Competitive Inhibition: Videos & Practice Problems

Competitive inhibition occurs when an inhibitor competes with the substrate for the enzyme's active site, decreasing the initial reaction velocity (v₀) and increasing the apparent K_m without affecting the V_max. The Michaelis-Menten equation in the presence of a competitive inhibitor is modified to reflect this change, where the apparent K_m is defined as α × K_m. In Lineweaver-Burk plots, the slope increases with more inhibitor, while the y-intercept remains unchanged, indicating that V_max is unaffected.

Competitive Inhibition

In this video, we're going to talk about our first type of reversible inhibition, which is competitive inhibition. And so, of course, with competitive inhibition, it's going to be caused by competitive enzyme inhibitors, which are actually the most common type of enzyme inhibitors in all of biochemistry, which is exactly why we're covering competitive inhibitors first before all of the other different types of reversible inhibitors. And so competitive enzyme inhibitors tend to be substrate analogs. And so the question is, what are substrate analogs? Well, substrate analogs are compounds that structurally similar to substrates. And so we'll be able to see that down below in our image as well where notice that the inhibitor takes on a shape that is structurally similar to the substrate, making competitive inhibitors substrate analogs. And so, when it comes to competitive inhibitors, it turns out that the key feature of them is actually the substrate for a position in the free enzyme's active site. In order to decrease the initial reaction velocity vno of the enzyme-catalyzed reaction. But really that's no surprise to us that the initial reaction velocity is decreased since we know from our previous lesson videos that by definition, all enzyme inhibitors regardless of what type are going to decrease the initial reaction velocity vno of the enzyme-catalyzed reaction. And so again, as I've already mentioned, the key feature of competitive inhibitors is actually this competition of how these competitive inhibitors can compete with the substrate. And as we'll see moving forward, none of the other types of reversible inhibitors will actually show this competition. And so this competition factor of competitive inhibitors is quite unique. And so, it also turns out it's important to emphasize that these competitive inhibitors will only bind to the empty active sites of free enzymes that are not bound to their substrates. And that's what defines free enzymes. And so, because competitive inhibitors only affect the free enzymes, that means that α, as well as kI, which are the degree of inhibition on the free enzyme and the inhibition constant of the free enzyme, are going to be the only ones that come into play for competitive inhibitors and we're not going to see α′ or k′I for competitive inhibitors. And so we'll be able to talk more about this idea, as we move along in our course. But for now, it's just important to know that these competitive inhibitors only affect free enzymes and not the enzyme-substrate complex. And so, as we'll see down below in our image, these competitive inhibitors will actually block the enzyme's active site and so the substrates cannot bind to the enzyme that's already bound to the competitive inhibitor, forming the EI complex. And so, if we take a look at our image down below on the left-hand side, we'll see that we've actually seen this image before in our previous lesson videos. And we can see that we have the same exact enzyme-catalyzed reaction that we've seen so many times before in our previous lessons. And notice that the competitive inhibitor here is represented by I and it's only affecting the free enzyme here and it's not affecting the enzyme substrate complex. And so the competitive inhibitor will bind to the free enzyme and form the EI complex and, of course, when the inhibitor is bound to the enzyme, the reaction is not going to be able to proceed, and so we'll get no reaction. And so over here on the right-hand side, essentially what we have is, the same exact, reaction just a different visual representation of that reaction. And so, you can notice here that the free enzyme, the active site is this little open circle part right here. And notice again that the competitive inhibitor is structurally similar to the substrate, making the competitive inhibitor a substrate analog. And so, what that means is that the competitive inhibitor is going to compete with the substrate for position in the enzyme's active site. If the substrate binds to the enzyme's active site first, then that means that the enzyme-substrate complex will form and that means that the substrate is going to be blocking the inhibitor from binding to the enzyme. And that means that if the substrate binds to the enzyme first, that the reaction is going to proceed as normal to form the product. However, if the inhibitor binds to the enzyme's active site before the substrate gets there, then the enzyme-inhibitor complex will form. And, of course, the inhibitor is going to be blocking the substrate from binding. And if the inhibitor is blocking the substrate from binding, then that is going to prevent the reaction from taking place. And so, again, the key feature about competitive inhibitors is that they directly compete with the substrate. A feature that we are not going to see again moving forward in our course when we talk about the other types of inhibitors. And so when the inhibitor competes with the substrate, what this means is that they are going to block each other from binding to the enzyme's active site. And again, you can see that the substrate will block the inhibitor from binding and the inhibitor will block the substrate from binding. And so, in our next lesson video, we'll introduce an analogy for you guys and some memory tools to help you guys memorize the effects that competitive inhibitors have on enzymes. So, I'll see you guys in our next video.

Competitive Inhibitor Effects

In this video, we're going to talk about the effects that competitive inhibitors have on enzymes. And so recall from our previous lesson videos, we mentioned that competitive inhibitors will actually make the apparent Km of an enzyme worse. And by worse, what we mean is that it's going to decrease the enzyme's affinity for the substrate. And, of course, a decreased affinity corresponds with an increase in the apparent Km. And so it's true that competitive inhibitors increase the apparent Km of an enzyme. But also recall from our previous videos that competitive inhibitors actually do not affect the apparent Vmax of an enzyme. And so the real question here is exactly how and why is it that competitive inhibitors have these particular effects on an enzyme.

So if we take a look at the left hand side over here, notice that this image represents an analogy for our enzyme catalyzed reaction. And notice that this adorably cute little puppy over here represents our free enzyme, and this cute little puppy's name is actually Zyme which is short for enzyme. And then over here what we have is a bone to represent our free substrate. Now of course recall from our previous lesson videos that the substrate will bind to the enzymes specifically at the enzyme's active site. And so in this analogy, the dog's mouth represents the enzyme's active site. And so when the bone binds to the dog's mouth, as it is over here in this image, it represents the enzyme substrate complex. And then, of course, the dog can eat the bone, catalyze the reaction unaltered while creating this poop over here on the ground.

Now I know talking about poop is a little weird and I can pretty much guarantee you that your professors are not going to relate adorably cute little puppies nor their poops to enzyme catalyzed reactions or these reversible inhibitors. However, this analogy right here is going to be incredibly helpful and incredibly useful for us as we move forward and talk about all of the different types of reversible inhibitors and the effects that those reversible inhibitors have, including helping us to understand the competitive inhibitor effects. And so speaking of competitive inhibitors, notice in our image down below we are using this soccer ball, which is the ball for a competitive sport, to represent our competitive inhibitor.

And so recall from our previous lesson videos that the soccer ball or this competitive inhibitor is unique in its ability to compete with the substrate. And again, this competition feature is incredibly unique to competitive inhibitors. And moving forward, as we talk about all of the other different types of reversible inhibitors, we are not going to see this competition feature again. And so we really want to emphasize this important competitive feature of the competitive inhibitor, to compete with the substrate for a binding position in the enzyme's active site, which is the dog's mouth here. And so notice when the competitive inhibitor or the soccer ball binds to the active site or the dog's mouth, it will create the enzyme inhibitor complex, and that will prevent the reaction from proceeding. And so, this is because the substrate simply cannot fit into the active site when the competitive inhibitor is bound. And so again they are competing for the same binding position in the dog's mouth.

Now over here we're showing you the Greek symbol alpha which you can recall from our previous lesson videos is representing the degree of inhibition on the free enzyme. And so this is just a reminder that the competitive inhibitor will only bind to the free enzyme, in the active site, again the dog's mouth. And once again, we do not want to lose sight of this competitive nature, that is so unique to competitive inhibitors. And so here, we're basically highlighting that feature by saying that the substrate and the competitive inhibitor are actually going to compete with each other to bind to the enzyme's active site. And so now that we have a general understanding of our analogy, let's go back up to our text to, again, determine exactly how and why competitive inhibitors have these particular effects.

Now, we're going to start with how competitive inhibitors increase the apparent Kilometers. And so the reason that competitive inhibitors increase the apparent Kilometers is because competitive inhibitors will actually decrease the concentration of free enzyme. And so notice down below, the competitive inhibitor is going to bind to the free enzyme and that's exactly why we have this alpha symbol right here. So recall that alpha is the degree of inhibition on the free enzyme. And so that reminds us that the competitive inhibitor will only bind to the free enzyme, and it will not bind to the enzyme substrate complex. And so, again, competitive inhibitors are going to decrease the concentration of free enzyme. And so when the competitive inhibitor binds to the free enzyme inhibitor complex, but in the meantime, it's decreasing the concentration of free enzyme. And of course, this equilibrium right here by Le Chatelier's principle is going to respond by shifting to the left. And so that's exactly what we're saying up above, that the decrease in the free enzyme concentration causes the k−1 reaction to shift to the left. And so, notice that here we have this green arrow to represent the shift to the left. And we also have Le Chatelier's principle right here, and Le Chatelier was a French scientist, which is why we have the French flag background behind it, blue, white, and red. And we have the one here that corresponds with the one in the text up above. So we know that there's going to be a shift to the left in this equilibrium, when the competitive inhibitor decreases the concentration of free enzyme. And so, if the reaction shifts to the left, that means that the enzyme substrate complex is going to break down into free enzyme and free substrate. And, of course, this is going to make it appear as if the free enzyme has a weakened affinity for the substrate. And that's what we're saying up above as well, that there's a weakened enzyme substrate affinity. And, of course, we know that the weakened enzyme substrate affinity corresponds with an increase in the apparent Kilometers. And that's exactly why competitive inhibitors increase the apparent Kilometers.

So now, moving on to the idea that competitive inhibitors do not affect the apparent Vmax. And this has to do with the unique feature of competitive inhibitors, which is their ability to compete. And so the reason that competitive inhibitors do not change the apparent Vmax is because the substrate can actually compete with the competitive inhibitor. And because the substrate can compete, that means that by sufficiently increasing the substrate concentration, it's possible for the substrate to outcompete the competitive inhibitor. And so, we can take a look at our image down below on the right-hand side to see how this idea works a little bit better. And so notice here that we have a large pile of bones, an increase in the substrate concentration. And when we increase the substrate concentration so much, we can essentially make the effects of the competitive inhibitor negligible, which is exactly why we have this big X through it. And so when there is this much substrate concentration, the competitive inhibitor is pretty much not even going to affect the enzyme catalyzed reaction, which means that the enzyme catalyzed reaction will be able to proceed normally when there is this much substrate. And of course, down below, the number 2 corresponds with the number 2 here. And so we can see that a sufficient increase in the substrate concentration allows the substrate to outcompete the competitive inhibitor and that's going to make the effects of the competitive inhibitor essentially negligible, which is why we crossed it out here, and if the enzyme catalyzed reaction can proceed as normal that means that the Vmax is gonna be kept exactly the same.

So let's take a moment to imagine if you were our little puppy in this analogy. Yeah, soccer balls are cool and all but, my gosh, look how many bones there are right here. Our little puppy simply cannot resist all of these bones, and so under saturating substrate concentrations when there are this many bones, all of these bones can outcompete the competitive inhibitor so that the effects of the competitive inhibitor are negligible. Meaning that our puppy is basically going to be ignoring the soccer ball competitive inhibitor and that means that our puppy will only be focused on this huge pile of bones and the enzymatic reaction will be able to proceed forward as if it weren't being affected at all by the competitive inhibitor. And so under saturating substrate concentrations, our enzyme can still keep the same Vmax and so the Vmax will be kept the same. And so what this means is that our enzyme is going to be able to convert the substrate into product at its maximal rate, and so it'll be able to create the maximum amount of poops over a specific period of time. Now, the good thing is neither you nor I need to deal with all of these poops that our puppy is producing. However, his owner will need to deal with all of these poops. However, the owner of our puppy is not going to come into play in our analogy until a little bit later in our course in different videos when we're talking about different reversible inhibitors. But for now, what we can say here about the competitive inhibitor is that they do not change the apparent Vmax.

And so the apparent Vmax is able to stay the same again because the substrate is able to outcompete the inhibitor under saturating substrate concentrations. And so since the apparent Vmax is not affected this also means that the catalytic constant or the kcat or the turnover number will also not be affected. And so recall from our previous lesson videos that the catalytic constant or the kcat is the maximum efficiency of an enzyme under saturating substrate concentrations. And so notice that number 3 right here is going to correspond with number 3 down below in our image. And so down below, we're going to recall the equation for the catalytic constant or the kcat. And so recall from our previous lesson videos that the kcat is equal to the ratio of the Vmax over the total enzyme concentration. And so recall that our memory tool for remembering the equation for the kcat is that even though the kcat is not a kitty cat, if it were a kitty cat every now and then we would have to take it to the vet. And so you can see the vet here highlighted can help us remember the equation for the kcat. Now again, because we've said that competitive enzyme inhibitors do not affect the Vmax and they also do not affect the total enzyme concentration, that means that this ratio will not change and the kcat or the catalytic constant also will not be affected.

And so this is really a lot of information to remember about the competitive inhibitor effects. And between you and me, who would have ever thought that these adorable cute little puppies and their nasty poops would help us get an A on our next test, but you learn something new every day. But, anyway, this is a lot of information. So how could we potentially memorize these important effects that competitive inhibitors have on enzymes? And so that's exactly what this next blue box down below is all about. And so when it comes to competitive inhibition, we know that the unique feature of competitive inhibition is again the ability for it to compete. And again, this is a feature that we're not going to see again moving forward. So we really want to appreciate and take into account this unique feature of competition right here, right now with competitive inhibition. And so, if we rewrite competitive with a K, it's literally telling us what happens to the Km. And so we can see that the Km is going to be increased. Now, in terms of the Vmax, again, we want to remind ourselves of the unique feature of competitive inhibition, the ability to compete. So we think about the soccer ball, and so we know that the substrate can compete with the soccer ball, and so it's possible for the substrate to outcompete the soccer ball. And so if the substrate can actually compete, then that means that it can keep the same Vmax. And of course, what this means is that if the substrate cannot compete, then it cannot keep the same Vmax. And so later in our course, we'll see that, again, this competition feature is unique to competitive inhibitors. So competitive inhibitors are going to be the only one that allow for t...

Competitive Inhibition & Michaelis-Menten Plots

So now that we know that competitive inhibitors increase the apparent Kilometers but do not affect the apparent V_max, in this video, we're going to talk about how competitive inhibitors affect the Michaelis Menten plot. And then later in a different video, we'll talk about how competitive inhibitors affect Lineweaver Burk Plots. And so recall from our previous lesson videos, we said that all enzyme inhibitors, regardless of what type they are, including competitive inhibitors, are all going to decrease the initial reaction velocity or the V₀ of an enzyme-catalyzed reaction. And so notice down below in our image on the left-hand side, this Michaelis Menten plot shows 2 different curves. This black curve here represents the enzyme-catalyzed reaction in the absence of inhibitor indicated by minus concentration of I, and this blue curve represents the enzyme-catalyzed reaction in the presence of a competitive inhibitor indicated by plus concentration of I. And notice that in the presence of a competitive inhibitor, if we focus on the initial reaction rate or the V₀, it's actually decreased as we mentioned in our previous lesson videos and up above. And so also recall from our previous lesson videos that competitive enzyme inhibitors only bind to the free enzyme, which means that alpha is going to measure the degree of inhibition on the free enzyme. And so, the degree of inhibition on the free enzyme alpha of a competitive inhibitor is only going to increase the apparent Kilometers as we already know from our previous lesson videos. And so in the presence of a competitive inhibitor, the apparent Kilometers is defined as the degree of inhibition on the free enzyme alpha times the Kilometers in the absence of a competitive inhibitor. And also, as we know from our previous lesson videos, competitive inhibitors do not change or affect the apparent V_max, which means that the apparent V_max is equal to the V_max in the absence of an inhibitor. And so notice down below on the left-hand side, we're showing you guys the Michaelis Menten equation that we already covered in our previous lesson videos. And so this is the Michaelis Menten equation in the absence of any inhibitors. And so if we want to get the Michaelis Menten equation in the presence of a competitive inhibitor, then all we need to do is take the Kilometers and substitute it with the apparent Kilometers, which is alpha times Kilometers, so that's what we have down below, and then take the V_max and substitute it with the apparent V_max, which is equal to just the V_max. So notice we didn't change it down below. And so this equation here represents the Michaelis Menten equation in the presence of a competitive inhibitor. And so again, taking a look at this Michaelis Menten plot right here, notice that even in the presence of a competitive inhibitor, the V_max is not changed. And so the blue curve and this black curve both have the same exact V_max. And so again, that shows that the V_max is not changed. And so, recall that the substrate concentration can be represented as bones from our previous lesson video. And so if we increase the concentration of bones, eventually, we'll get to a high enough concentration of bones where Scooby Doo is going to completely ignore that competitive soccer ball inhibitor and that means that Scooby Doo is gonna be able to produce the maximum amount of poops on the floor. And so, of course, this shows that, with competitive inhibitors, the substrate, the amount of bones can compete to keep that same V_max. However, even though the V_max is not changed, if we focus on the Kilometers, notice that the Kilometers is changed here. So the Kilometers, in the absence of inhibitor, with this black line here, notice is lower than the apparent Kilometers, for this blue curve here in the presence of a competitive inhibitor. And this means that the apparent Kilometers was increased in the presence of a competitive inhibitor And so pretty much what we're seeing is, the effects of competitive inhibition on the Michaelis Menten plot. And so over here on the right, this Michaelis Menten plot, all we're trying to show you is the effect of increasing the concentration of competitive inhibitor. And so notice that we have 3 curves here. We have this black curve here that again represents the enzyme-catalyzed reaction in the absence of an inhibitor, reaction in the presence of just plus 1 molar concentration of a competitive inhibitor. And notice that the Kilometers increases further here. And, notice that the green curve here represents the enzyme-catalyzed reaction in the presence of even more competitive inhibitors. So plus 2 molar concentration of competitive inhibitor. And so by increasing the concentration of competitive inhibitor even more, notice that the Kilometers is increased even further. And so you can see that the effect of competitive inhibitors is that they are going to increase the Kilometers and the more competitive inhibitor we add, the greater the Kilometers will be increased. But notice that no matter how much competitive inhibitor we add, the V_max is going to be unaltered. And again, that's because if we increase the amount of bones, eventually we'll get to a concentration of bones where Scooby Doo is going to completely ignore that competitive inhibitor and produce the maximum amount of poops on the floor. And so, this here concludes our lesson on how competitive inhibitors affect the Michaelis Menten plot. And in our next lesson video we'll talk about how competitive inhibitors affect the Lineweaver Burk plot

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Competitive Inhibition & Lineweaver-Burk Plots

In this video, we're going to talk about how competitive inhibitors affect Lineweaver-Burk plots. Recall from our previous lesson videos the discussion about shifting Lineweaver-Burk plots. If you don't remember much about this topic, make sure to check out those older lesson videos before continuing here. The slope of the line on a Lineweaver-Burk plot is represented by the ratio of K_MV_max. Competitive enzyme inhibitors do not affect V_max, but they do increase K_M. This leads to an increased slope on the plot when more competitive inhibitor is added. As more inhibitor is added, the line becomes steeper.

The Lineweaver-Burk plot, also known as a double reciprocal plot, plots the reciprocal of the initial reaction velocity (v₀) on the y-axis and the reciprocal of the substrate concentration on the x-axis. This plot is associated with the Lineweaver-Burk equation, resembling the equation of a line y=mx+b. Taking the reciprocal of the Michaelis-Menten equation in the presence of a competitive inhibitor gives the modified Lineweaver-Burk equation.

On a Lineweaver-Burk plot, crucial enzyme information is revealed through the intercepts. The y-intercept is the reciprocal of V_max, and the x-intercept is the negative reciprocal of K_M. Since competitive inhibitors do not affect V_max, the y-intercept remains unchanged. However, as K_M increases due to the inhibitor, its negative reciprocal, the x-intercept, shifts closer to zero, thereby appearing decreased in magnitude.

The accompanying diagram features two lines: a black line representing the enzyme-catalyzed reaction without inhibitor and a purple line indicating the presence of a competitive inhibitor. The y-intercept for both lines is the same, indicating no change in V_max. However, the x-intercept for the purple line is closer to zero, showing an increase in K_M. If the concentration of the competitive inhibitor were doubled, the slope of the purple line would become even steeper, with the x-intercept moving even closer to zero.

This lesson has delved into how competitive inhibitors affect Lineweaver-Burk plots. As we progress in our course, we'll practice these concepts further. I'll see you in our next video.

Which of the following would be altered on a Lineweaver-Burk plot in the presence of a competitive inhibitor?

• A. Slope of the line on the plot.
• B. Intercept on the 1/[S] axis.
• C. Intercept on the 1/V₀ axis.
• D. V_max.
• E. a and b only.
• F. a, b, c and d.

Answer: (E)

N-hydroxy-L-arginine (an intermediate in nitric oxide biosynthesis) can bind to the active site of arginase making its manganese reactive metal center unavailable for catalysis. How would an increased concentration of this intermediate be expected to affect the kinetic parameters of this enzyme?

• A. The apparent V_max will decrease.
• B. The apparent K_m for arginine will decrease.
• C. The apparent V_max will increase.
• D. The apparent K_m for arginine will increase.

Answer: (D)

An enzyme has a K_m of 8 μM in the absence of a competitive inhibitor and a K _?^??? of 12 μM in the presence of 3 μM of the inhibitor. Calculate the K_I.

• A. 1/6
• B. 1.5
• C. 3
• D. 6

Answer: (D)

Use the Lineweaver-Burk plot below to answer the following questions. Units of [S] are in nM.

A) Estimate the values of K_m & V_max as well as the K_?^??? & V_???^??? for the reactions in the absence and presence of the competitive inhibitor.
B) Would you expect the competitive inhibitor to be more effective under conditions of high or low [S]? Why?
C) If [I] = 10 nM, calculate the inhibition constant (K _I).

Alright. So here we have a 3-part practice problem broken up into part a, part b, and part c. It says to use the Lineweaver-Burk plot below to answer the following questions, and it tells us that the units of substrate concentration are in units of nanomolar. Taking a look at our Lineweaver-Burk plot down below, notice we actually have two lines. We have this black line here that corresponds to the enzyme-catalyzed reaction in the absence of inhibitors, so minus inhibitor, and then we also have this green line here that represents the enzyme-catalyzed reaction in the presence of competitive inhibitor. Looking at part a, it's asking us to estimate the values of the Michaelis constant (Km) and the Vmax, as well as the values of the apparent Michaelis constant and the apparent Vmax for the reactions in the absence and the presence of competitive inhibitor. What we need to recall in order to answer part a is that the intercepts in the Lineweaver-Burk plot are really what are going to reveal the Km and the Vmax. Recall that the x-intercept is just going to be a reciprocal. And because it falls on the negative region over here, we know it's going to be the negative reciprocal of the Km. And, of course, the y-intercept is also going to be a reciprocal. It's going to be the reciprocal of the Vmax. Starting with the absence of inhibitor, we can start with this black line that we see here, and we can also start with the x-intercept. The x-intercept is where this black line crosses the x-axis, which is right at this point right here, which corresponds with negative 0.2. So, we can say that the x-intercept in the absence of inhibitor is equal to negative 0.2 and that it is going to be equal to negative one over Km. Algebraically solving for the Km, we move it to this side of the equation by multiplying both sides of the equation by Km. What we get is the Km times negative 0.2 is equal to negative 1. Then we divide both sides of the equation by negative 0.2, and what we get is Km is equal to negative 1 divided by negative 0.2. If you take your calculator and do negative 1 divided by negative 0.2, you'll get an answer of 5. Since the units of the substrate concentration are in nanomolar, we can say that the units of the Km are also going to be in nanomolar. So, now that we've calculated the Km in the absence of inhibitor as 5 nanomolar, we can go ahead and do the same for the presence of inhibitor. In the presence of inhibitor, we know that the Km is going to be referred to as the apparent Km. We look at the x-intercept, which is where this green line is going to cross the x-axis, which corresponds with the value of negative 0.1. We can say that the x-intercept is going to be equal to negative 0.1, which is equal to the reciprocal of the Km, so negative one over Km. This is in the presence of inhibitor, so the Km is actually referred to as the apparent Km. We are going to solve for the apparent Km by moving the apparent Km to this side of the equation by multiplying both sides of the equation by apparent Km. What we get is that the apparent Km times negative 0.1 is going to be equal to negative 1, and then we divide both sides of the equation by negative 0.1. When we do that, what we get is that the apparent Km is equal to negative 1 divided by negative 0.1. If you take this in your calculator, what you'll get is 10, and the units are again going to be nanomolar. Now that we have the apparent Km as 10 nanomolar and, the Km as 5 nanomolar, we can move on to calculating the Vmaxes and the apparent Vmaxes. The Vmax is going to be indicated by the y-intercept, and the y-intercept is where these two lines cross the y-axis, which is at the same exact point. This indicates that in the presence and absence of competitive inhibitor, they have the same exact Vmax and we know that is definitely true when it comes to competitive inhibitors. This value here is approximately equal to a value of 0.075. It's the midway point between these two markers. So it's approximately 0.075, and this is going to be equal to 1 over the Vmax. Essentially, we're going to solve for the Vmax here, and we can move it to this side of the equation by multiplying both sides of the equation by Vmax. What we get is Vmax times 0.075 is equal to 1. We divide both sides of the equation by 0.075, and what we get is Vmax is equal to 1 over 0.075. If you type that in your calculator, what you'll get is an answer of about 13.33 repeating. Because the Vmax is the same in the presence and absence of inhibitor, we can say that this is also equal to the apparent Vmax as well. For part b, would you expect the competitive inhibitor to be more effective under conditions of high or low substrate concentration and why? Competitive inhibitors are going to compete with the substrate. The competitive inhibitor will be more effective if there is less substrate to compete with it. Therefore, the competitive inhibitor is going to be more effective at low substrate concentrations. Low substrate concentrations mean less competition between the substrate and the inhibitor, allowing the competitive inhibitor to be more effective. If we look at our Lineweaver-Burk plot on the right, we see that at low substrate concentration, which corresponds with further to the right of our x-axis here, the competitive inhibitor line starts to be higher than the absence of that competitive inhibitor line. Because it is higher up, it's further away from the 0 infinity marker, which means the initial reaction velocities are going to be smaller at lower substrate concentrations. For part c, if the concentration of inhibitors is equal to 10 nanomolar, calculate the inhibition constant of the free enzyme Ki. Recall that in the presence of competitive inhibitors, the apparent Km is defined as alpha, the degree of inhibition on the free enzyme, times the Km. We were able to determine the apparent Km as 10 nanomolar, and the Km in the absence of inhibitor as 5 nanomolar. So we actually have values to be able to solve for alpha, the degree of inhibition on the free enzyme. The apparent Km, which is 10 nanomolars, is going to be equal to alpha times the Km in the absence of inhibitor, which is 5 nanomolar. Solving for alpha, we divide both sides of the equation by 5 nanomolar. What we get is alpha is equal to 10 nanomolar divided by 5 nanomolar. Entering this into your calculator, you find that alpha is equal to 2. With the degree of inhibition on the free enzyme alpha, with a value of 2, we can go ahead and plug it into this equation where we know from our previous lesson videos that the degree of inhibition on the free enzyme alpha is defined as 1 plus the ratio of the concentration of inhibitor over the inhibition constant. Again, here we define alpha as 2, which equals this expression here. So, it's going to be equal to 1 plus the concentration of inhibitor, given to us as 10 nanomolar, over the inhibition constant Ki. We are going to solve for the Ki by subtracting 1 from both sides of the equation. When we do that, we get 2 minus 1, which is 1, and that's going to be equal to 10 nanomolar over the Ki. Algebraically solving for the Ki, we move it to this side of the equation by multiplying both sides of the equation by Ki, and then we divide both sides of the equation by 1. When we do that, the Ki is just going to be equal to 10 nanomolar, because 10 nanomolar over 1 is the same as 10 nanomolar. The inhibition constant is going to be 10 nanomolar if the concentration of inhibitor is 10 nanomolar. The Ki is going to be equal to 10 nanomolar. This concludes our 3-part practice problem. I'll see you guys in our next video.

Here’s what students ask on this topic:

What is competitive inhibition in biochemistry?

Competitive inhibition occurs when an inhibitor competes with the substrate for binding to the enzyme's active site. This type of inhibition is reversible and is characterized by the inhibitor being structurally similar to the substrate. As a result, the inhibitor can bind to the active site, preventing the substrate from binding. This decreases the initial reaction velocity (v₀) and increases the apparent Michaelis constant (K_m), but does not affect the maximum reaction velocity (V_max). The key feature of competitive inhibition is that the inhibitor only binds to the free enzyme, not the enzyme-substrate complex.

How does competitive inhibition affect the Michaelis-Menten plot?

In a Michaelis-Menten plot, competitive inhibition increases the apparent K_m without changing the V_max. This results in a rightward shift of the curve, indicating a higher substrate concentration is needed to reach half of V_max. The equation for the Michaelis-Menten plot in the presence of a competitive inhibitor is modified to include the apparent K_m (α × K_m). Despite the shift, the V_max remains the same because the substrate can outcompete the inhibitor at high concentrations.

What is the effect of competitive inhibitors on Lineweaver-Burk plots?

In Lineweaver-Burk plots, competitive inhibitors increase the slope of the line, which is the ratio of K_m to V_max. The y-intercept, which represents 1/V_max, remains unchanged because V_max is not affected by competitive inhibitors. However, the x-intercept, which is the negative reciprocal of K_m, shifts closer to zero, indicating an increase in K_m. This results in a steeper line as the concentration of the competitive inhibitor increases.

Why do competitive inhibitors increase the apparent K_m?

Competitive inhibitors increase the apparent K_m because they decrease the concentration of free enzyme available to bind with the substrate. When the inhibitor binds to the enzyme, it forms an enzyme-inhibitor complex (EI), reducing the number of free enzymes. According to Le Chatelier's principle, this shifts the equilibrium towards the formation of more free enzyme and substrate, making it appear as though the enzyme has a lower affinity for the substrate. This results in an increased apparent K_m.

How can the effects of competitive inhibitors be overcome?

The effects of competitive inhibitors can be overcome by increasing the substrate concentration. Since competitive inhibitors and substrates compete for the same active site on the enzyme, a higher concentration of substrate can outcompete the inhibitor. This allows the enzyme to bind more substrate molecules, effectively negating the impact of the inhibitor and allowing the reaction to proceed at its maximum velocity (V_max).