Mechanisms of Gene Regulation: Boolean Network Models of the Lactose Operon in Escherichia coli - Mathematical Concepts and Methods in Modern Biology - page 31

I=ideal(x1+x4*x5+x4,x1+x2,x1+x3,x4+1,x5+x6*x7+x6+x7
+1,x6+x3*x8,
x6+x7+x8+x9+x8*x9+x6*x8+x6*x9+x6*x8*x9, x8, x9+x8)
I.groebner_basis().
Evaluating, SAGE returns the following Groebner basis for the set of functions in the
left-hand sides of Eqs. ( 1.13 ):
[x1, x2, x3, x4 + 1, x5 + 1, x6, x7, x8, x9].
Thus, the system of equations from Eqs.( 1.13 ) has the same solution set as the
system of equations

x 1 = 0 x 4 + 1 = 0 x 7 = 0

x 2 = 0 x 5 + 1 = 0 x 8 = 0

x 3 = 0 x 6 = 0 x 9 = 0 .

This gives the steady state ( M , P , B , C , R , A , A l , L , L l ) = ( x 1 , x 2 , x 3 , x 4 , x 5 ,

x 6 0.

Exercise 1.20. Use SAGE to continue the computations from Example 1.10 and
show that the Boolean model of the lac operon from Eqs. ( 1.8 ) has the following fixed
points for the remaining combinations of parameter values: (1) For : a

, x 7 , x 8 , x 9 ) = ( 0 , 0 , 0 , 1 , 1 , 0 , 0 , 0 , 0 ) for L e = a = 0 , G e = g =

= 0 , g = 1 :

( M , P , B , C , R , A , A l , L , L l ) = ( 0 , 0 , 0 , 0 , 1 , 0 , 0 , 0 , 0 ) ;(2)For: a = 1 , g = 1 :

( M , P , B , C , R , A , A l , L , L l ) = ( 0 , 0 , 0 , 0 , 1 , 0 , 0 , 0 , 0 ) ;(3)For: ( a = 1 , g = 0 :

(

The results from Example 1.10 and Exercise 1.20 show that the Boolean model of
the lac operon from Eqs. ( 1.8 ) has the right qualitative behavior, predicting that the
operon is On only when external lactose is available and external glucose is not. In
this case all variables of the model, except for the repressor protein, are present. When
glucose is available, the operon is Off. All fixed points are biologically feasible.
Now that we understand how fixed points may be found as solutions of systems
of polynomial equations, we can use DVD again to analyze the model and compare
the results to those from Example 1.10 and from Exercise 1.20 . In DVD you can
either enter and run the model four times for the four different combinations of the
parameter values or use the approach we took in Example 1.5 (2) and include an
additional “variable” for each of the parameters with an update rule that keeps that
variable constant for all time steps. The analysis in DVD confirms the fixed points
from Example 1.10 and from Exercise 1.20 and affirms that the state space does not
contain limit cycles.

M , P , B , C , R , A , A l , L , L l ) = ( 1 , 1 , 1 , 1 , 0 , 1 , 1 , 1 , 1 ) .

1.5 CONCLUSIONS AND DISCUSSION
Utilization of lactose by the bacterium E.coli requires the production of two proteins,
the transporter lactose permease and the enzyme -galactosidase. Efficient use of cel-

lular energy and materials requires that the bacterium only make these two proteins
when needed. The regulatory mechanism of the lac operon ensures the repression
of these two proteins when glucose, the preferred energy source, is present and the

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