1. Go to The Na+ Action Potential Tutorial

2. Click on start the simulation.

3. In the Panel & Graph Manager click on Patch Parameters.

4. Turn off Na+ and K+ voltage gated channels ( set density of both to 0.0 S/cm2 ).

5. Set time base to 100 ms (Run Control, Total # (ms) , 100 ms )

• You now have a simplified situation with only one conductor in the membrane ( the so called leak conductance )

• This leak conductance has a reversal potential of -54.3 mV.

6. Change Run Control, Reset ( mV ) to -54.3 mV.

1. Make the current pulse duration 50 ms ( Stimulus Control , IClamp ).

2. Set delay to 10 ms.

3. Set amplitude to 0.01414 n

• This will cause a voltage change from -54 mV to -24 mV ( a 30 mV depolarization )

1A : Press "Reset and Run"

• right click to "Keep Lines"

  

1B : Predict what will happen when you double the current amplitude to 0.02828 nA

• $\mathrm{\Delta }V=30mV\ast 2=60mV$$\Delta V = 30\ mV * \textcolor{#61CC9E}{2} = 60\ mV$

• Peak = $-54\ mV + 60\ mV = 6\ mV$

  

1C : Predict what happens when you quadruple the current to 0.05656 nA

• $\mathrm{\Delta }V=30mV\ast 4=120mV$$\Delta V = 30\ mV * \textcolor{#61CC9E}{4} = 120\ mV$

• Peak = $-54\ mV + 120\ mV = 66\ mV$

1D : Calculate the input resistance of this cell

• $1c{m}^2={10}^8\mu m^2$$1\ cm^2 = 10^8\ \mu m^2$

1E : Measure the time constant of this cell

• Measure initial , and peak y-values

  • initial = -54.3

  • final = +65.724

• Calculate $\tau$$\tau$ graphically :

  • find difference in y-values :

    $$((65.724)-(-54.3))=120.024$$

  • scale by approximately 63% of the value :

    $$120.024\ast (1-\frac{1}{e})=75.8696379528388$$

  • add this number to initial y-value :

    $$(-54.3)+75.8696379528388=21.569637952838804$$

  • find corresponding x-value to this calculated y-value :

    

  • subtract off any experiment stimulus delay to get final value :

    $$13.4-10=3.4ms$$

( ( 65.724 ) - ( -54.3 ) ) * ( 1 - ( 1 / math.e ) ) + ( -54.3 )

• set amplitude to 0.04242 nA

• Press "Reset and Run"

2A : Predict what happens as you change capacitance ( Patch Parameters )

• try values of 0.5 , 1.0 , 2.0 , 4.0 $\mu F/c{m}^2$$\mu F / cm^2$

   

• green trace = $0.5\mu F/c{m}^2$$0.5\ \mu F / cm^2$

• black trace = $1.0\mu F/c{m}^2$$1.0\ \mu F / cm^2$

• red trace = $2.0\mu F/c{m}^2$$2.0\ \mu F / cm^2$

• blue trace = $4.0\mu F/c{m}^2$$4.0\ \mu F / cm^2$

• smaller capacitance = faster at charging and discharging

• higher capacitance = slower at charging and discharging

2B : Discuss how changing capacitance alters input resistance

• capacitance is not included in Ohm's law , $V = I * R$ or $V=\frac{I}{g}$$V = \frac{I}{g}$

• so it is not affected by input resistance

• steady-state ( final ) voltage = $\frac{I}{\text{total leak conductance ( g )}}$$\frac{I}{\text{total leak conductance ( g )}}$

• the total leak conductance input resistance is determined

2C : Discuss how changing capacitance alters the time constant

3A : Set the membrane capacitance at $2.0\mu F/c{m}^2$$2.0\ \mu F / cm^2$ and keep the time base at 100 ms

• press "Reset and Run"

3B : Double the leak conductance to $0.6mS/c{m}^2$$0.6\ mS/cm^2$ ( $0.0006S/c{m}^2$$0.0006\ S/cm^2$ )

• black trace = $0.3mS/c{m}^2$$0.3\ mS/cm^2$

• blue trace = $0.6mS/c{m}^2$$0.6\ mS/cm^2$

• Predict what happens to the steady state depolarization.

  • $\mathrm{\Delta }Vm$$\Delta Vm$ ( depolarization ) should be smaller ( half ) after doubling $g$

    • $V=\frac{I}{g}$$V = \frac{I}{g}$

• Discuss what happens to the time constant compared with the initial conductance level. ( This change may be hard to see easily )

  • time constant will decrease after increasing leak conductance

    • because time constant is directly proportional to resistance

      • $\tau \propto R$$\tau \propto R$

    • but resistance is inverse of conductance , so

      • $\tau \frac{1}{\propto }g$$\tau\ \frac{1}{\propto}\ g$

  

3C : Predict how much current you need to inject to make the depolarization the same , so you can directly compare time constants. ( Calculate what current you have to inject to get the same depolarization as when leak was only ½ as much ( original $0.3mS/c{m}^2$$0.3\ mS/cm^2$ ) )

• Original Current = $0.01414nA$$0.01414\ nA$

  • Produced a $30mV$$30\ mV$ depolarization

• Original Leak Conductance = $0.3mS/c{m}^2$$0.3\ mS/cm^2$

  • therefore Original Leak Resistance :

  $$\frac{1}{\frac{0.3mS}{c{m}^2}}=\frac{1\ast c{m}^2}{0.3\ast {10}^{-3}S}=3333\mathrm{\Omega }\cdot c{m}^2$$

  • for every $c{m}^2$$cm^2$ of membrane , the resistance is about 3333

    $$30mV=0.01414nA\ast R_{\text{original}}$$
  $$R_{\text{original}}=\frac{30mV}{0.01414\ast {10}^{-9}A}=2.1216407355021214\ast {10}^9\mathrm{\Omega }$$

   

• but now we want $R_{new}=\frac{R_{\text{original}}}{2}$$R_{new} = \frac{R_{\text{original}}}{2}$ :

• replace $R_{\text{original}}$$R_{\text{original}}$ :

• Measure the size of the time constant with the larger conductance.

  • initial = -54.3

  • final = -9.291

  • find difference in y-values :

    $$(-9.291)-(-54.3)=45.009$$

  • take around 63% :

    $$45.009\ast (1-\frac{1}{e})=28.451114232314552$$

  • add to initial y-value :

    $$(-54.3)+\left(28.451114232314552\right)=-25.848885767685445$$

  • find corresponding x-value :

    • 13.375

   

  

3D : Predict what happens if you double leak again to 1.2 mS/cm2 ( 0.0012 S/cm2 )

• red trace = $1.2mS/c{m}^2$$1.2\ mS/cm^2$

• Prediction = the resistance is lowered again by another factor of 2

• Discuss what happens to the level of steady state depolarization.

  • the depolarization will be halved

3E : If you go back to your original leak conductance of 0.3 mS/cm2 ( 0.0003 S/cm2 ) , predict what the membrane would depolarize to

• Calculate what current you have to inject to depolarize to +24 mV

  $$I=\frac{78.3\ast {10}^{-3}V}{2.1216407355021214\ast {10}^9\mathrm{\Omega }}=3.69054\ast {10}^{-11}A$$

• Measure the time constant now and discuss

  • initial = -54.3

  • final = -31.7955

  • find difference in y-values :

  $$(-31.7955)-(-54.3)=22.5045$$

  • take around 63% :

  $$22.5045\ast (1-\frac{1}{e})=14.225557116157274$$

   

  • add to initial y-value :

  $$(-54.3)+\left(14.225557116157274\right)=-40.07444288384272$$

  • find corresponding x-value :

    • 11.675

     

  • the time constant is smaller with higher leak conductance

    • because higher leak conductance = lower resistance

      • lower resistance ➡️ faster charging and discharging