Owen Lewis
Math 5110
Homework #5
12/10/2015
1 Read Chapters 1 and 2 in the textbook.
2 A biochemical reaction that is ubiquitous in metabolic systems is the following sequence:
PFK
glucose −−→ GGP −−→ FGP −−→ −−−−−−−−→ FDP −−→ products.
ATP −−→ ADP
Here, FGP is the abbreviation for fructose-6-phosphate and FDP is fructose-1,6diphosphate, two important metabolites. The conversion of FGP to FDP requires energy
(via the dephosphorylation of ATP to ADP), and is also catalyzed by the enzyme PFK. It
is generally known that PFK can have two states, one of which is more “active” and thus
increases the rate of conversion of FGP to FDP. It is also known that ADP can promote
the conversion of PFK to its active state. Thus, the products of the FGP/FDP reaction
increase the rate at which the reaction occurs. This small part of the reaction chain can
be represented schematically as
+
←-
substrates −−→ FGP −−→ ADP −−→ products.
+
Here, the symbol (←-) indicates the positive feedback of ADP on the consumption of
FGP. If we let x be the non-dimensional concentration of FGP and y be the concentration
of ADP, then the equations which have been derived to describe this reaction are
dx
= δ − kx − xy2
dt
dy
= kx + xy2 − y.
dt
(1)
(a) Find the single positive steady state of ADP and FGP.
(b) For what parameter values does the system behave as a positive feedback system
(as discussed in class) near the steady state?
(c) When is the steady state stable? What behavior does the system exhibit when it is
unstable?
(a) We wish to find the steady state which is defined by
0 = δ − kx − xy2 ,
0 = kx + xy2 − y.
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Rearranging the first equation we get kx + xy2 = δ. Plugging this into the second equation
we get δ − y = 0. Clearly, at steady state we have y∗ = δ. Now, plugging this back into
the first equation we get
δ − kx − xδ2 = 0.
A quick big of algebra gives x ∗ = δ2 /(k + δ2 ). Together with y∗ = δ defines the only
positive steady state.
(b) First, some notation. We will define the functions f ( x, y) = dx/dt and g( x, y) =
dy/dt. To identify a positive feedback system, we wish to determine when the jacobian
has the following sign
"∂f ∂f # −k − y2 −2xy
− −
∂x ∂y
∼
∂g ∂g =
+ +
k + y2 2xy − 1
∂x
∂y
Because x and y represent concentrations, we will assume they are always non-negative
quantities. Similarly, we will assume that because k is a rate constant, it is positive. We
see immediately that ∂ f /∂x < 0, ∂ f /∂y ≤ 0, and ∂g/∂x > 0. Therefore the only term we
need to worry about is the bottom right entry of the jacobian. We wish to find parameters
such that 2xy − 1 > 0. Obviously this cannot always be true, therefore we will focus on
making it true near the equilibrium. Plugging in x ∗ and y∗ we are left with the desired
inequality
δ
− 1 > 0.
2δ
k + δ2
A quick rearrangement gives
2δ2 > k + δ2 ,
which finally gives the condition
δ2 > k.
As long as this condition is met, the jacobian will have the desired sign pattern at the
equilibrium, and therefore (by continuity) in some neighborhood surrounding the equilibrium.
(c) Now, we wish to determine when the equilibrium is stable. First, lets evaluate the
jacobian at the equilibrium:
"
#
2
−k − δ2 − k2δ
2
+δ
.
J=
2
k + δ2 k2δ
−
1
2
+δ
For the purposes of this problem, we will assume that the system is a positive feedback
system and therefore from the previous part, 2δ2 /(k + δ2 ) > 1. To determine stability, we
can simply examine the slopes of the two null clines at the equilibrium. First, we must
calculate the slopes. From class, we know that the slope of the x-cline is given by
sf =
− ∂∂xf
∂f
∂y
=
2
−(k + δ2 )
2δ2
k + δ2
.
2
2
x’=d−kx−xy
y ’ = k x + x y2 − y
d=2
k=2
5
5
4.5
4.5
4
4
3.5
3.5
3
3
2.5
2.5
y
y
x’=d−kx−xy
y ’ = k x + x y2 − y
2
2
1.5
1.5
1
1
0.5
0.5
0
d = 0.5
k = 0.02
0
0
0.5
1
1.5
2
2.5
x
3
3.5
4
4.5
5
0
0.5
1
1.5
2
2.5
x
3
3.5
4
4.5
5
Figure 1: Phase plane of the ADP-FGP oscillator in the stable and unstable regimes
Similarly, we can calculate the slope of the y-cline by
sg =
∂g
− ∂x
∂g
∂y
=
−(k + δ2 )
2δ2
k + δ2
−1
.
By the assumptions on δ and k, it is clear that both of these are negative slopes, and s g
has a smaller denominator. Therefore it is a “more negative” slope, and we have s f > s g .
Therefore the system always passes this test for stability.
Finally, for stability to be ensured we must satisfy the condition
sf <
∂g
∂y
∂f
∂y
=
−2δ2
k + δ2
.
2δ2
−
1
k + δ2
Substituting our already calculated value of s f , we arrive at the inequality
−(k + δ2 )
2δ2
k + δ2
<
−2δ2
k + δ2
.
2δ2
−
1
2
k+δ
Rearranging a bit, we can simplify this as far as
−2δ2 (k + δ2 )2 + (k + δ2 )3 + 4δ4 < 0.
We can simplify this further, but I don’t really care to. In Figure 1, I have included plots
of the phase plane for the system for different parameter values. All parameters lie in the
positive feedback regime. One is stable and one is unstable. We can clearly see an stable
limit cycle surrounding the unstable equlibrium in the unstable regime.
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3 In 1977, Goldbeter and Segel proposed the following equations to model AMP signaling within the cell. cAMP (cyclic adenosine monophosphate) is derived from ATP and
is an important “second messenger” involved in signaling many biological processes.
The original equations are
da
= v − σΦ( a, c)
dt
db
= qσΦ( a, c) − k b b
dt
dc
k
= b b − k c c.
dt
h
(2)
Here, a, b, and c are the non-dimensional concentrations of intracellular ATP, intracellular cAMP, and extracellular cAMP respectively. The other parameters are dimensional.
The function Φ is given by
Φ=
a (1 + a ) (1 + c )2
L + (1 + a )2 (1 + c )2
,
and represents the allosteric kinetics of adenylate cyclase (BIG WORDS!). For the original analysis of this model and realistic parameter values, see “Unified mechanism for
relay and oscillation of cyclic AMP in Dictyostelium discoideum”, Goldbeter, A., and L. A.
Segel, PNAS, 74(4), 1977.
(a) Assume that the reaction for intracellular cAMP proceeds quickly to equilibrium
and reduce the system to two equations.
(b) Choose a value of L is very large, and a value of k c that is small. Show that for
moderate values of a, the system exhibits an “n-shaped” null-cline like we’ve seen
in class.
(c) The parameter v represents the input of ATP (energy) into the system. What is
the behavior of the system for small values of v? Does the system behaves like an
excitable system?
(d) What happens as more energy is put into the system?
(a) If we assume that the second equation proceeds quickly to equilibrium, then we may
make the assumption that
db
= qσΦ( a, c) − k b b ≈ 0.
dt
We rearrange these terms and solve for the quasi-steady state concentration b
qσ
b=
Φ( a, c).
kb
We can use this to eliminate b from the other two equations are arrive at the reduced
system:
4
da
= v − σΦ( a, c),
dt
qσ
da
=
Φ( a, c) − k c c.
dt
h
(b) The null-clines are shown in Figure 2. The purple line indicates the a null-cline where
σΦ( a, c) = v,
and the orange line indicates the c null-cline where
qσ
Φ( a, c) = k c c.
h
The parameter values used here are L = 106 , σ = 1.2, k c = 0.4, and q/h = 10. These
values were pulled (approximately) from the reference in the problem statement. As you
can see, the orange c null-cline has the desired n-shape which we have seen can give rise
to oscillations or excitable behavior.
a ’ = v − s (a (1 + a) (1 + c)2)/(L + (1 + a) (1 + a) (1 + c)2)
v = 0.04
c ’ = q s (a (1 + a) (1 + c)2)/(L + (1 + a) (1 + a) (1 + c) (1 + c)) − k c
s = 1.2
L = 10000
q = 10
k = 0.4
20
18
16
14
c
12
10
8
6
4
2
0
0
1
2
3
4
5
a
6
7
8
9
10
Figure 2: Null cline structure of the cAMP oscillator system.
(c) In Figure 3 I have shown the phase plane structure for the same values as in part
(b), with v = 0.04. The single fixed point shown is a stable spiral. As this is the only
equilibrium, and it is stable, we expect that all trajectories will eventually settle on the
fixed point. This implies a globally stable steady state concentration of intracellular ATP
and extracellular cAMP. However, as the figures show, a small deviation away from this
equilibrium results in a large “loop” through phase space. Examining the time course of a
and c, we see spikes that look like the action potential of neurons. If you look at c(t), you
will see that it spikes above the steady state value then “overshoots” to below the steady
state value, before settling back down to equilibrium. Thus, we can say that the system
exhibits excitable behavior.
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a ’ = v − s (a (1 + a) (1 + c)2)/(L + (1 + a) (1 + a) (1 + c)2)
v = 0.04
c ’ = q s (a (1 + a) (1 + c)2)/(L + (1 + a) (1 + a) (1 + c) (1 + c)) − k c
s = 1.2
L = 10000
q = 10
k = 0.4
2
2
a ’ = v − s (a (1 + a) (1 + c) )/(L + (1 + a) (1 + a) (1 +
v=
c)0.04
)
s = 1.2
2
c ’ = q s (a (1 + a) (1 + c) )/(L + (1 + a) (1 + a) (1 + c) (1 + c)) − k cL = 10000
q = 10
k = 0.4
20
9
a
18
c
8
16
14
7
6
10
a and c
c
12
8
6
5
4
4
3
2
0
2
0
1
2
3
4
5
a
6
7
8
9
10
1
0
20
40
60
80
t
100
120
140
160
Figure 3: Trajectory of the cAMP system showing excitability (v = 0.04).
(c) In Figure 4 I show the behavior of the system when v is increased to v = 0.08. The
pink a null-cline has moved up, and therefore the fixed point has gone across the “hip” of
the orange c null-cline. The system has undergone a Hopf bifurcation. The fixed point is
now unstable, and a limit cycle has arisen (illustrated in blue). This limit cycle appears to
be globally stable. I have included the time courses of two trajectories of the system. In
the middle panel, I show a trajectory which was initially close to the unstable fixed point.
We can see growing oscillations which quickly appear to settle on the periodic limit cycle.
In the third panel, I show a trajectory which was started far from the fixed point. We see
oscillations which initially decay and also appear to settle on the limit cycle. So it seems
that as we input more energy into the system, it transitions from excitable, to exhibiting
stable oscillatory behavior. This will lead to periodic spikes in both ATP and cAMP within
the cell.
a ’ = v − s (a (1 + a) (1 + c)2)/(L + (1 + a) (1 + a) (1 + c)2)
v = 0.08
c ’ = q s (a (1 + a) (1 + c)2)/(L + (1 + a) (1 + a) (1 + c) (1 + c)) − k c
s = 1.2
L = 10000
q = 10
k = 0.4
2
a ’ = v − s (a (1 + a) (1 + c)2)/(L + (1 + a) (1 + a) (1 +
v=
c)0.08
)
s = 1.2
c ’ = q s (a (1 + a) (1 + c)2)/(L + (1 + a) (1 + a) (1 + c) (1 + c)) − k cL = 10000
2
a ’ = v − s (a (1 + a) (1 + c)2)/(L + (1 + a) (1 + a) (1 +
v=
c)0.08
)
s = 1.2
c ’ = q s (a (1 + a) (1 + c)2)/(L + (1 + a) (1 + a) (1 + c) (1 + c)) − k cL = 10000
q = 10
k = 0.4
q = 10
k = 0.4
20
a
18
a
c
16
c
10
20
14
8
15
8
6
a and c
10
a and c
c
12
6
10
4
4
2
5
0
2
0
1
2
3
4
5
a
6
7
8
9
10
0
0
0
50
100
150
200
250
t
300
350
400
0
50
100
150
200
250
300
350
t
Figure 4: Trajectory of the cAMP system showing an apparently globally stable limit cycle
(v = 0.08).
In Figure 5, we show the behavior of the system when v is increased to v = 0.4. The
system has undergone another Hopf bifurcation. The limit cycle is gone, and the fixed
point is stable again (as it has gone to the other hip of the null cline). Again, we show
a time course of a solution which was started far away from the equilibrium. We see
decaying oscillations, indicating that this equilibrium is globally stable. However, now
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the system is not excitable. Trajectories decay directly to the steady state without first
moving way to take a long jaunt through phase space. This means that if a “lot” of energy
is put into the system, the cell will maintain a nice stable high level of ATP and cAMP. a ’ = v − s (a (1 + a) (1 + c)2)/(L + (1 + a) (1 + a) (1 + c)2)
v = 0.4
c ’ = q s (a (1 + a) (1 + c)2)/(L + (1 + a) (1 + a) (1 + c) (1 + c)) − k c
s = 1.2
L = 10000
q = 10
k = 0.4
a ’ = v − s (a (1 + a) (1 + c)2)/(L + (1 + a) (1 + a) (1 + vc)=2)0.4
s = 1.2
c ’ = q s (a (1 + a) (1 + c)2)/(L + (1 + a) (1 + a) (1 + c) (1 + c)) − k cL = 10000
q = 10
k = 0.4
20
45
18
a
c
16
40
14
35
12
a and c
c
30
10
8
6
25
20
4
15
2
10
0
0
1
2
3
4
5
a
6
7
8
9
10
5
0
0
50
100
150
200
250
t
Figure 5: Trajectory of the cAMP system showing an apparently globally stable equilibrium (v = 0.4).
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