Apparent Km and Vmax: Study with Video Lessons, Practice Problems & Examples

Inhibitors can alter the apparent Michaelis constant (K_m) and maximum reaction velocity (V_max) of enzymes. Competitive inhibitors increase the apparent K_m (weaker substrate binding) while leaving V_max unchanged. Uncompetitive inhibitors decrease both apparent K_m (stronger substrate binding) and V_max. Mixed and non-competitive inhibitors affect apparent K_m variably, depending on the degree of inhibition factors, α and α'. Understanding these dynamics is crucial for grasping enzyme kinetics and inhibition mechanisms.

Apparent Km and Vmax

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 km and or the theoretical maximal reaction velocity, vmax, of an enzyme. And so, really that's exactly what we mean by the apparent K̀ḿ and vₘₐₓ and so, the apparent kilometers and v max can be defined as these variables right here. And so you can see that the app in the superscript just means apparent, and so the apparent km is symbolized like this, and the apparent vmax is symbolized like this. And so all the apparent km and vmax are are the resulting km and vmax 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 km and the vmax 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 an 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 km and the vmax. And so we refer to the km and the vmax 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 an inhibitor, it's defined just as we've seen it before. And then in the presence of an 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.

Apparent Km and Vmax

Alright. So in this video, we're going to talk about how the apparent \( K_m \) and the apparent \( V_{\text{max}} \) of an enzyme are affected by the degree of inhibition factors, alpha and 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 alpha prime will indicate the degree at which the apparent \( K_m \) and the apparent \( V_{\text{max}} \) will be altered by the inhibitor. Depending on the type of inhibitor, alpha and alpha prime may affect the apparent \( K_m \) and the apparent \( V_{\text{max}} \) in different ways. And again, this is going to depend on the type of inhibitor.

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 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 going to 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. 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 \( K_m \), as we can see here, can be defined differently. The same goes for the apparent \( V_{\text{max}} \); it's possible for it to be defined differently depending on the type of inhibitor.

Looking at this first row right here, this is specifically for competitive inhibitors. And so we can see that the apparent \( K_m \) for 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 \( K_m \) can only increase in the presence of a competitive inhibitor. The effect that this has on the \( K_m \) is that it's going to have a weaker binding affinity for the substrate. If the enzyme has a weaker binding affinity for the substrate in the presence of a competitive inhibitor, the effect on the \( K_m \) is that it's going to be worse. The apparent \( V_{\text{max}} \) is actually equal to the normal \( V_{\text{max}} \), which means that in the presence of a competitive inhibitor, the apparent \( V_{\text{max}} \) is equal to the \( V_{\text{max}} \), which means that the \( V_{\text{max}} \) is not altered.

Now, moving on to our next row of inhibitors here, the uncompetitive inhibitors. And so you can see that with uncompetitive inhibitors, the apparent \( K_m \) is defined as \( \frac{K_m}{\alpha'} \). This means that the apparent \( K_m \) here is always going to be decreased in the presence of an uncompetitive inhibitor. A decreased \( K_m \) means a stronger binding affinity that the enzyme has for the substrate. However, looking at the apparent \( V_{\text{max}} \), notice that it is defined as \( \frac{V_{\text{max}}}{\alpha'} \), which means that the apparent \( V_{\text{max}} \) is going to decrease in the presence of an uncompetitive inhibitor.

In this last row here of inhibitors, what we have is both mixed and non-competitive inhibitors. And so you can see here that the apparent \( K_m \) is defined much differently than it is for both of these previous apparent \( K_m \)'s. It turns out that whether or not the apparent \( K_m \) will increase or decrease will depend on the exact value of alpha and alpha prime. And what it depends on is the value of alpha and alpha prime. And, of course, looking at the apparent \( V_{\text{max}} \), it is defined just as it was before with the uncompetitive inhibitor, which means that it is also going to be worse because it's going to decrease with alpha prime in the denominator.

We were able to make these conclusions on the effects just by interpreting the apparent \( K_m \)'s and the apparent \( V_{\text{max}} \)'s here. As we move forward in our course, we will talk more and more about each of these different types of inhibitors in their own separate videos. 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 \( K_m \) and the apparent \( V_{\text{max}} \) can be defined differently. This concludes our introduction to the apparent \( K_m \) and apparent \( V_{\text{max}} \), 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.

Apparent Km and Vmax Example 1

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 \( K_M \) is 10 micromolar. Calculate the apparent \( K_M \) when 4 micromolar 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. 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. It wants us to calculate the apparent \( K_M \), and so the apparent \( K_M \) is defined right here for the competitive inhibitor. We can go ahead and rewrite this down below.

What we're trying to solve for is the apparent \( K_M \). And, of course, this is going to be equal to \( \alpha \times K_M \). Now, we are given the value of \( K_M \) as 10 micromolar, but we're not really given the value of the degree of inhibition. So what we're gonna need to do is we're gonna need to calculate the degree of inhibition. 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 \( 1 + \frac{ [I] }{ K_I } \). 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 those values for \( \alpha \). So what we get is that \( \alpha = 1 + \frac{4 \, \text{micromolar}}{2 \, \text{micromolar}} \).

So if we do \( 4 \, \text{micromolar} \div 2 \, \text{micromolar} \) and then add 1, what we'll get is that \( \alpha = 3 \). 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 \( K_M \) is equal to \( 3 \times K_M \), which again we said earlier in our problem that the \( K_M \) is equal to 10 micromolar. And so, \( 3 \times 10 \, \text{micromolar} = 30 \, \text{micromolar} \). And so 30 micromolar is going to be the answer for our apparent \( K_M \) 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.

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

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.

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? 

• A. 3/2
• B. 2/3
• C. 1/3
• D. 3
• E. 1

Here’s what students ask on this topic:

What is the difference between apparent Km and Vmax in the presence of inhibitors?

Apparent Km and Vmax refer to the modified values of the Michaelis constant (Km) and maximum reaction velocity (Vmax) of an enzyme in the presence of inhibitors. Competitive inhibitors increase the apparent Km, indicating weaker substrate binding, while leaving Vmax unchanged. Uncompetitive inhibitors decrease both apparent Km, indicating stronger substrate binding, and Vmax. Mixed and non-competitive inhibitors affect apparent Km variably, depending on the degree of inhibition factors, α and α'. Understanding these changes is crucial for studying enzyme kinetics and inhibition mechanisms.

How do competitive inhibitors affect apparent Km and Vmax?

Competitive inhibitors increase the apparent Km, which means the enzyme has a weaker binding affinity for the substrate in their presence. This is because the inhibitor competes with the substrate for the active site of the enzyme. However, competitive inhibitors do not affect the apparent Vmax, meaning the maximum reaction velocity remains unchanged. The equation for apparent Km in the presence of a competitive inhibitor is given by:

$K_{m\text{'}}=K_m\times \alpha$

How do uncompetitive inhibitors affect apparent Km and Vmax?

Uncompetitive inhibitors decrease both the apparent Km and Vmax. The apparent Km decreases because the inhibitor binds to the enzyme-substrate complex, enhancing the enzyme's binding affinity for the substrate. The apparent Vmax also decreases because the inhibitor reduces the overall number of active enzyme-substrate complexes available for the reaction. The equations for apparent Km and Vmax in the presence of an uncompetitive inhibitor are:

$K_{m\text{'}}=\frac{K_m}{\mathrm{\alpha \text{'}}}$

$V_{\max \text{'}}=\frac{V_{\max }}{\mathrm{\alpha \text{'}}}$

What are the effects of mixed inhibitors on apparent Km and Vmax?

Mixed inhibitors can affect the apparent Km and Vmax in various ways, depending on the values of the inhibition factors α and α'. The apparent Km may increase or decrease, depending on whether α or α' is more dominant. The apparent Vmax, however, always decreases because the inhibitor affects the enzyme's ability to catalyze the reaction. The equations for apparent Km and Vmax in the presence of mixed inhibitors are:

$K_{m\text{'}}=\frac{K_m}{\mathrm{\alpha \text{'}}}\times \alpha$

$V_{\max \text{'}}=\frac{V_{\max }}{\mathrm{\alpha \text{'}}}$

Why is it important to understand apparent Km and Vmax in enzyme kinetics?

Understanding apparent Km and Vmax is crucial in enzyme kinetics because these parameters help elucidate how inhibitors affect enzyme activity. By studying these changes, researchers can determine the type of inhibition and the inhibitor's mechanism of action. This knowledge is essential for drug development, as many pharmaceuticals function as enzyme inhibitors. Additionally, understanding these dynamics aids in the design of experiments and the interpretation of kinetic data, providing insights into enzyme function and regulation.