Enzymes are often studied by using a range of concentrations of various substrates, products, and inhibitors in order to determine the reaction mechanism. The choice of particular experiments is made on the basis of the well-trodden path of standard enzyme kinetic practice, as described in many authoritative texts.(4-7)

As noted in Section 3.4.2, the function RateEquation provides a means of rapidly deriving the steady-state rate equations for virtually any enzyme mechanism. But for the present problem we must regroup the various terms involving unitary rate constants in the rate equation to couch the rate equation on a form that has the equivalent Michaelis-Menten constants, inhibition constants, and Fmax parameters. For arginase (Section 3.4.1) the relationships between the unitary rate constants and the steady-state parameters were readily apparent. However, for many enzyme mechanisms this is not the case.

The Vmax of an enzymic reaction is simply the rate obtained when the substrate concentration is (infinitely!) high. This situation ensures that the active site(s) of the enzyme are saturated with the substrate. When choosing a range of substrate concentrations for use in kinetic experiments the rule of thumb that is used is that "the highest substrate concentration must be at least 5 times the Km value" for the particular substrate. Recall that the Km is defined as the substrate concentration that gives half the maximum velocity. Then at the top of this range the enzyme will be operating at approximately 5/6 th of its Vmax.

For enzymes with two or more co-substrates, Vmax is the velocity when each co-substrate is at a saturating concentration, and the products are all at zero concentration. The corresponding Km values are now defined as the substrate concentration that gives the half-maximum velocity when all the other co-substrates are at saturating concentrations.

The biochemical literature is replete with kinetic data on enzymes with the maximal velocities often expressed in terms of the turnover number(s), and the steady-state parameters given as Michaelis constants and various types of inhibition constant. The challenge facing the metabolic modeler is to match these values with the appropriate terms in the rate equation that describes the appropriate enzyme mechanism. Hence, it is useful to have an automatic procedure for deriving the expressions for these parameters. This process is shown in the next example by using another urea cycle enzyme, ornithine carbamoyl transferase.

Q: Derive the steady-state rate equation and hence expressions for Vmax in the forward and reverse directions, as well as the respective Km values for each substrate, in the ornithine carbamoyl transferase reaction.

A: The reaction mechanism for this urea cycle enzyme is(8L

where CP, O, C, and P denote carbamoyl phosphate, ornithine, citrulline, and orthophosphate, respectively. There are two important points to note about this mechanism. First, the formation of a ternary complex occurs between the enzyme and both substrates; and second, the substrates and products bind/release to/from the enzyme in a specific order. These two features lead to the definition of this mechanism as a compulsory-ordered ternary-complex one.

We derive the overall steady-state expression for the reaction mechanism of Eqn [3.22] in a manner similar to that used in the example in Section 3.4.2. We begin by setting up the rate constant matrix and then using RateEquation, as follows:

Clear[Subscript];

{k_1, 0, k2o, 0}, {0, k-2, 0, k3}, {k4 , 0, k-3 c, 0}};

RateEquation[rcmOCT, {e, ecp, eocp, ep}]

The output is

RateEquation[rcmOCT, {e, ecp, eocp, ep}]

Enzyme |
Distribution Functions |

e/eo = |
(c k-3 k-2 k-1 + k-2 k-1 k4 + k-1 k3 k4 + o k2 k3 k4 ) |

ecp/eo |
= (cp k-4 k-3 k-2 + ccp k-3 k-2 k1 + cp k-2 k1 k4 + cp k1 k3 k4 ) |

eocp | |

/eo = |
(c p k-4 k-3 k-1 + cop k-4 k-3 k2 + ccpo k-3 k1 k2 + cp o k1 k2 k4 ) |

ep/eo = |
(p k-4 k-2 k-1 + p k-4 k-1 k3 + op k-4 k2 k3 + cp o k1 k2 k3 ) |

Steady-State Rate Equations d[cp]/dt =eo (cpk_4 k_3 k_2 k_i - cp o ki k2 k3 k4)/ Denominator d[p]/dt =eo (_ cpk_4k_3 k_2 k_1 + cp o k1 k2 k3 k4)/ Denominator d[o]/dt =eo (cpk_4 k_3 k_2 k_1 _ cp o k1 k2 k3 k4)/ Denominator d[c]/dt =eo (_ cpk_4k_3 k_2 k_1 + cp o k1 k2 k3 k4)/ Denominator

Denominator c k_3 k_2 k_1 + p (k_4 k_2 k_1 + c (k_4 k_3 k_2 + k _4 k_3 k_1 ) +

k_4 k_1 k3 + o (c k_4 k_3 k2 + k_4 k2 k3)) + k_2 k_1 k4 + k_1 k3 k4 + o k2 k3 k4 + cp (c k_3 k_2 k1 + k_2 k1 k4 + k1 k3 k4 + o (c k_3 k1 k2 + k1 k2 k3 + k1 k2 k4))

We can now apply the Limit function in various ways to determine the steady-state parameters.

Limitfexpr, x -> xo ] finds the limiting value of expr when x approaches xo

We start by choosing one of the rate expressions in the output generated above; it is usual practice to select an expression that describes the rate of change of concentration of a product (since we usually define the rate of product formation to be positive). We convert this particular expression to an input one by cutting and pasting it into an input Cell:

vel = (-cpk_4 k-3 k-2 k-i + cp o ki k2 k3 k4) / denom; denom =

c k-3 k-2 k-i + p (k-4 k-2 k-i + c (k-4 k-3 k-2 + k-4 k-3 k-i) + k-4 k-i k3 + o (c k-4 k-3 k2 + k-4 k2 k3)) + k-2 k-i k4 + k-i k3 k4 + o k2 k3 k4 + cp (c k-3 k-2 ki + k-2 ki k4 + ki k3 k4 + o (c k-3 ki k2 + ki k2 k3 + ki k2 k4)) ;

Now derive the expression for Fmax for the forward direction. This entails taking the limit of the rate equation as the two substrate concentrations go to infinity while also setting the product concentrations to zero.

Vmaxfl = Limit [vel, o ^Infinity]; Vmaxf2 = Limit[Vmaxfl, cp ^ Infinity]; Vmax,f = Vmaxf2 / . c ^ 0;

Print["Vmax in forward direction = ", Vmaxf]

The expression for the Km of ornithine is obtained by (1) solving the rate equation for the value of the ornithine concentration that yields a rate of Vmax/2; (2) setting the product concentrations to zero; and (3) taking the limit as the co-substrate(s) go to infinity. The analysis uses the powerful symbolic capability of Mathematica for this otherwise very tedious and error-prone task.

Km for ornithine - 1

Similarly, we obtain expressions for the other steady-state parameters. For example, Vmax for the reverse direction is obtained in the same manner as for the forward direction but with the products now viewed as substrates.

Vmaxr1 = Limit [vel, c ^Infinity]; Vmaxr2 = Limit [Vmaxr1, p ^ Infinity] ; Vmax,r = Vmaxr2 / . o ^ 0;

Print ["Vmax in reverse direction =", V^x^] ;

Vmax in reverse direction

The other Km expressions are determined as follows:

r Vmax,f 1

Print["Km for carbamoyl phosphate = ", Km/Cp] ;

kmcit2 = kmcitl / • {o 0 0, cp 0 0} ; Km,c = c / . Limit [kmcit2 , p 0 Infinity] ; Print["Km for citrulline = ", Km,c] ;

kmpi2 = kmpil / • {o 0 0, cp 0 0} ; Km,p = p / • Limit[kmpi2, c 0 Infinity] ; Print["Km for phosphate = ", K^p] ;

Km for citrulline = {

Having derived the relationships between the unitary rate constants and the experimentally measurable steady-state parameters, a consistent set of unitary rate constants can be calculated (see next worked example).

Q: Calculate a consistent set of unitary rate constants for the ornithine carbamoyl transferase mechanism shown in Eqn [3.22]. Use the three known values of the steady-state parameters and the equilibrium constant.(8)

A: In this mechanism there are eight unitary rate constants but only three well-known steady-state parameter values, plus the equilibrium constant of the overall reaction. Therefore it is necessary to assume values of four of the unitary constants. As in the example in Section 3.4.3, the program is also based on the Solve function.

The known parameter values are(8)

kcat,f = 1.4 x 103 ; Km,cp = 8.1 x 10-5 ; Km,o = 9.0 x 10-4 ; Keq = 1.0 x 105;

and we assume the following 'reasonable' values for four of the unitary rate constants:

k-2 = 1.0 x 103 ; k3 = 3.0 x 103 ; k-3 = 9.0 x 104; k-4 = 5.0 x 105;

We now Solve for the unitary rate constants:

Solve[{kcat,f == --— , Kn.cp == ——-— , Kn.o == ———-——

Keq == k1 k2 k3 k4 / (k-1 k-2 k-3 k-4)} , {k1 , k-1 , k2 , k4<]

{{k_1 ^ 62.7343, k1 ^ 1.7284 x 107, k2 ^ 2.07407 x 106, k4 ^ 2625.}}

In conclusion, the unitary rate constants that are consistent with the values of the steady-state parameters are {k1 = 1.73 x 107 mol-1 L s-1, k-1 = 62.73 s-1, k2 = 2.07 x 106mol-1 L s-1, k3= 3 x 103 s-1, k-3 = 9 x 104 mol-1 L s-1, k4 = 2.63 x 103 s-1, and k-4 = 2.07 x 106 mol-1 L s-1}. These are used in a model of the urea cycle that is described in Section 3.8.

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