Bistability in the Lactose Operon of Escherichia coli: A Comparison of Differential Equation and Boolean Network Models - Mathematical Concepts and Methods in Modern Biology - page 60

B , and A by half are calculated to be h M = 572 min, h B M , 1 . = 22 . 360 min, and

h A =

753 min.
Combining this information with the methodology outlined in Section 2.5 leads to
the following choice of variables for the Boolean model:

15 .

1. We will model the dynamics of the Boolean variables M , B , and A , denoting

-galactosidase, and allolactose.
2. We assume that glucose is absent and intracellular lactose is present at all times.
The intracellular lactose concentration L is a parameter for the model. From
Section 2.4 , we know that the model in Table 2.1 has multiple steady states
when the internal lactose concentration is in a certain intermediate (maintenance)
range, estimated numerically to be between (0.039, 0.055) mM (see Figure 2.4 ).
To distinguish between low, medium, and high lactose concentrations in the
Boolean case, we introduce an additional Boolean parameter L high .Thevalue
L

mRNA, β

1 implies intracellular lactose concentrations are within the maintenance
range or higher while L

=

0 implies intracellular lactose concentrations are
below the maintenance range. When L high =

=

1, internal lactose concentration is

higher than the maintenance range.
3. We select a discrete time step of about 10 min for the Boolean model. The delays
τ M ,τ B can then be ignored since they are much smaller than the time step.
Similarly, since h M

10 min, in the absence of new mRNA production, any
available amounts of mRNAwould be reduced below the discretization threshold
in one time step. Thus there is no need for introducing M old .
4. We define additional variables A old ,

B old ( 1 ) , and B old ( 2 ) to model the different

degradation rates of allolactose and β -galactosidase.

We emphasize that the choices regarding the size of the discrete time step and the
exact number of “old” variables chosen for this model represent just one of many pos-
sibilities. Alternative models are considered later in the chapter and in the exercises.
Combining these assumptions with the assumption that translation and transcrip-
tion happen in one time step, we obtain the following Boolean model:

f M = A f B old ( 2 ) = M ∧ B old ( 1 ) , (2.55)

f B = M ∨ ( B ∧ B old ( 2 ) ) f A = ( B ∧ L ) ∨ L hi gh ∨ (( A ∧ A old ) ∧ B ),

f B old ( 1 ) = M ∧ B f A old = (( B ∨ L ) ∧ L high ) ∧ A .

A justification for the transition functions in Eqs. ( 2.55 ) follows:
Transition Equation for M : In the presence of allolactose at time t , mRNA will be
produced and present at the next time step t + 1. Availability of M at time t does not

affect its availability at time t + 1 since M is assumed to degrade completely within

one time step.
Transition Equation for B : When mRNA is available at time t , translation and
transcription of

1. If amounts of
B produced earlier are still available in high enough concentrations to not fall below

β -galactosidase will take place and B = 1atstep t +