Purpose: This is based off of the week 2 homework, where students are beginning to understand what determines the
wave form of an AP. This combines the use of toxins, current wave forms, equilibrium potentials, and action potential
wave forms. Importantly, it addresses the kinetics of each channel (voltage gated delayed rectifier & voltage gated
sodium channels) and when they open or close. I’ve provided a .docx file at the bottom for ease of printing or use.
Discussion_3_group_activity_answers
Exercise: Split discussion sections into 5 groups, and provide each group a list of these questions. Assign each group 1
question they are responsible for, and provide ~5 minutes to discuss. If they have extra downtime, let them consider the
other questions as well. Re-group, and go over the answers for each scenario.
Assumptions: TTX blocks pore loop of voltage activated Na+ channel & TEA blocks pore loop of voltage activated
K+ delayed rectifiers. We’re assuming, for the “stimulation” questions, that we’re probably injecting current around 0.5
to 1 msec for less than a msec or less.
Normal AP and current wave forms: Source from McGraw Textbook.
AP & current response
1.
Administer TEA… stimulate to a little past threshold, what is the waveform of the AP?
Rather than a normal wave form, because TEA removes the delayed rectifiers, there is no outward K+ current from
delayed rectifiers. This leads to a slower return to resting membrane potential. The inactivation gates kick in from the
Na+ channels, so it shuts down the influx of Na+ ions… however, because there is no K+ outward current… the inside
of the cell remains positive until the Na+/K+ pump can bring the cell back to resting membrane potential. This will
take a long time, so it plateaus for a while and very slowly gets back to resting membrane potential.
Goal for #1: For students to understand that the delayed rectifiers are required to bring membrane potential back to a
negative voltage quickly.
Na+ channel inactivation kicks in, but no rapid decline back to rest due to no current from K+ delayed rectifiers.
After a very long period of time... will eventually return to rest. Note there is no undershoot!
2.
Administer tetradotoxin, voltage clamp at +40 mV for 1 minute… assume equilibrium potential for K+ is +60 mV
rather than -80 mV. What would the K+ current look like?
Tetradotoxin blocks the voltage gated Na+ channel… so the current response is only due to the K+ current from
delayed rectifiers. If you look at a normal K+ current, it’s a positive current with a delayed kinetics. Because your
patch clamping at +40 mV, and equilibrium potential for K+ is at +60 rather than -80, DF = +40 – +60 = -20 mV.
This means negative current. The wave form is still similar to a normal K+ current, except flipped and negative.
Goal for #2: For students to understand the usual K+ current wave form, and to understand that equilibrium potential
of K+ relative to the membrane potential is what determines the positivity or negativity of current. This tests their
understanding of what determines current, as many will want to believe that K+ current is still outwards, when in this
case, it would be inward.
K+ current over 1 min. Note plateau until after voltage clamp ends.
3.
Administer tetradotoxin, stimulate past threshold to a level that will activate K+ channels, what would K+ current
look like if:
a.
duration of stimulation is very short (not long enough for delayed kinetics).
b.
duration of stimulation is longer(long enough for delayed kinetics).
a. TTX blocks Na+ currents. If the stimulation is not long enough for delayed kinetics, the delayed rectifiers won’t kick
in, or will kick in only a little bit. This also depends on the probability of opening for the delayed rectifiers…
Probability of delayed rectifiers opening increases and the voltage becomes more and more positive. So you might see
a blip of positive current from K+ that returns back to no current shortly after stimulation.
b. Because you are stimulating longer, you’ll likely see a larger positive current from K+, it will rise slowly, and also
decrease slowly back to 0 after the stimulation is over.
Goal for #3: The goal for this question is to get them to understand the kinetics of K+ and the isolated wave form.
No current, or possibly a very brief small outward current from K+ delayed rectifiers.
A positive current from K+ delayed rectifiers opening... depending on magnitude and length of stimulation, will open
and then eventually close
4.
Administer tetradotoxin… clamp neuron at +40 mV for 1 minute, what would the K+ current look like during that
1 minute duration and a little past?
TTX blocks Na+, so the only current you see is from K+. The voltage clamp will pop the inner membrane voltage to
+40 mV, which means your delayed rectifiers will kick in. Once they kick in, you will get a slow rise of K+ positive
current… that will plateau at a certain point, and maintain at that level until the voltage clamp at +40 mV ends after 1
minute, then slowly decrease.
Goal for #4: The goal for this question is to get them to understand that the delayed rectifiers remain open for as long
as you are at a certain positive voltage… once you go back to a negative voltage will it finally close.
K+ current from delayed rectifiers remains open till +40 mV voltage clamp ends
5.
Administer TEA, clamp neuron at +40 mV for 1 minute… what would the Na+ current look like during that 1
minute duration and a little past?
TEA blocks K+, so the only current you see is from Na+. Clamping at +40 mV will stimulate the Na+ channels to
open. In response to depolarization pass threshold, the activation gate will pop open immediately, then after a delay,
the inactivation gate will close. This results in a normal Na+ current, which is negative… a rapid negative current
followed by inactivation. Reference the book for the Na+ current wave form. This occurs over ~2-4 msecs… for the rest
of the 1 minute, because you are still at +40 mV, your inactivation gate remains closed, and you don’t really have any
further Na+ current after that initial response.
Goal for #5: For students to understand that the Na+ channel activates rapidly, and then inactivates… and that
membrane potential must return to a negative enough potential before the inactivation gate reopens… thus, you have
that initial negative current, but for the rest of the 1 minute duration, there is no more Na+ current response.
Normal negative current from Na+ influx for a few msecs. For the rest of the voltage clamp, because you remain at +40
mV, inactivation gates remain closed, thus no current.