 Okay, we're going to start talking about enzymes. But first, this is the histogram for midterm two. It's very impressive. One of two things happened. Either you guys did really good or G. and Mark and Stephen got paid off by a large number of you to grade it easily. I don't know which it is, but these are all A's over here. It's only a smaller number of B's and hardly any C's. Okay, so this is the first time I've been a professor in 20 for 22 years. This is the first time the mean on the exam has been an A. The mean was like right here. It's never happened before in one of my classes where the mean on a midterm exam is an A? No. So you guys did really, really well. And I didn't think the exam was particularly easy. I really didn't. Now, what does all this mean for your future? This is the how am I doing score calculated a couple of days ago. All right? I dropped your two lowest quizzes to calculate this score. I dropped them. Okay, so your quiz average could still go up a little or down a little based on what you did today. But basically, it looks great. All right, look at all these A's over here. Look at all these B's. All right, all these people over here are doing pretty well. You folks, don't give up hope. All right, there's a 200 point final. All right, so far we have fixed, the quizzes are worth 200. Midterm one was worth 100. Midterm two is worth 100. So we fixed 400 points in the course. Nothing we can do about those 400 points right now. All right, we've taken those 400 points. We've engraved them in stone. But we haven't said anything about the last 200 points. That's one-third of all the points in the class are still going to be decided on that final exam. So this person right here can move down. This person right here can move up. These folks over here are getting an A minus. They might be able to pull out an A. These people over here are getting a B plus. They might be able to cross over this B barrier to get an A minus. Lots of movement within this histogram is still possible. All right, that's the message I want to send to you. So if you're getting worn down, it's the end of the school year, it's June 1 today. You know, can you possibly carry on for another week? I'm going to implore you to focus on the final exam and try to do well. Okay, now, maybe for most of you, it just doesn't matter. You've gotten into whatever grad school you're going to get into or professional school, medical school, pharmacy school, you've got a job lined up for the summer, whatever it is, maybe it doesn't matter to you. And for those of you who are in that situation, do whatever you feel like you need to do. I don't blame you. I personally did not work that hard. My spring quarter, my senior year, I didn't really work that hard. But for those of you who care, how many people really care what their grade is going to be in this class? For whatever reason, rational or irrational? All right, so you can still change your grade. Everybody who had their hands up and everybody who didn't too, but now, don't take my word for this. All right, I calculated this and somebody sent me an email today and said, you know, how am I doing scores? All messed up, blah, blah, blah. I went into the spreadsheet and I double checked and I don't think there's any mistakes in the spreadsheet, but let me show you what I did. I took your four highest quiz scores, multiplied by eight because the quizzes are graded from one to five so that four times five is 20 times eight is 160 quiz points, not including the one we took today, of course. 160 quiz points plus 100 plus 100 is 360, right? That's how I calculated your how am I doing score, right? So this is the percentage of your total points where each quiz is worth 40 points, okay? That's all it is. So don't take my word for it, check your own how am I doing score because this spreadsheet, I'm only going to add one column to it and I'm going to calculate your final grade from that, right? This spreadsheet is your final grade once I add the final exam to that and quiz seven also, okay? So it's important that it be exactly right. So if you don't think it's right, come and see me because we need to change the spreadsheet somehow. Maybe a quiz got left out or it got the wrong score, whatever it is, all right, it should be self-consistent with everything that's in the grade book already. When I construct this spreadsheet, I go to the grade book, I pull a column of numbers out and I paste it into the Excel spreadsheet. So what's in the grade book is what's in my spreadsheet, okay? Any questions on that? So, you know, as far as I'm concerned, this is exactly where we want to be going into the final exam. This is exactly the histogram I was hoping we would have at the beginning of the quarter, you know, the mean is in here somewhere, all right? You guys are looking like you're doing really well, all right, barring some sort of cataclysmic breakdown, all right? The histogram is going to look a lot like this at the end of the quarter, okay? But that doesn't mean that there won't be a lot of flux going on inside this histogram. There will be, there always is, right? Somebody's going to have them melt down for reasons that might be out of their control, you know, and someone's just going to have a breakout day and, you know, both of those things are going to happen. Okay, so half of the final exam is going to be on kinetics, half of it. There's going to be four problems. Two of them is going to be kinetics. One of them is going to be stat mech. One of them is going to be thermodynamics, all right? That's all we have time for. It's going to be four problems, four multi-part problems. So I'm going to spell that out next week. I'm going to tell you what those problems are going to be, more or less, like I always do, all right? So you're going to have an excellent idea. But this has always been on the final exam, enzyme kinetics. I just like this subject. I think if you understand enzyme kinetics, you understand some very important basic chemical kinetics principles. Okay? Now, I know most of you have seen this before. How many of you have seen enzyme kinetics before? Good. All right? The way you really learn stuff is by seeing it more than once, all right? The first time you see it, it sinks in millimeters. And then it doesn't go any further until you see it again the second time, and then it seeps in a little further, you know, and eventually it goes all the way to the core of your brain stem, and you understand it at a very deep level, all right? I'm still not at that point, myself, but you need to see topics like this more than once. And you'll take away something new, maybe. I'll say it a different way. You'll take away something new even if you've seen it before. Okay, so we're going back, talking about unimolecular reactions. You'll recall that I said that most unimolecular reactions are of two types. They're either decomposition reactions. You've got a reactant, a bond breaks, and it breaks into two particles or three particles, all right? If that's a decomposition reaction, or it's configuration changes. So it undergoes an isomerization, all right? Those are the two most common types of unimolecular reactions. They're both important. We want to understand how they occur, all right? Because it's unusual for a unimolecular reaction to just spontaneously happen because there's an activation energy that's required for most reactions, all right? There's an activation energy. In an isomerization reaction, the energy of those two states might be the same, but there's an energetic barrier between them, cis and trans, for example, all right? And in a decomposition reaction, usually a bond has to break. That's a lot of energy, okay? So reactions like this, in general, just don't happen, all right, A just reacts to give products. No. If it worked that way, there wouldn't be any A, all right? A would all react. It'd be gone, all right? You have to take A concentrated into a container at some partial pressure so that it can collide with itself and then this reaction happens, all right? There has to be an energetic collision somewhere. So the Langmuir-Hinshel-Wood method gives us the most primitive way to think about this, right? This is not the most advanced mechanism. This is the most primitive unimolecular reaction mechanism. What happens? Basically, A collides with itself. That's the key. That collision is going to impart the energy necessary for this reaction to move forward over this energy barrier, okay? But after that collision occurs, that's the step right here why this activated A star can collide with A again and get deactivated. That's the step right here, all right? So we form A star by a collision, but if A star undergoes a second collision, the energy can get transferred. The energy necessary for the reaction could be lost in the second collision. Now, it's a little surprising maybe that that can happen. You'd think the second collision would make it even more energized and that can happen too, all right? We don't talk about it, but that can happen too, all right? But what matters to us is that some fraction of these collisions are going to deactivate the A star. It's going to go back to A and that's what this shows right here. And finally, the last thing that can happen is this A star, if it's not deactivated, it can undergo its unimolecular reaction. Here I'm showing a decomposition, all right? But it could also be an isomerization. There could only be one B instead of two here. Okay, so the key is that this A star has enough energy to undergo this reaction. And so we can apply the steady-state approximation because A star is an intermediate. If there's an intermediate in the mechanism, you can apply the steady-state approximation. If there's no intermediate, you can't apply the steady-state approximation, right? There has to be an intermediate in the mechanism in order for the steady-state approximation to make any sense. What's its steady-state? It's the intermediate for goodness' sakes. Okay, so because A star is an intermediate, we can apply it, so we write an expression for the differential rate expression for A star. Here's the rate at which A star builds up and this reaction right here. Here's the rate at which it decays and this reaction right here and this reaction right here. And then if we're applying the steady-state approximation, we just set that equal to zero. These are just the differential rate expressions also for A and B. And then we can simplify this. We can move all the negative terms on one side, all the positive terms on the other side, set them equal to one another, and then solve for A star. And then finally we plug the steady-state A star concentration into the rate expression for product. Okay? And so A star steady-state is here. I'm just going to plug this whole thing in for that. I've got 2K2 from here and this is just the expression I derive for the steady-state concentration of A star. And I'm done. That's my Lindemann-Hinshelwood reaction rate for the reaction. What does it predict? Because these predictions turn out to be very general. We want to understand. If I make A big, so if I look at this expression, if I just look at it, I can already see there's going to be two limits to think about. All right? How do I know that? And by looking at it, there's a plus sign here. Right? In the denominator, there's an addition operation. All right? And so immediately it tells me that if this is huge, compared to this, I can neglect this. And I'm going to see one type of kinetics. Conversely, if this is huge, compared to this, these A's are going to cancel. I'm going to see completely different kinetics. Right? The key is I look at the expression and I see the addition operation. Whether it's in the numerator or the denominator, that tells me that there can be two limiting behaviors that are important in the kinetics. So the first one is let's say I make A huge and I blow this term up until K2 is small in comparison. Then, this A is going to cancel with this A right here. And so you see, I got rid of one of these A's and all of these constants then are going to be rolled up into some effective rate constant. And this thing is going to look like it's just a first order reaction in A. It's going to look like a first order reaction at high pressure or high concentrations of A. Now, what does this mean mechanistically? Well, look, if I make A big, what that means is that this deactivation reaction here happens with high efficiency. In other words, I'm creating this energized A star, all right? But if there's a high partial pressure of A, it's going to, A is going to be colliding into A star. A star is not going to live for very long and we're going to go back towards de-energized A. The de-energization of A is going to be very efficient. If that's true, it's just like this forward reaction is in rapid equilibrium. This is happening and then this is happening right away. And then this happens, you generate an A star, but A star deactivates it. Activation, deactivation. Activation, deactivation. It's happening rapidly. Okay? So, it's like the first step is in a pre-equilibrium. We can think about an equilibrium constant for this forward reaction right here, all right? It would just be A star over A because the A's cancel, of course, if I write an equilibrium constant expression. Okay? And so the rate of the reaction then would be 2 times K 2 times A star, right? Except that A star is just equal to big K times A and so that's my reaction rate. And hey, look at this. That equilibrium constant is equal to K1 over K minus 1. Yes. That's what you would expect it to be. Okay? So, if that first reaction is happening rapidly in the forward and reverse directions, which it will at high pressure, that's, we're going to see first order kinetics, all right? And there's two ways to think about it, all right? But basically, the deactivation process becomes very efficient. What about a small A? If I make A small, then this term goes away. Now, I've just got K2 in the denominator. All right? It's going to be overall second order. And what does this mean mechanistically? It means basically I'm turning this off. I'm turning off the back reaction. I make A star and at low pressures of A, it's not going to get deactivated. Once it forms, it's going to hang around until it reacts. So it's like there's no back reaction. And under those conditions, we're going to see second order kinetics. Okay? So this plot must not be mysterious to you, all right? If you're looking at this plot and you're going, it's only one slide and the presentation's got 85 slides. I don't need to understand that particular plot because it makes no sense. You do need to understand it. What am I plotting here? This is log of K effective, K effective, that K effective right there, all right? So what we're doing is we're saying, look, take this Lindemann-Hinshelwood rate expression, that mess right there, all right? And pretend that it has this form, all right? Take one of the A's out because there's an A squared in the numerator and roll everything else up into K effective, all right? And that's the K effective that I'm plotting here, all right? We're pretending that the reaction is a first order reaction, so it's the first order rate constant for the Lindemann-Hinshelwood mechanism. And on this axis, I'm plotting the partial pressure of A. Actually, I'm plotting the log of the partial pressure of A, all right? And why do we call it log? Because we want to convey the picture that we're plotting A over a huge range of pressures, all right? From very low to very high. That's why we're calling this the log, okay? And so what we want to notice here is that there's two regimes, two. The first one is one where this is where we've got second order kinetics, okay? And so K effective is going to be K, K1 times 2 times A, all right? I think you can see if I plug this in for K effective, I get second order reaction kinetics, right? I get A squared, all right? So here, the rate constant, all right, K effective depends on A, that's why we're seeing this linear behavior like this, but when we go to high concentrations of A, we get first order kinetics, okay? So we have first order kinetics assumed here already and so the rest of this is just a bunch of constants, okay? So what we should see in the lab is we see the reaction act like a second order reaction and then as we increase the pressure, increase the pressure, increase the pressure, it rolls over and starts to act like a first order reaction. That's the experimental behavior that we expect to see. It's surprising that that level of complexity exists. We're talking about a reaction that is A goes to products. Full stop, that's the whole reaction, all right? But we still see this complex behavior. We see the apparent reaction rate change, right? The parent mechanism change as we change the pressure of A even though all that can happen is A reacts to give products, right? There's still complexity to the behavior that we see in the lab, even though it's the world's simplest reaction, literally, okay? So we did some more algebra. We can take this Lindemann-Hinshelwood rate and here's that effective rate constant that we were just talking about. Now I've written down what it is, all right? Here's what it is mathematically, okay? And then I can take one over that and I get two terms because there's two terms in the denominator so if I put these guys in the numerator I'm going to have two terms corresponding to these two guys. There they are, all right? And then the normal way to diagnose whether Lindemann-Hinshelwood applies to the mechanism that you're talking about is to plot this one over K effective versus one over A. You should get a straight line. Just looking at this mathematics here you should get a straight line with a slope of one over 2K1, all right? And that should be the intercepts. We should see a positive intercept here, all right? And so qualitatively this blue line is doing what we expect it to do, all right? Now there's a breakdown that is occurring here. We haven't talked about why that happens. Hopefully we'll get to that before the end of the class. Okay, so now this looks a lot like this. I think you'll agree, all right? We've got a forward reaction and a reverse reaction. Forward reaction and a reverse reaction. And then we've got what looks like a unimolecular reaction step at the end here, just like we do right here, all right? So this looks similar to this. And we already understand Lindemann-Hinshelwood pretty well. So qualitatively we're going to expect that to do something similar. We can apply the steady-state approximation again. We've been through this a million times. We're not going to talk specifically about this reaction but if you do the mathematics you find out that it behaves exactly the same way as the Lindemann-Hinshelwood mechanism. All right, we're going to zoom through that. And you can look back on these slides and work through those equations if you want to. But basically what we want to do is we want to apply these mathematics now to this reaction. This is the enzyme kinetics that we're trying to get to today. And we're going to look at this carefully, all right? This looks like Lindemann-Hinshelwood. This looks like this, all right? Forward reaction and a reverse reaction, a unimolecular last step that's irreversible, all right? This is the classical enzyme kinetics mechanism. So this is the enzyme. This is the reactant. We call it the substrate enzyme kinetics. So one of the things that you should understand about enzyme kinetics is that it's been designed to confuse you. We haven't called this the substrate before. We've called it the reactant, all right? But now we're going to, these are both reactants. All right, but what do we call a reactant when it shows up as a product? A catalyst for gosh sakes, all right? So by golly, that's a catalyst. All right, it's going to react with s to form an enzyme substrate complex. And the reaction occurs in this enzyme substrate complex and then the product is released. And the enzyme is regenerated at the end of the process. All right, that's the picture. We don't call this guy a product because he started out here, all right? So he's a catalyst. That's what the role of the enzyme is to do. The enzyme is a very efficient biological catalyst. All right, it is this enormous thing in most cases. It's a giant molecule with a molecular weight of 100,000 or more, all right? And so it's got this really intricate architecture, all right? And its job is to reduce the activation energy of the reaction, which it can do in any of several ways. Hopefully, we'll be seeing more about that. So schematically, here's the cartoon. If you Google enzyme kinetics and you look at images, you see a lot of cartoons that look like this, all right? Here's the enzyme. This underestimates its relative size, all right? The enzyme is this thing that's 100,000 molecular weight. The substrate can be a tiny molecule like ethanol or water. All right? It could be a small biological molecule with a molecular weight of 50, okay? So it's not comparable in size to this, like we're drawing here, all right? It's tiny, it docks in the enzyme. The enzyme in general recognizes the substrate molecule and binds it, all right? There's not any substrate molecule that's swimming around in there, not, right? Even a chemically similar one will not bind to the enzyme. The enzyme has, thank you, yes, specificity. Okay, so here's this binding process. This is the so-called active site of the enzyme. It's, this cartoon is meant to convey the idea that it's recognizing this green substrate molecule. Now chemistry happens, all right? A bond between these two things is broken perhaps, all right? And the two products get spit out and the enzyme is regenerated, that's the picture, all right? That's the picture that we want to understand. Okay, so in five minutes, we're going to snap through some slides and we're going to look at how these kinetics work. And then we're going to come back and do it again on Monday so that we, we're all clear on this because this is important. Here's our kinetic scheme, all right? If we look at the differential rate law for the product, I think everyone will agree that it's K2 times the concentration of this enzyme substrate complex, okay? Nothing surprising about that. We're going to apply the steady-state approximation. When we do that, we've got the rate at which the enzyme substrate complex builds up, boom. We've got the rate at which it decays, boom. We've got the rate at which it reacts, boom. All right, here's, here's that reaction rate, that back rate there, here's that forward rate there, all right? This depletes the enzyme substrate complex, this depletes the enzyme substrate complex, so those are minus signs. And this builds up the enzyme substrate complex, so that's a plus sign. So it's just the standard steady-state approximation. See how important the steady-state approximation is? You see how we're applying it indiscriminately without thinking about what the rates of various, how big these rate constants are? Because we know darn good and well, it's the steady-state approximation doesn't always work. We're not paying any attention to that, are we? Okay, now it's hard to know what the enzyme concentration is because it's changing. Why? Because the enzyme reacts to form the enzyme substrate complex. Okay, so we don't know a priori what the free enzyme concentration is, but we do know the total enzyme concentration, because we put the enzyme in the stupid beaker. So we know, if there's one thing we know, it's how much enzyme we put into the beaker. Total enzyme, that's E0, all right? That's the amount we put in the beaker, all right? And it can have two forms. There can be free enzyme or enzyme substrate complex. There's no other options for the enzyme. In this simplest picture, we'll talk about complexities later. All right, so the enzyme can be in two states, free or bound by substrate. So now we can express the enzyme concentration as the difference, all right? This is the total enzyme concentration. This is the amount of it that's bound to the substrate. Okay, so then I can just plug that in for the enzyme. All right, I'm just going to plug that expression in for E here, boom. Okay, and then, when I distribute that across this K1, I'm going to get four different terms, three of them negative and one of them positive, and then I'm just moving that positive term over, so I'm putting all the negative ones on one side, the positive one on the other side, and then I'm solving for the enzyme substrate complex, boom. All right, there's the expression that I get if I just solve for ES in this expression right here. It turns out to be pretty easy to do. All right, don't forget that we're talking about the steady state enzyme substrate complex concentration. All right, it's not any, it's not, this is not a totally general expression. It applies to the steady state approximation. Okay, so earlier we said this is the rate of the reaction, right? This is the rate of that last step, enzyme substrate complex times K2, so now I can just plug that in for that, boom. That's the rate of the reaction. And finally, we're going to divide the numerator and the denominator both by K1. All right, when I do that, I'm going to lose K1 here. All right, I'm going to lose K1 here, and I'm going to put K1 under these guys, boom, there he is, now he's gone from there and he's gone from there and that's the McHales-Menton equation, the most important equation in enzyme kinetics. So we just derived it, it's pretty easy. This mess here is called the McHales constant. We'll have more to say about that on Monday. But, all right, see you on Monday.