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Biochemistry

Learn the toughest concepts covered in Biochemistry with step-by-step video tutorials and practice problems by world-class tutors

7. Enzyme Inhibition and Regulation

Uncompetitive Inhibition

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Uncompetitive Inhibition

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in this video, we're going to talk about our second type of reversible inhibition, which is uncompetitive inhibition. And so, of course, uncompetitive inhibition is going to be caused by uncompetitive enzyme inhibitors. And so, unlike competitive enzyme inhibitors, a NCCAM Pettitte of enzyme inhibitors will Onley bind to the enzyme substrate complex. And they do not buy into the free enzyme like again competitive inhibitors dio. And so, of course, when the uncompetitive inhibitor binds to the enzyme substrate complex, it's going to form the S I complex, and we already know that all enzyme inhibitors, regardless of what type they are, they're all going to decrease the initial reaction Velocity V not of the enzyme catalyzed reactions. So no surprise here. Now, what's important to know, as is already anticipated, uncompetitive in inhibitors do not compete, so there is no competition when it comes to uncompetitive enzyme inhibitors. And so what this means is that the uncompetitive inhibitor does not compete with the substrate for the same binding site and instead, uncompetitive inhibitors. Binding site is on Lee, created after the substrate binds to the free enzyme to form the enzyme substrate complex, and we'll be able to see this a little bit better down below in our image Now, also note that the binding of an uncompetitive inhibitor to the enzyme substrate complex is going to prevent the conversion of the substrate into the product. And so that is the nature of inhibitors as a whole. When they're bound to the enzyme, they're going to somehow prevent the reaction from taking place. And so if we take a look at our example image down below of uncompetitive inhibition notice on the left hand side, we have an image that we have seen before and our previous lesson videos. And so you can see that we have the same exact enzyme catalyzed reaction that we've seen in our previous lesson videos and noticed that the uncompetitive inhibitor here is not affecting the free ends. I'm like the competitive inhibitor did in our previous lesson video. Instead, the uncompetitive inhibitor is Onley binding to the enzyme substrate complex toe form, the S I complex. And of course, whenever the inhibitor is bound to the enzyme, that's going to prevent the reaction. So there will be no reaction when the inhibitors bound to the enzyme. And so over here on the right hand side. Essentially, what we have is the same exact image of the left hand side, except just a different visual representation, and so notice that the free enzyme over here is in red. The active site is this circular position here, and notice that the uncompetitive inhibitor here is not interacting with the free enzyme specifically because there's no binding site for the uncompetitive inhibitor on the free enzyme, and so notice that this allows the substrate to bind to the free enzyme Normally toe form. The enzyme substrate complex, however, notice that upon the formation of the enzyme substrate complex that induces the binding site for the uncompetitive inhibitor to form. And so now the uncompetitive inhibitor can bind to the enzyme substrate complex toe form, the E S I complex. And, of course, whenever the inhibitor is again bound to the enzyme, that's going to prevent the reaction from taking place. However, what's really important here again is this idea that with on competitive inhibitors, there's absolutely no competition between the uncompetitive inhibitor and the substrate. And so it's important to emphasize here that the uncompetitive inhibitor does not compete with the substrate and the reason that it doesn't compete with the substrate is because the substrate is actually already bound to the enzyme. When, um, the enzyme substrate, complex forms and again uncompetitive inhibitors will Onley bind to the enzyme substrate complex, which is again very, very different from competitive inhibitors. And so, in our next lesson, video will be ableto revisit our analogy, but change it up a little bit for uncompetitive inhibitors, and we'll also give you guys a little memory tool to help you guys memorize the effects that uncompetitive, uh, inhibitors have on enzyme, So I'll see you guys in those videos.
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Uncompetitive Inhibition

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in this video, we're going to talk about the effects that uncompetitive enzyme inhibitors have on enzymes and so recall from our previous lesson videos, we mention the fact that uncompetitive enzyme inhibitors will actually make the apparent K M oven enzyme appeared to get better. However, in the process, it makes the apparent V max of the enzyme worse. And so even though the apparent KM seems to get better in the presence of an uncompetitive inhibitor, uncompetitive inhibitors are still going to inhibit the enzyme catalyzed reaction because the better apparent K M is accompanied by the V Max getting worse, and so recall that a better apparent K M just means that the enzyme is going to appear to have a stronger affinity for the substrate. But of course, a stronger affinity, we know corresponds with a decreased K M. And of course, a worse apparent V Max is going to correspond with a decrease V max as well. And so, essentially, what we're saying here is that uncompetitive enzyme inhibitors will proportionally decrease both the apparent K M as well as the apparent V Max oven. Enza now also recall from our previous lesson videos that uncompetitive enzyme inhibitors do not compete, and so there's no competition between the substrate and the uncompetitive inhibitor. Now the real question is exactly how and why is it that uncompetitive enzyme inhibitors have these particular effects on an enzyme? And so, in order to understand that, let's take a look down below at our image and notice that we're using the same exact analogy that we used in our competitive inhibition lesson, where Scooby Doo here is binding to a Scooby snack bone and converting it into a Dukie over here on the floor and notice that instead of using our competitive inhibitor soccer ball here instead, what we're using is one of the most uncompetitive looking people that we know shaggy to represent the uncompetitive inhibitor. And that's because Shaggy is always so scared to compete, and he's got these skinny arms and skinny legs. And really, there's no way that Shaggy is gonna be able to compete with the football linebacker on a football field. And so that reminds us that Shaggy, the uncompetitive inhibitor here, is not going to compete with anything, including the substrate over here, and so also noticed that Shaggy are uncompetitive inhibitor has a very particular mood where he's upset with Scooby Doo for pooping on his living room couch and so noticed that shaggy here saying, You've had enough bone scoop and he's so upset with Scooby Doo that really he Onley cares about Scooby Doo when he's bound to that bone, and he's got that bone in his mouth. And really, when Shaggy is so upset with Scooby Doo in this mood, he's not going to care about Scooby Doo when he doesn't have that bone in his mouth. And again, all Shaggy wants to do is prevent Scooby Doo from pooping on his couch. And so, uh, this is exactly why we have this symbol right here, Alfa Prime. And so recall that Alfa Prime is the degree of inhibition on the enzyme substrate complex. And so that reminds us that uncompetitive enzyme inhibitors will Onley bind to the enzyme substrate complex, and they do not bind to the free enzyme. And so, of course, whenever Shaggy catches Scooby Doo with that bone in his mouth, he's gonna grab his dog leash and attach it to his color real quick, and they're going to form an enzyme substrate inhibitor complex like we see here and s I complex. And, of course, Shaggy is going to make sure that Scooby Doo does not eat that bone because if he does, it's gonna end up with a big mess. He's got to clean up. And so now that we understand our image a little bit better, let's go on back up to our text to revisit again exactly how and why Uncompetitive enzyme inhibitors have these particular effects on an enzyme, and we're going to start off with how uncompetitive inhibitors decrease the apparent K M. And so the reason that uncompetitive inhibitors decrease the apparent K M is because uncompetitive enzyme inhibitors also lower or decrease the concentration of enzyme substrate complex. And so, if we take a look at our image down below, notice that the uncompetitive inhibitor here is going to bind to the enzyme substrate complex to form the enzyme substrate inhibitor complex. But in the process, it's going to decrease the concentration of enzyme substrate complex. And of course, what that means is that this equilibrium over here is going to respond to the decreased enzyme substrate complex by shifting to the right. And so that's exactly what we're seeing up above. And so notice that by lush at lease principle, a lowered or decrease concentration of enzyme substrate complex causes the K one reaction to shift to the right and so notice that the number one up above corresponds with the number one down below. So we have low shot liaise principle here and notice that we have this pink arrow shooting to the right, reminding us that this equilibrium is going to shift to the right when uncompetitive enzyme inhibitors decrease the concentration of enzyme substrate complex. And so, of course, if this equilibrium shifts to the right, the enzyme substrate complex is going to form. And that's going to make it appears if the enzyme has a stronger affinity for the substrate, and so a strengthened enzyme substrate affinity, we know corresponds with a decreased K M. And so that's exactly how and why. Uncompetitive enzyme inhibitors decrease the apparent K M. And again, the decreased apparent cam appears to make the enzyme perform better because the enzyme is going to appear to have a stronger affinity for the substrate. But just because the apparent km appears to get better does not mean that uncompetitive inhibitors make the enzyme better. And that's because this decreased apparent km that appears to make the enzyme better is also accompanied by the apparent V max getting significantly worse. And so because the apparently max get significantly worse, uncompetitive enzyme inhibitors are still going to have an overall inhibitory effect on the enzyme catalyzed reaction. And so now we're going to move on to the fact that uncompetitive enzyme inhibitors also decrease the apparent V max. And so this has to do with the fact that uncompetitive enzyme inhibitors do not compete. And so again, we know that Shaggy down here are uncompetitive inhibitor. He is not very competitive. It also he's not gonna compete even with the substrate, which means that even if we increase the substrate here and added a huge pile of bones, it doesn't matter how many bones there are. Uh, Shaggy is not going to compete with those bones, and that means that the bones cannot out compete, shaggy. And that's exactly what we're saying up above in number two, that since the substrate can't out compete the uncompetitive inhibitor, this means that the effects of the uncompetitive inhibitor are not going to be reversed even when we increase the substrate concentration to saturating levels. Even if we were to make a huge pile of bones here, the uncompetitive enzyme inhibitor effects are not going to be reversed like what we saw with competitive inhibition. And so this means that the decreased initial reaction velocity of the uncompetitive inhibitor is going to translate to a decreased the max. And so notice here, down below. If we would apply this even if we had a huge pile of bones here, it doesn't matter how many bones there are. Shaggy. If he spots Scooby with a bone in his mouth, he's going to jump right on Scooby. And make sure that Scooby does not eat that bone preventing Scooby from making the maximum amount of poops regardless of how many bones there are. And that's why if the maximum amount of poops they're gonna be prevented, the V max is going to be decreased. And of course, uh, if uncompetitive inhibitors decrease the V max, that means that the catalytic constant or the K cat or the turnover number is also going to be decreased. And we know that the K cat represents the maximum catalytic efficiency under saturating substrate concentration. So even under saturating bone concentrations, uh, Scooby Doo is not gonna be able to create the maximum amount of poop poops per second. A zloty as shaggy here is ready to make sure that, uh, he's not going to poop on his couch. And so notice down below. Here, what we're saying is that the decreased catalytic constant or K cat or turnover number eyes going to be equal to the V max also being decreased on. Of course, the total enzyme concentration is not going to be affected by the uncompetitive inhibitor. And so the decreased the max is, of course, going to lead to the decrease K cat. And so this is a lot of effects to remember about uncompetitive enzyme inhibitors. And so how could you guys possibly go about memorizing all of these different effects that uncompetitive and, uh, inhibitors have? And so that's exactly why we have this box down below. And so when it comes to uncompetitive enzyme inhibitors, we really want to focus on this you here, which is quite unique for uncompetitive inhibition. And so this, you hear reminds me of a U turn And that just tells me that everything's going to be going down, including the K M and the V Max are both going to be decreased, as we said up above. And so notice that with uncompetitive and ambition, even though it says, uh, the K M is gonna be increased, we can't forget the you means a u turn. And so even though it says Cam is gonna be increased, the K M is actually going to be decreased when we consider the U turn. And, of course, with uncompetitive inhibition, there's absolutely no competition between the uncompetitive inhibitor and the substrate, and that means that the substrate cannot compete. And, of course, if the substrate can't compete, then it can't keep the same V max. And that means that the V Max is going to be decreased. And so again, all you need to remember with uncompetitive inhibition is that it represents a U turn and everything is going to be going down. So the K M being decreased and so is the V Max. And so this here concludes our introduction to the effects of uncompetitive enzyme inhibitors, and we'll be able to get some practice later in our course. But in our next lesson video, we'll talk about how uncompetitive enzyme inhibitors affect the McHale is meant and plot, so I'll see you guys there.
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Uncompetitive Inhibition

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So now that we know that uncompetitive enzyme inhibitors will proportionally decrease both the apparent K. M and the apparent V max oven enzyme in this video, we're gonna talk about how uncompetitive inhibitors affect met callous meant in plots. And so recall from our previous lesson videos that all enzyme inhibitors, regardless of what type they are, including uncompetitive inhibitors, they will all decrease the initial reaction velocity or the V, not of an enzyme catalyzed reaction. And so notice taking a look at our image down below. On the left hand side, we have, um, a callous, meant and plot showing two different curves. This black curve represents the enzyme catalyzed reaction in the absence of inhibitor. And this blue curve here represents the enzyme catalyzed reaction in the presence of uncompetitive inhibitor and notice that in the presence of uncompetitive inhibitor, the initial reaction rate is being decreased, as we indicated up above. And so also recall from our previous lesson videos that uncompetitive enzyme inhibitors will Onley bind to the enzyme substrate complex, which means that Alfa Prime is going to measure its degree of inhibition on the enzyme substrate complex. And so again, Alfa Prime the degree of innovation on the enzyme substrate complex. Oven uncompetitive inhibitor, as we already know, is going to proportionally decrease both the apparent km and the apparent V max oven enzyme. And so notice that the apparent K M is defined as came over Alfa Prime and the apparent V. Max is also defined as V Max over Alfa Prime. And so the Alfa Prime here is on Lee going to decrease the K M and the V Max. And so if we want to get the McHale is meant an equation in the presence of an uncompetitive inhibitor. All we need to dio is substitute the V Max with the apparent V Max, as we did down below and substitute the K m with the apparent cam, as we did down below as well. And so notice again Here, looking at this, McHale is meant and plot that clearly in the presence of inhibitor, the V Max is being decreased. Unlike with competitive inhibition so recall with competitive inhibition the blue curve went all the way up Eventually thio the V max so that it was unchanged. But here with uncompetitive inhibition again, the V max is being decreased. And that's because no matter how much we increase the substrate concentration, essentially no matter how many bones we add, of course, Shaggy is always going to be there to make sure that Scooby Doo does not convert the maximum amount of bones into poop. And so, Scooby Eyes only gonna be able to convert ah lowered amount of bones into poop indicated by a lowered V max, and also noticed that along with the V Max being decreased, uh, the K M is also being decrease of the apparent cam is also being decreased with respect Thio the K m in the absence of inhibitor. And so over here on the right, this McHale is mentioned plot is just trying to show us what happens if we increase the concentration of uncompetitive inhibitor even further and notice we have these three different curves. We have the black curve that again represents the enzyme catalyzed reaction in the absence of any inhibitor. And then we have this purple curve here that represents the enzyme catalyzed reaction in the presence of just one Moeller concentration of uncompetitive inhibitor, and noticed that the V Max is again being decreased. But so is the K M. And so the K M is decrease. The V max has decreased. But we have this green curve here that represents the enzyme catalyzed reaction in the presence of plus two Moeller concentration of uncompetitive inhibitor. And so notice that increasing the concentration of uncompetitive inhibitor even further will decrease the V Max even further. And it will decrease the K M even further as well. So, again, the K M is still going to be decreased here, along with the decreases in the V max. And so this year concludes our lesson on how uncompetitive inhibition effects McHale is mentioned plots. And in our next lesson video, we'll talk about how uncompetitive inhibition effects the line Weaver bird plot. So I'll see you guys in that video.
4
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Uncompetitive Inhibition

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So in this video, we're going to talk about how uncompetitive inhibitors affect line. Weaver Berg plots and so recall that uncompetitive inhibitors proportionally decrease both the apparent K M as well as the apparent V Max oven enzyme. And so, even though both the apparent cam and the apparent V Max are decreased, it turns out that the slope of the line on a line we were burke plot, which is the ratio of the K M over the V Max, is actually not changed. And so here, what we can say is that the slope is not going to change even in the presence of an uncompetitive inhibitor. And so this green background here behind proportionately decreased shows that, um, it corresponds with the slope of the line not changing. And of course, since the line with Robert Plot is also known as a double reciprocal plots, since it plots the reciprocal of the initial reaction velocity on the Y axis and the reciprocal of the substrate concentration on the X axis. Uh, what this means is that both the why intercept, which is the reciprocal of the V Max as well as the absolute value of the X intercept, which is the negative reciprocal of the K m. Both of these, because they are reciprocal. Is there going to be proportionally increased? And again, this has to do with the nature of the fact that the reciprocal Zehr plotted on these axes. And so if we take a look at our image down below over here on the left hand side, you can see that in the presence of uncompetitive inhibitor. This is how the line we were, Burke equation changes. And so notice that the slope of our line, which is again the ratio of the cam over the V Max eyes actually not being affected by the degree of inhibition factors. So there's no Alfa prime affecting um, this, uh, slope here and so that this means that the slope is not being changed. And again, this has to do with the fact that both the apparent K. M and the apparent V Max are proportionally being decreased. And so if we take a look at the line, we were Burke equation over here on the right notice, we have two different lines. We have this black line here that represents the enzyme catalyzed reaction in the absence of inhibitor indicated by negative concentration of I. And then we have this purple line here that represents the enzyme catalyzed reactions in the presence of uncompetitive inhibitor one Moeller concentration and so notice that even though the, uh, why intercept is being increased and the X intercept, the absolute value of the X intercept is also being increased. Um, the, uh, these values here are getting further and further away from our zero marker, which we know acts as the infinity marker for the V max and the K M themselves. And so you can see that the V max, um here is getting further away from the infinity marker, meaning it's being decreased. And three k. M here is also getting further away from the infinity market, which means it's being decreased as well, and they're being decreased proportionally, which is what allows the slope here of our line to say exactly the same. And so notice that the example here is asking us to draw the representative line for the enzyme activity if the concentration of uncompetitive inhibitor was doubled. And so here we have plus one concentration of I. So if we were to double this, it would be plus two concentration of I. And so, in the presence of even mawr uncompetitive inhibitor, the the Max eyes going to decrease even further. And so here we could say that the Y intercept is gonna get further away from the zero infinity marker, and the same goes for the K M is going to get further away, and it's gonna be proportionally decreased so that the slope stays exactly the same. And so here noticed that all of these slopes here are exactly the same. And that is a unique identifying feature that uncompetitive and inhibitors are present. And so what you'll notice is that when you think again about uncompetitive inhibitors, you think about that unique you and the unique you you think about the U turn. And so we know that both the K M and the V Max they're gonna be decreased because of this U turn. And so with this you turn. It can also help you remember that there's going to be parallel lines as well. And so you can see here that it here's our line for the, uh, in the absence of inhibitor and notice that it does a U turn here on it pretty much keeps the same line. Exactly. Parallel. And so you can see that this U turn has these two parallel lines. And so hopefully these little hints that helped me remember uncompetitive inhibitor effects online. We were Burke plots will also help you guys remember it a little bit as well. And so, uh, this year concludes our lesson, and we'll be able to get some practice as we move forward in our core. So I'll see you guys in our next video.
5
Problem

True or false: Increasing [S] in the presence of an uncompetitive inhibitor will lower the inhibition constant (KI).

6
Problem

In the presence of an uncompetitive inhibitor that binds __________ the substrate, the apparent Vmax ____________ and the apparent Km ____________ with respect to the Vmax and Km of the uninhibited enzyme.

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Problem

What is the effect of an uncompetitive inhibitor on the equilibrium between free enzyme & the ES-complex?

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Problem

Which of the following statements is true about uncompetitive inhibitors?

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