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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