in this video, we're going to finally introduce our last enzyme kinetics variable, the catalytic constant or the K cat. Now, as tempting as it might seem, the K cat does not actually stand for a kitty cat. Instead, it stands for the catalytic constant. And so the catalytic constant or the cake cat is really just the rate constant for the rate limiting step or the slowest step of an enzyme catalyzed reaction. And so what we need to note is that the rate limiting, or the slowest step oven enzyme catalyzed reaction ultimately is going to dictate the potential for the theoretical maximal reaction velocity or the V Max. And so, in other words, what we're saying is that a reaction cannot go any faster than its slowest step. And so it's the slowest step of the reaction that dictates the V max. And so because the catalytic constant is the rate constant for the rate limiting slowest step, that means that the catalytic constant is going to dictate the potential for the V Max and recall that the V Max can Onley occur at saturating substrate concentrations. So for that reason, we want to associate the catalytic constant K cat with saturating substrate concentration. Now what I want you guys to note is that for simple enzyme catalyzed reactions, which pretty much incorporate all of the enzyme catalyzed reactions that we're going to cover in this course, uh, they can be expressed by this image shown down below on the left, where we have three relevant rate constants K one K minus one and K two initially at the very, very beginning of the enzyme catalyzed reaction now also recalled from our previous lesson videos that of these three reactions that we see here in this expression, it is a K two that's actually going to be the rate limiting or the slowest step of the reaction for simple enzyme catalyzed reactions. And so because K two we're going to assume is the rate limiting, uh, the rate constant for the rate limiting step, that means that we can say that K two is going to be equal to K cat when it comes to simple enzyme catalyzed reactions. And that's exactly what we're saying up above is that the K cat we're going toe always assume is equal to the value of the K two and so we call that K two is just the rate constant for the product formation. And we can see that since K two here is directly involved with product formation. Now, if we were to specifically right the rate law for this reaction here that utilizes K to what we would get is this rate law right here. So recall that the rate law is just another way to write or express the reaction Velocity V. And this includes the initial reaction velocity. And so, essentially, the reaction velocity is gonna be equal to the rate constant K two times the concentration of the reactant or the substrate here, which is the enzyme substrate complex. Now, if we were to determine the rate law that uses the catalytic constant, then notice that we can determine that over here on the right. So what we need to recall is that the catalytic constant dictates the potential for the V max and is associated with saturating substrate concentrations. And so when the enzyme is completely saturated with substrate, then we could say that this initial reaction velocity is shown here, which can also be expressed as the change in the concentration of product over the change in time. Well, approximately equal the V max. And so it's totally okay to substitute the initial reaction velocity with the V max when the enzyme is saturated with substrate. Now also note that the concentration of the enzyme substrate complex will approximately equal the total concentration of enzyme when the enzyme is saturated with substrate. And so it's also okay to substitute the concentration of enzyme substrate complex with the total concentration of enzyme when it's saturated with substrate. And then, of course, because we know that it's the catalytic constant, this rate constant for the rate limiting slow step that dictates the V. Max. Whatever the V Max is incorporated like this, we know that it's okay to substitute in the K two that we see here with the K cat. And so we can say that K cat could go into the space here. And so really, we can see that this expression here is a way to express the theoretical maximal reaction velocity v Max. Now, if we wanted to define the K cat, all we need to do is take this total enzymes, uh, concentration and move it below the V Max, and that's what we get right here. So we can say that the K cat is also equal to this ratio of the theoretical maximum reaction velocity V max over the total enzyme concentration. And so, even though we said that the K cat is not a kitty cat, if it were a kitty cat, then you would imagine that we would have to take it to the vet every now and then. And so hopefully by remembering that that will help you remember how the K cat is the ratio of the V Max over the total enzyme concentration. And so this year concludes our introduction to the catalytic constant or the cake cat, and we're going to continue to define and express how to calculate the cake cat as we move forward in our next video. So I'll see you guys there.

2

concept

Kcat

Video duration:

10m

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So in our last lesson video, we introduce the catalytic constant or the K cat oven enzyme. And in this lesson video, we're going to continue to talk about and define the catalytic constant K cat specifically, how to calculate and interpret the K cat. And so recall from our last lesson video that in order to calculate the catalytic constant or the K cat, we're going to need both the total enzyme concentration and the theoretical maximal reaction velocity or the V max of the enzyme. And so, from our last lesson video, we define the K cat as the ratio of the V max over the total enzyme concentration. And so to further define and interpret the meaning of the catalytic constant or the K cat. Here we're defining the value of the K cat as the maximum amount of substrate that is being converted into product per second by just one single enzyme molecule, specifically under saturating substrate concentration. And so we'll be able to further explain exactly what this means down below when we get to our image. Now, what's also important to know is that the catalytic constant or the K cat is also commonly referred to as the turnover number and, uh, the catalytic constant or the K cat or the turnover number, which are all synonyms toe one another. Uh, they they have the unit of inverse seconds and so down below in our image notice on the left. What we have is an image to help you guys remember that the K cat is also known as the turnover number, and so I wouldn't recommend doing this. But if you were to invert a cat and then just drop it, it would probably turn over and land on its feet. So by remembering, uh, this image here, hopefully that'll remember help you guys remember that the K cat and the turnover number are the same exact thing. And so it's important to recognize for your practice problems moving forward. And so what we can say is that the turnover number is exactly the same thing. It's equal to the value of the K cat. And as we define already in our previous lesson video, uh, the K cat is going to be the ratio of the V max over the total enzyme concentrations. And so notice down below. What we have is this table and what we have are these four different enzymes in the first column and the second column. What we have is the function of each of these enzymes. And in the third column, what we have is the catalytic constant or the K cat, which is also equal to the turnover number. And we know from up above in our previous lesson, video that has units of inverse seconds. And so we also know from our last lesson video that the K cat and the turnover number are associated with saturating substrate concentrations. And so, in order to define exactly what does this number really mean? What we can say is that for this particular number here, 40 million. What this means is that one single molecule of the enzyme cattle ace, specifically under saturating substrate concentrations, can convert 40 million molecules of substrate into product per second. And that's why it has units of inverse seconds, which means per second. And so that is an incredibly large amount of substrate being converted into product per second. That is such a mind blowing number. I can't imagine anything happening 40 million times within a second, but cattle aces one single molecule of catalysts is able to convert this many molecules of substrate into product per second when the catalyst is saturated with substrate and so on the opposite and hear what we have is 15 for the enzyme DNA proliferates one. And so this means much smaller in comparison to the 40 million. And so what this 15 means again just to further clarify this idea of the interpretation of the cake can, What this 15 means is that one single molecule of the enzyme DNA polymerase one is, um, specifically under saturating substrate. Concentrations is capable of converting 15 molecules of substrate into product per second, and so that still seems that's pretty fast. 15 substrate into product per second is pretty quick, but in comparison to 40 million, it's, ah, lot slower and really DNA polymerase one is considered toe Have, ah, fairly slow, uh, catalytic efficiency, maximal catalytic efficiency. And so the reason for why that might be is recall from your previous biology courses that DNA proliferates one is involved with DNA replication, so essentially making mawr DNA and the DNA is incredibly important. So this enzyme wants to be ableto work at a much slower rate here in order to minimize the amount of errors during DNA replication. So you can see how, just by looking at the K cap, we can also get a sense for the functionality and the importance of either catalyzing very, very quickly lots of substrates into product per second or catalyzing slowly. Ah, small amount of substrate being converted into product per second. And so what? I also want you guys to notice in this chart on this table Is that in this column here? What we have is the inverse or the reciprocal of the catalytic constant. So what we have is one over the catalytic constant. And so what I want you guys to know is that the reciprocal of the catalytic constant or the cake cat or the turnover number is just going to be won over K cat. And because it is the reciprocal, it's also going to have the reciprocal units. So that means that instead of having uh, inverse seconds as the units, the reciprocal of the K cat or one over K, cat is gonna have just units of seconds. And if it has units of just seconds, that's just a unit of time. So that means that the inverse of the K cat represents some amount of time. And indeed, the inverse or the reciprocal of the K cat one over K cat represents the exact amount of time that's required for just one single catalytic event to take place by one molecule of the enzyme. And so, if we take a look down below at our, uh, column here with the reciprocal of the K cat, notice that it has units of second. So it represents the time for just one single catalytic event to occur by just one single enzyme molecule of the enzyme. And so, looking at this first number here, notice it says 2.5 times 10 to the negative eighth, and this has units of seconds. And when it's multiplied by 10 to the negative eighth, that makes it an incredibly incredibly small number. And so it only takes 2.5 times 10 to the negative eight seconds for one catalytic event to occur by one molecule of cattle lease. So that is incredibly, incredibly rapid. And when we compare the reciprocal of the K cat, um, down below with the the enzyme DNA polymerase noticed that it has, um a reciprocal of, uh, the K cat as 0.7 And of course, the units are seconds. And so what this means is that it takes 007 seconds for just one catalytic event to occur by the DNA poll, Emery's enzyme. And so this might seem pretty quickly 0.7 seconds. That's pretty fast for just one catalytic event. Now it seems pretty fast. But again, in comparison to this reciprocal of the cake cat, um, it's actually pretty slow because this is an incredibly small number and it happens fast. And this is, ah, much larger number than this one here, so it happens much slower over time. And so, essentially, what the reciprocal of the K cat represents is it represents the speed of catalysis, how fast one single catalytic event occurs and so ultimately hear what we're saying is that the catalytic constant or the K cat or the turnover number alone, is going to be used by biochemist as a measure of the maximal catalytic efficiency of an enzyme specifically under saturating substrate concentration. So this is one of the mawr important takeaways here is that the K cat is specifically under saturating substrate concentrations and it can Onley operate this quickly on have this fast of a catalytic event when it is saturated with substrate. And so this year concludes our lesson on the catalytic efficiency and the cake cat and, of course, the turnover number. And we'll be able to get some practice applying these concepts as we move forward. And I'll show you guys an example of, um, using the K cat and unexamined problem in our next video, so I'll see you guys there.

3

example

Kcat Example 1

Video duration:

2m

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All right. So here we have an example. Problem that's asking us what is the turnover number for carbonic and hydrates If the V max is equal to 60, mole arat e per second and the total enzyme concentration is equal to 0. Moeller. And so recall from our previous lesson videos that the turnover number is the same thing as the K cat. And so what we need to recall from our previous lesson videos is that the K cat can be defined by the ratio of the V Max for the enzyme over the total enzyme concentration. And so the cake cat is just a measure of the maximal catalog catalytic efficiency of an enzyme and so noticed that were given the V Max as 60,000 polarity per second. So we can plug that in 60,000 morality per second. And, uh, this is going to be over our total enzyme concentration, which were given as 0.1 Moeller so we can put that in down here 0.1 Moeller. And so when we do this, notice that the unit of molar ity cancels and we're just left with one over seconds here. So we're gonna have our answer in inverse seconds s O. That would be the same thing as this. And so if you take your calculator and do 60,000 divided by 0.1, you'll get the answer of 600,000. And again the units are inverse Second. So, uh, this answer here matches with answer option B so we can indicate here that be here is the correct answer for this example and essentially what this means, Uh, the fact that the K cat is equal to 600,000 and for seconds means that just one single molecule of the carbonic and hydrates enzyme is capable of converting 600,000 molecules of substrate into product per second under saturating substrate concentrations. And so that is ah, lot of molecules converted into product per second. And so again, this concludes our example and will be able to get some mawr practice utilizing these concepts in our practice problems. So I'll see you guys there

4

Problem

Problem

To calculate the turnover number of an enzyme, you need to know:

A

Total enzyme concentration [E]_{T}.

B

V_{0} of the catalyzed reaction when [S] >> [K_{m}].

C

V_{0} of the catalyzed reaction at low [S].

D

K_{m} of the enzyme.

E

Both a & b.

5

Problem

Problem

If 10 μg of an enzyme (MW = 50,000 g/mol) is added to a solution containing a [substrate] 100 times greater than the K_{m}, it catalyzes the conversion of 75 μmol of substrate into product in 3 min. What is the enzyme's turnover #?

a) 1.25 x 10^{5} min^{-1}

b) 2.5 x 10^{4} min^{-1}

c) 1.5 x 10^{2} min^{-1}

d) 3.5 x 10^{6} min^{-1}

A

1.25 x 10^{5}min^{-1}

B

2.5 x 10^{4}min^{-1}

C

1.5 x 10^{2}min^{-1}

D

3.5 x 10^{6}min^{-1}

6

concept

Kcat

Video duration:

3m

Play a video:

In this video, we're going to compare and contrast the catalytic constant or the K cat or the turnover number two, the Michaelis Constant or the K M oven enzyme. And so first, I want you guys to recall from our previous lesson videos that the catalytic constant or the K cat or the turnover number is a measure of the maximal catalytic efficiency of an enzyme, specifically under saturating substrate concentrations. And so, in comparison, I want you guys to also recall from some of our earlier lesson videos that the meticulous, constant or the K M is instead a measure of the binding affinity that an enzyme has for its substrate. And so we already knew these two pieces of information from our previous lesson video. So, really, the only thing that I want you guys to learn in this video is that the binding affinity, um, that an enzyme has for its substrate or the enzymes binding to its substrate is a completely separate event to enzyme ca Tallis, ISS, and so, essentially, what this means is that moving forward in our course, we want to think of enzymes binding to its substrate and enzyme catalysis as two completely separate events. And so what this means is that an enzyme could have a really, really high binding affinity to its substrate and bind really, really well. But that does not necessarily mean that the enzyme has a high capacity to convert the substrate and toe product efficiently. And so, essentially, what we're saying here is that the K M oven enzyme or the binding affinity that an enzyme has for its substrate is going to be completely separate. And it's not necessarily going to indicate the K cat that enzyme has or the maximal catalytic efficiency that enzyme has. And so even though the K cat and the K M both indicate two completely separate events, um, it's important that when a biochemist is studying an enzyme catalyzed reaction that they consider both the K cat and the K. M of the enzyme and so moving forward in our course, we're going to talk a little bit. Maura, about how the K cat and the K M um can be studied by biochemist toe reveal important information about an enzyme specifically when we talk about the specificity constant later in our course. But for now, this is the conclusion of our comparison of the K cat and the K M. And really, this is the Onley learning objective that I want you guys to know and so that here concludes our lesson and I'll see you guys in our practice videos.

7

Problem

Problem

Studies with mutated forms of an enzyme show that changing some active-site amino acids decrease the enzyme's turnover number (k_{cat}) but do not affect the K _{m} of the reaction. What is the best interpretation of these results?

A

The K_{m} of the enzyme for the substrate does not depend on amino acid side chains found in the active site.

B

The two terms, K_{m} and turnover number, are inversely proportional.

C

The transition state for this reaction is formed prior to the formation of the ES complex.

D

Amino acids involved in stabilizing transition-state complexes can be different than those affecting the ES-complex.

8

Problem

Problem

The turnover number for an enzyme is known to be 5000 min ^{-1}. From the following set of data, determine both the K_{m} and the total amount of enzyme E_{T}.