Solution of the Briggs-Haldane System
Exact Solution for Constant [S]
We consider the enzyme kinetics model with an intermediate enzyme-substrate and
three reaction rate constants, depicted by the reaction
E+S
k1
ES
k2
E+P
k-1
This reaction can be studies using the following system of differential equations:
d[E ]
 (k 1  k 2 )[ ES ]  k1 [ E ][ S ],
dt
d [ ES ]
 k1 [ E ][ S ]  (k 1  k 2 )[ ES ],
dt
d[S ]
 k 1 [ ES ]  k1 [ E ][ S ],
dt
d [ P]
 k 2 [ ES ].
dt
Assuming that [S] does not change much over small times, we let [ S ]  s0 . Then the
first two equations of the system become
d[E ]
 (k 1  k 2 )[ ES ]  k1 s 0 [ E ],
dt
d [ ES ]
 k1 s 0 [ E ]  (k 1  k 2 )[ ES ].
dt
This is a linear system of differential equations, which can be solved exactly for [E]
and [ES]. There are two methods that can be used, a matrix method and that of writing
this system as one linear, second order constant coefficient differential equation of the
type seen by undergraduates in a first course on differential equations. We employ the
latter method.
Differentiating the first equation, we obtain
d 2 [E]
d [ ES ]
d[ E ]
 (k 1  k 2 )
 k1 s0
.
2
dt
dt
dt
Now, replace
d [ ES ]
by the expression in the second equation of the system:
dt
d 2 [E]
d[ E ]
 (k 1  k 2 )( k1 s0 [ E ]  (k 1  k 2 )[ ES ])  k1 s0
.
2
dt
dt
This equation can then be turned into an equation in [E] alone by using the first
equation of the system. Solving for [ES] and substituting the new expression in the last
equation, we get
d 2 [E]
d[ E]
 (k 1  k 2  k1 s0 )
.
2
dt
dt
This equation can be integrated to yield
d[E ]
 Ae ( k 1  k 2  k1s0 ) t ,
dt
where A is an arbitrary constant. Another integration gives the result.
[E] 
Ae  ( k1  k2  k1s0 )t
 B,
k 1  k 2  k1 s 0
where B is a second integration constant.
From equation one, we can solve for [ES] to obtain
d[E]
 k1 s 0 [ E ]
Ae t /   k1 s 0 ( B  Ae t /  )
[ ES ]  dt

,
k 1  k 2
k 1  k 2
where we have introduced   k 1  k 2  k1 s 0 .
We can determine the integration constants from the initial conditions
[ E ](0)  e0 , [ ES ](0)  0.
These conditions determine a system of two algebraic equations for A and B:
[ E ](0)   A  B  e0 ,
[ ES ](0) 
A  k1 s0 ( B  A )
 0.
k 1  k 2
Solving this system, we find
A  e0 s0 k1 ,
B  e0  A  e0  e0 s0 k1 .
Inserting these expressions into the solutions for [E] and [ES], we obtain
[ E ]  e0 [k 1  k 2  s 0 k1e t /  ],
[ ES ]  e0 k1 s 0 [1  e  t /  ].
We can now find [P]. Integrating the fourth equation of the system yields
1
[ P]  e0 k1 k 2 s 0 [t  e t /   C ],

where the integration constant C can be determined from the initial condition [ P](0)  0.
1
One finds that C   , so

1
[ P]  e0 k1 k 2 s 0 [t  (e t /   1)].

Finally, we should note that there is a natural time scale in this problem. Namely, the
time   k 1  k 2  k1 s 0 .
Condition for Constant [S]
Finally, we need to check the third equation of the system. Since [S] is assumed to be
constant, then
d[S ]
 k 1 [ ES ]  k1 [ E ][ S ]  0
dt
for the time of observation. This gives a condition on the assumption that [S] does not
change very much. Using the solutions above and the typical magnitudes of the variables
used in the program, we can establish a bound on the right hand side of this equation.
We can easily bound this expression. Note that
| k 1 [ ES ]  k1 s 0 [ E ] | k1 k 2 s 0 e0  k1 s 0 e0 (k 1  k1 s 0 )e t / 
 k1k 2 s0 e0  k1 s0 e0 (k 1  k1 s0 )  k1 s0 e0 .
Typical values for the constants in this bound are
10 6  s 0  10 3 M 1 , e0  10 9 M 1 .
Then we find that
d[S ]
 k1 s 0 e0  10  4 M sec 1 .
dt
Scaling the System
In studying such systems, it is best to convert it to dimensionless quantities and
determine which terms are the dominant terms in the system. We will introduce the
following dimensionless variables:
[ E ]  a e( ), [ ES ]  a f ( ), [ P]  a p( ), [S ]  b s( ), t  T ,
where a and b are characteristics concentrations and T is a characteristic time.
Furthermore, we relate the rate constants by
k 2  k1 , k 1  k1 .
k 1  k 2
   .
k1
The system becomes
Then, K m 
e  [(   ) f  bes]k1T ,
f  [bes  (   ) f ]k T ,
1
ak T
s  [ f  bes] 1 ,
b
p  k1Tf ,
where x 
dx
. Typical values of the parameters are given by
d
a  10 9 M , 10 6 M  b  10 3 M , T  10 3 sec, k1  108 M 1 sec 1 ,
10 6 M    10 4 M ,   10 5 M , 1.1  10 5 M  K m  1.1  10 4 M .
Each equation has a factor of k1T  10 5 M 1 . In the first equation, the first term in the
first factor is of order K m and the second is of order b. These terms are generally of the
same order as indicated below. The overall contributions from each term in the first and
second equations are of order 0.1 to 10 for times on the order of T.
k 2 (sec 1 )
100
1000
10000
K m (M )
b (M)
K m k1T
1.1  10 5
2.0  10 5
1.1  10 4
1.0  10 6
1.0  10 5
1.0  10 4
1.1
2.0
11.
bk1T
0.1
1.0
10.
ab
k1T
10 3
0.1
1.0
10.
10 4
10 5
In the third equation, we see by a similar argument, that the terms are of the order of
a , or 10 5  a  10 3. This confirms that the change in [S] is small over this time scale.
b
b
Finally, the order of the term in the last equation is k1T , which varies on the orders of
10 1 to 10.
These considerations further confirm the need to keep all the terms in the system
studied above and confirm the assumption that [S] can be treated as a constant. Such an
analysis can also be used to search for systems in which the nonlinear terms in the first
three equation should be kept.
The Velocity
The velocity is given by
v
d [ P]
d[S ]

dt
dt
for the steady state solutions. Steady state solutions are obtained from the solutions
obtained above by seeking the asymptotic values of the solutions as t gets large. We find
that
d [ P]
 e0 s 0 k1 k 2 (1  e t /  )  e0 s 0 k1 k 2 ,
dt
and
d[S ]
 k 1 [ ES ]  k1 [ E ][ S ]  k1 k 2 s 0 e0  k1 s 0 e0 (k 1  k1 s 0 )e t /   e0 s 0 k1 k 2 .
dt
Therefore, we have
v  e0 s 0 k1k 2 
e0 s 0 k1k 2
.
k 2  k 1  k1 s 0
Thinking of v as a function of s 0 , this expression does not have an extremum, but it
does level off as s0 gets large. This asymptote is called vmax and is easily seen to be
vmax  e0 k 2 .
We now determine at what velocity s 0  K m , which is a well-known relation in the
study of this system. From the above expressions for v and v max , we have
v
e0 s0 k1k 2
k1 K m vmax
v

 max .
k 2  k 1  k1 s0 k 2  k 1  k1 K m
2
v max
.
2
Finally, the expression for v can be rewritten in terms of K m :
So, when s0  K m , v 
v
e0 s0 k1k 2
sv
 0 max .
k 2  k 1  k1 s0 K m  s0
In summary, we have obtained known expressions involving the velocity of this reaction.