Biology Reference
In-Depth Information
2.4
THE YILDIRIM-MACKEY DIFFERENTIAL EQUATION
MODELS FOR THE LACTOSE OPERON
In this section, we talk about two delay differential equation models [
8
,
9
] recently
developed to study the
lac
operon. Both models employ experimentally estimated
parameters and are compared against independently collected, published experimental
data from this genetic circuit. The first model is a 3 variable model and only takes
into account mRNA (
M
),
-galactosidase (
B
), and allolactose (
A
) dynamics. Internal
lactose (
L
) is assumed to be readily available and is included in the model as a
parameter [
8
]. In the second model, besides these three dynamic variables, the internal
lactose (
L
) and permease (
P
) are also explicitly modeled [
9
].
β
2.4.1
Model Justification
We need a few more preliminaries before presenting the differential equation models.
Modeling dilution in protein concentration due to bacterial growth:
When devel-
oping a mathematical model for bacterial systems like the
lac
operon, it is important
to know how fast the bacteria actually grow. Depending on the environmental con-
ditions, an
E. coli
population can double in size as fast as every 20 min [
10
]. This
increase in the cellular volume results in a dilution in the protein concentrations,
which can be critical in developing reliable mathematical models. Dilution in protein
concentrations due to bacterial growth can be modeled as follows: Suppose that
V
is
the average volume of a bacterial cell and
x
represents the number of molecules of a
protein
X
in that cell. Assume that the cell volume increases exponentially in time.
Then we have,
dV
dt
=
μ
V
(2.28)
where
0 denotes the growth rate. Also, assume that the degradation of X is
exponential. Then we can write,
μ>
dx
dt
=−
β
x
,
(2.29)
β>
where
0 represents this decay rate. Concentration of the protein
X
is equal to the
number of molecules
x
of
X
divided by volume of the cell, namely
x
V
,
[
x
]=
(2.30)
where
stands for concentration of the protein
X
. Differentiating both sides of
Eq. (
2.30
) with respect to time
t
results in
[
x
]
dx
dt
V
dt
x
1
d
[
x
]
dV
=
−
V
2
.
(2.31)
dt
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