Biology Reference
In-Depth Information
Figure 1.5 a, and that the transition functions are given by
x 1 (
t
+
1
) =
f x 1 (
x 1 (
t
),
x 2 (
t
),
x 3 (
t
)) =
x 2 (
t
)
x 2 (
t
+
1
) =
f x 2 (
x 1 (
t
),
x 2 (
t
),
x 3 (
t
)) =
x 1 (
t
)
x 3 (
t
)
x 3 (
t
+
1
) =
f x 3 (
x 1 (
t
),
x 2 (
t
),
x 3 (
t
)) =
x 1 (
t
)
x 2 (
t
)
x 3 (
t
).
(1.1)
With the understanding that the variables on the right-hand side are always evaluated
at time t and that the variables on the left-hand side stand for the values at time t
+
1
we can simplify the notation by removing t and t
+
1 from the equations:
x 1 =
f x 1 (
x 1 ,
x 2 ,
x 3 ) =
x 2
x 2 =
f x 2 (
x 1 ,
x 2 ,
x 3 ) =
x 1
x 3
x 3 =
f x 3 (
x 1 ,
x 2 ,
x 3 ) =
x 1
x 2
x 3 .
(1.2)
Assume next that at time t
=
0, the values of the variables x 1 ,
x 2 , and x 3 are x 1
=
0
0 are often referred
to as initial conditions . Using these values to evaluate the transition functions above,
we obtain the values of the variables at time t
,
x 2 =
0
,
x 3 =
1. The values of the model variables at time t
=
=
1:
x 1 =
f x 1 (
x 1 ,
x 2 ,
x 3 ) =
f x 1 (
0
,
0
,
1
) =
0
x 2 =
f x 2 (
x 1 ,
x 2 ,
x 3 ) =
f x 2 (
0
,
0
,
1
) =
0
1
=
1
x 3 =
f x 3 (
x 1 ,
x 2 ,
x 3 ) =
f x 3 (
0
,
0
,
1
) =
0
0
1
=
1
.
Now take the newvalues x 1 =
1. These values are used to evaluate the
transition functions f x i again, producing, for time t
0
,
x 2 =
1
,
x 3 =
=
2, the values x 1 =
1
,
x 2 =
1
,
=
x 3
1. Plugging these values into the functions again, returns the same values
=
,
=
,
=
x 1
1
x 2
1
x 3
1 which correspond to the values of the variables at time
t
=
3. Thus, for any future values of t , the values of the model variables will remain
x 1
=
1
,
x 2
=
1
,
x 3
=
1. We say that we have computed the trajectory of the state
(
. We say that (1,1,1) is a fixed point for
the Boolean network. Similar considerations show that (0, 0, 0) is also a fixed point
(see Exercise 1.4 ). Using different starting values for the Boolean variables will lead
to different trajectories. For instance, the initial state (0,1,0) generates the following
repeating pattern:
0
,
0
,
1
) : (
0
,
0
,
1
) (
0
,
1
,
1
) (
1
,
1
,
1
)
(
0
,
1
,
0
) (
1
,
0
,
0
) (
0
,
1
,
0
) (
1
,
0
,
0
) (
0
,
1
,
0
)
(
form a
cycle of length two . Fixed points can be considered to be cycles of length one.
Since the system is composed of three variables each of which can take values 0
and 1, there are 2 3
1
,
0
,
0
)...
(see Exercise 1.5 ). We say that the states
(
0
,
1
,
0
)
and
(
1
,
0
,
0
)
8 possible states for the system (see Exercise 1.2 ). Computing
the trajectories initiating from each of these eight initial states and plotting them as
in Figure 1.5 b visualizes all possible trajectories for the Boolean network, containing
all possible transitions between the eight states and, thus, forming the state space
transition diagram of the Boolean network. Clearly, for a much larger number of
variables, computing the trajectories by hand would be impossible and the use of
appropriate software is essential. The graphs in Figure 1.5 were created with the
=
 
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