Biomedical Engineering Reference
In-Depth Information
Note that Eq. (8.43) provides a nonlinear expression for
t
as a function of
q
S
and that, until
recently, it was thought impossible to write an expression of
q
S
as a function of
In 1997,
t.
Schnell and Mendoza used computer algebra to solve Eq. (8.43) for
q
S
as a function
of
t
using the Lambert
W
function. While Eq. (8.43) can be used to solve for
t
by substi-
tuting values of
q
S
from
q
S
ð
0
Þ
to zero, it is far easier to simulate Eq. (8.41) for
q
S
:
The product is given by
q
P
¼
K
2
q
ES
¼
K
2
q
E
ð
0
Þ
ð
8
:
44
Þ
þ
K
M
q
S
1
To eliminate
q
S
, we have
q
S
¼
q
S
ð
0
Þ
q
ES
q
P
:
Since
q
ES
is small compared to the other
quantities after the initial phase of the response, we have
q
S
¼
q
S
ð
0
Þ
q
P
and
q
P
¼
K
2
q
E
ð
0
Þ
K
2
q
E
ð
0
Þ
¼
ð
8
:
45
Þ
þ
K
M
q
S
K
M
q
S
ð
1
1
þ
0
Þ
q
P
Solving Eq. (8.45) using the same approach to find Eq. (8.43), we have
1
q
S
ð
Þ
q
P
q
S
ð
0
t
¼
q
P
K
M
ln
ð
8
:
46
Þ
q
E
ð
0
Þ
0
Þ
K
2
This approximation works well for the reaction, except during the initial quick phase when
both
q
P
are small.
Biochemists define the reaction rate as either the rate of disappearance of the substrate or
the appearance of the product. Using the Michaelis-Menten approximation, the reaction
rates are equal and are given by
q
ES
and
However, based on the true differential
equations in Eq. (8.34), these two definitions are not equal. Using the approximation in
Eq. (8.41), we have
V
¼
q
S
¼
q
P
:
V
¼
q
S
¼
K
2
q
S
q
E
ð
0
Þ
V
max
q
S
q
S
þ
K
M
V
max
1
Þ
¼
Þ
¼
ð
8
:
47
Þ
ð
q
S
þ
K
M
ð
þ
K
M
q
S
where
is the maximum velocity of the substrate disappearance. Equation
(8.47) is plotted in Figure 8.5. Note that at high substrate levels,
V
max
¼
K
2
q
E
ð
0
Þ
V
max
, since
all of the enzyme is engaged and the velocity saturates. In this region, the reaction rate is
independent of
V
approaches
At low substrate levels, the reaction rate is approximately linearly depen-
dent on substrate level,
q
s
:
V
¼
V
max
K
M
q
s
:
V
¼
V
max
2
Now consider when
:
Substituting into Eq. (8.47), we have
V
max
2
V
max
q
S
q
S
þ
K
M
¼
ð
Þ
