Question

Suppose a company develops two enzymes that degrade highly toxic compounds to non-toxic compounds. Your task is to degrade the greatest amount of toxic compound in the shortest amount of time.

A) Which enzyme is better to use when [S] = 0.167 mM? ____________

B) Which enzyme is better to use at saturating [S]? ____________

Accepted Answer

Alright. So here we have a 2 part practice problem broken up into part a and part b, and it says to suppose that a company develops 2 enzymes that degrade highly toxic compounds into non toxic compounds. And so you can imagine that this company really just wants to clean up a really polluted area. And so notice down below over here on the right hand side, what we have is a little table giving us information on both of these enzymes that the company has to choose from. And these enzymes are enzyme number 1 and enzyme number 2. And so, we're provided with the Michaelis constant or the kilometers for each of these enzymes, as well as with the theoretical maximal reaction velocity or the v max for both of these enzymes. Now, notice down below over here on the left hand side, what we have is the Lineweaver Burk plot or the double reciprocal plot showing the enzyme kinetic data for both of these enzymes. And so the line for enzyme number 1 is here in blue and the line for enzyme number 2 is here in red. And so the problem continues to say that your task is to degrade the greatest amount of toxic compound in the shortest amount of time, which suggests that we just wanna choose the enzyme with the higher reaction velocity. And so maybe your natural instincts tell you that if we want to degrade the greatest amount of toxic compound in the shortest amount of time, then all we need to do is choose the enzyme with the greatest maximal reaction velocity, or the greatest v max, which would of course be enzyme number 2. However, not so fast because it turns out that the answer depends on the concentration of substrate or the concentration of the highly toxic compound. And so we can see that down below in our Lineweaver Burk plot as well, that if we look at the substrate concentration values, notice that, at some substrate concentrations, the enzyme number 2 in red, which is closer to the infinity marker, will have a higher reaction velocity, whereas as at other substrate concentrations, enzyme number 1 is closer to the infinity marker, which means that enzyme number 1 will have a greater reaction velocity. And so the split here seems to be at a value that corresponds with, inverse substrate concentration of 2. And so, let's take a look at part a here to continue to answer this problem. So it says, which enzyme, either enzyme number 1 or enzyme number 2, would be better to use when the substrate concentration is equal to 0.167 millimolar? And so what we need to realize is that a substrate concentration equal to 0.167 millimolar is the same thing as a substrate concentration of 167 micromolar. And so notice that the value of the substrate concentration, 167 micromolar, is the same exact value as the Kilometers for enzyme number 1, which is a 167 micromolar, and this substrate concentration is also less than the Kilometers value for enzyme number 2, which is 500 micromolar. And so what this is suggesting is that when the concentration of substrate is equal to this value, like part a suggests, that neither of these two enzymes are going to be operating at their maximal reaction velocities or their v max, which means that the v max for part number a, when the substrate concentration equals this value, is irrelevant, and we're not going to be focusing on the v max to answer part number a. And so instead, what we're going to do is take a look at our Lineweaver Burk plot and apply this substrate concentration to determine which of these two enzymes has a higher reaction velocity at this particular substrate concentration. So in order to apply the substrate concentration to the Lineweaver Burk plot, we need to recall that a Lineweaver Burk plot is a double reciprocal plot, so it actually plots the reciprocal of the substrate concentration. And so if the substrate concentration is equal to 0.167 millimolar, then the reciprocal is gonna be 1 over substrate concentration, which is gonna be 1 over 0.167 millimolar. And if you type this into your calculator, 1 divided by 0.167, millimolar, what you'll get is an answer, that is about 5.99, and this is, of course, approximately equal to 6, and the units are going to be inverse millimolar. And so what we're saying is that when the substrate concentration is equal to 0.167 millimolar, like it indicates in part a, that corresponds with a value of, the inverse substrate concentration of 6. And so it's this value right here that we're specifically focused on, for part a. And so you can see that at this, particular substrate concentration of 6, that enzyme number 1, will actually have a initial reaction velocity that's closer to the 0 infinity marker, and, enzyme number 2 has a reaction velocity that's further away from the 0 infinity marker. So we can see that at this substrate concentration, enzyme number 1 actually has the higher initial reaction velocity, and so what that means is, at this substrate concentration, we're going to want to use enzyme number 1, and so we can go ahead and put that in as our answer. And so, part of the reason, that this is going to be true is because recall that a smaller value of the Kilometers, so a smaller value of the Kilometers, means that we actually have a stronger or a higher binding affinity. So the enzyme 1 will have a higher binding affinity to its substrate than enzyme number 2. And so that's why at these low substrate concentrations, enzyme number 1 ends up having a higher reaction velocity because it binds a lot better to the substrate at these lower substrate concentrations. And so now that we know the answer for part a, we can move on to part e b, which says, which asks which enzyme is better to use at saturating substrate concentrations? Now we also know from our previous lesson videos that at saturating substrate concentrations, max. So that means that for part b, we're going to want to focus specifically on the v max values. And of course, it's enzyme number 2 that has the higher v max at saturating substrate concentrations. So we're going to want to use enzyme number 2 specifically, for part b, at saturating substrate concentrations. And so this here concludes our answers to our 2 part practice problem, and I'll see you guys in our next video.

Struggling with Biochemistry?

Suppose a company develops two enzymes that degrade highly toxic compounds to non-toxic compounds. Your task is to degrade the greatest amount of toxic compound in the shortest amount of time.
A) Which enzyme is better to use when [S] = 0.167 mM?
B) Which enzyme is better to use at saturating [S]?

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