Math 4600: Homework 5 Solutions
Gregory Handy
[5.1] Look at the solution of the full model (bursting). What does s do during the silent and active
phase? Can you estimate the (lower) value of s where a saddle-node bifurcation happens just by
looking at this solution?
As Fig. 1 illustrates, during the silent phase of the model, s is decreasing, and during the active phase, s is
increasing. From class, we know that the saddle-node bifurcation should occur just before the active phase
begins (when s is a minimum). This value can be read from the graph to be ∼ 0.36.
−20
−30
V
−40
−50
−60
−70
0
1
2
3
4
t
5
4
x 10
0.5
s
0.45
0.4
0.35
0
1
2
3
4
t
5
4
x 10
Figure 1: Plot for Problem 1.
The figure above was produce with the files posted on class website (using default values):
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Script for HW 5 Prob1
clear;
close all;
% Full beta-cell model
%initial conditions
v0=-60;n0=0.0001;s0=.4;
% Conductances in pS; currents in fA; Ca concentrations in uM; time in ms
[t,Y]=ode45(@hw5_3_ode,[0 50000],[v0 n0 s0]);
v=Y(:,1); n=Y(:,2); s=Y(:,3);
figure(2)
subplot(2,1,1)
plot(t,v,’-’)
xlabel(’t’,’fontsize’,16)
ylabel(’V’,’fontsize’,16)
set(gca,’fontsize’,16)
subplot(2,1,2)
plot(t,s)
xlabel(’t’,’fontsize’,16)
1
ylabel(’s’,’fontsize’,16)
set(gca,’fontsize’,16)
and it requires the function
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% ODE file for HW 5 Prob1
function dy=hw5_3_ode(t,Y)
dy=zeros(3,1);
v=Y(1); n=Y(2); s=Y(3);
%Ik
gk=2700.0;vn=-16;sn=5.6;vk=-75;taun=20;
% Is
gs=200;vs=-52;ss=5;taus=20000;
% Ica
gca=1000;vca=25;vm=-20;sm=12;
%Ikatp
gkatp=120;
cm=5300;
%Functions
ninf=1./(1+exp((vn-v)/sn));
minf=1./(1+exp((vm-v)/sm));
sinf=1./(1+exp((vs-v)/ss));
ica=gca*minf.*(v-vca);
ik=gk*n.*(v-vk);
is=gs*s.*(v-vk);
ikatp=gkatp*(v-vk);
% Equations
dy(1)=(-ica-ik-is-ikatp)/cm;
dy(2)=(ninf-n)/taun;
dy(3)=(sinf-s)/taus;
end
[5.2] (a) In MATLAB plot nullclines of the two-dimensional V − n system for s = 0.8. In the same
plot, graph several solutions, with different initial conditions (using provided files) to explore
the stability of the steady state(s). What do you find? Do your findings match the bifurcation
diagram we did in class?
(b) Repeat part a) for s = −0.2
(c) Repeat part a) for s = 0.4
We must first solve for the nullclines of our reduced, two-dimensional system. We begin with the following
dv
= (−Ica (V ) − Ik (V, n) − Is (V ) − Ikatp (V ))/Cm
dt
n∞ (V ) − n
dn
=
.
dt
τn
Solving for the n-nullcline first, we find
n∞ (V ) − n
= 0 ⇒ n = n∞ (V )
τn
−1
⇒ n(V ) = [1 + exp((Vn − V )/sn )]]
.
We can solve for the V -nullcline by setting dv/dt = 0,
−Ica (V ) − Ik (V, n) − Is (V ) − Ikatp (V ) = 0
2
⇒ −gca m∞ (V )(V − Vca ) − gk n(V − Vk ) − gs s(V − Vk ) − gkatp (V − Vk ) = 0.
One may be tempted to solve this equation for V as a function of n (since we are solving for the V -nullcline,
but since m∞ is a function of V, this will be very difficult. Instead, we note that we can easily solve this for
n as a function of V
⇒ n(V ) = −
Ica (V ) + Is (V ) + Ikatp (V )
.
gk (V − Vk )
(a) As the left panel in Fig. 2 illustrates, when s = 0.8, we have one steady state, and the trajectories
chosen imply that this is a stable node.
(b) For s = −0.2, the middle panel shows that the v-nullcline has shifted up, and we now have one, higher
steady state. This time, the trajectories suggest that this is a stable spiral.
(c) Finally, for s = 0.4 we have three steady states. The trajectories show that the lower one is a stable
node, the middle is a saddle node, and high one is an unstable spiral.
0.15
0.15
n−nullcline
v−nullcline
n−nullcline
v−nullcline
n−nullcline
v−nullcline
0.25
0.1
0.1
0.15
n
0.05
0.05
n
n
0.2
0.1
0.05
0
0
0
−0.05
−80
−70
−60
−50
−40
−30
−20
−10
−0.05
−80
−70
−60
−50
V
−40
V
−30
−20
−10
−0.05
−80
−70
−60
−50
−40
−30
−20
−10
V
Figure 2: Plot for Problem 2. The left panel is for s = 0.8, middle for s = −0.2, and the left is for s = 0.4
with different trajectories (denoted by different line styles).
The code for part a) is given by the following:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%% Script for Prob 2a
clear;
close all;
% v-n fast subsystem for phase plane analysis and bifurcation diagram
% parameters for the system
%Ik
gk=2700.0;vn=-16;sn=5.6;vk=-75;taun=20;
% Is
gs=200;vs=-52;ss=5;
% Ica
gca=1000;vca=25;vm=-20;sm=12;
%Ikatp
gkatp=120;
cm=5300;
% s is a parameter for this 2D system
s=0.8;
% we will vary V to calculate our nullclines
v = [-80:.01:-10];
%Functions
ninf=1./(1+exp((vn-v)/sn));
minf=1./(1+exp((vm-v)/sm));
sinf=1./(1+exp((vs-v)/ss));
ica=gca*minf.*(v-vca);
is=gs*s.*(v-vk);
ikatp=gkatp*(v-vk);
% first, plot the n-nullclines
3
n_n_nullcline=1./(1+exp((vn-v)/sn));
plot(v,n_n_nullcline,’red’,’linewidth’,1.5)
hold on
% now lets plot the v-nullcline
n_v_nullcline = (is+ikatp+ica)./(-gk*(v-vk));
plot(v,n_v_nullcline,’g’,’linewidth’,1.5)
% now plot solution trajectories
%initial conditions
v0=-30;n0=0.1;
% Conductances in pS; currents in fA; Ca concentrations in uM; time in ms
[t,Y]=ode45(@hw5_2_ode,[0 4000],[v0 n0],odeset,s);
v=Y(:,1);
n=Y(:,2);
plot(v,n,’-’,’linewidth’,1.5)
%initial conditions
v0=-74;n0=0.05;
% Conductances in pS; currents in fA; Ca concentrations in uM; time in ms
[t,Y]=ode45(@hw5_2_ode,[0 4000],[v0 n0],odeset,s);
v=Y(:,1);
n=Y(:,2);
plot(v,n,’--’,’linewidth’,1.5)
%initial conditions
v0=-60;n0=-0.02;
% Conductances in pS; currents in fA; Ca concentrations in uM; time in ms
[t,Y]=ode45(@hw5_2_ode,[0 4000],[v0 n0],odeset,s);
v=Y(:,1);
n=Y(:,2);
plot(v,n,’.’,’linewidth’,1.5)
xlabel(’V’,’fontsize’,16)
ylabel(’n’,’fontsize’,16)
axis([-80 -10 -.05 .15])
legend(’n-nullcline’, ’v-nullcline’)
clear gca
set(gca,’fontsize’,16)
where it uses the ODE file
% ODE function for 2D system, where s is now a parameter
function dy=hw5_2_ode(t,Y,s)
dy=zeros(2,1);
v=Y(1);
n=Y(2);
%Ik
gk=2700.0;vn=-16;sn=5.6;vk=-75;taun=20;
% Is
gs=200;vs=-52;ss=5;
% Ica
gca=1000;vca=25;vm=-20;sm=12;
%Ikatp
gkatp=120;
cm=5300;
%Functions
ninf=1./(1+exp((vn-v)/sn));
minf=1./(1+exp((vm-v)/sm));
sinf=1./(1+exp((vs-v)/ss));
ica=gca*minf.*(v-vca);
ik=gk*n.*(v-vk);
is=gs*s.*(v-vk);
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ikatp=gkatp*(v-vk);
% Equations
dy(1)=(-ica-ik-is-ikatp)/cm;
dy(2)=(ninf-n)/taun;
end
Note: the code for parts b) and c) are achieved by changing the value of s.
[5.3] Using the MATLAB files for the full model, test the effect of changing the glucose level. The
effect of glucose is to BLOCK the ATPsensitive potassium current IK(ATP) , so increasing the glucose
level in the model corresponds to decreasing the conductance gK(ATP ) .
(a) Try gK(ATP ) equal 200, 120, and 60. Look at the solution V (t) and explain why the results
make sense in terms of amount of insulin that would be released.
(b) In the s-V plane, plot the s-nullcline, and the Z-shaped curve of the bifurcation diagram (steady
states vs. s as a parameter) for gK(ATP ) = 200 and gK(ATP ) = 120. Discuss the intersection of
the two curves in each case and how it corresponds to the solution that you found in part a)
(a) When gK(ATP ) = 200, corresponding to low levels of glucose, we see no bursting or spiking behavior
(Fig. 3, left column). This means that no insulin is release,which makes sense, since we have a very low
amount of glucose. Decreasing gK(ATP ) to 120 and then to 60 (middle and right column), corresponding
to increasing levels of glucose, we see bursting, following by steady spiking. Again, this makes sense,
since as glucose levels increase, we would expect more insulin to be released.
gkatp=200
gkatp=120
−60
gkatp=60
−20
0
−40
−20
V
V
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V
−62
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−40
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1
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5
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1
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t
4
x 10
−60
0
5
0.6
0.6
0.6
0.4
0.4
0.4
s
0.8
0
0
0.2
1
2
3
t
4
5
0
0
2
3
4
t
0.8
0.2
1
4
x 10
0.8
s
s
t
5
4
x 10
0.2
1
2
3
t
4
x 10
4
5
4
x 10
0
0
1
2
3
t
4
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4
x 10
Figure 3: Plot for Problem 3a.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Code for Prob 3a
clear;
close all;
% Full beta-cell model
%initial conditions
v0=-60;n0=0.0001;s0=.4;
for i = 1:3
% Conductances in pS; currents in fA; Ca concentrations in uM; time in ms
5
if i ==1
[t,Y]=ode45(@hw5_3_ode_v2,[0 50000],[v0 n0 s0],odeset,200);
elseif i == 2
[t,Y]=ode45(@hw5_3_ode_v2,[0 50000],[v0 n0 s0],odeset,120);
else
[t,Y]=ode45(@hw5_3_ode_v2,[0 50000],[v0 n0 s0],odeset,60);
end
v=Y(:,1);
n=Y(:,2);
s=Y(:,3);
subplot(2,3,i)
plot(t,v,’-’)
if i == 1
axis([0 50000 -70 -60])
elseif i == 2
axis([0 50000 -80 -20])
else
axis([0 50000 -60 0])
end
xlabel(’t’,’fontsize’,16)
ylabel(’V’,’fontsize’,16)
set(gca,’fontsize’,16)
subplot(2,3,i+3)
plot(t,s)
xlabel(’t’,’fontsize’,16)
ylabel(’s’,’fontsize’,16)
set(gca,’fontsize’,16)
axis([0 50000 0 0.8])
end
where we used the following modified ODE function
function dy=hw5_3_ode_v2(t,Y,gkatp)
dy=zeros(3,1);
v=Y(1);
n=Y(2);
s=Y(3);
%Ik
gk=2700.0;vn=-16;sn=5.6;vk=-75;taun=20;
% Is
gs=200;vs=-52;ss=5;taus=20000;
% Ica
gca=1000;vca=25;vm=-20;sm=12;
cm=5300;
%Functions
ninf=1./(1+exp((vn-v)/sn));
minf=1./(1+exp((vm-v)/sm));
sinf=1./(1+exp((vs-v)/ss));
ica=gca*minf.*(v-vca);
ik=gk*n.*(v-vk);
is=gs*s.*(v-vk);
ikatp=gkatp*(v-vk);
% Equations
dy(1)=(-ica-ik-is-ikatp)/cm;
dy(2)=(ninf-n)/taun;
dy(3)=(sinf-s)/taus;
end
(b) As Fig. 4 illustrates, when gK(ATP ) = 200 (left column), the Z-shaped bifurcation diagram has been
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shifted to the left, and the s-nullcline intersects it on the lower, stable branch, and we would expect no
bursting to occur. However, when gK(ATP ) = 120 (right column), we see that the Z-shaped bifurcation
diagram has shifted to the right, and the intersection is now in the middle, unstable branch. As we
discussed in class, the position of this intersection implies that bursting will now occur.
To plot the Z-shaped bifurcation diagram (as a function of s), we set dv/dt = 0,
−Ica (V ∗ ) − Ik (V ∗ , n) − Is (V ∗ ) − Ikatp (V ∗ ) = 0
⇒ −gca m∞ (V ∗ )(V ∗ − Vca ) − gk n∞ (V ∗ )(V ∗ − Vk ) − gs s(V ∗ − Vk ) − gkatp (V ∗ − Vk ) = 0,
where we use the fact that at steady state we have n = n∞ (V ∗ ). Similar to Prob 2, instead of solving
for V ∗ , we solve for the parameter s, which yields
s(V ∗ ) = −
g
Ica (V ∗ ) + Ik (V ∗ , n∞ (V ∗ )) + Ikatp (V ∗ )
.
gs (V ∗ − Vk )
=200
g
katp
=120
katp
−10
−10
Z−Shaped Bifucation
s−nullcline
Z−Shaped Bifucation
s−nullcline
−30
−30
−40
−40
V
−20
V
−20
−50
−50
−60
−60
−70
−70
−80
0
0.2
0.4
0.6
0.8
−80
0
1
0.2
s
0.4
0.6
s
Figure 4: Plot for Problem 3b.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%% Code for HW 5 Prob3b
clear;
close all;
% parameters for the system
%Ik
gk=2700.0;vn=-16;sn=5.6;vk=-75;taun=20;
% Is
gs=200;vs=-52;ss=5;
% Ica
gca=1000;vca=25;vm=-20;sm=12;
cm=5300;
% we will vary V to calculate our nullcline and bifurcation diagram
v = [-80:.01:-10];
for i = 1:2
if i == 1
gkatp=200;
else
gkatp=120;
gca=1000;
end
%Functions
7
0.8
1
ninf=1./(1+exp((vn-v)/sn));
minf=1./(1+exp((vm-v)/sm));
sinf=1./(1+exp((vs-v)/ss));
ica=gca*minf.*(v-vca);
ikatp=gkatp*(v-vk);
ik=gk*ninf.*(v-vk);
bif_dia = (ica+ik+ikatp)./(-gs*(v-vk));
% plots the bifurcation diagram
subplot(1,2,i)
plot(bif_dia,v,’linewidth’,1.5)
hold on
% plots the s-nullcline diagram
plot(sinf,v,’red’,’linewidth’,1.5)
xlabel(’s’,’fontsize’,16)
ylabel(’V’,’fontsize’,16)
axis([0 1 -80 -10])
legend(’Z-Shaped Bifucation’,’s-nullcline’)
clear gca
set(gca,’fontsize’,16)
end
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