 Hello. Hello, everybody. Hello. All right. We will turn that down. So a quick announcement. I'm not associated. I don't endorse anything, but if you do have announcements for class, especially for the students, I'd be happy to share them with you. So today we're going to switch gears. And if I go fast, ask me questions. Slow me down, because this lecture is very intense. This will be the most difficult lecture for this midterm, because there's a lot of topics. And we're going to be looking at those topics in a way that you might not be used to. So last lecture, we look at the binding of the oxygen ligand to myoglobin and hemoglobin. And today we're going to go one step further. So those are not chemical transformations. Proteins can also associate with small molecules and help to catalyze a chemical reaction. So today we're going to be looking at a class of proteins called enzymes. Those are the proteins that catalyze a chemical reaction. And we'll be thinking about enzyme-catalyzed reactions, both thermodynamically and kinetically. And this is a hallowed lecture in biochem. Every biochemistry student goes through this lecture. And so you might wonder, well, why are we thinking about enzyme kinetics and enzyme thermodynamics? Well, it turns out that it's a helpful thing for you to be able to mechanistically understand enzymes. And so enzymes, if we look at the enzymes under different conditions and look how the kinetic parameters are changing, because it's a little of a sense of how tightly the enzyme binds to its substrate, and how possessive the enzyme is. And this is useful to understand the mechanism. As you might know, many drugs are inhibitors to enzymatic activities. And so if we want to design a better drug, well, we should be able to really understand the molecular details about how inhibition of enzymes works. One useful thing that will fall out of today's lectures is that one of these kinetic parameters is helpful for you to estimate the physiological substrate concentrations. And we'll see how that works in a bit. Also, there are many diseases that are caused by abnormalities in enzymes. So if we understand how these abnormalities lead to defects in either affinity or the processivity of the enzyme, we might be able to design a new therapy. If you like to drink beer, enzymes are all over the brewing industry, and so there's a wide variety of applications to this stuff. It's not just in the medical field. Okay, so we're going to start from thermodynamics. And part of that is probably going to feel a lot like a review. So you've encountered chemical catalysts, these simple little catalysts in organic chemistry. We're just going to take that one step further, more complex type of catalyst. And then we're going to be looking at the kinetics of the enzyme-catalyzed reaction. I'll introduce you to some terms such as Km, Kcat, Kcat over Km. And I'll explain to you how these terms are derived experimentally using manipulations of equations that you'll see. We'll look at different types of inhibition, and if there's a minor miracle, we'll talk about high substrate reaction. So I will not rush. You will ask questions if I'm rushing. Please ask me to repeat anything that's confusing. So let's start with thermodynamics. So we can describe the equilibrium between two states of a molecule with this reaction coordinate. So in this case, instead of reactants, we say substrate, and then the substrate is transformed into a product. We can describe this equilibrium from an arbitrary contrived state called the standard state by these delta G knot terms. And so you remember these from organic chemistry? They describe how distant the standard state is from the equilibrium state. So starting from the standard state, which is a contrived concentration of molecules, how much do you have to change those concentrations to reach equilibrium? And so in biochemistry, so in that standard state in chemistry class was defined as a one molar concentration of substrates and products, or reactants and products. But in biochemistry, a one molar concentration of all reactants is not really physiologically relevant. So oftentimes protons are involved in enzymatic reactions. And a one molar concentration of protons would melt the enzyme. That's not useful. So typically physiological condition is at a pH 7, so that would be 10 to the minus 7 molar concentration of protons. So we introduced this new slight deviation of delta G knot, delta G prime knot. So that means a one molar concentration of everything is set for protons, which are at a 10 to the minus 7 molar concentration, or pH 7. And so between the transformation of substrates of product is a energetic barrier, a transition state, and this is a high energy configuration of the molecule through which the molecule must pass. So questions so far. This is probably familiar, so for this slight deviation with the prime notation. And remember there is a nonlinear relationship between the magnitude of delta G prime knot and the equilibrium constant. So remember we defined the equilibrium constant last time, concentration of products over reactants or products over substrates. And delta G prime knot is negative RT natural log of this equilibrium constant. So if there's a negative delta G, you attend to accumulate products. At equilibrium, this ratio will favor the formation of products. Now when you thought about this in chemistry class, you might have thought about exothermic type of reactions. So the enthalpy is the primary concern. But in biochemistry, the protein catalysts are going to precisely manipulate the entropy as well. So we're going to say a process is endergonic or exergonic. In other words, the delta G is positive or negative. Because there's a contribution both from the entropy and the enthalpy. We can't ignore the entropy. Enzymes have thousands of chiral centers. Every alpha carbon potentially is a chiral center. And these things are very precisely controlling the flexibility of the substrate. Also the enzyme affects the accessibility of substrate to water. So in solution, the substrate is floating around in water. But in an enzyme, the enzyme in many enzymes encase the substrate, exclude the water. So the entropy of the substrate is decreased. It's less moving, but the entropy of water is increased. So we can't just ignore that. We need to consider both the entropy and the enthalpy. Endergonic is when delta G is positively charged. And exergonic is when delta G is negatively charged. And so there's also this term of spontaneous. Very confusing. So a spontaneous exergonic reaction spontaneously forms product. When you think of a spontaneous person, you think of somebody that's acting very crazy and erratically. And you think that they're spontaneous. That means you do something crazy quickly. But in terms of thermodynamics, spontaneity just means the direction that things tend to go, not how quickly it occurs. So thermodynamics doesn't care if things take the age of the universe. It just tells you like for delta G is negative, things will spontaneously go in the direction towards product. It could take a very long time. For equilibrium to be achieved. Do you remember this from previous classes? Yes. That's right. That's right. So far it might be familiar. This is a nonlinear relationship between the equilibrium constant and delta G. Remember there's that natural log transformation. So as an equilibrium constant increases logarithmically, the delta G erratically decreases. So every tenfold change in the equilibrium constant, you add or subtract 5.7 kilojoules per mole. So this describes this equilibrium. So let's look in between. So the equilibrium is only defined by the free energy that's substituted in product. But how quickly the equilibrium is achieved is dramatically affected by this transition state. And this is a transient, unstable species. It's necessary to pass through this unstable species on the way to formation a product. But once a molecule has achieved the transition state configuration, there's an equal probability that it will return to substrate or product. But more substrate molecules achieve the transition state. So every time you get to the top of the hill, the ball could roll to the left or the right. But since more substrate molecules make it to the top of the hill, you equilibrium favors formation of the product. There's one way to think about this. And so there's two contributions to the high free energy of the transition state. And as you might imagine, it has to do with entropy and enthalpy. So in terms of entropy, if you want to cause a chemical transformation, oftentimes favorable bonds need to be broken in your substrate. And so there's... Oh, that's in terms of enthalpy. In terms of entropy, the substrate must adopt a particular very defined orientation in the transition state. So there's a high entropic cost of achieving that configuration. Whereas enthalpy, you might need to break stabilizing bonds. And so those two factors combine to make a high transition state energy. All right. So the equilibrium state is solely determined by the free energy of the substrate and the product, but the rate at which the equilibrium is achieved is determined by the height of this barrier. And so there is an equation. You can have this to your slide. I just added this. Some of you might be aware of the Renius equation. And so as the activation barrier increases, the rate decreases. So it's e to the minus height of the barrier divided by RT. And so the way enzymes are going to work is they're going to provide an alternate lower energy barrier through which to form products. And so enzymes do not affect that thermodynamics of this process, only the kinetics. And that makes sense because what does an enzyme do? It binds a free substrate and releases a free product. So just because the enzyme was touching it at one point, that doesn't change the free energy of this product floating around in solution. Okay. So here is a way in which a framework in which we're going to think about this. So an enzyme and a substrate in solution find each other and form an enzyme substrate complex. While bound to the enzyme, the substrate is chemically transformed through a transition state until you form an enzyme product complex. And then the enzyme and the product are released. And so what's happening here is the rate enhancement is coming from the reduction of the free energy barrier comparing the uncatalyzed process to the catalyzed process. That accelerates the rate. And these rate enhancements are spectacular. Look at the magnitude of some of these numbers. Here's one particular enzyme. You can pick out many enzymes by looking for the letters ace at the end of the name. And so this particular ace accelerates a chemical reaction 10 to the 17th fold above what the rate would be in the absence of the enzyme. And so it's hard to imagine. That's way beyond the U.S. debt. We're talking the number of grains of sand on the planet Earth. Earth is a big place. There's lots of grains of sand. That's how many are there. And the planet's worth of rate enhancement with this one enzyme. Just spectacular. Yes, I am a nerd. Okay, so this equilibrium, as I've said again and again, when I repeat things, I tend to think they're important. I tend to want you to think they're important. And I might ask you about this. So the equilibrium is not affected by lowering this activation energy. And so the way that they're providing this lowering of the activation energy is through binding energy. The enzyme makes more bonds to the transition state configuration of the molecule than the ground state. And those bonds are stabilizing the transition state, providing this lower energy pathway. Now, when you think about it, well, an enzyme in your cells is encountering a myriad of different substrates. And it's got to pick the right one. So you might imagine, okay, there's got to be a certain amount of specificity of the enzyme for its substrate, but not too much. Because when you think about things thermodynamically, if you formed a large number of bonds to your substrate, this free energy would be reduced. And you would still have a barrier between the molecule. What's important is the differential of the number of bonds to the substrate in its ground state configuration compared to the transition state. You make more bonds to the transition state than the ground state. It's necessary to make some bonds to the ground state because you have to find it with specificity. You don't want to transform the wrong substrate. Does that make sense so far? All right. Yes, it is a painful lecture. So one way that many people have thought about this in different ways. And so one of the earliest thoughts was the Lock and Key model. And so in this model, people are thinking mostly about specificity. In order for an enzyme to find its substrate, you might imagine the enzyme active site has to be very complementary to the shape of the substrate. And so in this, here's a picture of a pigly substrate binding to its enzyme. So it nestles down there. Do you see how it's less accessible when it's bound to the enzyme? Do you see some new favorable bonds being formed? So here we have a salt bridge, a positively charged functionality in the enzyme pairing up with a negatively charged functionality of the substrate. We maybe have some hydrogen bonds. But what we need to do is help to push this forward. So we're going to make more bonds to the transition state. And so in order for you to make more bonds to the transition states in the ground state, you might imagine, well, it's probably necessary for most enzymes to be flexible. So they need to adopt into an optimal configuration. They need to first bind their substrate. So they have to have some similarity. But then they need to adopt an optimal configuration for them to stabilize the transition state, to make the maximal number of bonds to the transition state. So many enzymes have this induced fit where they adopt one shape, and that shape is enough to make a specific interaction with the substrate, but then they change in configuration to make an optimal number of bonds with the transition state. So enzyme active sites are most complementary to the transition state configuration and not the ground state. Too fast, too slow. Just right. So the enzyme is providing binding energy. And this binding energy is reducing the free energy of the transition state. As you reduce this rate, we have the Arrhenius equation. So you reduce the barrier, the rate goes up directly. And you do not, as I mentioned before, want to make too many bonds to the substrate because that would also reduce the free energy in the enzyme substrate complex. Such that, for example, the curve would go down here, go back up, and go here. That's not as good. It's just making a few bonds to the substrate where you would have a smaller barrier. Okay, so binding energy is the magic. So there ends the lesson on the thermodynamics of enzymes. Any questions on that part? We're going to switch gears to the kinetic analysis of enzymes. I think the thermodynamics part is a little bit easier because it's very similar to ways you've thought about chemical transformations with small catalysts. But now let's look at the kinetics. And so remember I laid out this framework. An enzyme binds its substrate, makes an enzyme substrate complex while bound to the enzyme. The substrate is chemically transformed into a product. And the enzyme product complex dissociates. Each of these steps can be associated with a rate constant. They're in equilibrium, so there is a forward rate constant and a backwards rate constant. So we define a K1 rate constant. That's the rate of formation of enzyme substrate complex. And the rate of breakdown of enzyme substrate complex have to do with K-1 and K2. And the way that we're going to approach this kinetics is going to seem very contrived. I'm going to lay on this analysis to simplify things. A few assumptions. There's actually two. The first is the initial velocity. We're going to consider an enzymatic catalyzed reaction before a significant product is accumulated after equilibrium is reached in this step. So it's only the initial binding and before a significant product has accumulated. What does that assumption allow me to do here? Color-coded in the figure. Do you see it? Before significant product. So what is the rate? How can you define a rate in terms of rate constants? Do you see it? So remember, rate has to do with the rate constant and the concentration. At the initial state, what is the concentration of EP? Zero or very small, right? And so do you see how half these rate constants go away at the initial state when you're thinking about the rate of the process because K3 times EP is zero. K3 could be whatever it wants. So you multiply it by an infinitesimally small concentration, it goes away. So the rate of formation of product is K2 times ES. You buy that at the initial state, right? I'm not talking about the rate of formation of enzyme substrate complex. That would involve other things, right? So there's all kinds of stuff going away from enzyme substrate complex. I'm talking, what we care about is what is the rate of formation of product? And so that rate is just K2 ES, right? Not, you could say K2 ES minus K minus 2 EP, but K minus 2 times EP is zero. So we don't need to think about it. So it's very simple. The rate of the process at the initial state is K2 ES. Do you buy that? This is something that's hard to accept because either you want it to be the rate of formation of enzyme substrate complex or you're not thinking about it from the initial state. That's why it's confusing. Okay, so we define all these rate terms and now we need to think about what is the relationship between substrate concentration and the rate and this initial rate, abbreviated V0. So before we do that, we're going to introduce assumption two, steady state. The concentration of enzyme substrate complex is a constant. It's unchanging in the steady state. Okay, so we have two, in other words, the derivative of that concentration is zero. The slope of the line is zero. So here we have a substrate being converted to product. We have the steady state assumption in the initial rate assumption or limitation. So when you look at this picture, you see a shaded rectangle. Is this the zone of where this kinetics, both of these assumptions hold true? Any ideas? So initial state before a significant product has accumulated. Has significant product accumulated over here? Okay, so this shape is only telling you where steady state occurs. But we have two assumptions, steady state, which is all of this, and the initial state before a significant product has accumulated. So we refer to this region over here as pre-steady state because the concentration of enzyme substrate complex is increasing. You see that? We're just talking about this one little tiny part of the process. And you might think, useless. Surely enzymes are free ranging and they encounter all of life's challenges. But you'll see in a moment how that's not necessarily true and that these approximations are not necessarily limiting. So this is where we're talking about. Just this tiny little box. It reminds me of a scene in a particular movie. This is one little area where you can win the prize. Okay, so those are our only two assumptions. And now we're going to look at the, yes, the steady state assumption is one of two assumptions. The steady state assumption is represented by this entire rectangle. That's where you have the confluence of steady state and initial state before a significant product has accumulated. Okay, that's a good question. Okay. Okay, so here is the relationship between the rate of the enzyme catalyzed reaction and substrate concentration. You can see it makes a particular familiar shape that increases to a maximal value. The maximal value is called the Vmax, the maximal velocity. And then if you have a Vmax, you have a half of Vmax, and that half of Vmax has a certain substrate concentration that is necessary to achieve that half maximal rate. And so let's call it Km. Okay, so this is a hyperbolic relationship between the initial rate and the substrate concentration. Do you remember where we saw this? Myoglobin. Ding, ding, ding, ding. So here is the equation that describes the relationship between rate and substrate concentration. So you have rate equals some constant times s divided by some constant plus s. So it's very familiar to this equation. Y, V0, equals some constant times x over A, some constant plus x. That's the equation of a hyperbola. So there's some relationship between the fractional occupancy of binding and rate. If all enzymes are bound to a substrate, that enzyme is working at its maximal rate. If half the enzyme molecules are bound to substrate, it's working at its half maximal rate. If 10% of enzymes are bound to substrate, it's working at its 10% of its maximal rate. Think of it in terms of assembly lines in Detroit, making automobiles. You have robots. If a robot, only half of the time, is doing what it needs to do, the whole assembly line would be half as efficient. Okay, so these are just little molecular machines turning out widgets, stamping out products. And so there's a logical relationship to transform a molecule and enzyme must bind to that molecule. So the more that binds, the more that can be transformed. So it makes sense that both of these are hyperbolic relationships. Okay, so this is restating that. Any questions so far? No. All right, but look at this equation. Do you see, remember with theta, we had theta, fractional occupancy in myoglobin or any protein, is equivalent to L over L plus 1 over KA. What does that tell you about this particular constant? Intuitively, without going through an extensive analysis. But there's got to be some sort of connection, right, between a 1 over KA and a KM. What's another way to express 1 over KA? KD. So there's some sort of similarity between a KD and a KM. So KD is inversely related to binding affinity. So this KM has something to do in an inverse sort of way with how tightly an enzyme binds its substrate. Okay, but we're not using calculation, we're just using intuition to make this statement. So we have KM and Vmax, and those are useful. If we know KM, we know the concentration of substrate necessary to fill half the binding sites on the enzyme. If we know Vmax, we know the maximal processivity of the enzyme. And so we want to be able to determine these constants because they tell us about something about our enzyme. And if we alter the conditions and see how these constants change with different conditions, we can maybe get some insights into the mechanism of the enzyme. So we have KM, Vmax, and I'll be introducing later the more generic term, Kcat. And so now we're going to derive an equation. This is a proud tradition in biochemistry. I just plot this thing down. V0 equals Vmax times concentration of substrate over KM plus S as if it was an epiphany. But it came from somewhere. It was directly derived from this equation in our two assumptions that we've made. And so we have that here is the framework. I don't think I need to go over that again. Remember I introduced this concept. The rate of formation of product is equal to K2ES when our two assumptions hold true. Do you still believe it? Work it out on, think it through. Make sure you believe that. It's hard to swallow that pill. Now an enzyme can either be bound to substrate or not. So ET phone home, the total amount of enzyme is either in a configuration where it's not bound to a substrate or it's bound to substrate. At the initial state, it's pretty likely that the concentration of substrate will be much slower than the concentration of enzyme substrate or much smaller than the concentration of enzyme substrate complex. You with me so far? This one's hard to swallow, but the rest of this makes some amount of sense. So what we're going to do here is we're going to think about how can I figure out the rate of formation of the enzyme substrate complex? There's two sides to this. How quickly it's made and how quickly it's degraded or it comes apart. The rate of formation of enzyme substrate complex is equal to K1 times ES. Do you see that? It has nothing to do with these great out terms. The rate of breakdown of enzyme substrate complex, we've got this term and we've got this term. So the rate of breakdown is the concentration of ES times that term plus the concentration of ES times that term. You with me so far? So that's how fast it's made, how fast it's broken down. If we're at steady state where the concentration of enzyme substrate complex is a constant, what do we do next? You see it on the slide, but why can we do that? So if the concentration of enzyme substrate complex is a constant, it must be being made as quickly as it's being broken down. So if these rates are equivalent, we can set the two sides of these equations equal to each other. Here I've made just a little substitution here where I solve for E. E equals ET phono minus ES. So we substitute that in there. That'll be handy later on. So we're just setting the two sides of these equations equal to each other. And now it's going to be a parade of arithmetic manipulations of this equation. I'm about to move it up into the side corner. There it is. I didn't do anything. This is the same. I want you to keep track. So here we just have the same equation. We've collected terms, I think, is all we've done here. And now what we want to do is solve for enzyme substrate complex concentration and cluster the constants. See if we can simplify this a bit. So here we're moving things around, collecting constants. Now we're solving for enzyme substrate complex, dividing through. Now we've solved for enzyme substrate complex, but we still have a bit of a mix of things over here. So we want to condense those constants. So here we have all the constants together. You're like, so what? Well, if they're all constant, I can just call it a new constant. The person that did this work was Michaelis Menten. So he said, I think we need a Michaelis Menten constant. So he called that a KM. So forever, his name is associated with this odd assortment, this collection of rate constants. So we define, Michaelis define Menten, defined KM, a new constant. In terms of these rate constants, K2 plus K minus 1 divided by K1. When I draw a box around something, I'm giving you a hint that I feel it's a valuable piece of information that you should remember. So this is how you calculate the KM. You just K2 plus K minus 1 divided by K1. So what I'm going to do now is substitute that back in. No sorcery here. I'm just putting KM into this equation. But the problem is, well, I did solve for enzyme substrate complex concentration, but it's a bit difficult in the lab to go and measure that. You have a huge enzyme. It finds a tiny little substrate, and it's not analytically that different. It has just a little bit more stuck to it. So we want to rearrange this equation a bit and substitute it in. So this was our initial rate, V0 equals K2ES. Do you remember that? So it's the initial state, concentration of EP, and all these other things, very low, so we just forget the rest of that. V0 equals K2ES. We already solved for ES. You can stick that in there. It's just a simple substitution here to make it more interesting. Move the K2 to the other side, and then they put it back. All right? So V0 equals KT, ET thrown home, times S divided by KM plus S. Do you see how we're almost there? So when you think about it, K2ET, well, if all the enzyme is bound to substrate, it's working at its maximal rate, right? So we're going to do one last... I moved it, I just moved it up here. It's always a little panic. So V0 equals K2ES, but the maximal rate is K2ET. So if all the enzyme is bound to a substrate, it'll be working at its maximal rate. So Vmax equals K2ET, we can substitute green for Vmax, getting our final equation, the McHale-Listman equation. Now is that so bad? A little bit. It's a bit unnatural. You want to say, why did we do that? Well, one thing that we gained, we get a sense of how we calculate a KM. What is a KM? Where did this come from? But let's poke in probably this equation to get more of a sense of its nature. And so if you want to determine Vmax and KM, you have to determine the shape of this curve. So you have to go in the lab, and you have a reaction rate detector meter, maybe you measure the rate of formation of product, and you add continually increasing concentrations of substrate and measure the reaction rate, and you can figure out this curve. Once you've figured out this curve, you could use computers to calculate what the asymptote of the hyperbola is. That would be Vmax. Once you know Vmax, you know half Vmax. So you can have graph paper. Do you guys ever use this? I'm very dated. So you have graph paper, and you can determine, okay, what is the corresponding substrate concentration necessary to fill half those binding sites? Well, that's KM. You say, well, really? Why do you say that? I don't believe what you say, man. So if you don't believe me, you can substitute in. So let's figure out what is the substrate concentration necessary to achieve half the maximal rate? So if V0, the rate, is equal to half its maximal value, substituting that in and rearranging, rearranging, yada, yada, yada. Whoa! Look at this. Isn't this spectacular? So you have Vmax over 2 equals all these other things, and you just shift things around at a half maximal rate. KM is the substrate concentration. Isn't that confusing? So KM is both an odd assortment of rate constants, and it's a measurable substrate concentration. It has units of concentration. Right? So KM is the number of substrate molecules, the concentration of substrate molecules, necessary to fill half the enzyme binding sites. Okay? So if we need more of substrate molecules to fill half the binding sites, this curve would be shifted to the right, and if we need less substrate molecules to fill the active site, it would be shifted to the left. Okay? It's weird. Rate constants and a concentration. What's up with that? So KD is equal to a ligand concentration. Remember this? When half the binding sites are occupied. So there's this equivalence between the number of occupied binding sites and this KM. When half are full, you're at a concentration of KM. Okay? With me so far, we're going to poke and prod just a little bit more to get a sense of this. Well, one question about one of the assumptions. How do we know the rate of formation of ES is equal to the rate of breakdown of EP? It doesn't matter what the rate of breakdown of EP is, because the concentration of EP is negligible under our set of assumptions. Do you see that? That's a great question. Have me stop. All right. Okay, so let's consider two scenarios. What about high substrate concentrations or low substrate concentrations? Well, it's high substrate concentrations. This denominator is dominated by substrate concentration. So at substrate concentrations, much higher than a KM concentration, you can just neglect this. Just say substrate concentration is 10 KM. You add one more. Big deal. It doesn't make much of a difference. So at very high substrate concentrations, V naught equals V max S over S. So at high substrate concentrations, V naught equals V max. As you increase the amount of substrate, you approach a maximal rate. That makes a certain amount of sense. What about at low substrate concentrations? Well, in that scenario, KM would be much higher than substrate concentrations, so we can ignore substrate concentrations. So at low substrate concentrations, V naught equals V max S over KM. I think that's written here. Or not yet. And so that's the equation of a line, right? Y equals some constant times another variable, substrate concentration. The slope of the line, M, so Y equals Mx plus B. We're going to use this fundamental concept again and again today. Y equals Mx plus C. So M, the slope, is V max divided by KM. So at low substrate concentrations, there's a linear relationship between the concentration of substrate and the rate of the reaction. So we're just examining this. But how do I determine these constants? They're very useful in comparing different substrates for one enzyme, enzyme under different conditions, or different enzymes for one substrate. So I want to calculate these so I can compare enzymes, but how do I calculate them? Well, we need to think about what they are. So KM, so we'll get to that in a moment. But the Michaelis-Minton constant is an apparent dissociation constant, right? So it's not really an equilibrium constant. It's calculated by a odd assortment of rate terms. And KM, remember KM equals K2 plus K minus 1 divided by K1. Dissociation constant, KD, is K minus 1 divided by K1. So the Michaelis-Minton constant is a dissociation constant only when K2 is rate limiting. So one very good question to go over this lecture today, is it necessary that the K2 be rate limiting for this whole framework to hold true? And what do you think about that? Okay, so it's apparent because it's not calculated in the same way that other dissociation constants are calculated. It's a weird assortment of rate terms. And it only is a dissociation constant if K2 is much smaller than K minus 1. And so the maximal velocity is achieved. The more enzyme is bound to some strain. So as we saw from poking around in that curve. Okay, so these are KM and Dmax. These are very useful terms. And here's what I just said. I'm more conveniently shown to you. And so if K2 is much smaller than K minus 1, KM would equal to K minus 1 over K1. So what does that mean? If this step is rate limiting compared to these other steps, then KM is a dissociation constant. But maybe K2 is of similar magnitude to K minus 1. Then it would not be a dissociation constant. Now, for the majority of enzymes, yes, it's true that this step is rate limiting. And in that case, there would be an inverse relationship between the KM and the binding affinity because KD equals 1 over KA. But that is only true. It's only an approximation of binding affinity if this term is much smaller than this term, which is generally the case. Does that make sense so far? So this lecture is very useful to go over the video, pause it, like what in the world is this guy saying? Get caught up and watch it again and again until it makes sense. Read other textbooks. Okay, so that's great. It's very conceptual. Let's use a real-world example. Very useful. So here is one of our favorite substrates. Makes us happy. We can oxidize happy substrate to form acid aldehyde. Acid aldehyde can be additionally oxidized to full carboxylic acid acetate by acid aldehyde dehydrogenases. We'll learn about dehydrogenases or things that change oxidation state of molecules. So here we have one oxidation, two oxidation. So as it turns out in this room, there's a diversity of different types of this enzyme. So some of you have a mutant form of this enzyme that has a different Km than others of you in the room. So some people are luckily endowed with an acid aldehyde dehydrogenase with a low Km. So that means a smaller amount of this material allows you to achieve the maximal rate. So that means that you're able to more efficiently clear this acid aldehyde to convert it to product. Whereas other of you consuming a same amount of this substrate, well, you have unfortunately the high Km variant of this enzyme. In that variant, it takes a larger amount of the substrate for you to achieve the half-maximal rate. So the amount of acid aldehyde begins to build up in your bodies. And so you might go to a party and you look around and you say, the acid aldehyde actually causes a physiological response. Your face gets blushed and your heart starts quivering. I think I might have one of this, the high Km form. And you can go and tell people just by looking at them what kind of acid aldehyde dehydrogenase they have. That's not nerdy enough. So that's a real-world, very useful example of how genetic diversity leads to different forms of enzymes that have different parameters. So it's useful to calculate Km. And so, as I've been saying, once you calculate this Km and Vmax, it's useful for comparing enzymes under different scenarios. So for example, you can measure the Km and the Vmax at different temperatures in pH. If you change the pH and there's not a change in Vmax or in Km, well then probably the enzymatic catalyzed reaction doesn't involve the transfer of protons. You've just been able to come up with a mechanistic insight based on the calculation of these parameters. And you might also say, well, that's really, you remember Steve Martin, this is one little section of where our assumptions hold true. And you say, that doesn't make sense. At Brown, we're liberated. We could do whatever we want, right? But it makes sense in the context of one enzyme in a test tube with some substrate. It does make sense when an enzyme is cooperating with other enzymes. So one enzyme's product is the next enzyme substrate. So product tends to not accumulate because enzymes occur in pathways. They manipulate metabolites in pathways. So it seems contrived, but it actually is often the case that the product does not accumulate. And this consideration of the initial state covers a wide range of enzymes, the real world performance of enzymes. But if we went into lab and we tried to calculate this, think about what you would have to do. You'd have to keep increasing substrate concentration. You have your reaction rate, oh meter, detecting the rate of formation of product, for example, and you would start to collect some data. But you see that to really determine the shape of this curve, you're going to need, you know, if you just collect data down here at low substrate concentrations, you might say, well, this is sort of a linear relationship. Maybe Professor Salman is a liar. It's not real. So really what you need to get the shape of this curve is you need to collect some data up here. Well, that can be a problem because in order for an enzyme to chemically transform a substrate, it has to be in solution. So as you increase the concentration of substrate higher and higher, some substrates reach their solubility limit. They crash out of solution, right? So experimentally, it can sometimes be difficult to determine the shape of this curve. So now what we're going to do is to transform the McHale-Smithson equation so that we can be able to have a more convenient form of this equation that allows us to determine these rate constants. And so what we're going to do here is very simple. We invert both sides of the reaction. Remember, so multiply through, multiply through. Have I violated any natural laws of science? No, we're just inverting. We're doubly inverting this equation. Once we've inverted, we're going to rearrange it. And again, you might think, why would I have done this out of the blue? They tried every manipulation of this equation until they got one. It was useful to overcome this experimental difficulty. So now we have this equation. 1 over V naught is Km over V max s, plus s over V max s. Well, we can simplify that, obviously, one more step. And then we get to this double reciprocal equation. Look at the nature of this. Do you see how we've converted a hyperbola into a line? You're like, no, you haven't. Yes, I have. So it is a line if you consider y to be 1 over V naught and x to be 1 over s. So if you plot this, so y equals mx plus b, y 1 over V naught equals Km over V max m times x 1 over s plus a constant. That's exactly, you've converted a hyperbola by double reciprocating it. You've created a linear relationship. The slope is Km over V max. And the y intercept, remember that b is the y intercept. That's 1 over V max. So I've double reciprocated this thing. And here we can plot some data. So the slope is Km over V max. But we can also calculate the y and the x intercepts. So if you fundamentally disagree with my perception that b is a y intercept, you could prove it to yourself. So at the y intercept, let's prove it. Maybe you're skeptical. At the x intercept, y equals 0. Remember, y is 1 over V naught, so we just set that equal to 0. That's going to be the x intercept. To set that equal to the rest of this equation, rearrange things. And we find that at the x intercept, 1 over substrate concentration equals negative 1 over Km. Oh, that's nice. What about if we do the same thing for the y intercept, where x is equal to 0? So first, we have to solve for x. We have to solve for 1 over substrate concentration. So rearrange this equation, solve for x. And then set that equal to 0. Rearrange things. Ah, at the y intercept, it is true that b is a y intercept. 1 over V naught is 1 over V max. That's a little bit confusing. Why is it confusing? Where are you collecting data? How much data are you collecting right on this line? Or over here? Can you say extrapolate? You are not collecting any data here. I went into the test tube and applied a vacuum force that pulled the substrate out. And that's my negative substrate concentration. No, that's ridiculous. You have positive substrate concentration. So your data you're collecting is over here. But what you can do is draw a line through the data points so you increase the substrate concentration, right? Collect some measurements of the rate. And you can obviously invert those easily. Draw a line through those points. Extrapolate through both the y-intercept and the x-intercept. The y-intercept is 1 over V max. So if you know the y-intercept, you know V max. The x-intercept is negative 1 over Km. If you know the x-intercept by extrapolation, you know the Km. You're like, that seems a little easier, but I don't quite understand why you're doing this to us. Well, let me show you some data. So here we have equally spaced data points. So I've gone into the lab. I measured out on a balance or whatever, half a Km worth of substrate. So that's enough to fill half of those enzyme active sites. And then I measured out a Km. And I measured out 1.5 Km to Km. And so then I measured the rates corresponding to those. And so we're collecting a fair amount of data here, but not much data over here. What happens to the distribution of the corresponding points on the double reciprocal pot? Where are low substrate concentrations occurring on this plot? The ones that are easier to measure. From your perspective, do you see that? So this point, and this is 1.5 Km. This is 1 Km. Those are very handy points, because those really allow you to determine the slope of that line. Whereas really high substrate concentrations are all clustered here. So if you just measured high concentrations, you wouldn't be able to tell the slope of the line. So the experimentally easier measurements occur spaced out in such a way that you can more easily extrapolate the line. So it's just a simple manipulation. And now we have in hand an ability to calculate Km and Vmax. I often ask questions about this type of manipulation of the equation. And I might ask you, why is it useful? Why would you do this to yourself? You're with me so far? Yes. Lower substrate concentrations are less likely to cause a problem with solubility. And the lower concentration when the hyperbola, it's sort of, remember, linear in that earlier region. So you don't get a real good shape of the curve. Now, perhaps if you collected lots of data and you were able to make tight measurements, you could maybe be able to tell the shape. But it's a lot easier. Here's a point draw line between them instead of minute calibrations of some polynomial curve fit. Any other questions? Okay. So we've talked about V0 equals K2ES. Remember, that's the rate of formation of product. We're not concerned with the rate of formation of enzyme substrate complex, but the rate of formation of products, the rate of overall process is K2ES. But some enzymes aren't simple. So I started with a very simple example. Substrate product. Substrate, it's like a seesaw between substrate and product. But you'll see lots of examples of enzymes throughout this semester that have intermediates. So substrate gets a little bit transformed, a little bit more, a little bit more. So we can use McKay-Lux-Minton kinetics for those more complex enzymes if we can figure out which step is rate limiting. And so for those types of enzymes, we can define a Kcat. A Kcat is whatever step is rate limiting in the process. Another way of thinking of Kcat is in terms of the maximal rate. So if all enzyme is bound to substrate, the enzyme, remember we've agreed that the enzyme is working at its maximal rate, and the Kcat gives you a sense of how many widgets the robot is putting out per second. So Kcat is referred to as the turnover number. So it sort of has two definitions. It's an abstraction of a complex process down into a K2. Or it's the number of substrate molecules converted to product per enzyme active site per second. So a Kcat of one million means that one single enzyme is able to convert one million substrate to product per second. So Kcat gives you a sense of the maximal processivity of the enzyme. Working all out, steams, things falling off, going all out for it. If all the enzymes are bound to substrate and they're cranking out widget after widget, that's the maximal processivity of the enzyme, the Kcat. So we call it a nice turnover time. Does that make sense? This is also often confusing. It has multiple definitions. Okay. So let's look at some examples of these measurements. So here is Kcat for a variety of enzymes, and Km for a variety of enzymes. Do you see how there's a massive dynamic range in the maximal processivity of these enzymes, the Kcat? So this catalyst enzyme converts 40 million substrate molecules to product in one second. Isn't that spectacular? Can you think of it in Detroit? GM is like one million manipulations of the car per second. This is an incredible machine. It blows away anything that we've created. Whereas other enzymes are like, oh, there's a substrate here. And I'm going to convert you. Wait for it. Wait for two seconds. So that's only sitting there. Very, very slow. And there's what? Almost eight orders of magnitude and difference in the maximal processivity. But think of Km. Look at the range of that. So if K2 is slow rate limiting, Km is a dissociation constant universally related to the binding affinity. So if that is the case for these enzymes, then there's a large range of binding affinity of the enzyme for its substrate. Maximum processivity and the binding affinity varies dramatically. But the combination of both factors gives you the only real world perception of how an enzyme works. As it turns out, it makes sense. Enzymes don't typically work at their maximal rate. Stuff starts falling apart if you're going really, really fast. Enzymes typically work at or below a Km concentration of their substrate. Evolution has evolved these enzymes so that they're not working at their maximal rate so that intermediates don't accumulate. If it's working at its maximal rate, you put a little bit more substrate in, it'll just start to build up because it's like, come on, man, I can't go any faster. So nature has evolved these machines to work at or slightly below their Km. But really, to get a sense for this, we need to combine these two terms, how tightly it binds and how processive it is. And so this gives you a little sense of why it makes sense. You can think of it in terms of intermediates building up, but you can also think of it in terms of stabilization. Something is going to work very fast. It's going to bind, or if the Km is going to be pretty low, it's going to bind that substrate really tightly. But thermodynamically, we've thought about this. That's not ideal. You don't want to bind it too tightly because if you bind it too tightly, if the Km has to be really small, then this would be thermodynamically overstabilized. There would be a barrier between the steps. Do you see that? Okay, so how can we combine Kcat and Km to get a sense of the real-world performance of an enzyme? Well, let's consider the physiological state. The physiological state is where the substrate concentration tends to be lower than the Km. So if the concentration of substrate is much lower than Km, we can neglect this term from the denominator, and so the McKay-Lis-Minton equation is transformed into this. V-naught equals Kcat ETS divided by Km. So Kcat over Km is a second-order rate constant for the enzyme encountering the physiological normal set of conditions. Another way, and Kcat over Km is confusing. So think through this a lot. Read a lot of text. The other way I like to think about Kcat over Km is Kcat over Km is processivity normalized for binding affinity. That I find less confusing, but it might not help others. So it's either you can think of it as a second-order rate constant under physiological conditions, or you can think of it as the combination of processivity and affinity of the enzyme to get the best estimate of the real-world performance of the enzyme. So if we can calculate Kcat over, we can calculate Kcat, we can calculate Km, and then we can look at some real-world examples. Look at these collection of enzymes. They catalyze their reaction at a certain maximal processivity, and they also have a certain apparent affinity for their substrates. Do you see how there's what? Seven orders of magnitude. If I say it well better, I could actually say. But there's a huge orders of magnitude of variation in Kcat, and a corresponding huge variation in Km. But all these enzymes are equally as efficient in the real-world. Because when you divide, they might say, some of these have huge Kcats and some have slower Kcats. Some have really tight binding affinities, some have looser binding, apparent binding affinities. But when I divide the two by each other, you get about the same number. And these particular sets of enzymes are nearly perfection in evolution. So these are achieving, the only limiting factor with these enzymes is the diffusion of the substrate into the enzyme. Once it's there, it just pops right into product. So diffusion can occur at this rate, right, in terms of the enzyme and its substrate. And these are approaching what's called catalytic perfection. This is the mastery of evolution here. So these have 10 to the 7 order of magnitude for Kcat over Km. So I can go in the lab. So if I didn't know, if I didn't look at Kcat or Km, I would have naively thought that these enzymes had different abilities. But when I combine these and under the real-world conditions of the enzyme, they're just as good as each other. So it's useful when you calculate Kcat's Km, Kcat over Km to compare things, compare different enzymatic reactions, for example. Does that make sense? So one question from the audience. How is Kcat affected by the physiological environment like changes in pH? Ah, that's a good question. I can... changes in pH affect Kcat. Does anybody have ideas of a real-world example? I cannot... I'll have to think about that. Most of you guys can jump on Wikipedia and find some specific examples of Kcat being affected by that. Okay, I'll have to come back to that. You stumped me. I have no idea. All right. Any other questions? This is a long, freaking lecture, so we just have one more topic. Bear with me. And that's inhibition. So we're going to look at the mechanism of action of inhibition. There's different types of inhibitors. An inactivator is an irreversible inhibitor. Once this thing binds to an enzyme, that enzyme is poisoned. It can no longer process substrates. But then there's reversible inhibitors. And we can categorize these reversible inhibitors in terms of different mechanisms of action. Competitive, uncompetitive, mixed in the ultra-different non-competitive. So this is torturous. Non-competitive is different than uncompetitive. Semantically, let's say, but biochemically different. Sorry about that. I didn't do it. I did not come up with this. So let's look at competitive inhibitions. Think of the word competitive. Something is competing. We have two possibilities, either an enzyme binds a substrate or an enzyme binds an inhibitor. If an enzyme binds the inhibitor, it is precluded from binding a substrate. It can't do it. Sometimes, but not always, in a competitive inhibition, the inhibitor binds at the active site and physically blocks the substrates from binding. Doesn't necessarily have to be that way. You can think about how that could be. I'll let you think about that one. I'm not going to answer all questions. But let's think for each of these types of inhibitors. We have an experimental method to determine KM and Vmax. Remember these double reciprocal plots? We can add different amounts of inhibitors and see how it affects KM and Vmax. Let's think about these things logically. If it's competitive inhibition, what do you think is going to happen to Vmax? The maximal rate giving an infinite amount of substrate concentration. It's not going to change it because you can saturate this out. If you just keep adding more substrate, you're going to tend to favor the equilibrium because this is all in equilibrium. You're going to tend to favor the equilibrium of formation ends on substrate product, which then gets converted to product. Vmax is unchanged by competitive inhibition just by logical thinking processes. What about the KM? In other words, what happens to the substrate concentration that's necessary to fill half the binding site? It increases. If K2 is rate limiting, what does that say about the binding affinity? Decreases. You see this so far? You can go in the lab and do an experiment. You can independently vary the substrate concentration and the inhibitor concentration and make a bunch of rate observations. We vary substrate concentration. We vary inhibitor concentration. As the inhibitor concentration increases, the y-intercept does not change, so 1 over Vmax is unchanged. Negative 1 over KM, as we go up, negative 1 over KM decreases, so 8 KM increases. Do you see it? All you have to do, there's definitely nothing about the mechanism of action of this inhibitor. You do this experiment, collect this data. If the curved line-weaver berkplot looks like this, it's competitive inhibition. That curve is diagnostic that the inhibitor and the substrate cannot bind into the enzyme at the same time. So it's very useful at getting at the mechanism of action there. Throughout this we'll have various mathematical ways to calculate these, but we're not going to concern though there's alpha and alpha prime. Don't worry about those. Just think about these logically. Uncompetitive inhibition is the opposite case where the inhibitor can only bind once substrate is already bound to the enzyme. What would that do to Vmax? It would decrease it. Do you see that? What about KM? Yeah, so that's what the slide says, but what about why? Does that wait a minute? An inhibitor reduces KM, increasing the binding affinity. So when you think about KM, it's this equilibrium. We tend to favor the formation of enzyme substrate complex when we trap the enzyme substrate complex by binding to an inhibitor. So this is tending to make more enzyme substrate complex present. So it has the counterintuitive effect of increasing the parent binding affinity but decreasing Vmax. If I just keep adding infinite amounts of substrate concentration, everybody's like, yeah, party time, baby. There's lots of things for me to bind to. You can't saturate. These are not competing. You can't just keep adding more substrate and the inhibitor is more happy. And so Vmax is decreased. Do you see that? What if you have perhaps a more realistic scenario where both possibilities exist? This is called mixed inhibition. Mixed inhibition. So these are the curves that you get and you can go through it on your own time. Vmax is decreased as you increase inhibitor concentration and KM is also decreased. So that's diagnostic. That would be a guess at the mechanism. What if we have both possibilities? A tad of competitive and a tad of uncompetitive. Think about it. How might this affect Vmax? Competitive doesn't affect Vmax. Uncompetitive decreases Vmax. Unchanged plus decreased equals what? Decreased. Uh-oh. What about KM? KM is going in opposite directions for competitive and uncompetitive. So whether KM is increased or decreased would depend entirely on the nature of this equilibrium. If it's more like the competitive inhibition, then the KM would tend to be decreased. No, KM would be increased. If it's more like uncompetitive inhibition, KM would be decreased. This is thinking about it intuitively. You can do this nasty math and you could figure out perhaps using calculus that this is true as well. But common sense sometimes is useful. So if we do a curved line weaver-burke plot, this is the kind of plot you get, then that allows you to determine Vmax and KM. And noncompetitive inhibition is a special case of mixed inhibition where KM is unchanged. So we have one process pushing KM in one way, the other process pushing it in the other way. If the forces are equal, KM is unchanged and these lines would intersect at the x-axis. All right, so let's hit the clicker. This is a summary table. Very useful for exams. So if you forget the logical part of this, you can just memorize this table. I'm going to give you a little extra time. If you can't stay, I'll announce it on the video. You can watch the answer on the video. I think this actually requires a paper and a pencil. So I'll let you guys go a little bit longer. You seem to be having an active conversation. Is everybody done voting? Okay, I would like to give the answer. Has everybody done voting? If you haven't, did you turn in a paper? Stick now or ever hold your voice. See, almost all of you got it right. This is some noxious question. So Micaela Smithen has all kinds of these types of questions. So it's just, we have the equation up there. We've substituted in a arbitrary thing to get the answer. So the process is easy. It's just the calculation is difficult in your head.