Kinetics: Enzyme Catalysis

In this experiment, our focus will be on the kinetics of enzyme catalysis. Despite the vast number of enzyme structures and the equally vast number of reactions catalyzed, most enzyme-catalyzed processes exhibit essentially the same kinetics. Considering the complexity of the molecules and reactions, the kinetics is relatively simple. We approach it first from the experimental standpoint, which produces an observed rate law; we then address the question of mechanism. Equation (1) shows a typical enzyme catalyzed reaction in generic form.

Here, E stands for the enzyme, and is written over the arrow to indicate that the reaction is catalyzed. The reactant molecule on which the enzyme works is called the substrate. Although there may be more than one reactant molecule, in most cases only one of them interacts directly with the enzyme. We will condense (1) to the form in (2).

Kinetics studies of enzyme catalyzed reactions usually involve the measurement of initial rates, for reasons that need not concern us here. Throughout the discussion that follows, it will be assumed that the word, rate, stands for initial rate. The general rate law, applicable to most enzyme catalyzed processes for which studies have been done, is in (3).

[E]_T stands for the total concentration of enzyme present in the solution, and k_obs is the observed first-order rate constant. The observed rate constant exhibits an interesting dependence on the concentration of substrate, S. When [S] is small, k_obs increases nearly linearly with [S]; in other words, the rate is first order in S. However, as larger and larger concentrations of S are used, the increases in k_obs become smaller and smaller; k_obs eventually levels off to a constant value that is independent of further increases in [S]. At this point, the rate is zero-order in S. The relationship between k_obs and [S] is shown. To an experienced kineticist, this behavior is known as saturation kinetics, and can be explained by a rate law of the form in (4).

Here A and B are constants that are presumably related to rate constants of the elementary steps in the reaction mechanism. We interpret the observed kinetics in terms of (4) as follows. When [S] is small, B is the dominant term in the denominator. Under these conditions, (4) simplifies to

Thus the rate is first order in E, first order in S, with k_obs = A/B. When [S] is large, it is the dominant term in the denominator, and (4) simplifies to

The rate is first order in E, zero-order in S, with k_obs = A. The form of the expression in 4 is consistent with the observed kinetics. For historical reasons, enzyme-substrate systems that obey the kinetic form in 4 are said to exhibit Michaelis-Menten kinetics, after two early investigators in the field. The constant, B, is called the Michaelis constant, and is usually given the symbol, K_M. We will not use this symbol, however, because it implies that B is an equilibrium constant. Instead, we will use the symbol M. We make the switch at this point:

The plot of Rate versus substrate concentration nicely shows the saturation behavior of the kinetics; unfortunately, it is not possible to obtain accurate values for A and M from this plot. However, there are two rearrangements of 8 that allow the experimental data to be plotted in linear form. The first is obtained by reciprocating both sides of 8:

A plot of 1/k_obs versus 1/[S] should be linear with slope M/A and intercept 1/A. Such a plot is shown for the earlier data. A reciprocal plot of this type is called a Lineweaver-Burk plot. The values of A and M obtained from the plot are A = 0.01 s^-1, M = 0.0055 M.

The second linear form is obtained from 8 by, first, multiplication of both sides by k_obs;

second, multiplication by A;

and third, subtraction of the first term on the right from both sides.

A plot of k_obs versus k_obs/[S] should be linear with slope -M and intercept A. This plot, called an Eadie-Hofstee Plot, is shown. It produces M = 0.0055 M, A = 0.01 s^-1.

The physical interpretation of Michaelis-Menten (saturation) kinetics goes something as follows. First, the rate is first order in both enzyme and substrate when [S] is small. This implies that 1 molecule of enzyme and 1 molecule of substrate must collide and interact in order to form products. Second, the rate is first order in E and zero-order in S when S is large. The implication of this observation is that the substrate forms an adduct of some type with the enzyme, which we might symbolize as ES. This adduct then reacts further to form products. When [S] becomes large enough, Le Chatelier's Principle requires that all of the free enzyme is converted to the adduct. Once this happens, the rate no longer increases with increases in [S], because [ES] has reached its maximum possible value. This type of reasoning led biochemists to the simplest mechanism consistent with the observations:

We must now develop the rate law implied by this mechanism, to see that it matches what is observed. It is not legitimate to make assumptions at the outset about the relative magnitudes of the rate constants, because we have no information about that. The development of the rate law must be carried out as generally as possible, then examined for limiting behavior. We use the so-called steady state approximation. The essence of this approximation is that the concentration of the intermediate adduct, ES, stays essentially constant as reaction proceeds. This idea is fairly simple to implement. First, we recognize that the rate of formation of products, P, is given by

This rate law is strictly correct and always true within the context of the mechanism in 13. However, it is difficult to use because [ES] is not accessible. As usual, we need to express the rate law in terms of measurable concentrations: those of E and S. Further work is required. Now we invoke the steady state assumption, which is that

This is a mathematical way to say what we said above: the concentration of ES remains constant as reaction proceeds. We now use the elementary steps to obtain an expression for [ES]:

Solving for [ES] gives

which can be substituted for [ES] in the rate law, 14.

We are close, but not quite done. Recall that the experimental rate law is written in terms of the total enzyme concentration, [E]_T. Interpreted in terms of our mechanism,

In words, the enzyme is partitioned between the unbound and substrate bound forms. We cannot at the outset expect to know how much enzyme is in each form. However, we do know the total amount; it is the amount we put into the solution. We must therefore obtain an expression for [E] in terms of [E]_T, which can be done using 17 and 19:

Solving for [E] gives

Substituting this expression into the rate law, 18, and doing some simplifying arithmetic gives

This has a familiar look. In fact, if we identify k_cat with A, and (k_cat + k_-1)/k₁ with M, we obtain a form that is identical with the experimental rate law:

Within the context of mechanism 13, we can draw the following conclusions:

• The value of A gives a direct measure of the rate constant k_cat, which governs the rate at which the enzyme substrate adduct reacts to form products.
• The Michaelis constant, M, is seen to be (k_cat + k_-1)/k₁. This gives the ratio of the rate of consumption of ES to its rate of production. Importantly, it is clear that M is not an equilibrium constant, because it involves the sum of rate constants from two different elementary processes.

Glucose oxidase catalyzes the oxidation of glucose by oxygen according to the following equation.

The hydrogen peroxide produced in this process is a good oxidizing agent, and in the presence of the reducing agent, 2,2'-azino-bis(3-ethylbenzthiazoline-6-sulfonic acid), abbreviated ABTS, it is catalytically reduced to water by the enzyme, peroxidase.

The second reaction occurs much more rapidly than the first. The oxidized form of ABTS, ABTS⁺, absorbs strongly at 725 nm. The molar absorptivity of the dye at this wavelength is 1.9*10⁴ M^-1cm^-1. Thus by monitoring the absorbance at 725 nm, the kinetics of glucose oxidase catalyzed glucose oxidation can be readily studied.

1. During an enzyme assay run, does the reaction rate change? How?
2. Construct a plot of initial rate versus glucose oxidase concentration. What is the order of the reaction in glucose oxidase?
3. Construct a plot of initial rate versus glucose concentration. What is the order of the reaction in glucose?
4. What is the purpose of ABTS in the reaction mixture?
5. What is the purpose of horseradish peroxidase in the reaction mixture?
6. Do you think reaction rate would depend on the concentration of horseradish peroxidase? ABTS? Why or why not?

• UV-visible spectrometer
• glass or quartz spectrometer cell
• Small volume pipettors
• Stopwatch (optional)
• Stock solution of glucose oxidase (0.5 Sigma units/mL)
• Stock solutions of sugars, 1.0 M
• Cocktail Solution #1 (contains pH 7.00 phosphate buffer, ABTS, horseradish peroxidase, and 0.090 M glucose).
• Cocktail Solution #2 (contains pH 7.00 phosphate buffer, ABTS, horseradish peroxidase, and glucose oxidase).

Note to instructor: Click here for recipes for preparation of solutions.

Safety glasses must be worn at all times in the laboratory. ABTS is thought to be non-toxic; however, avoid contact with the skin anyway.

Record all data in your notebook. Obtain the necessary equipment and clean the glassware thoroughly using brushes and Alconox detergent. Rinse with distilled water and dry thoroughly.

Part 1--Dependence of Initial Rate on Enzyme Concentration. In this part of the experiment, you will determine the order of the reaction rate in glucose oxidase by varying the initial concentration of enzyme while keeping the initial concentration of substrate (glucose) fixed. Kinetics runs (called enzyme assays) will be carried out on solutions of 1.0-mL total volume. To prepare these solutions, you will mix a precise volume of a stock solution of glucose oxidase with a precise volume of "cocktail" solution #1, which contains the reagents below:

Obtain about 5 mL of "cocktail" solution #1 and about 0.50 mL of glucose oxidase stock solution. Prepare each of the following solutions in turn and carry out a kinetics run according to the instructions following the table.

Measure the reagents directly into a spectrometer cuvette in the order given. Upon addition of the glucose oxidase aliquot to the solution, shake the solution vigorously to mix and start timing. Read the absorbance of the solution at 725 nm at 1-minute intervals for a total of 6 minutes, and record the absorbance/time points.

Part 2--Dependence of Initial Rate on Glucose Concentration. In this part of the experiment, you will determine the order of the reaction rate in glucose by varying the initial glucose concentration while keeping the initial concentration of enzyme fixed. Enzyme assays will be carried out on solutions of 1.0-mL total volume. Solutions will be prepared by mixing a precise volume of glucose stock solution with a precise volume of "cocktail" solution 2, which contains the following reagents:

Obtain about 5 mL of Cocktail Solution #2 and about 1.00 mL of glucose stock solution. Prepare each of the following solutions in turn and carry out a kinetics run according to the instructions following the table.

Measure the reagents in the order given. Upon addition of the glucose aliquot to the solution, shake the solution vigorously to mix and start timing. Read the absorbance of the solution at 725 nm at 1-minute intervals for a total of 6 minutes, and record the absorbance/time points.

Clean-up. When you have finished all of your work:

• Clean Labkit glassware by recommended procedures, shake off excess water, and return to the Labkit.
• Clean Pasteur pipets, graduated pipets, and vials by recommended procedures, shake off excess water, and place in the drying oven.
• Clean spectrometer cells by rinsing with tap water. Aspirate dry.
• Clean and return all borrowed equipment to the instructor before leaving lab.
• Clean up your work area before leaving lab.

All reaction solutions can be flushed down the sink with water.

1. This experiment.
2. The appropriate sections of your textbook.
3. Concepts of Chemistry Chapter 15

1. The initial rate of an enzyme-catalyzed reaction varies with total enzyme concentration as shown in the table. What is the order of the reaction with respect to enzyme?

   total enzyme concentration, M	Initial Rate, M/min
   1.0*10-8	0.0032 M/min
   1.8*10-8	0.0060 M/min
   3.2*10-8	0.0099 M/min
   4.4*10-8	0.0141 M/min

2. The initial rate of an enzyme-catalyzed reaction varies with substrate concentration as shown in the table. Determine the Michaelis constant and the maximum reaction velocity from the data.

   [Substrate], M	Initial Velocity, M/min
   0.01	5
   0.025	12
   0.04	22
   0.1	47