KinAB234-30-08.nb (in DeterminKinPar on Desktop)  [97]

Determination of the Rapid-Equilibrium Kinetic Parameters for the Enzyme-Catalyzed Reaction A + mB -> Products with the Minimum Number of Velocity Measurements

Robert A. Alberty
Department of Chemistry
MIT 6-215
77 Mass. Ave.
Cambridge, MA 02139
alberty@mit.edu

1.  Determination of kinetic parameters for A + 2B -> products

1.1  Ordered mechanism for A + 2B -> products when A is bound first and the two B molecules are bound by EA in a single reaction

1.1.1  Derivation of the rate equation

1.1.2  Determination of kinetic parameters with the minimum number of velocity measurements

1.2  Ordered mechanism for A + 2B -> products when the two B molecules are bound first by E in a single reaction and then A is bound

1.2.1  Derivation of the rate equation

1.2.2  Determination of kinetic parameters with the minimum number of velocity measurements

1.3  Random mechanism for A + 2B -> products

1.3.1  Derivation of the rate equation

1.3.2  Determination of kinetic parameters with the minimum number of velocity measurements

1.3.3  Application of the random program to velocities for the ordered mechanism when A is bound first

1.4  Alternative random mechanism for A + 2B -> products

1.4.1  Derivation of the alternative rate equation

1.4.2  Determination of kinetic parameters with the minimum number of velocity measurements

1.4.3  Application of the calckinpars141alt to velocities for the ordered mechanism when 2B is bound before A

2.  Determination of kinetic parameters for A + 3B -> products

2.1  Ordered mechanism for A + 3B -> products when A is bound first and the three B molecules are bound by EA in a single reaction

2.1.1  Derivation of the rate equation

2.1.2  Determination of kinetic parameters with the minimum number of velocity measurements

2.2  Ordered mechanism for A + 3B -> products when the three B molecules are bound by E in a single reaction

2.2.1  Derivation of the rate equation

2.2.2  Determination of kinetic parameters with the minimum number of velocity measurements

2.3  Random mechanism for A + 3B -> products

2.3.1  Derivation of the rate equation

2.3.2  Determination of kinetic parameters with the minimum number of velocity measurements

2.3.3  Application of random program calckinpars231 to velocities v211fab for the ordered mechanism when A is bound first

2.4  Alternative random mechanism for A + 3B -> products

2.4.1  Derivation of the rate equation

2.4.2  Determination of kinetic parameters with the minimum number of velocity measurements

2.4.3  Application of the random program calckinpars231alt to velocities v221ab for the ordered mechanism when 3B is bound first

3.  Discussion

4.  References

1.  Determination of kinetic parameters for A + 2B -> products

1.1  Ordered mechanism for A + 2B -> products when A is bound first and the two B molecules are bound by EA in a single reaction

There are many enzyme-catalyzed reactions of the type A + mB -> products, but their kinetics have not received much attention.  These reactions are generally oxidoreductase reactions, and m can be as large as 6 or more [1].  Reductase reactions may consume large numbers of hydrogen ions, and so their kinetics may involve very large pH effects.  Since the rate equations may be quite complicated, it is a good idea to start with the rapid-equilibrium assumption because this yields the simplest possible rate equations.  The discussions here are on reactions with m = 2, and m = 3.  The apparent equilibrium constants and changes in the binding of hydrogen ions are known as functions of pH and ionic srength for a dozen reactions A + 2B -> products, and two reactions A + 3B -> products [2].  These two types of reactions are each discussed for ordered mechanisms in which A is bound first, mB is bound first, and for the random mechanism.  The effects of pH on these reactions have been described [3-6].  This package backs up the article R. A. Alberty, "Determination of rapid-equilibrium kinetic parameters for the enzyme-catalyzed reactions A + 2B -> products and A + 3B -> products with the minimum number of velocity measurements."

1.1.1  Derivation of the rate equation

The rapid-equilibrium mechanism has 2 equilibrium expressions, one expression for the velocity, and 3 kinetic parameters

E + A = EA		kA = e*a/ea                  kA = 10   
EA + 2B =(Subscript[EAB, 2])	(kB^2) = ea*(b^2)/eab2         kB = 20
(Subscript[EAB, 2]) -> products	v = (10^npH)kf*eab2

Note that kB has the units of a concentration.  
	In deriving the rate equation for a rapid-equilibrium mechanism, it is necessary to be careful about the number of equilibrium expressions used in calculating the equilibrium concentration of the enzyme-substrate complex that yields products.  The reactions that are assumed to be in equilibrium must be independent.  A convenient way to check on this is to use N' = C' + R', where N' is the number of reactants, C' is the number of components (the things that are conserved), and R' is the number of independent reactions [2].  For this mechanism, N' = 5, C' = 3 (E, A, B), R' = 2, so that 6 = 3 + 3.  This mechanism involves 4 kinetic parameters, and so 4 velocities have to be measured.

Solve[{kA==e*a/ea,(kB^2) ==ea*(b^2)/eab2,et==e+ea+eab2},{eab2},{e,ea}]

The reaction velocity is obtained by multiplying eab2 with (10^npH)*kf.

((a (b^2) et*(10^pH)*kf)/(a (b^2)+a (kB^2)+kA (kB^2)));

Replace (10^pH)*kf*et with vfexp.

v111=((a (b^2) vfexp)/(a (b^2)+a (kB^2)+kA (kB^2)))

This is the general equation for the velocity that is used in the computer program calckinpars111 to calculate 3 kinetic parameters from 3 measured velocities.  When the kinetic parameters are specified, the function of [A] and [B] is given by v111fab.  To demonstrate the calculation of kinetic parameters from measured velocities, the following values are specified for the kinetic parameters: vfexp = 1, kA = 10, and kB =20.

v111fab=v111/.kA->10/.kB->20/.vfexp->1

This rate equation is used to calculate velocities at selected pairs of substrate concentrations.  An overview of the velocities for this mechanism is obtained by plotting v111fab versus [A] and [B].

plot1=Plot3D[v111fab,{a,0,50},{b,0,100},PlotRange->{0,1},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(a)"];

plot2=Plot3D[Evaluate[5*D[v111fab,a]],{a,.0001,50},{b,.0001,100},PlotRange->{0,1},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(b)"];

plot3=Plot3D[Evaluate[D[v111fab,b]],{a,.0001,50},{b,.0001,100},PlotRange->{0,.04},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(c)"];

plot4=Plot3D[Evaluate[D[v111fab,a,b]],{a,.0001,50},{b,.0001,100},PlotRange->{-.01,.05},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(d)"];

fig1=Show[GraphicsArray[{{plot1,plot2},{plot3,plot4}}]]

Fig. 1.  (a) Plot of v111fab versus [A] and [B].  Notice the sigmoid character of the curves at constant [A] and variable [B].  (b) Plot of 5*dv111fab/d[A] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [A] when [A] is low and [B] is high.  (c)  Plot of dv111fab/d[B] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [B] when [B] is about equal to kB and [A] is high.  (d)  Plot of (d^2)v111fab/d[A]d[B] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [A] and [B] together when [B] is high and [A] is low.  This will be important when the random mechanism is considered in Section 1.3.

1.1.2  Determination of kinetic parameters with the minimum number of velocity measurements

The kinetic parameters for this mechanism are obtained by solving 3 equations for 3 measured velocities.  The following program has been written to make it convenient to introduce the measured velocities and the pairs of concentrations {[A],[B]} that yielded these velocities.  The choice of substrate concentrations has been discussed in two previous papers.

calckinpars111[v1_,a1_,b1_,v2_,a2_,b2_,v3_,a3_,b3_]:=Module[{},(*This program calculates vfexp, kA, and kB from three experimental velocities for A + 2B -> products at three pairs of substrate concentrations on the assumption that A is bound first and two B molecules are bound to EA at the same time.  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 medium [B].*)
Solve[{v1==a1*b1^2*vfexp/(a1*b1^2+a1*kB^2+kA*kB^2),v2==a2*b2^2*vfexp/(a2*b2^2+a2*kB^2+kA*kB^2),v3==a3*b3^2*vfexp/(a3*b3^2+a3*kB^2+kA*kB^2)},{vfexp,kA,kB}]]

Rate equation v111fab is used to calculate 3 velocities that will be treated as experimental data to test the program calckinpars111 to calculate the 3 kinetic parameters.

Calculate kinetic parameters using  {200,200}, {0.1,200}, {200,3}.
The effect of [B] on the velocity is very large because it depends  on (([B])^2).

v111fab/.a->200/.b->200//N

v111fab/.a->.1/.b->200//N

v111fab/.a->200/.b->3//N

calckinpars111[.98961,200,200,.49751,.1,200,.020979,200,3]

The following Mathematica operations make it possible to make a table.

line5={vfexp,kA,kB}/.calckinpars111[.98961,200,200,.49751,.1,200,.020979,200,3]

line5[[2]]

kB can be negative because only kB^2 is involved, but only the positive value is of interest in enzme kinetics.  These are the expected values of the 3 kinetic parameters, but the effects of experimental errors in velocity measurements and substrate concentrations.  The effects of errors in substrate concentrations are simple becaue they cause the corresponding errors in Michaelis constants.

Calculate the effects of 5% errors in the measured velocities, one at a time.

line6={vfexp,kA,kB}/.calckinpars111[1.05*.98961,200,200,.49751,.1,200,.020979,200,3];

line7={vfexp,kA,kB}/.calckinpars111[.98961,200,200,1.05*.49751,.1,200,.020979,200,3];

line8={vfexp,kA,kB}/.calckinpars111[.98961,200,200,.49751,.1,200,1.05*.020979,200,3];

Table 1.  Effects of 5% errors in velocity measurements on the 3 kinetic parameters for A + 2B -> products when A is bound first and the following pairs of substrate concentrations are used: {200,200}, {0.1,200}, {200,3}

TableForm[{line5[[2]],line6[[2]],line7[[2]],line8[[2]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3"},{"vfexp","kA","kB"}}]

1.2  Ordered mechanism for A + 2B -> products when the two B molecules are bound by E in a single reaction

1.2.1  Derivation of the rate equation

The rapid-equilibrium mechanism has 2 equilibrium expressions, one expression for the velocity, and 3 kinetic parameters


E + 2B =(Subscript[EB, 2])		(((kB2))^2) = e*(b^2)/eb2    kB2 = 20
(Subscript[EB, 2]) + A = (Subscript[EAB, 2])	kA2 = eb2*a/eab2    kA2 = 10
(Subscript[EAB, 2]) -> products	v = (10^npH)kf*eab2

The following calculations are based on kA2 = 10 and kB2 = 20.

Solve[{(kB2^2) ==e*(b^2)/eb2,kA2==eb2*a/eab2,et==e+eb2+eab2},{eab2},{e,eb2}]

v121=((a (b^2) vfexp)/(a (b^2)+(b^2) kA2+kA2 (((kB2))^2)))

v121fab=v121/.kA2->10/.kB2->20/.vfexp->1

Note that this rate equation has a (b^2) term in the denominator, rather than an a term.  This shows that this mechanism can be distinguished from the mechanism with A bound first.

Construct 3D plots.

plot5=Plot3D[v121fab,{a,0,50},{b,0,100},PlotRange->{0,1},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(a)"];

plot6=Plot3D[Evaluate[5*D[v121fab,a]],{a,.0001,50},{b,.0001,100},PlotRange->{0,1},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(b)"];

plot7=Plot3D[Evaluate[D[v121fab,b]],{a,.0001,50},{b,.0001,100},PlotRange->{0,.04},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(c)"];

plot8=Plot3D[Evaluate[D[v121fab,a,b]],{a,.0001,50},{b,.0001,100},PlotRange->{-.005,.005},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(d)"];

fig2=Show[GraphicsArray[{{plot5,plot6},{plot7,plot8}}]]

Fig. 2.  (a) Plot of v121fab versus [A] and [B].  Notice the sigmoid character of the curves at constant [A] and variable [B].  (b) Plot of 5*dv121fab/d[A] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [A] when [A] is low and [B] is high.  (c)  Plot of dv121fab/d[B] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [B] when [B] is about equal to kB and [A] is high.  (d)  Plot of (d^2)v121fab/d[A]d[B] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [A] and [B] together when [B] is high and [A] is low.  This will be important when the random mechanism is considered in Section 1.3.

Note (d) indicates {1,20}

1.2.2  Determination of kinetic parameters with the minimum number of velocity measurements

calckinpars121[v1_,a1_,b1_,v2_,a2_,b2_,v3_,a3_,b3_]:=Module[{},(*This program calculates vfexp, kA2, and kB2 from three experimental velocities for A + 2B -> products when 2B are 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 medium [B].*)
Solve[{v1==a1*b1^2*vfexp/(a1*b1^2+b1^2*kA2+kA2*kB2^2),v2==a2*b2^2*vfexp/(a2*b2^2+b2^2*kA2+kA2*kB2^2),v3==a3*b3^2*vfexp/(a3*b3^2+b3^2*kA2+kA2*kB2^2)},{vfexp,kA2,kB2}]]

Velocities are calculated at 3 pairs of substrate concentrations: {200,200}, {1,200, and {200,3}.

v121fab/.a->200/.b->200//N

v121fab/.a->1/.b->200//N

v121fab/.a->200/.b->3//N

Calculation of kinetic parameters assuming 2B molecules are bound by the enzymatic site before A.

calckinpars121[.95193,200,200,.090090,1,200,.30560,200,3];

line21={vfexp,kA2,kB2}/.calckinpars121[.95193,200,200,.090090,1,200,.30560,200,3];

line21[[2]]

Calculate the effects of 5% errors in the velocities, one at a time.

calckinpars121[1.05*.95193,200,200,.090090,1,200,.30560,200,3];

line22={vfexp,kA2,kB2}/.calckinpars121[1.05*.95193,200,200,.090090,1,200,.30560,200,3];

calckinpars121[.95193,200,200,1.05*.090090,1,200,.30560,200,3];

line23={vfexp,kA2,kB2}/.calckinpars121[.95193,200,200,1.05*.090090,1,200,.30560,200,3];

line24={vfexp,kA2,kB2}/.calckinpars121[.95193,200,200,.090090,1,200,1.05*.30560,200,3];

Table 2.  Effects of 5% errors in velocities for the ordered mechanism A + 2B -> products when 2B is bound first.  Velocities are calculated at {200,200}, {1,200, and {200,3}.

TableForm[{line21[[2]],line22[[2]],line23[[2]],line24[[2]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3"},{"vfexp","kA2","kB2"}}]

1.3  Random mechanism for A + 2B -> products

1.3.1  Derivation of the rate equation

The random mechanism is

E + A = EA		kA = e*a/ea                   kA = 10
E + 2B = (Subscript[EB, 2])		(kB2^2) = e*b^2/eb2      kB2 = 30
EA + 2B =(Subscript[EAB, 2])	(kB^2) = ea*(b^2)/eab2          kB = 20
(Subscript[EAB, 2]) -> products	v = (10^npH)kf*eab2

There are 4 kinetic parameters, and so the minimum number of velocities that have to be measured is 4.  This mechanism involves 6 reactants, 3 components, and 3 equilibrium relations, and so N' = C' + R' is  6 = 3 + 3.  It is also represented by

             kA
E  +  A  =  EA
+                +
2B             2B
||   (kB2^2)      ||  (kB^2)
(Subscript[EB, 2])          (Subscript[EAB, 2])-> products

Section 1.4 deals with an alternative mechanism that is represented by

             kA
E  +  A  =  EA
+                
2B             
||   (kB2^2)      
(Subscript[EB, 2])+ A = (Subscript[EAB, 2])-> products
             kA2

The difference between these two mechanisms is that they use different equilibrium constant expressions.  The first mechanism involves kA, kB2, and kB.  The alternative mechanism involves kA, kB2, and kA2.  The term "alternative" is not meant to indicate one mechanism is superior to the other, but they do have different uses that will be discussed.  These two mechanisms are related thermodynamically.  This relation is shown by 

             kA
E  +  A  =  EA
+                +
2B             2B
||   (kB2^2)      ||  (kB^2)
(Subscript[EB, 2])+ A = (Subscript[EAB, 2])-> products
             kA2
             
This mechanism is not suitable for representing rapid-equilibrium kinetics because the four reactions are not independent.  The relation between the 4 Michaelis constants is kA (kB^2) = (kB2^2)kA2, and so kA2 = kA (kB^2)/(kB2^2) =  10*(20/30)^2 because kA = 10, kB = 20, and kB2 = 30.  Thus the experimental determination of the kinetic parameters kA, kB, and kB2 discussed in this section makes it possible to calculate kA2 in addition. 
	The rate equation is derived from an independent set of reactions.

Solve[{kA==e*a/ea,kB^2 ==ea*b^2/eab2,kB2^2==e*b^2/eb2,et==e+ea+eb2+eab2},{eab2},{e,ea,eb2}]

v131=((a (b^2) vfexp (kB2^2))/((b^2) kA (kB^2)+a (b^2) (kB2^2)+a (kB^2) (kB2^2)+kA (kB^2) (kB2^2)))

v131fab=v131/.vfexp->1/.kA->10/.kB->20/.kB2->30

An overview of the kinetics of A + 2B -> products for this random mechanism is obtained by making 3D plots.

plot9=Plot3D[v131fab,{a,0,50},{b,0,100},PlotRange->{0,1},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(a)"];

plot10=Plot3D[Evaluate[5*D[v131fab,a]],{a,.0001,50},{b,.0001,100},PlotRange->{0,1},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(b)"];

plot11=Plot3D[Evaluate[D[v131fab,b]],{a,.0001,50},{b,.0001,100},PlotRange->{0,.04},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(c)"];

plot12=Plot3D[Evaluate[D[v131fab,a,b]],{a,.0001,50},{b,.0001,100},PlotRange->{-.005,.005},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(d)"];

fig3=Show[GraphicsArray[{{plot9,plot10},{plot11,plot12}}]]

Fig. 3.  (a) Plot of v131fab versus [A] and [B].  Notice the sigmoid character of the curves at constant [A] and variable [B].  (b) Plot of 5*dv131fab/d[A] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [A] when [A] is low and [B] is high.  (c)  Plot of dv131fab/d[B] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [B] when [B] is about equal to kB and [A] is high.  (d)  Plot of (d^2)v131fab/d[A]d[B] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [A] and [B] together when [B] is high and [A] is low.  

	These plots are helpful in choosing pairs of substrate concentrations that yield the smallest effects of errors n the measured velocities.

1.3.2  Determination of kinetic parameters kinetic parameters with the minimum number of velocity measurements

calckinpars131[v1_,a1_,b1_,v2_,a2_,b2_,v3_,a3_,b3_,v4_,a4_,b4_]:=Module[{},(*This program calculates vfexp, kA, kB, and kB2 from 4 experimental velocities for the random mechanism of A + 2B -> products in which A is bound first, and then 2B are bound.  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 medium [B], and the fourth velocity is at low concentrations of A and B, but not too low.*)
Solve[{v1==((a1 (b1^2) (kB2^2) vfexp)/((b1^2) kA (kB^2)+a1 (b1^2) (kB2^2)+a1 (kB^2) (kB2^2)+kA (kB^2) (kB2^2))),v2==((a2 (b2^2) (kB2^2) vfexp)/((b2^2) kA (kB^2)+a2 (b2^2) (kB2^2)+a2 (kB^2) (kB2^2)+kA (kB^2) (kB2^2))),v3==((a3 (b3^2) (kB2^2) vfexp)/((b3^2) kA (kB^2)+a3 (b3^2) (kB2^2)+a3 (kB^2) (kB2^2)+kA (kB^2) (kB2^2))),v4==((a4 (b4^2) (kB2^2) vfexp)/((b4^2) kA (kB^2)+a4 (b4^2) (kB2^2)+a4 (kB^2) (kB2^2)+kA (kB^2) (kB2^2)))},{vfexp,kA,kB,kB2}]]

v131fab/.a->200/.b->200//N

v131fab/.a->1/.b->200//N

v131fab/.a->200/.b->5//N

v131fab/.a->5/.b->5//N

calckinpars131[.96832,200,200,.18004,1,200,.056110,200,5,.0200445,5,5];

line31={vfexp,kA,kB,kB2}/.calckinpars131[.96832,200,200,.18004,1,200,.056110,200,5,.0200445,5,5];

line31[[4]]

Change velocities 5%, one at a time.

line32={vfexp,kA,kB,kB2}/.calckinpars131[1.05*.96832,200,200,.18004,1,200,.056110,200,5,.0200445,5,5];

line33={vfexp,kA,kB,kB2}/.calckinpars131[.96832,200,200,1.05*.18004,1,200,.056110,200,5,.0200445,5,5];

line34={vfexp,kA,kB,kB2}/.calckinpars131[.96832,200,200,.18004,1,200,1.05*.056110,200,5,.0200445,5,5];

line35={vfexp,kA,kB,kB2}/.calckinpars131[.96832,200,200,.18004,1,200,.056110,200,5,1.05*.0200445,5,5];

Calculation of kA2 = kA (kB^2)/(kB2^2) =  10*(20/30)^2

10*(20/30)^2//N

9.97*(20.55/29.83)^2//N

10.02*(19.99/30.97)^2//N

10.85*(19.44/30.42)^2

9.20*(20.04/28.81)^2

Append[line31[[4]],4.44444]

Table 3.  Effects of 5% errors in measured velocities on the determination of kinetic 
parmeters for the random mechanism of A + 2B -> products.  Velocities are calculated at {200,200}, {1,200}, {200,5}, and {5,5}.

TableForm[{Append[line31[[4]],4.44],Append[line32[[4]],4.73],Append[line33[[4]],4.17],Append[line34[[4]],4.43],Append[line35[[4]],4.45]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3","1.05*v4"},{"vfexp","kA","kB","kB2","kA2"}}]

The values of vfexp, kA, kB, and kB2 were calculated from velocities, and the values of kA2 were calculated using kA2 = kA (kB^2)/(kB2^2).

1.3.3  Application of the random program calckinpars131 to velocities for the ordered mechanism when A is bound first

A program for a random mechanism can be applied to data for a non-random mechanism.  This is useful when a reaction is being studied for the first time.  Since calckinpars131determines kA and kB, it can be applied to velocities for the first mechanism discussed in Section 1.1.  When this is done, a fourth velocity is needed.

v111fab/.a->5/.b->5//N

Section 1.1.2 showed that

v111fab/.a->100/.b->100//N

v111fab/.a->1/.b->100//N

v111fab/.a->100/.b->5//N

calckinpars131[.95785,100,100,.69444,1,100,.053763,100,5,.020408,5,5];

line41={vfexp,kA,kB,kB2}/.calckinpars131[.95785,100,100,.69444,1,100,.053763,100,5,.020408,5,5];

line41[[4]]

This shows that kB2 is infinity or negative infinity.  This is right because there is no kB2 in the ordered mechanism when A is bound first.

v131

v111

v131 can be written as

((a (b^2) vfexp)/((b^2) kA (kB^2)/(kB2^2)+a (b^2) +a (kB2^2)+kA (kB^2)))

When (kB^2) is set equal to infinity, v111 is obtained.

Calculations of effects of 5 % errors in measured velocities.

line42={vfexp,kA,kB,kB2}/.calckinpars131[1.05*.95785,100,100,.69444,1,100,.053763,100,5,.020408,5,5];

line43={vfexp,kA,kB,kB2}/.calckinpars131[.95785,100,100,1.05*.69444,1,100,.053763,100,5,.020408,5,5];

line44={vfexp,kA,kB,kB2}/.calckinpars131[.95785,100,100,.69444,1,100,1.05*.053763,100,5,.020408,5,5];

line45={vfexp,kA,kB,kB2}/.calckinpars131[.95785,100,100,.69444,1,100,.053763,100,5,1.05*.020408,5,5];

Table 4.  Effects of 5% errors in velocities when calckinpars131 is applied to velocity meaurements for ordered A + 2B -> products when A is bound first.

TableForm[{line41[[4]],line42[[4]],line43[[4]],line44[[4]],line45[[4]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3","1.05*v4"},{"vfexp","kA","kB","kB2"}}]

i is the square root of -1.  This shows the advantage of using a more general program when studying a reaction for the first time.

1.4  Alternative random mechanism for A + 2B -> products

1.4.1  Derivation of the alternative rate equation

The alternative rapid-equilibrium mechanism is

E + A = EA		kA = e*a/ea                   kA = 10
E + 2B = (Subscript[EB, 2])		(kB2^2) = e*b^2/eb2      kB2 = 30
(Subscript[EB, 2]) + A =(Subscript[EAB, 2])        kA2 = eb2*a/eab2         kA2 = 4.44444
(Subscript[EAB, 2]) -> products	v = (10^npH)kf*eab2

The first two reactions are the same as the mechanism in the preceding section.  The important change is that EAB2 is formed from (Subscript[EB, 2]) + A.   The number of equilibrium expressions is 3, and so  N' = 6, C' = 3, R' = 3 and 6 = 3 + 3.  This mechanism involves 4 kinetic parameters, and so 4 velocities have to be measured to determine the 4 kinetic parameters.  This mechanism is represented by

             kA
E  +  A  =  EA
+                
2B             
||   (kB2^2)      
(Subscript[EB, 2])+ A = (Subscript[EAB, 2])-> products
             kA2

As shown in Section 1.3, the relation between the 4 equilibrium constants is kA (kB^2) = (kB2^2)kA2, and so kB = Sqrt[kA2 (kB2^2)/kA].  Thus kB is given by

Sqrt[4.44444*30^2/10]

The rate equation for the alternative mechanism is obtained by use of Solve.

Solve[{kA==e*a/ea,kA2 ==eb2*a/eab2,kB2^2==e*b^2/eb2,et==e+ea+eb2+eab2},{eab2},{e,ea,eb2}]

v141=((a (b^2) vfexp kA)/(a (b^2) kA+(b^2) kA kA2+a kA2 (kB2^2)+kA kA2 (kB2^2)))

This can be compared with v131:

v131

v141 is different from v131 and involves different symbols.

v141fab can be derived by putting in the kinetic parameters given above that are consistent with the parameters in v131.

v141fab=v141/.vfexp->1/.kA->10/.kB2->30/.kA2->4.44444

This equation can be compared with v131fab that was calculated using v131/. vfexp -> 1 /. kA -> 10 /. kB -> 20 /. kB2 -> 30

v131fab

This is a different function of a and b than v141fab, but calculate some velocities.

v131fab/.a->100/.b->100//N

v141fab/.a->100/.b->100//N

v131fab/.a->5/.b->5//N

v141fab/.a->5/.b->5//N

Although the equations are different, these are exactly the same velocities obtained with v131fab in Section 1.3.2.  The reason these two rate equations yield the same velocities is that there is a thermodynamic relation between the kinetic parameters in the two mechanisms.  Since v131fab and v141fab yield the same velocities, there is no need to construct 3D plots or to calculate velocities to test calckinpars141.

1.4.2  Determination of kinetic parameters with the minimum number of velocity measurements

calckinpars141[v1_,a1_,b1_,v2_,a2_,b2_,v3_,a3_,b3_,v4_,a4_,b4_]:=Module[{},(*This program calculates vfexp, kA, kB2, and kA2 from 4 experimental velocities for the alternative random mechanism of A + 2B -> products.  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 medium [B], and the fourth velocity is at low [A] and low [B].*)
Solve[{v1==((a1 (b1^2) kA vfexp)/(a1 (b1^2) kA+(b1^2) kA kA2+a1 kA2 (kB2^2)+kA kA2 (kB2^2))),v2==((a2 (b2^2) kA vfexp)/(a2 (b2^2) kA+(b2^2) kA kA2+a2 kA2 (kB2^2)+kA kA2 (kB2^2))),
v3==((a3 (b3^2) kA vfexp)/(a3 (b3^2) kA+(b3^2) kA kA2+a3 kA2 (kB2^2)+kA kA2 (kB2^2))),
v4==((a4 (b4^2) kA vfexp)/(a4 (b4^2) kA+(b4^2) kA kA2+a4 kA2 (kB2^2)+kA kA2 (kB2^2)))},{vfexp,kA,kB2,kA2}]]

calckinpars141[.96832,200,200,.18004,1,200,.056110,200,5,.0200445,5,5];

line51={vfexp,kA,kA2,kB2}/.calckinpars141[.96832,200,200,.18004,1,200,.056110,200,5,.0200445,5,5];

Calculate the effects of 5% errors in the measured velocities, one at a time.

line52={vfexp,kA,kA2,kB2}/.calckinpars141[1.05*.96832,200,200,.18004,1,200,.056110,200,5,.0200445,5,5];

line53={vfexp,kA,kA2,kB2}/.calckinpars141[.96832,200,200,1.05*.18004,1,200,.056110,200,5,.0200445,5,5];

line54={vfexp,kA,kA2,kB2}/.calckinpars141[.96832,200,200,.18004,1,200,1.05*.056110,200,5,.0200445,5,5];

line55={vfexp,kA,kA2,kB2}/.calckinpars141[.96832,200,200,.18004,1,200,.056110,200,5,1.05*.0200445,5,5];

The values of kA, kA2, and kB2 can be used to calculate values of kB by use of thermodynamics: kB = Sqrt[kA2 (kB2^2)/kA].

Sqrt[4.44444*30^2/10]

Sqrt[4.173*29.83^2/9.97]

Sqrt[4.439*30.97^2/10.02]

Sqrt[4.452*30.42^2/10.86]

Sqrt[4.728*28.81^2/9.20]

Table 5.  Effects of 5% errors in measured velocities, one at a time, on determinations of kinetic parameters for the alternative mechanism using calckinpars141 for A + 2B  -> products

TableForm[{Append[line51[[2]],20.],Append[line52[[2]],19.30],Append[line53[[2]],20.61],Append[line54[[2]],19.48],Append[line55[[2]],20.65]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3","1.05*v4"},{"vfexp","kA","kA2","kB2","kB"}}]

The values of vfexp, kA, kA2, and lB2 were calculated from velocities.  These values were used to calculate kB.  These results are the same as those in Table 3.

1.4.3  Application of the calckinpars141 to velocities v121 for the ordered mechanism when 2B is bound before A

The more general program calckinpard141 can be applied to experimental velocities calculated using v121.

It is necessary to determine one more velocity with v121fab

v121fab/.a->200/.b->200//N

v121fab/.a->1/.b->200//N

v121fab/.a->200/.b->3//N

v121fab/.a->5/.b->5//N

calckinpars141[.95193,200,200,.090090,1,200,.30560,200,3,.028571,5,5]

line61={vfexp,kA,kA2,kB2}/.calckinpars141[.95193,200,200,.090090,1,200,.30560,200,3,.028571,5,5];

line61[[2]]

Calculate the effects of 5% errors in the measured velocities.

line62={vfexp,kA,kA2,kB2}/.calckinpars141[1.05*.95193,200,200,.090090,1,200,.30560,200,3,.028571,5,5];

line63={vfexp,kA,kA2,kB2}/.calckinpars141[.95193,200,200,1.05*.090090,1,200,.30560,200,3,.028571,5,5];

line64={vfexp,kA,kA2,kB2}/.calckinpars141[.95193,200,200,.090090,1,200,1.05*.30560,200,3,.028571,5,5];

line65={vfexp,kA,kA2,kB2}/.calckinpars141[.95193,200,200,.090090,1,200,.30560,200,3,1.05*.028571,5,5];

The following table is not used in the article.

Effects of 5% errors in measured velocities, one at a time, on determinations of kinetic parameters applying calckinpard141  to experimental velocities calculated using v121 for A + 2B  -> products

TableForm[{line61[[2]],line62[[2]],line63[[2]],line64[[2]],line65[[2]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3","1.05*v4"},{"vfexp","kA","kA2","kB2"}}]

These values can be compared with the values in Table 3.

2.  Determination of kinetic parameters for A + 3B -> products

2.1  Ordered mechanism for A + 3B -> products when A is bound first and the three B molecules are bound by EA in a single reaction

2.1.1  Derivation of the rate equation

The rapid-equilibrium mechanism has 3 reactions and 3 kinetic parameters

E + A = EA		kA = e*a/ea 
EA + 3B =(Subscript[EAB, 3])	(kB^3) = ea*(b^3)/eab3
(Subscript[EAB, 3]) -> products	v = (10^npH)kf*eab3

kA is assumed to be 10.  kB is assumed to be 20.  Note that kB has the units of a concentration.  vfexp = (10^pH)kf*et is assumed to be 1.00.

Solve[{kA==e*a/ea,(kB^3) ==ea*(b^3)/eab3,et==e+ea+eab3},{eab3},{e,ea}]

The rate of reaction is obtained by multiplying eab3 with (10^npH)*kf.

((a (b^3) et*(10^npH)*kf)/(a (b^3)+a (kB^3)+kA (kB^3)))

Replace (10^pH)*kf*et with vfexp.

v211=((a (b^3) vfexp)/(a (b^3)+a (kB^3)+kA (kB^3)))

This is the general form for the velocity that is used in the computer program calckinpars211 to calculate 3 kinetic parameters from 3 measured velocities.  When the kinetic parameters are specified, the function of [A] and [B] is given by v211fab.

v211fab=v211/.kA->10/.kB->20/.vfexp->1

The rate equation v211fab is used to explore the properties of this mechanism.

plot5=Plot3D[v211fab,{a,0,50},{b,0,100},PlotRange->{0,1},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(a)"];

plot6=Plot3D[Evaluate[5*D[v211fab,a]],{a,.0001,50},{b,.0001,100},PlotRange->{0,1},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(b)"];

plot7=Plot3D[Evaluate[D[v211fab,b]],{a,.0001,50},{b,.0001,100},PlotRange->{0,.04},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(c)"];

plot8=Plot3D[Evaluate[D[v211fab,a,b]],{a,.0001,50},{b,.0001,100},PlotRange->{-.01,.05},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(d)"];

Show[GraphicsArray[{{plot5,plot6},{plot7,plot8}}]]

Fig. 4.  (a) Plot of v211fab versus [A] and [B].  Notice the sigmoid character of the curves at constant [A] and variable [B].  (b) Plot of 5*dv211fab/d[A] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [A] when [A] is low and [B] is high.  (c)  Plot of dv211fab/d[B] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [B] when [B] is about equal to kB and [A] is high.  (d)  Plot of (d^2)v211fab/d[A]d[B] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [A] and [B] together when [B] is high and [A] is low.  This will be important when the random mechanism is considered in Section 2.3.

2.1.2  Determination of kinetic parameters with the minimum number of velocity measurements

When A is bound first, the general equation for the velocity is

v211=((a (b^3) vfexp)/(a (b^3)+a (kB^3)+kA (kB^3)))

calckinpars211[v1_,a1_,b1_,v2_,a2_,b2_,v3_,a3_,b3_]:=Module[{},(*This program calculates vfexp, kA, and kB from three experimental velocities for A + 3B -> products at three pairs of substrate concentrations on the assumption that three B molecules are bound to EA in a single reaction.  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 medium [B].*)
Solve[{v1==a1*b1^3*vfexp/(a1*b1^3+a1*kB^3+kA*kB^3),v2==a2*b2^3*vfexp/(a2*b2^3+a2*kB^3+kA*kB^3),v3==a3*b3^3*vfexp/(a3*b3^3+a3*kB^3+kA*kB^3)},{vfexp,kA,kB}]]

Calculate velocities at the following pairs {a, b} of substrate concentrations: {100,100},{0.1,100},{100,5}.

v211fab/.a->100/.b->100//N

v211fab/.a->.1/.b->100//N

v211fab/.a->100/.b->5//N

Note reducing [B] to 1 or 0.1 yields too low a velocity.

calckinpars211[.99128,100,100,.55310,.1,100,.014006,100,5];

Mathematically, there are 3 solutions to this set of 3 equations, but only the last is of interest in biochemistry.

line41={vfexp,kA,kB}/.calckinpars211[.99128,100,100,.55310,.1,100,.014006,100,5];

line41[[3]]

Calculate effects of 5% errors in velocities, one at a time.

calckinpars211[1.05*.99128,100,100,.55310,.1,100,.014006,100,5];

line42={vfexp,kA,kB}/.calckinpars211[1.05*.99128,100,100,.55310,.1,100,.014006,100,5];

calckinpars211[.99128,100,100,1.05*.55310,.1,100,.014006,100,5];

line43={vfexp,kA,kB}/.calckinpars211[.99128,100,100,1.05*.55310,.1,100,.014006,100,5];

calckinpars211[.99128,100,100,.55310,.1,100,1.05*.014006,100,5];

line44={vfexp,kA,kB}/.calckinpars211[.99128,100,100,.55310,.1,100,1.05*.014006,100,5];

Table 6.  Effects of 5% errors in measured velocities on the determination of kinetic parameters for the ordered mechanism for A + 3B -> products when A is bound first using {100,100}, {0.1,100}, {100,5}.

TableForm[{line41[[3]],line42[[3]],line43[[3]],line44[[3]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3"},{"vfexp","kA","kB"}}]

2.2  Ordered mechanism for A + 3B -> products when A is bound first and the three B molecules are bound by E in a single reaction

2.2.1  Derivation of the rate equation

This rapid-equilibrium mechanism has 3 reactions and 3 kinetic parameters

E + 3B =(Subscript[EB, 3])		(((kB3))^3) = e*(b^3)/eb3  
(Subscript[EB, 3]) + A = (Subscript[EAB, 3])	kA3 = eb3*a/eab3  
(Subscript[EAB, 3]) -> products	v = (10^npH)kf*eab3

Solve[{(kB3^3) ==e*(b^3)/eb3,kA3==eb3*a/eab3,et==e+eb3+eab3},{eab3},{e,eb3}]

v221=((a (b^3) vfexp)/(a (b^3)+(b^3) kA3+kA3 (((kB3))^3)))

This function is used in the program to determine kinetic parameters.

v221fab=v221/.kA3->10/.kB3->20/.vfexp->1

Note that this rate equation has a (b^3) term in the denominator, rather than an a term.  This shows that this mechanism can be distinguished from the mechanism with A bound first.

3D plots of v221fab

plot9=Plot3D[v221fab,{a,0,50},{b,0,100},PlotRange->{0,1},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(a)"];

plot10=Plot3D[Evaluate[5*D[v221fab,a]],{a,.0001,50},{b,.0001,100},PlotRange->{0,1},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(b)"];

plot11=Plot3D[Evaluate[D[v221fab,b]],{a,.0001,50},{b,.0001,100},PlotRange->{0,.05},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(c)"];

plot12=Plot3D[Evaluate[D[v221fab,a,b]],{a,.0001,50},{b,.0001,100},PlotRange->{-.001,.005},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(d)"];

Show[GraphicsArray[{{plot9,plot10},{plot11,plot12}}]]

Fig. 5.  (a) Plot of v221fab versus [A] and [B].  Notice the sigmoid character of the curves at constant [A] and variable [B].  (b) Plot of 5*dv221fab/d[A] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [A] when [A] is low and [B] is high.  (c)  Plot of dv221fab/d[B] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [B] when [B] is about equal to kB and [A] is high.  (d)  Plot of (d^2)v221fab/d[A]d[B] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [A] and [B] together when [B] is high and [A] is low.  This will be important when the random mechanism is considered in Section 2.3.

2.2.2  Determination of kinetic parameters with the minimum number of velocity measurements

calckinpars221[v1_,a1_,b1_,v2_,a2_,b2_,v3_,a3_,b3_]:=Module[{},(*This program calculates vfexp, kA3, and kB3 from three experimental velocities for A + 3B -> products at three pairs of substrate concentrations on the assumption that three B molecules are bound to E at the same time.  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 medium? [B].*)
Solve[{v1==((a1 (b1^3) vfexp)/(a1 (b1^3)+(b1^3) kA3+kA3 (kB3^3))),
v2==((a2 (b2^3) vfexp)/(a2 (b2^3)+(b2^3) kA3+kA3 (kB3^3))),
v3==((a3 (b3^3) vfexp)/(a3 (b3^3)+(b3^3) kA3+kA3 (kB3^3)))},
{vfexp,kA3,kB3}]]

Calculate velocities at the following pairs {a, b} of substrate concentrations: {100,100},{1,100},{100,5}.

v221fab

v221fab/.a->100/.b->100//N

v221fab/.a->1/.b->100//N

v221fab/.a->100/.b->5//N

line61={vfexp,kA3,kB3}/.calckinpars221[.90843,100,100,.090253,1,100,.13333,100,5];

line62={vfexp,kA3,kB3}/.calckinpars221[1.05*.90843,100,100,.090253,1,100,.13333,100,5];

line63={vfexp,kA3,kB3}/.calckinpars221[.90843,100,100,1.05*.090253,1,100,.13333,100,5];

line64={vfexp,kA3,kB3}/.calckinpars221[.90843,100,100,.090253,1,100,1.05*.13333,100,5];

Table 7.  Effects of 5% errors in velocities for the ordered mechanism A + 3B -> products when 3B is bound first.

TableForm[{line61[[3]],line62[[3]],line63[[3]],line64[[3]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3"},{"vfexp","kA3","kB3"}}]

2.3  Random mechanism for A + 3B -> products

2.3.1  Derivation of the rate equation

There are two possible rapid-equilibrium mechanisms for this reaction.  An alternate mechanism is discussed in Section 2.4.
	The rapid-equilibrium mechanism is

E + A = EA		kA = e*a/ea  
E + 3B = (Subscript[EB, 3])		(kB3^3) = e*b^3/eb3 
EA + 3B =(Subscript[EAB, 3])	(kB^3) = ea*(b^3)/eab3 
(Subscript[EAB, 3]) -> products	v = (10^npH)kf*eab3

This mechanism is also represented by
  
             kA
E  +  A  =  EA
+                +
3B             3B
||   (kB3^3)      ||  (kB^3)
(Subscript[EB, 3])          (Subscript[EAB, 3])-> products

Section 2.4 deals with an alternative mechanism that is represented by

             kA
E  +  A  =  EA
+                
3B             
||   (kB2^3)      
(Subscript[EB, 3])+ A = (Subscript[EAB, 3])-> products
             kA3

These two mechanisms can be combined to discuss thermodynamics

            kA
E  +  A  =  EA
+                +
3B             3B
||   (kB3^3)      ||  (kB^3)
(Subscript[EB, 3])+ A = (Subscript[EAB, 3])-> products
              kA3
              
Note that kA (kB^3) = (kB3^3)kA3. The following values have already been assigned: kA = 10, kB = 20, and kB3 = 30.  Therefore,
kA3 = kA (kB^3)/(kB3^3). 

Have to be careful about the number of equilibrium expression to use in the equilibrium concentration of (Subscript[EAB, 3]).  N' = 6, C' = 3, R' = 3, so that 6 = 3 + 3.  The reaction (Subscript[EB, 3]) + A = (Subscript[EAB, 3]) is at equilibrium when the first three reactions in the mechanism are at equilibrium, even though it is not included in the mechanism.  This mechanism involves 4 kinetic parameters, and so 4 velocities have to be measured.   vfexp = 1, kA = 10, kB = 20, and kB3 = 30.

Solve[{kA==e*a/ea,kB^3 ==ea*b^3/eab3,kB3^3==e*b^3/eb3,et==e+ea+eb3+eab3},{eab3},{e,ea,eb3}]

v231=((a (b^3) vfexp (kB3^3))/((b^3) kA (kB^3)+a (b^3) (kB3^3)+a (kB^3) (kB3^3)+kA (kB^3) (kB3^3)))

This general rate equation is used in the computer program.

v231fab=v231/.vfexp->1/.kA->10/.kB->20/.kB3->30

3D plots of v231fab.

plot13=Plot3D[v231fab,{a,0,50},{b,0,100},PlotRange->{0,1},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(a)"];

plot14=Plot3D[Evaluate[5*D[v231fab,a]],{a,.0001,50},{b,.0001,100},PlotRange->{0,1},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(b)"];

plot15=Plot3D[Evaluate[D[v231fab,b]],{a,.0001,50},{b,.0001,100},PlotRange->{0,.05},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(c)"];

plot16=Plot3D[Evaluate[D[v231fab,a,b]],{a,.0001,50},{b,.0001,100},PlotRange->{-.001,.005},ViewPoint->{-2,-2,1},AxesLabel->{"[A]","[B]",""},Lighting->{{"Ambient",GrayLevel[1]}},PlotLabel->"(d)"];

Show[GraphicsArray[{{plot13,plot14},{plot15,plot16}}]]

Fig. 6.  (a) Plot of v231fab versus [A] and [B].  Notice the sigmoid character of the curves at constant [A] and variable [B].  (b) Plot of 5*dv231fab/d[A] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [A] when [A] is low and [B] is high.  (c)  Plot of dv231fab/d[B] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [B] when [B] is about equal to kB and [A] is high.  (d)  Plot of (d^2)v231fab/d[A]d[B] versus [A] and [B].  This shows that the velocity changes the fastest with respect to [A] and [B] together when [B] is high and [A] is low.

2.3.2  Determination of kinetic parameters with the minimum number of velocity measurements

calckinpars231[v1_,a1_,b1_,v2_,a2_,b2_,v3_,a3_,b3_,v4_,a4_,b4_]:=Module[{},(*This program calculates vfexp, kA, kB, and kB3 from 4 experimental velocities for the random mechanism of A + 3B -> products.  An example of 4 choices of [A] and [B] are {200,200},{1,200},{200,5},{5,5}.*)
Solve[{v1==((a1 (b1^3) (kB3^3) vfexp)/((b1^3) kA (kB^3)+a1 (b1^3) (kB3^3)+a1 (kB^3) (kB3^3)+kA (kB^3) (kB3^3))),v2==((a2 (b2^3) (kB3^3) vfexp)/((b2^3) kA (kB^3)+a2 (b2^3) (kB3^3)+a2 (kB^3) (kB3^3)+kA (kB^3) (kB3^3))),v3==((a3 (b3^3) (kB3^3) vfexp)/((b3^3) kA (kB^3)+a3 (b3^3) (kB3^3)+a3 (kB^3) (kB3^3)+kA (kB^3) (kB3^3))),v4==((a4 (b4^3) (kB3^3) vfexp)/((b4^3) kA (kB^3)+a4 (b4^3) (kB3^3)+a4 (kB^3) (kB3^3)+kA (kB^3) (kB3^3)))},{vfexp,kA,kB,kB3}]]

v231fab/.a->200/.b->200//N

v231fab/.a->1/.b->200//N

v231fab/.a->200/.b->5//N

v231fab/.a->5/.b->5//N

{1,40} suggested

line41={vfexp,kA,kB,kB3}/.calckinpars231[.98438,200,200,.25164,1,200,.014660,200,5,.0051655,5,5];

line41[[9]]

This is kA,kB,kB3

Calculate effects of 5% errors in velocity measurements, one at a time.

line42={vfexp,kA,kB,kB3}/.calckinpars231[1.05*.98438,200,200,.25164,1,200,.014660,200,5,.0051655,5,5];

line43={vfexp,kA,kB,kB3}/.calckinpars231[.98438,200,200,1.05*.25164,1,200,.014660,200,5,.0051655,5,5];

line44={vfexp,kA,kB,kB3}/.calckinpars231[.98438,200,200,.25164,1,200,1.05*.014660,200,5,.0051655,5,5];

line45={vfexp,kA,kB,kB3}/.calckinpars231[.98438,200,200,.25164,1,200,.014660,200,5,1.05*.0051655,5,5];

When kA, kB, and kB3 have been determined with velocity measurements, kA3 can be calculated using kA3 = kA kB^3/(kB3^3), even though (Subscript[EB, 3])+ A = (Subscript[EAB, 3]) was not included in the mechanism.

10*(20/30)^3//N

10*(20.34/30)^3//N

10*(20/30)^3//N

10*(19.64/30)^3//N

10*(20.2/30)^3//N

Table 8.  Effects of 5% errors in velocities for the random mechanism A + 3B -> products

TableForm[{Append[line41[[9]],2.96],Append[line42[[9]],3.11],Append[line43[[9]],2.96],Append[line44[[9]],2.81],Append[line45[[9]],3.05]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3","1.05*v4"},{"vfexp","kA","kB","kB3","kA3"}}]

2.3.3  Application of the random program calckinpars231 to velocities v211fab for the ordered mechanism when A is bound first

calckinpars231 involves kA, kB, and kB3, and so it can be applied to velocities calculated using v211fab that involves kA ans kB.
In order to apply the random program (calckinpars231) to data from v211fab for the ordered mechanism A + 3B -> products when A is bound first, it is necessary to calculate a fourth velocity using v211fab:  that can be done at {5,5}.

Calculate velocities at the following pairs {a, b} of substrate concentrations: {100,100},{0.1,100},{100,5},{5,5}.

[A different choice might be {100,100},{1,100},{100,30},{5,30}.]

v211fab

v211fab/.a->100/.b->100//N

v211fab/.a->.1/.b->100//N

v211fab/.a->100/.b->5//N

v211fab/.a->5/.b->5//N

calckinpars231[.99128,100,100,.55310,.1,100,.014006,100,5,.0051814,5,5];

Dimensions[calckinpars231[.99128,100,100,.55310,.1,100,.014006,100,5,.0051814,5,5]]

line91={vfexp,kA,kB,kB3}/.calckinpars231[.99128,100,100,.55310,.1,100,.014006,100,5,.0051814,5,5];

line91[[9,1]]

line91[[7,4]]

line91[[7]]

This is kA, kB, kB3 in random.  I expected kB3 to be infinite.  kB3 is missing in odered mechansm, BUT there is a kB3.

Calculate effects of 5% errors in measured velocities.

line92={vfexp,kA,kB,kB3}/.calckinpars231[1.05*.99128,100,100,.55310,.1,100,.014006,100,5,.0051814,5,5];

line92[[9]]

line93={vfexp,kA,kB,kB3}/.calckinpars231[.99128,100,100,1.05*.55310,.1,100,.014006,100,5,.0051814,5,5];

line93[[7]]

line94={vfexp,kA,kB,kB3}/.calckinpars231[.99128,100,100,.55310,.1,100,1.05*.014006,100,5,.0051814,5,5];

line94[[7]]

line95={vfexp,kA,kB,kB3}/.calckinpars231[.99128,100,100,.55310,.1,100,.014006,100,5,1.05*.0051814,5,5];

line95[[9]]

Table 9.  Effects of 5% errors in velocities when the random program is applied to data for the ordered mechnism A + 3B -> products when A is bound first.

TableForm[{line91[[7]],line92[[9]],line93[[7]],line94[[7]],line95[[9]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3","1.05*v4"},{"vfexp","kA","kB","kB3"}}]

This shows that the expected values are obtained for vfexp, kA, and kB.  The erratic values of kB3 shows that the rate equation for the ordered mechanism with A bound first does not have a term for E + 3B = (Subscript[EB, 3]) ((kB3^3) = e*b^3/eb3).

2.4  Alternative random mechanism for A + 3B -> products

2.4.1  Derivation of the rate equation

An alternative random mechanism  that involves vfexp, kA, kB3, and kA3 is

E + A = EA		kA = e*a/ea                 10
E + 3B = (Subscript[EB, 3])		(kB3^3) = e*b^3/eb3     30
(Subscript[EB, 3]) + A =(Subscript[EAB, 3]) 	kA3 = eb3*a/eab3        2.9630
(Subscript[EAB, 3]) -> products	v = (10^npH)kf*eab3

The first two reactions are the same as the mechanism above.  The relation between the two mechanisms was shown in Section 2.31.
              
Note that kA (kB^3) = (kB3^3)kA3. The following values have already been assigned: kA = 10, kB = 20, and kB3 = 30.  Therefore,
kA3 = kA kB^3/(kB3^3)

kB = (kB3^3*kA3/kA)^(1/3)

kB is given by

(kB3^3*kA3/kA)^(1/3)

(30^3*2.9630/10)^(1/3)

Derivation of the rate equation.

Solve[{kA==e*a/ea,kB3^3 ==e*b^3/eb3,kA3==eb3*a/eab3,et==e+ea+eb3+eab3},{eab3},{e,ea,eb3}]

v231alt=((a (b^3) vfexp kA)/(a (b^3) kA+(b^3) kA kA3+a kA3 (kB3^3)+kA kA3 (kB3^3)))

80/27//N

v231altfab=v231alt/.kA->10/.kB3->30/.kA3->2.96296/.vfexp->1

v231altfab/.a->200/.b->200//N

v231altfab/.a->1/.b->200//N

v231altfab/.a->200/.b->5//N

v231altfab/.a->5/.b->5//N

v231altfab=v231alt/.kA->10/.kB3->30/.kA3->2.96296/.vfexp->1

These are exactly the velocities obtained with the mechanism above.  To obtain the same coefficients as before multiply numerator and denominator by 2700.

2700*{8*10^5,8*10^4,29.6296,10}

Since the rate equations for the two mechanisms are identical, the two mechanisms are simply different ways to describe the kinetics of random A + 3B -> products, but they describe it in terms of different Michaelis constants.  The first mechanism uses vfexp, kA, kB3, and kB.  The alternative mechansm uses vfexp, kA, kB3, and kA3.  
The rate equation for the ordered mechanism in which A is bound first involves vfexp, kA, and kB.  This is a special case of the first mechanism when there is no kB3 reaction.  This means that calckinpars231 can be applied to data on the ordered mechanism, as shown by Table 9. 
The rate equation for the ordered mechanism in which 3B is bound first involves vfexp, kA3, and kB3.  This is a special case of the alternative mechanism when there is no kA reaction.  This means that calckinpars231alt can be applied to data on the ordered mechanism when A is bound first, as shown by Table 11.

2.4.2  Determination of kinetic parameters with the minimum number of velocity measurements

calckinpars231alt[v1_,a1_,b1_,v2_,a2_,b2_,v3_,a3_,b3_,v4_,a4_,b4_]:=Module[{},(*This program calculates vfexp, kA, kB3, and kA3 from 4 experimental velocities for the random mechanism of A + 3B -> products.  An example of 4 choices of [A] and [B] are {200,200},{1,200},{200,5},{5,5}.*)
Solve[{v1==((a1 (b1^3) kA vfexp)/(a1 (b1^3) kA+(b1^3) kA kA3+a1 kA3 (kB3^3)+kA kA3 (kB3^3))),v2==((a2 (b2^3) kA vfexp)/(a2 (b2^3) kA+(b2^3) kA kA3+a2 kA3 (kB3^3)+kA kA3 (kB3^3))),
v3==((a3 (b3^3) kA vfexp)/(a3 (b3^3) kA+(b3^3) kA kA3+a3 kA3 (kB3^3)+kA kA3 (kB3^3))),v4==((a4 (b4^3) kA vfexp)/(a4 (b4^3) kA+(b4^3) kA kA3+a4 kA3 (kB3^3)+kA kA3 (kB3^3)))},
{vfexp,kA,kB3,kA3}]]

The velocities calculated above can be used because the results from the rate equation are the same as in the first mechanism.

calckinpars231alt[.98438,200,200,.25164,1,200,.014660,200,5,.0051655,5,5]

line111={vfexp,kA,kA3,kB3}/.calckinpars231alt[.98438,200,200,.25164,1,200,.014660,200,5,.0051655,5,5];

Calculate the effects of 5 % errors in velocities.

line112={vfexp,kA,kA3,kB3}/.calckinpars231alt[1.05*.98438,200,200,.25164,1,200,.014660,200,5,.0051655,5,5];

line113={vfexp,kA,kA3,kB3}/.calckinpars231alt[.98438,200,200,1.05*.25164,1,200,.014660,200,5,.0051655,5,5];

line114={vfexp,kA,kA3,kB3}/.calckinpars231alt[.98438,200,200,.25164,1,200,1.05*.014660,200,5,.0051655,5,5];

line115={vfexp,kA,kA3,kB3}/.calckinpars231alt[.98438,200,200,.25164,1,200,.014660,200,5,1.05*.0051655,5,5];

The values of kB can be calculated by use of

(30^3*2.9630/10)^(1/3)

(29.84^3*3.17/9.99)^(1/3)

(30.67^3*2.77/10)^(1/3)

(30.25^3*2.96/10.8)^(1/3)

(29.34^3*2.96/9.22)^(1/3)

Effects of 5% errors in velocities on the kinetic parameters for the alternative random mechanism for A + 3B -> products

TableForm[{Append[line111[[3]],20.00],Append[line112[[3]],20.35],Append[line113[[3]],19.99],Append[line114[[3]],19.64],Append[line115[[3]],20.09]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3","1.05*v4"},{"vfexp","kA","kA3","kB3","kB"}}]

These are the expected results.  See Table 8.

2.4.3  Application of the random program calckinpars231alt to velocities v221ab for the ordered mechanism when 3B is bound first

The velocities for the ordered mechanism when 3B is bound first are given by

v221

v221fab

Earlier calculated the following velocities at the following pairs {a, b} of substrate concentrations: {100,100},{1,100},{100,5}.

v221fab

v221fab/.a->100/.b->100//N

v221fab/.a->1/.b->100//N

v221fab/.a->100/.b->5//N

To use calckinpars231alt an additional velocity is needed: calculate it at {5,10}.

v221fab/.a->5/.b->10//N

calckinpars231alt[.90843,100,100,.090253,1,100,.133333,100,5,.052632,5,10]

This is correct because v221 does not have a kA term.

line121={vfexp,kA,kA3,kB3}/.calckinpars231alt[.90843,100,100,.090253,1,100,.133333,100,5,.052632,5,10]

line121[[3]]

Calculate the effects of 5% errors in velocities.

line122={vfexp,kA,kA3,kB3}/.calckinpars231alt[1.05*.90843,100,100,.090253,1,100,.133333,100,5,.052632,5,10]

line123={vfexp,kA,kA3,kB3}/.calckinpars231alt[.90843,100,100,1.05*.090253,1,100,.133333,100,5,.052632,5,10]

line124={vfexp,kA,kA3,kB3}/.calckinpars231alt[.90843,100,100,.090253,1,100,1.05*.133333,100,5,.052632,5,10]

line125={vfexp,kA,kA3,kB3}/.calckinpars231alt[.90843,100,100,.090253,1,100,.133333,100,5,1.05*.052632,5,10]

Table 10. Effects of 5% errors in velocities, one at a time, when calckinpars231alt is applied to data for the ordered mechanism for A + 3B --> products when 3B is bound first

TableForm[{line121[[3]],line122[[3]],line123[[3]],line124[[3]],line125[[3]]},TableHeadings->{{"No errors","1.05*v1","1.05*v2","1.05*v3","1.05*v4"},{"vfexp","kA","kA3","kB3"}}]

This is an example of a more general program yielding the correct kinetic parameters for a simpler mechanism and showing which parameter is missing in the rate equation for the smpler mechanism.

3.  Discussion

This package has been concerned with the determination of kinetic parameters for the enzyme-catalyzed reactions A + 2B -> products and A + 3B -> products.  For these two biochemical reactions, the rapid-equilibrium rate equation is derived with Solve, and a program has been written to calculate the kinetic parameters with the minimum number of measured velocities.  These calculations are made for ordered and random mechanisms.  The mechanisms are the simplest possible because 2B or 3B is assumed to be involved in a single reaction.  More terms could be included by adding B, one molecule at a time.  More terms could also be added from steady-state rate equations.  
	The general principles for the selection of [A] and [B] for velocity measurements on A + B -> products have been discussed earlier, but this discussion has been extended in this article, but no attempt has been made to find the optimum concentrations of the substrates because that depends on the method for determining velocities for a specific set of kinetic parameters.  In each case discussed here, a table is given to show the effects of 5% errors in the velocity measurements, one at a time. 
	Six rapid-equilibrium rate equations are given.  They are obtained by typing v111, v121, v131, v211, v221, and v231.  This package provides six programs for calculating kinetic parameters for the six mechanisms discussed.  These programs can also be used to calculate the effects on the kinetic parameters of experimental errors in velocity measurements and substrate concentrations.  The point is made that in studying the kinetics of a reaction for the first time, the calckinpars program for the random case should be used because it can yield the kinetic parameters for the ordered mechanism with A or B bound first as well as for the random mechanism.  
	When the kinetic parameters for a reaction have been determined with the minimum number of velocity measurements, a second set of substrate concentrations should be used to be sure that the kinetic parameters are independent (within expected experimental uncertainties) of the choices of substrate concentrations.  The accuracy of the values of the kinetic parameter can be improved by making replicate measurements of velocities.  A complete search for the best choices of substrate concentrations have not been made in calculating the tables in this paper.
	These calculations have all been concerned with mechanisms in which 2B or 3B are bound in a single reaction, but it is possible to make calculations on mechanisms in which B is bound one at a time.  There the issue will be how different the dissociation constants have to be so that it will be clear that the addition is stepwise, rather than in a single reaction.

4. References

[1]  Enzyme Nomenclature: http://www.chem.qmal.ac.uk/iubmb/enzyme/
[2]  Alberty, R. A., 2006.  Biochemical thermodynamics: Applications of Mathematica, Wiley, Hoboken, N. J. 
[3]  Alberty, R. A., 2007.  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.
[4]  Alberty, R. A., 2007.  Three mechanisms and rapid-equilibrium rate equations for a type of reductase reaction.  Biophys. Chem., 131, 71-91.
[5]  Alberty, R. A., 2007.  Effects of pH in rapid-equilibrium enzyme kinetics, J. Phys. Chem. B, 111, 14064-14068.
[6]  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.