Biomedical Engineering Reference
In-Depth Information
These equilibriums are subjected to the law of mass action:
k
1
k
−1
[ES]
[E]
=
=
K
a
[S]
K
a
is the association equilibrium constant and is inverse propor-
tional to the dissociation equilibrium constant K
D
:
·
[S]
[ES]
[E]: enzyme concentration; [S]: substrate concentration; [ES]: con-
centration of the enzyme-substrate complex; [P]: product concen-
tration; k
1
: association equilibrium rate constant; k
−1
:dissociation
equilibrium rate constant (rate constant of the back reaction).
It is stated that during an in vitro enzymatic reaction the con-
centration of the enzyme shall not change during the test, and that
the substrate concentration exceeds the enzyme concentration in
orders of magnitude: in a first approximation the substrate con-
centration is practically constant, too. Both of these assumptions
transform a reaction of 2
nd
order into the much simpler reaction of
0
th
order. If the concentrations of enzyme and substrate are similar,
we get a reaction of 1
st
order. The reaction rate v for the association
reaction
1
K
a
k
−1
k
1
[E]
·
=
=
=
K
D
→
E+S
ES
is in the case of large substrate excess (0
th
order)
d[S]
dt
=
=
v
k
1
and
−
d[S]
dt
d[P]
dt
=
=
=
v
k
1
·
[S]
=
respectively. Integration of the differential quotients over time t
0
to t gives
=
ln[S]
ln[S
0
]−k
1
·
t
=
[S
0
]: substrate concentration at t
0
This equation allows the determination of the rate constant
of the association reaction, and analogously, by measuring the
product forming, the dissociation rate constant.
A further important value is the time of half-change
t1
2
, i.e.,
that time at which half the amount of substrate is converted ([S]
/
=
0.5
·
[S
0
]):
0.69315
k
In steady state, i.e., when association and dissociation occur at the
same speed, the change of [ES] and [E] are the same in the time
ln2
k
=
=
t
1/2
