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Home , Voltage clamp

1. Draw a schematic set-up for a clamp experiment including , voltage electrode, ground electrode, voltage meter, holding voltage, amplifier, current meter, and current electrode.

a. What is the goal of a voltage clamp? i. To manipulate membrane voltage independently of ionic currents, so that current-voltage relationships of membrane channels can be studied b. Briefly, how does a voltage clamp work? i. A voltage clamp holds constant, allowing membrane currents to fluctuate and be recorded through a system c. What is the role of each component of the voltage clamp? i. Voltage electrode 1. Records charge concentration inside ii. Voltage meter 1. Measures voltage inside cell (relative to ground) iii. Command Generator 1. Programs the “holding” (i.e. “desired”) voltage iv. Difference Amplifier 1. Subtracts actual voltage from programmed voltage to generate an ‘error signal’. Using Ohm’s law, the amplifier then generates a current to alter the membrane potential and reduce the error signal to zero - thus holding the membrane potential at the programmed voltage. v. Current Meter 1. Measures current (columbs/second) vi. Current electrode 1. Transmits current into axon

2. a. Draw a graph of the time course of the change in membrane potential during an .

b. Draw a second graph showing the time course of the change in total membrane current during an action potential. Include the voltage step used to depolarize the cell (e.g. -70mV to 0mV and back to -70mV). c. On the same graph draw separate curves for the Na and K currents during an action potential.

3. A depolarizing voltage clamp pulse is delivered to an axon being held at -70mv and the inward Na+ and outward K+ currents are recorded. If you want to eliminate the Na+ current and record only the outward K+ current, you have three ways of accomplishing this aim. Name all three. Assume you have control over the external solution and the voltage of the clamp pulse. Make certain that your procedures make the Na+ current zero. a. Block Na channels with TTX. b. Voltage clamp to ENa+.

c. Make [Na+ ]out equal to [Na+ ]in

4. Now that you have isolated the Na and K currents you begin holding the cell at different from -60mV to +70mV and you notice something interesting. Voltage-gated K currents are always outward, and voltage-gated Na currents inward at some values and

outward at others. Why is does this occur? (Hint: EK= -80mV and ENa=50mV) a. For the K currents, you are stepping from -60mV to +70mV and never cross the reversal potential thus as you go more positive the currents will only increase outward. However, for the Na current, we cross the reversal potential so the electrical driving force overpowers the chemical driving force and Na flows outward.

5. Using the following equation, which is a derivation of Ohm’s Law (INa = gNa (Vm – ENa)): a. Estimate and draw o b. n a graph the magnitude of the current for voltage dependent Na+ channels when the membrane potential is voltage clamped at -70 mV and then is stepped up to -20mV, 0mV, 20mV, 50mV, and 70mV. Specifically indicate if the current is very large, large, small, very small or none and indicate whether or not it is an inward or an outward current. Assume gNa reaches its maximum value at all voltages more positive than -25 mV. Assume a standard ENa of +50mV. i. Notice that, as Vm is changed by the researcher, the value of the driving force (Vm – ENa) changes as well. Therefore: 1. Vm = -20mV; Very Large inward current of Na+ 2. Vm = 0mV; Large inward current of Na+ 3. Vm = 20mV; Small inward current of Na+ 4. Vm = 50mV; No current flow of Na+ 5. Vm = 70mV; Small Outward current of Na+ ii. Further explanation: When the membrane potential is clamped at a voltage below the equilibrium potential for Na+, Na+ will flow into the cell. As the membrane potential approaches the equilibrium potential of Na+, no current will flow. You have reached equilibrium. iii. Once the membrane potential has exceeded that of the equilibrium potential of Na+, Na+ will flow out of the cell towards its concentration gradient. To calculate this out mathematically use: INa = gNa (Vm – ENa). iv. At voltages more positive than -25mV, gNa is maxed out, so this is negligible (this value will not affect INa ); the units, however, are still necessary to properly calculate INa, so do not ignore gNa in your calculation.

6. Given these values, what is the of this cell? EK = -80mV, ENa = +50mV. Assume K+ conductance is 10 times the Na+ conductance. a. gK = 10gNa b. Use V = IR → I = gV → I = g(Vm-Ex); for each use the equation Ix=gx(Vm- Ex) c. Since we are at rest the Na current and K currents are equal and opposite: gNa(Vm-ENa) = -gK(Vm -EK) d. Substitute: gNa(Vm-ENa) = -10gNa(Vm-EK) e. Solve for Vm = -68 mV