Facilitated diffusion through biological membranes is driven by a difference in solute concentrations. If it were driven by ATP, what you'd have is primary active transport. And it is most certainly specific with regard to substrate. Now, the specificity of a potassium channel for potassium over sodium is mainly the result of the differential interaction of the selectivity filter of the protein. Considering this is an ion channel, it's unlikely that it would be hydrophobic, that it would have a lot of phospholipids or cholesterol for that matter. Because it's going to want to be hydrophilic so it can interact with the ions better. Now, let's do a bit of a throwback problem here to our friend, the enzyme kinetics problem, except now we're masquerading it as a transport problem and looking for \(K_T\) instead of \(K_M\). So, everything has changed. Right? But still, it's good to practice this because it could come up. So, you're going to want to solve this using a Lineweaver-Burk type plot, but we don't actually need to graph it. I'm just drawing a representation of it here to talk about a couple of things in case you forgot. So, this point right here, the y-intercept, that is equal to \( \frac{1}{{V_{\text{max}}}} \), right? And this point right here, the x-intercept is equal to \( -\frac{1}{{K_M}} \) or in this case \( K_T \). So, we don't actually need to graph this, but we do need to use some of the data and a little bit of math. And we're going to have to get the reciprocal for some of these values. So, this is going to become 5, this is going to remain 1, this will be 17.5, and this will be 13. And we're going to need to use our \( y = mx + b \) equation for a line. Alright. So first things first, we need to find \( m \), the slope, and we're going to do that by determining the change in rise over change in run. So that is going to be \( \frac{17.5 - 13}{5 - 1} \) which is equal to \( \frac{4.5}{4} \) or 1.125. Now, to find \( b \), we need to plug in some values to our equation. So let's do \( 13 = 1.125 \times 1 + b \). So \( b \) is equal to \( 13 - 1.125 \) which is equal to 11.875 and remember that equals \( \frac{1}{V_{\text{max}}} \). It's not exactly what we want. What we want is \( V_{\text{max}} \). So \( V_{\text{max}} \) is going to be \( \frac{1}{11.875} \) which turns out is equal to 0.084 micromoles per second. Now to find \( K_T \), we need to substitute 0 in for \( y \), so we're going to say \( 0 = 1.125x + 11.875 \) which is \( b \). And we need to rearrange to get \( x \) by itself, so we get \( \frac{-11.875}{1.125} = x \). And let me just scroll a little bit here so we can see this better. My head doesn't cut it off. Alright. So that is equal to \( x \) and that comes out to about -10.6. Remember that's equal to \( -\frac{1}{{K_T}} \). So \( K_T \) is going to be equal to \( -\frac{1}{{-10.6}} \) which comes out to 0.094 millimolar. Alright, moving on to question 40. The type of membrane transport that uses ion gradients as the energy source is secondary active transport. Primary active transport uses ATP. Secondary active transport uses these gradients often established due to primary active transport. These gradients are usually there because pumps are pumping things actively. Passive transport relies on concentration gradients and includes simple diffusion and facilitated diffusion. Alright. Last problem. And this one is kind of an interesting one, so bear with me on the explanation but here is how to think about it. If you, let's say, have a membrane with 3 transporters in it, right? And one transporter transports all amino acids including lysine and arginine. And then you have another transporter that's specific to arginine and another one that's specific to lysine. And it turns out that this arginine transporter is inhibited by lysine. And this lysine transporter is inhibited by arginine. And by the way, this little symbol that I'm drawing right here, it's kind of like a perpendicular symbol, that's what we use in neuroscience to mean inhibition. So I'm just saying that those are inhibiting those transporters. So anyways, if the arginine transporter is inhibited by lysine, well then only half the lysine is going to get through because it's going to have to go through this other transporter. Likewise, if only, if the arginine is inhibiting this lysine transporter, only half of the lysine is going to get through because it's going to have to just use that other one. So that's why the answer to this question is 3. That's all I have, for this video. If you have any questions, please leave comments on the videos, and I will answer them and good luck on your exam.
Review 1: Nucleic Acids, Lipids, & Membranes
Practice - Membrane Transport 2