#11 Biochemistry Enzymes Iii Lecture for Kevin Ahern's Bb 450/550

Captioning provided by Dissability Access Services at Oregon State University. Dr. Ahern: Okay, folks, let's get started. We are rapidly making our way through enzymes and we're in very good shape with stuff so I will finish where I finish in the lecture is where the material will stop. That will likely be down in here. I'll repeat that. Where I stop today, that's where the material for the exam 1 will stop. I have tentatively put in a request for a review of session time for ALS4001 for Saturday at 3:00. That's my tentative time. I see frowns. I always see frowns. I don't see smiles but I see frowns. As I said, I will videotape that so if you can't make it, hopefully you'll be able to watch the videotape. As always, I recommend you come get your questions answered. That's why I have a review session. Watching other people's questions I don't think is necessarily the best way to review. If you have questions, come see me. I'll be more than happy to meet you and hopefully try to answer your questions. Last time I finished talking about perfect enzymes. Perfect enzymes are pretty remarkable things. We see, at least I hope you see why not all enzymes are perfect, that driving too fast can cause problems and that's certainly the case. We can imagine the perfect enzymes. We will see increasingly that enzymes are such powerful catalysts that cells really have to put a throttle on them. Cells need to control enzymes. We can't just let enzymes just go crazy and there are some very interesting strategies that cells use to control enzymes. We're going to mention one of those today. We'll see some others as we get going through the term. But enzymes, because they're so fast and they are so powerful, cells really do need to keep a handle on them. That's important for them to do. Turn that off, alright. So the first thing I want to talk about today is again related to some mechanisms of enzymatic action. These are for enzymes that bind to multiple substrates. Not all enzymes bind to multiple substrates. So a prime example is there's an enzyme in glycolysis called phosphoglucose isomerase and it binds glucose six phosphate and it rearranges it and makes it fructose six phosphate. No other factors involved. It's only one substrate bound by that enzyme. There's a substrate that comes in, there's a product that's released, but it doesn't have two things. Many and in fact most enzymes have at least two substrates that they bind to. A + B goes to C + D. We'll see some examples of those today. So when we have multiple substrates, one of the questions that arises is, "well is there an order or is there a specific way "in which these substrates bind?" So I will give you some examples of some interesting strategies that exists for enzymes. The first of these are what are called sequential displacement. Sequential displacement is a mechanism that really can be divided into two categories. One in which the order of binding of the substrates matters. The other the order doesn't matter. They're both called sequential displacement. So let's look at an example of an enzyme for which the order in which the substrates bind matter. We can see here that here's an enzyme that binds these guys over here. We'll specifically focus on pyruvate and NADH and if the enzyme doesn't bind the substrates in the proper order, the reaction will not go. My first question to you is can you imagine a scenario why that might be, based on things we have talked about with respect to enzymes? Why would this class of enzymes need to bind things in order? What have we learned so far about enzymes that would suggest that order might be important for an enzyme? Student: When it binds one specific molecule, it changes the shapes so that it combines to the other one Ahern: His answer is exactly right. The model in which the binding of one changes the shape of the enzyme so that now a binding set for the second one becomes possible. What was the model that we talked about that had that enzyme changing shape like that? The induced fit, right? So this model is very consistent with the induced fit when we had the induced fit, we had A, B and C. We saw A binding and that created a binding site or a better binding site for B and then for C and the enzyme could do its thing. Well in the case of this particular enzyme, and by the way, I'm going to show you a couple of diagrams. You're not responsible for the diagrams. I only show them to you so you can see something about how the enzyme works. So in this case, the enzyme binds to NADH first. The binding by the enzyme of NADH causes a shape change that now favors the binding of pyruvate. And after pyruvate has bound, the reaction is catalyzed. So very, very simple example of an induced fit happening when we have ordered binding of sequential displacement. The other mechanism that's involved or the other possible mechanism of sequential displacement is called random binding. And as its name would suggest, it doesn't really matter which order in which these bind. So if I had this reaction, creatine + ATP, it means that ATP can bind first, or creatine can bind first. It doesn't really matter which one is the first one to hit the enzyme. Now you might look at that and say, "does that mean the sequential displacement doesn't happen?" The answer is no. Does that mean that induced fit doesn't happen? And of course the answer is no. Induced fit happens essentially with any enzyme but the induced fit doesn't have to solely involve the binding sights of the substrates. In the first example, it did involve the binding sights for the substrates. You might say well how does the induced fit work on an enzyme like this? And my answer to that is that virtually every catalytic action is the product of induced fit. So when we see catalysis happening, that alone is pretty good evidence that the induced fit is occurring. So random binding and there are many enzymes that have random binding, it doesn't matter the order in which they bind. We'll talk about this enzyme later in the term. It's very interesting. It's a very important enzyme in our muscles. It's called creatine kinase. Creatine kinase makes this very high energy intermediate called phosphocreatine that as we will see has implications for us as we exercise. Yes? Student: Can you say one more time the relationship to the random model in regards to reduced fit? Ahern: So how does induced fit work with the random model? The induced fit works with the random model as it works with any enzyme model in the fact that catalysis is occurring. Catalysis depends on that induced fit. There are slight changes produced in the enzyme by the binding in the substrate. This may cause, for example, the active site to have a slightly different configuration. And that slightly different configuration might put different molecules together. I'll give you an example. So here's an enzyme that we'll talk about later in the term called hexokinase. Hexokinase I believe is a random, I'm not sure about that but we'll say for the moment it is because it gives a good example. Hexokinase is the first enzyme in the pathway of glycolysis and it's a really cool enzyme when we look at its structure. So hexokinase takes two molecules. It takes glucose and it takes ATP. You don't need to know that right now. So it has to bind two substrates. It binds one substrate up here, let's say this is glucose. It binds the other substrate down here. We'll say that is the ATP. And the binding of the two of them causes a conformational change in the enzyme such that these jaws literally close. So either one can bind. It didn't matter if they were actually change the shape of the binding site. But what changed was the shape of the enzyme. And so those jaws closing now bring the ATP very close to the glucose that's in the enzyme and a phosphate is able to move from ATP onto the glucose. That change then induce the enzyme's jaws to open up and let go. So now instead of having glucose up here, I have glucose six phosphate. And instead of having ATP down here, I have ADP. So we see that the shape of the enzyme changed as a result of the binding of the substrates. The binding of the substrate didn't affect the binding of the other substrate but in fact the enzyme. Does that make sense? So we'll see other examples of enzymes changing shape as a result of binding of substrates and as I say, virtually every catalysis we can think of, there's some changes happening in that enzyme that's making that possible and again, this is consistent with the idea that this is how enzymes are able to catalyze reactions so much faster than chemical catalysts can because the enzymes are flexible. So that's two models that are fairly straight forward to understand for binding of two different substrates. There's a third model and it doesn't really fit into sequential displacement. It also involves binding of different, of two substrates but in this case, it binds the substrates separately. One after the other. And in the process, the enzyme is continually changing its state. It's continually changing its state. What does that mean? Well, this class of enzymes are called double displacement enzymes. And in a double displacement enzyme, the enzyme is actually grabbing something from one of the substrates and taking it and exchanging it with another substrate. I'll explain this as I get going through it. But the enzyme exists in two states. So the enzyme that catalyzes this reaction is called by the general name of transaminase, transaminase. [spelling out transaminase] Transaminases are interesting enzymes in that they catalyze reactions that move an amine from one molecule and put it onto, in the place of an oxygen on another molecule. And the oxygen on the other molecule is moved back onto the amine of the first molecule. So there's literally a swapping that's happening of the amine and the oxygen. So I need to explain to you now how this enzyme works. How does this enzyme accomplish this? Because it does not bind both substrates and the same time. It does bind both substrates however. Well, let's look start out our enzyme in the state I'll call O. The O state, the enzyme has bound to it an oxygen. It's just carrying out here. So it's got this oxygen on itself. When it's got its oxygen on itself, it's looking to swap that oxygen for a nitrogen. So in this case, it binds to aspartate. When it binds to the aspartate, what the enzyme does is it catalyzes a swap. The oxygen that it's carrying, it gives to the aspartate and the amine that is a nitrogen that is on the aspartate, passes it off to the enzyme. Now the enzyme is in a different state. It's in what I would think of as the N state. The enzyme lets go of the aspartate except for it's no longer aspartate, aspartate has got an oxygen and that turned it into oxaloacetate. So we've taken a nitrogen and we've made it into an oxygen. That's the first part of our reaction. Now the enzyme is in the N state and it wants to bind to an oxygen containing molecule. It binds to alpha-ketoglutarate and it takes that nitrogen, I'm sorry that nitrogen that it has, passes it off and takes the oxygen away from alpha-ketoglutarate. That in turn converts alpha-ketoglutarate into glutamate or glutamic acid. So in essence what's happened is we have started with something that had a nitrogen, it ended up with an oxygen. We start with something that had an oxygen and it became a nitrogen. Now what I've essentially done, is I've taken aspartic acid, made oxaloacetate. I've taken alpha-ketoglutarate and I've made glutamic acid. So why would a cell want to do this? Well a cell might want to do this, imagine if you will, if it has way too much aspartate but not enough glutamate. This provides a nitrogen source that we can convert non-nitrogen containing alpha-ketoglutarate into glutamate and then the byproduct of that is oxaloacetate. I don't want to focus so much on the reaction as I do want to focus on the enzyme. The enzyme is existing in two states. It started in an O state, where it donated oxygen and grabbed nitrogen that converted it into the N state. And in the N state, it did the reverse. It grabbed an oxygen and gave up the nitrogen that it had and it moved back into the O state. So the exam is doing something that we call ping pong kinetics. It's going back and fourth between two different states. Once it's in the O state, it's ready to go back and do another one of these. So it goes back and fourth between N and O and N and O all the time. Does that make sense? Student: Ping pong kinetics. Ahern: Ping pong kinetics, yeah. Like a ping pong ball going back and fourth Student: Why doesn't it return the amine to the aspartic acid? Ahern: Why doesn't it do what? Student: Instead of carrying that ping pong ball over to the side, why doesn't it just... Ahern: Why doesn't it just put it right back on aspartic acid? Because once it's put on aspartic acid, it lets go of it. Aspartic acid goes out into the solution. So it literally will go from one molecule to the other molecule, back and fourth and back and fourth. Yes sir? Student: These productive reactants generally yield one specific chirality like in this case? Ahern: You always get the same chirality with this. All the enzymes that work on, essentially all the enzymes that work on amino acids are very chiral specific. You'll always get the same thing, yes. Yeah? Student: What about the transfer of the associated hydrogen. Is that actually... Ahern: Say that again now? Student: On the aspartate on the far left, on the reactant side, you have a hydrogen going back. Ahern: Yeah, I think that, that's a good question. That would be picked from a section I haven't shown you, but that would be picked up from the solution. Student: So does it pull the oxygen or the amine group off right away? Or does it wait for the other partner? Ahern: I'm not sure I understand... are you saying... Student: As soon as it binds to an aspartate, does it pull off the amine group? Ahern: Yes. So it's not waiting for alpha-ketoglutarate to come along. So that's the point here is it's not binding to both substrate. That's different from what we saw before. Because if it were binding both substrates, then we can imagine there might be an order or randomness to that. That's not what's happening here. So the enzyme is binding to one or the other. Once it's bound to one and does its thing, it lets it go. And that now leaves it a candidate to bind something else. These enzymes, by the way, tend to be very broad in their specificities. This could be aspartic acid, but this could also be asparagine. There's quite a wide variety of substrates they will accommodate and the idea is that these are very important in passing nitrogens from one amino acid to another. And nitrogen balance and nitrogen movement in the body is very important. Nitrogen is a precise resource. Yes ma'am? Student: So is it possible for the enzyme to bind [inaudible] acetate and invert it back Ahern: Okay her question basically is, "is this reverse action possible?" and the answer is absolutely. Any enzymatic reaction is reversible and what will be driving it is concentration. Student: Let's say an enzyme [inaudible] Ahern: It can happen with the same enzyme, yeah. It's independent of the enzyme itself. Student: What's the name of this enzyme again? Ahern: The name of this enzyme is called transaminase. Student: So just to make sure I understand, it binds to aspartate and removes the amine group. Ahern: Yup, binds to the aspartate and takes the amine off. Student: It kicks off what's left of the aspartate. Ahern: It puts the oxygen on aspartate. Student: Where did it get the oxygen from? Ahern: It's carrying it. That's in the O state. When the enzyme is in the O state, it has an oxygen on it. Student: which it got from ... Ahern: A previous oxygen. It's always been back and fourth between N and O. Student: Is there not sort of a chicken-egg question a little bit? Ahern: Not really because the first time the enzyme worked, it had to get one or the other. Then the rest of its life it's going to go back and fourth, back and fourth. So it's not really a chicken or egg question, though I see why you might think it is. We could debate that I suppose, right? Did the oxygen come first or the nitrogen come first? Extra credit, right? Other questions? Alright, so that mechanism is called as I said, ping pong kinetics. It's also called double displacement and both are equivalent. Now I mentioned in the beginning of the lecture that enzymes are, because they are so powerful and so able to make tremendous changes in very short time, cells have to control them. And there are about three different control mechanisms that I will talk about this term and we're getting ready to encounter the first of them and the first one is called allosterism. Allosterism. [Spelling allosterism] An enzyme that exhibits allosterism is called allosteric. And that's what you see right there. Right there. Now, what does this mean? I need to tell you what this control mechanism is. We'll talk next term about the synthesis of cholesterol. Cholesterol synthesis is very complicated. There are about 25 enzymes that are involved in converting a very small molecule ultimately by putting a lot of them together into cholesterol. Making cholesterol is very energetically intensive. It takes a lot of ATP to do that and so cells don't want to be making cholesterol if they don't need cholesterol. So they have something called a feedback mechanism and we'll talk about another feedback mechanism next week. But they have a mechanism called feedback in which cholesterol which is the end product of this very long pathway, binds to one of the first enzymes in the pathway and when it binds to that enzyme, it causes the enzymes shape to change very slightly such that it's much less active. Now I'm going to give you a definition of what I've just told you. Allosterism is a property where a small molecule interacts with an enzyme and affects its activity. It's a property where a small molecule interacts with an enzyme and affects its activity. In the example I just gave you, cholesterol is the small molecule. An early enzyme in the pathway is the enzyme and the effect is to reduce the enzyme's activity. Some enzymes will have activity reduced. Some enzymes will have activity increased. And we talk about this, we commonly talk about it as if we are turning on the enzyme or turning off the enzyme. You need to realize that we don't have absolutes like that usually in the cell. Usually we think of it more like the volume. We're turning up the volume or we're turning down the volume. And that's what happens usually with allosterism. Did you have a question? Student: Naw. Ahern: Or you just got tired? Student: Well here's the question. So it doesn't have to be the product to allosterically interact with... Ahern: So his question is "does it have to be a product to allosterically interact?" The answer is no, it does not. We will see an example where another molecule affects an enzyme and it's not a product in any way. Yes? Connie Student: When you say turn off and turn on vs. turn the volume off and the volume on, is each enzyme completely turned off? Ahern: Her question is a good one, you're getting a little ahead of me. "Can each enzyme be turned on or off?" and the answer is "no." Most enzymes, this may surprise you based on what I've been telling you, but most enzymes are not controlled. "That doesn't make any sense. "You just said enzymes are powerful, "they can cause problems, they could do too much "and now you're saying that most enzymes are not controlled?" And the reason that most enzymes are not controlled is because cells are very picky in the controls that they have. When we study metabolism, what we will see is that metabolism occurs in what we call pathways. Enzyme A converts molecule 1 into molecule 2. Then molecule 2 becomes a substrate for enzyme B that converts it into molecule 3. We can see a connectedness of these pathways. 3 goes to 4, 4 goes to 5, 5 goes to 6, etc. Cells are efficient in that they will frequently control the first enzyme in the pathway. If they control the first enzyme in the pathway, it doesn't matter how much of the other enzymes that you have because there's not going to be hardly an 2, or 3 or 4 or 5 or 6 there. So by controlling the major enzymes of the pathway, which are usually the first ones, cells are efficient in being able to regulate what enzymes do. You had a question back here? Student: What does it mean to say small molecules? Ahern: "What does it mean when I say small Molecule?" It's a non-protein. Molecules, we can think of these as substrates. Substrates in general are small in comparison to proteins. Cholesterol is a pretty good sized molecule but it pales in comparison to the size of a protein. You have a question here? Student: [Inaudible] Ahern: The question about what I mean by control. The answer is I mean they are not being able to have their activity regulated. On or off, up or down, whatever we want to say. Student: So a cell can turn off like you said but it won't necessarily turn off the first one, it can also turn off the third one because it wants to make whatever the first two enzymes are making? Ahern: Basically, her question is "does it always have to be the first enzyme that's regulated?" and the answer is it doesn't always. It's not always that case. We'll see an example when we talk about glycolysis. Glycolysis is the break down of sugar. It's a very odd pathway in its regulation. There are reasons why it is that way. But it doesn't solely regulate the first enzyme in the pathway. Yes sir? Student: When talking about cholesterol regulating one of its earlier precursor enzymes, would the terminology be negative feedback loop or down regulation or how is that generally described? Ahern: For cholesterol regulation? Student: Yeah for the one you described. Ahern: For cholesterol's regulation, the term I use is called feedback inhibition. But we'll talk more, actually, let me say, just to say that, I will talk about a specific feedback inhibition next week that will be a bit more relevant for us because we haven't talked about cholesterol. Good questions. Good thinking about this. I bring this up because when we look at the kinetics of allosteric enzymes, we see something very interesting. If we plot V vs S and we have an allosteric enzyme, here's what we see. What does this look like to you? Have you seen a curve like this before? You saw it for hemoglobin, right? You saw it for hemoglobin and, "but Kevin, hemoglobin isn't an enzyme!" What were we plotting when we were doing hemoglobin? This is why the axes are important. What were we plotting when we were doing hemoglobin? We're plotting oxygen concentration so that would behave like a substrate. That's similar. What are we plotting on the Y axis? You won't remember, that's fine. We're plotting what percentage of the enzyme is saturated is what percentage of the enzyme is bound with oxygen. Here we're plotting reaction velocity. Those are very different things, so why does this curve look like the curve for hemoglobin? Is it a coincidence? I'll give you a hint. It's not a coincidence. Any thoughts? Elizabeth? Student: [Inaudible] Ahern: Is it similar to the cooperative binding? There is some parallel to the cooperative binding. That's not really the answer to the why it is sigmoidal. There is similar behavior to cooperative binding. We'll see shape changes, we'll see that next week. I tell you what I'm going to do. I'm not going to answer that question. I'm going to leave you guys to think about that and that will be your extra credit on the exam. What was it? Why does this plot resemble the plot of hemoglobin? And I'm giving you a hint. It's not just the cooperativity. So that will not be the answer. I'll give you another hint. The answer lies in what the Y axes are telling us. In this case, we're looking at velocity. In the case of hemoglobin, we're looking at percent bound with oxygen. I've already told you your extra credit question for the exam. You can consult. I will not tell you the answer if you consult with me. I would ask you to not consult with each other. You want to consult with people, that's fine. But I want each person doing it on their own. Not a group project for example. It's kind of a fun activity. Now, the last thing I want to talk about relevant to our enzymes is a very important consideration for drugs and enzymes. So when we think about medical applications or medical implications of enzymes, we think about, "well how do I inhibit enzymes?" Because inhibiting enzymes is a way that many drugs will actually work. If we examine the different types of inhibition that can occur, there are three main ones that people talk about. And we're only going to talk about 2 of them in this class. We're going to talk about what's competitive inhibition and non-competitive inhibition. We're not going to talk about un-competitive inhibition because I find that students find it confusing I don't think we really need to go into that to understand the basic principles of inhibition. Let's think about what happens with competitive inhibition. The word competitive tells us something. Something is competing with something else. In the case of competitive inhibition, we have a substrate that looks like this in green. It's shown here in green. The competitive inhibitor is a molecule that in this case resembles the normal substrate but it has something different. It has something different about it. Now, it's similar enough to the normal substrate that the enzyme binds it without even thinking about it. It binds it. But when it binds it, it's just kind of stuck there. It doesn't do anything on it. Whereas the case of a normal substrate, the enzyme catalyzes a reaction on it. Now I want to emphasize that these two methods that you see on the screen are reversible. They are not covalent, they are reversible. That is the enzyme can bind the competitive inhibitor but it will also let go of it. If it bounded and did not let go of it, we would have a different kind of inhibition that I'll talk about later. But both competitive and non-competitive involve reversible inhibitors. The enzyme can bind them, the enzyme can let go of them. Let's think, so that's a general thing. I'll talk about the kinetics in a second. By contrast, there's a very different mechanism called non-competitive inhibition. And it's depicted on the bottom of the screen. In this case we have a, actually this is uncompetitive, in this case we have an enzyme that binds to a normal substrate, here's the same enzyme, but instead of the inhibitor resembling the substrate as it happened with competitive, the non competitive inhibitor binds to a different portion of the enzyme. And you can see what's happened with this non-competitive inhibitor. It is bound to a different portion of the enzyme. It didn't resemble the substrate at all and it caused the enzyme to change shape and the enzyme no longer functions. That's a very fundamental difference between competitive and non-competitive inhibition. In competitive inhibition, the inhibitors will almost always resemble the normal substrate. In non-competitive inhibition, there will almost always not resemble the normal substrate. And as a consequence, they work in different places. Competitive inhibitors work at the active site, non-competitive inhibitors work at other sites on the enzyme. Yes, sir? Student: Does the substrate have to be in the enzyme for the non-competitive inhibitor to work? Ahern: Does the what? Student: Does the substrate have to be in the enzyme for the non-competitive inhibitor to work? Ahern: Does the substrate have to be in the enzyme for the non-competitive inhibitor to work? No it does not. It may actually prevent in some cases. In this case, we see it bound. But it could for example prevent the normal substance from binding. Yes, sir? Student: [inaudible] Ahern: I'm not going to talk about the un-competitive because it actually is very different and involves action of catalysis. I can't do that here, no. I can tell you separately if you'd like to talk about. Student: Oh yeah, I had a question about it to. I wondered, does it require energy to reverse non-competitive than it would require to reverse competitive? Ahern: Her question is "does it require energy to reverse these?" And the answer is energy is not involved. Energy is not involved. So we're thinking of either competing for the active site or binding something else. That's the only two possibilities that we're going to consider here. Let's think about this. Here's an example of a competitive inhibitor. Methotrexate is a man-made drug. Dihydrofolate is a normal substrate for an enzyme that's involved in nucleotide metabolism. Our cells need to use dihydrofolate ultimately to make nucleotides. If I give cells methotrexate, which resembles dihydrofolate, that same enzyme that uses dihydrofolate will bind to methotrexate and will not function. They're competitive inhibitors. They both bind to the active site of the enzyme. The enzyme can catalyze the reaction using this guy. It cannot catalyze a reaction using this guy. If I treat cells with methotrexate, these cells will ultimately die. If I don't take it away. If I don't take it away, these cells are going to die. Methotrexate is used in some types of chemotherapy. Because cells that can't make nucleotides can't divide and die. If I just gave it to a person and I didn't give them anything else, what's going to happen? They're going to die and that's the end of it. However, if I give them methotrexate for a short period of time, ala chemotherapy, and then take it away by flooding the cells with a normal substrate, then what happens is the cells that divide the most rapidly are the most sensitive. They have the greatest need for nucleotide, this might be a cancer cell for example, and so I have selectively killed rapidly dividing cells like cancer cells. I may also kill other cells that are rapidly dividing like hair cells or intestinal cells. And so use of this drug may have very nasty side effects for people taking chemotherapy. There's many other examples of chemotherapy but you can see how and why people might lose their hair, why people might feel very nauseous because they're losing intestinal lining and it's not being replaced properly. But this drug is very effective on some types of rapidly growing cells. Methotrexate is used for other purposes as well so I don't want to say it's only a chemotherapy. Dose is going to be important. Dose is going to be important. Small quantities that's used to treat certain problems. That's a competitive inhibitor. Let's think about what happens with a competitive inhibitor. This figure I don't like. It's too confusing in my opinion, so we're going to make it simple. We're going to focus on this green line and this black line. And we're not going to focus on these other two green lines. We're also not going to pay any attention to this up here. I'm going to talk you through what happens with competitive inhibition. Alright? You've seen that enzymes that are not allosteric enzymes behave with a hyperbolic plot when I do V versus S. Velocity versus substrate concentration. Relative rate and velocity. If I take, and I'm going to go back and describe that reaction that I described to you before where I did my experiment. I said if I do a V versus S plot, what did I do? I took 20 tubes, I put the exact same amount of buffer, the same amount enzyme and I put varying amounts of substrate. Each tube had a different amount of substrate. And why did I put only the substrate varying? What did I say? I only want to have one variable. If I had 2 variables, I've got a problem. So I want to measure the inhibition of this enzyme using an inhibitor. Am I going to put varying amounts of inhibitor or am I going to put the same amount of inhibitor? I'm going to put the same amount of inhibitor, right? Each tube is going to get the same amount of inhibitor and the only variable is going to be my substrate concentration, right? Let's think about this for a second. Let's imagine when I start my experiment, my tube number 1 has, let's say, 10,000 molecules of substrate. And into each tube that I'm doing of my 20 tubes, I put in a hundred thousand molecules of inhibitor. That tube #1 was going to be the most likely thing that the enzyme is going to see. It's going to see the inhibitor because I've got 10 to 1 times the inhibitor, right? Tube #2 I increase the S. I increase the S and now I've got 20,000 molecules of substrate and 100,000 molecules of inhibitor again. It's going to be twice as likely as the first tube that the enzyme's going to find substrate but it's still going to be much more likely it's going to find inhibitor, right? I keep adding, I get to the 20th tube, and by the 20th tube, I've got 2 million molecules of substrate and 100,000 molecules of inhibitor. Now what's going to be the most likely thing the enzyme's going to find? It's going to be the substrate by a factor of 20 to 1. Right? If I go far enough out, I might have a thousand or ten thousand times as much substrate as I have inhibitor. 99.9% of the time, the enzyme, when I get to high substrate concentration, is going to be finding substrate and .1% of the time, it might be finding inhibitor. Everybody follow that? At very high concentrations, that's what's going to happen. Just a second, okay? What's going to happen to the V max? It turns out because I'm competing, when I'm competing, my substrate is winning the race. It's out competing the inhibitor because the concentration factor is beating it out. Everybody follow? In essence, 99.9% of the enzyme out here is active. I can't tell them apart from 100%. I can't measure that accurately. So when I'm measuring competitive inhibition, V max stays the same. It stays the same. It might take awhile to get out here but ultimately it will be out here. Did you have a quick question? Student: [Inaudible] Ahern: "Do we have to account for the different affinities between the inhibitor and the enzyme?" At one level yes, but not for a V vs S plot, no. If I said to you, if I look at tube #1, what percent of the enzyme would you say is inactive most of the time? I had 10,000 versus 100,000. I might only have about 10% of the enzyme active. Right? Because 90% of it's going to bound with an inhibitor and when it's bound to an inhibitor it's not doing anything. Everybody follow? I might expect that that velocity would be lower because velocity is going to depend upon how much enzyme they got active. Out here, I got 99.9% of the enzyme active. It doesn't matter. It's essentially 100%. I've out competed the inhibitor and that's the important part of competitive inhibition. Competitive inhibition, the substrate outcompetes the inhibitor ultimately at high concentration. The substrate wins. So V max does not change if I compare uninhabited vs inhabited enzymes. What happens to KM? Good thing to think about KM. What's going to happen to KM? Is it going to be the same also? Nobody wants to stick their neck out for that one. Let's think about what this means. Here is our V max of 100 let's say. That means half of V max is going to be right here at 50, right? Because, yeah, half of V max can be there because KM, I have to measure the velocity of V max over 2. So there's V max over 2 right there. What is the substrate, what is the KM for the uninhibited reaction? It's right here. What's the KM for the inhibitor reaction? It's over here. What has happened to the KM? It's gone up. It's gone up. So when we have competitive inhibition, KM increases. V max does not change. Everybody with me there? Questions? With competitive inhibition, V max does not change. KM increases. Now I think you should be able to logic that out in your heads. I think that's important to be able to think through because it saves you having to memorize something else. Now, this stands in contrast to non-competitive inhibition. Let's look at non-competitive inhibition. Non-competitive inhibition, they're not competing. I told you in the case of competitive inhibition that the reason that the substrate won was because the concentration of the substrate was outcompeting the inhibitor. In non-competitive inhibition, the name tells us they're not competing. Nothing stops the non-competitive inhibitor from binding to the enzyme site in the first case. The substrate could outcompete the inhibitor. They were fighting for the same place. In non-competitive inhibition, we don't see that fight going on. Let's imagine, if you will, in non-competitive inhibition, that I have 100,000 enzyme molecules in each tube. Every tube has the same amount of enzyme. And I put in my 10,000 molecules of inhibitor. What's going to be the percentage of the enzyme that's going to be inhabited in tube A? What percentage will be inhabited? 10%, right? One in 10 will be inhabited. 10,000 of those are going to bind to those 100,000 enzymes. 10,000 enzymes are going to be inactive and 90,000 are going to be active, right? What's going to be the case in tube #2? Exactly the same thing. Because I had the same amount of enzyme, I had the same amount of inhibitor and then tube #2, I've got 10% of my enzyme is inhabited. And tube #3, tube #4, tube #5, and all the way up to the highest concentration of substrate that I use, I always have 10% of my enzyme knocked out. Now we talked about V max, I said something about V max that it wasn't specific for an enzyme. What did I say about V max? What did it depend on? The amount of enzyme that I had. Right? So if I did a reaction, I did a series of V vs. S with a given amount of enzyme, I would get my plot. I would get a V max. If I did a different set of reactions where I only used 90% of the enzyme, would my V max be the same? It would be lower. Non-competitive inhibition, the V max is lower because you have a fixed percentage of the enzyme that is inhabited. The other side of this will surprise you. And I'm not going to go through it here, but the other side of it is that non-competitive inhibition, the KM stays the same. How can that be the case? Well, let's think about it for a second. Do you suppose what I said about KM, is it a constant for an enzyme or not? Yeah it is. Yeah it is. KM is a characteristic of an enzyme. For a given set of conditions, the KM will be the same independent of how much enzyme I use. That was one of the beauties of KM. It was a characteristic of an enzyme. I can compare KM values between enzymes. In fact, we did that the other day. We compared KM values of enzymes. So the KM I get if I have 100% of the enzyme will be exactly the same KM I would get if I had 90% of the enzyme. KM does not change in non-competitive inhibition. Now the last thing I want to say and I know it's a little rushed but I want to get it in here for this exam. Student: On that graph... [inaudible] How would you... Ahern: Well, I can point, this one has got a lot of inhibition that's gone with it. So I don't like the way this graph is drawn. Come see me separately and I'll show you. But the argument is that the KM independent of the amount of enzyme that I use. Student: Yeah, I understand. It was just the graph. Ahern: Now, you've learned about Lineweaver-Burk plots. I want you to think about what those mean relative to these types of inhibition. Here is a Lineweaver-Burk plot that shows an uninhabited reaction in black and an inhabited reaction in red. I've plotted the same data, taken the inverse of everything, and look what's happened. They cross at the Y axis. If you recall, the place where a Lineweaver-Burk plot crosses on the Y axis is one over Vmax. We're looking at competitive inhibition because Vmax doesn't change. Competitive and non-competitive will have exactly the same value at Vmax. What does change is KM. KM gets larger and that means minus 1 over KM gets closer to zero. This plot is what we would see for competitive inhibition if we did a Lineweaver-Burk plot. By contrast, if we did a non competitive inhibition reaction and compared it to a normal reaction, we see that the two lines cross at minus 1 over KM. That's not surprising because the KM values for those two are the same. Why is the red higher? We're plotting one over Vmax, not Vmax. One over a smaller number gives you a larger number. So this is what the plot looks like when I do that reaction for a non-competitive inhibitor. Questions on that? Shannon? Student: Can you say where that crossover is? Ahern: Minus 1 over KM right there. Because KM is the same for the two. That's where the material for the exam will stop and I will see you guys on Friday. I will announce the time of the review session on Friday for sure also. [END]