Mechanisms of Gene Regulation: Boolean Network Models of the Lactose Operon in Escherichia coli - Mathematical Concepts and Methods in Modern Biology - page 20

Table 1.2 DVD output for the update functions in Eqs. ( 1.6 ) See the text for
details.

ANALYSIS OF THE STATE SPACE [m = 2, n = 5]
There are 4 components and 4 fixed point(s)

Components Size Cycle Length

1 8 1

2 8 1

3 8 1

4 8 1

2 5 nodes
Printing fixed point(s).
[ 0 0 0 0 0 ] lies in a component of size 8.
[ 0 0 0 0 1 ] lies in a component of size 8.
[ 0 0 0 1 1 ] lies in a component of size 8.
[ 1 1 1 1 0 ] lies in a component of size 8.

TOTAL: 32 =

G e = 0, we should enter the model into DVD as

f 1 = (( ∼ 0 ) ∗ ( x 3 + 1 ))

f 2 = x 1

f 3 = (( ∼ 0 ) ∗ (( x 2 ∗ 1 ) + ( x 3 ∗ ( ∼ x 2 )))).

2. Introduce two new variables, x 4for L e and x 5for G e . The update rules for these

variables are f 4 = x 4 and f 5 = x 5 and the entire model is

f 1 = (( ∼ x 5 ) ∗ ( x 3 + x 4 ))

f 2 = x 1

f 3 = (( ∼ x 5 ) ∗ (( x 2 ∗ x 4 ) + ( x 3 ∗ ( ∼ x 2 ))))

f 4 = x 4

f 5 = x 5 . (1.6)

Taking the second approach and running the model in DVD produces the output
in Table 1.2 and the state space diagram in Figure 1.9 . There are four fixed points,
each one of which corresponds to a different combination of the parameter values for
L e and G e . Each such combination corresponds to a component in the state space
transition graph in Figure 1.9 . The last two values of each state encode the values of
the parameters L e and G e , respectively. Viewed this way, Figure 1.9 is identical to
the analysis of the lac operon model presented in Figure 1.7 .

Exercise 1.11. Use DVD to analyze the model of the lac operon from Eqs. ( 1.4 ),
running it four times for the four different values of the parameters. Compare the
results with those in Table 1.2 , Figure 1.7 and Figure 1.9 .