How membrane potential varies with stimulus amplitude

1. Choose tutorial "Introduction to Neurons in Action"

2. Click on Start the Stimulation

3. Click on Stimulus Control ( in Panel & Graph Manager window )

4. Click on IClamp ( in Stimulus Control window )

5. Click on Voltage vs. Time Plot ( in Panel & Graph Manager window )

6. Increase Total # ( ms ) to 8 ms ( in Run Control window )

7. Click on Reset & Run ( in Run Control window )

8. Right Click on the Graph , hold, scroll down to Keep Lines start stimulating at 0.06 nA

1. Vary the amplitude of stimulation ( change amplitude , nA , in Stimulus Control window )

2. Click on Reset and Run

3. Describe what happens ( sketch representative voltages traces to illustrate )

   • 0.0300 nA = blue = halving = still no action potential

   • 0.0600 nA = black = default

   • 0.0658 nA = red = first action potential

   • 0.0900 nA = green = action potential happening quickly and not delayed

   

4. Use the mouse to move over plot , left-click to read amplitude. ( write down each Istim and Vpeak pair )

   • 0.0300 nA = blue = -62.3641 mV

   • 0.0600 nA = black ( original ) = -59.5748 mV

   • 0.0658 nA = red = +36.8368 mV

   • 0.0900 nA = green = +43.5287 mV

5. Determine whether action potential amplitude changes with stimulus amplitude.

   • yes , action potential amplitude is direction proportional to stimulus amplitude

6. Determine the threshold amplitude.

   • 0.0658 nA = +36.8368 mV

7. Determine whether threshold changes with stimulus amplitude

   • the time it takes to reach threshold is inversely proportional to the stimulus amplitude

   • threshold is reached quicker with larger stimulus amplitudes

1. Erase Lines

2. Change Stimulus amplitude to 0.18 nA

3. Keep Lines

4. Reset and Run to get control trace

5. Click on Patch Parameters

a.) Extracellular Na⁺ Concentration

6. Change Extracellular Na. Start by halving & doubling.

1. Describe what happens and the underlying processes at work , measuring Vm at key points during the time course ( sketch representative voltages traces to illustrate ) :

   • $[Na{]}_{\text{extracellular}}=70.0mM$$[Na]_{\text{extracellular}} = 70.0\ mM$ = blue

     • $V_{\text{max}}=+29.4385mV$$V_{\text{max}} = +29.4385\ mV$ @ $1.475\ ms$

   • $[Na{]}_{\text{extracellular}}=140mM$$[Na]_{\text{extracellular}} = 140\ mM$ = black ( original )

     • $V_{\text{max}}=+45.3089mV$$V_{\text{max}} = +45.3089\ mV$ @ $1.375ms$$1.375\ ms$

   • $[Na{]}_{\text{extracellular}}=280mM$$[Na]_{\text{extracellular}} = 280\ mM$ = red

     • $V_{\text{max}}=+61.0205mV$$V_{\text{max}} = +61.0205\ mV$ @ $1.3ms$$1.3\ ms$

   • $[Na{]}_{\text{extracellular}}=500mM$$[Na]_{\text{extracellular}} = 500\ mM$ = green

     • $V_{\text{max}}=+74.0768mV$$V_{\text{max}} = +74.0768\ mV$ @ $1.25ms$$1.25\ ms$

   

2. Determine which parts of the AP changes , if any

   $$[Na{]}_{\text{extracellular}}\propto V_{\text{max}}$$
   $$[Na{]}_{\text{extracellular}}\frac{1}{\propto }\text{time to peak}$$

3. Determine which parts of the AP do not change.

   • threshold potential

   • time to reach $V_{\text{min}}$

4. Discuss your interpretation of these results.

   • Driving Force = $\left(V_m\right)-\left(E_{\text{ion}}\right)$$\left(\ V_{m} \ \right) - \left(\ E_{\text{ion}} \ \right)$

     • Driving Force $\frac{1}{\propto }$$\frac{1}{\propto}$ $E_{ion}$$E_{ion}$

   • $E_{\text{cation}}\propto \frac{[\text{extracellular}]}{[\text{intracellular}]}$$E_{\text{cation}} \propto \frac{[\ \text{extracellular}\ ]}{[\ \text{intracellular}\ ]}$

     • nernst is more positive the more extracellular concentration is increased

     • nernst is more negative the more inside concentration is increased

   • $[Na{]}_{\text{extracellular}}\propto$$[\ Na\ ]_{\text{extracellular}} \propto$

5. Contrast these results with what happened when you changed stimulus amplitude.

   • modifying stimulus amplitude determines if we reach threshold or not , and how quickly

   • modifying extracellular sodium concentration doesn't change threshold

   • changes the depolorization phase , steeper slope

b.) Intracellular Na⁺ concentration

7. Erase Lines

8. Set extracellular Na+ concentration back to default ( click on red check )

9. Reset & Run to get control trace

10. Change Intracellular Na+

1. Describe what happens ( by measuring Vm ) and the underlying processes at work ( sketch representative voltages traces to illustrate ) :

   • $[Na]_{\text{intracellular}} = 7.0\ mM$ = blue

     • $V_{\text{max}}=+61.0205mV$$V_{\text{max}} = +61.0205\ mV$ @ $1.3ms$$1.3\ ms$

   • $[Na]_{\text{intracellular}} = 14\ mM$ = black ( original )

     • $V_{\text{max}}=+45.3089mV$$V_{\text{max}} = +45.3089\ mV$ @ $1.375ms$$1.375\ ms$

   • $[Na]_{\text{intracellular}} = 28\ mM$ = red

     • $V_{\text{max}}=+29.4163mV$$V_{\text{max}} = +29.4163\ mV$ @ $1.5ms$$1.5\ ms$

   • $[Na{]}_{\text{intracellular}}=56mM$$[Na]_{\text{intracellular}} = 56\ mM$ = green

     • $V_{\text{max}}=+13.3703mV$$V_{\text{max}} = +13.3703\ mV$ @ $1.65ms$$1.65\ ms$

   

2. Determine which parts of the AP changes , if any

   $$[Na{]}_{\text{intracellular}}\frac{1}{\propto }{V}_{\text{max}}$$
   $$[Na{]}_{\text{intracellular}}\propto \text{time to peak}$$
   $$[Na{]}_{\text{intracellular}}\propto \text{time to reach }{V}_{\text{min}}$$

3. Determine which parts of the AP do not change.

   • threshold voltage

4. Discuss your interpretation of these results.

   • everything is flipped compared to changing proportions of extracellular concentrations

5. Contrast these results with what happened when you changed stimulus amplitude.

   • no difference between this and extracellular

1. Erase Lines

2. Set intracellular Na concentration back to default ( if not already )

3. Change Stimulus amplitude to 0.18 nA

4. Keep Lines

5. Reset and Run to get control trace

a.) Extracellular K⁺ Concentration

6. Change Extracellular K. Start by halving & doubling.

1. Describe what happens and the underlying processes at work , measuring Vm at key points during the time course ( sketch representative voltages traces to illustrate ) :

   • $[K{]}_{\text{extracellular}}=2.5mM$$[K]_{\text{extracellular}} = 2.5\ mM$ = blue

     • $V_{\text{max}} = +43.2048\ mV$ @ $1.775\ ms$

     • $V_{\text{min}}=-92.4244mV$$V_{\text{min}} = -92.4244\ mV$ @ $4.3ms$$4.3\ ms$

   • $[K{]}_{\text{extracellular}}=5mM$$[K]_{\text{extracellular}} = 5\ mM$ = black ( original )

     • $V_{\text{max}}=+45.3089mV$$V_{\text{max}} = +45.3089\ mV$ @ $1.375ms$$1.375\ ms$

     • $V_{\text{min}}=-76.5002mV$$V_{\text{min}} = -76.5002\ mV$ @ $4.375ms$$4.375\ ms$

   • $[K{]}_{\text{extracellular}}=10mM$$[K]_{\text{extracellular}} = 10\ mM$ = red

     • $V_{\text{max}} = +46.6865\ mV$ @ $1.225ms$$1.225\ ms$

     • $V_{\text{min}}=-60.2096mV$$V_{\text{min}} = -60.2096\ mV$ @ $5.375ms$$5.375\ ms$

   • $[K{]}_{\text{extracellular}}=20mM$$[K]_{\text{extracellular}} = 20\ mM$ = green

     • $V_{\text{max}} = +47.9175\ mV$ @ $1.125\ ms$

     • $V_{\text{min}}=-38.5307mV$$V_{\text{min}} = -38.5307\ mV$ @ $6.15ms$$6.15\ ms$

   

2. Contrast with what happened when you changed extracellular Na+ concentration :

    

   $$[K{]}_{\text{extracellular}}\propto V_{\text{max}}$$
   $$[K{]}_{\text{extracellular}}\frac{1}{\propto }\text{time to peak}$$
   $$[K{]}_{\text{extracellular}}\propto V_{\text{min}}$$
   $$[K{]}_{\text{extracellular}}\propto \text{time to reach}{V}_{\text{min}}$$

b.) Intracellular K⁺ concentration

7. Erase Lines

8. Set extracellular K+ concentration back to default ( click on red check )

9. Reset & Run to get control trace

10. Change Intracellular K+

1. Describe what happens ( by measuring Vm ) and the underlying processes at work ( sketch representative voltages traces to illustrate ) :

   • $[K]_{\text{intracellular}} = 62.0\ mM$ = blue

     • $V_{\text{max}} = +46.6865\ mV$ @ $1.225ms$$1.225\ ms$

     • $V_{\text{min}}=-60.2095mV$$V_{\text{min}} = -60.2095\ mV$ @ $5.4ms$$5.4\ ms$

   • $[K]_{\text{intracellular}} = 124\ mM$ = black ( original )

     • $V_{\text{max}}=+45.3089mV$$V_{\text{max}} = +45.3089\ mV$ @ $1.375ms$$1.375\ ms$

     • $V_{\text{min}}=-76.5002mV$$V_{\text{min}} = -76.5002\ mV$ @ $4.375ms$$4.375\ ms$

   • $[K]_{\text{intracellular}} = 248\ mM$ = red

     • $V_{\text{max}} = +43.2048\ mV$ @ $1.775\ ms$

     • $V_{\text{min}} = -92.4275\ mV$ @ $4.275ms$$4.275\ ms$

   • $[K{]}_{\text{intracellular}}=500mM$$[K]_{\text{intracellular}} = 500\ mM$ = green

     

     $$[K{]}_{\text{intracellular}}\frac{1}{\propto }{V}_{\text{max}}$$
     $$[K{]}_{\text{intracellular}}\propto \text{time to peak}$$
     $$[K{]}_{\text{intracellular}}\frac{1}{\propto }{V}_{\text{min}}$$
     $$[K{]}_{\text{intracellular}}\frac{1}{\propto }\text{time to reach }{V}_{\text{min}}$$

    

2. Try increasing and decreasing extracellular Na+ concentration by 3 mM , then extracellular K+ concentration by 3 mM :

   • increasing $[Na{]}_{\text{intracellular}}$$[Na]_{\text{intracellular}}$ and increasing $[K{]}_{\text{intracellular}}$$[K]_{\text{intracellular}}$ :

     

   • increasing $[Na{]}_{\text{intracellular}}$$[Na]_{\text{intracellular}}$ and decreasing $[K{]}_{\text{intracellular}}$$[K]_{\text{intracellular}}$ :

     • very similar to above , just earlier in time

        

   • decreasing $[Na{]}_{\text{intracellular}}$$[Na]_{\text{intracellular}}$ and decreasing $[K{]}_{\text{intracellular}}$$[K]_{\text{intracellular}}$ :

   

   • increasing $[K{]}_{\text{intracellular}}$$[K]_{\text{intracellular}}$ while decreasing $[Na{]}_{\text{intracellular}}$$[Na]_{\text{intracellular}}$ :

     

1. Go to the "The Na Action Potential" Tutorial

2. Start the Simulation

3. Increase the total ms to 6 ms

4. In Stimulus Control , increase the stimulus amplitude to 0.22 nA

5. Keep lines

6. In Patch Parameters , half and double the Na Channel Density ( halving is like blocking Na channels )

1. Discuss your observations of what happens.

   • $\frac{0.06\ S}{cm^2}$ = blue :

     • $V_{\text{max}} = +32.5236\ mV$ @ $1.6\ ms$

     • faster to reach $V_{\text{min}}$

   • $\frac{0.12\ S}{cm^2}$ = black ( original )

     • $V_{\text{max}}=+45.5878mV$$V_{\text{max}} = +45.5878\ mV$ @ $1.225ms$$1.225\ ms$

   • $\frac{0.24\ S}{cm^2}$ = red

     • $V_{\text{max}} = +51.1379\ mV$ @ $1.0\ ms$

     • faster to reach $V_{\text{max}}$$V_{\text{max}}$

1. First make a prediction as to what halving the K+ channel density will do :

   • sodium will have less competition , so it will reach $V_{\text{max}}$$V_{\text{max}}$ faster

2. Return Na Channel Density to default

3. Repeat exercise described above , but now change K+ Channel Density ( halving is like blocking K+ channels )

4. Discuss your observations of what happens

   • $\frac{0.018S}{c{m}^2}$$\frac{0.018\ S}{cm^2}$ = blue :

     • $V_{\text{max}}=+49.7401mV$$V_{\text{max}} = +49.7401\ mV$ @ $1.15ms$$1.15\ ms$

     • slower time to reach $V_{\text{min}}$

     • faster time to reach $V_{\text{max}}$$V_{\text{max}}$

   • $\frac{0.036S}{c{m}^2}$$\frac{0.036\ S}{cm^2}$ = black ( original )

     • $V_{\text{max}}=+45.5878mV$$V_{\text{max}} = +45.5878\ mV$ @ $1.225ms$$1.225\ ms$

   • $\frac{0.072S}{c{m}^2}$$\frac{0.072\ S}{cm^2}$ = red

     • $V_{\text{max}}=+36.9994mV$$V_{\text{max}} = +36.9994\ mV$ @ $1.5ms$$1.5\ ms$

     • faster to reach $V_{\text{min}}$

     • slower to reach $V_{\text{max}}$$V_{\text{max}}$

   

5. Compare with your prediction

   • slower time to reach $V_{\text{min}}$

   • faster time to reach $V_{\text{max}}$$V_{\text{max}}$