Mixed Inhibition - Online Tutor, Practice Problems & Exam Prep

Mixed inhibitors can bind to either the free enzyme or the enzyme-substrate complex, affecting enzyme kinetics. They always decrease the apparent V_max, while the apparent K_m can either increase or decrease based on the degree of inhibition on the free enzyme (α) versus the enzyme-substrate complex (α'). The relationship is defined by the equations: $(\alpha K)/\mathrm{\alpha \text{'}})$ for K_m and $(VV)/\mathrm{\alpha \text{'}})$ for V_max.

Mixed Inhibition

In this video, we're going to talk about our 3^rd type of reversible inhibition, which is mixed inhibition. Now, of course, mixed inhibition is going to be caused by mixed enzyme inhibitors. And so these mixed enzyme inhibitors actually have mixed binding since they can either bind to the free enzyme or they could bind to the enzyme-substrate complex. And, of course, we know that all enzyme inhibitors, regardless of what type they are, are all going to decrease the initial reaction velocity or the v₀ of an enzyme-catalyzed reaction. So no surprise there. And so, really, you can think of mixed inhibitors as having some of the mixed features of competitive and uncompetitive inhibitors that we already talked about in our previous lesson videos since they can bind to either the free enzyme like competitive inhibitors and they can bind to the enzyme-substrate complex like uncompetitive inhibitors. Now, ultimately, the binding of a mixed inhibitor to either the free enzyme or the enzyme substrate complex is going to prevent the conversion of the substrate into the product. And, of course, we kind of already knew this as well. Whenever an enzyme inhibitor is bound to the enzyme, it's going to inhibit the enzyme by preventing the reaction from taking place.

Now, what's important to also recognize here is that mixed inhibitors do not even mention the word "competition" at all. And so, with mixed inhibitors, there's absolutely no competition that takes place between the mixed inhibitor and the substrate. And so, part of this is because mixed inhibitors are actually going to bind to allosteric sites on the enzyme. An allosteric site is just an alternative site on the enzyme other than the active site. And since the substrate is binding to the active site and the mixed inhibitor is binding to other sites other than the active site, there's no competition between the two. Now, what's also important to take into account for mixed inhibitors is that they can actually bind with different affinities to either the free enzyme or the enzyme-substrate complex. And recall from our previous lesson videos that the inhibition constant, K_I, is the inhibition constant for the free enzyme that describes the affinity that the free enzyme has for the inhibitor. And the inhibition constant of the enzyme-substrate complex or K_I' describes the affinity that the enzyme-substrate complex has for the inhibitor. And so when it comes to mixed inhibitors, K_I cannot be equal to K_I', meaning that, the mixed inhibitor will have different affinities to the free enzyme and the enzyme-substrate complex. And so if it turns out that the K_I does equal the K_I', then that is what defines a noncompetitive inhibitor. But we're going to talk about noncompetitive inhibitors later in our course in a different video. For now, we're gonna focus on these mixed inhibitors here where the K_I cannot equal K_I'. And so, if we take a look at our example down below of mixed inhibition, on the left-hand side over here, notice what we have at the top is the same exact enzyme-catalyzed reaction that we've seen so many times before in our previous lesson videos. And notice that when it comes to the mixed inhibitor, that the mixed inhibitor can actually bind to the free enzyme or to the enzyme-substrate complex to form the E-I complex or the E-S-I complex. And so in either case, whenever the inhibitor is bound to the enzyme, the reaction is not going to be able to take place. And so notice that the inhibitor binding to the inhibitor, for a mixed inhibitor, does not compete with the substrate for binding and so what this means is that the substrate is free to bind to the enzyme as it pleases regardless of if the inhibitor is bound or if the inhibitor is not bound, the substrate is always free to bind. And the same goes for the inhibitor. The inhibitor is free to bind to the enzyme, regardless of if the substrate is bound or not. And so that's partly what we mean here by no competition between the mixed inhibitor and the substrate for binding. And so over here on the right-hand side, what we have is essentially the same exact thing that we have on the left-hand side, just a different visual representation. And so notice that our free enzyme here actually has two binding sites. It has the active site over here and then it has the mixed inhibitor binding site over here. And so notice that the substrate is free to bind to the active site, whenever it is open and available. And the inhibitor is also free to bind to the mixed inhibitor binding site whenever it is open and available. So this allows the mixed inhibitor to bind to the free enzyme or it allows the mixed inhibitor to bind to the enzyme-substrate complex. So here when the mixed inhibitor is bound to the free enzyme, notice that the substrate is still capable of binding to the active site. So, this equilibrium down here is just showing how there's no competition for binding between the mixed inhibitor and the substrate. However, the most important thing to recognize is that the mixed inhibitor, if it is bound to the enzyme, regardless in what form, the reaction is not going to be able to take place. And so, again, it's important to note that the mixed inhibitor is going to have a different binding affinity to the free enzyme and to the enzyme-substrate complex. Meaning that K_I will not equal K_I'. And so in our next lesson video, we're going to revisit our "Scooby Doo" analogy and apply mixed inhibition to it as along with giving you guys a memory tool for memorizing, the effects that mixed inhibitors have. And so, this concludes our introduction to mixed inhibitors and I'll see you guys in our next video.

Mixed Inhibitor Effects

In this video, we're going to talk about the effects that mixed inhibitors have on enzymes. It turns out that mixed inhibitors may either increase or decrease the apparent Michaelis constants. There are some mixed inhibitors that will increase the apparent Michaelis constants, and there are other mixed inhibitors that will decrease the apparent Michaelis constants. Later in this video, we'll discuss the exact factors that determine whether a mixed inhibitor will either increase or decrease the apparent Michaelis constants. But all mixed inhibitors will always decrease the apparent Vmax. The real question is exactly how and why mixed inhibitors have these particular effects on an enzyme. To understand that, let's look at our analogy in this image. We're using the same image from our previous lesson videos to represent the enzyme-catalyzed reaction, where we have an adorably cute little puppy named Zyme binding to its substrate bone and converting it into a poop on the ground. Notice we're using the owner of this puppy to represent the reversible inhibitor, and we know from our previous lessons that the owner is not competitive. He's wearing a shirt that says, "We are all winners," and he has no interest in competing with anyone or anything, including the substrate. This reminds us that mixed inhibitors do not compete with the substrate for a binding position to the enzyme's active site. The enzyme's active site is the dog's mouth, and we're saying that the owner will not be binding to the dog's mouth, which makes perfect sense. One important way that these mixed inhibitors differ from the uncompetitive inhibitors from our previous videos is that the owner in the mixed inhibitors will display mixed emotions. Notice on the right-hand side, he displays an angry upset emotion, which can be told by his facial expression. On the left-hand side, he displays the opposite emotion: a very happy and loving emotion. He is upset on the right-hand side because he is upset at the enzyme-substrate complex version of his dog with the bone in his mouth. He knows that when his dog has the bone in his mouth, there will soon be poop on his living room floor that he has to clean up. The angry upset version of our mixed inhibitor tells the dog to drop the bone when he finds that his dog has the bone in his mouth. The mixed angry version of the inhibitor will quickly get the leash and take the dog outside so that the dog does not poop on the floor. The mixed inhibitor, the owner, will therefore be preventing the dog from pooping, saying "No pooping." This would be the enzyme-substrate-inhibitor complex where you have the enzyme, the substrate (which is the bone), and the inhibitor (which is the owner), all bound together. With the happy loving version of the mix inhibitor, notice that he says "I love you, Zyme," and he is going to love the free enzyme version of the dog, where the dog does not have the bone in its mouth. Notice that this version, when it binds to the free enzyme version, will form the enzyme-inhibitor complex, which is really just this cute little portrait of the owner and his puppy. The owner has these loving emotions towards his dog. It's important to remember that the symbol α is the symbol used to represent the degree of inhibition on the free enzyme. We have this α symbol next to the happy version of our owner because the owner is associated with the free enzyme version of the dog. Recall that the symbol α' represents the degree of inhibition on the enzyme-substrate complex. This reminds us that the mixed inhibitor version with the angry upset mood is going to bind to the enzyme-substrate complex. Mixed inhibitors can bind to both the free enzyme and the enzyme-substrate complex. If α is greater than α', it means that the mixed inhibitor will have a stronger affinity for the free enzyme. If α' is greater than α, that means that the mixed inhibitor will have a stronger affinity for the enzyme-substrate complex. Regardless of whether the mixed inhibitor is bound to the free enzyme forming the EI complex or if the mixed inhibitor is bound to the enzyme-substrate complex forming the ESI complex, the mixed inhibitor will prevent the dog from pooping on the floor. The mixed inhibitor is going to prevent the enzyme-catalyzed reaction from proceeding, regardless of whether it's binding to the free enzyme or the enzyme-substrate complex. Another important note is that the enzyme, when in the enzyme-inhibitor complex, can still bind to the substrate, but when it does, the inhibitor will already be there to prevent the dog from pooping on the ground. This is a reversible binding. Now that we understand our analogy better, let's revisit how and why mixed inhibitors have these particular effects. The effect on the apparent Michaelis constants turns out to be governed by Le Chatelier's principle. What influences this reaction shift is the magnitude of the degree of inhibition on the free enzyme, α, as well as the degree of inhibition on the enzyme-substrate complex, α'. α and α' dictate whether this equilibrium is going to shift left or right. If α is greater than α', or, in other words, if the degree of inhibition on the free enzyme is greater than the degree of inhibition on the enzyme-substrate complex, this equilibrium right here in the middle will respond by shifting to the left as indicated by this arrow and the color coding of the green. When α is greater than α\prime, that means that our owner is going to have greater loving emotions towards the free enzyme than he will have upset emotions towards the enzyme-substrate complex. That means that the mixed inhibitor is going to have a greater affinity for the free enzyme than it will for the enzyme-substrate complex. If α is greater than α\prime, that means that the concentration of the free enzyme will decrease more than that of the enzyme-substrate complex. According to Le Chatelier's principle, this equilibrium right here will respond to the decrease in free enzyme by shifting to the left and again, that's why we have the green arrow shifting to the left right here. When this equilibrium shifts to the left, that means that the enzyme-substrate complex is going to break down into a free enzyme and free substrate. This will make it appear as if the enzyme has a weakened affinity for the substrate, and a weakened affinity for the substrate will correspond with an increase in the Michaelis constant, or an increase in the Kilometers. That is, when α is greater than α\prime, there will be a shift to the left in this equilibrium, and that corresponds with an increase in the Michaelis constant or an increase in the Kilometers. Of course, when α is less than α\prime, the opposite will be true. There will be a shift to the right in this equilibrium. That will correspond with a decrease in the Michaelis constant or a decrease in the Kilometers. When α is less than α\prime, the equilibrium shifts to the right. When this equilibrium shifts to the right, the enzyme will form a complex with the substrate, making it appear as if the enzyme has a stronger affinity for the substrate. An increased affinity for the substrate corresponds with a decrease in the Michaelis constant, which is exactly what we said earlier: when α is less than α\prime, the Michaelis constant will decrease. To remember these effects caused by α and α', imagine mixing a unique xi symbol, shaping it like a greater than or a less than symbol. Notice that in "mix", the xi represents an increase and the delta in "mixed" represents a decrease, specifically referring to the Michaelis constant, Km. We always think of α before α'. If α is greater than α\prime, like what this box says, then the result is an increase in the Kilometers. If α is less than α\prime, like what we see over here, then the result is a decrease in the Kilometers. Now that we understand a bit better exactly how the Kilometers can be increased or decreased, let's move on to the fact that the apparent Vmax is always decreased. We know that this is related to the fact that mixed inhibitors are not competitive. Mixed inhibitors cannot compete with the substrate, meaning it's impossible for the substrate to outcompete the mixed inhibitor. Since the substrate cannot outcompete the mixed inhibitor, the effects of the mixed inhibitor are irreversible, even if we increase the substrate concentration to saturating levels. This decrease in the initial reaction velocity, which accompanies all enzyme inhibitors, is not reversible and corresponds with the Vmax also being decreased. Since mixed inhibitors decrease the apparent Vmax, this also means that the catalytic constant, or the kcat or turnover number, will also be decreased. The Vmax is decreased because if the substrate can't compete, which, with mixed inhibitors, we know is the case, then the Vmax can't remain the same, so it's decreased. The only reversible inhibitor that allows the Vmax to remain the same is the competitive inhibitor where the substrate can compete. This concludes our introduction to the effects of mixed inhibitors. In our next lesson video, we'll talk about how mixed inhibitors affect the Michaelis-Menten plot. I'll see you in that video.

Mixed Inhibition & Michaelis-Menten Plots

So now that we know that mixed inhibitors can either increase or decrease the apparent \(k_m\), but they always decrease the apparent \(v_{\text{max}}\), in this video, we're going to talk about the effects that mixed inhibitors have on the Michaelis-Menten plot. And so, recall from our previous lesson videos that mixed inhibitors have mixed enzyme binding. This is because mixed inhibitors can either bind to the free enzyme or they could bind to the enzyme substrate complex. And since they can do either of these bindings, it's \(\alpha\) and \(\alpha'\) that will both measure the degree of inhibition.

Now, \(\alpha'\) is always going to lead to the decrease of the apparent \(v_{\text{max}}\). You can see here that the apparent \(v_{\text{max}}\) is just defined as this ratio of the \(v_{\text{max}}\) divided by \(\alpha'\). Now, it's the ratio of \(\alpha\) to \(\alpha'\) that's actually going to either lead to the increase or decrease of the apparent \(k_m\). This is how the apparent \(k_m\) is defined: \(\alpha \times k_m / \alpha'\). As we discussed in our previous lesson video, the greater the degree of inhibition to the free enzyme relative to the enzyme-substrate complex, essentially, if \(\alpha\) is greater than \(\alpha'\), the apparent \(k_m\) is going to be increased. Whereas, if the reverse is true, if \(\alpha\) is less than \(\alpha'\), then that means the apparent \(k_m\) will be decreased.

Now, if it turns out \(\alpha\) is exactly equal to \(\alpha'\), that will mean the apparent \(k_m\) is not changed at all, and at that point, the inhibitor is referred to as a noncompetitive inhibitor. But we're going to talk more about noncompetitive inhibitors a little bit later in our course in a different video.

For now, let's take a look at our image down below of these Michaelis-Menten plots. Notice, over here in the presence of a mixed inhibitor, the Michaelis-Menten equation changes as so: where the \(v_{\text{max}}\) is substituted with the apparent \(v_{\text{max}}\), which is \(v_{\text{max}}/\alpha'\), and the \(k_m\) is substituted with the apparent \(k_m\), which is this ratio of \(\alpha/\alpha' \times k_m\).

Now, over here on the right, what we can see is that all enzyme inhibitors, regardless of what type of enzyme inhibitor they are, they're all going to decrease the initial reaction velocity. It's clear here that in the presence of these inhibitors, these mixed inhibitors, which are the colored curves here, it's very clear that the initial reaction velocity is being decreased. Mixed inhibitors always decrease the apparent \(v_{\text{max}}\). And so, it's very clear that the \(v_{\text{max}}\) for these enzymes in the presence of mixed inhibitors is clearly decreased, whereas, if we compare it to the curve here in the absence of inhibitor, it has a much higher \(v_{\text{max}}\).

Also, notice that we have these two curves here: this curve here in pink is when \(\alpha\) is less than \(\alpha'\), and this curve here in green is when \(\alpha\) is greater than \(\alpha'\). And as mentioned above, when \(\alpha\) is greater than \(\alpha'\), the apparent \(k_m\) is increased. So looking at this green curve right here, looking at the \(k_m\), notice that it's actually being increased in comparison to this black \(k_m\) here in the absence of inhibitor. And because the \(k_m\) is being increased, we know it's true that \(\alpha\) will be greater than \(\alpha'\) for that inhibitor. Whereas, with this pink curve here, where \(\alpha\) is less than \(\alpha'\), notice that there's actually a decrease to the \(k_m\) here with respect to the absence of inhibitor.

This here concludes our lesson on how mixed inhibitors can affect Michaelis-Menten plots. In our next lesson video, we'll talk about how mixed inhibitors affect the Lineweaver-Burk plot. So I'll see you guys there.

Mixed Inhibition & Lineweaver-Burk Plots

In this video, we're going to talk about how mixed inhibitors affect Lineweaver-Burk plots. Recall that way back in some of our older lesson videos, we talked about shifting Lineweaver-Burk plots. The skills we developed in those older lesson videos are actually going to be very useful here in this lesson. If you don't remember much about shifting Lineweaver-Burk plots, make sure to go back and check out those older lesson videos before you actually continue here.

Now, that being said, recall from our last lesson video, we said that mixed enzyme inhibitors will always decrease the apparent v_max, but they can either increase or decrease the apparent K_m depending on the values of the degree of inhibition on the free enzyme, alpha, and the degree of inhibition on the enzyme substrate complex, alpha prime. What this means is that mixed enzyme inhibitors can actually change the slope of the line on the Lineweaver-Burk plot, which is this ratio of K_m over V_max in multiple ways.

We know that enzyme, mixed enzyme inhibitors, are going to decrease the V_max, but the K_m can either be increased or decreased and not necessarily proportional to the decrease with the V_max. This means it's possible for mixed inhibitors to increase the slope, but they can also decrease the slope, again depending on the values of alpha and alpha prime. Of course, since mixed enzyme inhibitors always decrease the apparent V_max, this means that the reciprocal of the apparent V_max, which is the y-intercept, is always going to be increased. But, since the apparent K_m can either increase or decrease, this means that the X-intercept, the negative reciprocal of the K_m, will either be decreased or increased.

If we take a look at our image down below, notice over here on the left-hand side, what we have is how the Lineweaver-Burk equation changes in the presence of mixed inhibitors. Essentially, all we need to do is take the reciprocal of the Michaelis-Menten equation in the presence of mixed inhibitors, and this is what we end up getting. Over here on the right, in our Lineweaver-Burk plot, notice that this black line right here, corresponds with the enzyme-catalyzed reaction in the absence of inhibitors (minus concentration of inhibitor).

If we were to draw in the lines for a mixed inhibitor, we know that mixed enzyme inhibitors are always going to decrease the V_max, and decreasing the V_max leads to an increase in the y-intercept. We know that the y-intercept is going to increase and get further away from our 0 marker, which is the infinity marker. That means that really even though the y-intercept is increasing, the V_max is decreasing. We can go ahead and plot a point up here. That will be the new Y-intercept. When it comes to the K_m, it can either increase or it can decrease. That means, of course, that the x-intercept will either decrease or increase. If the x-intercept goes in this direction here, that means that it's getting closer to the 0 infinity marker. Because it's closer to the 0 infinity marker, that means that when it goes in this direction, the K_m itself is actually being increased.

Here's one potential possibility for a mixed inhibitor, it could look something along this line. The other possibility for the mixed inhibitor would be if the K_m, instead of being increased, was being decreased in the opposite direction, getting further away from the 0 infinity marker. We could put a point over here to represent that, and we could go ahead and say that if it had the same decrease in V_max, it would be there. Here is another possibility for the mixed inhibitor. You can see how the mixed inhibitor can actually change the slope in multiple ways. It can either increase the slope, but it could also potentially decrease the slope. So, you can imagine another situation if we were to draw a red line here, where the V_max is only decreased to this point and the K_m is decreased to this point, and so you get a line something like this, and that would be a decrease in the slope and another possibility for a mixed inhibitor.

There are a lot of possibilities when it comes to mixed inhibitors and how they affect Lineweaver-Burk plots. But the main point here is that the K_ms can either increase or decrease, but the V_max is always going to decrease. This concludes our lesson on how mixed inhibitors can affect Lineweaver-Burk plots, and we'll be able to get some practice moving forward in our next couple of video videos. I'll see you guys there.

Draw the representative lines for enzyme activity for an inhibitor with α > α' and a separate line with α < α'.

Alright. So this practice problem wants us to draw the representative lines for enzyme activity for an inhibitor where alpha is greater than alpha prime, and then it wants us to draw a separate line where the reverse is true and alpha is less than alpha prime. And so recall that when it comes to mixed inhibition, the unique feature that we want to focus on is the x. And so if we write out mixed inhibition like this, we can see that the x can represent a greater than as well as a less than sign. And so we know that when alpha is greater than alpha prime, as in our x down below, that is suggesting that the kilometers is going to be increased where the I here represents an increase in the Kilometers. And so, an increased Kilometers is actually going to shift the x intercept towards the 0 marker, which we know acts as the infinity marker on a Lineweaver-Burk plot. And so, an increased Kilometers with respect to, the enzyme catalyzed reaction in the absence of an inhibitor is going to be shifted towards the 0. So we can go ahead and shift it, right here towards the 0. And then, of course, we know that because mixed inhibition does not have any competition, with the substrate, that means that the Vmax can't stay the same. And if the Vmax can't compete, it can't keep the same Vmax and so it's going to be decreased. And a decreased Vmax is actually going to move the y intercept further away from the 0 infinity marker. So it's going to be somewhere up here. So we can go ahead and draw in our line here for our mixed inhibitor, where alpha is greater than alpha prime. So now that we've got our first line here, our first line where alpha is greater than alpha prime, we can go on to our next line where alpha is less than alpha prime. And so when alpha is less than alpha prime, as indicated by the x here, notice that, that is going to cause a decrease in the km, and the d here represents decrease. And a decreased km is going to get further away from the 0 infinity marker. So it's going to move in this direction here. And so we can go ahead and put in our decreased kilometers here. And, of course, the Vmax is still also going to decrease, so we're going to have the same position right here. And so, this line right here represents where alpha is less than alpha prime. And so now that we've drawn our two lines, we've completed our practice problem and so I'll see you guys in our next video.

When a mixed inhibitor favors binding to the enzyme-substrate complex (ES) over the free enzyme (E), the apparent substrate affinity (apparent K_m) is:

• A. Greater than the substrate affinity for E (K_m)
• B. Less than the K_m
• C. Equal to the K_m

Complete the chart by indicating with an 'x' which kinetic parameters are affected by each factor.

Alright. So this practice problem wants us to complete the chart down below by indicating with an "x" which kinetic parameters are affected by each factor listed in the far right column. And so looking at competitive inhibitors, we already know from our previous lesson videos that competitive inhibitors will increase the Km. So, there's an increase in the Km with competitive inhibitors. And because competitive inhibitors can actually compete with the substrate, if the substrate can compete, then it can keep the same Vmax. And so the Vmax is going to be kept exactly the same whereas the Km is going to be increased. So what this means is that we can mark in this column that the Km will be affected and only the Km will be affected. Now, moving on to mixed inhibitors, we already know that mixed inhibitors cannot compete with the substrate. And so what that means is that the Vmax is going to be altered for sure. And we also know that mixed inhibitors, depending on the value of alpha and alpha prime, will either increase or decrease the Km, and so mixed inhibitors are going to affect both the Km and the Vmax. So we can go ahead and mark an x here for both being affected. Now recall from our previous lesson videos, we talked about the effect of urea and 6 molar urea on proteins. And we know that urea is an agent that will denature the proteins. And so if the protein is going to be denatured then all of its kinetic parameters are going to be affected, including the Km and the Vmax. So, we can go ahead and mark an x for both since again, 6 molar urea leads to protein denaturation. Now, doubling the substrate concentration might increase the initial reaction velocity so that it approaches the Vmax. But again, the Km and the Vmax themselves are not going to be affected by doubling the substrate concentration. So we can mark neither for this right here. So, now that our table has been filled out, this completes our practice problem and I'll see you guys in our next video.

Here’s what students ask on this topic:

What is mixed inhibition in enzyme kinetics?

Mixed inhibition occurs when an inhibitor can bind to either the free enzyme or the enzyme-substrate complex. This type of inhibition affects enzyme kinetics by always decreasing the apparent V_max and either increasing or decreasing the apparent K_m. The degree of inhibition on the free enzyme (α) and the enzyme-substrate complex (α') determines the changes in K_m. If α > α', K_m increases; if α < α', K_m decreases. Mixed inhibitors bind to allosteric sites, not the active site, so there is no competition with the substrate.

How do mixed inhibitors affect the Michaelis-Menten plot?

Mixed inhibitors alter the Michaelis-Menten plot by decreasing the apparent V_max and either increasing or decreasing the apparent K_m. The plot shows a lower V_max in the presence of mixed inhibitors, indicating a reduced maximum reaction rate. Depending on the relative values of α and α', the K_m can either shift to the right (increase) or to the left (decrease). This results in different curves on the plot, reflecting the mixed inhibitor's effect on enzyme affinity and catalytic efficiency.

What is the difference between mixed inhibition and noncompetitive inhibition?

Mixed inhibition and noncompetitive inhibition both involve inhibitors binding to sites other than the active site. However, in mixed inhibition, the inhibitor has different affinities for the free enzyme and the enzyme-substrate complex, leading to different effects on K_m. In noncompetitive inhibition, the inhibitor binds equally well to both the free enzyme and the enzyme-substrate complex, resulting in no change in K_m but a decrease in V_max. Thus, K_i = K_i' in noncompetitive inhibition, while K_i ≠ K_i' in mixed inhibition.

How do mixed inhibitors affect the Lineweaver-Burk plot?

In a Lineweaver-Burk plot, mixed inhibitors cause an increase in the y-intercept, indicating a decrease in V_max. The x-intercept can either move closer to zero (indicating an increase in K_m) or further from zero (indicating a decrease in K_m), depending on the relative values of α and α'. This results in lines with different slopes, reflecting the mixed inhibitor's varying effects on enzyme kinetics. The plot helps visualize how mixed inhibitors alter both the maximum reaction rate and the enzyme's affinity for the substrate.

Why do mixed inhibitors always decrease the apparent V_max?

Mixed inhibitors always decrease the apparent V_max because they bind to allosteric sites on the enzyme, preventing the enzyme from catalyzing the reaction efficiently. This binding reduces the number of active enzyme molecules available to convert substrate to product, thus lowering the maximum reaction rate. Since the substrate cannot outcompete the mixed inhibitor, even at high concentrations, the decrease in V_max is a consistent effect of mixed inhibition.