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QUESTION (AP EXTENDED) Suppose you have neuron wlth Kt channels that open In response to a neurotransmitter: Ifyou add the neurotransmitter to a neuron at rest (Vo ...

Question

QUESTION (AP EXTENDED) Suppose you have neuron wlth Kt channels that open In response to a neurotransmitter: Ifyou add the neurotransmitter to a neuron at rest (Vo = -65 mV), how will membrane potential change? Will this trigger the generation of an action potential?QUESTION (AP EXTENDED) Suppose you inject K+ ions into an individual, which dramatically increases extracellular K concentration s0 that the Kt equilibrium potential is about -20 mV: How might this affect resting membrane potential?

QUESTION (AP EXTENDED) Suppose you have neuron wlth Kt channels that open In response to a neurotransmitter: Ifyou add the neurotransmitter to a neuron at rest (Vo = -65 mV), how will membrane potential change? Will this trigger the generation of an action potential? QUESTION (AP EXTENDED) Suppose you inject K+ ions into an individual, which dramatically increases extracellular K concentration s0 that the Kt equilibrium potential is about -20 mV: How might this affect resting membrane potential? How do you think this would affect neuron 5 ability to generate action potentials?



Answers

The membrane potential of a cell is determined by the relative permeability of the membrane to various ions. When acetylcholine binds to its receptors on the postsynaptic muscle membrane, it causes a massive opening of channels that are equally permeable to sodium and potassium ions. Under these conditions, $$V_{m}=\left(V_{K^{*}}+V_{N_{A}^{*}}\right) / 2$$ If $\left[\mathrm{K}^{+}_{\text {in }}\right]=140 \mathrm{mM}$ and $\left[\mathrm{Na}_{\text {in }}^{+}\right]=10 \mathrm{mM}$ for the muscle cell, and $\left[\mathrm{Na}_{\text {out }}^{+}\right]=150 \mathrm{mM}$ and $\left[\mathrm{K}_{\text {cut }}^{+}\right]=5 \mathrm{mM},$ what is the membrane potential of the neuromuscular junction of an acetylcholine-stimulated muscle?

This question is asking us to think about what would happen if a signal was originated in a neuron in the middle of an ax on instead of in its normal manner, which, as a reminder, is normally from the cell body and then running along the axon. That's the normal direction for a signal to go. But in this case or in this question is asking what would happen if a signal started here, right in the middle of the ax on? And the answer is, it would radiate out in both directions. Normally, normally sell signals. Air neuron signals are directional because the influx of sodium and potassium happens in a kind of wave like manner down along the ax on and by originating closer to the cell body. The re polarization process prevents signals from going backwards, but because it's actually happening right here in the middle of the Axiron, no such mechanism would be in place, and so the signal would transfer or move in both directions.

Hey, I'm gonna be walking you through problem number one from chapter 48 of Candle Abolish Textbook. And this question is asking us about deep polarization. Now, what is deep polarization? Deep polarization occurs when we have a neck influx of positive ions leading thio uh, the membrane potential becoming more positives if we're thinking about, um, action potentials, uh, we see that membranes. I mean, the membrane de polarizes until it reaches a threshold point at which point sodium voltage gated ion channels open, allowing for more influx of sodium to the cell, allowing for even greater polarization. So to answer the question of what is deep polarization, it is see the membrane potential becoming more positive? This video was helpful, and I'll see you in the next one.

Hi, My name is Priya, Mona's and I will be answering my interview question. So this question consists of two parts. The first part, part A is dealing with patch clamp diagrams. So four patch clamp diagrams the peaks that we see something like this shows that the ion channel ion channel is open and when we do not see a peak, that means the ion channel is closed. So if we're looking at the figure, we can actually see two distinct channels based on the differences in current. So the 1st, 3rd and last peak, which are kind of bigger are approximately equivalent. So we can say that that is one ion channel or one distinct type of ion channel. In contrast, the 2nd, 3rd or the second and middle three peaks are also approximately equivalent. But they're a lot smaller and so we can conclude that this is also a separate but one type of ion channel. So in conclusion, we have two distinct ion channels based on our patch clamp diagram. Alright. So Part B is actually looking at how maney ions cross the open ion channel within one millisecond so part me, and were given a lot of helpful information. So for this question, we actually need to know from the patch clamp diagram how much current is flowing through these two distinct channels and were given a pretty helpful legend. So the legend says that about one third of the larger peak is too fecal lamps. So if it's one third, then the larger channel is likely, or PICO amperes is likely going to be six PICO and pierce for the larger channel, the smaller channel looks to be about double the amounts of are given legend. So we can say that this current is likely for PICO and Pierce, so knowing that we can move on to the givens provided in the question. So the first given is that we know one ampere is equal to one cool, um per second. That's our first given on, then the second given is one. PICO and peer is equal to 10 to the negative 12th amperes. All right, and then next we know that sodium is going thio, so our sodium ion is going to carry a charge of 1.6. Let's see, 1.6 times 10 to the negative 19th columns. All right, So what does this all mean? And why do we care? So we know from earlier that I can show up Here is we have different currents, right? We have four PICO amperes for one and six PICO amperes for the other. So for four PICO em peers, we need Thio convert it to amps. Right? So are just amperes, right? So how do we do that? We multiply four PICO amperes by one times 10 to the negative, 12th and pierce And then this way over one peek a lamp. Here we can cancel out our PICO in Pierce and we are left with four times 10 to the negative, 12th and Pierce. Okay, so we're just going based off of are given here. Next, we want to convert our amperes into columns. So using our first given, we're going to do for times 10 to the negative. 12th Can pierce. And we're going to do times one cool own per second. And so we are going to get we can cross out our amperes and we're going to get four times ton to the negative 12. Cool. OEMs per second. All right, so now we have that now we need Thio Move on to our last given, which is feeling with our sodium ions. So we know that one sodium ion is 1.6 times 10 to the 19th columns. So we take our four times 10 cool ums per second and we multiply that by one sodium ion from which is what were given 1.6 times 10 to the negative 19th columns. So here we can cross out our columns and we're left with 2.5 times. Yeah, let's see times 10 to the seventh, sodium sodium ions over one second. Okay, but the question is asking for ions per millisecond. So our last step is to just multiply this by one second over 10 to the three milliseconds. Uh, so if we do this 10 to the three milliseconds, we get to 0.5 times 10 to the fourth ions her one No, a second. Okay, so we follow the same pattern, remember for our other channel here for our six PICO am's. So PICO amperes. So going on here, getting some room, we're going to start with six PICO amperes. We're gonna multiply this, remember, we have to convert to amperes times one times 10 to the negative. 12th and piers over one PICO ampere. Next, we have to convert to columns over second over second. So remember, that's pretty easy. It's just one. Cool. Um, for a second, then we're gonna multiply this. We have to convert the ions, so we're gonna have one sodium I on over 1.6 times 10 to the negative 19th. Cool on. And we can go back and make sure that our units are canceling properly. Right? So we cancel our PICO amperes, we're left with amperes. We cancel our cool ums so we're left with, uh, we're gonna be left with our Dampier's and our ions over seconds. But remember, we still need to get to milliseconds. So we're going to dio times one second over 10 to the third milliseconds, and so we can cancel our seconds out. Okay, so now that we have all of this, if we multiply through, we're left with three 0.8 times, 10 to the fourth. And this is going to be sodium ions. I'll just put it here over not seconds.

For this question, we need to look at a diagram. So for patch clamp figures, the peaks that we see show when current is passing through an open ion channel. So looking at the diagram, we can actually see two distinct ion channels. And I'll tell you why. So the 1st, 3rd and last peak are relatively large and relatively equivalent to one another. Okay, so we consider this one type of ion channel now, the second, and many of the middle peaks are again equivalent to each other, but smaller than the first peak that we talked about. So this is another distinct ion channel. So based on this information, we can answer the first part of the the question where we have to distinct Okay, ion channels shell. The next part of the question is asking how many ions pass through each channel. So we are given quite a bit of helpful information in the question. But first we're gonna look at the patch clamp diagram. So according to the legend, we know about how much PICO amperes is based on the peaks given. So looking at the legend, we can approximate that the larger peak is about six PICO amperes on the smaller peak is approximately for PICO amperes. Next we look at the givens provided in the question, so we know that one PICO Ampere is equal to approximately 10 to the negative 12 amperes. We also know that one ampere is equal to one Coolum per second. We also know that the Sony um ion is equal Thio about 1.6 times 10 to the negative 19th cool lumps. So let's start with our smaller channel. We know that are smaller. Channel is approximately four PICO amperes. We also know that one PICO AMP. E er is 10 to the negative 12th Dampier's so four PICO amperes would just be equal to four times 10 to the negative 12th. Dampier's using simple unit conversion. We also know that one in fear is equal to one Coolum per second. So plugging in our four, we know that four four times 10 to the negative 12th amperes is going to be equal to four times 10 to the negative 12th cool ums per second. Next, the question is asking how Maney sodium ions are going to pass through per milliseconds. So the last or the second last step is converting our cool OEMs per second into sodium ion. So how do we do that? We take four times ton to the negative. 12. Oops, negative 12th Cool ums per second on DWI Multiply with are given one sodium ion because that's what we want over 1.6 times 10 to the negative 19th cool lumps. Looking at our equation here, we can cancel out our columns and we are left with two 0.5 times 10 to the seventh ions sodium ions per second. However, the question is asking us for ions per millisecond. So we take 2.5 times 10 to the seventh ions for a second on, we multiply by 1/10 to the three milliseconds on, we get to 0.5 times ton to the fourth ions per millisecond. Next we repeat the process. So we also take into account our six PICO amperes. So following the same process, we take six pickle amperes Onda We multiply it by one times 10 to the negative 12 amperes because we need to convert to amperes. Oh, your mind PICO ampere Continue crossing out. Then we multiply it by one. Cool. Um per second on. Then we can cross out our amperes going along. We dio one sodium ion. Uh huh. Over 1.6 times ton to the negative 19th. Cool ums. So Oh, my God. Just at I on to keep track of our units. So now we're gonna cross out our cool ums. So I'm just gonna rewrite down here because I'm running out of space a little bit, So just rewriting Here we have one sodium. I am over 1.6 times 10 to the negative. 19th Coolum on Our last step is we just need to make sure that we cancel out our seconds and convert to milliseconds. Okay, So and we're just following the exact same process as we did here with four PICO amperes. And so this is our answer for the first part. Andi, for the second part, once we multiply everything through, we're going to get 3.8 times 10 to the fourth ions per millisecond. Yeah,


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