Biology Reference
In-Depth Information
The state space graph of the PDS from Example
3.2
is shown in Figure
3.1
(b).
For example, there is an arrow from state
(
2
,
1
)
to state
(
2
,
0
)
because
F
((
2
,
1
))
=
(
f
1
(
2
),
f
2
(
1
))
=
(
2
,
0
)
using modulo 3 arithmetic (since the field is
Z
3
). The graph
consists of 3
2
=
9 vertices arranged into three connected components. Notice that
. Such points are called
fixed points
and
these two are the only fixed points for this PDS. Also notice that
F
F
((
0
,
0
))
=
(
0
,
0
)
and
F
((
1
,
2
))
=
(
1
,
2
)
◦
F
((
0
,
2
))
=
. This is called a
limit cycle
and its length is two. Analogously, for any positive integer
m
,a
limit cycle of
length m
arises whenever
F
m
F
(
F
((
0
,
2
)))
=
F
((
1
,
0
))
=
(
0
,
2
)
, or graphically,
(
0
,
2
)
(
1
,
0
)
(
v
)
=
v
(and
m
is the smallest positive integer with this
n
. Fixed points can be considered as limit cycles of length
property) for some
v
∈ F
one. For this PDS,
is the only limit cycle of length larger than one.
Exercise 3.1.
Construct by hand the state space graphs and wiring diagrams for the
following PDSs. Can a vertex in the state space graph have an out-degree greater than
one, that is, can the vertex have more than one directed edge coming out of it? How
about a vertex in the wiring diagram?
(
0
,
2
)
(
1
,
0
)
2
2
1.
F
=
(
f
1
,
f
2
)
: Z
2
→ Z
2
, where
f
1
=
x
1
+
x
2
,
f
2
=
x
1
x
2
.
3
3
2.
F
=
(
f
1
,
f
2
,
f
3
)
: Z
3
→ Z
3
, where
f
1
=
x
1
+
x
2
,
f
2
=
x
1
x
3
+
1
,
f
3
=
x
2
x
3
.
Exercise 3.2.
Compute the fixed points of the PDS in Exercise
3.1
(1) algebraically.
Compare your answers to what you found in the previous exercise. Hint: A fixed point
will satisfy
x
1
=
Exercise 3.3.
In Exercise
3.1
(2), you saw that the PDS has no fixed points. If instead
we consider the system over
x
1
+
x
2
and
x
2
=
x
1
x
2
.
Z
5
, would you expect to see fixed points? Verify your
answer using the web tool Analysis of Dynamic Algebraic Models (ADAM) at
dvd.vbi.vt.edu/adam.html
.
Exercise 3.4.
Find three different PDSs over
Z
3
that have the wiring diagram as in
Figure
3.5
.
Exercise 3.5.
Z
2
that
fits the data as in Figure
3.6
. Is the PDS unique? What if the entire state space was
given?
Let's reverse engineer! Find a PDS on three variables over
Exercise 3.6.
[
14
] Phosphorylation is the addition of a phosphate group to a protein
or any other organic molecule. It turns many protein enzymes on and off, thereby
altering their function and activity. Suppose proteins A and B interact. Let
x
1
reflect
the absence (0)/presence (1) of protein A, whereas
x
2
indicates if the protein B has
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