klA = e*a/ea kIB = e*b/eb

kA = eb*a/eab v = kf*eab

K^]A = [E][A]/[EA]

tf

= [E][B]/[EB]

KA = [EB][A]/[EAB]

Vfexp = kf [E],

(6) (7) (8) (9)

AB is not involved, but note K

kinetic parameters K\

and Vf

. Another investigator could use mechanism II and report values for the kinetic

parame-ters for K\

and Vf

. This chapter makes the point that both investigators should report values for K\

and Vfexp. This is true at each pH, and so the pH dependencies of the five kinetic parameters can be determined using either

mechanism.

Solve[vars,eqs,elims] in Mathematica can be used to derive rapid-equilibrium rate equations. To calculate the equilib-rium concentration eab, there are three equilibequilib-rium equations and one conservation equation for each mechanism. Note that e, ea, and eb have to be eliminated in using Solve.

The rapid-equilibrium rate equation for mechanism I is derived as follows:

Solve[{klA == e * a / ea, kIB == e * b / eb, kB == ea * b / eab, e t == e + ea + eb + e a b }, {eab} , {e, ea, eb}]

a b e t kI B

ea b - > b k B kl A + a b kI B + a k B kI B + k B kl A kI B

Since v = kf*eab, it is necessary to replace et with vfexp = kf*et = kfet to obtain the rate equation. At a specified pH, vfexp = kfet.

a b vfex p kI B vrandABl =

b k B kl A + a b kI B + a k B kI B + k B kl A kI B a b kI B vfex p

b kB klA + a b kIB + a kB kIB + kB klA kIB

The expression for vrandABl can be converted to the usual form by dividing the numerator and denominator by a b kIB.

Divide the denominator by a*b*kIB, one term at a time.

{bkBkIA, a b k l B , a k B k l B , kBklAklB} / (a * b * kIB) , kB klA kB kB klA,

1 akIB b a b ^J

This yields the following rate equation for mechanism I:

kB klA kB kB klA vfexp / 11 + + — +

a kIB b a b vfexp

' / (

kB kB klA kB klÄ 1 + b a b a kIB

The rapid-equilibrium rate equation for mechanism II is derived as follows:

Solve [{kl A == e * a / e a , kIB == e * b / eb, kA == eb * a / eab, e t == e + ea + eb + e a b }, {eab} , {e, ea, eb}]

a b e t klA

eab a b klA + b kA klA + a kA kIB + kA klA kIB -

;})

Since v = kf*eab, it is necessary to replace et with vfexp to obtain the rate equation.

a b vfexp klA vrandAB2 =

a b klA + b kA klA + a kA kIB + kA klA kIB a b klA vfexp

a b klA + b kA klA + a kA kIB + kA klA kIB

The expression for vrandAB2 can be converted to the usual form as follows:

Divide the denominator by a*b*kIA one term at a time.

{abkIA, bkAkIA, akAklB, kAklAklB} / (a * b * klA) , kA kAklB kAklB,

{i, —, , )

L a b klA a b '

This yields the following rate equation for mechanism II:

/ I kA kAklB kAkIB\

vf exp / 1 + — + + ' \ a bkIA a b )

vfexp

, kA k A k l B k A k l B 1 + — + + a a b b klA

These two rate equations can be interconverted using klA kB = kIB kA.

■ 4.1.2 Use of the two mechanisms to calculate velocities

In order to make numerical calculations, the values of the four kinetic parameters in mechanism I are arbitraily taken to be vfexp = 1, klA = 5, kIB = 40, and kB = 20. kA is given by

kA = klA kB/klB = 5*20/40 = 2.5 (10) The two rate equations can be tested by arbitrarily choosing [A] = 10 and [B] = 20.

vrandABl / . vfexp -» 1 / . klA -» 5 / . kIB -► 40 / . k B - » 2 0 / . a-» 1 0 / . b -» 20 / / N 0.363636

vrandAB2 / . vfexp -» 1 / . klA -> 5 / . kIB -> 40 / . kA -> 2.5 / . a -» 10 / . b - > 2 0 / / N 0.363636

When mechanism I is used to interpret velocities for random A + B -» products, the values of vfexp, klA, kIB, and kB are obtained directly, and they can be used to calculate kA = klA kB/klB. This calculation should be made, and the value of kA should be reported because it is the equilibrium constant for EB + A = EAB. When more complicated enzyme-catalyzed reac-tions are studied, more equilibrium constants can be obtained in this way. For example, when A + B + C -» products has a completely random mechanism, seven equilibrium constants can be obtained using the rapid-equilibrium rate equation, but five more equilibrium constants can be obtained using equations like kA kIB = klA kB (see Chapter 6 and [34]).

There are two specialized forms of the rate equations that are useful; one is obtained by specifying the kinetic parameters, and the other is obtained by specifying substrate concentrations. When the values of the kinetic parameters are specified, the velocity is a function of the substrate concentrations. To demonstrate this using mechanism I, the following values are arbitrarily assigned to the four kinetic parameters: vfexp = 1, klA = 5, kB = 20, and kIB = 40. The rate equation is given by

vrandABl / . vfex p - » 1 / . kl A - » 5 / . k B - » 2 0 / . kI B - » 4 0 40 a b

4000 + 80 0 a + 100 b + 4 0 a b

This form of the rate equation will be used in this section to make plots of velocities for various [A] and [B].

The other form of the rate equation is obtained by substitution of specified values of a and b.

vrandABl / . a - > 10 0 / . b - » 8 0 800 0 kI B vfex p

80 k B kl A + 800 0 kI B + 10 0 k B kI B + k B kl A kI B

Since vrandABl/. vfexp -> 1 /. klA -» 5 /. kB -> 20 /. kIB -> 40 is a function of [A] and [B], the velocity can be plotted versus [A] and [B]. More information about the shape of this surface can be obtained by taking partial derivatives with respect to [A] and [B], and the mixed second derivative with respect to [A] and [B]. Plots of these derivatives are useful in choosing pairs of substrate concentrations {[A],[B]} to use in estimating the kinetic parameters with the minimum number of velocity measure-ments.

piot i = Plot3 D [Evaluat e [vrandAB l / . vfex p - » 1 / . kl A - » 5 / . k B - » 2 0 / . kI B - > 40 ] , {a , .0001 , 15} , {b , .0001 , 60} , PlotRang e - » {0 , 1} , Viewpoint- » {-2 , - 2 , 1} ,

AxesLabel - » {"[A]" , "[B]" , " " } , Lightin g - > {{"Ambient" , GrayLevel[1]}} , PlotLabe l - > "v"] ; plot J = Plot3 D [Evaluat e [1 0 * D [vrandAB l / . vfex p - » 1 / . kl A - » 5 / . k B - > 2 0 / . kI B -»40 , a ] ] ,

{a , .0001 , 15} , {b , .0001 , 60} , PlotRange- » {0 , 1.5} , Viewpoint- » {-2 , - 2 , 1} , AxesLabel- » {"[A]" , " [B ] " , " " } ,

Lightin g - > {{"Ambient" , GrayLevel[1]}} , PlotLabe l - > "10dv/d[A]"] ;

plot K = Plot3 D [Evaluat e [3 0 * D [vrandAB l / . vfex p - > 1 / . kl A - » 5 / . k B - » 2 0 / . kI B -»40 , b ] ] , {a , .0001 , 15} , {b , .0001 , 60} , PlotRange- . {0 , 1.3} ,

Viewpoint- » {-2 , - 2 , 1} , AxesLabel- » {"[A]" , " [B ] " , " " } ,

Lightin g - > {{"Ambient" , GrayLeve l [ 1]} } , PlotLabe l - > "30dv/d[B]"] ;

plot L = Plot3D[Evaluate[75 0 *D[vrandAB l / . vfex p - » 1 / . kl A - » 5 / . k B - » 2 0 / . kI B - 40 , a , b ] ] , {a , .0001 , 15} , {b , .0001 , 60} , PlotRange- » {-1 , 7} ,

Viewpoin t - { - 2 , - 2 , 1 } , AxesLabe l - » { " [A ] " , " [B ] " , " " } ,

Lightin g - > {{"Ambient" , GrayLevel[1]}} , PlotLabe l - > "750d(dv/d[A]/db"] ;

fig22ran d = GraphicsArray[{{pioti , plotJ} , {plotK , plotL}} ]

v 10dv/d[A ]

30dv/d[B] 750d(dv/d[A]/db

00 ω

Fig. 4.1 (a) v versus [A] and [B] for random mechanism I with V

= 1.00, K

10dv/d[A]. (c) 30dv/d[B]. (d) 750d(dv/d[A]/db.

The plot of 10dv/d[A]shows that the velocity is most sensitive to [A] at high [B] and low [A]. The plot of 10dv/d[B]

shows that the velocity is most sensitive to [B] at high [A] and low [B]. The plot of 1000d(dv/d[A]/d[B] shows that the velocity

is most sensitive to [A] and [B] together at low [A] and low [B]. d(dv/d[A]/db shows a distinguishing feature of the random

mechanism, as can be seen by comparing this figure with Fig. 3.6 in the preceeding chapter.