) describes the kinetic behavior of an enzyme as modeled by the kinetic scheme inequation (3). Looking at equation (17) at very low [S], when [S] KM, V [S]Vmax/KM, thatis, the rate is proportional to [S] (describes the linear region of the plot in figure 4). At high [S],

7when [S] KM, V Vmax and hence independent of [S] (the “plateau” region of the plot infigure 4).Equation (17) can be re-arranged as:111K MV Vmax Vmax [S](18)If we were to plot 1/V vs. 1/[S] we would obtain a straight line with a y-intercept 1/Vmaxand a slope KM/Vmax (see figure 5). This plot is called a Lineweaver-Burke plot.y 2.055E 05x 1.284E 051/[S]Figure 5Significance of KMFrom equation 18, when [S] KM, then V Vmax/2. Hence KM is equal to the substrateconcentration at which the reaction rate is half its maximum value. In other words, if anenzyme has a small value of KM, it achieves its maximum catalytic efficiency at low substrateconcentrations. Hence, the smaller the value of KM, the more efficient is the catalyst. Thevalue of KM for an enzyme depends on the particular substrate . It also depends on the pH ofthe solution and the temperature at which the reaction is carried out. For most enzymes KMlies between 10-1 and 10-7 M.

8Determining KM and Vmax experimentallyTo characterize an enzyme-catalyzed reaction KM and Vmax need to be determined. The waythis is done experimentally is to measure the rate of catalysis (reaction velocity) for differentsubstrate concentrations. In other words, determine V at different values of [S]. Then plotting1/V vs. 1/[S] we should obtain a straight line described by equation (18). From the y-interceptand the slope, the values of KM and Vmax can be determined. For example, use EXCEL to plotthe data shown below. Fit the data to a straight line, and from the equation of the straight linedetermine the values of KM and Vmax.[S] (mM)8.335.552.771.380.83V [S] (mM-1)0.120.180.360.721.21/V (sec/mM)2.76E 052.95E 053.64E 055.02E 056.73E 05You should obtain a plot as shown in figure 6 below. From the fit to the data show thatKM 1.6 for this data and Vmax is 4.32 x 10-6 mM/sec8.00E 05y 3.700E 05x 2.311E 057.00E 056.00E 055.00E 054.00E 053.00E 052.00E 051.00E 050.00E 0000.20.40.61/[S]0.811.21.4

9Figure 6

10EXERCISESUse Excel to answer the questions given below.1) In a typical experiment to characterize an enzyme, KM and Vmax need to be measured. Todo this, first a series of experiments are run, where for a given initial substrate concentration,the rate at which the product is formed is monitored. The rate of product formation ismonitored by measuring the product concentration as a function of time. A plot ofconcentration of product vs. time will be linear at short time (as long as most of the substratemolecules have not reacted) and the slope of the linear fit to the data is the rate of productformation for a given initial substrate concentration. This is the rate of the reaction (V). Theexperiment is repeated for different initial substrate concentrations (keeping the enzymeconcentration fixed) and the rate of product formation at each of these substrate concentrationsdetermined. This series of experiments results in a set of data: reaction rates (V) as a functionof substrate concentration [S]. Using this data a plot of 1/V vs. 1/[S] is constructed – this plotshould be linear as dictated by equation (18), and from the linear fit KM and Vmax aredetermined.You will use Excel to analyze data from an experiment of the type just described to determinethe KM and Vmax that characterizes the enzyme used. Open the file “exercise1” in Excel. (Givethe file a new name [File - SAVE AS], save to DISK, and work from this saved file.)This file contains the data from five runs (RUN 1 to RUN 5) monitoring the productconcentration as a function of time for different initial substrate concentrations (all carried outwith the same enzyme concentration). Plot each data set as product concentration vs. time. Fitthe data to a straight line and determine the slope. The slope is the rate of the reaction, V, for aparticular value of [S].Note: In the “Add Trendline” dialog box (accessed from the Chart menu), first select “Linear”(for a linear fit to the data) and then click on Options and select the “Display equation onchart” option. This should show the equation of the straight line fit to the data. Double clickon the equation. This will bring up another dialog box. Click on the “Number” option andchoose Scientific and increase the number of decimal places to 3.You should end up with five values of V for each of the five values of [S]. Plot 1/V vs. 1/[S]and determine KM and Vmax from the straight line fit to the data and equation 18.

11Submit a printout of the plot of 1/V vs. 1/[S] and report the values of KM and Vmax.2) Open file “exercise2” in Excel. (Give the file a new name [File - SAVE AS], save to DISK,and work from this saved file.)This file contains a series of data from three experiments. In each experiment, the reactionrates (V) were monitored as a function of substrate concentration [S], but in each of the threeexperiments a DIFFERENT ENZYME was used. The concentration of the enzymes was keptfixed, as well as the pH of the solution and the temperature of the reaction. In addition, theV’s for each experiment were determined for the same substrate concentration.From the data given, plot Lineweaver-Burke plots for each experiment, determine whichenzyme (X, Y or Z) is catalytically most efficient, and explain why. Submit printouts of theplots you created along with your calculations for KM.3) Auxiliary questions:(i)What are the units for KM and why does this constant have these units?(ii)In the Michaelis-Menten Scheme, what would happen to V (equation 15) if, insteadof making the steady-state approximation, you assumed that the enzyme-substratecomplex (ES) were in equilibrium with E S? Show what approximation you wouldneed to make in the steady-state result for V to get the equilibrium result for V.