in this video, we're going to introduce the Michaelis Constant or the K M oven enzyme. So the McHale is constant is an enzyme kinetics variable that's abbreviated as just K M. And so the K M or the McHale is constant as well. See, moving forward in our course can actually be defined in several different ways. And so again, as we move forward in our course, we'll talk about all of those different ways to define The McHale is constant or the K M. But as we can see in this lesson video, one of the ways that we can define the McHale is constant or the K M is as an exact substrate concentration. In fact, the K M is the exact substrate concentration at which the initial reaction velocity or the V not is exactly equal to one half of the theoretical maximal reaction velocity or the V Max. And so again, because the McHale is constant or the K M is an exact substrate concentration. Then we can say that when the value of the substrate concentration is exactly equal to the value of the K M. That exactly half of all of the available enzyme active sites are going to be full or occupied with substrate to form the enzyme substrate complex. And so let's take a look at our image down below to clear some of this up. And so notice on the right. What we have is our enzyme kinetics plot that we've seen so many times in our previous lesson videos. So notice on the Y Axis, what we have is the initial reaction velocity. And on the X axis, what we have is the substrate, concentration and units of malaria. He and of course, we can see that are green curve. Right here is the same curve that we've seen so many times before in our previous lesson videos. And so we already know from our previous lesson videos on the V max oven enzyme that the V Max acts as a horizontal Assam tote toe limit. The initial reaction velocity of an enzyme catalyzed reaction and so we can see that this curve right here will get really, really, really close to the V max. But it's not actually attained by any enzyme. And so that's why we said that the V Max is better to find as a theoretical maximal reaction velocity. Now, in our previous lesson videos, we said that the V max can Onley be attained under saturating substrate concentrations where the substrate concentration is really, really high. And, uh, that is the only way where the initial reaction velocity can approach the V max. So the question is, how do we determine the substrate concentration that's needed to get half of all of the available enzyme active sites occupied? Well, it's pretty easy. All we need to do is find the value of the V max and then cut that value in half. Or essentially take the V max over to. And so, essentially, the point on our curve that corresponds with half of the V max is going to, uh, correlate with the substrate concentration that equals the value of the K M. And so what we can see here is that the K M is really just expressed as an exact substrate concentration, and it's the exact substrate concentration that's needed to get the initial reaction velocity equal to exactly half of the V max. And so what's really important to note here is that the K M is on Lee associated with ah, half of the V max. But the K M is not equal to the value of half of the V Max. The K M instead is an exact substrate concentration. So it has units of substrate, concentration or units of concentration. And so another way to look at this K m here is over here in this box. And so, as we mentioned already up above in our lesson that when the value of the substrate concentration is equal to the value of the K. M, we can say that exactly half or 50% of all of the available enzyme active sites are going to be full or occupied with substrate. And so notice over here, what we have are enzymes as these brown structures and these little dots here are our substrate. And so notice that we have a total of six enzymes here. And so when the value of the substrate concentration is equal to the value of the K M, notice that exactly half or 50% of all of these enzymes here are actually of occupied or full with substrate to form the enzyme substrate complex. So notice that we have three enzymes here uh, in the form of the enzyme substrate complex, which is exactly half of all of our enzymes. And so what we can say again is that when the substrate concentration value is equal to the value of the K M, that exactly half of all of the available enzymes, or half of the total concentration of enzyme, is going to equal the concentration of the enzyme substrate complex. And so what's also important to note, Um, as we move forward in our course, we'll talk more about this idea later. But it's important to note now that the K M or the McHale is constant, is actually an intrinsic property of an enzyme, which means that it does not get affected by the amount of enzyme that's present. It's a property of the enzyme that won't change. Uh, regardless of how much enzyme is present. However, the K M oven enzyme can actually change depending on the conditions. And so what we can say is that the K M will vary under different conditions such as different. PH is different temperatures or even different solvents. And so this concludes, our introduction to the McHale is constant and again as we continue to move forward in our course, we'll talk. Maura, Maura, about this. McHale is constant, so I'll see you guys in our next video.

2

Problem

Problem

What is the initial velocity of a reaction when the concentration of substrate is set equal to the K_{m}?

A

V_{0} is equal to V_{max}.

B

V_{0} is equal to one-half [S].

C

The [S] is so low that V_{0} is negligible.

D

Not enough information to determine V_{0}.

E

V_{0} = one-half V_{max}.

3

concept

Km Enzyme

Video duration:

8m

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So we already know at this point that the McHale is constant or the K M can actually be expressed in multiple ways. And in our last lesson video, we literally said that the K M waas a substrate concentration and more specifically, what we said was that the K M. Was the exact substrate concentration that allows for the initial reaction velocity Thio equal half of the V Max. Now, in this lesson video, we're gonna focus on how the McHale is constant. K M can also be defined as a substrate concentration when it is expressed with rate constants. And so what I want you guys to recall from our previous lesson videos is that reaction rates or reaction velocity symbolized with the can be expressed by rate laws. And we know that rate laws incorporate rate constants which are abbreviated with the lower case, letter K. And of course, we also already know from our previous lesson videos that initially at the very, very, very beginning oven enzyme catalyzed reaction, there are Onley three relevant rate constants and so down below in our image here of our enzyme catalyzed reaction, we can see that this is the reaction initially at the very, very beginning of the enzyme catalyzed reaction. And we know that because there are only three relevant rate constants K one K minus one and K two and the reverse rate constant here, uh, for K minus two is negligible at the very, very beginning of the enzyme catalyzed reaction. And that's why we have this initial reaction velocity symbol here. Don't remind you guys that this is at the very, very beginning. Now all of this is pretty much review from our previous lesson video. So what I want you guys to know in this video is that the meticulous, constant K M can also be defined as just a compilation of these three rate constants that were already familiar with from our previous lesson videos. And so you can see that the K M is also defined by this ratio right here. Now I'll admit, at first glance, this ratio might look pretty random, but this ratio is actually not random at all. In fact, this ratio is defined specifically under conditions known as steady state conditions, which will talk more about later in our course. But for now, all I want you guys to know is that this ratio right here that defines the McHale is constant. K M is not random, and it's actually defined under steady state conditions. And so again, as random as this ratio might appear, if we take an even closer look at this ratio, we'll see that it's just the ratio of the some of the two dissociation rate constants for the enzyme substrate complex over the association rate constant for the enzyme substrate complex. And that's exactly what we're saying. Down below is that the Michaelis constant K M is defined as the enzyme substrate complex dissociation over the enzyme substrate complex association. And of course, if we take a look at our enzyme catalyzed reaction over here on the left, notice that the enzyme substrate complex can actually disassociate in two different ways, it can disassociate backwards via K minus one to form the free enzyme in the free substrate. Or the enzyme substrate complex could also dissociate forward via K two to form the free enzyme and the free product. And so that's why we can take a minus one and K two and some them together to get our rate, uh, to get our dissociation of the enzyme substrate complex. And then, of course, we know that the enzyme substrate complex can on Lee associate in this reaction here via K one. And that's why the for the association on the bottom, we place in K one now again from our last lesson video, we said that the K M. Was an exact substrate concentration. And so here, even when the K M is expressed as this ratio of rate constants, it still represents that same substrate concentration. And I can prove it to you guys just by looking at the units. And so it turns out that the McHale is constant or the K M. Even when it's expressed as this ratio of rate constants, it still has units that are equal to molar ity. And polarity is, of course, a unit of concentration, which shows that the K M is still a substrate concentration. And so we could have actually determined this by using the previous knowledge that we acquired from our previous lesson video. So we know that both K minus one and K two Onley have one reactant, which is the enzyme substrate complex, and because they only have one reactant. They are both first order rate constants. And we know that first order rate constants have units off inverse seconds from our previous lesson videos. And so, if we were to just focus on the units down below and say that K minus one and K two have units of inverse seconds, so we could say in verse seconds plus inverse seconds. If we're just focusing on the units now, of course, K one, on the other hand, instead of having one reacted, it actually has to react. It's. And because it has to react, it's K one is a second order rate constant. And we know from our previous lesson videos that second order rate constants have units of inverse polarity in verse seconds. And so again, I'm proving to you guys how the K M still has units of morality. So this units here must somehow, uh, convert or simplify two units of polarity. So essentially inverse seconds plus inverse seconds. Uh, the final units on the top number, regardless of what the value is the units, if we're only focusing on the units three units air still just gonna be inverse seconds on the top. And then, of course, the bottom is still gonna have inverse polarity, Inverse seconds. And so here, what we can see is that the units of inverse seconds cancels out. And then what you end up getting is one over in verse, molar ity. And, of course, one over in verse. Polarity is the same exact thing as just saying molar ity. And so this goes to show how this compilation of rate constants in this ratio here still has units of mole aren t which means that K M is still a substrate concentration just like what we said in our last lesson video. Now, what I also want you guys to note is that, uh, it has to be the dissociation over the association of the enzyme substrate complex and it can't be the reverse. It can't be association over dissociation because if it were like that, then the units would turn out to be inverse mole, aren t And that would mean that KM has units of inverse similarity. An inverse mole Arat e is not a unit of substrate concentration. So that's how we know it has to be the dissociation over the association and not vice versa. And so later, through our course, will be able to talk more about exactly how, uh, the K M is not just the substrate concentration, but an exact substrate concentration that allows for the initial reaction velocity to equal half of the V. Max. We'll talk more about that later in our course. But for now, all I want you guys to really know from this video is that the K M can also be expressed with rate constants in this format. Right here. So that concludes his lesson video, and I'll see you guys in our next video.

4

Problem

Problem

Select the best description of the K _{m}.

A

Equal to the product concentration at initial reaction conditions.

B

Equal to the substrate concentration when the reaction rate is half its maximal value.

C

Equal to the ratio of the sum of the ES dissociation rate constants over the ES association rate constant.

D

More than one of the above are true.

5

concept

Km Enzyme

Video duration:

7m

Play a video:

So from our last lesson video, we know that the McHale is constant or the K M is just a substrate concentration, and it can be expressed with rate constants. Now, in this lesson video, one of the things that we're going to do is defined. Yet another way to express the McHale is constant. K M. And so the K M can also be defined or expressed as the ratio of the free enzyme concentration times, the free substrate, concentration over the enzyme substrate, complex concentration. And so notice down below. We can refresh our memories that the McHale is constant. K M is just expressed as the enzyme substrate complex dissociation over the enzyme substrate complex association. And we know that this ratio of the rate constants defines the K M from our last lesson video. And so here. What we're saying is that the dissociated form of the enzyme substrate complex is really just going to be the free enzyme and the free substrate. And of course, the associated form of the enzyme substrate complex on bottom is just going to be the enzyme substrate complex. And so it's this ratio here that also defines the McHale is constant. K m. And again, we already know from our previous lesson videos that even though these ratios here might seem random, they're actually not random at all. And they're derived under conditions known as steady state conditions, which again, we're going to talk about a little bit later in our course. But for now, all I want you guys to know is that the McKell is constant. K M is defined by these ratios now, because the K M is defined by these ratios, it means that the K M measures and enzymes binding affinity for its substrate and by affinity. What we mean is the strength of its attraction. Affinity is essentially the same as the attraction. And so it turns out that the McHale is constant. K M actually has an inverse relationship with an enzymes binding affinity for its substrate. And so what I mean by that is that the greater the larger the value of the K M is, the smaller the binding affinity and enzyme has for its substrate, or the weaker the binding affinity and enzyme has for its substrate. And so just to clear that up, let's take a look at our image on the bottom left, uh, down below. And so notice that when we have a large value of the K m indicated by this large up arrow and large size of the K m. Notice that that means that the dissociation of the enzyme substrate complex is going to be quite large. And of course, this means that the association of the enzyme substrate complex is going to be quite small. And so if the enzyme substrate complex is dissociating ah, lot, that means that it's breaking apart a lot. And that must mean that the enzyme does not have a very strong affinity for its substrate. And so this means that when we have a large K M that the enzymes affinity for its substrate will be weak. And so we will have a week affinity. Now, with the reverse scenario, if we have a small K M value of the K M indicated by the small, uh, downwards arrow in the small size of the K M notice. This means that this time the dissociation of the enzyme substrate complex is going to be quite small, and instead the association of the enzyme substrate complex is going to be quite large. And so, of course, if there's ah lot of association of the enzyme substrate complex going on, that means that the enzyme in the substrate will be associating together to form the enzyme substrate complex. And that means that the enzyme must have a strong affinity for the substrate. And so that is exactly what a small km indicates is a strong affinity that the enzyme has for its substrate. And so notice that in this example, problem here it's asking us which enzyme has a stronger affinity for its substrate. Is it enzyme a or is it enzyme B and down below were given this, uh, enzyme kinetics plot here where we have the initial reaction rate, or V not on the Y axis in the substrate, concentration and units of polarity on the X axis. And so, just like we indicated before, it's actually going to be the magnitude of the K M. That dictates the strength of the affinity. And so what we're going to want to do is determine the K M for both of these enzymes, so notice that we have to curves here. We have this red curve here for enzyme B, And then we have this blue curve here for enzyme s. So we need to determine the K M in order to determine which one has a stronger affinity for the subject. And so if we start with enzyme, be, uh, noticed that enzyme B has a V max that's showing up somewhere horizontally over here. And so we know that the K M from our previous lesson videos is a substrate concentration. In fact, it's the substrate concentration that allows the initial reaction velocity to equal half of the V max. And so, if here we have our V max half of the V, max is gonna be somewhere over here. And so that means that the point here on our curve that corresponds with half of the V Max is going to be really, really low and associate, uh, somewhere over here. So the red and enzyme B has a K M. That is really, really low and corresponds somewhere over here. Now, if we take a look at enzyme a notice that it's V Max is somewhere way up here and so half of the V Max is going to be somewhere about that corresponds somewhere around here, and that corresponds with this point. And this point here corresponds with the K M. That's right around this region right here. And so by doing this, we can clearly see that the K M of enzyme A is actually greater than its larger than the K M of enzyme be and so remember that a larger K M means a weaker affinity. So this means that enzyme A because it has a larger value for the K M. It has a weaker affinity. And that means that, of course, K. M of enzyme B is smaller and has a stronger affinity. So that means that clearly the struck the enzyme that has the stronger affinity for its substrate is going to be enzyme be. And so we could go ahead and indicate that it's enzyme be here. That is the correct answer for this problem. And so this is, uh, concludes our lesson on how the K M the McHale is constant. K M can be used as a measure of enzymes affinity for its substrate, and we'll be able to get practice utilizing these concepts here in our next couple of practice videos. So I'll see you guys there

6

Problem

Problem

According to the chart below, which one of the following enzymes has the strongest affinity for its substrate?

A

Lysozyme.

B

Penicillinase.

C

Î²-Galactosidase.

D

Chymotrypsin.

E

Carbonic anhydrase.

7

concept

Km Enzyme

Video duration:

3m

Play a video:

So in one of our previous lesson videos, we mentioned the fact that the Michaelis Constant or the K M, is an intrinsic property of an enzyme. And so all this means is that the K M is completely independent of the enzyme concentration. And so what this means is we can say that when the value of the substrate concentration is equal to the value of the K M. That exactly half of all of the available enzyme active sites are going to be occupied with substrate, and this will be true regardless of the total enzyme concentration. And really, that's the main take away of this video. The fact that the total enzyme concentration does not affect the Michaelis constant K M. And so recall from our previous lesson videos on the V max oven enzyme that the total enzyme concentration does actually affect both the initial reaction velocity and the maximal reaction velocity. But what we're learning in this video, however, is that altering the total enzyme concentration does not affect the K M oven enzyme. And so if we take a look at our image down below notice, we have this enzyme kinetics plot where we have the initial reaction velocity on the Y axis and the substrate concentration on the X axis and notice that we have these two different curves and our plot. We've got this red curve right here and we've got this green curve right here and notice the Onley difference between these two curves in our plot is the total amount of enzyme. So notice that the green curve has double the total amount of enzyme in comparison to the red curve below. And so again, as we already know from our previous lesson, videos on the V max oven enzyme doubling or changing the total enzyme concentration is also going to alter both the initial reaction velocity shown by the change in the curve. And it's going to change the maximal reaction velocity in comparison to this one that's down below. And so what we're learning again in this video is that even though varying the total enzyme concentration changes both the initial and the maximal reaction velocity, it does not change the K M. And so if we take a look at the K M here, notice that it has the same exact value for both of these curves the red and the green curve, despite the fact that they have different amounts of total enzyme concentration. And so I noticed that the K M is exactly the same, however, noticed that half the V Max is going to be different for both of these curves. And that's because the V Max itself is different for both of these curves. However, because the K M stays the same, that tells us that the K M is an intrinsic property of the enzyme that's independent of the total enzyme concentration. So that concludes our lesson, and we'll be able to apply these concepts as we move forward in our course, so I'll see you guys in our next video.

8

Problem

Problem

Indicate which region of the Enzyme Kinetics plot below best corresponds to each statement.

A) Initial reaction velocity is limited mainly by the [S] present: ______

B) Initial reaction velocity limited mainly by the [E] present: ______

C) The active site of an enzyme is most likely free/unoccupied: ______

D) The active site of an enzyme is most likely occupied by substrate: ______

E) This region includes the points corresponding to Km & Â½Vmax: ______

Video duration:

5m

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9

Problem

Problem

Use the data in the following chart to determine the Km of the enzyme.