WEEK 2 DISCUSSION QUESTIONS & ANSWERS FALL 2017

1a. Using Voltage Clamp, you systematically hold the membrane potential at increasing amounts in a normal neuron. Design an experiment to isolate the Na and K currents.

Simple answer is to add in TTX (tetrodotoxin) to isolate K currents (blocks Na channels), and TEX (tetraethylammonium) to isolate Na currents (blocks K channels). Or change the ion bathing solutions. To isolate K currents, replace Na with with a larger, impermeant cation (Choline Chloride)

1b. Now that you have isolated the Na and K currents you begin holding the cell at different voltages 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)

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.

2. Given these values, what is the resting potential of this cell? EK = -80mV ENa = +50mV + + K conductance 10 times the Na conductance (gK = 10gNa)

Using the equation Ix=gx(Vm -Ex) for each ion. Since we are at rest the Na current and K currents are equal and opposite, we can solve for Vm. So gNa(Vm-ENa)=-gK(Vm -EK) > gNa(Vm-ENa) = -10gNa(Vm-EK), Solve for Vm =-68 mV

3. Using the following equation, which is a derivation of Ohm’s Law (INa = gNa (Vm – ENa)), estimate 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? 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, and a standard ENa of +50mV.

ENa is constant at +50 mV gNa is maxed out at all proposed holding potentials.

Vm is changed by the researcher (to -20mV, 0mV, 20mV, 50mV & 70mV), changing the value of (Vm – ENa) Vm = -20mV; Very Large inward current of Na+ Vm = 0mV; Large inward current of Na+ Vm = 20mV; Small inward current of Na+ Vm = 50mV; No current flow of Na+ Vm = 70mV; Small Outward current of Na+

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. 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) 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.

4. In the same cell as in Question 3, you change the extracellular Na+ concentration to 15mM (inside the cell remains at 18mm). What happens to the magnitude of current through voltage gated Na+ channels when you do the same set of experiments, voltage clamping the membrane potential at -70mV then stepping to -20mV, 0mV, 20mV, 50mV, and 70mV?

Since the concentration of Na+ outside the cell has changed, we must calculate a new ENa:

ENa = 58*log(13mM/18mM)

→ ENa = -5mV

As ENa is now -5mV, the flow of Na+ will change from inward to outward as the membrane potential goes from <-5mV to >-5mV. Vm = -20mV; small inward current of Na+ Vm = 0mV; Very small outward current of Na+ Vm = 20mV; Small outward current of Na+ Vm = 50mV; Large outward current of Na+ Vm = 70mV; Very large outward current of Na+

5. Describe an experiment using voltage-clamp to find out how long it takes voltage-gated Na+ channels to de-inactivate (i.e., transition from its inactivated state back to its closed-but-ready-to-open state)

We know that it takes time for the voltage gated Na+ channels to de-inactivate. To start, we would hold the cell for several seconds at a membrane potential where we know all the voltage-gated Na+ channels are open (e.g., -25mV). Then, we would move the holding potential to -70mV for a period of time (T), then back to - 25mV, and measure the amount of current flowing. We would gradually increase the time T during which the cell was held at -70mV, and measure the current flowing after return to -25mV.

During extremely short intervals of T, we would expect the current measured to remain constant, as no Na+ channels have de-inactivated. However, as we increase T, we expect more and more channels to be de- inactivated while at -70mV, meaning that they could open again when the cell is voltage-clamped to -25mV, meaning that we would see increasing conductance on the return to -25mV.

At some point, T will be long enough that ALL voltage-gated Na+ channels are able to de-inactivate while at -70mV, and any longer durations of T will not results in a greater current at -25mV. When the amount of current flowing asymptotes, we have discovered the time T at which all channels have de-inactivated. The curve we find, as we go from short T to long T, describes the probability that a channel de-inactivates given a certain amount of time at -70mV.

6) It is 1986 and you developed the patch clamp technique. You obtained the following data from your first experiment.

What does this tell you about the ion channel in this recording? Provide a rationale for your answer.

The downward deflection of current indicates that this is an inward current. This recording also suggests that the time a channel is open is variable in duration (some are open for a short period of time while others are open for a longer period of time). This is shown by the variation in the duration of the current events. Each response has two conductance states, open or closed. This is shown by the sharp rise and sharp decline of the traces. They all pass the same amount of current because all of the first 3 traces are of the same magnitude. The fourth depolarization suggests that two channels have opened simultaneously since the size is twice the size of the first three. The conclusion based on this recording is that this channel could be either a sodium, chloride, or calcium channel.