The problems just about some calculations and graph plotting. The equations for calculation are...

Question

The problems just about some calculations and graph plotting. The equations for calculation are provided. No other source can be used. The reference material will be provided1. 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  membrane voltage 3 membrane containing K+ channels Cell VinVout [K+]in [K+]out Vmembrane = Vm = (Vin – Vout)  = Nernst potential for K+ ≡ Ek say [K+]in  = 0.4M and  [K+]out  = 0.01M Will Vmembrane = Ek be zero, or  positive, or negative?  Why? also negative ions for charge  balance (they are not permeable) Thought experiment: cell with membrane only permeable to K+ - in that case  membrane voltage is set by [K+].   4 membrane containing K+ channels Cell VinVout [K+]in [K+]out Vmembrane = Vm = (Vin – Vout)  = Nernst potential for K+ ≡ Ek say [K+]in  = 0.4M and  [K+]out  = 0.01M !" = $% &' () *+ , *- . !/ = -96mV Thought experiment: cell with membrane only permeable to K+ - in that case  membrane voltage is set by [K+].   ~ 26mV valence: +1 for K+ also negative ions for charge  balance (they are not permeable) 5 [ ] [ ]i o x X X zF RT E ln= [ ] = concentration R = ideal gas constant = 8.314 J/K°/Mole T = absolute temperature in K° (C+273~300) z = algebraic charge of the ion F = Faraday’s constant 96,500 Coulomb/mole Nernst potential ~ 26mV ENa ~ +37 mV EK ~ -90mV ECl ~ -67mV Based on these E values, is [K]  greater inside or outside the cell?  What about [Na], [Cl]?   cell with Na+, Cl-,  K+ intracellular and  extracellular   concentrations  Cell Each ion (X) will have its own E value, based on  its own concentrations and its valence. E values  are not based on permeability.  Vmembrane is  based on E values and permeabilities.  6 Cell Vin Vout [K+]in [Cl-]in [Na+]in [Cl-]out [Na+]out [K+]out Current (I) can be applied (in hair cells,  this occurs naturally via stereocilia  channels, in an experiment current can  be applied with an electrode) Cell membrane is lipid  bilayer with capacitance  ~ 1uF/cm2 I = #$ %&$ %' + )*+ &$ − -*+ + )./ &$ − -./ + )0 &1 − -1 Note:  concentrations [ ] and  thus E values are fixed and  unchanging – they are not  affected by applied current,  voltage changes etc.   g values do change.  7 at rest (d/dt =0), and with no injected current (I=0): 0 = #$% &' − )* + #,- &' − ),- + #. &/ − )/ resting membrane voltage ≡ &' = 012345067367508380125067508 I = :' ;&' ;

Vijay Kumar · Accepted Answer

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;
beta_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;
beta_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;
beta_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 referring 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;
beta_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;
r=-1/(gk*vk*4*((n0)^3)*dn0);
L=r/(double(alp_n(0))+double(beta_n(0)));
R_inv=gl+(gk*(n0)^4);
x1=@(f) (r+(2*pi.*f*L)*i)
x2=@(f) ((1-(4*pi*pi.*f.*f)+r*R_inv)+((2*pi.*f*L*R_inv)+(r*2*pi.*f*C))*i)
Z=@(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 membrane so,

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...

Answer To: 1. Show that you can reproduce the plots below, and be aware of unites-ohms, henrys, farads are in...

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