RateAB12 - 27 - 07 (in DeterminKinPars on Desktop)  [94]

Rapid - Equilibrium Rate Equations for A + B -> Products and Determination of Kinetic Parameters

Robert A. Alberty

Department of Chemistry, Massachusetts Institute of Technology, Cambridge, MA 03139 alberty@mit.edu

1.  Derivation the rate equation for ordered A + B -> products and calculations of kinetic parameters

1.1 Introduction

1.2  Calculation of kinetic parameters from velocities at {100, 100}, {1, 100}, and {100, 1}

1.3  Calculation of kinetic parameters from velocities at {150,150}, {.1,150}, {150,.1}

1.4  Calculation of kinetic parameters from velocities at {20,20},{5,20},{20,5}

2.  Justification of the recommendations for choices of substrate  concentrations for the ordered mechanism

3.  Derivation the rate equation for random A + B -> products and calculations of parameters

3.1  Introduction

3.2  Calculation of velocities at specified values of kinetic parameters and substrate concentrations  {100,100},{1,100},{100,1},{5,5}

3.3  Calculation of kinetic parameters from measured velocities for the random mechanism

3.4  Calculation of kinetic parameters from measured velocities at {100,100},{1,100},{100,1},{5,5}

3.5  Calculation of kinetic parameters from velocities at {150,150}, {1,150}, {150,1},{1,1}

3.6  Try a narrower range of substrate concentrations: {50,50}, {10,50}, {50,10},{10,10}.

4.  Justification of the recommendationns for substrate concentrations for the random mechanism

5  What happens when the wrong program is used?

5.1  Application of calckinparsrandAB to velocities for the ordered mechanism

6.  Competitive inhibition of ordered A + B -> products

6.1  Introduction

6.2  Program for the calculation of the competitive inhibition consatant

7.  Uncompetitive inhibition of ordered A + B -> products

7.1 Introduction

7.2  Derive the rate equation for uncompetitive inhibition

7.3  Program for uncompetitive inhibition

7.4  Determination of kinetic parameters

8.  Mixed inhibition of ordered A + B -> products

8.1 Introduction

8.2 Program for the determination of mixed inhibition constants (Subscript[K, Ic]) and (Subscript[K, Iu])

8.3 Determination of kinetic parameters

9.  Application of calckinparsordABIm to competitive inhibition data

10.  Application of calckinparsordABIm to uncompetitive inhibition data

11.  Discussion

12.  References

1.  Derivation the rate equation for ordered A + B -> products and calculations of kinetic parameters

1.1 Introduction

This package provides programs and calculations used in R. A. Alberty, Determination of Kinetic Parameters of Enzyme-Catalyzed Reactions with a Minimum Number of Velocity Determinations, J. Theor. Biol., 2008, in press.

Mathematica can be used to derive the rapid-equilibrium rate equation from the mechanism of an enzyme-catalyzed reaction.  The concentrations of reactants are represented by e, a, b, eq, eab because names starting with capital letters are operations in Mathematica.  The biochemical equilibrium constants are represented by kIA and kB.  The ordered mechanism is

E + A = EA	        kIA = e*a/ea												(1)
EA + B = EAB         kB = ea*b/eab												(2)
EAB -> products      v= (10^npH)*kf*eab												(3)

The velocity v of reaction is proportional to (10^npH), kf, and eab.  In (10^npH), the n is the number of hydrogen ions consumed in the rate-determining reaction.  When no hydrogen ions are consumed, n = 0.  If one hydrogen ion is consumed, n = -1, etc Alberty, 2007a,2007b, 2007c, 2008).
	In using biochemical thermodynamics to obtain the expression for eab, it is important to be careful about the number of equilibrium expressions that are used.  It is important that these reactions be independent; that is, none of them can be obtained by adding or subtracting other reactions in the set.  There is a simple test, and that is N' = C' + R', where N' is the number of reactants, C' is the number of components, and R' is the number of independent reactions (Alberty, 2006).  The number of kinetic parameters in this case is equal to the number of terms in the denominator of the rate equation.  The denominator terms arise from E, EA, and EAB.  It is also important to understand that the equilibrium composition that is calculated using thermodynamics satisfies all possible equilibrium constant expressions that can be written with the N' reactants.   
	To calculate the equilibrium concentration eab, there are two equilibrium equations and one conservation equation.  The number N' of reactants is 5 (a, b, ea, eab, and e), the number C' of components components is 3 (a, b, and e), and the number R' of  independent reactions is 2, so that N' = C' + R' is 5 = 3 + 2.  Note that there is only one way to form ea and one way to form eab in this mechanism.
	Solve[eqs,vars,elims] can be used to derive the expression for the equilibrium concentration of EAB.  This operation solves for the variables when the elims are eliminated.

Solve[{kIA==e*a/ea,kB==ea*b/eab,et==e+ea+eab},{eab},{e,ea}]

The rapid-equilibrium velocity of reaction is given by (10^npH)*kf*eab, where n is the number of hydrogen ions consumed in the rate-determining reaction.  When the expression for eab is multiplied by (10^npH)*kf,  (10^npH)*kf*et can be replaced by vfexp, the limiting velocity in the forward direction. The velocity of the forward reaction is given by

vordAB=((a b vfexp)/(a b+a kB+kB kIA));

An advantage of the derivation of the rate equation in Mathematica is that the rate equation is in computer-readable form.  The three kinetic parameters are calculated by solving 3 simultaneous equations.  As a trial, three velocities are calculated for specified vfexp, kIA, and kB and specified pairs of concentrations of the substrates by use of the ReplaceAll operation (/.x->).  Since there are 3 kinetic parameters, 3 velocities have to be determined.  The following pairs {a,b} of concentrations are used: {100,100}, {1,100}, and {100,1}.

vordAB/.vfexp->1/.kIA->5/.kB->20/.a->100/.b->100//N

vordAB/.vfexp->1/.kIA->5/.kB->20/.a->1/.b->100//N

vordAB/.vfexp->1/.kIA->5/.kB->20/.a->100/.b->1//N

The following program was written in Mathematica to calculate the values of vfexp, kA, and kB from three measured velocities at three pairs of substrate concentrations on the assumption that the mechanism is ordered (that is reactions 1-3), with A being bound first.

calckinparsordAB[v1_,a1_,b1_,v2_,a2_,b2_,v3_,a3_,b3_]:=Module[{},(*This programm calculates vfexp, kIA, and kB from three experimental velocities for A + B -> products at three pairs of substrate concentrations on the assumption that the mechanism is ordered, with A bound first.  The first velocity is at high [A] and high [B], the second velocity is at low [A] and high [B], and the third velocity is at high [A] and low [B].*)
Solve[{v1==a1*b1*vfexp/(a1*b1+a1*kB+kIA*kB),v2==a2*b2*vfexp/(a2*b2+a2*kB+kIA*kB),v3==a3*b3*vfexp/(a3*b3+a3*kB+kIA*kB)},{vfexp,kIA,kB}]]

1.2  Calculation of kinetic parameters from velocities at {100, 100}, {1, 100}, and {100, 1}

The program calckinparsordAB is used to calculate the kinetic parameters from the 3 velocities calculated above (now considered to be measured velocities) at {100,100}, {1,100}, and {100,1}.

calckinparsordAB[.826446,100,100,.4544545,1,100,.0454545,100,1]

line11={vfexp,kIA,kB}/.calckinparsordAB[.826446,100,100,.4544545,1,100,.0454545,100,1]

The right-hand sides of the simultaneous rate equations solved by the program are

vordAB/.a->100/.b->100//N

vordAB/.a->1/.b->100//N

vordAB/.a->100/.b->1//N

The values of the 3 kinetic parameters calculated using calckinparsordAB are correct, but the question is "How sensitive are they to experimental errors in the measured velocities and substrate concentrations?"

Introduce 5% errors in velocities, one at a time.

calckinparsordAB[1.05*.826446,100,100,.4544545,1,100,.0454545,100,1]

line12={vfexp,kIA,kB}/.calckinparsordAB[1.05*.826446,100,100,.4544545,1,100,.0454545,100,1]

calckinparsordAB[.826446,100,100,1.05*.4544545,1,100,.0454545,100,1]

line13={vfexp,kIA,kB}/.calckinparsordAB[.826446,100,100,1.05*.4544545,1,100,.0454545,100,1]

calckinparsordAB[.826446,100,100,.4544545,1,100,1.05*.0454545,100,1]

line14={vfexp,kIA,kB}/.calckinparsordAB[.826446,100,100,.4544545,1,100,1.05*.0454545,100,1]

TableForm[{line11[[1]],line12[[1]],line13[[1]],line14[[1]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3"},{"vfexp","kIA","kB"}}]

Calculate the effects of 5% errors in [A].

calckinparsordAB[.826446,1.05*100,100,.4544545,1.05*1,100,.0454545,1.05*100,1]

Calculate effects of 5% errors in [B].

calckinparsordAB[.826446,100,1.05*100,.4544545,1,1.05*100,.0454545,100,1.05*1]

Note that the effects of errors in concentrations are what you would expect.  They are independent of the choices of velocities.  The limiting velocity is not affected by errors in the substrate concentrations, but the two calculated Michaelis constants are each increased by 5%.  These calculations are summarized in Table 1 in the manuscript.

1.3  Calculation of kinetic parameters from velocities at {150,150}, {.1,150}, {150,.1}

This shows the effects of widening the range of substrate equations.  The 3 velocities are given by

vordAB/.vfexp->1/.kIA->5/.kB->20/.a->150/.b->150//N

vordAB/.vfexp->1/.kIA->5/.kB->20/.a->.1/.b->150//N

vordAB/.vfexp->1/.kIA->5/.kB->20/.a->150/.b->.1//N

The simultaneous rate equations solved by the program are

vordAB/.a->150/.b->150//N

vordAB/.a->.1/.b->150//N

vordAB/.a->150/.b->.1//N

Calculate the kinetic parameters.

calckinparsordAB[.878906,150,150,.128205,.1,150,.00481541,150,.1]

line21={vfexp,kIA,kB}/.calckinparsordAB[.878906,150,150,.128205,.1,150,.00481541,150,.1]

Introduce 5% errors in the 3 velocities, one at a time.

calckinparsordAB[1.05*.878906,150,150,.128205,.1,150,.00481541,150,.1]

line22={vfexp,kIA,kB}/.calckinparsordAB[1.05*.878906,150,150,.128205,.1,150,.00481541,150,.1]

calckinparsordAB[.878906,150,150,1.05*.128205,.1,150,.00481541,150,.1]

line23={vfexp,kIA,kB}/.calckinparsordAB[.878906,150,150,1.05*.128205,.1,150,.00481541,150,.1]

calckinparsordAB[.878906,150,150,.128205,.1,150,1.05*.00481541,150,.1]

line24={vfexp,kIA,kB}/.calckinparsordAB[.878906,150,150,.128205,.1,150,1.05*.00481541,150,.1]

TableForm[{line21[[1]],line22[[1]],line23[[1]],line24[[1]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3"},{"vfexp","kIA","kB"}}]

Note that the errors are a little smaller than for {100,100},{1,100},{100,1}.
Calculate the effects of 5% errors in [A].

calckinparsordAB[.878906,1.05*150,150,.128205,1.05*.1,150,.00481541,150,.1]

Calculate the effects of 5% errors in [B].

calckinparsordAB[.878906,150,1.05*150,.128205,.1,1.05*150,.00481541,150,1.05*.1]

These calculations are summarized in Table 2 of the manuscript.

1.4  Calculation of kinetic parameters from velocities at {20,20},{5,20},{20,5}

Try a narrower range of substrate concentrations.

vordAB/.vfexp->1/.kIA->5/.kB->20/.a->20/.b->20//N

vordAB/.vfexp->1/.kIA->5/.kB->20/.a->5/.b->20//N

vordAB/.vfexp->1/.kIA->5/.kB->20/.a->20/.b->5//N

Calculate the kinetic parameters.

calckinparsordAB[.444444,20,20,.333333,5,20,.166667,20,5]

line31={vfexp,kIA,kB}/.calckinparsordAB[.444444,20,20,.333333,5,20,.166667,20,5]

This is correct, as expected, but now calculate the effects of 5% errors in the measured velocities.

calckinparsordAB[1.05*.444444,20,20,.333333,5,20,.166667,20,5]

line32={vfexp,kIA,kB}/.calckinparsordAB[1.05*.444444,20,20,.333333,5,20,.166667,20,5]

calckinparsordAB[.444444,20,20,1.05*.333333,5,20,.166667,20,5]

line33={vfexp,kIA,kB}/.calckinparsordAB[.444444,20,20,1.05*.333333,5,20,.166667,20,5]

calckinparsordAB[.444444,20,20,.333333,5,20,1.05*.166667,20,5]

line34={vfexp,kIA,kB}/.calckinparsordAB[.444444,20,20,.333333,5,20,1.05*.166667,20,5]

TableForm[{line31[[1]],line32[[1]],line33[[1]],line34[[1]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3"},{"vfexp","kIA","kB"}}]

Calculate the effects of 5% errors in [A].

calckinparsordAB[.444444,1.05*20,20,.333333,1.05*5,20,.166667,1.05*20,5]

Calculate the effects of 5% errors in [B].

calckinparsordAB[.444444,20,1.05*20,.333333,5,1.05*20,.166667,20,1.05*5]

These effects are summarized in Table 3 in the published article.  When a narrower range of substrate concentrations are used, the kinetic parameters are determined less accurately.  The effects of errors in [A] and [B] are unchanged.

2.  Justification of the recommendations for choices of substrate  concentrations for the ordered mechanism

First consider the velocity for the ordered mechanism as a function of the concentrations of the substrates.  Since the velocity is a function of [A] and [B], a 3D plot can be used to show this dependence.  The rate equation used above is

vordAB

vord2=vordAB/.vfexp->1/.kIA->5/.kB->20

It is of interest to inquire into the effects that small changes in the concentrations of A and B have on the reaction rate.   This can be done by taking partial derivatives of vord2.  3D plots can be constructed to display vord2, dvord2/da, dvord2/db, and d(dvord2/da)db.

plotA2=Plot3D[Evaluate[vord2],{a,.0001,15},{b,.0001,60},PlotRange->{0,1},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(a)"];

plotB2=Plot3D[Evaluate[5*D[vord2,a]],{a,.0001,15},{b,.0001,60},PlotRange->{0,1.5},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(b)"];

plotC2=Plot3D[Evaluate[30*D[vord2,b]],{a,.0001,15},{b,.0001,60},PlotRange->{0,1.3},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(c)"];

plotD2=Plot3D[Evaluate[1000*D[vord2,a,b]],{a,.0001,15},{b,.0001,60},PlotRange->{-1,7},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(d)"];

fig1ord2=Show[GraphicsArray[{{plotA2,plotB2},{plotC2,plotD2}}]]

Fig. 1 (a) ν versus [A] and [B] for the ordered mechanism with (Subscript[V, fexp]) = 1.00, (Subscript[K, IA]) = 5.00, and (Subscript[K, B]) = 20.00.  (b) 10dν/d[A] for the ordered mechanism.  (c)  10dν/d[B] for the ordered mechanism.  (d)  1000d(dν/d[A]/db for the ordered mechanism.

(*Export["fig1ord2.tif",fig1ord2,ImageResolution->300,ImageSize->700]*)

Figure 1 (a) shows superimposed Michaelis plots at constant [A] and constant [B].  Figure 1 (b) shows that the velocity is most sensitive to [A] at high [B] and low [A].  Figure 1 (c) shows that the velocity is most sensitive to [B] at high [A] and low [B].  Figure 1 (d) shows that the velocity is most sensitive to [A] and [B] together at low [A] and low [B].  Figure 1 (d) will be impotant in considering the random mechanism.

3.  Derivation the rate equation for random A + B -> products and calculations of parameters

3.1  Introduction

The random mechanism is

E + A = EA	       kIA = e*a/ea												(4)
E + B = EB             kIB = e*b/eb												(5)
EA + B = EAB         kB = ea*b/eab												(6)
EAB -> products     v= kf*eab													(7)

To calculate the equilibrium concentration eab, there are three equilibrium equations and one conservation equation.  There are 6 reactants (a, b, ea, eb, eab, and e), 3 components (a, b, and e), and three biochemical equilibrium expressions.   N' = C' + R' is 6 = 3 + 3.  Note that there is only one way to form ea, one way to form eb, and one way to form eab in this mechanism.

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

The velocity vfordAB is proportional to eab, and the limiting velocityis given by vfexp = (10^npH)*kf*eab.

vrandAB=((a b vfexp kIB)/(b kB kIA+a b kIB+a kB kIB+kB kIA kIB))

As a trial, the velocity can be calculated for arbitrary vfexp, kIA, and kB and arbitrary pairs of concentrations of the substrates:

The velocities are readily calculated at {100,100},{1,100},{100,1},{5,5}.

vrandAB/.kIA->5/.kIB->10/.kB->20/.vfexp->1/.a->100/.b->100//N

vrandAB/.kIA->5/.kIB->10/.kB->20/.vfexp->1/.a->1/.b->100//N

vrandAB/.kIA->5/.kIB->10/.kB->20/.vfexp->1/.a->100/.b->1//N

vrandAB/.kIA->5/.kIB->10/.kB->20/.vfexp->1/.a->5/.b->5//N

These velocities are considered to be experimental velocities.

3.2  Calculation of kinetic parameters from measured velocities for the random mechanism

The following program was written in Mathematica to calculate the values of vfexp, kIA, kB, and kIB from four measured velocities at four pairs of substrate concentrations on the assumption that the mechanism is random.

calckinparsrandAB[v1_,a1_,b1_,v2_,a2_,b2_,v3_,a3_,b3_,v4_,a4_,b4_]:=Module[{},(*This programm calculates vfexp, kIA,kIB and kB from four experimental velocities for A + B -> products at four pairs of substrate concentrations on the assumption that the mechanism is random.  The first velocity is at high [A] and high [B], the second velocity is at low [A] and high [B], the third velocity is at high [A] and low [B], and the fourth velocity is at low [A] and low [B].*)
Solve[{v1==a1*b1*kIB*vfexp/(b1*kB*kIA+a1*b1*kIB+a1*kB*kIB+kB*kIA*kIB),v2==a2*b2*kIB*vfexp/(b2*kB*kIA+a2*b2*kIB+a2*kB*kIB+kB*kIA*kIB),v3==a3*b3*kIB*vfexp/(b3*kB*kIA+a3*b3*kIB+a3*kB*kIB+kB*kIA*kIB),v4==a4*b4*kIB*vfexp/(b4*kB*kIA+a4*b4*kIB+a4*kB*kIB+kB*kIA*kIB)},{vfexp,kIA,kIB,kB}]]

3.3  Calculation of kinetic parameters from measured velocities at {100,100},{1,100},{100,1},{5,5}

vrandAB/.a->100/.b->100/.vfexp->1/.kIA->5/.kB->20/.kIB->10//N

vrandAB/.a->1/.b->100/.vfexp->1/.kIA->5/.kB->20/.kIB->10//N

vrandAB/.a->100/.b->1/.vfexp->1/.kIA->5/.kB->20/.kIB->10//N

vrandAB/.a->5/.b->5/.vfexp->1/.kIA->5/.kB->20/.kIB->10//N

calckinparsrandAB[.763359,100,100,.0819672,1,100,.0452489,100,1,.0909091,5,5]

Introduce 5% errors in the velocities, one at a time.

calckinparsrandAB[1.05*.763359,100,100,.0819672,1,100,.0452489,100,1,.0909091,5,5]

calckinparsrandAB[.763359,100,100,1.05*.0819672,1,100,.0452489,100,1,.0909091,5,5]

calckinparsrandAB[.763359,100,100,.0819672,1,100,1.05*.0452489,100,1,.0909091,5,5]

calckinparsrandAB[.763359,100,100,.0819672,1,100,.0452489,100,1,1.05*.0909091,5,5]

Make a list of the numbers only.

values=calckinparsrandAB[.763359,100,100,.0819672,1,100,.0452489,100,1,.0909091,5,5]

{vfexp,kIA,kB,kIB}/.values

The following obtains the numbers without calculating values.

line1={vfexp,kIA,kB,kIB}/.calckinparsrandAB[.763359,100,100,.0819672,1,100,.0452489,100,1,.0909091,5,5]

line2={vfexp,kIA,kB,kIB}/.calckinparsrandAB[1.05*.763359,100,100,.0819672,1,100,.0452489,100,1,.0909091,5,5]

line3={vfexp,kIA,kB,kIB}/.calckinparsrandAB[.763359,100,100,1.05*.0819672,1,100,.0452489,100,1,.0909091,5,5]

line4={vfexp,kIA,kB,kIB}/.calckinparsrandAB[.763359,100,100,.0819672,1,100,1.05*.0452489,100,1,.0909091,5,5]

line5={vfexp,kIA,kB,kIB}/.calckinparsrandAB[.763359,100,100,.0819672,1,100,.0452489,100,1,1.05*.0909091,5,5]

To eliminate the outer curly brackets, the first element has to be selected with {{1}}.

line5[[1]]

{line1[[1]],line2[[1]],line3[[1]],line4[[1]],line5[[1]]}

Table of determinations of kinetic parameters for random A + B -> products from four velocity measurements at four pairs of substrate concentrations (a,b): {100,100}, {1,100}, {100,1},  {5,5}.

TableForm[{line1[[1]],line2[[1]],line3[[1]],line4[[1]],line5[[1]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3","1.05*v4"},{"vfexp","kIA","kB","kIB"}}]

This is Table 4 in the published article.

3.4  Calculation of kinetic parameters from velocities at {150,150}, {1,150}, {150,1},{1,1}

Increase the range of substrate concentrations and calculate the velocities.

vrandAB/.kIA->5/.kIB->10/.kB->20/.vfexp->1/.a->150/.b->150//N

vrandAB/.kIA->5/.kIB->10/.kB->20/.vfexp->1/.a->1/.b->150//N

vrandAB/.kIA->5/.kIB->10/.kB->20/.vfexp->1/.a->150/.b->1//N

vrandAB/.kIA->5/.kIB->10/.kB->20/.vfexp->1/.a->1/.b->1//N

calckinparsrandAB[.830258,150,150,.0847458,1,150,.0460123,150,1,.00763359,1,1]

line51={vfexp,kIA,kB,kIB}/.calckinparsrandAB[.830258,150,150,.0847458,1,150,.0460123,150,1,.00763359,1,1]

Introduce 5% errors in the velocities, one at a time.

calckinparsrandAB[1.05*.830258,150,150,.0847458,1,150,.0460123,150,1,.00763359,1,1]

line52={vfexp,kIA,kB,kIB}/.calckinparsrandAB[1.05*.830258,150,150,.0847458,1,150,.0460123,150,1,.00763359,1,1]

calckinparsrandAB[.830258,150,150,1.05*.0847458,1,150,.0460123,150,1,.00763359,1,1]

line53={vfexp,kIA,kB,kIB}/.calckinparsrandAB[.830258,150,150,1.05*.0847458,1,150,.0460123,150,1,.00763359,1,1]

calckinparsrandAB[.830258,150,150,.0847458,1,150,1.05*.0460123,150,1,.00763359,1,1]

line54={vfexp,kIA,kB,kIB}/.calckinparsrandAB[.830258,150,150,.0847458,1,150,1.05*.0460123,150,1,.00763359,1,1]

calckinparsrandAB[.830258,150,150,.0847458,1,150,.0460123,150,1,1.05*.00763359,1,1]

line55={vfexp,kIA,kB,kIB}/.calckinparsrandAB[.830258,150,150,.0847458,1,150,.0460123,150,1,1.05*.00763359,1,1]

TableForm[{line51[[1]],line52[[1]],line53[[1]],line54[[1]],line55[[1]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3","1.05*v4"},{"vfexp","kIA","kB","kIB"}}]

Effect of 5% error in [A]

calckinparsrandAB[.830258,1.05*150,150,.0847458,1.05*1,150,.0460123,1.05*150,1,.00763359,1.05*1,1]

Effect of 5 % error in [B]

calckinparsrandAB[.830258,150,1.05*150,.0847458,1,1.05*150,.0460123,150,1.05*1,.00763359,1,1.05*1]

The effects of these errors are shown in Table 5.  The velocity at {1,1} is pretty small, but the enzyme concentration can be increased.

3.5  Calculation of kinetic parameters from velocities at {50,50}, {10,50}, {50,10},{10,10}.

[Not included in MS.]

vrandAB/.kIA->5/.kIB->10/.kB->20/.vfexp->1/.a->50/.b->50//N

vrandAB/.kIA->5/.kIB->10/.kB->20/.vfexp->1/.a->10/.b->50//N

vrandAB/.kIA->5/.kIB->10/.kB->20/.vfexp->1/.a->50/.b->10//N

vrandAB/.kIA->5/.kIB->10/.kB->20/.vfexp->1/.a->10/.b->10//N

calckinparsrandAB[.609756,50,50,.384615,10,50,.294118,50,10,.2,10,10]

Introduce 5% errors in the velocities, one at a time.

calckinparsrandAB[1.05*.609756,50,50,.384615,10,50,.294118,50,10,.2,10,10]

calckinparsrandAB[.609756,50,50,1.05*.384615,10,50,.294118,50,10,.2,10,10]

calckinparsrandAB[.609756,50,50,.384615,10,50,1.05*.294118,50,10,.2,10,10]

calckinparsrandAB[.609756,50,50,.384615,10,50,.294118,50,10,1.05*.2,10,10]

This shows that when the substrate concentrations do not cover a wide range (with respect to the unknown Michaelis constants), less accurate kinetic parameters are obtained.  Thus the objective of the velocity measurements is to cover as wide a range of substrate concentrations as possible.

4.  Justification of the recommendationns for substrate concentrations for the random mechanism

The rate equation when vfexp = 1.00, kIA = 5, kB = 20, and kIB = 10 is the following:

vrandAB152010=vrandAB/.vfexp->1/.kIA->5/.kB->20/.kIB->10

plotI=Plot3D[Evaluate[vrandAB152010],{a,.0001,15},{b,.0001,60},PlotRange->{0,1},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(a)"];

plotJ=Plot3D[Evaluate[10*D[vrandAB152010,a]],{a,.0001,15},{b,.0001,60},PlotRange->{0,1.5},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(b)"];

plotK=Plot3D[Evaluate[30*D[vrandAB152010,b]],{a,.0001,15},{b,.0001,60},PlotRange->{0,1.3},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(c)"];

plotL=Plot3D[Evaluate[750*D[vrandAB152010,a,b]],{a,.0001,15},{b,.0001,60},PlotRange->{-1,7},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(d)"];

fig22rand=Show[GraphicsArray[{{plotI,plotJ},{plotK,plotL}}]]

(*Export["fig22rand.tif",fig22rand,ImageResolution->300,ImageSize->700]*)

Fig. 2 (a) ν versus [A] and [B] for the random mechanism with (Subscript[V, fexp]) = 1.00, (Subscript[K, IA]) = 5.00, (Subscript[K, IB]) = 10.00, and (Subscript[K, B]) = 20.00.  (b) 10dν/d[A] for the random mechanism.  (c)  30dν/d[B] for the random mechanism.  (d)  750d(dν/d[A]/db for the random mechanism.

Fig. 2 (d) makes clear why the fourth experimental velocity for the rndom mechanism should be measured at the lowest possible concentrations of A and B.  This figure also shows a distinguishing feature of the random mechanism.

5  What happens when the wrong program is used?

5.1  Application of calckinparsrandAB to velocities for the ordered mechanism

Earlier three velocities were calculated for the ordered mechanism using vordAB: 0.826446 at {100,100}, 0.454545 at {1,100}, and 0.0454545 at {100,1}.   These calculations are repeated here.

vordAB/.vfexp->1/.kIA->5/.kB->20/.a->100/.b->100//N

vordAB/.vfexp->1/.kIA->5/.kB->20/.a->1/.b->100//N

vordAB/.vfexp->1/.kIA->5/.kB->20/.a->100/.b->1//N

To use calckinparsrandAB, one more velocity is needed, and that is calculated at {5,5}.

vordAB/.vfexp->1/.kIA->5/.kB->20/.a->5/.b->5//N

Now calckinparsrandAB is used to calculate the kinetic parameters assuming the mechanism is random.

calckinparsrandAB[.826446,100,100,.454545,1,100,.0454545,100,1,0.111111,5,5]

This yields the correct values for kIA and kB, and it indicates that kIB is infinite.  This is in accord with the rate equation for the random mechanism that is

vrandAB

vrandAB/.a->5/.b->5

This rate equation can also be written as

((a b  vfexp)/(b kB kIA/kIB+a b+a kB+kB kIA))

Thus when kIB = ∞, this rate equation reduces to the rate equation for the ordered mechanism.

vordAB

Thus when the mechanism is ordered, the use of calckinparsrandAB shows that the mechanism is not random by giving a very large value for kIB, and it gives the correct values of vfexp, kIA, and kB for the ordered mechanism.  This is an important conclusion.

6.  Competitive inhibition of ordered A + B -> products

6.1  Introduction

When a competitive inhibitor I is present, the following reaction is added to the ordered mechanism in equations 1-3.

E + I = EI   kIc = e*i/ei													(8)

To calculate the equilibrium concentration eab, there are 3 equilibrium equations and one conservation equation.  There are 7 reactants (a, b, i, ei, ea, eab, and e), 4 components (a, b, i, and e), and so N' = C' + R' is 7 = 4 + 3.  Note that there is only one way to form ea, one way to form eab, and one way to form ei in this mechanism.  There are 4 kinetic parameters, and so 4 velocities have to be measured.  The rate equation is derived as follows:

Solve[{kIA==e*a/ea,kB==ea*b/eab,kIc==e*i/ei,et==e+ea+eab+ei},{eab},{e,ea,ei}]

To obtain the expression for the velocity, et is replaced with vfexp.

vordABIc=((a b vfexp kIc)/(i kB kIA+a b kIc+a kB kIc+kB kIA kIc))

The rate equation with specified kinetic parameters is

vordABIc/.vfexp->1/.kIA->5/.kB->20/.kIc->30

The 4 velocities can be measured at {a,b,i}: {100,100,0}, {1,100,0}, {100,1,0}, and {5,5,30}.

vordABIc/.vfexp->1/.kIA->5/.kB->20/.kIc->30/.a->100/.b->100/.i->0//N

vordABIc/.vfexp->1/.kIA->5/.kB->20/.kIc->30/.a->1/.b->100/.i->0//N

vordABIc/.vfexp->1/.kIA->5/.kB->20/.kIc->30/.a->100/.b->1/.i->0//N

vordABIc/.vfexp->1/.kIA->5/.kB->20/.kIc->30/.a->5/.b->5/.i->30//N

6.2  Program for the calculation of the competitive inhibition consatant

calckinparsordABIc[v1_,a1_,b1_,i1_,v2_,a2_,b2_,i2_,v3_,a3_,b3_,i3_,v4_,a4_,b4_,i4_]:=Module[{},(*This programm calculates vfexp, kIA, kB, and kIc from four experimental velocities for A + B -> products in the presence of a competitive inhibitor at four triplets of substrate concentrations and the inhibitor concentration on the assumption that the mechanism is ordered, with A bound first.  The first velocity is at high [A], high [B], and zero [I], the second velocity is at low [A], high [B], and zero [I], the third velocity is at high [A] and low [B], and zero [I], and the fourth velocity is at medium [A], medium [B], and moderate [I].*)
Solve[{v1==((a1 b1 kIc vfexp)/(i1 kB kIA+a1 b1 kIc+a1 kB kIc+kB kIA kIc)),v2==((a2 b2 kIc vfexp)/(i2 kB kIA+a2 b2 kIc+a2 kB kIc+kB kIA kIc)),
v3==((a3 b3 kIc vfexp)/(i3 kB kIA+a3 b3 kIc+a3 kB kIc+kB kIA kIc)),v4==((a4 b4 kIc vfexp)/(i4 kB kIA+a4 b4 kIc+a4 kB kIc+kB kIA kIc))},{vfexp,kIA,kB,kIc}]]

6.3  Determination of kinetic parameters for competitive inhibition

The kinetic parameters are calculated as follows:

calckinparsordABIc[.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,.076923,5,5,30]

Now calculate kinetic parameters when there are 5% errors in the four measured velocities one at a time.

calckinparsordABIc[1.05*.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,.076923,5,5,30]

calckinparsordABIc[.82645,100,100,0,1.05*.454545,1,100,0,.0454545,100,1,0,.076923,5,5,30]

calckinparsordABIc[.82645,100,100,0,.454545,1,100,0,1.05*.0454545,100,1,0,.076923,5,5,30]

calckinparsordABIc[.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,1.05*.076923,5,5,30]

Since there are rather large errors in kIc, raise the inhibitor concentration.

The 4 velocities can be measured at {a,b,i}: {100,100,0}, {1,100,0}, {100,1,0}, and {5,5,100}.

vordABIc/.vfexp->1/.kIA->5/.kB->20/.kIc->30/.a->5/.b->5/.i->100//N

calckinparsordABIc[.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,.044776,5,5,100]

Now calculate kinetic parameters when there are 5% errors in the four measured velocities one at a time.

calckinparsordABIc[1.05*.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,.044776,5,5,100]

calckinparsordABIc[.82645,100,100,0,1.05*.454545,1,100,0,.0454545,100,1,0,.044776,5,5,100]

calckinparsordABIc[.82645,100,100,0,.454545,1,100,0,1.05*.0454545,100,1,0,.044776,5,5,100]

calckinparsordABIc[.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,1.05*.044776,5,5,100]

These errors are a little smaller, and so try {100,100,0}, {1,100,0}, {100,1,0}, and {1,1,100}.   This is not used to make a Table in the article.

vordABIc/.vfexp->1/.kIA->5/.kB->20/.kIc->30/.a->1/.b->1/.i->100//N

calckinparsordABIc[.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,.00220103,1,1,100]

calckinparsordABIc[1.05*.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,.00220103,1,1,100]

calckinparsordABIc[.82645,100,100,0,1.05*.454545,1,100,0,.0454545,100,1,0,.00220103,1,1,100]

calckinparsordABIc[.82645,100,100,0,.454545,1,100,0,1.05*.0454545,100,1,0,.00220103,1,1,100]

calckinparsordABIc[.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,1.05*.00220103,1,1,100]

Make the last triplet {1,100,100} so that there are more enzymatic sites for the competitive inhibitor to bind to.

vordABIc/.vfexp->1/.kIA->5/.kB->20/.kIc->30/.a->1/.b->100/.i->100//N

calckinparsordABIc[.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,.180723,1,100,100]

line61={vfexp,kIc,kIA,kB}/.calckinparsordABIc[.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,.180723,1,100,100]

calckinparsordABIc[1.05*.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,.180723,1,100,100]

line62={vfexp,kIc,kIA,kB}/.calckinparsordABIc[1.05*.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,.180723,1,100,100]

calckinparsordABIc[.82645,100,100,0,1.05*.454545,1,100,0,.0454545,100,1,0,.180723,1,100,100]

line63={vfexp,kIc,kIA,kB}/.calckinparsordABIc[.82645,100,100,0,1.05*.454545,1,100,0,.0454545,100,1,0,.180723,1,100,100]

calckinparsordABIc[.82645,100,100,0,.454545,1,100,0,1.05*.0454545,100,1,0,.180723,1,100,100]

line64={vfexp,kIc,kIA,kB}/.calckinparsordABIc[.82645,100,100,0,.454545,1,100,0,1.05*.0454545,100,1,0,.180723,1,100,100]

calckinparsordABIc[.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,1.05*.180723,1,100,100]

line65={vfexp,kIc,kIA,kB}/.calckinparsordABIc[.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,1.05*.180723,1,100,100]

TableForm[{line61[[1]],line62[[1]],line63[[1]],line64[[1]],line65[[1]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3","1.05*v4"},{"vfexp","kIC","kIA","kB"}}]

Table 6 in the article summarizes this calculation of kinetic parameters with {1,100,100} as the last triplet.  Note that when the products are P and Q, adding them separately yields a competitive inhibition term just like I.

7.  Uncompetitive inhibition of ordered A + B -> products

7.1 Introduction

Uncompetitive inhibition adds the following reaction to the mechanism in equations 1-3.

EAB + I = EABI   kIu = eab*i/eabi												(9)

N' = C' + R' is the same as for competitive inhibition, and 4 velocities have to be determined.

7.2  Derive the rate equation for uncompetitive inhibition

Solve[{kIA==e*a/ea,kB==ea*b/eab,kIu==eab*i/eabi,et==e+ea+eab+eabi},{eab},{e,ea,eabi}]

To obtain the expression for the velocity, et is replaced with vfexp.

vordABIu=((a b vfexp kIu)/(a b i+a b kIu+a kB kIu+kB kIA kIu))

This function is used in writing the computer program.
The rate equation with specified kinetic parameters is

vordABIu/.vfexp->1/.kIA->5/.kB->20/.kIu->40

In studing uncompetitive inhibition, high [A] and high [B] have the advantage that there is a lot of EAB for I to bind on.  Therefore, use {100,100,0}, {1,100,0}, {100,1,0}, and {100,100,100}.

vordABIu/.vfexp->1/.kIA->5/.kB->20/.kIu->40/.a->100/.b->100/.i->0//N

vordABIu/.vfexp->1/.kIA->5/.kB->20/.kIu->40/.a->1/.b->100/.i->0//N

vordABIu/.vfexp->1/.kIA->5/.kB->20/.kIu->40/.a->100/.b->1/.i->0//N

vordABIu/.vfexp->1/.kIA->5/.kB->20/.kIu->40/.a->100/.b->100/.i->100//N

7.3  Program for uncompetitive inhibition

calckinparsordABIu[v1_,a1_,b1_,i1_,v2_,a2_,b2_,i2_,v3_,a3_,b3_,i3_,v4_,a4_,b4_,i4_]:=Module[{},(*This programm calculates vfexp, kIA, kB, and kIu from four experimental velocities for A + B -> products in the presence of an ucompetitive inhibitor at four triplets of substrate concentrations and the inhibitor concentration on the assumption that the mechanism is ordered, with A bound first.  The first velocity is at high [A], high [B], and zero [I], the second velocity is at low [A], high [B], and zero [I], the third velocity is at high [A] and low [B], and zero [I], and the fourth velocity is at high [A], high [B], and high [I].*)
Solve[{v1==((a1 b1 kIu vfexp)/(a1 b1 i1+a1 b1 kIu+a1 kB kIu+kB kIA kIu)),
v2==((a2 b2 kIu vfexp)/(a2 b2 i2+a2 b2 kIu+a2 kB kIu+kB kIA kIu)),
v3==((a3 b3 kIu vfexp)/(a3 b3 i3+a3 b3 kIu+a3 kB kIu+kB kIA kIu)),
v4==((a4 b4 kIu vfexp)/(a4 b4 i4+a4 b4 kIu+a4 kB kIu+kB kIA kIu))},{vfexp,kIA,kB,kIu}]]

7.4  Determination of kinetic parameters

The kinetic parameters are calculated as follows:

calckinparsordABIu[.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,.269542,100,100,100]

line71={vfexp,kIu,kIA,kB}/.calckinparsordABIu[.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,.269542,100,100,100]

calckinparsordABIu[1.05*.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,.269542,100,100,100]

line72={vfexp,kIu,kIA,kB}/.calckinparsordABIu[1.05*.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,.269542,100,100,100]

calckinparsordABIu[.82645,100,100,0,1.05*.454545,1,100,0,.0454545,100,1,0,.269542,100,100,100]

line73={vfexp,kIu,kIA,kB}/.calckinparsordABIu[.82645,100,100,0,1.05*.454545,1,100,0,.0454545,100,1,0,.269542,100,100,100]

calckinparsordABIu[.82645,100,100,0,.454545,1,100,0,1.05*.0454545,100,1,0,.269542,100,100,100]

line74={vfexp,kIu,kIA,kB}/.calckinparsordABIu[.82645,100,100,0,.454545,1,100,0,1.05*.0454545,100,1,0,.269542,100,100,100]

calckinparsordABIu[.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,1.05*.269542,100,100,100]

line75={vfexp,kIu,kIA,kB}/.calckinparsordABIu[.82645,100,100,0,.454545,1,100,0,.0454545,100,1,0,1.05*.269542,100,100,100]

TableForm[{line71[[1]],line72[[1]],line73[[1]],line74[[1]],line75[[1]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3","1.05*v4"},{"vfexp","kIu","kIA","kB"}}]

These results are summarized in Table 7.

8.  Mixed inhibition of ordered A + B -> products

8.1 Introduction

When a mixed inhibitor I is present, the following two reactions are added to the ordered mechanism:

E + I = EI                    kIc = e*i/ei  											(10)
EAB + I = EABI         kIu = eab*i/eabi											(11)

To calculate the equilibrium concentration eab, there are 4 equilibrium equations and one conservation equation.  There are 8 reactants (a, b, i, ei, ea, eabi, eab, and e), 4 components (a, b, i, and e), and so N' = C' + R' is 8 = 4 + 4.  Note that there is only one way to form ea, one way to form eab, one way to form ei, and one way to form eabi in this mechanism.  There are 5 kinetic parameters, and so 5 velocities have to be measured.  The rate equation is derived as follows:

Solve[{kIA==e*a/ea,kB==ea*b/eab,kIc==e*i/ei,kIu==eab*i/eabi,et==e+ea+eab+ei+eabi},{eab},{e,ea,ei,eabi}]

To obtain the expression for the velocity, et is replaced with vfexp.

vordABIm=((a b vfexp kIc kIu)/(a b i kIc+i kB kIA kIu+a b kIc kIu+a kB kIc kIu+kB kIA kIc kIu))

This equation is used in writing the program to calculate the kinetic parameters. 
The rate equation with specified kinetic parameters is

vordABIm/.vfexp->1/.kIA->5/.kB->20/.kIc->30/.kIu->40

The 5 velocities can be measured at {a,b,i}: {100,100,0}, {1,100,0}, {100,1,0}, {1,100,100}, and {100,100,100}.
						vfexp		kIA	   kB		kIc		kIu

vordABIm/.vfexp->1/.kIA->5/.kB->20/.kIc->30/.kIu->40/.a->100/.b->100/.i->0//N

vordABIm/.vfexp->1/.kIA->5/.kB->20/.kIc->30/.kIu->40/.a->1/.b->100/.i->0//N

vordABIm/.vfexp->1/.kIA->5/.kB->20/.kIc->30/.kIu->40/.a->100/.b->1/.i->0//N

Note these three velocities are the same because I is not present.

vordABIm/.vfexp->1/.kIA->5/.kB->20/.kIc->30/.kIu->40/.a->1/.b->100/.i->100//N

vordABIm/.vfexp->1/.kIA->5/.kB->20/.kIc->30/.kIu->40/.a->100/.b->100/.i->100//N

8.2 Program for the determination of kinetic parameters

calckinparsordABIm[v1_,a1_,b1_,i1_,v2_,a2_,b2_,i2_,v3_,a3_,b3_,i3_,v4_,a4_,b4_,i4_,v5_,a5_,b5_,i5_]:=Module[{},(*This programm calculates vfexp, kIA, kB, kIc, and kIu from four experimental velocities for A + B -> products in the presence of a mixed inhibitor at five triplets of substrate concentrations and the inhibitor concentration on the assumption that the mechanism is ordered, with A bound first.  The first velocity is at high [A], high [B], and zero [I], the second velocity is at low [A], high [B], and zero [I], the third velocity is at high [A] and low [B], and zero [I], the fourth velocity is at low [A], high [B], and high [I], and the fifth velocity is at high [A], high [B], and high [I].*)
Solve[{v1==((a1 b1 kIc kIu vfexp)/(a1 b1 i1 kIc+i1 kB kIA kIu+a1 b1 kIc kIu+a1 kB kIc kIu+kB kIA kIc kIu)),v2==((a2 b2 kIc kIu vfexp)/(a2 b2 i2 kIc+i2 kB kIA kIu+a2 b2 kIc kIu+a2 kB kIc kIu+kB kIA kIc kIu)),
v3==((a3 b3 kIc kIu vfexp)/(a3 b3 i3 kIc+i3 kB kIA kIu+a3 b3 kIc kIu+a3 kB kIc kIu+kB kIA kIc kIu)),v4==((a4 b4 kIc kIu vfexp)/(a4 b4 i4 kIc+i4 kB kIA kIu+a4 b4 kIc kIu+a4 kB kIc kIu+kB kIA kIc kIu)),v5==((a5 b5 kIc kIu vfexp)/(a5 b5 i5 kIc+i5 kB kIA kIu+a5 b5 kIc kIu+a5 kB kIc kIu+kB kIA kIc kIu))},{vfexp,kIA,kB,kIc,kIu}]]

8.3 Determination of inhibition constants (Subscript[K, Ic]) and (Subscript[K, Iu])

The kinetic parameters are calculated as follows:

calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.124481,1,100,100,.267142,100,100,100]

line81={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.124481,1,100,100,.267142,100,100,100]

calckinparsordABIm[1.05*.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.124481,1,100,100,.267142,100,100,100]

line82={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[1.05*.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.124481,1,100,100,.267142,100,100,100]

calckinparsordABIm[.82645,100,100,0,1.05*.45455,1,100,0,.045455,100,1,0,.124481,1,100,100,.267142,100,100,100]

line83={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[.82645,100,100,0,1.05*.45455,1,100,0,.045455,100,1,0,.124481,1,100,100,.267142,100,100,100]

calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,1.05*.045455,100,1,0,.124481,1,100,100,.267142,100,100,100]

line84={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,1.05*.045455,100,1,0,.124481,1,100,100,.267142,100,100,100]

calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,1.05*.124481,1,100,100,.267142,100,100,100]

line85={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,1.05*.124481,1,100,100,.267142,100,100,100]

calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.124481,1,100,100,1.05*.267142,100,100,100]

line86={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.124481,1,100,100,1.05*.267142,100,100,100]

TableForm[{line81[[1]],line82[[1]],line83[[1]],line84[[1]],line85[[1]],line86[[1]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3","1.05*v4","1.05*v5"},{"vfexp","kIA","kIc","kIu","kB"}}]

These results are summarized in Table 8.

9.  Application of calckinparsordABIm to competitive inhibition data

This program requires 5 velocities, and so the velocity at {100,100,100} is calculated using vordABIc.

vordABIc/.vfexp->1/.kIA->5/.kB->20/.kIc->30/.a->100/.b->100/.i->100//N

The velocities calculated earlier for competitive inhibition at {100,100,0}, {1,100,0}, {100,1,0}, and {1,100,100} are 0.82645, 0.454545, .0454545  and 0.180723.  With this fifth velocity 0.80429 at {100,100,100}, calckinparsordABIm can be used

calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.180723,1,100,100,0.80429,100,100,100]

line91={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.180723,1,100,100,0.80429,100,100,100]

This indicates that (Subscript[K, Iu]) is infinite.  The rate equation for mixed inhibition is the following

((a b vfexp kIc kIu)/(a b i kIc+i kB kIA kIu+a b kIc kIu+a kB kIc kIu+kB kIA kIc kIu))

that can be rearranged to clarify the role of (Subscript[K, Iu]).  Divide the numerator and the denominator of the rate equation for mixed inhibition with kIu.

((a b kIc vfexp)/(a b i kIc/kIu+i kB kIA+a b kIc+a kB kIc+kB kIA kIc))

When kIu is infinite, the rate equation for competitive inhibition is obtained.  The effects of 5% errors in the 5 velocities are as follows:

calckinparsordABIm[1.05*.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.180723,1,100,100,0.80429,100,100,100]

line92={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[1.05*.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.180723,1,100,100,0.80429,100,100,100]

calckinparsordABIm[.82645,100,100,0,1.05*.45455,1,100,0,.045455,100,1,0,.180723,1,100,100,0.80429,100,100,100]

line93={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[.82645,100,100,0,1.05*.45455,1,100,0,.045455,100,1,0,.180723,1,100,100,0.80429,100,100,100]

calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,1.05*.045455,100,1,0,.180723,1,100,100,0.80429,100,100,100]

line94={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,1.05*.045455,100,1,0,.180723,1,100,100,0.80429,100,100,100]

calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,1.05*.180723,1,100,100,0.80429,100,100,100]

line95={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,1.05*.180723,1,100,100,0.80429,100,100,100]

calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.180723,1,100,100,1.05*0.80429,100,100,100]

line96={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.180723,1,100,100,1.05*0.80429,100,100,100]

TableForm[{line91[[1]],line92[[1]],line93[[1]],line94[[1]],line95[[1]],line96[[1]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3","1.05*v4","1.05*v5"},{"vfexp","kIA","kIc","kIu","kB"}}]

This is Table 9 in the paper.  Thus 5% errors in the 5 measured velocities do not affect the conclusion that kIu is infinite.

10.  Application of calckinparsordABIm to uncompetitive inhibition data

This program requires 5 velocities, and so the velocity at {1,100,100} is calculated using vordABIu.

vordABIu/.vfexp->1/.kIA->5/.kB->20/.kIu->40/.a->1/.b->100/.i->100//N

The velocities calculated earlier for uncompetitive inhibition at {100,100,0}, {1,100,0}, {100,1,0}, and {100,100,100} are 0.826545, 0.454545, .0454545  and 0.269542.  With this fifth velocity 0.212766 at {1,100,100}, calckinparsordABIm can be used

calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.269542,100,100,100,0.212766,1,100,100]

line101={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.269542,100,100,100,0.212766,1,100,100]

This indicates that (Subscript[K, Ic]) is infinite.  The rate equation for mixed inhibition is the following

((a b vfexp kIc kIu)/(a b i kIc+i kB kIA kIu+a b kIc kIu+a kB kIc kIu+kB kIA kIc kIu))

that can be rearranged to clarify the role of (Subscript[K, Ic]).  Divide the numerator and the denominator of the rate equation for mixed inhibition with kIu.

((a b vfexp kIu)/(a b i+i kB kIA kIu/kIc+a b kIu+a kB kIu+kB kIA kIu))

When kIc is infinite, the rate equation for uncompetitive inhibition is obtained.

((a b vfexp kIu)/(a b i+a b kIu+a kB kIu+kB kIA kIu))

The effects of 5% errors in the 5 velocities are as follows:

calckinparsordABIm[1.05*.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.269542,100,100,100,0.212766,1,100,100]

line102={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[1.05*.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.269542,100,100,100,0.212766,1,100,100]

calckinparsordABIm[.82645,100,100,0,1.05*.45455,1,100,0,.045455,100,1,0,.269542,100,100,100,0.212766,1,100,100]

line103={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[.82645,100,100,0,1.05*.45455,1,100,0,.045455,100,1,0,.269542,100,100,100,0.212766,1,100,100]

calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,1.05*.045455,100,1,0,.269542,100,100,100,0.212766,1,100,100]

line104={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,1.05*.045455,100,1,0,.269542,100,100,100,0.212766,1,100,100]

calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,1.05*.269542,100,100,100,0.212766,1,100,100]

line105={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,1.05*.269542,100,100,100,0.212766,1,100,100]

calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.269542,100,100,100,1.05*0.212766,1,100,100]

line106={vfexp,kIA,kIc,kIu,kB}/.calckinparsordABIm[.82645,100,100,0,.45455,1,100,0,.045455,100,1,0,.269542,100,100,100,1.05*0.212766,1,100,100]

TableForm[{line101[[1]],line102[[1]],line103[[1]],line104[[1]],line105[[1]],line106[[1]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3","1.05*v4","1.05*v5"},{"vfexp","kIA","kIc","kIu","kB"}}]

Thus 5% errors in the 5 measured velocities do not affect the conclusion that kIc is infinite.  These results are summarized in Table 10.

	These inhibition calculations have been made for ordered A + B -> products, but the same calculations can be made for random A + B -> products, but the rate equations will each have one more term in the denominator.

11.  Discussion

The application of calckinparsrandAB to velocities for the ordered mechanism showed that (Subscript[K, IB]) is infinite; in other words, the mechanism does not have the reaction E + B = EB.  Similarly, the application of calckinparsABIm to velocities for competetive inhibition showed that (Subscript[K, Iu]) is infinite; in other words, the mechanism does not have the reaction EAB + I = EABI.  The application of calckinparsABIm to velocities for uncompetive inhibition showed that (Subscript[K, Ic]); in other words, the mechanism does not have the reaction E + I = EI.  This can be generalized by saying that a more general program can be used to determine whether an enzyme-catalyzed reaction A + B -> products is ordered or random.  A more general computer program can also be used to determine whether an inhibtor is competitive, uncompetitive, or mixed.  
	In using this method, it is important to use more than one set of substrate concentrations to be sure that the calculated values of the kinetic parameters are not changing with the choice of of substrate concentrations.  The fisher information matrix provides a more general way to estimate kinetic parameters (Van Trees, 1968).

12.  References

Alberty, R. A., 2006.  Biochemical thermodynamics: Applications of Mathematica, Wiley, Hoboken, N. J. 
Alberty, R. A., 2007a.  Two different ways that hydrogen ions are involved in the thermodynamics and rapid-equilibrium kinetics of the enzyme-catalysis of S = P and S + (Subscript[H, 2])O = P. Biophys. Chem., 128, 204-209.
Alberty, R. A., 2007b.  Three mechanisms and rapid-equilibrium rate equations for a type of reductase reaction.  Biophys. Chem., 131, 71-91.
Alberty, R. A., 2007c.  Effects of pH in rapid-equilibrium enzyme kinetics, J. Phys. Chem. B, 111, 14064-14068.
Alberty, R. A., 2008. Rapid-equilibrium rate equations for the enzymatic catalysis of A + B = P + Q over a Range of pH.  Biophys. Chem, 132, 114-126.
Alberty, R. A. Determination of Kinetic Parameters of Enzyme-Catalyzed Reactions with a Minimum Number of Velocity Measurements, J. Theor. Biol., in press.
Cook, P. F., Cleland, W. W., 2007.  Enzyme kinetics and mechanism, Garland Science, London and New York.
Cornish-Bowden, A., 2004.  Fundamentals of enzyme kinetics, 3rd ed. Portland Press, London.
Duggleby, R. G., 1979.  Experimental designs for estimating kinetic parameters for enzyme-catalyzed reactions.  J. Theor. Biol. 81, 672-684.
Van Trees, H. L. 1968. Detection, Estimation, and Modular Theory, Part 1. Wiley, New York.