Learn the toughest concepts covered in Biochemistry with step-by-step video tutorials and practice problems by world-class tutors

6. Enzymes and Enzyme Kinetics

1

Calculating Vmax

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in this video, we're going to begin our discussion on some of the tools that we have to calculate the theoretical maximal reaction velocity or the V Max. And so it turns out that the V max can actually be calculated in multiple ways. And we're really going to talk about two primary ways to calculate the V Max. And so the first way that the V Max can be calculated is just by algebraic rearrangement. It can be calculated by algebraic rearrangement specifically of the meticulous meant in, or the line Weaver Burke equations and so down below. In our example, we're going to algebraic Lee. Rearrange. The McHale is meant in equation to solve for the V Max, and so notice down below. What we have is the meticulous mental equation right here on the left and on the right. What we have is the line Weaver Burke equation. And so what we need to recall is that the line Weaver Burke equation is literally just the reciprocal of the McHale is meant an equation. And so that means that the relationship between the McHale is meant in, and the line Weaver Burke equations is that they are reciprocal of one another. And so they could be pretty easily, uh, inter converted between each other. And so, essentially, starting over here with the McHale is meant an equation we can solve for the V, max. So we wanna isolate and solve for this variable, and we can start off by essentially just moving this entire denominator here. And so what we could do is just move this denominator up into this side right here. And we could do that by multiplying both sides of our equation here by the denominator k M plus substrate concentration. And what that does is it gets rid of the denominator on the right side. So we're just left with the V Max times the substrate concentration on the right. And then on the left side, all we have is the initial reaction velocity times this denominator. So essentially the initial reaction velocity times, the k m plus the substrate concentration. So now again, we want to continue to isolate and solve for this v Max, just like we're being asked to do in the example. And so what we could do is get rid of this substrate concentration by dividing both sides of the equation by the substrate concentration and so that will get rid of the substrate concentration on the right so that we have V Max all by itself on the right. And of course, we're going to have the same, uh, expression that we had up above. So we're gonna have the initial velocity times, the K M plus the substrate concentration on the top and all divided by the substrate concentration, uh, that we needed to do to get rid of this on the right. And so what, you can see years we've algebraic Lee rearranged. The McHale is meant in equation to solve for the V max. And so this is actually in expression that we can use to solve for the to calculate the V max and some of our practice problems, and so essentially will be able to utilize some of this moving forward in our course. But in our next lesson video, we're going to talk about a second major way that we can use to calculate the V Max. So I'll see you guys in that video

2

Calculating Vmax

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So, in addition to calculating the V Max through algebraic rearrangement of the McHale is meant in and the line Weaver Burke equations. We can also calculate the V Max through rate laws. And so what we need to recall from our previous lesson videos is that the V max is actually directly proportional to both the product formation rate constant, which is K two, and the total enzyme concentration. And so we did cover this in some of our previous lesson videos. But just to refresh our memories, let's recall some information from our previous lesson videos and so recall that biochemist mainly focus on measuring and plotting the initial reaction velocity, or the V not of enzyme catalyzed reactions. And so we know that the initial reaction velocity or the V not is also expressed as the change in the concentration of product over the change in time. And so because it's the product concentration that is relevant to biochemist, they actually focus on measuring the initial reaction velocity of the product formation step in our typical enzyme catalyzed reactions, and so notice down below. What we have is our typical enzyme catalyzed reaction, and we know that This is the initial reaction because notice that the rate constant K minus two here is being ignored. It's negligible. So we know that this is the initial enzyme catalyzed reaction and notice that with this initial enzyme catalyzed reaction, there's only one rate constant that effects the change in the product concentration. And that is the rate constant. K two. And so that's why biochemists are mainly focused on the product formation step oven enzyme, catalyzed reaction and the rate law for the product formation step and so down below. What we have is the rate law for the product formation step. And so we know that the rate law is really just another way to express reaction velocities. And so it's going to be the rate law is going to be the initial reaction velocity, which is equal to the rate constant, which is going to be K two for the product formation step so we can say its equal to K two, uh, times the concentration of the reactant. And so, for this product formation reaction right here, the reacting of this reaction is actually the enzyme substrate complex. So it's gonna be times the concentration of the enzyme substrate complex. And then we know it's going to be raised to the power of the reaction order. And because this is, ah, simple enzyme catalyzed reaction, we know that the coefficient, which is essentially one here, is going to be the reaction order, so we can assume that it's a one. And so what we're saying is that it's this expression right here that is the rate law for the product formation step. And again, this is all review information from our previous lesson video. So no new information in this video. And so the next thing that we need to recall from our previous lesson videos is that this initial reaction velocity So this initial reaction velocity V not is really the velocity that has the best chance at approaching the theoretical maximum velocity v max. And again, that's what we're interested in. Calculating in this video is the V Max, And so what we can say is that there is this possibility of variable substitution where we can substitute the initial reaction Velocity V, not with the V max, and that's exactly what we're going to do with this rate law down below. So recall that it's really, um, conditions that are under saturating substrate concentrations, where the initial reaction velocity can actually approach and be approximately equal to the theoretical maximum velocity. And so, under saturating substrate concentrations, we can actually take the initial reaction velocity in the rate law for the product formation step and substitute it with the, uh v max. And so that's exactly what we've done here. Substituted the initial reaction velocity with the V Max and also recall that under saturating substrate concentrations all of the enzyme, the total enzyme concentration is going to be associated with substrate. There's so much substrate because it's saturating. And so that means that the total enzyme concentration is going to approximately equal the concentration of the enzyme substrate complex. And so what that means is we could take the enzyme substrate complex here and substituted with the total enzyme concentration. And so that's exactly what we're doing here. We're substituting the, uh, concentration of the enzyme substrate complex with the total enzyme concentration. And of course, this rate law here is still going to be the rate law for the product formation steps. So it's still going to be K two here And so essentially, what we're saying is that this here is another way to calculate the V Max of a reaction through this rate law and so moving forward with our practice problems, we're going to need to identify which of the two methods do we need to use to calculate for the V Max. And so we'll be able to get some practice making these decisions and our practice problems, so I'll see you guys there.

3

Calculating Vmax Example 1

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All right. So here we have an example problem that wants us to calculate the maximum reaction velocity or the V max oven enzyme. If the Michaelis constant K M is equal to seven million Euler and the initial reaction velocity or V not is equal to 86.71 micro Mueller per second when the substrate concentration is equal to 25 million Euller. And so in our previous lesson videos, we talked about two predominant ways of calculating the maximum reaction Velocity v Max and so of these two ways. We can see that in this problem that were neither given the product formation rate, constant K two or the total enzyme concentration, which means that we're not gonna be able to use this method here to calculate the V Max. And that means that we're going to need to algebraic Lee rearrange. The McHale is meant in equation to solve for the V Max. And so I recall from our previous lesson videos that we can do that by multiplying both sides of the equation by the denominator to move it up here and then dividing both sides of the equation by the substrate concentration to move that away over here So that V Max is all by itself. And so when we do that, what we end up getting is an equation that says that V Max is going to be equal to the initial reaction velocity times, the k M plus the substrate concentration, um, all divided by the substrate concentration. And so, essentially, all we need to do to calculate the V Max in this problem is to plug in the values that we have for these variables. However, what we need to note is that these variables here are in different units. So the K M is in units of Millie Mueller. The substrate concentration is also in units of Millie Mueller. But the initial reaction velocity is in units of micro Moeller per second, and all of our answer options have units of molar ity. And so what this means is we might as well convert all of these units of concentration into units of mole arat e so that it matches one of our answer options. And so we need to do that first before we try plugging in the values into our expression. And so if we take this initial reaction velocity the 86.71 micro molars eso if we take 86 point first off, let's write. This is for the initial reaction velocity. And, uh, we said that it's 86.71 micro Moeller per second. Uh, let's exclude the second. We'll put the seconds back in afterwards, but we want to convert this micro Moeller into mole aren t so that it matches our answer option. So, uh, what we know is that there are 10 to the Sixth Micro Moeller in one mole arat e. So all we need to do is take 86.71 and divided by 10 to the six, and we'll get an answer of 8.671 times 10 to the negative fifth, and this is going to be in units of molar ity. So now that we've converted into units of polarity weaken, bring back in this seconds that we had before, so we can just bring that in. So that's our initial reaction velocity that we're going to plug in here. Um, for the k m. We're told that the K M is equal to seven million Mueller. And so for seven million Mueller, we want to convert that into molar ity as well so that they're all matching the units on DSO. We know that there are 1000 million Mueller in one Moeller and so this allows our units to cancel. So if we take seven and divided by 1000 we'll get the answer of 0.7 Moeller And so this is our new K M. And then, of course, last but not least, we have our substrate concentration, which we're told is 25 million Mueller. So again we want to convert that all into units of mole aren t so that their matching. So again we know that there are 1000 million Mueller in one Moeller one Mueller so that allows our units to cancel. And so 25 divided by 1000 is going to be equal to 0.25 Moeller. And so now we have all of our units matching And so now we can plug in these values into our expression over here. So if we rewrite this will have the V max, which is what we're calculating for is going to equal our V, not which we converted into these units here, So it's gonna be 8.671 times 10 to the negative fifth. And this is gonna be units of morality per second, and we'll put this in parentheses altogether. So that's our initial reaction velocity. And this is gonna be multiplied by our K M, which is 0.70 point Mueller and, uh, plus our substrate concentration, which is now in units of malaria's wealth. 0.25 Mueller. And so this is all going to be divided by these substrate concentration, which again is 0.25 Moeller. And so don't forget that you always want to do the parentheses first here. So what you want to do in your calculators is add up 0.7 plus 0. and take the answer of that and multiply it by 8.671 times 10 to the negative fifth more clarity per second and then take the answer of all of that and divided by 0.25 Mueller. And when you do that What you'll get is that the V Max is going to be equal to approximately 1.11 times 10 to the negative fourth and the units are going to be mole Arat e per second. And so notice that this answer option here 1.11 times 10 to the negative fourth polarity per second matches with answer Option C and so we can indicate that. See, here is the correct answer for this problem and that concludes our example here. So I'll see you guys and our practice problems where we could get more practice applying these concepts.

4

Problem

Suppose an enzyme (MW = 5,000 g/mole) has a concentration of 0.05 mg/L. If the *k*_{cat} is 1 x 10^{4} s^{-}^{1}, what is the theoretical maximum reaction velocity for the enzyme?

A

1050 µM/s.

B

100 µM/s.

C

150 µM/s.

D

105 µM/s.

5

Problem

For a Michaelis-Menten enzyme, what is the value of V_{max} if at 1/10 Km, the V_{0} = 1 μmol/min.

A

1.5 μmol/min.

B

11 μmol/min.

C

19 μmol/min.

D

103 μmol/min.

E

μmol/min.

6

Problem

Carbonic anhydrase catalyzes the hydration of CO_{2}. The Km of carbonic anhydrase for CO_{2} is 12 mM. The initial velocity (V_{0}) of the enzyme-catalyzed reaction was 4.5 μmole*mL^{-1}*sec^{-1} when [CO_{2}] = 36 mM. Calculate the V_{max} of carbonic anhydrase.

A

8.1 x 10^{2} M s^{-1}.

B

6 x 10^{-3} M s^{-1}.

C

2.5 x 10^{-4} M s^{-1}.

D

7.3 x 10^{-5} M s^{-1}.

7

Problem

Triose phosphate isomerase catalyzes the conversion of dihydroxyacetone phosphate (DHAP) to glyceraldehyde-3-phosphate (G3P) during glycolysis; however, this is a reversible reaction. The K_{m} of the enzyme for G3P is 1.8 x 10 ^{-5} M. When [G3P] = 30 μM, the initial rate of the reaction (V_{0}) = 82.5 μmole*mL^{-1}*sec^{-1}. Calculate the V_{max}.

A

0.493 M s^{-1}.

B

1.201 M s^{-1}.

C

0.067 M s^{-1}.

D

0.132 M s^{-1}.

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