Apparent Km and Vmax - Video Tutorials & Practice Problems

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concept

Apparent Km and Vmax

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So now that we've covered the degree of inhibition on the free enzyme and the enzyme substrate complex essentially Alfa and Alfa prime in this video, we're going to introduce the apparent K M and V Max. And so, in the presence of inhibitors, it's possible that the inhibitor could cause an apparent change to either the Michaelis constant K M and or the theoretical maximal reaction Velocity V Max of an enzyme. And so really, that's exactly what we mean by the apparent Cam and V. Max. And so the apparent K M and V Max can be defined as these variables right here and so you can see that the A P p in the superscript just means apparent. And so the apparent K M is symbolized like this, and the apparent V Max is symbolized like this. And so all the apparent Came and V Max are are the resulting K M and V Max. That enzyme has specifically in the presence of an inhibitor and so down below in our image, we can clear some of this up and so notice on the left hand side in the absence of an inhibitor. Essentially, when there's no inhibitor present. The cam in the V max are expressed exactly as we've seen them in all of our previous lesson videos. And so we can see. Uh, this is in the absence of inhibitor, however, as soon as we start to add some inhibitor. So in the presence of inhibitor, notice that depending on the type of inhibitor, it's possible that the inhibitor could cause an apparent change to the K M and the V Max. And so we refer to the K M and the V Max in the presence of inhibitor as the apparent K M and the apparent V Max. And so it's pretty simple. In the absence of inhibitor. It's defined just as we've seen it before and then in the presence of inhibitor. We refer to it as the apparent camp and the apparent V Max. And so, in our next lesson video, we'll talk about how the apparent Cam and the apparent V Max are really modified by the degree of inhibition factors that we talked about in our previous lesson videos Alfa and Alfa Prime. And so I'll see you guys in those videos

2

concept

Apparent Km and Vmax

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7m

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all right. So in this video, we're going to talk about how the apparent K M and or the apparent V Max oven enzyme are affected by the degree of inhibition factors Alfa and or Alfa Prime. And so we know from our previous lesson videos that Alfa and Alfa Prime are the degree of inhibition factors on the free enzyme and the enzyme substrate complex, respectively. And so Alfa and or Alfa Prime will indicate the degree at which the apparent km and or the apparent V max will be altered by the inhibitor. And so, depending on the type of inhibitor, Alfa and or Alfa Prime may affect the apparent km and or the apparent V Max in different ways. And again, this is going to depend on the type of inhibitor. And so if we take a look at our table down below, notice that each row here represents different types of reversible inhibitors and so this table actually has ah lot of information, and I want you guys to know that I definitely do not expect you guys to memorize all of the information in this table right now. In fact, as we move forward in our course. We're pretty much gonna break down everything that's in this table and more details. So we're gonna see everything all over again. So again, not necessary for you to memorize this right now. But that being said, however, there are a few major takeaways that I want you guys to get from this table. And one of them is that depending on the type of inhibitor, the apparent km, as we can see here, can be defined differently on, uh, the same goes for the apparent V max. It's possible for it to be defined differently depending on the type of inhibitor. And so, looking at this first row right here, this is specifically for competitive inhibitors and so we can see that the apparent K M for competitive inhibitors is equal to Alfa Times, K m. And we know again that Alfa is the degree of inhibition on the free enzyme and from our previous lesson videos. We know that Alfa is always going to be greater than or equal toe one, which means that the apparent km can Onley increase in the presence of a competitive inhibitor. And so the effect that that's gonna have on the K M. We know that if the cam is increased, that means that it's gonna have a weaker, binding affinity for the substrate. And so, if the enzyme has a weaker, binding affinity for the substrate in the presence of a competitive inhibitor, that means that the effect on the K M is that it's going to be worse in the presence of a competitive inhibitor. Now notice here for the apparent V max that we actually have it blink in our table. And that's because the apparent V max is actually equal to the normal V max. And so what this means is that in the presence of a competitive inhibitor, the apparent V max is equal to the V. Max means that the V max is not altered. And so the effect on the V Max in the presence of a competitive inhibitor is that there's no effect or no change. Now, moving on to our next row of inhibitors here, it's actually the uncompetitive inhibitors, and so you can see that with uncompetitive inhibitors, the apparent K M is defined a little bit differently than it was with the competitive inhibitor. And again, that's because it depends on, uh, it all depends on the type of inhibitor that we're using. And so the apparent km for the uncompetitive inhibitor is K M, divided by Alfa Prime. And so what this means again? We know that Alfa Prime is the degree of inhibition on the enzyme substrate complex, and Alfa prime must always be greater than or equal toe one, which means that the apparent km here is always going to be decreased in the presence of uncompetitive inhibitor, and a decreased K M actually means a stronger, binding affinity that the enzyme has for the substrate. And a stronger binding affinity here means that it's actually going to have a better KM because it's gonna have a stronger, binding affinity for the substrate. However, if we take a look at the apparent V, Max noticed that it is defined as V Max over Alfa Prime, which means that Alfa I'm sorry, the apparent V Max is going to decrease in the presence of uncompetitive inhibitor, and that means that the effect on the V Max is that it's going to be worse. And so, even though the apparent KM seems to be getting better in the presence of an uncompetitive inhibitor. The V max is getting worse. And so you can see that the inhibitor is always gonna have some something that makes the enzyme perform worse than normal. And so in this last row here of inhibitors, what we have is both mixed as well as non competitive inhibitors. And again, the reason that we're grouping mixed in non competitive is because noncompetitive inhibitors, as we'll see moving forward in our course, is just a specific type of mixed inhibitor. And so you can see here that the apparent K M is defined much differently than it is for both of these previous apparent cams. And so you can see that it turns out that whether or not the apparent km will increase or decrease will depend on the exact value of Alfa and, uh, Alfa Prime. And so if, uh, depending on the value of Alfa Alfa Prime, the K M, uh could get better or worse. So what we're gonna right here is that it depends and what it depends on is the value of Alfa and Alfa Prime. And of course, we can see here looking at the apparent V Max that it's defined exactly as it was before with the uncompetitive inhibitor. And so what that means is that it's also going to be worse because it's going to decrease with Alfa Prime here in the denominator. And so we were able to make these conclusions on the effects just by interpreting the the Apparent Kms and the apparent V. Max is here, and so again, as we move forward in our course, we're going to talk more and more details about each of these different types of inhibitors, um, in their own separate videos. And so definitely no need for you guys to memorize all of this information. But one of the major takeaways is that again, depending on the type of inhibitor, the apparent km and the apparent V max can be defined differently. And so this concludes our introduction, uh, to the apparent cam and apparently Max and we'll be able to get some practice problems applying these concepts that we've learned as we move forward. So I'll see you guys there

3

example

Apparent Km and Vmax Example 1

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3m

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So here we have an example problem that says that the value of the in division constant on the free enzyme K I for a certain competitive inhibitor is to micro Mueller when no inhibitor is present, the value of the McKell is constant. K m is 10 Micro Mueller calculate the apparent cam when four micro Mueller of inhibitor is present. And so, in order to solve for this example problem here, we're going to need to use some of the information that's in our table up above. And so, uh, notice here that in our example problem We're specifically looking at a competitive inhibitor. So we're gonna want to focus on this first row here on our table, and it wants us to calculate the apparent km. And so the apparent K M is defined right here for the competitive inhibitor and so we could go ahead and rewrite this down below. So what we're trying to solve for is the apparent K. M. And of course, this is gonna be equal to Alfa Times, the K M. Now we are given the value of the K M as Micro Mueller, but we're not really given the value of the degree of inhibition. And so what we're going to need to do is we're gonna need to calculate for the degree of inhibition and so recall from our previous lesson videos that the degree of inhibition on the free enzyme is defined by this equation right here where the degree of inhibition on the free enzyme Alfa is equal to one plus this ratio of the concentration of inhibitor over the inhibition constant. Now we are given the concentration of inhibitor as four Micro Mueller and were also given the inhibition constant as to micro Mueller so we could go ahead and plug in those values for Alfa. So what we get is that Alfa is equal to one. Plus the concentration of inhibitor is for Micro Mueller So one plus four micro Moeller over theme inhibition constant which is given to us as to micro Mueller so we can put a to micro Mueller here. And so if we do four Micro Mueller divided by two Micro Mueller and then add one, what we'll get is that Alfa is equal to three. And so now that we have our Alfa which is equal to three, we can take it and we can essentially plug it into our expression over here. And so if we do that, what we get is that the apparent K M is equal to three times the k m, which again we said earlier in our problem that the K M is equal to 10 Micro Moeller. And so three times 10 Micro Mueller is going to be equal to 30 Micro Mueller and so 30 Micro Moeller is going to be the answer for our apparent km under the's conditions that air described in our problems. And so this is the answer to our example problem and we'll be able to get some practice on our next couple of videos, so I'll see you guys there.

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Problem

Problem

Competitive inhibitor A at a concentration of 2 μM doubles the apparent K _{m} for an enzymatic reaction, whereas competitive inhibitor B at a concentration of 9 μM quadruples the apparent K_{m}. What is the ratio of the K _{I} for inhibitor B to the K _{I} for inhibitor A?

A

1.5

B

3

C

4

D

2/3

E

1/4

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Problem

Problem

The K_{I} value for a certain competitive inhibitor is 10 mM. When no inhibitor is present, the K_{m} value is 50 mM. Calculate the apparent K_{m} when 40 mM inhibitor is present.

A

20 mM.

B

10 mM.

C

100 mM.

D

150 mM.

E

250 mM.

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Problem

Problem

Uncompetitive inhibitor A at a concentration of 4 mM cuts the K _{m}^{app} in half for an enzymatic reaction, whereas the K_{m}^{app} is one-fourth the K_{m} in the presence of 18 mM uncompetitive inhibitor B. What is the ratio of the K'_{I} for inhibitor A to the K'_{I} for inhibitor B?