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6. Enzymes and Enzyme Kinetics

1

Calculating Km

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in this video, we're going to begin our discussion on the tools that we have to calculate the Michaelis constant or the K M oven enzyme. And so, just like the V Max, the K M can also be calculated in multiple ways. And so we're actually going to talk about two predominant ways to calculate for the K M of the enzyme. And in this video we're going to focus on the very first way. And then in our next listen video, we'll talk about the second way to calculate for the K M. And so the K M, just like the V Max, can also be calculated by algebraic rearrangement of both. The McKell is meant in and the line Weaver Burke equations. And so notice down below. In our example, it's asking us toe algebraic Lee rearrange. The McHale is meant in equation in order to solve. For the McHale is constant or the K M. And so notice down below we have The McHale is meant in equation over here, on the left and on the right. What we have is the line Weaver Burke equation, and we know from our previous lesson videos that these two equations are reciprocal of one another so they can be converted between each other just by taking the reciprocal. And so if we solve for the K M in the McHale is meant in equation over here. That just means that we want to isolate this K m variable all by itself. And so we can start by getting this k m out of the denominator. So essentially just moving it out of the bottom of this fraction. And we could do that by essentially just taking the entire denominator and moving it up here. And the way that we could do that is by multiplying both sides of the equation by the entire denominator. And when we do that, we get rid of the denominator on the right side so that we're on Lee left with the top, which is just V Max times the substrate concentration. And then, of course, on the left side, what we've done is we've taken this expression and we've moved it up here. So we're gonna have RV not times the denominator, which is K m plus the substrate concentration. And again, what we're trying to do is solve for the K m So we wanna isolate for this variable. So what we can easily do is get rid of this V not. And we can do that by dividing both sides of the equation to move it to the other side. Eso essentially. What we'll do is we'll still have our v, Max, Um, time substrate concentration over here. But now we've got the V not, uh, being dividing this entire expression. And so on the left, what we're left with is just this expression right here. So we still have k m plus the substrate concentration. So now all we need to dio is again we're solving for this k m. So all we need to do is get rid of this substrate. And we can do that by subtracting the substrate concentration from both sides of the equation. And so what we end up with is the same exact expression over here on the left. So we still have our V, Max, uh, times the substrate concentration, And this is going to be over the initial reaction velocity. But now we're subtracting off the substrate concentration. And so what we're left with on the left hand side of the equation is just the K M. And so now we can see that R. K M is essentially isolated. However, we can further simplify this expression here. We can actually, uh, take out a common factor here. So we can actually factor out both of these substrate concentrations here, and we can factor them both out to the front and eso When we do that, Essentially, what we get is the substrate concentration out in the front on then, of course, we just have V Max over the initial reaction velocity. So v Max over the initial reaction velocity. And then, uh, this when it's factored out, it turns into, ah one. And so, really, this here is the reduced form for rearranging the Michaelis Menton equation to solve for the K M. Just like we wanted to dio And so recall from our previous lesson videos, we had defined the k m, uh, in several ways. And one of the ways that we define the K M was the exact substrate concentration that was needed to get the initial reaction velocity to half of the V max. And so essentially, what that saying is if we were to rewrite this expression over here, and we were to try to rewrite this expression. We could say that the k M actually, let's move this up. We would say that the K M is going to be equal to the substrate concentration, which is gonna be multiplied by the V max. And then, um, it's gonna be over the initial reaction velocity minus one. So here we have the initial reaction velocity. But we said that the K M is the exact substrate concentration where the initial reaction velocity is half of the V max. So that means that in the bottom, what we really have is half the max for the initial reaction velocity. And so if you take uh, v Max divided by half V max, what you end up getting is you end up getting to and so essentially what you end up getting is to, uh, minus one here, time substrate, times the substrate concentration, and this is going to be still equal to the K M. And so to minus one is just going to be one, and then one times the substrate concentration is just going to be the substrate concentration, and so you can see that the K M will be equal to the substrate concentration and so you can see that the K M is literally just a substrate concentration. It's the substrate concentration that allows for the initial reaction velocity to be equal to half of the V max. And so that is how we define the K M in one of our ways. So that concludes, our first method for calculating the McHale is constant or the K M oven enzyme. And in our next lesson video, we'll talk about the second predominant way to calculate the K M. So I'll see you guys there.

2

Calculating Km

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So in our last lesson video, we talked about one way to be able to calculate the McHale is constant or the K M oven enzyme. And that was to algebraic Lee rearranged the meticulous meant an equation. Now, in this video, we're gonna talk about a second alternative way to calculate the K M oven enzyme. And so recall in our previous lesson videos where we talked about the K M oven enzyme, we said that the K M oven enzyme can be defined in multiple ways, and it could be expressed with rate constants. And so recall that the K M is really just the ratio of the some of the enzyme substrate complexes to dissociation rate constants, which our K minus one, uh and, uh, K two over its association rate, constant K one. And so notice down below and our image on the left. Here we have our typical enzyme catalyzed reaction at the very, very, very beginning of the reaction. So we have the initial V not here to remind us of that, and that's because we can see that the rate constant K minus two is being ignored here. That's why it's not being shown and so recall that, uh, the enzyme substrate complex can disassociate in two different ways. It can associate backwards via K minus one to form the free enzyme unfree substrate. And it can also dissociate forwards via K two to form the free enzyme and free product. And so the enzyme substrate complex dissociation rate constants are just going to be K minus one backwards and K two forwards. And so we know that this is going to be over the enzyme substrate complex association rate constant. So that's going to be the free enzyme and the free substrate associating with each other via K one to form the enzyme substrate complex. So that would be K one down below in the bottom. And so this is one way to be ableto or another way to be able to calculate the K M of the enzyme is to just take the some of the dissociation rate constants over the association rate concert as stated up above and so also recall from our previous lesson videos. We we additionally had said that this also correlates with the ratio of the free enzyme concentration times, the free substrate concentration over the concentration of the enzyme substrate complex. And so that's because the free enzyme in free substrate is a measure of the enzyme substrate complex dissociation, whereas the concentration of enzyme substrate complex is a measure of the enzyme substrate complex association. So over here we can say that it's a measure of the free enzyme concentration times, the free substrate concentration over the concentration of the enzyme substrate complex. And so this is really correlating, uh, thio other ways that we are calculating r k m. And so again, all of this is review from our previous lesson video. So no new information here. It's just a refresher. And so also recall from our previous lesson videos that a small value of the K M actually means that there's going to be less dissociation of the enzyme substrate complex. And there's going to be mawr enzyme substrate, complex formation or association. So, essentially, if the K M value is really, really small, Aziz, this is indicating over here. Uh, the only way that we could get the K M value small is if we say that there is, uh, a little bit of enzyme substrate complex free enzyme in free substrate. So if the K M is small, we could say that there's a little bit of free enzyme in free substrate and a lot of the enzyme substrate complex form or of the enzyme substrate complex. So, again, this is all refresher from our previous lesson videos, and we'll be able to utilize thes different methods of calculating the K M and our practice problems, so I'll see you guys in our next video.

3

Calculating Km Example 1

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All right. So here we have an example problem that says the following rate constants were measured for a simple enzyme catalyzed reaction determined the McHale is constant or the k m for the enzyme and noticed that were given the values for the three relevant rate constants for a simple and I'm catalyzed reaction. And then we've got four potential answer options for the K M of the enzyme. So recall from our previous lesson videos, we talked about two predominant ways to calculate the K m of an enzyme. And so notice the first way, which is algebraic rearrangement of the meticulous man equation, uh, noticed that we're not actually given the values for the initial reaction velocity, the V max or the substrate concentrations. Which means that we cannot use the first method here off algebraic Lee rearranging the McHale Ismet in equation to determine the K M. And so, for that reason, we can actually just go ahead and just remove this completely. And, um, essentially, what that means is we're going to need to use our second method of calculating the K M and recall that the second method of calculating the K M is related to the rate constants and expressing km with rate constants. And so we know that a simple enzyme catalyzed reaction, uh, shown here is going to have three relevant rate constants. And we know that the K M is going to be the rate constants for the enzyme substrate complex dissociation over the association of the enzyme substrate complex. So essentially, what we need to recall is that the enzyme substrate complex can disassociate in two different ways. It can associate backwards via K minus one to form the free substrate and free enzyme. And it could associate Ford's Via K two to form the free enzyme and the free product. And so it's going to be the dissociation over the association. So these two dissociation is here. It's gonna be the some of them. So it's gonna be K minus one plus K two, and then it's gonna be over the rate constant for the association of the enzyme substrate complex. So that would be via K one to form the enzyme substrate complex. And so what we could do is we can put the K one on the bottom, and so we know that these rate constants here. Uh, the dissociation of the enzyme substrate complex is associated with the free enzyme and the free substrate and the association of the enzyme substrate complex is, of course, associated with the concentration of the enzyme substrate complex. And so this is another way to express this equation. Now, notice that we're not actually given the values for the concentrations of the enzyme or the substrate or the enzyme substrate complex. So we're not gonna be able to use that portion of the equation. And we're gonna be focusing specifically on this portion right here. And so notice that were given K one a Z two times 10 to the, uh, two times 10 to the eighth Inverse similarity in verse seconds. And then we're given the value of K one and K two as well. So all we need to do is plugging these values into our expression here to calculate for the K M. So essentially, what we see here is that the K M is going to be equal to, uh, K minus one, which is given to us as one times 10 to the third, So it will be one times 10 to the third inverse seconds plus the concentrate or I'm sorry. Plus the value of the K two, which is five times 10 to the third inverse seconds on then. Of course, uh, this is all going to be divided by K one Theus Association rate constant, which is two times 10 to the eighth and units of inverse molar ity Inverse seconds. And so, essentially, if you take your calculators and you do one times 10 to the third, plus five times 10 to the third and divide that answer by two times 10 to the eighth, which will get is the answer for R K M, which is equal to three times 10 to the negative fifth Moeller. And so we can see that three times 10 to the negative fifth. Moeller matches with answer option A so we could go ahead and indicate that a Here is the correct answer for this example problem and that, um, concludes this example problem and we'll be able to apply these concepts moving forward and our practice problems. So I'll see you guys there

4

Problem

To determine the Km from a Lineweaver-Burk plot you would:

A

Multiply the reciprocal of the x-axis intercept by -1.

B

Multiply the reciprocal of the y-axis intercept by -1.

C

Take the reciprocal of the x-axis intercept.

D

Take the reciprocal of the y-axis intercept.

5

Problem

The V_{max} for an enzyme is 9 mg/min. Calculate the K_{m} if the [S] = 5 mM when the V_{0} = 3 mg/min.

A

100 M.

B

10 M.

C

10 mg.

D

10 mM.

6

Problem

Calculate the Km of an enzyme using Michaelis-Menten kinetics if the forward rate constant for ES formation is 4.3 x 10^{6} M^{-1}s^{-1}, the reverse rate constant for ES dissociation into E + S is 2.4 x 10^{2} s^{-1}, and the forward rate constant for ES dissociation into E + P is 1.2 x 10 ^{3} s^{-1}.

A

3.35 x 10^{-4} M^{-1}.

B

3.58 x 10^{3} M.

C

3.85 x 10^{3} M^{-1}.

D

3.35 x 10^{-4} M.

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