Apparent Km and Vmax - Video Tutorials & Practice Problems
On a tight schedule?
Get a 10 bullets summary of the topic
1
concept
Apparent Km and Vmax
Video duration:
2m
Play a video:
Video transcript
So now that we've covered the degree of inhibition on the free enzyme and the enzyme substrate complex, essentially alpha and alpha prime, in this video, we're going to introduce the apparent Kilometers 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 kilometers 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 k m and v max are are the resulting k m and v max that an 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 k m and the v max are, expressed exactly as we've seen them in all of our previous lesson videos. And so we can see, 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 an 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 an inhibitor as the apparent Kilometers 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 Kilometers and the apparent v max. And so, in our next lesson video, we'll talk about how the apparent Kilometers and the apparent v max are really modified by the degree of inhibition factors that we talked about in our previous lesson videos, alpha and alpha prime. And so I'll see you guys in those videos.
2
concept
Apparent Km and Vmax
Video duration:
7m
Play a video:
Video transcript
Alright. So in this video, we're going to talk about how the apparent k m and or the apparent v max of an enzyme are affected by the degree of inhibition factors, alpha and or alpha prime. And so we know from our previous lesson videos that alpha and alpha prime are the degree of inhibition factors on the free enzyme and the enzyme substrate complex respectively. And so, alpha and or alpha prime will indicate the degree at which the apparent Kilometers and or the apparent v max will be altered by the inhibitor. And so depending on the type of inhibitor, alpha and or alpha prime may affect the apparent Kilometers 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 a lot of information and 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 in more detail. So, we're going to 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 Kilometers, as we can see here, can be defined differently. And, 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 Kilometers for enzyme and from our previous lesson videos, we know that alpha enzyme. And from our previous lesson videos, we know that alpha is always going to be greater than or equal to 1, which means that the apparent Kilometers can only increase in the presence of a competitive inhibitor. And so the effect that that's gonna have on the Kilometers, we know that if the Kilometers 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 Kilometers is that it's going to be worse in the presence of a Vmax that we actually have it blank 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 Vmax is equal to the Vmax, means that the Vmax 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 un uncompetitive inhibitors. And so you can see that with uncompetitive inhibitors, the apparent Kilometers is defined a little bit competitive inhibitor. And, again, that's because it depends on, it all depends on the type of inhibitor that we're using. And so the, apparent Kilometers for the uncompetitive inhibitor is Kilometers divided by alpha prime. And so what this means, again, we know that alpha prime is the degree of inhibition on the enzyme substrate complex, and alpha prime must always be greater than or equal to 1, which means that the apparent k m 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 k m because it's gonna have a stronger binding affinity for the substrate. However, if we take a look at the apparent V Max, notice that it is defined as V Max over alpha prime, which means that alpha 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 Vmax is that it's going to be worse. And so even though the apparent Kilometers 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 and non competitive is because non competitive 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 Kilometers is defined much differently than it is for both of these, previous apparent KMs. And so you can see that, it turns out that whether or not the Apere Kilometers will increase or decrease will depend on the exact value of alpha and, alpha prime. And so if, depending on the value of alpha and alpha prime, the kilometers, could get better or worse. So what we're gonna write here is that it depends. And, what it depends on is the value of alpha and alpha prime. And, of course, we can see here looking at the, parent 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, alpha 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 Maxes 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, 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 Kilometers and the apparent v max can be defined differently. And so this concludes our introduction, to the, apparent cam and apparent v 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
Video duration:
3m
Play a video:
Video transcript
So here we have an example problem that says that the value of the inhibition constant on the free enzyme k I for a certain competitive inhibitor is 2 micromolar. When no inhibitor is present, the value of the Michaelis constant kilometers is 10 micromolar. Calculate the apparent cam when 4 micromolar of inhibitor is present. And 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, notice here that in our example problem, we're specifically looking at a competitive inhibitor. So, we're going to want to focus on this first row here on our table. And it wants us to calculate the apparent Kilometers, and so the apparent Kilometers is defined right here for the competitive inhibitor. And so we can go ahead and rewrite this down below, so And, so, we can go ahead and rewrite this down below. So, what we're trying to solve for is the apparent Kilometers. And, of course, this is going to be equal to alpha times the kilometers. Now, we are given the value of the kilometers as 10 micromolar, but we're not really given the value of the degree of inhibition. And so what we're gonna 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 alpha is equal to 1 plus this ratio of the concentration of inhibitor over the inhibition constant. Now we are given the concentration of inhibitor as 4 micromolar, and we're also given the inhibition constant as 2 micromolar. So we can go ahead and plug in the as 2 micromolar. So we can go ahead and plug in those values for alpha. So what we get is that, alpha is equal to 1 plus the, concentration of inhibitor is 4 micromolar. So 1+4micromolar over, the inhibition constant, which is given to us as 2 micromolar. So we can put in, 2 micromolar here. And so if we do 4 micromolar divided by 2 micromolar and then add 1, what we'll get is that alpha is equal to 3. And so now that we have our alpha, which is equal to 3, 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 Kilometers is equal to 3 times the Kilometers, which again we said, earlier in our problem that the Kilometers is equal to 10 micromolar. And so, 3 times, 10 micromolar is going to be equal to 30 micromolar. And so 30 micromolar is going to be the answer for our apparent Kilometers under, these conditions that are described in our problem. And so, this is the answer to our example problem, and we'll be able to get some practice, in our next couple of videos. So I'll see you guys there.
4
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 Km. 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
5
Problem
Problem
The KI value for a certain competitive inhibitor is 10 mM. When no inhibitor is present, the Km value is 50 mM. Calculate the apparent Km when 40 mM inhibitor is present.
A
20 mM.
B
10 mM.
C
100 mM.
D
150 mM.
E
250 mM.
6
Problem
Problem
Uncompetitive inhibitor A at a concentration of 4 mM cuts the K mapp in half for an enzymatic reaction, whereas the Kmapp is one-fourth the Km 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?