1. Choose tutorial "The Neuromuscular Junction"

2. Click on Start the Stimulation

3. Increase Total # ms to 4 ms

4. Click on Reset & Run

5. Keep Lines

1. Closely examine the three traces.

   

2. Compare and contrast the differences between them.

   Trace	Units	Shape	Details
   Top ( Voltage )	mV	rises , then decays	depolarization of the membrane due to ACh binding and causing ion movement
   Middle ( Current )	nA	sharp dip , then rises	influx of positive ions ( Na⁺ , K⁺ ) through ACh receptor channels
   Bottom ( Conductance )	μS	rises and falls symmetrically	opening and closing of ACh receptor ion channels

3. Explain what might account for some traces going up and others going down.

   • Voltage : increase = depolarization

   • Current : inward = negative ( goes down )

   • Conductance : always positive ( channel opening )

4. Explain what might account for the differing time courses the voltage and current responses.

   • Conductance changes first , driven by ACh binding to receptors

   • Current changes next , closely tracking conductance

   • Voltage changes last , because its the integration of current over time

     • "RC properties" of the cell

       • membrane’s capacitance and resistance

6. Increase receptor conductance to 5 μS and then to 10 μS ( g max in Alpha Synapse window )

7. Continue increasing in steps of 10 μS at least up to 60 μS

5. Determine whether voltage and current continue to increase proportionally. Explain briefly.

   • qualitatively :

     • $g_{\text{max}}\propto V_m$$g_{\text{max}} \propto V_m$

     • $g_{\text{max}}\propto I$$g_{\text{max}} \propto I$

   • but the rate at which they increase levels off after 50 μS

   • once it gets near the reversal potential , further increasing conductance doesn't really change current and voltage.

6. Estimate the half time for current decay. ( try decreasing Total # ms to 1.0 ms )

   • when $g_{\text{max}}=60\mu S$$g_{\text{max}} = 60\ \mu S$ , peak current = -1854.54 nA at 0.5125 milliseconds

   • half of that = -927.27 nA , at around 0.57 milliseconds

8. Erase traces

9. Change total # ms to 25

10. In Alpha Synapse panel, change onset to 20 ms

11. Set g max back to 2 μ Siemens

12. Keep lines

13. Reset & Run

14. Click on Stimulus Control , IClamp

15. Increase the amplitude of the current pulse in 10 nA increments, at least to 50 nA

7. Look at each graph carefully. As you change the injected current , observe what happens to conductance , voltage , and current changes induced by ACh

   • Conductance :

     • stays the same for all current injections

   • Voltage :

     • at the 20 ms mark , ACh is released , and attempts to move towards the reversal potential of the ACh channel

     • when we are more negative than the reversal potential , ACh as a positive cation enters the cell , making the membrane potential more positive.

     • when we are more positive than the reversal potential , ACh as a positive cation leaves the cell , making the membrane potential more negative.

   • Current Changes Induced by ACh :

     • $\text{Injected Current}\frac{1}{\propto }\text{ACh Current}$$\text{Injected Current}\ \frac{1}{\propto}\ \text{ACh Current}$

     • smaller ( less negative ) with more injected current

16. Repeat the experiment in voltage clamp.

17. Set total # ms back to 2.5 and onset to 1 ms.

18. Select voltage clamp on the stimulus control

8. Create an IV plot to determine the reversal potential ( you will need to design the voltage clamp protocol )

   

   • Data

   $$0=2\ast x+30$$
   $$x=\frac{-30}{2}=-15$$

   • Reversal Potential $\approx -15mV$$\approx -15\ mV$

9. If the [Na+ ]o = 120mM and [Na+ ]i = 12mM and [K+ ]o = 4mM and [K+ ]i = 110mM, calculate the gNa/gK of the AChR.

   $$E_{Na}=60\ast lo{g}_{10}\left(\frac{120}{12}\right)=60mV$$
   $$E_K=60\ast lo{g}_{10}\left(\frac{4}{110}\right)=-86.360mV$$

   • so we know $E_{rev}=-15mV$$E_{rev} = -15\ mV$

   $$-15=\frac{(g_{Na}\ast E_{Na})+(g_K\ast E_K)}{g_{\text{total}}}$$

   • let $x=\frac{g_{Na}}{g_K}$$x = \frac{g_{Na}}{g_K}$

     • therefore , $g_{Na}=x\ast g_K$$g_{Na} = x * g_K$

     • and then $g_{\text{total}}=x\ast g_K+gK$$g_{\text{total}} = x * g_K + gK$

       • or $g_K\ast (x+1)$$g_K * \left(\ x + 1 \ \right)$

   • substituting stuff back in :

   $$-15=\frac{(x\ast g_K\ast 60mV)+(g_K\ast -86.36mV)}{g_K\ast (x+1)}$$

   • factoring :

   $$-15=\frac{g_K\ast ((60\ast x)-86.36)}{g_K\ast (x+1)}$$

   • canceling out $g_K$$g_K$ :

   $$-15=\frac{(60\ast x)-86.36}{x+1}$$

   • algebra :

   $$\frac{g_{Na}}{g_K}\approx 0.951467$$

10. Based on the IV plot , determine whether the conductance of AChR is voltage dependent.

    • its NOT voltage dependent

    • the conduction fit / trend line is linear

      • constant conduction

19. Close stimulus control

20. Erase traces

21. Click “Add HH channels”

22. In Alpha Synapse panel, begin with g max at 8 μSiemens and gradually decrease it to 2 μS

11. Explain what happens. ( threshold )

    • action potential only occurs when the postsynaptic depolarization reaches threshold

    • this requires a minimum $g_{\text{max}}=3.533$$g_{\text{max}} = 3.533$

12. Indicate the disease that resembles what you have modeled.

    • Myasthenia Gravis

    • autoimmune destruction of AChRs

    • with fewer AChRs , $g_{\text{max}}$$g_{\text{max}}$ decreases , making it hard to fire an action potential , and muscle weakness