Lecture 25. Enzymes (second & final attempt)
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Transcript: English(auto-generated)
00:05
Okay, we're going to give enzymes another try today. So there's a mistake in the way I calculated the how
00:24
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
00:43
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
01:01
to check your how am I doing score and found this error. Okay, quiz 7 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.
01:23
Quiz 6 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 re-graded all
01:41
of these and reposted them and so the quiz scores that are up there for quiz 6 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
02:01
and a few of you have worked them out for me, but so far just 16 percent 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.
02:26
Let me explain the situation as follows. I teach general chemistry, analytical chemistry, physical chemistry. When I teach general chemistry and I get evaluations
02:45
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
03:01
or an intermediate guy, they just know they can't hold their feet up for 30 seconds. They have no experience with drill instructors and so when I get feedback from them 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,
03:24
something wrong with your pants or you guys are like these guys right here, your grizzled veterans. Most of you are graduating in a week
03:41
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. I know there's issues with this class and it would really help me out if you went through
04:01
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.
04:21
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,
04:41
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. It turns out all of this stuff is in Chapter 21 of your book and a chapter that's called Catalysis.
05:00
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, thermal, kinetics that are the most important topics I think. So this is one of those topics.
05:20
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
05:41
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
06:02
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
06:21
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
06:42
of bond breaking reaction occurs. 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.
07:01
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. 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,
07:22
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
07:41
but hopefully it will get us through this lecture. Here's the mechanism I showed on the previous slide. Here's the rate of the reaction. I'm just depicting here the rate at which P is formed. I think you can see that it's a unimolecular reaction from the enzyme substrate complex with a rate constant K2.
08:06
We're going to apply the steady state approximation again. It's another example of that and to do that we set the time rate of change of the intermediate ES equal to 0. So I think you can see there's a rate at which ES builds up
08:23
and two ways at which ES is consumed. 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
08:40
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. Presumably the enzyme is going to get bound, form the enzyme substrate complex
09:01
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. At least if we're studying this reaction
09:21
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. Now in a natural system 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
09:43
anything, we can't study enzyme kinetics under those conditions in general. 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.
10:01
So presumably we know this E zero 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
10:22
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 and when I distribute this K1 over these two terms I'm now going to get four terms instead of three, one, two, three,
10:42
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, 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
11:04
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. We're assuming that the steady state approximation is correct.
11:23
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
11:44
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. So this is the new expression for the rate of the reaction
12:02
that I get, still subject to the steady state approximation and now I'm going to roll all these constants up and call them KM, K big M. All of those guys together are going to be the Michaelis constant and this equation is the Michaelis-Menten equation.
12:24
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
12:42
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. We need to do some more work.
13:03
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.
13:22
If there's an addition operation in the expression for the rate, there's going to be limiting cases. 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 the denominator. What are the corresponding limiting cases?
13:41
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.
14:04
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
14:21
to be K1 times E0, the initial concentration of enzyme times the substrate concentration. Yes. Did everyone hear that?
14:41
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.
15:20
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
15:42
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?
16:01
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
16:21
to be quasi-constant because it's very low. This also means that E is approximately equal to E zero because ES is approximately zero, right? So essentially what this means is that the formation
16:40
of the ES in this first reaction here is the rate limiting step in this sequence of reactions. 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
17:01
in substrate, right? In this limit of high K2 the reaction is first order in 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 parenthesis here, right, and I just end up with this expression right here and S is going to cancel
17:22
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, right? It only depends on the initial enzyme concentration, all right? In the limit of large S, large concentrations
17:42
of substrate, the reaction rate is constant, right? You don't see any change in the, 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
18:04
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?
18:20
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?
18:40
This means that the reaction of ES 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.
19:00
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
19:23
substrate, equilibrium lies way over here, all right? And it's a bad enzyme. It's not a very good enzyme. Enzyme is not recognizing the substrate. Maybe the substrate is not the normal substrate for that enzyme. The reaction rate is just slow, okay?
19:43
Enzyme, 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.
20:03
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,
20:21
so we know what this reaction is going to do as a function of S. All 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,
20:43
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.
21:03
In that limit there's a maximum reaction rate V max that's equal to K2 times the total concentration of enzyme that I started out with in this reaction. So this is what any enzyme will do as a function
21:21
of the substrate concentration. That's what this equation predicts. It's also what's experimentally observed. Now, there's some other things that are indicated here that I'll explain to you.
21:42
In the limit of large S we obtain the maximum rate V max. I already showed that to you. Here's the V max 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 V max.
22:02
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 E is a V max, yes.
22:25
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. V max over E0 is given by this expression right here
22:41
but we also call V max over E0 the turnover number. I think you can see V max over E0 is obviously equal to K2. What are the units of K2 going to be?
23:01
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 and so that's okay. So the turnover number is going to have units of per second.
23:22
Reactions per second is what it represents. Okay, so V max over E0, I mean I just, so if we work out the units, this just shows that, V max is moles, molar per second if you will.
23:41
That's the reaction rate. The concentration of enzyme is molar so the units of this quotient here are 1 over seconds. We're going to call that K catalysis or K2 or the turnover number. All three of those things are the same.
24:04
If someone says the turnover number they're just talking about K2. If someone says K cat 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.
24:22
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 how I was calling it DP DT?
24:42
No difference. I'm going to just call it V, the velocity. Divide that by V max. Remember V max is just K2 times E0. 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
25:04
and if I divide by V max 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
25:21
and do some more algebra on it to get an equation that is not called the Lineweaver-Burk equation. It should be called the Lineweaver-Burk equation but it's not but if you make a plot using this equation it is called the Lineweaver-Burk plot.
25:41
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,
26:02
that's the vertical axis, horizontal axis rather. That'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 and I've never noticed that.
26:23
To be clear we're talking about the initial substrate concentration and the initial velocity. So you take your enzyme solution in your pH controlled buffer, you add some substrate
26:42
and you measure usually spectroscopically how fast the products are formed as a function of time and you extrapolate to zero time and you measure the reaction rate at time zero. Of course you're measuring initially the reaction rate over a range of times but you extrapolate to zero
27:03
to get the initial rate. We call that V zero and it applies of course to the initial substrate concentration. That's usually what we're plotting in a Lineweaver-Burk plot. Here's what it looks like. 1 over S, horizontal axis.
27:21
1 over V, vertical axis. Notice it says V zero. That means the initial velocity. This should really say S zero. That's the initial substrate concentration. 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
27:41
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,
28:00
I get minus 1 over KM from the not Lineweaver-Burk equation. So this is an extremely useful equation. The Michaelis-Menten equation, no, not so much.
28:23
Not useful except that I can derive the not Lineweaver-Burk equation from it and that's really useful. Here's a problem from last year's final exam. The following results are obtained from the substrate concentration reaction rate.
28:45
This should say VS zero. This should say V zero. Which substrate concentration? Is it constant during this whole reaction? Not necessarily. It's going to be consumed as a function of time. Is the reaction rate going to be constant as a function
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of time during this reaction? Not necessarily. We're really talking about the initial substrate concentration and the initial velocity. Here's some data. Tell me about this enzyme. I want to know everything about its kinetic behavior.
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Well, we can get everything. In other words, we can get KM and we can get E zero and we can get V max by making a Lineweaver-Burk plot. So to make a Lineweaver-Burk plot we don't want S. We
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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. Here are the data points you're going to plot. You're going to draw this axis because you know you don't want it here.
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You're making a Lineweaver-Burk plot. You want it in the middle somewhere 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.
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So we can grade it. 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. So maybe you can use the edge of your notebook
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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. So now that I've got this I can determine V max, Km, K cat,
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all of these things, everything that you can possibly determine in terms of the observational kinetics of this enzyme we can extract from this Lineweaver-Burk plot. Getting the right answer starts
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with getting the right Lineweaver-Burk 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
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so far in another class? Now what could possibly mess this up? Well, an inhibitor, a molecule that inhibits this reaction could mess it up.
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Inhibitors are extremely important in enzyme kinetics. This is the beauty of the Lineweaver-Burk plot. How the Lineweaver-Burk plot changes with and without inhibitor is diagnostic of how the inhibitor is acting on the enzyme.
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It's a thing of beauty. You can measure the Lineweaver-Burk plot without the inhibitor, add the inhibitor and look at what happens to the Lineweaver-Burk plot, add a little more inhibitor, add a little more inhibitor, of course you have to measure the initial velocity
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of the reaction at a series of substrate concentrations for added inhibitor 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.
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Is this another example of confusing enzyme terminology? Yes, non and un are not the same. I don't know who came up with this stuff. Okay, so what's the difference?
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It's very simple. In competitive enzyme inhibition the inhibitor is binding at the same active site that the substrate wants to bind to. If the inhibitor is in there the substrate can't bind. It's blocking it. That's competitive inhibition. The word perfectly matches what's happening.
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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.
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Finally, in uncompetitive inhibition the inhibitor is binding away from the active site but only at the enzyme substrate complex, right? So there's a subtle difference, all right? Non-competitive and uncompetitive. In non-competitive the inhibitor combined either
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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.
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Good luck. It's hard to remember. All right? Here's a cartoon. Here's the inhibitor. Here's the substrate. The inhibitor's blocking the active site. That's this piece of the pie here. Once the inhibitor's in there, the substrate can't bind. Obviously the reaction slows down, but in a very characteristic way.
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It slows down 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
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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.
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If there's a thousand substrates for every inhibitor 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
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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
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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.
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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.
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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 1 over V max. It does not change for competitive inhibition. So if I add in, here's no inhibition.
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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 Kms. There's Km, 1 over Km is down here. Minus 1 over Km, 1 over V max. All right? And I can look at this line weaver Burke plot and say boom.
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It's competitive inhibition obviously. 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? Well in this limit, here's the inhibitor.
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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.
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And in that limit what happens is Km is unaffected but V max gets reduced. All right? So this is 1 over V max. Remember? This intercept is 1 over V max. And so as I increase the inhibitor concentration, Km doesn't change but the intercept changes
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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 Burke plot allows us to diagnose immediately which of these two cases is operating.
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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?
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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.
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That slope is preserved and I get a different intercept for 1 over V max, a different intercept for 1 over KM but the ratio stays the same. Isn't that amazing? So I can classify if I am willing to do the work
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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.
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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.
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It's not in your book. It's not even in Leninger. It's value added that you can't get very easily anywhere else. All right are you ready for this because I'm not going to laboriously do all this math.
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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, reversible formation of the enzyme substrate complex, reaction of the enzyme substrate complex to give products. Now we're going to add two other possibilities. The enzyme can reversibly form a complex
41:26
with the inhibitor that's I and the enzyme substrate complex can form a complex with the inhibitor as well. So there's two new possibilities that we have to consider. These two new possibilities encompass all of the things
41:42
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,
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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.
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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, right, the equilibrium constant for the inhibitor interacting with free enzyme, that equilibrium constant.
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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
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that long before we get to your equation 21.8B. All right, here's the non-Leaver-Burk equation. Here's the non-Leaver-Burk equation for enzyme inhibitors.
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All right, check this out. I've got an alpha prime instead of a one. I've got a K times the Michaelis-Menten 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.
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All right, this equation explains those three behaviors that we just talked about. How does it do that? First of all, notice that if K alpha, sorry, if alpha prime is one and alpha is one,
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then the inhibitor is exerting no effect. I just get the non-Leaver-Burk 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
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than one, alpha prime equals one. V max does not change, but KM gets bigger. This table is a thing of beauty. It's also not found in Leininger. 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.
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V max 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
45:06
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
45:24
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.
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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
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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.
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All right, that's the succinate after it gets dehydrogenated. Succinate dehydrogenase, succinate dehydrogenase will form fumarate, form this double bond. All right? The classic competitive inhibitor is melonic acid.
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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.
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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?
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Because on Wednesday we're going to talk about something else. Okay, look at chapter 21 and then see
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if you have questions about this stuff.