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 63

we argue that this is unnecessary.

f M = A 1 ∨ ( M ∧ M old ) f B old ( 1 ) = M 2 ∧ B

f M 1 = M f B old ( 2 ) = M 2 ∧ B old ( 1 )

(2.57)

f M 2 = M 1 f A = ( B ∧ L ) ∨ L high ∨ (( A ∧ A old ) ∧ B )

f M old = A 1 ∧ M f A 1 = A

f B = M 2 ∨ ( B ∧ B old ( 2 ) ) f A old = (( B ∨ L ) ∧ L high ) ∧ A .

Analysis of the long-term behavior of the models from Eqs. ( 2.56 ) and ( 2.57 )
leads to results similar to those for the Boolean model from Eqs. ( 2.55 ). Since the
state space for the model defined by Eqs. ( 2.56 ) is relatively small, we have presented
it in Figure 2.9 (see Exercise 2.6 ). Regardless of the initial conditions, for low lactose
(

L = 0 , L high = 0 ) the system settles in a state where M = 0 , A = 0, and B = 0

and the operon is Off. For high lactose ( L = 1 , L high = 1 ) the system settles in a

state where M = 1 , A = 1, and B = 1 and the operon is On. For intermediate levels

0001 0011 0101 0111 1001 1011 1101 1111

0110

0010

1000 1110 1010 1100

0000 0011

0001

0100

1000 1010

0110

0100

0010

(A)

0111

0000

1011

1001

1110 1000 0110 0100

1100 1010 0000 0010 1100 1110

0011 0001 0111

0101

0101

1001 1011

1111

1111

(B) (C)

1101

1101

FIGURE 2.9
The state space diagram displaying the points (M,B,B old , A) for the Boolean model from
Eqs. ( 2.56 ). Panel A: Low level of internal lactose, corresponding to L 0.

Single fixed point, the operon is Off; Panel B: High level of lactose, corresponding to
L

= 0 ;L high =

1. Single fixed point, the operon is On; Panel C: Medium level of lactose,
corresponding to L

= 1 ;L high =

= 1 ;L high = 0. Two possible fixed points exist, the operon can be On

or Off.