Biology Reference
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−
d
[
S
]
d
[
P
]
Differentiating both sides of this equation with respect to
t
yields
=
.
dt
dt
Hence, Eq. (
2.14
) becomes
d
[
S
]
V
max
[
S
]
=−
]
.
(2.18)
dt
K
m
+[
S
The evolution of the substrate concentration can be simulated by solving this differen-
tial equation after fixing
V
max
and
K
m
and assigning a value to the initial concentration
of the substrate. Then, from Eq. (
2.17
), one can compute the progress curve for the
concentration of the product.
Exercise 2.1.
Consider the reactions where two substrates
S
1
and
S
2
compete
for binding to an enzyme
E
to produce two different products
P
1
and
P
2
. Under the
assumption that each reaction follows theMichaelis-Menten kinetics, derive rate equa-
tions for
P
1
and
P
2
in this system and explain the effects of the competition occurring.
k
1
k
2
k
3
−→
E
+
S
1
ES
1
P
1
+
E
.
p
1
p
2
p
3
−→
E
+
S
2
ES
2
P
2
+
E
.
2.3.2
Multi-Molecule Binding and Hill Equations
Some enzymes react with more than one substrate molecule, e.g., the reaction in
Eq. (
2.19
), in which
n
-molecules of a substrate react with an enzyme
E
. For simplicity,
let's assume that the binding sides on the enzyme are all identical, the bindings take
place simultaneously and the bindings of
n
-molecules are independent. Otherwise,
the influence of each binding step to other binding steps has to be considered.
nS
k
1
k
3
→
E
+
k
2
ES
n
P
+
E
.
(2.19)
k
1
−→
This reaction consists of three individual reactions. In the first reaction,
E
+
nS
n
-molecules of a substrate
S
interact with one molecule of an enzyme
E
to form
an enzyme-substrate complex
ES
n
with a rate constant
k
1
.
ES
n
ES
n
,
k
2
−→
nS
is the
second reaction, in which the enzyme-substrate complex
ES
n
can break down into the
enzyme
E
and
n
-substrate
S
molecules with a rate constant
k
2
. In the third reaction,
ES
n
E
+
k
3
−→
P
, the enzyme
E
releases unchanged from the complex and a product
P
is produced with a rate constant
k
3
. The differential equation modeling dynamics
of the concentration of the enzyme-substrate complex
ES
n
is the difference between
the production and consumption rates of this complex.
ES
n
is produced with the first
reaction and it is used up with both the second and the third reactions. Hence,
E
+
[
ES
n
]
dt
d
n
=
k
1
[
][
]
−
(
k
2
+
k
3
)
[
ES
n
]
.
E
S
(2.20)
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