In reality, it is uncommon to encounter enzymic reactions that can be simply described by the Michaelis-Menten mechanism. Those enzymes that appear most commonly to conform to this simple model are hydrolytic ones. As this name suggests water is a co-reactant with the substrate, so the enzyme has in reality two substrates and not one. Because the concentration of water in a reaction mixture is so large (55.55 mol L-1), relative to the other substrate(s), it is assumed to be constant and therefore it is not usual to include the concentration of water in the rate equation. On the other hand, it is not possible to ignore the fact that two products are formed in most hydrolytic reactions. A good example of this is again provided by arginase, an enzyme of the urea cycle that we first considered in Section 3.2.4.

The steady-state rate equation for any enzymic reaction mechanism can, in principle, be derived in the same way as that for the Michaelis-Menten mechanism (Section 2.3.2).

Q: Derive an expression for the steady-state rate equation for arginase in terms of its unitary rate constants.

A: The complete reaction mechanism of arginase is

where A, U, and O denote arginine, urea, and ornithine, respectively.

A rate equation for this mechanism can be derived by using the steady-state assumption described in Section 2.3.2. First we write expressions that describe the rate of change of the concentrations of all but one enzyme form.

Clear[k, Subscript] ;

eqn : = e'[t] % -ki ea + k_i ea - k-3 eo + k3 eo; eqn2 : = ea' [t] % k1 ea - k-1 ea - k2 ea;

By applying the steady-state assumption, the left-hand side of each equation is set equal to zero.

Expressions describing the steady-state concentrations of each of the enzyme forms can then be derived by solving the above simultaneous equations and the conservation of mass equation for the enzyme.

solution = Solve [{eqn1, eqn2 , e0 % e + ea + eo}, {e, ea, eo}]

o k_3 k_1 + o k_3 k2 + aki k2 + k_1 k3 + ak1 k3 + k2 k3 ' e0 (k_i + k2) k3

_a k1 (k_1 _ k3 ) _ (k_1 + k2) (_o k_3 _ ak1 _ k3) '

ea ^ - a k1 ( k _ 1 _ k 3 ) _ ( k _1 + k 2 ) (_ o k_ 3 _ a k _ k 3) = =

The rate of production of urea is given by varg : = u' [t] % k2 ea and hence the steady-state rate equation is varg / . solution

From the last question, the steady-state rate equation for arginase is v = dEL =_ki k2 h [A] [E] 0_

v dt k3 (k- 1 + k2) + k-3 (k_ 1 + k2) [O] + k1 (k2 + k3 ) [A] , [ ]

and the correspondences between the unitary rate constants and the steady-state parameters are readily derived.

Note the use of nomenclature to denote the forward catalytic steady-state rate constant, k£at, and its corresponding maximum velocity. Also, since there are Michaelis and inhibition constants for each reactant, it is necessary to use double subscripts such as Km,A and KiO, respectively.

Another way to derive steady-state rate equations is to use the RateEquation function that is fully described in Appendix 1. It automatically derives a steady-state rate equation for almost any enzyme mechanism with up to 14 enzyme forms. Before this function is used for the first time you must evaluate all the cells in Appendix 1 so that the appropriate .m file is created. This allows the function to be called using the following commands.

<< rateequationderiver RateEquationfrcrn, el]

loads in the function RateEquation derives the steady state rate equation for an enzyme mechanism defined in the rate constant matrix (rcm - see below and Appendix 1). The argument el is optional and is a list of user defined names for the enzyme forms of the reaction mechansim.

Deriving rate equations.

Q: Use the function RateEquation to derive the rate expression given in Eqn [3.18]. A: First it is necessary to load in the function RateEquation.

<< rateequationderiver

Then, the only 'thinking task' required of you is to set up the rate constant matrix based on the reaction scheme in Eqn [3.17]. The matrix is

Hence, in Mathematica the matrix is entered as rcmArginase = {{0, k1 a, k-3 o} , {k-1, 0 , k2 } , {k3 , 0, 0}} ;

The function RateEquation is applied as follows:

RateEquation[rcmArginase, {e, ea, eo}] The output is

Enzyme Distribution Functions e/eo = (k-i k3 + k2 k3 ) ea/eo = ak1 k3

Steady-State Rate Equations d[a]/dt =-aeok1k2k3/ Denominator d[o]/dt =aeok1k2k3/ Denominator

Denominator o (k-3 k-1 + k-3 k2) + k-1 k3 + k2 k3 + a (k1 k2 + k1 k3)

Thus, it can be seen that the functional form of the output is the same as Eqn [3.18], as required by the question.

3.4.3 Calculating a consistent set of unitary rate constants

Having determined the steady-state rate equation, as well as the relationships between the unitary rate constants and the steady-state kinetic parameters, it is possible to calculate a consistent set of unitary rate constants.

Q: Calculate a consistent set of unitary rate constants for the arginase mechanism shown in Eqn [3.17] by using the three well-known(8) values of the steady-state parameters for the enzyme, KmA, KijO and the turnover number.

A: Use the Mathematica function Solve as follows:

The known parameter values are(8)

kcat,f = 4.5 x 103; (*mol L-1 s-1*) Km,a = 5.0 x 10-3; (*mol L-1 *) Ki/o = 3.0 x 10-3; (*mol L-1 *)

Only three steady-state parameter values are known, so two of the five unitary rate constants must be assigned values. Select the two second-order rate constants and set them to 1/10th of the diffusion limit value (Section 3.3.1).

Now use Solve to determine the values of the unitary rate constants from the nonlinear algebraic equations that relate the steady-state parameters to the unitary rate constants (Section 3.4.1).

Km,a % k3 (k-1 + k2) / (k1 (k2 + k3)) , Ki,o % k3 / k-3} , {k-1 , k2 , k3 }]

Thus the set of unitary rate constants whose values are consistent with the values of the steady-state parameters are {k1 = 1 x 107 mol-1 L s-1, k-1 = 5.35 x 104 s-1, k2 = 5.29 x 103 s-1, k3 = 3 x 104 s-1, k-3 = 1 x 107 mol-1 L s-1}.

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