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6. Enzymes and Enzyme Kinetics

1

Michaelis-Menten Equation

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all right, So now that we've covered the McHale is meant in assumptions. In this video, we're going to talk about The McHale is meant in equation, and so the McHale is meant in equation were commonly going to abbreviate as just the M M equation in our lesson. And so the McHale is meant in equation is just a mathematical description of the initial reaction rates or the V, not of an enzyme catalyzed reaction. And so, really, that should be no surprise to us, since we know from our previous lesson videos that biochemist mainly focus on the initial reaction rates or the V not of enzyme catalyzed reaction. And so now we're talking about an equation that allows us to calculate these initial reaction rates. And so, really, the McHale is meant in equation mathematically relates the initial reaction velocity Veena to the substrate concentration, and it does that via the theoretical maximum reaction Velocity V Max and the Michaelis Constant or the K M. And so, essentially, what we're saying is that the McHale is meant in equation mathematically relates all of these enzyme kinetics variables that we already talked about in our previous lesson videos and we'll be able to see that down below in our example on the left hand side. Now, if we take a quick look at our example down below at this plot that we have on the right notice that this is our enzyme kinetics plot, which is also commonly known as just a meticulous meant in kinetics plot. And so again, we've seen this plot so many times before and our previous lesson videos. So we're already familiar that on the y axis we have the initial reaction rate or the V not. And on the x axis we have the substrate concentration. And of course, we know that it's typical for enzymes to show this curve that we see here on this plot. Now, what you may not have known is that the shape of the curve that we see here is, um, actually has a name to it, and the name of the shape is actually called a rectangular hyperba. And so, essentially, what we'll see is that the McHale is meant in equation is actually describing the rectangular hyperbole a shape that we see in these enzyme kinetics plot eso often these, um, typical curves that we see in these enzyme kinetics plots. And so again, the enzyme kinetics plot that we see down below plots the initial reaction velocity on the Y axis and it plots the substrate concentration on the X axis. And so what you'll notice is that the rectangular hyperba um just like a line we know has a an equation. Why equals M x plus B, a rectangular hyperba or this shape right here also has an equation. And so the equation for a rectangular hyperbole curve is actually this equation that's shown here. And so what we'll see is that the McHale is meant in equation. Really? All it does is it simply substitutes. It's substitutes, enzyme kinetics variables into this rectangular hyperbole equation that we see up above. So all we do is we take enzyme kinetics variables and we plug them in to this equation that we see here. So to see what I mean, let's take a look at our image down below on the left hand side here. And so notice on the left. What we have is the rectangular hyperbole, a equation, the same one that we have up above here. And so it's exactly the same. And so notice that, uh, the rectangular hyperbolic equation is color coded so that we can see how we can substitute enzyme kinetics variables into this equation to get our meticulous mental equation. And so the why here and the rectangular hyperbole equation is going to be what we have on the Y axis of our enzyme kinetics plot. And what we have on the Y axis is our initial reaction rate V Not so you can see that we take the why and we plug it in with V Not since that's what's on the Y axis, then what we have is this a here in this A This read A is going to be our V max and the B and the denominator. The bottom here is going to be our McHale is meant in K. M. And so what can help us remember? That is, if we take a look at our enzyme kinetics plot. Notice that the V Max is towards the top of our plot, whereas the K M is on the X axis towards the bottom of our plot. So you can see K m is on the bottom and the V Max is on the top of our plot. So that's what helps us remember to substitute these variables accordingly. And then, of course, we know that the X is going to be exactly what we have on the X axis, which is the substrate concentration. And so, essentially, what we could do is, uh, plug into these X is this substrate concentration over here? So we'll have substrate concentration. And so, essentially, this is our meticulous meant in equation and moving forward in our course, we're going to be able to use this equation to solve for the initial reaction velocity and for other variables as well. Asl Ong as were given the appropriate variables to solve for the missing variables. And so, uh, you can see how really the McHale is meant in equation includes all of these variables that we, uh, discussed before in our previous lesson video so we can see that the V max is included in theory, the equation. So is the k M, um, down below? And we have the substrate concentration included in the equation as well as the initial reaction velocity. And so this concludes our lesson on the introduction to the McHale is meant in equation. And And our next video, I'll show you guys an example of how to utilize. Uh, this McHale is meant in equation. And so I'll see you guys in that video.

2

Michaelis-Menten Equation Example 1

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All right, So now that we formally introduce the McHale is meant in equation. In our last lesson video in this video, I'm going to show you guys how to utilize the McHale is meant an equation in a two part example problem. And so this example problem wants us to consider the following enzyme kinetics data down below for the enzyme catalyzed reactions shown right here where we have substrate a being converted into product be. And so looking at this enzyme kinetics data down below notice in the first column On the left hand side, what we have is the concentration of our substrate A and units of micro moles. And in the second column on the right, what we have is the V not, or the initial reaction velocity of the enzyme catalyzed reaction in units of micro moles per minute. And so, looking at party, it's asking us what is the meticulous constant or the K m of the enzyme. And so what we need to recall from our previous lesson videos is that the cam can be defined as notice that the cam can be defined as the exact substrate concentration. When the initial reaction velocity is exactly equal to half of the V max. And so notice that we are given the initial reaction velocity on our data over here. But we're not directly given the V, Max. And so what we need to do is use this data on the right to help us determine what the V max is. And so what we need to realize over here with this data is that in this first column, the substrate concentration, it starts really, really low, and it's steadily increases. And then the increases get greater and greater and greater toe a point where the concentrations are so great that the initial reaction velocity doesn't even change. And so this is suggesting that the substrate concentrations are so high that it's completely saturating the enzyme and recall from our previous lesson videos that when an enzyme is completely saturated with substrate, the initial reaction velocity can approach the V max. And so what we're saying is that these initial reaction velocities of 80 must be approaching and getting really close to the V max. And so we can pretty much say that R V max is equal to a value of 80 micro moles per minute. And so now that we have our V Max, if we want to get half the V Max, then all we need to do is take half of the value of the V Max and half of our V Max, which is 80 is going to be 40 of course, and notice that 40 actually shows up here in our data. And so this is a velocity that's equivalent to half of the V Max and notice that the substrate concentration when the initial reaction velocity is half the V max is going to be the K. M. And so this substrate concentration here of 50 micro moles corresponds with half of the V max. And so we can say that this substrate concentration right here is equivalent to RK mm. And so that is exactly what we're going to put as our answer. 50 Micro Mueller is r k m. And that is the answer to part at now moving on to Part B. It's asking what is the value of the V not or the initial reaction velocity when the concentration of substrate is equal to 43 we'll assume the units are micro Mueller. And so, in order to solve this, we're going to need to utilize our Michaelis Menton equation or the MM equation which is provided down below in this box here. And so we want to solve for the initial reaction velocity. So all we need to do is plug in all of these other values here and calculated using our calculator. So let's go on and do that. So we have the initial reaction Velocity V, not eyes equal to the V Max here and the V Max we previously identified as being 80 micro moles per minute. And so we're going to multiply this by the concentration of our substrate, which is given to us as 43 Micro Mueller. So we'll go ahead and multiply this by 43 Micro Mueller and then all of this is going to be divided by the K M, which we already previously identified as being 50 Micro Mueller. So we can put that in down below and then this is going to be added to the substrate concentration again, which we know is 43 Micro Mueller given to us is 43 Micro Mueller And so all we need to do is plug all of this into our calculator. So if you do 80 micro moles per minute times 43 micro moles and take that answer and divided by 50 micro mold plus 43 micro molds, you'll get the answer, which is approximately equal to 36. which rounds off to about 37. And the units of the initial reaction velocity are gonna equal the units of the initial reaction velocity that we have in our chart. So it's gonna be micro moles per minute. And so this here is the answer that we are looking for. So 37 micro moles per minute. And so we can say that when the initial, uh when the concentration of our substrate is equal to 43 Micro Mueller three initial reaction velocity will be 37 micro moles per minute. And so we could have kind of check to look at our date over here to see if our answer makes sense and so noticed that a concentration of 43 Micro Mueller is just below the concentration of 50 Micro Mueller just a little bit smaller. And so we expect that the velocity that corresponds with 43. It's just gonna be a little bit smaller than the velocity that corresponds with 50. And so when we see that 37 is just a little bit smaller than this 40 here, we can see that our velocity is in the right ballpark. So we know that our answer is in the right ballpark. And so also with this example shows us is that as long as we have three out of the four total variables that are present in our McHale is meant in equation, then we can solve for that fourth and missing variable. And so that shows us the power of the McHale is meant an equation, and we'll be able to continue to utilize it as we move forward in our next video, so I'll see you guys there.

3

Problem

A) Suppose the [S] = 10 K_{m}. Use the Michaelis-Menten equation to determine what percentage of the V_{max} will be equal to the value of V _{0}.

B) Now suppose the [S] = 20 K_{m}. Use the Michaelis-Menten equation to determine what percentage of the V_{max} will be equal to the value of V _{0}. What conclusion can be made from these calculations?

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4

Problem

Which of the following statements about a V_{0} vs. [S] plot for a Michaelis-Menten enzyme is false?

A

As [S] increases, V_{0} also increases.

B

At very high [S], the curve becomes a horizontal line that intersects the y-axis at K_{m}.

C

K_{m} is the [S] at which V_{0} = ½ V_{max}.

D

The shape of the curve is a hyperbola.

5

Problem

What is the ratio of [S] to K_{m} ( [S] / K_{m }) when the V_{0} of an enzyme-catalyzed reaction is 80% of the V_{max}?

A

1.

B

2.

C

3.

D

4.

E

5.

6

Problem

An enzyme-catalyzed reaction was carried out with a [substrate] initially 1000 times greater than the Km for that enzyme. After 9 minutes, 1% of the total substrate was converted into 12 μmoles of product. If in a separate experiment, one-third as much enzyme and twice as much substrate had been combined, how long would it take for the same amount of product (12 μmoles) to be formed?

A

1.5 min.

B

13.5 min.

C

27 min.

D

3 min.

E

6 min.

7

Problem

An enzyme catalyzes a reaction at a velocity of 10 μmol/min when all enzyme active sites are occupied with substrate. The K_{m} for this substrate is 1 x 10^{-5} M. Assume that Michaelis-Menten kinetics are followed, calculate the initial reaction velocity (V_{0}) when:

A) [S] = 1 x 10^{-5} M. V_{0} = ___________

B) [S] = 1 x 10^{-2} M. V_{0} = ___________

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