Submit 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:11What are the units for KM and why does this constant have these units?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 youneed to make in the steady-state result for V to get the equilibrium result for V.wouldBiochemistry* (1)☆(ii)Please help. EXERCISESUse Excel to answer the questions given below.1) In a typical experiment to characterize an enzyme, Kè 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.10You 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.

The general equation for initial velocity (V) of a reaction involving a single substrate on an enzyme with a single binding site for the substrate is given below. V=Vmax[S][S] + a here 'a' is a constant. In steady state assumption we get 'Km' as 'a' , while in equilibrium assumption we get 'Ks' as 'a'. Lets have a look at this in detail. A reaction involving a single substrate on an enzyme with a single binding site for the substrate can be represented as shown below. E + S ⇌k-1k1 ES ⇌k-2k2 E+ P The Michaelis-Menten enzyme kinetics is concerned with only the initial period of the reaction and that's why we measure the initial rate of the reaction. Since there aren't enough product present at this initial period of the reaction, we can neglect the formation of ES from E and P. So, the above representation can be modified as shown below. E + S ⇌k-1k1 ES →k2 E + P The steady state assumption is that, in the initial period of the reaction, the rate of change of [ES] is negligible i.e. rate of formation of ES is equal to the rate of breakdown of ES. Rate of formation of ES= Rate of breakdown of ESk1[E][S] = k-1[ES] + k2[ES][E][S][ES] = k-1 + k2k1 = Km ----(eq1)We know, [E] =[E0] -[ES] ----(eq2)here, [E0] is total enzyme concentrationSubstituting (eq2) on (eq1);[E0] -[ES][S][ES] = Km[ES]=[E0][S][S]+Km ----(eq3)We know, V=k2[E0]----(eq4)here 'V' is initial reaction rateSubstituting (eq4) on (eq3) ;V=k2[E0][S][S]+KmSince, Vmax = k2[E0]V=Vmax[S][S]+Km Now, lets consider the equilibrium assumption. The equilibrium assumption considers that an equilibrium is quickly set up between the reaction that forms ES from E & S and the reaction where ES gets broken down to E and S. And it further assumes that under the initial conditions, the rate of formation of P from ES does not influence or affect this equilibrium. Thereby, the difference between the steady state and equilibrium assumptions is that, the steady state assumption considers the reaction ES→k2 E + P , while the equilibrium assumption does not consider this reaction. So, according to the equilibrium assumption; Rate of formation of ES = Rate of breakdown of ESk1[E][S] = k-1 [ES][E][S][ES] =k-1k1 = Ks 'Ks' is the dissociation constant of ES. So, under steady state assumption, [E][S][ES] gives us the Michaelis-Menten constant, Km . While the under the equilibrium assumption, [E][S][ES] gives us another constant called the dissociation constant of ES, Ks . The rest of the steps to find 'V' can be done similar to the way we did it in Steady state assumption. At the end, we get V in equilibrium assumption as, V=Vmax[S][S]+Ks So, the Km in steady state assumption gets replaced by another constant called Ks in the equilibrium assumption , because we do not consider the reaction ES→k2 E + P in the equilibrium assumption. The units of Km are the different units of concentration like molar (M), millimolar (mM), micromolar (μM) , etc. Km has the units of concentration because Km is effectively the substrate concentration when the initial rate of reaction is equal to Vmax2 .