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Learn the toughest concepts covered in Biochemistry with step-by-step video tutorials and practice problems by world-class tutors

6. Enzymes and Enzyme Kinetics
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Reaction Orders 8m
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in this video, we're going to begin our discussion on reaction orders. So recalling our previous lesson videos, we talked about the rate law and we said that the rate law is just another way to express the reaction velocity or V here. And so the rate law says that the reaction velocity is equal to the rate constant k times each initial reactant concentration raised to the power of the order. And so this reaction order here is exactly what we're gonna be focusing on in this video. And so the reaction order is a proportionality relationship between the reaction rate V and each initial substrate concentration or each initial reactant concentration. Remember that substrates are the reactant of enzyme catalyzed reactions now recall from our previous lesson videos, we said that the reaction orders are frequently going to be equal to the coefficients of the substrates, especially in our course moving forward. However, we said that the coefficients of the substrates do not always equal the reaction orders, and so the reaction orders actually need to be experimentally determined. Now it turns out that the substrate coefficients will Onley equal the reaction order for elementary reactions and elementary reactions air just those reactions with Onley one transition state. And so if the reaction on Lee has one transition state, then it's considered an elementary reaction, and we know that the substrate coefficients will equal the reaction order. However, if it's a non elementary reaction, meaning that it's not elementary, then we know that this will not be true and that the substrate coefficients will not equal the reaction order now. What's important to note is that the overall reaction order, regardless of if the reaction is non elementary or elementary, will always be equal to the sum of the individual reaction orders for all of the substrates. And we'll be able to see how that works better down below in our example. Now, in this example, we're going to determine the overall reaction order for the following reactions. And the reaction over here on the left notice of what we have is ozone gas plus oxygen gas being converted into two moles of oxygen dioxide gas, and so notice that it's telling us that this is an elementary reaction. So we know that this reaction here on Lee has one transition state and that one transition state is being represented by this one arrow here in this reaction. Now, if we were to draw the rate law for this particular reaction, notice that the rate law is expressed right here, and the rate law is just the reaction velocity, which is equal to the rate constant k times the concentration of the initial, uh, substrates or reacting. So we have one substrate here, ozone gas, and we have another substrate here just oxygen. And then because this is an elementary reaction, we know that the substrate coefficients are going to be equal to the substrate reaction order. And so we noticed that the reaction orders that air replaced here do match the substrate coefficients of one. And so what's important to note is that the overall reaction order is always going to be the sum of the individual reaction orders. And so because we have the first reaction order as one and the second reaction orders one we know that one plus one will give us the overall reaction order. So we know that the overall reaction order is second order for this reaction. Now, if we take a look at this non elementary reaction over here on the right notice. What we have is nitrogen dioxide gas plus carbon monoxide gas will give us nitrogen monoxide gas and carbon dioxide gas. And so notice that this is a non elementary reaction, which means that it has more than one transition state. And in this reaction, we can see that's true because we have more than one reaction arrow, which means that there's some kind of intermediate that forms that's not being shown in this overall reaction here. And just because there are two reaction arrows here, uh, doesn't mean that this reaction will always be expressed with two reaction arrow. Sometimes we'll see that this will be expressed with one reaction arrow, and it will make it look like it's an elementary reaction when in reality it's actually not in elementary reaction. And it's really broken up into two smaller reactions. And so what that means is we can't really rely on the arrows toe. Let us know if this is a non elementary reaction or not. However, moving forward in our course, we know that all of the enzyme catalyzed reactions we're going to assume our elementary reactions with Onley one transition state. However, taking a look at this non elementary reaction if we were to draw out the rate law for this reaction, which is shown here, notice that the rate law is equal to the reaction velocity, which is equal to the rate constant K times, the concentration, the initial concentration of the reactant. So we have nitrogen dioxide here and carbon monoxide here, and then notice that because this is a non elementary reaction, the substrate coefficients do not necessarily equal the substrate reaction order, and we can see that here as well. So notice that the reaction order for this nitrogen dioxide gas is, too. But the coefficient here is not to the coefficient is actually one, so they don't match each other. And that's okay, because this is a non elementary reaction, and the same applies for the carbon monoxide gas. Notice that it's reaction order zero. Even though it's coefficient is one and so because they don't match, we know that's OK because that's a non elementary reaction. But still, even though this is a non non elementary reaction, the overall reaction order is always going to be the sum of the individual reaction orders for all the substrates and So we have a reaction order of two plus a reaction order of zero. So to play zero is still too. So that means that the overall reaction order is still going to be second border. And so notice that the reaction order the overall reaction order for this elementary reactions, Second order. And so is the overall reaction order for this non elementary reaction order. So the overall reaction order does not necessarily tell us information about the individual reaction orders for each reactant. And, uh, that is why we say that the individual reaction orders must be experimentally determined. And so again, uh, our course, our biochemistry course. Most of the professors are not going to expect us to use data to determine the reaction orders. And instead, what we're going to do is on Lee, look at reactions and assume that reactions are elementary reactions for the particular enzyme catalyzed reactions that we'll be looking at. And so what we're going to talk about next are the three common overall reaction orders, and those are zero order reactions, first order reactions and second order reactions. And so moving forward in our course, we're gonna talk about each of these common overall reaction orders and their own separate videos. And so we'll start off by talking about zero order reaction, so I'll see you guys in that video.
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Problem

The rate law for an elementary and/or nonelementary reaction is _________________________:

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Reaction Orders 5m
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All right, So now that we know a little bit about reaction orders in this video, we're going to talk about reactions that have an overall reaction order of zero or zero order reactions. And so zero order reactions are reactions where the substrate concentration has zero effect on the reaction rate or the reaction velocity. And so if the substrate concentration has zero effect on the reaction rate, that means that even when we change the substrate concentration, the reaction rate will not change. And so enzyme catalyzed reactions will actually exhibit zero order kinetics when an enzyme is completely saturated with substrate and we'll be able to see an example of that down below in our image. Now, the zero order rate constant units, our, uh, specifically molar ity times inverse seconds, which is express right here. But I really don't want you guys to focus on the units of the rate constant just yet, because later in our course and a different video, we're gonna focus specifically on all of the rate constant units. And so if these units are a bit confusing to you right now, that's okay, because again, we're gonna talk about the units later in a different video now down below. In our image notice here we have a specific reaction where we have reacting a being converted into product B, and we're told that this is a zero order reaction, which means that the overall reaction order is zero. And so if we take a look at the rate law for this reaction, we know that the rate law is just another way to express the reaction rate or the reaction velocity. And it's equal to the rate constant k times the concentration of the reactant or the substrate on. We only have one substrate here it's substrate s O. It's the rate constant k times the concentration of substrate. And then it's gonna be raised to the reaction order. And because we're told that this is a zero order reaction, we know that the reaction order must be zero. And so any number raised to the power of zero is going to be one. Try it in your calculators, take any number and raise it to the power of zero. You'll get an answer of one. And so essentially, what we have here is one times K and one times K is going to be K. And so essentially what we're saying is that 40 order reactions, the reaction rate or the reaction velocity will be equal to the rate constant K. And so, if we take a look at this graph that we have over here, notice that we have the reaction rate on the Y axis and the substrate concentration on the X axis, and this is specifically a zero order reaction. So that means that the substrate concentration has zero effect on the reaction rate. So even when we change the substrate, concentration toe, low substrate concentrations and high substrate concentrations, it has zero effect on the reaction rate. So notice that the reaction rate remains constant and the value of this reaction right here, we know is going to be exactly equal to the rate constant value. And so, essentially, what we have over here is another graph that we're gonna be analyzing Ah, lot as we move forward throughout our course. And this is just showing a typical curve for a typical enzyme catalyzed reaction. And so on the Y axis, what we have is the initial reaction rate and on the X axis, what we have is the substrate concentration. And again, this is a typical curve that we see for any kind of enzyme catalyzed reaction when we plot this data here. And so what you'll notice is that there comes a point in the reaction where the reaction is actually displaying zero order kinetics. And we can kind of see that here, because notice we have this horizontal line just like we have the horizontal line over here. And so we can see that by changing the substrate concentrations in this area where it's displaying zero or two kinetics. The reaction velocity does not change. And so what this means is that within this region where zero ordered kinetics are being displayed, the reaction rate does not depend on the substrate concentration. And again, the reason that an enzyme catalyzed reaction reaches ah, point where it displays zero order kinetics is because the enzyme is being saturated with substrate. And so at this point, uh, here on our graph coming down, the enzyme becomes saturated with substrate. And so even by increasing the substrate here, uh, substrate concentrations were not able to affect the reaction rate and so that's important to note that, uh, enzymes. When they're saturated with substrate, they display zero order kinetics. And so that concludes our lesson here on zero order reactions. And we'll be able to take a look at an example of this in our next video, so I'll see you guys there.
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Reaction Orders Example 1 2m
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all right. So in our last lesson video, we said that zero order reactions are reactions where the substrate concentration has zero effect on the reaction rate or the reaction velocity. And so here we're showing you guys an example of zero order kinetics and notice on the left. Over here we have a bunch of cars that are lined up, and these cars represent our substrate. And on the right, what we have here is a one lane bridge that represents our enzyme. And so the cars crossing over to the other side represents the reaction. And so up above, we have a question that's asking. Does increasing the number of cars or does increasing the substrate concentration also increase the rate that the cars cross the one lane bridge? And so recall that here we have so many cars that are lined up to cross this one lane bridge, and essentially, what we're saying is that this one lane bridge this ends I'm here is pretty much already saturated with substrate. And so when it's saturated with substrate, we know that it's going to display zero order kinetics. And that means that increasing the substrate concentration will not increase the rate that the cars will cross this bridge. And so, even if we were toe, add 10 mawr lanes of cars over here, that would not increase the rate that all of these cars would cross this one lane bridge. Because remember, this one lane bridge is already operating at its maximum rate and adding Mork, ours will not increase the rate that the cars will cross the bridge. So essentially, what we're saying is that the answer to this problem is no that, uh, increasing the substrate concentration or the number of cars will not increase the rate that the cars will cost the wrong lane bridge. And that is because the enzyme or the bridge is already saturated with substrate. Now, one of the only ways that we can actually increase the rate of a reaction when the enzyme is saturated with substrate is by adding mawr enzyme. And so if we were to ADM or one lane bridges, then that would increase the rate that the cars would cross the bridge because there's now more lanes and there's MAWR cars that can be crossed within ah, smaller amount of time. And so this concludes our example on zero order kinetics, and I'll see you guys in our next video
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Reaction Orders 3m
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All right. So now that we've covered zero order reactions in this video, we're going to talk about first order reactions. And so first order reactions are just reactions where the rates of the reaction are directly proportional to Onley one substrate concentration. And this includes uni molecular reactions or reactions that only have one molecule uni meaning one. And so these thes reactions involve Onley a single substrate, for example, Converting substrate a into product be now the first order rate. Constant units are specifically inverse seconds. But again, I don't want you guys to focus on the units of the rate constant just yet because later in our course, we're gonna have ah specific video dedicated to helping you guys understand rate constant units. So if this is confusing to you guys right now, that's okay, because again, later we'll talk more about it to clear it up so down below. In our example, what we have is an example of first order kinetics, and so notice that we're looking at a uni molecular reaction that involves only one molecule, and we're converting substrate a into product B. And we're told that this is a first order reaction And so when we, uh, consider the rate law, which is another way to express the reaction rate or the reaction velocity, we know that it's gonna be equal to the rate constant k times the concentration of the Onley Substrate year, which is substrate A. And of course, because we're told this is a first order reaction, we know that the reaction order must be one for this substrate. And so, if we take a look at this graph over here, where we have the reaction rate on the Y axis and the substrate concentration on the X axis, notice that instead of getting a horizontal line like what we saw for zero order reactions instead, this time what we're seeing is a diagonal line that's going up like this. And that's because the rate of the reaction is directly proportional to the substrate concentration. So we can say here that the rate of the reaction here is directly proportional to the substrate concentration. This is a symbol that means directly proportional. And all that means is that as the substrate concentration, uh, increases, so does the reaction rate. And so if we take a look at this other graph over here of a typical enzyme catalyzed reaction where we have the initial reaction rate on the Y axis and the substrate concentration on the X axis notice again from our last lesson video. We already know that if we increase the concentration toe a point where it's saturating the, uh, enzyme, then the enzyme will be displaying zero order kinetics where the rate of the reaction does not respond depend on the substrate concentration on. We already know this again from our last lesson video. And so here in this video we can see that early on where we have ah, lower substrate concentration and the reaction notice that the reaction the enzyme catalyzed reaction is displaying first order kinetics, where the rate of the reaction actually is directly proportional and depends on the substrate concentration. So you can kind of see that it's going up in this diagonally in this fashion, similar to how we can see. It's going diagonally over here in this fashion. And so we'll be able to apply these concepts here on first order reactions as we move along through our course. But for now, this concludes our lesson here on first order reactions, and I'll see you guys in our next video
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Reaction Orders 8m
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Problem

What is the overall reaction order for the following rate law? v = k [A] 1 [B]1 [C]0

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Reaction Orders 3m
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all right. So now that we've covered both 1st and 2nd order reactions in this video, we're going to briefly talk about pseudo first order reactions. And so sudo is just a word that pretty much just means fake. And so pseudo first order reactions are just fake first order reactions. And really, all they are are reactions that appear to be first order reactions. But in reality, they're actually second order reactions, and really other than that, there's nothing special about them. And so pseudo first order reactions can occur when the concentration of one substrate is much greater than the concentration of the other. And so, for instance, uh, if the concentration of B is much, much, much greater than the concentration of a, then we could have ourselves a pseudo first order reaction. And so remember that when the concentrations of a substrate are really, really, really high, like the concentration of be here, then they're going to be at saturating concentrations. And when the enzyme is saturated with substrate, it's going to have zero order kinetics, and it's going to appear that the concentration of B does not affect the reaction rate when indeed it might actually do that in a second order reaction. And when the concentration of A is so small in comparison to be, this makes the substrate a the limiting re agent and as the limiting re agent, it's going to appear that Onley the concentration of a alone will dictate the rate of the reaction when in reality, it's not just the concentration of a it's also the concentration of B. And so if we take a look up above at the graph from our second order kinetics, uh, reaction that we talked about in our previous lesson video. What we're saying is that a pseudo first order reaction is really just it's actually just a second order reaction, and it's nothing special about that. However, when we graph a pseudo first order reaction, it appears toe look just like a first order reaction. And that's because, again, the concentration of one substrate is much greater than the other. And so, even though with a pseudo first order reaction, the reaction appears to be first order. We know that it's actually a second order reaction, and so this can sometimes, uh, cause a little bit of complications and challenges for a biochemist when they're trying to study a reaction. So they have to vary the concentrations to make sure that these conditions here don't exist, because otherwise it could fake out a biochemist. And so that concludes our brief lesson on pseudo first order reactions, and we'll be able to get some practice as we move along with our practice video, so I'll see you guys there.
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Problem

Which of the following options is true for a reaction with the provided rate law: v = k [NO]2 [O2]

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Problem

Consider the nonenzymatic elementary reaction from A → B. When the initial [A] = 20 mM, the reaction velocity is measured as 5 μM/min. Determine the reaction order and calculate the rate constant for the reaction.

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Problem

Consider the nonenzymatic elementary reaction A  B.  When the [A] = 20 mM, the reaction velocity is measured as 5 µM of “B” produced per minute.  Calculate the rate constant for the reaction. Hint: Consider the rate law.

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Problem

The hypothetical elementary reaction 2A --> B + C has a rate constant of 10-6 M-1s-1.  What is the reaction velocity when the concentration of A is 10 mM? 