Collected Data : https://docs.google.com/spreadsheets/d/1UArGLeBwguDwpFXvkdP19wLGHYiIc84Duz7DeYf39ck/edit?usp=sharing

• Determine how $I_K$$I_K$ varies with stimulus voltage

1. Choose tutorial "Voltage Clamping a Patch"

2. Click on "Start the Simulation"

3. Click on VClamp

4. Click on Membrane Current Plots

5. Increase Total ms to 12 ms

6. Increase Stimulus Duration ( for testing level ) to 10 ms

7. Remove Na+ conductance , by setting to 0.0 S/cm^2

1. Vary the test amplitude of stimulation ( change voltage , mV in Stimulus Control window ) , starting at −100 mV and stepping in 10 mV increments to +100 mV. Measure the largest current for each voltage step ( check beginning and end of testing level ). Left clicking on trace to get a reading for x,y ( y is the amplitude in mA/cm2 ). Plot the current vs the voltage step, on the graph provided ( next page )

   $V_m\left(mV\right)$$V_m\ \left(\ mV \ \right)$	$I\left(\frac{mA}{c{m}^2}\right)$$I\ \left(\ \frac{mA}{cm^2} \ \right)$
   -100	-0.00800178
   -90	-0.00450186
   -80	-0.000954157
   -70	0.00274448
   -60	0.0137072
   -50	0.0762199
   -40	0.25128
   -30	0.56275
   -20	0.971343
   -10	1.42375
   0	1.88685
   10	2.34651
   20	2.79808
   30	3.24056
   40	3.67417
   50	4.09954
   60	4.51741
   70	4.92852
   80	5.33359
   90	5.73329
   100	6.12823

   

   • we get the conductance value for free from the slope

   $$g=0.0342\frac{\frac{nA}{c{m}^2}}{mV}=0.0342\frac{\frac{{10}^{-9}A}{{\left({10}^{-2}\right)}^2{m}^2}}{{10}^{-3}V}=0.0342\frac{\frac{{10}^{-5}A}{m^2}}{{10}^{-3}V}=0.0342\ast {10}^{-2}\frac{A}{V\cdot m^2}=\frac{0.000342A}{V\cdot m^2}$$

   • Siemens $\left(S\right)$$\left(\ S \ \right)$ is $\frac{Amperes}{Volts}$$\frac{Amperes}{Volts}$

   • so here I guess its $\frac{Siemens}{m^2}$$\frac{Siemens}{m^2}$

   • now to find approximated nernst value for sodium :

   $$y=(0.0342\ast x)+\left(2.28\right)$$

   • let $y=0$$y = 0$ :

     $$0=(0.0342\ast x)+\left(2.28\right)$$
     $$x\approx -66.6667mV$$

2. Calculate chord conductance ( G ) for these K+ currents , then construct a plot of GK versus Vm

   $$I=g\ast (\left(V_m\right)-\left(E_{\text{ion}}\right))$$

   • solving for $g$$g$ :

   $$g=\frac{I}{\left(V_m\right)-\left(E_{\text{ion}}\right)}$$

   $V_m\left(mV\right)$$V_m\ \left(\ mV \ \right)$	$g\left(\frac{{10}^{-2}A}{V\cdot m^2}\right)$$g\ \left(\ \frac{10^{-2}\ A}{V \cdot m^2} \ \right)$
   -100	0.0003528277261
   -90	0.0003550642795
   -80	0.0003561616275
   -70	0.0003748777489
   -60	0.0007913630853
   -50	0.002789791735
   -40	0.00673293856
   -30	0.01189218317
   -20	0.01694567436
   -10	0.02114867575
   0	0.02440281424
   10	0.02687223005
   20	0.02875104037
   30	0.03019502241
   40	0.03131724073
   50	0.03219845901
   60	0.03289671645
   70	0.03345429369
   80	0.03390259406
   90	0.03426521477
   100	0.03456009159

   

   • the curve is sigmoidal , and you could fit it with a Boltzman equation to find the slope

     • a large slope = channel requires a broader voltage range to switch from closed to open state

     • a small slope = channel opens very quickly over a narrow voltage range

   • but we just can take $g_{max}$$g_{max}$ , like $0.03456009159$$0.03456009159$ , and divide it by 2

     $$\frac{g_{max}}{2}=\frac{0.03456009159}{2}=0.017280045795$$

   • now go to the G / V plot , and lookup corresponding voltage for where $\frac{g_{max}}{2}$$\frac{g_{max}}{2}$ is :

     • $[??,0.017280045795]$$[\ ??\ ,\ 0.017280045795\ ]$

     • from the table , its somewhere between $-20mV$$-20\ mV$ and $-10mV$$-10\ mV$

     • the actual value we call now $V_{1/2}$$V_{1/2}$

     • $V_{1/2}$$V_{1/2}$ = the voltage at which 50% of the channels are open

       • if $V_{1/2}$$V_{1/2}$ is more negative , the channel activates at lower voltages

         • probably a sodium channel

       • if $V_{1/2}$$V_{1/2}$ is more positive , the channel requires a stronger depolarization to activate

         • probably a potassium channel

3. Current reverses ( changes from inward to outward: negative to positive ) at the Nernst potential for the permeant ion. Locate that reversal point for the K+ current and compare with the anticipated Nernst value.

   • looking at the table of values we collected , current reverses polarity between $-80mV$$-80\ mV$ and $-70mV$$-70\ mV$

   • our predicted value of $-66.6667mV$$-66.6667\ mV$ is somewhat near this range

4. Describe any changes of the kinetics with voltage ( kinetics refers to the time course of the current: rate of rise and rate of fall )

   • constant slope on G / V plot between $[-50mV,-10mV]$$[\ -50\ mV\ ,\ -10\ mV\ ]$

   • delay in opening , then open for long period of time

• Determine how $I_{Na}$$I_{Na}$ varies with stimulus voltage

1. Choose tutorial "Voltage Clamping a Patch"

2. Click on "Start the Simulation"

3. Click on VClamp

4. Click on Membrane Current Plots

5. Increase Total ms to 6 ms

6. Increase Stimulus Duration ( for testing level ) to 4 ms

7. Set Na+ conductance to default value

8. Remove K+ conductance , by setting to 0.0 S/cm^2

1. Vary the amplitude of stimulation (change voltage, mV in Stimulus Control window), starting at −100 mV and stepping in 10 mV increments to +100 mV. Measure the maximal current for each voltage step. Plot the current vs the voltage step, on the graph provided on next page.

   $V_m\left(mV\right)$$V_m\ \left(\ mV \ \right)$	$I\left(\frac{mA}{c{m}^2}\right)$$I\ \left(\ \frac{mA}{cm^2} \ \right)$
   -100	-2.57E-09
   -90	-1.37E-07
   -80	-6.34E-06
   -70	-0.000237441
   -60	-0.00635058
   -50	-0.0879799
   -40	-0.441195
   -30	-0.947073
   -20	-1.33333
   -10	-1.55554
   0	-1.61346
   10	-1.52326
   20	-1.31529
   30	-1.01759
   40	-0.653983
   50	-0.242015
   60	0.2097
   70	0.692602
   80	1.20084
   90	1.72929
   100	2.27465

   

2. Locate the reversal potential for the Na+ current and compare with the anticipated Nernst value

   $$0=(6.79\ast {10}^{-3}\ast x)+(-0.22)$$
   $$x=32.4006$$

   • fitted linear equation is obviously very wrong , as we don't have a straight line for the entire trace

   • looking at the collected data in the table , it crosses zero between $+50mV$$+50\ mV$ and $+60mV$$+60\ mV$

   • the programmed $E_{Na}$$E_{Na}$ in the simulation is at $+55.449$$+55.449$

3. Describe any changes of the kinetics with voltage ( also note time of peak INa )

   

   • it reaches the peak time for $I_{Na}$$I_{Na}$ quicker and quicker as you raise $V_m$$V_m$

4. Repeat the stimulation sequence with both Na+ and K+ conductances at default values ( 10 ms duration testing level ). Click on Membrane Current Plots ( in Panel & Graph Manager ) and VClamp.i graph ( in Stimulus Control Panel ).

   

   

1. Choose tutorial "The Na Action Potential"

2. Start the Simulation

3. Click on IClamp

4. Change total # ms to 10 ms

5. Reset and Run

1. Using your results from the voltage dependence of INa and IK ( review your plots from previous pages ) , and the dependence of INa and IK on driving force $I_i=g_i\ast (V_m-E_i)$${\ I_i = g_i * \left(\ V_m - E_i \ \right)\ }$ , predict the INa and IK that occur during the action potential.

   • inward sodium current , outward potassium current

2. Draw the time course of your predicted Na+ and K+ currents along with the aligned action potential. ( Pick key time points along the action potential to note the conductance and driving force. )

    

   

6. Now open Membrane Current Plots along with Voltage vs Time Plot

7. Change total # (ms) to 10 ms

8. Reset and Run

3. Compare your prediction with the simulation and note differences. If different , redraw the currents along with the aligned action potential.

4. Briefly explain the time course of INa and IK

   • sodium opens early and closes quickly for inactivation

   • potassium delay in opening , but then stay open for longer

9. Now open the membrane conductance plots

10. Assure that the axis extends to 10 ms

11. Reset and Run

5. Add the conductance plots to your drawing above

   

6. Describe any differences between the time course of the conductance plots and the current plots.

   

   • the lag is much more obvious for potassium in the conductance plot

   • you can more clearly when sodium is conducting

7. Examine the peak of the action potential and the Na+ Nernst potential , and briefly explain the relationship.

   • action potential peak = $+45.3934mV$$+45.3934\ mV$

   • sodium Nernst = $+55.442$$+55.442$

   • the action potential peak approaches the nernst values as you conduct ions of that type

8. Examine IK and INa at the time points when Vm is not changing ( maximums and minimums ) , and Vm when either IK or INa are not changing. Briefly explain the relationship between these currents and Vm.

   • when $V_m$$V_m$ is not changing ( at min and max ) , current = zero

   • when current is not changing , the channels are inactivated or closed