10 0223 BME402HW3

10 0223 BME402HW3 - BME
402
Spring
2010
 
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Unformatted text preview: BME
402
Spring
2010
 
 Question
1
 @t=0,
gNA=5nS
 C=12
pF
 Vm@t=5
ms?
 
 Remember
for
a
differential
equation
of
the
following
form:
 
 dV/dt=
(‐V
+
V∞)/
Tau
 
 the
solution
is:
 
 V(t)=
De‐t/Tau+A
 
 We
will
solve
the
KCL
equation
for
dV/dt
 
 (Vm‐El)gl
+
(Vm‐ENa)gNa
+
C(dVm/dt)
=
0
 
 dVm=

‐Vm
+
(El*gl/(gl+gNa))
+
(ENa*gNa/(gl+gNa))

 dt

 
 
 (C/(gl+gNa))

 

 
 Vm(t=0‐)=
D+A=
‐75
mV
 Vm(t=∞)=
A
=
El*gl
+
ENa*gNa

=
33.3333
 
 
 gl
+
gNa
 
 ‐75
mV=
D
+
33.33333
 D=‐108.33
 
 Vm(t)=
‐108.33*
exp(‐t/.002
sec)
+33.33
 
 Vm(t=5ms)=
24.441
mV
(+/‐
0.005
was
accepted)
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Homework
3
Solutions
 
 BME
402
Spring
2010
 
 Question
2
 Homework
3
Solutions
 
 
 i=1;
 v(i)=
‐100;
 vmax=25;
 
 %%whole
itteration
 
 while
(v(i)
<=
vmax)
 
 %%
m_inf
 if
(abs(v(i)
+
45)
<=0.01)
 



a_m(i)=1;
 else
 



a_m(i)
=
(‐0.1*(v(i)
+
45))/(exp(‐(v(i)
+
45)/10)
‐
1);
 end;
 
 %%
n_inf
 if
(abs(v(i)
+
60)
<=0.01)
 



a_n(i)
=
.1;
 
 BME
402
Spring
2010
 
 Homework
3
Solutions
 
 else
 



a_n(i)
=
(‐0.01*(v(i)
+
60))/(exp(‐(v(i)
+60)/10)
‐
1);
 end;
 
 %%
 
 a_h(i)
=
(0.07*(exp(‐(v(i)+70)/20)));
 b_n(i)
=
(0.125*exp(‐(v(i)+70)/80));
 b_m(i)
=
(4*exp(‐(v(i)+70)/18));
 b_h(i)
=
1/(exp(‐(v(i)+40)/10)+1);
 
 
 n_inf(i)=
a_n(i)/(a_n(i)+b_n(i));
 Tau_n(i)=
1/(a_n(i)+b_n(i));
 m_inf(i)=
a_m(i)/(a_m(i)+b_m(i));
 Tau_m(i)=
1/(a_m(i)+b_m(i));
 h_inf(i)=
a_h(i)/(a_h(i)+b_h(i));
 Tau_h(i)=
1/(a_h(i)+b_h(i));
 
 v(i+1)=v(i)+1;
 i=i+1;
 
 end
 
 subplot(2,1,1),
plot(v(1:126),n_inf,v(1:126),m_inf,v(1:126),h_inf);
 xlabel('Voltage
(mV)'),
ylabel('Asymptote');
 title('Infinity
values'),
legend('n_i_n_f','m_i_n_f','h_i_n_f');
 
 subplot(2,1,2),
plot(v(1:126),Tau_n,v(1:126),Tau_m,v(1:126),Tau_h)
 xlabel('Vm
(mv)'),
ylabel('t
(ms)');
 title('Tau
values'),
legend('Tau_n','Tau_m','Tau_h');
 
 Question
3
 Using
your
formulas
that
you
have
implemented
in
Matlab,
evaluate
m∞,
h∞
and
the
 corresponding
τ
values
during
P1
and
P2
(4
numbers
total).
Round
to
the
nearest
 thousandth.
 
 For
period
P1:
 m∞
=
[a]
=
0.992
 h∞
=
[b]
=
0.001
 τm
=
[c]
=

0.180
 τh
=
[d]
=
1.005
 
 For
period
P2:
 m∞
=
[e]
=
0.004
 h∞
=
[f]
=
0.966
 BME
402
Spring
2010
 
 Homework
3
Solutions
 
 τm
=
[g]
=
0.082
 τh
=
[h]
=
5.077
 
 Question
4
 m(t)
=
[i].e‐t/[j]
+
[k]
 
 During
P1,
m
is
growing
exponentially
from
m∞1
=
0.094
to
m∞2
=
0.992
with
taum
=
 0.180
ms.
 
 m(t)
=
(m∞1
‐
m∞2).
e‐t/taum
+
m∞2
 
 Question
5
 gNa_bar
=

0.0012
*
4π
*
102
=
1.5080
μS
 m
and
h
at
0.5
ms
can
be
calculated
from
your
formulas,
and
then
 gNa
=
gNa_bar.
m(at
t=0.5ms)3.
h(at
t=0.5ms)
=
0.315
μS
 
 Question
6
 BME
402
Spring
2010
 
 Homework
3
Solutions
 
 clear;

 clc;

 


 %Plotting
m,
h,
m3h

 


 %Period
1

 t
=
[0:.0001:1];
%time
in
ms

 


 m_1
=
‐0.8980.*exp(‐t/0.1795)+0.9916;

 h_1
=
0.4169.*exp(‐t/1.0054)+0.0013;

 m3h_1
=
(m_1.^3).*h_1;

 


 %Period
2

 t2
=
[1.0001:.000001:4];
%time
in
ms

 


 m_2
=
0.9841.*exp((‐t2+1)/0.0820)+0.0041;

 h_2
=
‐0.8105.*exp((‐t2+1)/5.0768)+0.9660;

 m3h_2
=
(m_2.^3).*h_2;

 


 subplot(2,2,1),
plot(t,m_1,'blue',t2,m_2,'blue')

 BME
402
Spring
2010
 
 Homework
3
Solutions
 
 subplot(2,2,2),
plot(t,h_1,'green',t2,h_2,'green')

 subplot(2,2,3),
plot(t,m3h_1,'red',t2,m3h_2,'red')

 subplot(2,2,4),

 plot(t,m_1,'blue',t2,m_2,'blue',t,h_1,'green',t2,h_2,'green',t,m3h_1,'red',t2

 ,m3h_2,'red');

 


 pause

 


 %plotting
Na+
current

 


 %Period
1

 


 gNa_1
=
(0.0012*4*3.14159*100).*m3h_1;
%in
micro
Siemens

 INa_1
=
(10‐55).*gNa_1;

 


 %Period
2

 


 gNa_2
=
(0.0012*4*3.14159*100).*m3h_2;
%in
micro
Siemens

 INa_2
=
(‐90‐55).*gNa_2;

 


 figure


 plot(t,INa_1,'magenta',t2,INa_2,'magenta')
 ...
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This note was uploaded on 04/04/2010 for the course BME 302L taught by Professor Maarek during the Spring '08 term at USC.

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