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Enzyme Kinetics: Inversion of Sucrose

Abstract
Enzymes are a class of proteins that catalyze practically all biochemical reactions. The enzymatic reactions were looked at by the use of two comparable reactions of enzyme invertase and a acidic form of the reaction. The order of the nonacidic enzymatic reaction was zero order due to substrate concentration and due to the fact that the reaction was time dependent. It is also due to having large R2 value at 0.9932 for D run and a smaller R2 value at 0.9902 for C run. The acidic runs had a first order reaction, which had a lower R2 values at 0.9028 for run D and 0.9028 for run C. Also, the percent error in run C at 33.31% was found to be much lower than the percent error in run D at 55.77%, which mean the concentration of run C is more effective than run D.

I. Introduction In this experiment the chemical kinetics of the enzyme catalyzed inversion of sucrose was studied. The reaction that we will study in this experiment is the inversion of sucrose, catalyzed by the enzyme invertase that is derived from yeast: The rate of reaction of this reaction was compared to the same reaction that is to be catalyzed by hydrogen ions. Enzymes make up an important class of proteins that are used to catalyze a wide array of biochemical reactions. The enzyme that was used to catalyze the reactions in this lab experiment was the enzyme invertase (β-fructofuranidase). The first to propose the basic mechanism for enzyme-catalyzed reactions was Michaelis and Menten, this was later confirmed by a study of the kinetics of the sucrose inversion. This mechanism is given by Eq. (1) where E is the enzyme, S is the reactant, ES is an intermediate, and P is the product, and K as the rate constant. The constant K’s which are shown for each of the backward and forward reactions are the equilibrium constants for each particular reaction. The first step that leads to the formation of the enzyme-substrate complex ES is a rapid reaction which both has large equilibrium constants (K1 and K-1). The decomposition of this complex to form products is relatively slow. The back-reaction to which the E and the P reunite to create the complex ES is ignored. Do to the fact that the rate constant for the reverse reaction (K-2) is relatively small and the concentration of the P will also be small since we are only concerned with the initial stages of product formation. When the rate of formation of the products are ignored, the back-reaction of the reformed E and P are produce and the complex ES is ignored this gives rise to Eq. (2). Note that the expression (ES) indicates the concentration of the enclosed species in units of mol L-1. r=+(d(p))/dt=K_2 (ES) (2) The steady state approximation was intended to be used by Michaelis and Menten for the complex ES. This approximation will give us an expression of the concentration (ES) in terms of enzyme and substrate concentrations. This assumption was based from assuming that the change of the concentration of our complex (ES) with respect to time is very small when compared to the rates of formation and destruction of ES. This shows that the change in the concentration of the complex with respect to time is given by Eq. (3). By simple manipulation Eq. (3) then yields the expression for Eq. (4). (d(ES))/dt=K_1 (E)(S)-K_(-1) (ES)-K_2 (ES)≅0 (3) (ES)=(K_1 (E)(S))/(K_(-1)+K_2 )=((E)(S))/K_m (4)
The Michaelis-Menten constant Km is defined by the expression in Eq. (5). 〖 K〗_m=(K_(-1)+K_2)/K_1 (5) The instant concentration of the enzyme E is not a known function but since the enzyme is conserved we will not need it. The total concentration in the form of the enzyme-substrate complex and free enzyme is equal to the initial enzyme concentration, which is denoted as (E)0. By combining this statement along with Eqs. (2) and (4) we get the following. r=(K_2 (E)_0 (S))/(K_m+(S)) (6) In this experiment we measured the initial rate r0 as a function of the initial enzyme concentrations (E)0 and (S)0. Eq. (6) can be rewritten in the following two ways to yield Eqs. (7) and (8). 1/r_0 =1/(K_2 〖(E)〗_0 )+K_m/(K_2 〖(E)〗_0 )*1/〖(S)〗_0 (7) r_0/〖(S)〗_0 =(K_2 〖(E)〗_0)/K_m -r_0/K_m (8) When 1/r0 is plotted against 1/(S)0 this is called a Lineweaver – Burk(L-H) plot which gives a intercept that can be used to determine K2(E)0 and a slope that can be used to determine Km Eq. (7). The Eadie – Hofstee (E-H) plot is when r0/(S)0 is plotted against r0 and allows us to calculate Km from the slope and K2(E)0 from the intercept Eq(8). A acid-catalyzed reaction rate can be carried out by using a less efficient catalyst H+ ion which gives the following rate. r=+(d(P))/dt=-d(S)/dt=K_H 〖(H〗^+)(S) (9) KH is the rate constant for this acid catalyzed reaction. This rate is studied at hydrogen ion concentrations ranging from 0.1M to 0.5M in the absence of the enzyme. The activation energy of a reaction is the energy that the reaction must overcome in order for the reaction to take place. Rate constants are occasionally represented by the Arrhenius equation, k=Ae^(-E_a/RT) (10) where Ea is the activation energy, R is the gas constant, and T is the temperature. A is a constant which can be determined by the given initial conditions of the reaction. The use of an enzyme can lower the activation barrier by lowering the activation energy and allowing for a faster reaction. The activation energies of both the enzyme catalyzed and hydrogen ion catalyzed reactions will be determined in this lab experiment. Data analysis will be performed on the data obtained in this lab experiment. The data analysis will help calculate the product concentrations. Beers law will be used to convert the obtained absorbance values to the product concentration. The equation for Beers law is A=log⁡(I_0/I)=εcd (11)

where A is the absorbance, I/I0 is the fraction of the light transmitted through an absorbing solution, ε is the absorption coefficient constant, c is the concentration, and in this case d is the path-length constant of the cuvettes.

IV. Discussion
In this experiment we look at the rates of duplicate runs such as D5 and E3 to determine whether the initial rate measurements are reproducible. It was determined that the rates were 9.78 x 10-6 M/s and 9.51 x 10-6 M/s. Considering that these values differ by an order of magnitude, it is unlikely that the initial rate measurements can be reproduced. The enzyme-catalyzed reaction is compatible with the Michaelis-Menten mechanism because the kinetics of the reaction between sucrose and the enzyme can be predicted by the constant Km and the concentration of the substrate sucrose. If Km is significantly smaller than our substrate concentration, the reaction proceeds with zero-order kinetics. However, if Km is significantly larger than the substrate, the reaction proceeds with first-order kinetics. The Km in our experiment was smaller than our sucrose concentration which leads us to believe that our enzyme-catalyzed reaction was predicted to be zero-order. If we look at figures 3 and 5, we can see that both runs C and D follow zero-order kinetics due to the fact that their R2 values are near to 1. This lines up with our prediction based on Km and the sucrose concentrations. On the other hand, if we look at figures 4 and 6, we can see that the reactions apparently proceed as first order because the R2 are near to 1as well. At the beginning the substrate concentration is high leading to zero-order kinetics as expected but overall, the substrate concentration decreases and the reaction proceeds with first order kinetics.
While analyzing our Lineweaver-Burk(L-B) and Eadie-Hofstee(E-H) graphs by observing that the values determined by the L-B is lower than the maximum rate observed after 20 minutes for run D while the E-H is slightly higher which makes it appear more accurate. The L-B graph has a larger R2 value whereas the E-H graph has a much smaller one, which makes L-B appear to be correct. However, it turns out that the E-H is more consistent with the data.
Due to the fact that an absorbable enzyme we were not found and time elapsed we were unable to make a comparison between the enzyme-catalyzed and acid-catalyzed reactions. Also, we were unable to finish runs F-H which entailed nonenzymatice hydrolysis of sucrose and the dependence of the rate on temperature.
In conclusion in many chemical reactions, the speed of the inversion of sucrose depends upon temperature. Hence the reaction must be studied at a controlled temperature which requires the use of a constant-temperature bath for the reacting mixture. It should be noted also that the enzyme catalysts, which are proteins, are sensitive to many microorganisms found in the laboratory and special care should be taken to exclude contaminants from the enzyme preparations during experiments

References Enzyme Kinetics: Inversion of Source Chapter X from Experiments in Physical Chemistry, Eighth Edition by Garland, Nibler, Shoemaker, 2009; pp 109/271-121/283.

Appendix A: Copy of Students’ Carbonless Lab Book Pages

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Physical Chem

...STK 2023 PHYSICAL CHEMISTRY 2 ASSIGNMENT 1 Question 1 (a) Rate of formation = 13d[O2]dt Rate of disappearance = - 12d[O3]dt (b) Rate of formation = 12d[HF]dt = d[O2]dt Rate of disappearance = - 12d[HOF]dt (c) Rate of formation = 12d[BrNO]dt Rate of disappearance = - 12d[NO]dt= - d[Br2]dt Question 2 2NO(g) + H2(g) N2O(g) + H2O(g) (a) Rate = k[NO]2[H2] (b) 0.005 = k(0.075)2(0.400) 0.005 = k(2.25 x 10-3) k = 2.22 M-2 s-1 Question 3 Temperature (K) | Rate constant (M/s) | 1/T (K-1) | ln k | 573 | 2.91 x 10-6 | 0.00175 | -12.75 | 673 | 8.38 x 10-4 | 0.00149 | -7.08 | 773 | 7.65 x 10-2 | 0.00130 | -2.57 | Plot of ln k against 1/T (K-1) Slope of graph, m = -7.08-(-12.75)0.00149-0.00175 = 5.67- 0.00026 = 21807.69 Using Arrhenius equation, m = - Ea8.314 J/Kmol 21807.69 = - Ea8.314 J/Kmol Ea = 181 kJ/mol Question 4 (a) Rate of reaction of NO is second order. Rate of reaction of H2 is first order. Total rate of reaction = third order. (b) If [NO] is doubled, the rate of reaction quadruples. (c) If [H2] is halved, the rate of reaction is halved. Question 5 Initial concentration = Ao Final concentration = A Time, t = 65s – 45s = 20s Use ln A = -kt + ln Ao ln (8.70 x 10-3M) = - (k)(20s) + ln (7.30 x 10-2M) -4.744 = - (k)(20s) + (-2.617) 2.127 = (k)(20s) k = 0.106 s-1 Question 6 The rate law is determined based on the slow reaction since the fast reactions are...

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