Biology Reference
In-Depth Information
IV. OSCILLATIONS DRIVEN BY A SINGLE-SYSTEM
FEEDBACK LOOP
In this section, we revisit models of formal networks similar to the model
in Example 6, focusing on networks describing a feedback interaction
between two hormones, A and B. The concentration of one, hormone A,
regulates the secretion of another, hormone B, which in turn controls the
release of A, as shown in Figure 10-14:
(
+
)
(
−
)
A
A
Elimination
Elimination
(
+
)
(
−
)
D
D
B
B
Elimination
Elimination
FIGURE 10-14.
Formal networks of two-node/one-feedback oscillators. The left panel depicts a network in which
the main Hormone B is stimulated, while the other shows a model in which B is inhibited. D denotes
a delay in the corresponding interconnection. In both networks, A and B are subject to elimination.
(Reprinted from Farhy, L. S. [2004]. Modeling of oscillations in endocrine networks with feedback,
Methods in Enzymology, 384, 54-81. Copyright 2004, with permission from Elsevier.)
Notice that in both constructs shown in Figure 10-14, any one of the
hormones suppresses, indirectly, its own secretion. Therefore, these
networks contain a negative feedback loop. Such networks are quite
common in endocrinology. For example, in the male rat, GH stimulates
the release of somatostatin with a lag of 60 to 120 minutes, and
somatostatin, in turn, suppresses the release of GH.
As before, we assume the two hormones, A and B, are continuously
secreted (driven by nonrhythmic excitatory inputs) in certain pool(s),
such as the systemic circulation, where they are subject to elimination.
The release of hormone B is regulated by hormone A. Hormone B
itself exerts a delayed effect on the secretion of A. The interactions
between A and B are assumed to be dose-responsive. For simplicity, we
assume no delay in the action of A on B and no basal secretions.
The equations corresponding to the schemes in Figure 10-14 could be
written following the pattern in Example 10-6. More specifically, the
system of ODEs describing the schematic diagram in the left panel is:
dC
A
dt
¼a
1
C
A
ð
t
Þþ
a
n
B
ð
C
B
ð
t
D
B
Þ=
T
B
Þ
þ
1
(10-14)
n
A
dC
B
dt
¼b
ð
C
A
ð
t
Þ=
T
A
Þ
C
B
ð
t
Þþ
b
1
:
n
A
ð
C
A
ð
t
Þ=
T
A
Þ
þ
