Part 1 : Where is the site of action potential initiation ?

1. Choose tutorial "Site of Impulse Initiation"

2. Impulse initiation in response to a depolarizing synaptic input

3. Click on Reset & Run

   • that delivers a depolarizing synaptic input to the middle of the dendrite at t = 0.1

4. To study the movie , click on Continue for ( ms ) in steps of 0.2 ms

5. Look carefully at the sequence of voltage changes in dendrite , soma , and axon

• Note : This simulation uses unmyelinated axon of the squid

• In this simulation , Na+ channels are uniformly distributed along the axon

  • ( no Na+ channel clusters at the axon initial segment )

• Vertebrates have axon initial segment in both myelinated and unmyelinated axons

1. Where is the site of action potential initiation ?

   • $200\mu m$$200\ \mu m$ @ $1.72$$1.72$ , $-51mV$$-51\ mV$

   

2. How does the action potential propagate ?

   • continuously

6. Change the Na channel densities in the axon

7. Run the stimulation and use Keep Lines in the Voltage vs Time window

8. In the Active Axon Parameter panel , halve the Na chan , then run the stimulation

9. Restore the channel densities to default values

10. Next , double the Na chan density , then run the stimulation

3. What happens to the action potential ?

   • halving = no action potential

   • doubling = bigger , earlier , faster conduction

Part 2 : How does axon diameter affect conduction velocity ?

1. Choose tutorial "The Unmyelinated Axon"

2. Click on Voltage vs Time , Dual Traces

3. Increase the time base ( Total # ( ms ) ) to 6 ms

4. Click on Reset & Run

• The responses are for the squid giant axon ( diameter 500 μm )

• The red trace is the action potential near the site of stimulation and the distance between red and black electrodes is 8 mm

1. Use the crosshairs to measure the action potential conduction velocity ( m/s )

   • Red Peak = 1.025 , 38.0981

   • Black Peak = 1.375 , 41.8389

   

   $$v=\frac{8mm-0mm}{1.375ms-1.025ms}=\frac{22.857meters}{second}$$

2. Calculate the action potential conduction velocity ( m/sec ) , and plot the value in the graph below

   • asdf

Squid thinner axon , unmyelinated axon

5. Decrease the diameter ( Axon Parameters ) and current amplitude ( IClamp ) as follows :

6. Click on Reset & Run for each

• Reducing current amplitude will offset the changes made to passive properties induced by the reduction in axon diameter so that the stimulus artifact isn’t too big

3. Calculate the conduction velocity ( m/s ) for each , and plot the value in the graph below

   Diameter	Amperes	Red $\left(ms\right)$$\left(\ ms \ \right)$	Black $\left(ms\right)$$\left(\ ms \ \right)$	$Velocity\left(m/s\right)$$Velocity \left(\ m/s \ \right)$
   400	15000	1	1.45	$\frac{8}{1.45-1}=17.778$$\frac{8}{1.45 - 1} = 17.778$
   300	10000	1	1.55	$\frac{8}{1.55-1}=14.545$$\frac{8}{1.55 - 1} = 14.545$
   200	5000	1.05	1.825	$\frac{8}{1.825-1.05}=10.323$$\frac{8}{1.825 - 1.05} = 10.323$
   100	2500	0.95	2.2	$\frac{8}{2.2-0.95}=6.4$$\frac{8}{2.2 - 0.95} = 6.4$
   50	1000	0.975	2.9	$\frac{8}{2.9-0.975}=4.1558$$\frac{8}{2.9 - 0.975} = 4.1558$

   

Part 3 : How does myelin thickness affect conduction velocity ?

Typical frog myelin

1. Choose tutorial "The Myelinated Axon"

2. Click on Voltage vs Time , Dual Traces

3. Increase the time base ( Total # ( ms ) ) to 3.0 ms

4. Click on Reset & Run

• The responses are for a frog axon with a diameter of 10 μm

• This simulated axon has ten myelinated regions with 150 wraps of myelin

• Each myelinated region is 1 mm long

• The nodes of Ranvier between each myelinated region are 3.2 μm long

• The red trace is the action potential at the first node and the black trace is the action potential at the 9th node , approximately 8 mm further along the axon

1. Calculate the action potential conduction velocity ( m/sec ) , and plot the value in the graph below

   • Red Peak = 0.325 , 35.0446

     • electrode-1 is at $1000\mu m$$1000\ \mu m$

   • Black Peak = 0.7375 , 36.6867

     • electrode-9 is at $9000\mu m$$9000\ \mu m$

   $$v=\frac{8mm}{0.7375ms-0.325ms}=\frac{19.394meters}{second}$$

Thinner Myelin

5. Decrease the number of myelin wraps to 100 , 50 , 20 , and then 10

2. Calculate the conduction velocity ( m/s ) for each , and plot the value in the graph below

   • 100 = $\frac{8mm}{0.7375ms-0.325ms}=\frac{16meters}{second}$$\frac{8\ mm}{0.7375\ ms - 0.325\ ms} = \frac{16\ meters}{second}$

   • 50 = $\frac{8mm}{1.1125ms-0.4ms}=\frac{11.228meters}{second}$$\frac{8\ mm}{1.1125\ ms - 0.4\ ms} = \frac{11.228\ meters}{second}$

   • 20 = $\frac{8mm}{1.7ms-0.5625ms}=\frac{7.033meters}{second}$$\frac{8\ mm}{1.7\ ms - 0.5625\ ms} = \frac{7.033\ meters}{second}$

   • 10 = $\frac{8mm}{2.3625ms-0.5625ms}=\frac{4.4444meters}{second}$$\frac{8\ mm}{2.3625\ ms - 0.5625\ ms} = \frac{4.4444\ meters}{second}$

   

3. What happened with 10 wraps of myelin? What might have caused this to occur ?

   • not enough resistance provided , therefore very slow

Thicker Myelin

6. Increase the number of myelin wraps to 200 , 250 , 300 , 400 , 500 , 600 , and then 800

4. Calculate the conduction velocity ( m/s ) for each , and plot the value in the graph below

   • 200 = $\frac{8mm}{0.675ms-0.3ms}=\frac{21.333meters}{second}$$\frac{8\ mm}{0.675\ ms - 0.3\ ms} = \frac{21.333\ meters}{second}$

   • 250 = $\frac{8mm}{0.6375ms-0.3ms}=\frac{23.704meters}{second}$$\frac{8\ mm}{0.6375\ ms - 0.3\ ms} = \frac{23.704\ meters}{second}$

   • 300 = $\frac{8mm}{0.6125ms-0.2875ms}=\frac{24.615meters}{second}$$\frac{8\ mm}{0.6125\ ms - 0.2875\ ms} = \frac{24.615\ meters}{second}$

   • 400 = $\frac{8mm}{0.575ms-0.2875ms}=\frac{27.826meters}{second}$$\frac{8\ mm}{0.575\ ms - 0.2875\ ms} = \frac{27.826\ meters}{second}$

   • 500 = $\frac{8mm}{0.55ms-0.275ms}=\frac{29.091meters}{second}$$\frac{8\ mm}{0.55\ ms - 0.275\ ms} = \frac{29.091\ meters}{second}$

   • 600 = $\frac{8mm}{0.5375ms-0.275ms}=\frac{30.476meters}{second}$$\frac{8\ mm}{0.5375\ ms - 0.275\ ms} = \frac{30.476\ meters}{second}$

   • 800 = $\frac{8mm}{0.525ms-0.2625ms}=\frac{30.476meters}{second}$$\frac{8\ mm}{0.525\ ms - 0.2625\ ms} = \frac{30.476\ meters}{second}$

   

   

5. How does the slope of the graph change ?

   • decreased slope

Part 4 : How does demyelination affect nerve conduction ?

• The preparation is :

  • left half = demyelinated

  • right half = myelinated

  • 5 node / myelinated internode pairs ( figure )

• For demyelinated axon , this simulation uses ion channel densities in normal unmyelinated axon.

• Actual demyelinated axon properties ( ion channel isoforms , distribution , density , etc. ) are different from this simulation , and are not fully understood

Stimulate the left end of the demyelinated ( bare axon )

1. Choose tutorial "Partial Demyelination"

2. Click on Reset & Run

   • This will stimulate left end of the bare half of the axon.

3. In the Voltage-vs-Time graph , look carefully at the traces of voltage changes at the center of the bare half of the axon ( red ) , at the first node of myelinated portion ( blue ) , and the right end of the preparation ( black )

4. To slow down the process , click on Reset ( mV ) , then repeatedly click on Continue for ( ms ) in steps of 0.2 ms

1. How does the action potential propagate ?

   • as soon as it gets to the internode , the propagation stops

Stimulate the right end of the myelinated axon

• This simulation is more relevant to the actual conditions in demyelinating diseases

5. Close the Stimulus Control ( Axon ) panel to avoid generating impulses at both ends

6. Press "Stimulus 'Trode in Node[4]" in Panel & Graph Manager to insert stimulating electrode at the right most node of the myelinated region.

7. Click on Reset & Run

2. What happens to the action potential ?

   • won't propagate going left into the bare axon region

   • conduction block

What changes in demyelinated axons will improve conduction? ( Na channels )

• This simulation modifies axon parameters in demyelinated region.

8. Continue from simulation 4.2

9. Launch the Bare Axon Parameters panel

10. Change the density of functional Na channels.

3. By how much must you change the density to enable invasion of the bare axon ? Increase or decrease ?

   • increase by $\frac{0.0007S}{c{m}^2}$$\frac{0.0007\ S}{cm^2}$

4. What changes in demyelinated axons will improve conduction? ( K channels ) Similarly , change the density of functional K channels.

   • decrease by $\frac{0.00001S}{c{m}^2}$$\frac{0.00001\ S}{cm^2}$