Practice - Membrane Transport 2 - Online Tutor, Practice Problems & Exam Prep

Facilitated diffusion across biological membranes relies on solute concentration differences, while primary active transport uses ATP. Ion channels, like potassium channels, exhibit substrate specificity due to their selectivity filters. Secondary active transport utilizes ion gradients established by primary active transport. Understanding enzyme kinetics through Lineweaver-Burk plots aids in calculating V_max and K_T. Inhibition can occur between transporters, affecting the transport efficiency of amino acids. Key equations include the slope-intercept form for determining transport rates and constants.

Practice - Membrane Transport 2

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.

Here’s what students ask on this topic:

What is the difference between facilitated diffusion and primary active transport?

Facilitated diffusion and primary active transport are both mechanisms for moving substances across cell membranes, but they differ significantly. Facilitated diffusion relies on the concentration gradient of the solute and does not require energy input. It uses specific transport proteins to move substances from an area of higher concentration to an area of lower concentration. In contrast, primary active transport requires energy in the form of ATP to move substances against their concentration gradient, from an area of lower concentration to an area of higher concentration. This process involves transport proteins known as pumps.

How do potassium channels achieve specificity for potassium over sodium?

Potassium channels achieve specificity for potassium (K⁺) over sodium (Na⁺) primarily through the selectivity filter of the protein. The selectivity filter is a narrow region within the channel that interacts differently with K⁺ and Na⁺. The filter is designed to stabilize the dehydrated K⁺ ion through specific interactions, such as coordination with carbonyl oxygen atoms. Sodium ions, being smaller, do not fit as well into the selectivity filter and are not stabilized in the same way, thus preventing their passage through the channel.

What is secondary active transport and how does it differ from primary active transport?

Secondary active transport uses the energy stored in ion gradients, which are typically established by primary active transport, to move substances across the membrane. Unlike primary active transport, which directly uses ATP to pump ions against their concentration gradient, secondary active transport relies on the energy from the movement of one ion down its gradient to drive the transport of another substance against its gradient. This process often involves symporters or antiporters, which move ions and other molecules in the same or opposite directions, respectively.

How can Lineweaver-Burk plots be used to determine V_max and K_T in membrane transport studies?

Lineweaver-Burk plots are double reciprocal plots used to linearize the hyperbolic relationship between substrate concentration and reaction rate in enzyme kinetics. In membrane transport studies, these plots can be adapted to determine V_max (maximum transport rate) and K_T (transport constant). The y-intercept of the Lineweaver-Burk plot is equal to 1/V_max, and the x-intercept is equal to -1/K_T. By plotting 1/transport rate (1/v) against 1/substrate concentration (1/[S]), one can determine these constants from the intercepts and slope of the resulting straight line.

What role do ion gradients play in secondary active transport?

Ion gradients play a crucial role in secondary active transport by providing the energy needed to move substances across the membrane. These gradients are typically established by primary active transport, which uses ATP to pump ions like Na⁺ or H⁺ out of the cell, creating a high concentration outside and a low concentration inside. Secondary active transporters, such as symporters and antiporters, use the energy released when ions move back down their concentration gradient to transport other molecules against their gradient. This coupling allows cells to efficiently import or export essential substances without directly using ATP.