Voltage Clamp Explained (Tetrodotoxin And Tetraethylammonium) | Clip

[@shortfutiball7023]

excellent video for a complex topic, clear with beautiful diagrams and graphics !!! you deserve much more recognition for the work you produce... you have my subscription and my like !!

[@sciencewithtal]

Thank you very much for the kind words! I am glad you enjoyed it and found it helpful!

[@cluckoothechicken5721]

I have been learning this all semester, and never quite fully understood it. I watched your video and I understand the difference between the clamp current and ionic current. Thank you so much!

[@sciencewithtal]

Wonderful, glad it helped! Thank you!

[@jokingbat5]

Great video!

[@sciencewithtal]

Thanks!

[@gbi42bce21]

great video!! Only a couple of things, why at minuti 1:20 at -50mV we have no current but at -20mV we have, what exactly happens between that two values? what is the factor that triggers the channels to open? and what happens at 0mV to have such a high inward current respect to -20mv and +20mv? Sorry, I'm bit confused. Thanks a bunch!!!

[@sciencewithtal]

Hi, good question! This is something that I look into a bit deeper in this video about the Hodgkin-Huxley model (Hodgkin-Huxley Model of Voltage-Gated Channels Explained). At 11:39 to 13:34, what I try to illustrate with the sequential steps is that the current and conductance of sodium and potassium (which can be isolated from voltage clamp experiments) are drastically different. The reason why the sodium current happens first and closes quickly whereas the potassium current opens later and is sustained has everything to do with the kinetics of the voltage gated channels (which can be explained by the Hodgkin-Huxley Model). To answer your question more directly, the main factor that makes the current increase between -50, -20 and 0 mV for sodium is that since they are voltage-gated (VG) channels, the channels open a bigger pore with higher voltage and let more sodium ions enter, thus, leading to a higher current. For the +20 mV condition, recall that the equilibrium potential of sodium is about +60 mV so as the command voltage approaches that value, the sodium current diminishes since there is less net movement into the cell (let me know if you need clarifications on the equilibrium potential). On the other hand for potassium, its equilibrium potential is at about -80 mV so its current keeps rising. Let me know if this helps, thanks for the feedback!

[@gbi42bce21]

@@sciencewithtal thanks a bunch!! all clear now, I will also give a look at the video you’ve mentioned

[@黄蟮]

It is said that oxytocin can make people loyal. Have you ever extracted oxytocin?​@@sciencewithtal

[@黄蟮]

​@@sciencewithtalAre you planning to extract oxytocin from ergot?

[@eymendediler5357]

I didn't fully understand why there is an inward current during hyperpolarization, such as when Vcommand is set to -70 mV. Because typically, inward current indicates depolarization. Could you explain this?"

[@sciencewithtal]

Good question, that was something confusing for me as well! Basically, if I reword what I mention in the video: we have a cell at rest at -60 mV, and now we clamp it to -70 mV. To do so, the voltage clamp will send negative charges to push the membrane potential towards -70 mV. Given that the neuron wants to stay at -60 mV, the neuron will open leak channels (IL) that will send positive charges inside to get back to rest and this is what you see on the readout as the inward/depolarizing current. In other words, with voltage clamp you are measuring the current response of the neuron and since it gets hyperpolarized, its response will be to open channels and depolarize to come back to the resting potential. Hope that clarifies your question, let me know if I can help further!

[@eymendediler5357]

​Thank you very much! Crystal clear now.

[@adhiraj2461]

@@sciencewithtal well i think to do clamp it send charge based on the feedback(which is a typical property of an OPAMP - used as an feedback amplifier here) i.e.- in the above example as the clamp voltage is -70mv and is lesser than the membrane potential which is -60v i.e the negative feedback is there which means it takes positive charge from the membrane or say gives negative charge to the membrane - which is okay and the example is inward but based on the definition of inward it says the negative current is the movement of positive charge in the cell. but here no positive charge is going to the cell even its not depolarizing.. what i guess is that inward and outward is basically depending on what cell wants to do like if we are hyperpolarizing to -70mv and but the cell wants to be at -60v which is its resting potential so what it will do - it will basically take positive charge to- well i know finally it has to be at -60 that's why i am also confused with the thing i thought about it as.

[@adhiraj2461]

please reed the cut part i am unable to remove it but it's not done intentionally

[@hiroboy7506]

Nice video, also I don’t understand Trial 1. Vm = -60 and Vc = -70, so we want a negative current to come-in (a.k.a positive current OUT of the cell, making Vm more negative). But you mentioned in Trial 1 that it’s an "inward current" (movement of positive ions INTO the cell) ? Isn't this contradictory to what you explained earlier? I see someone else asked exactly the same question, meaning this part is indeed unclear. I read your answer explanation in the comments, but it’s still confusing. How does that even make sense? Your explanation of inward/outward currents is based on positive ion movement, so it should naturally refer to the capacitive current (Ic), but suddenly you mention it’s due to the leak current (Il). Leak current isn't even in the the previous equation. Please help us to understand !

[@wubwub616]

I don't understand. I thought that in an operation amplifier, the potential between the (+) and (-) ends must be 0. But in that case, the potential between extracellular and intracellular space AKA membrane potential, should be 0

[@sciencewithtal]

From the way I understand, the offset between the (+) and the (-) terminals has to be set to 0 to give the corresponding difference in voltage between the two. Basically, it sets the ground/reference electrode to 0 mV and makes it so that the voltage picked up by the electrode inside the neuron is the membrane potential (e.g. -70 mV). If the (+) and (-) terminals are offset by -10 mV then the reading will give -80 or -60 mV depending on the arrangement of the apparatus. Let me know if that makes sense!