 Okay. We're going to give enzymes another try today. So there's a mistake in the way I calculated the how am I doing scores. And the mistake is that if you missed a quiz, not if you got a zero, but if you missed a quiz, then it was calculating your how am I doing score incorrectly. So a few of you pointed this out to me and I fixed it. And so I posted a new how am I doing score on Friday. It's separate from the earlier one. And so if you missed a quiz, your how am I doing score should be higher than it was on Thursday. All right. Just check that if you would. Thank goodness a few of you went to the trouble to check your how am I doing score and found this error. Okay. Quiz seven scores I think are already posted. All right. So that should be all of the quizzes for the quarter. You should see all the scores. The key is posted as well on the results page. Quiz six scores have been updated. We made an error in the way we graded that. Excuse me. Completely my fault. There was an error in the notation that we used on the quiz. And so G and Mark went through and regraded all of these and reposted them and so the quiz scores that are up there for quiz six are updated. If you want to double check your quiz score, that would be a good thing to do. Now, the electronic evaluations got turned on last week. And a few of you have worked them out for me. But so far just 16% of you, I need, and there's no way I can force you to do this, but I want to ask you if you would please to take the course evaluation. Let me explain the situation as follows. I teach general chemistry, analytical chemistry, physical chemistry. When I teach general chemistry and I get evaluations from the students, they're almost useless. And the reason is it's not really their fault. I mean these students right here, they don't know if the drill instructor is a good guy, a bad guy, or an intermediate guy. They just know they can't hold their feet up for 30 seconds. All right? They have no experience with drill instructors. And so when I get feedback from them, you know, I can feel like I've done the best job in the world of teaching this class and I will get all of this extraneous feedback, you know, your pants are something wrong with your pants or, you guys are like these guys right here, all right? You're grizzled veterans. Most of you are graduating in a week and you've seen everything. You've seen bad teaching, you've seen good teaching, you've seen intermediate teaching, you're calibrated. You know what exactly, what issues exist with the class. All right? I know there's issues with this class and it would really help me out if you went through and did these evaluations and not only just scoring them because I know that's the easiest thing because that's just radio buttons but if you actually wrote some comments to help us make the course better. I read every one of these comments. I get a printout of a list of all the comments and I read those all, all right? I don't get to see them for about two weeks because they don't allow me to see what your comments are before I issue the final grades but after I issue the final grades a week after that they let me see what your comments are, okay? So please if you would, I know it takes time, probably takes about 10 minutes, 15 minutes to do this. I'd appreciate it. All right, so we're going to talk about enzyme kinetics. Turns out all of this stuff is in chapter 21 of your book, all right, in a chapter that's called catalysis. So what we want to do and once again all quarter we've been doing this, we've been cherry picking certain topics from stat mech, thermo, kinetics that are the most important topics I think. And so this is one of those topics. The basic idea is we want to understand how enzymes catalyze reactions. Now we're not really learning how enzymes do this, we're studying the phenomenology of enzyme catalysis. In other words we're looking at the rates of enzyme reactions, we're trying to understand how we can break the mechanism of enzyme substrate catalysis down and turn it into a modular thing that we can assign constants to and make measurements on and compare enzymes against one another and so forth. We're trying to really understand the phenomenology of enzyme substrate catalysis. So we've got an enzyme, we've got a substrate. When the substrate docks in the enzyme what this schematic diagram here is trying to depict is that there is a recognition event that has to occur. In other words the enzyme is not going to catalyze this reaction whatever it is for any substrate. There has to be a recognition of the substrate by the enzyme at the active site in order for the substrate to dock and once this docking occurs then the enzymatic reaction can proceed. And in this case it looks like some sort of bond breaking reaction occurs. Okay? So this enzyme substrate complex is meant to represent this entity here and then products are produced and released from the enzyme at that point. Once this reaction occurs the affinity of the products for the enzyme is lower than the affinity of the substrate for the enzyme. If that was not true the products would just stay bound to the enzyme and it would be game over. Okay? So the enzyme has to release the products once they're formed in the active site. If that didn't happen the enzyme would be pretty useless wouldn't it? So that's what this event is depicting here. Okay. So we want to work out some equations that are specific to enzyme catalysis that help us to understand these reactions. The batteries in my laser pointer are dying so I'm going to use it sparingly but hopefully it'll get us through this lecture. Here's the mechanism I showed on the previous slide. Here's the rate of the reaction. All right? I'm just depicting here the rate at which P is formed. All right? I think you can see that it's a unimolecular reaction from the enzyme substrate complex with a rate constant K2. We're going to apply the steady state approximation again. All right? It's another example of that. And to do that we set the time rate of change of the intermediate ES equal to zero. And so I think you can see there's a rate at which ES builds up and two ways at which ES is consumed. And so that equation we could have written down earlier because it's just the steady state approximation and now we have to say some things that are specific to enzyme reactions. We're going to make some substitutions into this equation and one of the substitutions we want to make is for the enzyme concentration because we don't know a priori what it is as a function of time, all right? Presumably the enzyme is going to get bound from the enzyme substrate complex and the free enzyme concentration is going to go down and the enzyme substrate concentration is going to go up, presumably that's what has to happen. And so we don't know what that is, what E is but we do know how much total enzyme we've got, right? At least if we're studying this reaction in the laboratory we added a certain amount of enzyme at the beginning of the experiment to study the enzyme kinetics that we're trying to study in the lab, right? Now in a natural system, you know if there are cells around and we've got some extract from a liver and some enzymatic chemistry is going on, we don't know anything, all right? We can't study enzyme kinetics under those conditions in general, right? To study enzyme kinetics we've got to take the enzyme control, the pH, put some substrate in contact with it and measure the reaction rate as a function of time somehow, we've got to do that. Okay, so presumably we know this E0 at least if we're doing the experiment in the lab we know it. So the total enzyme can only exist in two forms, free enzyme and enzyme substrate complex and so now I can solve for the free enzyme in terms of the total concentration of enzyme and of course it's just equal to the total concentration minus the enzyme substrate complex. So now I can plug that into this expression, boom, here's my new steady-state approximation expression, all right? And when I distribute this K1 over these two terms I'm now going to get four terms instead of three, one, two, three, four terms and I can move the one generation term over to the left-hand side, put all the minus signs on the right side, all right? So all of that has got to be equal to that if this is equal to zero and now I just solve this for ES and that's easy to do because I've got ES, ES, ES, I just factor that out and this is the expression that I get. Oh, I want to remember that this ES concentration that I'm using now refers to the steady-state enzyme substrate concentration, all right? We're assuming that the steady-state approximation is correct. So this is the equation that I get for that and earlier we said that the rate at which P is formed is just equal to K2 times the enzyme substrate concentration. And so now if I just plug ES into here I get this equation right here, looks like a mess, okay? But I'm going to divide the numerator and the denominator by K1, so K1 is going to go away here and K1 is going to go away here and K1 is going to end up in the denominator here, all right? So this is the new expression for the rate of the reaction that I get, still subject to the steady-state approximation and now I'm going to roll all these constants up in column KM, K big M, all right? All of those guys together are going to be the McHale-List constant and this equation is the McHale-List-Menten equation. It is the most important equation in enzyme catalysis, okay? But it's moderately useless. In other words, we've derived this equation, it's very important, it contains all of the kinetic rate information for enzymatic reactions but we can't extract information from this equation very easily in the form that it's in. Ironically, it's the most important equation in enzyme kinetics but we can't really use it, all right? We need to do some more work. What do we need to do? Well, first of all, let's think about this equation. Let's see what it's telling us. Notice that in the denominator, there's an addition operation. What does that mean? There's going to be limiting cases, all right? If there's an addition operation in the expression for the rate, all right, there's going to be limiting cases but why? Because one half of this addition operation could be large compared to the other half or vice versa, okay? So we've got an addition operation in denominator. What are the corresponding limiting cases? Well, if K2 is big, remember K2 is part of KM, if K2 is big, how big? Big compared to K minus 1 and K2 over K1, if that's big compared to S, then this thing simplifies quite a bit. Notice if K2 is large, then K minus 1 can be neglected and S can be neglected. So I've just got K2 over K1 and K2 is going to cancel and so the rate of the reaction is going to simplify very substantially, right? The rate of the reaction is just going to be K1 times E0, the initial concentration of enzyme, times the substrate concentration. Yes. Did everyone hear that? Shouldn't it always be true that K2 is large if the steady state approximation is correct? The answer is yes, right? But we're going to abuse this equation on a routine basis, all right? So this limiting case is not always going to be observed. It turns out it is useful. And the reason is that we're going to apply it to the initial, in other words, we're going to measure a rate at the beginning of the enzyme substrate reaction using an initial concentration of substrate and an initial concentration of enzyme and under those conditions, this equation is going to work pretty well. Okay, so what does this mean? If K2 is big, all right, we've derived this simplified equation, but conceptually what does that mean? What does it mean if K2 is big? What it means is that the reaction doesn't even know ES exists, right? If K2 is big, as soon as ES is formed, boom, it reacts immediately, right? So its concentration is very low, which is good. That's in compliance with what the steady-state approximation is assuming, right? The enzyme substrate concentration is going to be quasi-constant because it's very low. This also means that E is approximately equal to E0 because ES is approximately zero, right? And so essentially what this means is that the formation of the ES in this first reaction here is the rate limiting step in this sequence of reactions, all right? The rate at which the ES is formed because as soon as the ES is formed, boom, it reacts like a shot, all right? Notice also that the reaction is first order and substrate, all right? In this limit of high K2, the reaction is first order and substrate. We'll come back to that. What if S is big, all right? If S is big, then I can neglect this whole guy in the parentheses here, all right? And I just end up with this expression right here and S is going to cancel and so the reaction rate is just, in that case, just given by this expression here. It doesn't even depend on S. Not only that, it doesn't even depend on time, all right? It only depends on the initial enzyme concentration, all right? In the limit of large S, large concentrations of substrate, the reaction rate is constant, all right? You don't see any change in the rea- So conceptually what we expect to see is that low substrate concentration, well, we'll get to that, okay? What does this mean? What's happening? It means in this limit when S is big, all of the enzymes tied up as enzyme substrate complex, all right? We've got very high concentration of S. We've driven this reaction forward because it's first order in S and we've basically saturated all the enzyme. All the enzymes got a substrate on it, right? There's so much substrate around that this reaction has been driven so far to the right based on Le Chatelier's principle that we've tied up all of the enzyme as enzyme substrate complex. The enzyme substrate complex concentration is approximately equal to the total enzyme concentration, okay? This means that the reaction of S is rate limiting. This second step is now rate limiting, okay? And that's what it looks like, all right? I'm using that rate constant for the total reaction rate. E0 is just equal to ES, all right? So in this limit, the reaction doesn't care about the substrate concentration. What if K minus 1 is big? Not very interesting, it happens. Reaction will be slow, not an interesting limiting case, but one that could happen, right? K minus 1 could be large compared to K2 and then K minus 1 over 1 could be large compared to S. But if that's true, the equilibrium lies far towards the substrate equilibrium lies way over here, all right? And it's a bad enzyme, right? It's not a very good enzyme. The enzyme's not recognizing the substrate. Maybe the substrate is not the normal substrate for that enzyme. The reaction rate's just slow, okay? The substrate doesn't want to dock with the enzyme to form the enzyme substrate complex. Okay, what if S is small? What happens at low concentrations of substrate? Check it out, all right? Here's the Michaelis-Menten equation, all right? If I make S small, it disappears from the denominator. I end up with this expression right here. The reaction is first order in S. Okay? So we said when S is big, the reaction rate becomes constant. When S is slow, the reaction rate is first order in S. Okay, so we know what this reaction's going to do as a function of S, right? It's going to be first order at low S. So this is a plot of the reaction rate versus the concentration of S. At small S, we see first order kinetics for S. In other words, the reaction rate goes up linearly as a function of the concentration of S. I think you can see this looks like a straight line down here, right? But as S gets larger, it starts to curve and at high concentrations of S, it becomes concentration independent, all right? And in that limit, there's a maximum reaction rate, Vmax, that's equal to K2 times the total concentration of enzyme that I started out with in this reaction, all right? So this is what any enzyme will do as a function of the substrate concentration, all right? That's what this equation predicts, all right? And it's also what's experimentally observed. Now, there's some other things that are indicated here that I'll explain to you. In the limit of large S, we obtain the maximum rate, Vmax. I already showed that to you, right? Here's the Vmax and we're not quite there yet. We would have to go to higher and higher and higher concentrations of substrate. But eventually, this reaction rate will asymptotically approach this dashed line which represents Vmax. That's the maximum reaction rate for the enzyme. That's a reaction rate that is characterized by the fact that the enzyme, every enzyme's got a substrate stuck on it. So at that point, the reaction can't go any faster. Okay. So K2 times theses of Vmax, yes. Okay. Now, there's lots of things in enzyme kinetics that are confusing, but here's one of them, all right? So let's get rid of one thing that's confusing here. Vmax over E0 is given by this expression right here, but we also call Vmax over E0 the turnover number. I think you could see Vmax over E0 is obviously equal to K2, all right? What are the units of K2 going to be? Seconds to the minus 1. Why? Because ES reacts to give products, right? So it's a unimolecular reaction. And so the rate constant is going to have units of 1 over seconds, okay? And so that's okay. So the turnover number is going to have units of per second. Reactions per second is what it represents. Okay? So Vmax over E0, I mean, I just, so if we work out the units, this just shows that. Vmax is molar per second, if you will. That's the reaction rate. The concentration of enzyme is molar. So the units of this quotient here are 1 over seconds, right? We're going to call that K catalysis or K2 or the turnover number. All three of those things are the same, right? If someone says the turnover number, they're just talking about K2. If someone says Kcat, they're just talking about K2. If someone says K2, they're just talking about the turnover number because they're all the same thing. They all have units of 1 over seconds. Isn't that confusing? Why have three names for the same thing? I don't know. Okay. Now, let's take the ratio between the reaction rate. Now I'm just going to call the reaction rate V. Remember, I was calling it DPDT? No difference. I'm going to just call it V, the velocity. Divide that by Vmax. Remember, Vmax is just K2 times Z0. If I divide these two things, obviously that's going to cancel. And so I'm just left with S over S plus Km. And if I take the reciprocal of that, I get this. And if I divide by Vmax and cancel these S's over here, I get this, which is the same as this. So all we've done is to take the Michaelis-Menten equation and do some more algebra on it to get an equation that is not called the Line-Weaver-Berke equation. It should be called the Line-Weaver-Berke equation. But it's not. But if you make a plot using this equation, it is called the Line-Weaver-Berke plot. How are you going to make a plot? Take 1 over S. That's going to be your vertical axis. No. That's supposed to be X. That's supposed to be, sorry. That's the vertical axis, horizontal axis rather. That's such, it's a typo. This should be Y. 1 over the velocity is the vertical axis. The slope is this, the intercept is that. I think I've used this slide for three years. I've never noticed that. All right. To be clear, we're talking about the initial substrate concentration and the initial velocity. All right. So you take your enzyme solution in your pH-controlled buffer. All right. You add some substrate and you measure, usually spectroscopically, how fast the products are formed as a function of time. And you extrapolate to zero time, all right. And you measure the reaction rate at time zero. Of course, you're measuring initially the reaction rate over a range of times, all right. But you extrapolate to zero to get the initial rate. We call that V zero, all right. And it applies, of course, to the initial substrate concentration. That's usually what we're plotting in a line weaver-berk plot. Here's what it looks like. All right. 1 over S, horizontal axis. 1 over V, vertical axis. Notice it says V zero. That means the initial velocity. All right. This should really say S zero. That's the initial substrate concentration. All right. We should get a straight line from that. The slope of the line is going to be KM over V max. The intercept on the y-axis is going to be 1 over V max and the intercept on the x-axis is going to be minus 1 over KM. Really? Yes? Check it out. If I take 1 over V and I make it zero and I solve for what 1 over S is going to be at 1 over V max equals zero, I get, minus 1 over KM, all right, from the not line weaver-berk equation. Okay? So, this is an extremely useful equation. The Michaelis-Menten equation, no, not so much. All right? Not useful except that I can derive the not line weaver-berk equation from it and that's really useful. Here's a problem from last year's final exam. All right? The following results are obtained by the not line weaver-berk equation. Substrate concentration, reaction rate, this should say VS zero. This should say V zero. Which substrate concentration is it constant during this whole reaction? Not necessarily, all right? It's going to be consumed as a function of time, right? Is the reaction rate going to be constant as a function of time during this reaction? Not necessarily, all right? We're really talking about the initial substrate concentration and the initial velocity. Here's some data. Tell me about this enzyme, all right? I want to know everything about its kinetic behavior, all right? Well, we can get everything. In other words, we can get KM and we can get E zero, all right? And we can get V max by making a line weaver-berk plot. So, to make a line weaver-berk plot, we don't want S. We want 1 over S. We don't want V. We want 1 over V. I want to plot 1 over V versus 1 over S. Here's a piece of graph paper that you're going to have on your answer sheet, all right? Here are the data points that you're going to plot. You're going to draw this axis because you know you don't want it here, all right? You're making a line weaver-berk plot. You want it in the middle somewhere, all right? Because you want to extrapolate this thing to the 1 over V max equals zero. This dash line here is meant to represent the maximum possible errors that any student could possibly make in plotting these data points, isn't it? So, we can grade it, okay? Now, what's helpful to you is to have a straight edge so that you can plot these data points here and then draw a straight line through them, all right? So, maybe you can use the edge of your notebook, but if I was you, I would buy one of those cheap plastic rulers that are about that long. That's very handy for this purpose, a straight edge, okay? So, now that I've got this, I can determine V max, Km, Kcat, all of these things, everything that you can possibly determine in terms of the observational kinetics of this enzyme we can extract from this line weaver-berk plot, okay? Getting the right answer starts with getting the right line weaver-berk plot and the right intercepts and the right slope, okay? You with me so far? This is straight out of Chapter 21. Now, how many people have heard everything I've said so far in another class? Now, what could possibly mess this up? Well, an inhibitor, all right? A molecule that inhibits this reaction could mess it up, all right? And inhibitors are extremely important in enzyme kinetics. And this is the beauty of the line weaver-berk plot, all right? How the line weaver-berk plot changes with and without inhibitor is diagnostic of how the inhibitor is acting on the enzyme. It's a thing of beauty, all right? You can measure the line weaver-berk plot without the inhibitor, add the inhibitor and look what happens to the line weaver-berk plot, add a little more inhibitor, add a little more inhibitor, all right? Of course, you have to measure it as the initial velocity of the reaction at a series of substrate concentrations, all right, for added inhibitor, all right? But once you've done that, you can determine exactly what type of enzyme inhibition is occurring. There's three possibilities. The inhibition could be competitive, non-competitive or uncompetitive. Is this another example of confusing enzyme terminology? Yes! None and on are not the same. All right, I don't know who came up with this stuff. Okay? So, what's the difference? It's very simple. In competitive enzyme inhibition, the inhibitor is binding at the same active site that the substrate wants to bind to, all right? If the inhibitor is in there, the substrate can't bind. It's blocking it, all right? That's competitive inhibition. The word perfectly matches what's happening. In non-competitive inhibition, the inhibitor is binding away from the active site and it can bind either to the free enzyme or to the enzyme substrate complex. But it's binding away from the active site. It's not binding at the active site. So, the inhibition is not competitive. And finally, in uncompetitive inhibition, the inhibitor is binding away from the active site but only at the enzyme substrate complex. All right? So, there's a subtle difference, all right? Non-competitive and uncompetitive. In non-competitive, the inhibitor can bind either to the free enzyme or the enzyme substrate complex. Non-competitive. In uncompetitive, it can only bind to the enzyme substrate complex. You see the difference? Free enzyme or enzyme substrate complex. That's non-enzyme substrate complex only. That's un. Good luck. It's hard to remember. All right? Here's a cartoon. Here's the inhibitor. Here's the substrate. The inhibitor is blocking the active site. That's this piece of the pie here. Once the inhibitor is in there, the substrate can't bind. Obviously, the reaction slows down but in a very characteristic way. All right? It slows down, all right? And so, here's what the Michaelis-Menten data looks like. All right? Initial velocity, initial substrate concentration. All right? Here's what happens in the absence of the inhibitor. Now I add the inhibitor, the whole curve shifts to the right, but if I make the substrate concentration high enough, I get the same V max. What's happening? I'm out competing the inhibitor. I've got a certain inhibitor concentration and if I add enough substrate, eventually this guy wins over this guy. All of the enzymes get tied up by substrate if there's 1,000 substrates for every inhibitor. All right? I think you'll agree the substrate's going to start to win and in that limit, I get the same V max, right? The enzyme is maxing out during the reaction that I care about. The inhibitor is having a relatively small effect on that. Okay? So, if it's a competitive inhibition situation, you can always out-compete the inhibitor by making the substrate concentration enormous. In that limit, you get the same V max. What does that mean? Oh, so let me point out that in competitive inhibition, you see how this is half V max here, this dash line right here? What happens at half V max? Why is that interesting? Well, it turns out at half V max, if you do the math, Km is equal to the substrate concentration. I put it on this slide so you can check it out later on. All right? Here's the substrate concentration. That's equal to Km at V max over 2. All right? So, if you're just looking at your raw data and you haven't made a line weaver Burke plot yet, you can look at this plot right here. You can go to half V max, that's equal to Km right there. And so, you can see the effect of the inhibitor is to increase Km, all right? But it doesn't affect V max. What influence does that have on our line weaver Burke plot? This is one over V max. It does not change for competitive inhibition. So, if I add in, here's no inhibition, now I add inhibitor, now I add some more, now I add some more, or I get a series of straight lines that have the same intercept but different Km's. There's Km, one over Km is down here, minus one over Km, one over V max, all right? And I can look at this line weaver Burke plot and say, boom, it's competitive inhibition, obviously, all right, same intercept, different slope. And the slope is changing in this, it's getting, the slope is getting larger as I increase the inhibitor concentration. What about non-competitive? While in this limit, here's the inhibitor, it's binding away from the active site now, you see that? It's binding, it can bind, in this case what I'm showing here is binding to free enzyme but can also bind to the enzyme substrate complex, all right, if it's non-competitive inhibition. And in that limit, what happens is Km is unaffected but V max gets reduced, all right? So this is one over V max, remember? This intercept is one over V max and so as I increase the inhibitor concentration, Km doesn't change but the intercept changes and the slope gets bigger, I can look at this and I can easily tell the difference between these two cases right here, all right? This guy is preserving this intercept, this guy is preserving this intercept, all right? So the line weaver-berk plot allows us to diagnose immediately which of these two cases is operating. I mean, I know when I add inhibitor the reaction is slowing down, I can see that, all right? But I want to understand where the inhibitor, how the inhibitor is acting on the enzyme, is it plugging up the active site, is it binding away from the active site, is it doing that for the enzyme substrate complex only or for the free enzyme as well, all right, I can tell all that just by doing a few experiments. Finally, uncompetitive amazingly gives us a third case that we can easily tell the difference. In non-competitive, this slope which is Km over V max turns out that slope is preserved and I get a different intercept for one over V max, a different intercept for one over Km but the ratio stays the same, isn't that amazing? So, I can classify, if I am willing to do the work because there's a non-trivial amount of work here, I mean, you've got to do four or five experiments to map out every one of these lines with different initial substrate concentrations, different initial enzyme concentrations to get different initial velocities. Now, if you buy this book which is the Bible of Biochemistry, even it does not contain what I'm about to show you which is the mathematical derivation of where these straight lines come from. It's not in your book, it's not even in Leninger, it's value added but you can't get it very easily anywhere else, all right, are you ready for this because I'm not going to laboriously do all this math, I'm just going to click through these slides and show you because it's a thing of beauty. Here's the mechanism that we've been talking about so far, all right, reversible formation of the enzyme substrate complex, reaction of the enzyme substrate complex to give products, all right. Now, we're going to add two other possibilities. The enzyme can reversibly form a complex with the inhibitor, that's I, all right, and the enzyme substrate complex can form a complex with the inhibitor as well, all right. So, there's two new possibilities that we have to consider. These two new possibilities encompass all of the things that can happen, all three types of enzyme inhibition, competitive, non-competitive, and uncompetitive, okay, and all we do is we write a conservation of mass equation, the total enzyme that I'm putting in my beaker has to exist in one of four states, free enzyme, enzyme substrate complex, enzyme inhibitor complex, enzyme substrate inhibitor complex. There's only four possibilities, right? Now, I do the math on this, and this is what that looks like. I have to create some definitions. I'm going to derive something called, I'm going to define rather something called alpha, which is one plus the inhibitor concentration over big K1, what's big K1? That guy, all right, the equilibrium constant for the inhibitor interacting with free enzyme, that equilibrium constant. What's K1 prime? That's the equilibrium constant for the enzyme substrate complex. That's going to be A prime, one plus that over, okay, and so I do math and I do a little algebra and we can write equations, and it doesn't take us that long before we get to your equation 21.8B, all right? Here's the non-Liever-Berberk equation. Here's the non-Liever-Berberk equation for enzyme inhibitors, all right? Check this out. I've got an alpha prime instead of a one. I've got a K times the Michaelis meton constant, all right? Instead of Km, all right? So I've just inserted alpha prime here and alpha here and now I've got an equation that is completely general for all types of enzyme inhibition, all right? This equation explains those three behaviors that we just talked about, all right? How does it do that? First of all, notice that if K alpha, sorry, if alpha prime is one and alpha is one, then the inhibitor is exerting no effect. I just get the non-Liever-Berberk equation, all right? That's this case, alpha equals one, alpha prime equals one, zilch, nothing happens, all right? For competitive inhibition, alpha is greater than one, alpha prime equals one. Vmax does not change, but Km gets bigger. This table is a thing of beauty. It's also not found in Leninger. Non-competitive inhibition inhibitor binds to both ES and E away from the active site, yes, yes, yes, both alpha and alpha prime are greater than one. Vmax gets smaller, Km doesn't change, that's what that actually means, what these two alpha means. And finally, uncompetitive inhibition, alpha doesn't change, but alpha prime is greater than one. Both of these two things are smaller, but the ratio is preserved, so the slope stays the same for uncompetitive inhibition. These are those three cases I just showed you. I just put the slide in here so you can look at it next to this, yes. So one thing that should be obvious is that for competitive inhibition, right, the inhibitor looks like the substrate. If it doesn't, it's not going to inhibit the reaction. This is the classic competitive inhibition reaction, succinate dehydrogenase, dehydrogenating succinate, all right? This is the mitochondrial membrane. Here's what that enzyme looks like. It's got a transmembrane component that plugs into the mitochondrial membrane. This is the lipid bilayer, all right? We've got these alpha helical peptide regions that are very hydrophobic that like to insert into the lipid bilayer, all right? And then we've got this more globular, extracellular part. Here's the reaction. Here's the succinate. That's this guy, all right? Here's the fumarate, that's the product of the reaction, all right? That's the succinate after it gets dehydrogenated, succinate dehydrogenase, succinate dehydrogenase. We'll form fumarate, form this double bond, all right? The classic competitive inhibitor is melonic acid. Look at this, and look at this, all right? If I rotate about that single bond, I think you're going to agree that this looks a lot like that. There's an extra carbon here, all right? But melonic acid inhibits succinyl dehydrogenase, succinate dehydrogenase, inhibits the heck out of it, all right? Why? Because the active site gobbles this stuff up. It looks just like this. It's very similar in size, okay? Did I finish early? Are there any questions about this? Because on Wednesday, we're going to talk about something else. Okay, look at chapter 21, and then see if you have questions about this stuff.