1. Show that you can reproduce the plots below, and be aware of unites-ohms, henrys, farads are in MKS units (volts, amps, seconds etc). In particular, when we find L in terms of and , which are...

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1. Show that you can reproduce the plots below, and be aware of unites-ohms, henrys, farads are in MKS units (volts, amps, seconds etc). In particular, when we find L in terms of and , which are expressed in units of msecs, we need to convert to seconds.
Hints:
2. For this question, do not include the Na+ channel. Evaluate the system at V=0 (Vm= Vr)
From the equations for L, r and the equations for Q and 0, one way to have a higher 0 and Q is to increase [dn∞/dV]v. Play with the values in the and functions to produce such an effect and show that it is possible to increase Q and 0 slightly in this way.
1) Write out your new and functions, and plot out your new and , n∞ and dn∞/dV (similar to the plots from problem 1 but with your new values, the values can be any what you want, just explain or show what you did)
2) Find your L and r values at V=0 (C will uncharged, R is uncharged)
3) Calculate the new values for Q, 0 and f0
4) Show Zmag and Zphase (similar to the plots from problem 1 with your new values), and compare measured Q and 0 and f0 values to the calculated values
Hints for this question:
3. Return the K+ channel parameters to their original values (from problem 1). Now include the Na+ channel, with the approximations (so that it boils down to a negative resistance, see attached materials for reference). Evaluate the system at V=0
1) Find the value for R’ [R’= (R*rm)/(R+rm)]
2) Calculate the value for Q, 0 and f0
3) Show Zmag and Zphase and compare measured Q and 0 and f0 values measured off the graph to the calculated values
4) Turtle hair cells have Q values between 2 and 5. Without changing r, L and C, calculate the value of R’ that will result in a Q value of 3. Plot Zmag and confirm the Q value measured off the graph is around 3

II. Hodgkin Huxley equations: active
mem
ane voltage
3
mem
ane containing K+ channels
Cell
VinVout
[K+]in
[K+]out
Vmem
ane = Vm = (Vin – Vout)
= Nernst potential for K+ ≡ Ek
say [K+]in = 0.4M and
[K+]out = 0.01M
Will Vmem
ane = Ek be zero, or
positive, or negative? Why?
also negative ions for charge
alance (they are not permeable)
Thought experiment: cell with mem
ane only permeable to K+ - in that case
mem
ane voltage is set by [K+].
4
mem
ane containing K+ channels
Cell
VinVout
[K+]in
[K+]out
Vmem
ane = Vm = (Vin – Vout)
= Nernst potential for K+ ≡ Ek
say [K+]in = 0.4M and
[K+]out = 0.01M
!" =
$%
&' ()
*+ ,
*- .
!/ = -96mV
Thought experiment: cell with mem
ane only permeable to K+ - in that case
mem
ane voltage is set by [K+].
~ 26mV
valence: +1 for K+
also negative ions for charge
alance (they are not permeable)
5
[ ]
[ ]i
o
x X
X
zF
RT
E ln=
[ ] = concentration
R = ideal gas constant = 8.314 J/K°/Mole
T = absolute temperature in K° (C+273~300)
z = alge
aic charge of the ion
F = Faraday’s constant 96,500 Coulom
mole
Nernst potential
~ 26mV
ENa ~ +37 mV
EK ~ -90mV
ECl ~ -67mV
Based on these E values, is [K]
greater inside or outside the cell?
What about [Na], [Cl]?
cell with Na+, Cl-,
K+ intracellular and
extracellular
concentrations
Cell
Each ion (X) will have its own E value, based on
its own concentrations and its valence. E values
are not based on permeability. Vmem
ane is
ased on E values and permeabilities.
6
Cell
Vin
Vout
[K+]in
[Cl-]in
[Na+]in
[Cl-]out
[Na+]out
[K+]out
Cu
ent (I) can be applied (in hair cells,
this occurs naturally via stereocilia
channels, in an experiment cu
ent can
e applied with an electrode)
Cell mem
ane is lipid
ilayer with capacitance
~ 1uF/cm2
I = #$
%&$
%' + )*+ &$ − -*+ + )./ &$ − -./ + )0 &1 − -1
Note: concentrations [ ] and
thus E values are fixed and
unchanging – they are not
affected by applied cu
ent,
voltage changes etc.
g values do change.
7
at rest (d/dt =0), and with no injected cu
ent (I=0):
0 = #$% &' − )* + #,- &' − ),- + #. &/ − )
esting mem
ane voltage ≡ &' = 0123450 XXXXXXXXXX
I = :'
;&'
;< + #* &' − )* + #,- &' − ),- + #. &/ − )
What is a simple way of thinking about
the relationship between Vm and the
individual conductance (g) values?
At rest Vmem
ane ~ -78mV. What does that tell
us about the g values?
Recall: ENa ~ +37 mV, EK ~ -90mV, ECl ~ -67mV
Explain in words how a change in g values
could lead to an action potential.
8
The conductance of a given ion is modulated by having some of the channels
open and others closed.
The probability of voltage-gated channels being open depends on the voltage
across the mem
ane, Vm.
Opening occurs with a delay. Therefore, the gated conductances are both voltage
and time dependent. gk(V,t) gNa(V,t)
C
m
Vm
Extracellula
Intracellula
E
Na
g
Na
E
K
g
K
L
g
l
E
9
ecall: g is for “conductance” g=1/R
g units are mhos (1/ohms)
Hodgkin & Huxley Model
XXXXXXXXXXKmKNamNaLmLmm EVngEVhmgEVgdt
dVCI -×+-××+-+= 43
)1()( nn
dt
dn
nn -+-= a
)()1(
)()1(
hh
dt
dh
mm
dt
dm
hh
mm
a
a
--=
--=
1
1
07.0
4
1
251.0
10
30
20
18
10
25
+
=
×=
×=
-
-
=
÷
ø
ö
ç
è
æ -
-
-
÷
ø
ö
ç
è
æ -
n
n
n
n
a
na
e
e
e
e
h
h
m
m
( )
÷
ø
ö
ç
è
æ -
-
×=
-
-
×=
80
10
10
125.0
1
1001.0
n
n
na
e
e
n
n
These terms prescribe how the conductances for K+ and Na+ vary
with mem
ane voltage and with time:
{{
gNa(V,t) gK(V,t)
Note V is expressed in units of mV,
⍺ and β are expressed in units of ms.
Time & voltage dependence of K+ channel
Differential equation for n:
αn and βn are functions of voltage:
)1()( nn
dt
dn
nn -+-= a
Note, here V = (Vm-Vr) =
actual mem
ane voltage - resting voltage.
• Probability of one gate being open = n
• Four gates in the K channel
• All must be open for conduction
• Probability that a channel is open = n*n*n*n = n4
• Conductance
4ngg KK ×=
11
V = (Vm-Vr) = mem
ane voltage - resting voltage.
ewriting the
main HH
equation in
terms of V:
12
Talk about what
would produce an
offset from resting
voltage.
)1()( nn
dt
dn
nn -+-= a
We will “linearize” the Hodgkin-Huxley model to demonstrate frequency
tuning. Na+ conductance is not necessary to demonstrate the basic
frequency tuning.
Hodgkin Huxley model, excluding Na channel;
focusing on K+ channel.
α and β are functions of V
13
III. Derive electrical resonance -- voltage oscillations and tuned
electrical impedance -- in squid axon by linearizing Hodgkin
Huxley equations. (Mauro et al. paper.)
-65
-64
-63
-62
-61
-60
-59
XXXXXXXXXX
Time (msec)
M
em
an
e
Vo
lta
ge
(m
V)
Ringing response in squid axon Tuned impedance in squid axon
Same behavior as in
the tuned LRC circuits
from last week:
14
Basic concept of “linearization”:
f = f(x) (f is a function of x)
Taylor series around some x point, xo:
22
2
( )( ) ( ) ( ) .........
2
( ) ( ) ( )
o
o o
o o
o o
o
x xdf d f
f x f x x x
dx dx
df
f x f x x x
dx
-
= + - + +
» + -
f
xo x
f(xo)
Slope = df/dx|o
Linear approximation: f (x) is ~ a straight line as
long as you don’t move too far from point where
you find the slope of the line.
o
df
f x
dx
d d=
15
Familiar linear equations: V=IR, dV/dt=I/C, V=LdI/dt
Voltage and Cu
ent (V and I) are linearly related, through R, L, C.
(There is no nonlinear term like I2, sin(I) etc.)
Linear equations often good approximation for “small enough” departures from starting
point even if for larger departures there is a nonlinearity (consider example, y=sin(x)).
-65
-64
-63
-62
-61
-60
-59
XXXXXXXXXX
Time (msec)
M
em
an
e
Vo
lta
ge
(m
V)
Voltage “scales linearly” with
cu
ent when linear
approximation holds – linear
approx was good for voltage
departures up to 3 mV here. Voltage response
with cu
ent pulse
of 1,2,3 and 4 nA
Simulation of full nonlinear Hodgkin Huxley
equations displayed linear behavior with “small
enough” cu
ent pulses. When cu
ent pulses got
too large, runaway nonlinear response (action
potential) occu
ed.
16
Back to Hodgkin-Huxley focusing on the K+ channel
This is a nonlinear equation because of the voltage dependence of the
probability of channel opening, n.
)1()( nn
dt
dn
nn -+-= a
To linearize, consider small pertu
ations ! around starting voltage, V
17
XXXXXXXXXXn n
dn n n
dt
a= - + -
XXXXXXXXXXm L K v v K
d VI C g V g n V n n V V
dt
dd d d dé ù= XXXXXXXXXXë û
[ ]( ) n n nn n v v
v v v
d d dn i n V
dV dV dV
a b ad w b a d
é ùé ùæ ö æ ö æ ö+ + = - +ê úê úç ÷ ç ÷ ç ÷
è ø è ø è øë ûë û
(1 )n nnv v nv
Answered 4 days AfterApr 03, 2022

Solution

Vijay Kumar answered on Apr 08 2022
13 Votes
Question 01:
1. Show that you can reproduce the plots below, and be aware of unites-ohms, henrys, farads are in MKS units (volts, amps, seconds etc). In particular, when we find L in terms of and , which are expressed in units of msecs, we need to convert to seconds.
Hints:
Solution 01:
Please type the following commands in MATLAB.
% Please note for both plots above variable V needs to be varied from -40 at step of 10 to +60 or etc.
clc
clear all
v= -40:1:10;
for i=1:1:length(v)

k=0.01;
h=(10-v(i));
l=exp((10-v(i))/10);
alp_n(i) = (k.*h)./l;
end
plot(v,alp_n)
xlabel("Voltage in mv");
ylabel("Values of alpha, beta and n_inf");
xlim([-40 10])
ylim([0 1])
hold on
for i=1:1:length(v)

k1=0.125;
eta_n(i) = k1*exp(-v(i)/80);
end
plot(v,beta_n)
hold on
for i=1:1:length(v)

k=0.01;
h=(10-v(i));
l=exp((10-v(i))/10);
alp_n(i) = (k*h)/l;
k1=0.125;
eta_n(i) = k1*exp(-v(i)/80);
n_inf(i)=alp_n(i)/(alp_n(i)+beta_n(i));
end
plot(v,n_inf);
legend('alpha_n','beta_n', 'n_inf')
% this is differential plot of .

Vs dV needs to be plotted.
clc
clear all
syms f(v)
k=0.01;
h=(10-v);
l=exp((10-v)/10);
alp_n(v) = (k*h)/l;
k1=0.125;
eta_n(v) = k1*exp(-v/80);
f(v) = alp_n(v)/(alp_n(v)+beta_n(v));
Df = diff(f,v)
v=-40:10:60;
fplot(Df, [-40 60])
legend(‘dn_inf/dV’);
This results into this
By refe
ing to page three for the values.
%define them in MATLAB.
% now
%type the following command in MATLAB.
clc
clear all
C=10^-6;
syms f(v)
k=0.01;
h=(10-v);
l=exp((10-v)/10);
alp_n(v) = (k*h)/l;
k1=0.125;
eta_n(v) = k1*exp(-v/80);
f(v) = alp_n(v)/(alp_n(v)+beta_n(v));
Df = diff(f,v)
n0=double(f(0));
dn0=double(Df(0));
gk=30*10^3;
gl=0.3*10^3;
vk=-12*1e-3;
vl=10.6*1e-3;
=-1/(gk*vk*4*((n0)^3)*dn0);
L=
(double(alp_n(0))+double(beta_n(0)));
R_inv=gl+(gk*(n0)^4);
[email protected](f) (r+(2*pi.*f*L)*i)
[email protected](f) ((1-(4*pi*pi.*f.*f)+r*R_inv)+((2*pi.*f*L*R_inv)+(r*2*pi.*f*C))*i)
[email protected](f) (r+(2*pi.*f*L)*i)/(@f x2);
f = 0:10:200;
figure
subplot(2,1,1)
plot(f, abs(Z(f)))
title('Amplitude')
subplot(2,1,2)
plot(f, (180*angle(Z(f))))
title('Phase')
Question 02:
For this question, do not include the Na+ channel. Evaluate the system at V=0 (Vm= Vr)
From the equations for L, r and the equations for Q and 0, one way to have a higher 0 and Q is to increase [dn∞/dV]v. Play with the values in the and functions to produce such an effect and show that it is possible to increase Q and 0 slightly in this way.
1. Write out your new and functions, and plot out your new and , n∞ and dn∞/dV (similar to the plots from problem 1 but with your new values, the values can be any what you want, just explain or show what you did)
1. Find your L and r values at V=0 (C will uncharged, R is uncharged)
1. Calculate the new values for Q, 0 and f0
1. Show Zmag and Zphase (similar to the plots from problem 1 with your new values), and compare measured Q and 0 and f0 values to the calculated values
Hints for this question:
Solution
NA, CL, K+ all three needs to be consider in mem
ane...
SOLUTION.PDF

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