11. Biological Membranes and Transport

Thermodynamics of Membrane Diffusion: Charged Ion

11. Biological Membranes and Transport

# Thermodynamics of Membrane Diffusion: Charged Ion

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concept

## Thermodynamics of Membrane Diffusion: Charged Ion

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So now that we've covered the thermodynamics of membrane diffusion for uncharged molecules in our previous lesson videos in this video we're going to introduce the thermodynamics of membrane diffusion, but for charged ions. And so if we take a look at our image down below, over here on the left hand side, we'll see a little reminder that in this video we're specifically focusing on charged ions. And so notice here in yellow, what we have is our charged ion, more specifically, our positively charged cat ion. And again, in this video, we're going to be focusing on the thermodynamics as charged ions like this one diffuse across a membrane through a membrane channel from its initial side of the membrane to its final side of the membrane. And so it's important to note is that unlike uncharged molecules, when charged ions diffuse across a membrane, the trans membrane potential or the trans membrane voltage delta Sigh must be considered. And so recall from our previous lesson videos that we already introduced the trans membrane potential or the trans membrane voltage Delta Sigh. And so recall from those previous lesson videos that the trans membrane potential or the Trans membrane voltage. Delta Sigh is really just equal to the difference in electrical charge across the membrane. Now, if we were to Onley, consider the trans membrane potential or the electrical Grady int Onley or alone. Then the Delta G transport of ions would be given as the following equation right here, where Delta G transport would be equal to the variable z times the variable f times Delta side the trans membrane potential. Now the variable Z is really just equal to the Net charge of the defusing ion and the variable F is known as Faraday's Constant, named after the scientist Michael Faraday, and really fair days, Constant F is just equal to the magnitude of the charge of one mole of electrons. And despite electrons being negatively charged, F fair days constant is always a positive value. In fact, F Faraday's constant is equal to positive. 96,485 with units of jewels, times inverse voltage divided by most or with units of cool ums per mole, depending on your textbook. Now it's important to note is that whenever you have a number or variable raised to the negative one, all we need to do is take the reciprocal of it. And so when we take the reciprocal of that, what we can get is Jules over volts times moles. And so, uh, this unit right here is the same thing as this unit up above. And that's important to keep in mind as we move forward and tackle out our practice problems. Now again, this equation right here would be true if Onley, the trans membrane potential or electrical Grady int alone was considered. But what we need to recall, however, from our previous lesson videos is that when ions diffuse across the membrane, it does not Onley depend on the electrical Grady int. It actually depends on the electro chemical Grady int. And you might recall that the electro chemical Grady int is really just a combination of the chemical ingredient and the electrical Grady int. And so if we take a look at our image down below right here on the right hand side, notice that this blue shaded region here is the portion of the equation explained by the chemical Grady int and this blue portion alone is the same exact equation that we introduced in our previous lesson videos when we talked about the thermodynamics of membrane diffusion for uncharged molecules. And that's because uncharged molecules do not depend on electrical ingredients. They Onley depend on chemical ingredients, however, because ions charged ions depend on the electrochemical radiant, they require both the chemical Grady int portion of the equation and the addition of the electrical Grady int portion of the equation, which is this part of the equation here. And so recall that it is, as we indicated up above, it's going to be easy times. The variable F, which you'll see sometimes, is stylized to this style right here for fair days, constant and then times Delta sigh the trans membrane potential. And so, really, when we're considering the thermodynamics of membrane diffusion for charged ions, all we need to do is tag on this electrical Grady int portion of the equation. And then we're set, and we can follow the same exact steps as before and get our answer. And so this year concludes our lesson and introduction to the thermodynamics of membrane diffusion for charged ions, and and our next video will be able to show you guys an example of how to apply this entire equation here to calculate the thermodynamics of membrane diffusion for charged ions. And so I'll see you guys in that next video.

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example

## Thermodynamics of Membrane Diffusion: Charged Ion Example 1

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all right. So here we have an example problem that wants us to calculate the energy costs or the Delta G transport of pumping calcium ions. Or see a two plus from the side assault to the extra cellular space. If the temperature is equal to 37 degrees Celsius, the trans membrane potential Delta Sigh is equal to 0.5 volts, where the inside of the cell is still negative. The site is solid. Calcium concentration is equal to 1.0 times 10 to the negative seventh Mueller and the extra cellular calcium concentration is equal to 1.0 times 10 to the negative third Moeller. And so, in order to solve problems like this one, we're actually going to follow a four step process that's very similar to the four step process that we covered in our previous lesson videos with just a little bit of differences. Now in step number one, it's the same a step number one from our previous lesson video. So all we need to do is determine the net charge of the defusing molecule. And if we take a look at our example problem, of course, the defusing molecule is calcium because that's what's being pumped across the membrane and calcium. Here is clearly a charged ion, so we could go ahead and check off this box. And, of course, by the two plus here we know that it's charges, actually, plus two. And so we can write that here in this blank. And so because this is a charged ion, of course, what this means is that we are going to use the same equation that we introduce from our previous lesson video that includes both the Chemical Grady Int and the Electrical Grady Int since again charged ions diffuse based on the electrochemical Grady int the combination of both of these. And so now that we know that we're going to use the equation from our previous lesson video to solve this problem, we can move on to step number two, which is actually slightly different from the step number two in our previous lesson videos. And so here in this step number two, we're still going to determine the direction of diffusion by establishing the initial and the final sides. But we're also going to determine the sign of the trans membrane potential. Delta sigh whether or not, it's going to be positive or negative when we include it into our equation. And so drawing a sketch is sometimes going to be very helpful for step number two. So notice down below. We have our cell membrane down below, and you can see that we have the ion channel embedded. And so, of course, to determine the direction of diffusion. Looking back at our example problem notice that we're pumping calcium from the site us all to the extra cellular space. And so if we call the left side over here, the inside of ourselves or the side, us all. And of course, the right side of ourselves over here, the outside of ourselves or the extra cellular space. Then we know that calcium is going to be pumped in this direction across the membrane again from the inside to the outside, from the site us all to the extra cellular space. And so, of course, that means that this left side over here, the inside of the cell is going to be the initial side of the membrane, and the outside of the cell on this end of the arrow is going to be the final side of the membrane. And so now we've established our initial and final sides. We can actually rewrite the concentrations of calcium that are given to us on each side. And so the Salafist Solich, the site of solid calcium concentration, is given to us as 1.0 times 10 to the negative seventh Moeller and again side of Solich means it's going to be on the inside so we could go ahead and write that concentration over here 1.0 times, 10 to the negative seventh Moeller. And then, of course, the extra cellular calcium concentration is given to us as 1.0 times 10 to the negative third Mueller. So we can write that over here on the outside of the cell, 1.0 times, 10 to the negative third Moeller. All right, so now to moving on to determine the sign of the trans membrane potential, whether or not it should be positive or negative when we include it into our equation. And so, of course, what we need to recall from our previous lesson videos is that the delta symbol eyes actually equal to final minus initial. And so, of course, because our final side over here is the outside of the cell and our initial side is the inside of the cell. What this means is that the trans membrane potential Delta Sigh is going to be equal to the final, which is going to be sigh out since again, Final is on the outside of the cell. And, of course, it's going to be minus the initial. And again the initial is on the inside of the self, so it's gonna be minus sigh. And and so because we know that the inside of the cell ISMM or negative with respect to the outside of the cell, which we know is more positive. And that's also indicated here in our example problem, the inside is more negative. We know that if you take something that is mawr positive and you subtract something that is mawr negative, you're going to get a value that is positive. And so what this means is that we want to make sure that our trans membrane potential, when we input it into our equation, that it's going to have a positive value and so notice that our trans membrane potential is actually provided as a positive value. So really, we didn't have to make any changes here. However, if this trans membrane potential were provided as a negative value, then we would have to switch the sign to make sure that it has a positive value before we included into our equation. And so really, this is it for step number two. We've determined the direction here, and we've determined the sign of Delta Side to be a positive value when we included into our equation. So now we can move on to step number three. And step number three is actually the same as step number three in our previous lesson videos. So all we need to do is check all of the units on all of our numbers and, if necessary, we need to convert the units to ensure compatibility with our equation. And so what this means is that we should check the temperature, make sure our temperatures and units of Kelvin we need to check our trans membrane potential. Make sure it's in units of volts, since I recall that the A fair day's constant eyes going toe have units of volts and it as well, and so what? You'll notice is that our trans membrane potential is given to us in units of volts, so we don't need to make any changes there. Uh, notice that our temperature is actually given to us in units of degrees Celsius. So before we can use our temperature here, we know we need to convert it into units of Calvin. So that's something that we're going to do on. Then notice that the calcium concentration on on the side of Solich side is given to us in units of polarity and the extra cellular calcium concentration is also given to us in units of polarity. So the units of polarity here, uh, match each other, and when we put them into this ratio, we know that they're going to cancel out. So really, there's no changes that we need to make here in terms of the units. So the only thing that we really need to change again is the temperature, and so here. What we have is that the temperature is 37 degrees Celsius and if we want to convert it to units of Calvin, we need to add 273.15 to it and When we do that, we get a value of 310 0. Calvin. And so this is the temperature that we want to use in our equation. And so really, that is it for step number three. So now we can move on to step number four, which is actually the most straightforward step, because all we need to do is plug in all of the given values with the appropriate units. Of course, into the correct equation which we've already established are correct equation. Is this equation up above, Um, and then all we need to do is algebraic Lee saw for the missing variable, which, of course, is going to be Delta G transport. And so we can go ahead and start to fill in our equation over here. So what we'll have is Delta G transport is going to be equal to the If we go back up to our equation, you can see it's gonna be our times t times the natural log of the concentration on the final side over the concentration on the initial side. So are here recall or actually you can see up above has a value of 8.315 jewels per mole times Calvin. So we can fill that in down below 8. jewels per mole times Kelvin. So that's our gas. Constant are times the temperature in Calvin, which again is 310.15 Calvin. So we can put that in here. 310.15 Calvin times the natural log of the concentration on the final side of the membrane. Again, concentration on the final side over the concentration on the initial side. And again, the concentration on the final side over here is 1.0 times to the negative. Third Mueller. So 1.0 times 10 to the negative third Mueller over the concentration on the initial side, which is 1.0 times 10 to the negative seventh. Mueller. Yeah, And so, um, this here all represents the chemical Grady int portion of our equation. So now we can go ahead and add the electrical Grady int portion of our equation. So now we can go ahead and add, so plus the electrical portion. And of course, we know that it's going to be Z, which recall is the net charge of the defusing molecule. And we've already determined that the net charges plus two here so, uh, down below as e. We can go ahead and say that the net charge is going to be positive, too. And this is going to be multiplied by F. Which recall is Faraday's constant and recall fair days. Constant has a value of 96, jewels per volt times mole and the fair days constant. Whether or not you need to memorize it is going to be something that you should ask your professor. Um, some professors may expect you to memorize this value. Other professors will provide this value for you. And so we have fair days constant in there now. And so now we need to multiply that by the trans membrane potential Delta sigh, ensuring that it has the appropriate units and so are Delta Sai. We determine is gonna have a positive value and so we can plug in our delta sigh as a positive value with units of volts 0.5 volts. So time 0. Volz here for our trans membrane potential and So this portion right here represents the electrical Grady int. Whereas this portion over here represents the chemical ingredient from up above in this equation chemical radiant and electrical radiant. So now is the easiest part of all, which is just a plug all of this into our calculators. And so when we do that, what we get is Delta G Transport is going to be equal to if you take 8.315 and multiply it by 310.15 which will get is an answer of about 2578. And the units are gonna be jewels per mole since the units of Calvin are going to cancel out. And so that was just multiplying these two right here. And then, Of course, if you take the natural log of if you take the natural log of 1.0 times 10 to the negative third, Mueller divided by 1.0 times 10 to the negative seventh Mueller and type that into your calculator, which will get is an answer of about 9. And then, of course, we're going to be adding this to this number over here. So if you take your calculator and do two times 96,485 times 0.5 what you'll get is an answer of 96,004. I'm sorry, 9648 0. jewels per mole. And again, that's because the volts air going to cancel out. And we're just left with units of jewels Permal. And so this is what this is here, our electrical Grady in what's left of it. And this is what's left of our chemical ingredient. So now we can condense this further by multiplying 2578.897 times 9. and then add on 96th out or I'm sorry, 9648.5. And so when you do that, you get the answer of Delta G. Transport is equal to a value of 33, 0.917 and the Onley units that are left are jewels per mole. So that's the units on this Jules per mole. And so really, this is the answer. The Delta G transport is equal to 33,400 0.917 jewels per mole. Now, if we wanted to convert the units to kill a Jules per mole, then all we need to do is multiply this and we know that there are 1000 jewels and one killer Jewell. And so this allows the jewels to cancel out. And we're just left with killing joules per mole. And so if you take 33,400 0.917 and divided by 1000, which will get is an answer of about 33.4, um, rounded off. It's about 33.4 and the only units that are left are killing jewels promote, so we can add that end killer jewels per mole. And so this would be our answer in units of killing joules per mole, 33.4 killed, jewels promote. And so this here is the answer. This is the Delta G transport, the energy costs of pumping calcium under these particular conditions right here. And so that concludes this example problem and moving forward in our course, we're gonna be able to get some practice applying each of these four steps to solve problems that are similar to this one. So again, that concludes this example problem. And I'll see you guys in our next video.

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Problem

ProblemCalculate the free energy change (Î”G _{transport}) for the movement of Na ^{+} into a cell when its concentration outside is 150 mM and its cytosolic concentration is 10 mM. Assume that T = 20Â°C and Î”Î¨ = â€“50 mV (inside negative).

A

â€“1.7 KJ/mol.

B

â€“11.4 KJ/mol.

C

â€“11,600 KJ/mol.

D

11.4 KJ/mol.

E

14.3 KJ/mol.

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Problem

ProblemCalculate the Î”G_{transport} required to move 1 mole of Na ^{+} ions from inside the cell ([Na^{+}] inside = 5 mM) to the outside of the cell ([Na ^{+}] outside = 150 mM) when Î”Î¨ = â€“70 mV (inside negative) & the temperature is 37Â°C.

A

â€“15.5 KJ/mol.

B

2.0 KJ/mol

C

15.5 KJ/mol.

D

â€“2.0 KJ/mol.

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Problem

ProblemCalculate the Î”G_{transport} when Ca^{2+} ions move from the endoplasmic reticulum ([Ca^{2+}] = 1 mM) to the cytoplasm ([Ca^{2+}] = 0.1 Î¼M). Assume that Î”Î¨ = 0 and T = 25Â°C.

A

23 KJ/mol.

B

â€“23 KJ/mol.

C

â€“17 KJ/mol.

D

17 KJ/mol.