3.2.1 Progress curve of a Michaelis-Menten reaction

Consider the irreversible Michaelis-Menten scheme - it is the simplest enzyme mechanism of all (see also Eqn [2.31]).

ki k2

Eqn [3.2] expresses the fact that in a conventional steady-state kinetic analysis of such an enzyme only two parameters are estimated, Km and Vmax, and these encapsulate all the information that is required to describe the progress of the reaction after the establishment of a steady state of the EA concentration. In other words, if we assume that [EA] is constant, which involves ignoring the rapid pre-steady-state phase of the time course, the progress of the reaction is described by the solution of the MichaelisMenten differential equation.

Q: Simulate the decline of substrate in a reaction that is catalyzed by a simple MichaelisMenten enzyme; let the Km and Vmax be 5 mmol L-1 and 100.0 |imol L-1 min-1, respectively, and let the initial concentration of the substrate be 10 mmol L-1. Simulate a time course of 400 min.

A: The relevant differential equation is (Eqn [2.1])

and a suitable Mathematica program entails the use of NDSolve (see Section 1.4.6), as follows:

Clear[Subscript];

NDSolve[{a' [t] % -v1, a[0] % a0 },a[t], {t, 0, 400}]

{{a[t] ^InterpolatingFunction[{{ 0., 400.}}, <>][t]}}

By using the resulting InterpolatingFunction (Section 1.4.6) we can Plot the time course.

AxesLabel -> {"Time (min)", "Concentration (M)"}];

Concentration (M)

Q: In the previous example we only simulated the decline of A. How do we simulate the appearance of P?

A: The conservation of mass condition for A specifies that

where the subscript t denoted any given time, hence

So, using our solution from the previous example, we must Plot Eqn [3.5] to solve this problem.

Plot [Evaluate [{a [t] , a0 - a[t]} /. solution], {t, 0, 400}, AxesLabel -> {"Time (min) ", "Concentration (M) " }] ;

Concentration (M)

Concentration (M)

We note that the two progress curves are mirror images of each other with the horizontal line of reflection being at half the starting concentration of A. This outcome is a direct consequence of the conservation relationship in Eqn [3.4]. For the MichaelisMenten reaction scheme this is not strictly true since there will be some buildup of the enzyme-substrate complex EA (see Eqn [2.12]) prior to it attaining a steady state. But as long as [A] is much greater than the concentration of the enzyme, this will usually be an exceptionally good approximation for times beyond a second or less.

3.2.2 Pre-steady-state Michaelis-Menten scheme

Eqn [3.2] contains the definition of the steady-state kinetic parameters of a simple Michaelis-Menten enzyme:

where [E]0 is the molar concentration of the active sites of the enzyme.

If we wish to simulate the pre-steady-state phase of the reaction, it is necessary to specify the values of the unitary rate constants. In the absence of any experimental studies that give these values, or at least some of them, we must solve the nonlinear algebraic equations that are expressed in Eqns [3.6] and [3.7] for the three values (k1, k-1, and k2).

It is usually possible to make a reliable estimate of the turnover number of an enzyme (Section 2.3.2) from an estimate of Vmax, hence k2 is readily evaluated by rearranging Eqn [3.7]. This outcome is possible provided that the enzyme has been purified and its molecular weight, and order of oligomerization (see Section 3.2.3), are known. Solving Eqns [3.6] and [3.7] for the three unknowns clearly means that the value of one of the unitary rate constants must be arbitrarily specified. Suppose we give k1 the value aa mol-1 L s-1, then symbolic expressions for the other two unitary rate constants can be obtained by using the Solve function.

Solve[/hs == rhs, vars] attempts to solve an equation or set of equations for the variables vars

Q: What is the procedure for determining expressions for the unitary rate constants for a Michaelis-Menten enzyme in terms of the steady-state parameters Vmax and Km ?

A: This problem, which is one of solving a pair of simultaneous nonlinear algebraic equations, is solved as follows:

Clear[Subscript];

The analysis for this problem is rather elementary and the equations could easily have been done accurately without symbolic computation. However, the basic approach is applicable to sets of much more complex expressions of steady-state parameters so it is a strategy well worth understanding.

3.2.3 Enzyme oligomerization and the turnover number

There is a complication with the definition of [E]0; it pertains to the concentration of active sites since many enzymes are oligomeric. If the enzyme is oligomeric then [E]0 must be related to the concentration of the holoenzyme (or whole oligomer), [E']0, by the expression

where n is the number of subunits in the oligomeric holoenzyme and m is the number of active sites on each monomer (subunit). If we know the concentration of the holoenzyme, the value of the forward catalytic breakdown rate constant is related to the turnover number of each active site by the expression

where m is almost invariably 1, and the maximum velocity is given by

3.2.4 Specific examples of enzyme mechanisms

Consider an example of the above analysis for a simple hydrolytic enzyme, arginase. Arginase catalyses the hydrolysis of arginine to ornithine and urea.

Although we will return to this enzyme a little later, in this instance let us ignore the fact that there are two products of the reaction and assume that the kinetics of arginase are well described by the Michaelis-Menten equation. Then it is simple to relate the unitary rate constants to the steady-state kinetic parameters.

Q: Determine a set of unitary rate constants that are consistent with the known steady-state parameters for a 1 ng L-1 solution of arginase assuming it to be a simple MichaelisMenten enzyme with a Km for arginine of 5.0 mmol L-1 and a turnover number of 4.5 x 103 s-1. The molecular mass of arginase is ~105,000 Da and it exists in solution as a trimer.

A: First, we use Eqn [3.8] to convert the gram concentration to a molar concentration of enzyme.

9.52381 x 10-12

By applying Eqn [3.8] we obtain the concentration of subunits in the trimer.

2.85714 x 10-11

Therefore, by using Eqns [3.6] and [3.7] we obtain the following values for the Michaelis-Menten parameters:

Finally, by arbitrarily selecting a value of 1 x 107 mol-1 L s-1 for k^we can use Solve with Eqns [3.6] and [3.7] to determine the values of k-1 and k2.

The answer to the question is that the unitary rate constants that are consistent with the values of the steady-state parameters are {k1 = 1 x 107 mol-1 L s 1, k-1 = 4.55 x 10 s-1, k2 = 4.5 x 103 s-1}.

Q: Use the unitary rate constants determined in the previous example, in conjunction with the relevant differential equations, to simulate a time course of the arginase-catalyzed reaction. Compare the results with those obtained by solving the time course using the Michaelis-Menten equation (Eqn [3.3]).

A: With the parameters and solution still defined from the last example and assuming that the initial concentration of arginine is 10 mM, we solve the time course by using the unitary rate constants and rate equations with the following input:

preSSTimecourse = NDSolve[{

a'[t] % - k1a[t] e [t] + k-1ea[t], ea' [t] % k1a[t] e[t] - (k-1 + k2) ea[t] , e'[t] % -k1a[t] e[t] + (k_1 + k2) ea[t] , p'[t] % k2 ea[t] , a[0] % a0, ea[0] % 0, e[0] % e0,p[0] % 0.0} /. solution, {a, ea, e, p} ,

Note the application of the /. solution operator to the list of differential equations. This introduces the values of k-1 and k2 that were obtained in the previous example; k1 is still an assigned value. We have also specified the option AccuracyGoal ^ 12 (see Section 1.6.4). The reason for this is that we are interested in the concentration of the product in the 'pre-steady-state' phase of the reaction that occurs within the first few milliseconds when the concentration of product is very low.

Similarly, we solve for the time course using the Michaelis-Menten rate equation (Eqn [3.3]).

ssTimecourse=

By graphing the two time courses we obtain Plot [

{Evaluate[p[t] /. preSSTimecourse], Evaluate[(a0 - a[t]) / . ssTimecourse]}, { t, 0.0, 1} ,

AxesLabel ^ {"Time (min)", "Concentration (M) " } , PlotRange -> All] ;

Concentration (M)

2x10 f

4x10 f

8x10 f

6x10 f

2x10 f

4x10 f

8x10 f

6x10 f

Time (min)

Figure 3.3. Time course of the arginase-catalyzed reaction. Comparing pre-steady-state reaction kinetics to Michaelis-Menten kinetics.

It is evident that under the conditions used and in the timescale considered, there is no perceptible difference between the pre-steady-state and steady-state solutions for the time course. However, if we plot just the first 100 ms of the timecourse,

{Evaluate[p[t] /. preSSTimecourse], Evaluate [(a0 - a[t]) / . ssTimecourse]} , {t, 0.0, 0.0001} ,

AxesLabel ^ {"", "Concentration (M)"}, PlotRange -> All];

Concentration (M)

2x10

4x10

6x10

8x10

2x10

4x10

6x10

8x10

0.000 02 0.00004 0.000060.00008 0.0001

Figure 3.4. Time course of the arginase-catalyzed reaction. Comparing pre-steady-state reaction kinetics (lower curve) to Michaelis-Menten kinetics (upper curve) within the first 100 ms.

Figure 3.4. Time course of the arginase-catalyzed reaction. Comparing pre-steady-state reaction kinetics (lower curve) to Michaelis-Menten kinetics (upper curve) within the first 100 ms.

Thus, we see a small time lag before the pre-steady-state time course has the same slope as that described by the solution of the steady-state equation.

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