tinywow_video_to_text_65331326.txt

All right, so let's talk a little bit about enzyme connect, forget the math of it and stuff like that just a little bit. But before we get there, we've got two models of enzyme connect we want to talk about. So the lock and key model, which is kind of the older, more established model and then the i...

All right, so let's talk a little bit about enzyme connect, forget the math of it and stuff like that just a little bit. But before we get there, we've got two models of enzyme connect we want to talk about. So the lock and key model, which is kind of the older, more established model and then the induced fit model, which is the now more accepted model that makes up for some shortcomings that we get in the lock and key model. Now on the lock and key model, the enzyme and the substrate are complementary. They bind each other. The substrate fits into the active side of the enzyme perfectly, so which is why the enzyme would want to bind to that substrate. So when it binds the substrate, then it converts that substrate into a product and releases it. So now that it's free again, it could find another substrate bind it and convert it to product over and over again. Again, those enzymes being catalyst are not consumed in the reaction. Now turns out that's the lock and key model. Now the problem with this model is it turns out that the process of binding should actually lower the energy. So and that would actually, if you lower the energy of this complex, then it would make the conversion and the next step actually have a higher activation energy, which is obviously defeating the purpose. We want to lower activation energy overall, not a higher activation energy. And so this enzyme substrate complex, as we call this, so we shouldn't want to lower its energy, which is what the lock and key model would predict. And so instead they come up with this induced fit model. And so the enzyme in the substrate, well, it turns out in this model the active side of the enzyme is not a perfect fit for the substrate. There's some complementarity, so it's not a perfect fit, but the process of binding causes the enzyme to undergo a conformational change so that it now is induced to have a perfect fit. And so the binding of the substrate by the enzyme actually causes the enzyme to change its shape just a little bit. And so that now the active site is perfectly complementary to that substrate. So and then it converts it to product and releases it and again, it's free now to bind more substrate and catalyzes reaction over and over again. And so it turns out this idea that the substrate enzyme aren't a perfect fit, but are made to be a perfect fit or induced to be a perfect fit, it avoids this idea that you know, this enzyme substrate complex would lower the energy as much, which would then raise the activation energy for the catalytic stuff and stuff like that, it avoids that entirely. So and is generally a more accepted model. Now the truth is all the enzyme kinetics, you're typically going to study in your undergraduate weather biology course or undergraduate biochemistry course. They're usually a little bit simplistic and these get way more complicated, but we're going to keep it simple. We're not going to go any further than you need to for the DAT or the OAT. All right. So now we're going to talk about Michaelis met in kinetics, the most common model presented. It's not the only model. There are other ones out there. There are more complex ones out there and stuff like this, but it is kind of the kind of leaping off point, good places to start in talking about enzyme kinetics and the one most commonly presented both in biology and biochemistry in your undergraduate experience. And it breaks up catalysis into two steps. So in the first one, here step one is binding of the substrate. So the enzyme is going to bind the substrate forming what's called an enzyme substrate complex. So and then you've got the catalysis step to step two. So and that's where the enzyme substrate complex, that substrate is going to be converted into a product and be released from the enzyme as a process. And so it breaks up any kind of enzymatic reaction that follows Michaelis met in kinetics is broken up into these two steps. We look at the first one as being reversible, you can bind and you can unbind. So but we view the second step as being irreversible. It's only going to go in the forward direction again for more complex models. It turns out that's not necessarily always going to be true, but it's a great model for many enzymes. So so one that's worth studying. All right. So if you take a look at the equation involved in Michaelis met in kinetics. So the velocity of catalysis is equal to the maximum velocity times your substrate concentration all over Km, which is called the Michaelis constant plus the substrate concentration. And so in this case, one thing you should know about the Michaelis constant, that Michaelis constant is inversely related to binding. And so it turns out a larger value for the Michaelis constant Km is going to mean lower affinity for the enzyme to the substrate. So and vice versa. So a smaller value of Km would actually imply a larger affinity for the enzyme for the substrate titer binding, if you will. So definitely want to know that about your Michaelis met in constant Km there. All right. So if you plot this and you plot velocity here versus your substrate concentration, you end up with a graph that looks like this here. And so initially at low substrate concentrations on the X axis here, so your velocity increases linearly. So we can see why that is so it's low substrate concentrations. The substrate might be a lot lower than Km and in the denominator, if you're adding a much smaller number over here than Km, then it's not really going to affect much. And so your denominator is essentially going to just reduce down to Km plus a really tiny number, which is still Km. But in the numerator, so we can see that your velocity is proportional to your substrate concentration. If you double the substrate concentration, the numerator there would double the value of the velocity under these conditions. And so as a result, that's why you get this kind of linear graph, because velocity and substrate are proportional to the other at low concentrations. So but as your substrate concentration gets higher and higher and higher, then all of a sudden this substrate concentration denominator becomes significant. So but you can get to some point where actually you get so much substrate that now substrate concentration is much bigger than the Km value, in which case now it would be Km, that's going to be insignificant. So it really high substrate concentrations way over here on the graph. It's now Km that's insignificant. And so the denominator is just going to be a small number plus a much larger number. And so it's just going to reduce down to being just plain old substrate concentration that Km would now be insignificant. And so as a result, you'd have substrate concentration numerator, substrate concentration denominator and they cancel and you'd just be left with V equaling Vmax. And that's why the graph asymptotically approaches this max velocity right here for that enzyme. That's kind of it. So it's a typical velocity versus substrate concentration curve. You can approximate Vmax right where it's asymptotically approaching. So there's another important piece of information you can elucidate as well. And that's the point where you reach half the maximum velocity. And so it turns out that Km itself knows if you're going to add Km and Fs, they must have the same units. Well, substrate concentration has units of concentration. Well, it turns out that the Km value also has units of concentration. The only way they can be out of it. And it turns out if you choose the substrate concentration that happens to so be the Km value for whatever enzyme substrate combination you're working with, all of a sudden now you end up with Km plus Km and then you have a Km right there as well. Well, Km plus Km would be 2Km and then you have a Km numerator and a Km denominator and those cancel. And you just end up with Vmax over 2. And so it turns out when your substrate concentration is equal to the Km value, that special value, so special value of your substrate right here on the graph, that's when your velocity is going to equal half the max velocity, also something you should be able to infer. So here's our max velocity, exactly halfway up to that point would be here and we can always go over to the curve and drop down and be like, oh, that's where the substrate equal to Km value. That's how you can approximate the Km value from just such a graph. Cool. Odds are you're probably not going to be doing any calculations with this, with like this Michaelis-Mett equation. And it's probably not worth memorizing and stuff like this. But from a curve like this, you should be able to approximate both Vmax. So as well as the Km value. So just as we've just done. Okay. So here we've got one additional plot here. So we've still got the velocity versus substrate curve right here, same one we had in the last slide. So we've got another new one. This is the double reciprocal plot. It's also called the line weaver Burke plot. So same depth, you're probably more likely to see this is the double reciprocal, but I'll put both names up there. So, and what's nice about this graph right here, so when they call it the double reciprocal plot, because if you take them a K-less-Mett equation and inverse both sides. So you've got V equal. So instead of V, you're going to have one over V. And instead of equaling Vmax times S over Km plus S, you're going to invert that side and have Km plus S over Vmax times S. And once you do that, you can actually rearrange this a little bit to get a linear plot of one over V versus one over S, instead of V versus S. And because it gives you a linear plot, well, conveniently, we can find the slopes and y-intercepts and x-intercepts, technically, for linear plots. And it turns out the lovely constants that are relevant here are related to these plots. And this is which is the part you want to know. So again, you're probably not doing any crazy calculations here and stuff like that. But what you want to do know is that when you do a double reciprocal plot of one over V versus one over substrate concentration, you will get a straight line. So a couple of key points. So in that first one, is that y-intercept right there. So it turns out the y-intercepts related to Vmax, and technically it's equal to one over Vmax. But it's related to Vmax. So whereas the x-intercept right there is related to the km value. It turns out it's equal to negative one over km. So notice I've kind of dashed this point in because the truth is you can't have a negative value on this axis because you can't have a negative value of substrate concentration. So we just kind of have to extrapolate it out, but we'll find out when we start playing with inhibitors and stuff like that, and these become important. So the last part, and this probably is a little bit less important, but the slope of this graph is equal to km over Vmax. And so the idea is we take that invert in the Kalis-Ment equation. We fit it to a y equals mx plus b equation, and we can find out that m value is the slope of km over Vmax, and the b value of the x, I'm sorry, the y-intercept, is equal to one over Vmax. And so by doing a least squares analysis and stuff like that, getting the y-intercept and the slope, we can figure out both km and Vmax in such a case. So that's what that gets its utility. So the big thing for you though on the DAT and the OAT is you just want to understand that when you're given a double reciprocal plot that you can use the y-intercept to get Vmax and the x-intercept to get to km. All right, we also want to talk about two types of inhibition. So many enzymes actually have inhibitors that will slow down the rate of catalysis in one way, shape or form. So the first type of inhibition we want to talk about is competitive inhibition. So in a competitive inhibitor is going to compete with the substrate for the active site. And so instead of the substrate binding of the active site, this form of inhibitor can bind there instead. So in the enzyme can't bind both the substrate and the inhibitor at the same time to the active site, it's the same site. So you can only bind one or the other. And so that's why they call it competitive inhibitor. It's competing with the substrate. And so obviously the more of this inhibitor you have, well then the more of the enzymes in your solution are going to be blocked from binding substrate because they're bound to this inhibitor instead. So big thing you want to take away though is that it is binding at the active site. So and then you want to know how it's going to influence things. Well it turns out it's going to affect km but it is not going to affect Vmax. So we say that km increases and we can make this make sense. We said earlier that the Michaelis constant km had an inverse relationship between the affinity for an enzyme and substrate. And so if km is going up, well that means there's lower affinity now for the substrate. And the reason there's lower affinity for the substrate is that the enzyme can say like about the substrate, about the inhibitor. Wow, but now there's this great inhibitor over here so I'm less likely to now bind the substrate. So this competing inhibitor is effectively lowering the affinity we now have for the substrate. So that's why km is increasing in this case with a competitive inhibitor. So it turns out that Vmax is not going to change. Now you have to add a lot more substrate to get to that point. That way you have enough substrate to outcompete all the inhibitor that's there. But if you add enough, then eventually you're going to get some point of saturation again where there's just so much substrate that effectively all the enzymes now are binding substrate rather than the little bit of inhibitor that you still have in your solution in comparison. So we kind of take a look at what this does with the graph here, velocity versus substrate. So you can see the regular graph is still in red here. So and again approaching saturation here with no inhibitor. So and you can see now with a competitive inhibitor you're still going to reach that same point of saturation but it's going to take a lot higher substrate concentration to get there. Now if we've really come really close to the point of saturation all the way at this point right here so with the with no inhibitor but we're not quite there when there's a competitive inhibitor. So you can see that the Km is increasing. So again you reach Km or you can a ballpark Km when you reach half vmax. Well these have the same vmax and so half vmax is going to occur on both of these curves right around here. So but for the uninhibited version. So we see Km right here and then for the inhibited version we see a larger value of Km. It occurs in a higher substrate concentration. And so that's how you can kind of tell. And so if you saw a graph like this you should be able to look at this graph and be like, oh look at that. They have the same vmax. They definitely don't have the same Km. That is going to be the hallmark of a competitive inhibitor. Now it turns out if you take a look at the double reciprocal plot and you're going to look at that. So if you recall it was the y-intercept that was related to vmax. And it was the x-intercept that was related to Km. Well again we're saying that vmax is not changing at all with a competitive inhibitor. And that's why we've got three different curves here for three different concentrations of the inhibitor. We're increasing this way but every one of them is going to pass through that same y-intercept because vmax is not changing. And so when you get a double reciprocal plot and you have the same y-intercept that again is also another way to recognize that you've got a competitive inhibitor. Now if you notice though these do not have the same x-intercept. Which makes sense because again it's that x-intercept we use to get Km. And so if Km is changing then your x-intercept should be changing amongst these curves as well. So you can also see the slope is changing. We said the slope is equal to Km over vmax. So and again vmax is not changing but Km was and so the slope should definitely be changing as well. Cool big thing though is that if you saw a plot of different inhibitor concentrations on the double reciprocal here. So you should recognize from just such a plot with the same y-intercept that's the hallmark of competitive inhibition. So the second type of inhibition we're going to study is called non-competitive. Now turns out there's a third type called uncompetitive that we're just going to ignore completely. If you're in a biochemistry course it'd probably be relevant. We'll talk about non-competitive inhibitors. So you should know that a non-competitive inhibitor does not bind at the active site. It's binding somewhere else on the enzyme. So in the case of non-competitive it can either bind the free enzyme or the enzyme substrate complex. Either one is fair game. It's got some complementary site other than the active site where it can bind in either case. That's the first little piece of trivia to take away. Combine either the free enzyme or the enzyme substrate complex. So and then now instead of the vmax not changing it's the vmax that's actually going to change and now it's the Km that's going to stay the same. So kind of the opposite situation to what we saw with a competitive inhibitor. And so it turns out this is going to cause the enacts to go down but the Km is going to be unchanged. Now if you recall with a competitive inhibitor we said that Km was changing because affinity for the enzyme or affinity for the substrate by the enzyme was changing due to the competition at the active site. Well now we don't have any competition at the active site because the non-competitive inhibitor is not binding at the active site and so we don't have to worry about any change in affinity. So binding is unchanged therefore Km is going to be unchanged. So however though what the non-competitive inhibitor does is it causes a conformational change in the enzyme that makes it not as good at catalysis now. And so it's affecting the catalytic step in the caliphantic kinetics. As a result your vmax is going down. So you can see the velocity versus substrate concentration curve. So Hallmark here is that no matter how much substrate you add you're not going to get to the same vmax here. So you can see the inhibitor is plotted in blue there. You're not getting the same vmax as you were without the inhibitor and that will be the hallmark of your non-competitive inhibitor instead. So also note though that whether I look at vmax when there's no inhibitor and go to half vmax the plot over and look down there's our Km value. So but also if you go to vmax with the inhibitor and go halfway up to that go over and go down. It's at exactly the same value and so again Km is not changing here with non-competitive inhibitor. And so it occurs at exactly the same point here on both curves. Now if we take a look at the double reciprocal plot so now it's the Km that's not changing. And if you recall it was the x-intercept that we were going to use to get that Km value. It's equal to negative one over Km but when we used to get that Km and since Km is not changing it's the x-intercept where they would appear as if they were going to intercept instead. So but now the y-intercept is going to be different because vmax is going down. And if vmax is going down well then one over vmax is going to be going up. And you're going to get three different y-intercepts in the case of non-competitive inhibition. So and the greater the concentration of inhibitor the lower the vmax and the lower the vmax the higher the one over vmax would be on that y-intercept. Notice the slope is also changing because again that slope is equal to the ratio of Km over vmax. And in this case Km is not changing but vmax is causing that slope to change. So once again though the big takeaway here is on this double reciprocal plot by noticing that they have the same x-intercept that would be the key to recognizing that this is an example of a non-competitive inhibitor.