Biology Reference
In-Depth Information
population sizes. For instance, in the first of Eq. (7-3), we see that the rate
of increase of [C] is proportional to the product of [A] and [B], while the
rate of decrease of [C] is proportional to [C.] The second and third
equations follow from conservation of quantities arguments:
þ
þ
Because [A]
[C] and [B]
[C] are constants, we obtain:
d
dt
ð½
0, and
d
A
þ½
C
Þ ¼
dt
ð½
B
þ½
C
Þ ¼
0, which shows that
d
dt
½
d
dt
½
and
d
d
dt
½
A
¼
C
dt
½
B
¼
C
:
In equilibrium, the rates of change of the concentrations are zero;
that is,
d
½
C
d
½
A
and
d
dt
¼
½
B
¼
0
;
¼
0
;
0. From Eq. (7-3), we obtain
dt
dt
k
1
½
A
½
B
k
2
½
C
¼
0, which yields the law of mass action
k
1
k
2
½
k
1
k
2
.
½
C
¼
A
½
B
¼
K
a
½
A
½
B
with K
a
¼
When the law of mass action is applied to a balanced biochemical
equation of the type
mA
þ
nB
$
C
;
where m and n are the stoichiometric coefficients, the mathematical for-
mulation of the law of mass action is:
m
n
½
C
¼
K
a
½
A
½
B
:
(7-4)
We shall use this formulation in creating some of the mathematical
models that follow.
E
XERCISE
7-1
(a) Write the differential equations representing the rates of change
for the reaction mA
C, and use them to obtain the law of
mass action in the form of Eq. (7-4).
þ
nB
$
(b) Explain why the stoichiometric coefficients m and n appear as
exponents in the formulation (7-4) of the law of mass action.
Binding reactions of the type in Eq. (7-1) are saturable because of the
fixed number of receptors available. As the unbound drug concentration
increases, the mass action equilibrium between the unoccupied
receptors [Receptor] and the receptors with the drug bound
[Drug-Receptor] will shift towards the latter. This follows directly from
Eq. (7-2), which implies that the ratio:
½
Drug
Receptor
¼
K
a
½
Drug
½
Receptor
