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6. Enzymes and Enzyme Kinetics

1

Rate Constants and Rate Law

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in this video, we're going to begin our discussion on rate constants and the rate law. And so I'm sure most of you guys have already covered rate constants and the rate law in your previous chemistry courses. And so a lot of the information in this video might sound familiar to you guys. And so in this video, we're on Lee going to talk about the rate constants. But later, in a different video, we'll talk about the rate law. And so what I want you guys to recall from your previous chemistry courses is that every reaction has what's known as a rate constant, which can be abbreviated with lower case variable K. And so all the rate constant is is a constant positive value for a reaction that indicates the reaction rate, efficiency or probability under very specific set conditions. And so the higher the value of the rate constant the mawr likely it is that the reaction will be faster. And so what's important to note is that when the rate constant is equal to zero, that means that no reaction will occur or take place. And again, the reaction the rate constant is a positive value, which means that the rate constant will never be negative. And so when we're considering a typical or a standard enzyme catalyzed reaction, there are four rate constants that we will be considering. And those are K one K minus one, K two and K minus two. And so, starting with K one K one is the free enzyme and the free substrate association rate constant that forms the enzyme substrate complex. And so, looking down at our typical enzyme catalyzed reaction here, notice that the free enzyme and the free substrate associating forwards to form the enzyme substrate complex is a reaction that has a rate constant of K one. And then, of course, K minus one will be the exact opposite reaction or essentially, the enzyme substrate complex dissociation rate, constant backwards toe reform, the free enzyme and the free substrate. So essentially going from the enzyme substrate complex and dissociating backwards to reform the free substrate and the free enzyme. And this is going to be K minus one. Now K two is going to be the enzyme substrate complex dissociation rate, constant forwards, toe form, the product and so down below. You can see that here we have the enzyme substrate complex and it can associate backwards. But it can also disassociate forwards toe form the product and the free enzyme. And this rate constant here will be k to. And then, of course, K minus two is gonna be the exact opposite or essentially taking the enzyme, the free enzyme and the free product and associating them together to reform the enzyme substrate complex. So, essentially taking the free product and the free enzyme and going backwards to reform this enzyme substrate complex will be K minus two. And so what I want you guys to recall is that V is the variable that represents the reaction rate or the reaction velocity which we already covered in our previous lesson videos. And we know that typically when we're measuring the reaction velocity, we're looking at the change in the product concentration over the change in time. And so what you can see here is that this here represents the enzyme catalyzed reaction. At any given time, these four rate constants are gonna be the ones that we need to consider. However, when we're looking to measure the reaction velocity at any given time, we're going to look at the rate constants that directly affect the change in the product concentration. Now notice that both K two and K minus two directly affect the change in the product. Concentration. Que two will increase the product concentration, whereas K minus two will decrease the product concentration directly. Now K minus one and K one do not directly affect the product concentration. They only directly affect the concentrations of the enzyme substrate complex and the free enzyme and free substrate. And so what that means is that because we're focusing on the change in the product concentration when it comes to reaction velocity. Were Onley very interested in measuring these two rate constant right here? And so when there are two to rate constants to consider that makes calculating the reaction velocity. Ah, lot more complicated. And so if we focus on a reaction when it's at its initial stages very, very, very early on, then were able to eliminate one of these rate constants K minus two and on Lee have one rate constant that effects the product concentration. And so, in our next lesson video, we're gonna talk more about this idea of the rate constants that are going to be considered during a, uh, initial stages of an enzyme catalyzed reaction. So I'll see you guys in that video.

2

Rate Constants and Rate Law

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So in our last lesson video, we said that there are four rate constants to consider at any given time during a typical enzyme catalyzed reaction. And those four rate constants are K one K minus one, K two and K minus two. Now, instead of focusing on the enzyme catalyzed reaction at any given time, if we instead focus on one specific period of time at the very, very beginning of the enzyme catalyzed reaction essentially at initial stages of the enzyme catalyzed reaction. Then there are Onley three rate constants for us to consider instead of four. And so which of these four rate constants here can we ignore during the initial stages of an enzyme catalyzed reaction? Well, it turns out that it's actually the K minus two rate constant that we can ignore during the initial stages of an enzyme catalyzed reaction. And so, essentially, over here, we can put that K minus two can be ignored. And so the reason that K minus two can be ignored during the initial stages of an enzyme catalyzed reaction is because when biochemists are studying enzyme catalyzed reactions in a laboratory, they Onley ad free enzyme and free substrate to the reaction mixture, but they do not add any enzyme substrate complex or any product. And so what we can say is that initially, at the very, very beginning of an enzyme catalyzed reaction, the product concentration is going to be zero, and so there's going to be no product. And so if there is no product, then that means that there is no product available to associate with the enzyme for this reverse reaction to occur. And so that means that, uh, at the initial stages of an enzyme catalyzed reaction, K minus two is ignored. And so here, what we can indicate is that biochemist will focus on the initial reaction velocity or the V, not oven enzyme catalyzed reaction for this reason. And so the initial reaction velocity is just the velocity at the very, very, very beginning of an enzyme catalyzed reaction during the initial stages, specifically where the rate constant K minus two is, uh, negligible and we can ignore it. And so notice down below. What we have is the initial enzyme catalyzed reaction at the very, very beginning of the reaction, essentially, where the time is approximately equal to zero. Since we're just a few microseconds into the reaction. And so again we can say that the product concentration just a few microseconds into the reaction is going to be equal to zero. And so notice that the reverse rate constant K minus two eyes not being included down below notice. It's not present here. And so that's because that's the one that is negligible. And instead we have here is that K one, uh, K minus one and K two are not negligible. It's on Lee K minus two. That is negligible here. And so another point that I want to make here is that K to moving forward in our course, we're always going to assume is going to be the slowest step of the enzyme catalyzed reaction. So K two is going to be the rate limiting step, and K one and K minus one will be much, much faster than K two. And so I recall from your previous chemistry courses that the slowest step, or the rate limiting step is always going to be the one to dictate the rate of the reaction. So the overall reaction here cannot occur any faster than its slowest step so it's the slowest step that actually dictates the rate. And so the K two variable here is really what biochemists are measuring. And so because with the initial reaction velocity, the K minus two variable is being ignored. What this means is that the change in the product concentration over the change in time is the change in the product. Concentration is on Lee going to be affected by one rate constant instead of being affected by two rate constants like it is at any given time during the reaction. And so that's why moving forward in our course, we're going to talk about how biochemist, uh, tend to focus on the initial reaction velocity at the very beginning of a reaction. And so it's this reaction velocity here that is, uh, measurable, uh, to biochemist, because again, it's on Lee affected by one rate constant, which is a lot easier. Thio consider, uh, than measure considering to rate constants. And so this here essentially concludes our lesson and what I want you guys notices Over here on the right, what we have is our second star. And so this star recall is just a reminder that we're going to see these concepts come up again later in our course, when we're talking about McHale is meant an enzyme kinetics and the assumptions. And so, for now, all I want you guys to recognize is that these stars are going to come back up again later in our course. But for now, this concludes our lesson on the rate constants, and in our next lesson video, we're going to talk about the rate law, so I'll see you guys there.

3

Problem

Which of the following rate constants is negligible for the initial velocity (V _{0}) of an enzyme-catalyzed reaction?

A

k_{1}.

B

k_{-1}.

C

k_{2}.

D

k_{-2}.

4

Rate Constants and Rate Law

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So now that we've brushed up a little bit on rate constants in this video, we're gonna refresh our memories on the rate law from our previous chemistry courses. And so recall that the rate law uses rate constants toe help us calculate the reaction rate, veep and so recall from our previous lesson videos that the reaction rate or the reaction Velocity V is literally just equal to the change in the product. Concentration over the change in time and the change in product concentration really just means the final product concentration minus the initial product concentration and the change in time just means the final time, minus the initial time. And so imagine a scenario where the final concentrations of reacting some products are unknown. Well, in that scenario, we could not use this method here toe. Help us calculate the reaction rate or the reaction velocity because we're missing the final product concentration. And so what we can do is we can actually still calculate the reaction rate or the reaction velocity, but we need to use the rate law in order to do that. And so, really, the rate law is just a mathematical relationship between the reaction rate v the rate constant K and each initial reactant concentration. And so we Onley need initial concentrations here and we do not need final concentrations. And typically we know the initial concentrations of reactions because when biochemists are studying reactions in the lab, they Onley add the reactant and so they know exactly how much of the reactions that they're adding. And so for the rate law, all we need to do is multiply the rate constant K by all of the initial reactant concentrations raised to the order, the reaction order. And we'll talk a lot more about this reaction order later in our course in a different topic now down below, we have an example of the rate law and notice we have the reaction velocity or the reaction rate, which we know is symbolized with the variable the and in our previous lesson videos, we said that this reaction rate of reaction velocity is just to find as the change in the product concentration over the change in time. But again, if we do not know the final concentration of product, then that means that we cannot use this method right here. If we don't know the final concentrations, but we can still use the rate law to help us get the reaction rate or the reaction velocity. And the rate law is in this bracket that we see here. And so all the rate law is is the reaction. Velocity V is equal to the rate constant que here times the initial concentration of the reactant raised to their reaction order. And so, if you have to reactant, then you'll have the rate constant times reactant number one concentration raised to the reaction order times reactant number two concentration race to the reacting order. So this is all multiplication right here. And really, that is the rate law. So again, we're going to talk a lot. Maura, about these reaction orders, uh, later in our courts in different videos. But it's important to know that these reaction orders must be experimentally determined. But in our course moving forward, we're going to see that frequently. The reaction orders are just equal to the coefficients of the react. It's and so if we take a look at our example and try to determine the rate law for each of the simple reactions that air shown uh, you can see over here on the left, we have a simple reaction where we're converting, reacting a into product be. And if we're trying to determine the rate law, remember the rate law is always gonna be the reaction. Velocity V is going to be equal to the rate constant K okay, times the concentration of the reactant. And so in this reaction, we only have one reacted. So it's gonna be times the concentration of a and then it's gonna be raised to the power of the order. And as we said previously, frequently, we'll see that the coefficients of the reactant will equal the reaction order. And so we could see that here, the reacting A does not have a coefficient or a number in front of it. So that means that we assume this number is just one. And that means that the reaction order will be one. And so this is the rate law for this simple reaction. Now, if we do the same for this reaction over here, notice that this time if we want to get the rate law, we have, uh, to reactant to consider, so it's gonna be the rate law will be the reaction Velocity V is equal to still the rate constant k times the concentration of the first reacting a raised to the order which will be the coefficient of one times the second reactant be the concentration of the second reacted be raised to the coefficient as well, and the coefficient is going to be a one. And so this is the rate law for this simple reaction. And so, in our next lesson in our next practice, video will be able to get some practice, uh, applying the rate law concepts that we refresh our memories on in this video. And then in our next lesson, video will be able to apply the rate law to enzyme catalyzed reactions, So I'll see you guys in those videos.

5

Problem

Calculate the reaction rate for the following simple reaction if k = 1.3 x 10 ^{-1} M^{-1}s^{-1}, initial [A] = 4.0 x 10^{-3} M, and the initial [B] = 6.0 x 10 ^{-3} M (for simple reactions, assume coefficients are reaction orders):

**A + B → 2C**

A

2.13 x 10^{-6} M.

B

3.12 x 10^{-6} M/s.

C

1.32 x 10^{-6} s.

D

3.12 x 10^{-6} M^{-1}s^{-1}.

6

Rate Constants and Rate Law

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all right. So now that we brush up a bit on rate constants and the rate law and this video, we're gonna talk a little bit about how to apply rate laws to enzyme catalyzed reactions. And so recall from our previous lesson videos that when biochemists are studying the reaction velocity or the reaction rate which could be expressed with the variable V, they tend to express the reaction, velocity or rate as the change in product concentration over the change in time. But we also know from our previous lesson videos that the reaction rate or the reaction velocity here can also be expressed and rewritten as the rate law. And so we already covered the rate law in our last lesson video. So let's take a look at our example down below at our typical enzyme catalyzed reactions shown here. And we know that this enzyme catalyzed reaction is specifically at the initial stages at the very, very beginning of the reaction. And we know that because notice that the reverse reaction here expressed by K minus two is not being included. And so because K minus two is negligible, we know that this is specifically the um, going to be the initial enzyme catalyzed reaction. And so let's right the rate law for the product formation step. And so the product formation step is going to be expressed with the rate constant K two. Since that's the one that leads to the product formation and so the product formation rate Law, we already know that the rate law is literally just going to be the reaction velocity. And in this case, because this is the very beginning of the reaction, this is going to be the initial reaction. Velocity is going to be equal to the rate constant, which is going to be K two here times the reactant concentration and the reactant of this particular reaction right here is actually the enzyme substrate complex. And so what we can say is that the enzyme substrate complex is our reactant for this particular product formation step. And then, of course, because we're going to assume that all of our enzyme catalyzed reactions are simple reactions. Um, we're going to assume that the coefficient in front of our reactant is going to be the reaction order. And so because it has no coefficient listed here, we'll assume that the coefficient is one. And so this here is the rate law for the product formation, uh, step and so we can apply all of these rate laws. Each of these reactions here have their own particular rate laws. And our next practice problem will be able to, uh, apply the rate laws to these rate constants as well. Now what? I want you guys, uh, to realize in addition to applying the rate laws thio enzyme catalyzed reactions is that there's Onley one rate constant in this expression right here that directly affects the change in the product concentration. And that is K two and so four enzyme catalyzed reactions. This formation of the product will Onley depend on K two? And that's because K minus two again is negligible early on, at the very beginning of the reaction, and K one and K minus one alone do not affect the product concentration. So if we were to look at K one alone, notice that it doesn't change the product concentration k one would Onley change the enzyme substrate complex concentration and the enzyme free enzyme and free substrate concentration and the same applies for K minus one And so these two do not affect the product concentration. And so, essentially, what we can say is that when biochemists are studying and measuring and plotting the reaction velocities of an enzyme catalyzed reaction, they're really gonna be plotting the initial reaction velocities or the V, not for this particular product formation step. So when they're plotting the velocities, they're Onley plotting the velocity of this step right here. And these steps over here are not going to be the ones that the Biochemist tend to focus on. And remember that this step right here we're always going to assume is the slowest step. So these two steps here are going to occur much faster than this step right here. And so over here, essentially, what we can see is a graph where we have the product concentration on the Y axis and the time on the X axis. And we've seen this graph before in our previous lesson videos. So we know that the curve is going to look like this where the product concentration levels off and we know that the slope of the line that's tangent toe, Any point on this curve is going to represent the reaction velocity. And so if we take a point that's very, very early on in our reaction here and we take the slope of the line, that's tangent to this point, we will get the reaction velocity. And so this is early early on in our reaction with a time of zero. And so that's why we, uh this slope here is going to represent the initial reaction velocity and really, because we're measuring product concentration and on Li Ke to affect the product concentration. This reaction velocity that we see here is really just the reaction velocity for this product formation Step K two step here. So that's something important to keep in mind as we move along in our course. But again, we're gonna be able to get some more practice applying rate laws to enzyme catalyzed reaction in our next practice video. So I'll see you guys there

7

Problem

Write out the rate law equations for each association/dissociation indicated below.

a) Rate law for ES dissociation into E + P: V = ______________

b) Rate law for E + S association: V = ______________

c) Rate law for ES dissociation back into E + S: V = ______________

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8

Problem

In a typical enzyme-catalyzed reaction, when & why is the rate constant k _{-2} negligible?

A

At the very beginning of a reaction because the [S] & [P] are at equilibrium and not yet disturbed by the enzyme.

B

Initially towards the beginning of a reaction because enzymes are getting off to a slow start.

C

At the start of a reaction when [S] are at their highest, [P] are at their lowest, & the reverse reaction is unlikely.

D

At the end of a reaction when the substrate and product are at equilibrium with each other.

E

As soon as the reaction begins when the reaction rate is at its lowest.

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