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.

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

1. First you will examine the influence of increasing current injection into the cell.

  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"

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

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

image-20250117195156172

1D : Calculate the input resistance of this cell

image-20250117195946370

V=IR
ΔV=VfinalVinitial
ΔV=( 24.3 )( 54.3 mV )=30 mV
3.0102 V=0.01414109 AmperesR
R=3102 V0.01414109 Amperes=2121640735.5021214 Ohm

1E : Measure the time constant of this cell

image-20250117195430479

τ=RC

2. Now you will examine the influence of changing cell capacitance

image-20250118071551810

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

Capacitance Cell’s Total Time Spent Charging and Discharging

image-20250118071734089

2B : Discuss how changing capacitance alters input resistance

2C : Discuss how changing capacitance alters the time constant

Time Constant( τ )=Resistance ( R ) Capacitance ( C )
τC

3. Now look at the responses of the cell to changing resistance ( 1 / conductance )

3A : Set the membrane capacitance at 2.0 μF/cm2 and keep the time base at 100 ms

3B : Double the leak conductance to 0.6 mS/cm2 ( 0.0006 S/cm2 )

image-20250118074618897

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 0.3 mS/cm2 ) )

30 mV=InewRoriginal2
Inew=230 mVRoriginal
Inew=230 mV30 mV0.01414109 A
Inew=2( 0.01414109 A )=0.02828109 A

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

image-20250118075315561

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

ΔV=( +24 mV )( 54.3 mV )=78.3 mV