Use the Lineweaver-Burk plot below to answer the following que...

Question

Use the Lineweaver-Burk plot below to answer the following questions. Units of [S] are in nM.

A) Estimate the values of K_m & V_max as well as the K_?^??? & V_???^??? for the reactions in the absence and presence of the competitive inhibitor.

B) Would you expect the competitive inhibitor to be more effective under conditions of high or low [S]? Why?

C) If [I] = 10 nM, calculate the inhibition constant (K _I).

Accepted Answer

Alright. So here we have a 3 part practice problem broken up into part a, part b, and part c. And it says to use the Lineweaver Burk plot below to answer the following questions, and it tells us that the units of substrate concentration are in units of nanomolar. And so units of nanomolar. And so taking a look at our Lineweaver Burk plot down below, notice we actually have two lines. We have this black line right here that corresponds to the enzyme catalyzed reaction in the absence of inhibitor, so minus inhibitor, And then we also have this green line right here that represents the enzyme catalyzed reaction in the presence of competitive inhibitor. And so looking at part a, it's asking us to estimate the values of the k m and the v max as well as the values of the apparent k m and the apparent v max for the reactions in the absence and the presence of competitive inhibitor. And so, what we need to recall in order to answer part a is that the intercepts in, the Lineweaver Burk plot are really what are going to reveal the k m and the v max. And so, recall that the x intercept is just going to be a reciprocal and because it falls on the negative region over here, we know it's going to be the negative reciprocal of the k m. And, of course, the y intercept is also going to be a reciprocal. It's going to be the reciprocal of the v max. And so, starting with the, absence of inhibitor, we can start with this black line that we see here, and, we can also start with the x intercept. And so the x intercept is gonna be where this black line crosses the x axis, which is right at this point right here, which corresponds with negative 0.2. So what we can say is that the x intercept in the absence of inhibitor is equal to negative 0.2. And, of course, the x intercept is going to be equal to negative one over kilometers, so we can make this equal to negative one over the kilometers. And, of course, we're going to add algebraically solve for the kilometers, so we can move it to this side of the equation by multiplying both sides of the equation by kilometers. And when we do that, what we get is the kilometers times negative 0.2 is equal to negative 1. And then we can divide both sides of the equation by negative 0.2, and what we get is kilometers is equal to negative 1 divided by negative 0.2. And, of course, if you take your calculator and do negative 1 divided by negative 0.2, you'll get an answer of 5. And because the units of the substrate concentration are in nanomolar, we can say that the units of the kilometers are also going to be in nanomolar. And so now that we've calculated the Kilometers in the absence of inhibitor as 5 nanomolar, we can go ahead and do the same, for the presence of inhibitor. And so in the presence of inhibitor, we know that the Kilometers is going to be referred to as the apparent Kilometers, And so, that's important to keep in mind. And, of course, we are going to look at the x intercept, which is where this green line here is going to cross the x axis, which is right at this point right here, which corresponds with a value of negative 0.1. And so what we can say is that the x intercept is gonna be equal to negative 0.1, and that is going to be equal to the negative reciprocal of the k m, so negative one over k m. And, again, because this is in the presence of inhibitor, the Kilometers is actually referred to as the apparent Kilometers. And so we're going to algebraically solve for the apparent Kilometers, and so we can move the apparent Kilometers up to this side of the equation by multiplying both sides of the equation by apparent kilometers. And when we do that, what we get is that the apparent kilometers times negative 0.1 is gonna be equal to negative 1, and then we can divide both sides of the equation by negative 0.1. And so when we do that, what we get is that the apparent kilometers is equal to negative 1 divided by negative 0.1. And, of course, if you take this in your calculator, what you'll get is 10, and the units are again going to be nanomolar. And so now that we have, the apparent Kilometers as 10 nanomolar and, the Kilometers as 5 nanomolar, we can move on to, calculating the Vmaxes and the apparent Vmaxes. Now, again, the v max is going to be indicated by the y intercept, and the y intercept is gonna be where these two lines cross the y axis, which, notice, is at the same exact point. And so, this indicates that in the presence and absence of competitive inhibitor, they have the same exact v max, and we know that is definitely true when it comes to competitive inhibitors. And so, essentially, this value right here is approximately value of 0.075. It's essentially, the midway point between this, these, 2, markers. So it's approximately 0.07 5, and, this is going to be equal to 1 over the v max. And so, essentially, we're going to solve for the v max here, and, essentially, we can move it up to this side of the equation by multiplying both sides of the equation by v max. And when we do that, what we get is v max times 0.075 is equal to 1. And, of course, we can divide both sides of the equation by 0.075, And what we get is v max is equal to 1 over 0.075. And if you type that in your calculator, what you'll get is an answer of about 13.33 repeating. And, of course, because the v max is, the same in the presence and absence of inhibitor, what we can say is that this is also equal to the apparent v max as well. And so, essentially, this is going to be, the apparent v max in both the absence as well as the presence of inhibitor. Now, now that we've determined that, we can move on to part b, which asks, would you expect the competitive inhibitor to be more effective under conditions of high or low substrate concentration and why? Now we know that competitive inhibitors are going to compete with the substrate. And so, the competitive inhibitor will be more effective if there is less substrate to compete with it. And so the competitive inhibitor is going to be, more effective, at low substrate concentrations. And so what we can indicate is that it will be more effective at low substrate concentrations, and the reason is because there's less competition, between, the, substrate. So between the substrate and the inhibitor, when there's low substrate concentration. And that allows, again, the, competitive inhibitor to be more effective. And so if we take a look at our line, we have a Burke plot over here on the right. What we can see is that a low substrate concentration actually corresponds with further to the right of our x axis here, and that's because, again, the substrate concentration is the in, the inverse of the substrate concentration that's being plotted. So, smaller substrate concentrations correspond further to the right. And notice as we move further to the right, the, competitive inhibitor actually starts to, be, higher than the, absence of that competitive inhibitor. And so recall that, because it is higher up, it's further away from the 0 infinity marker, which means the initial reaction velocities are going to be smaller as the substrate concentration gets smaller as well. And so that allows the competitive inhibitor to be more effective and inhibit the reaction even more at lower substrate concentrations. So now moving on to part c, it says that if the concentration of inhibitors is equal to 10 nanomolar, calculate the inhibition constant of the free enzyme k I. And so, in order to do this, we're going to need to recall some previous equations from, our previous lesson videos. And so recall that in the presence of, competitive inhibitors, the apparent Kilometers is defined as alpha, the degree of inhibition on the free enzyme, times the kilometers. And so, previously, up above in part a, we were able to determine the apparent kilometers as 10 nanomolar and we were able to determine the, Kilometers in the absence of inhibitor as 5 nanomolar. So we actually have values to be able to solve for alpha, the degree of inhibition on the free enzyme. So, what we can say is that the apparent Kilometers, which is, again, 10 nanomolars, is going to be equal to alpha times, the Kilometers in the absence of inhibitor, which is, again, going to be 5 nanomolar. And so we're going to solve for alpha, and so, we can divide both sides of the equation by 5 nanomolar. And when we do that, what we get is alpha is equal to 10 nanomolar divided by 5 nanomolar. And, of course, if you type that in your calculator, what you'll get is that alpha is equal to 2. Now, now that we have the degree of inhibition, on the free enzyme alpha, with a value of 2, we can go ahead and plug it into this equation over here where we know from our previous lesson videos that the degree of inhibition on the free enzyme alpha is defined as 1 plus the ratio of the concentration of inhibitor over the inhibition constant. And so we have the concentration of inhibitor, so we can go ahead and plug that in into this expression and then algebraically solve for the k I. And so, again, here, what we can do is put a little separator line here. And, again, we define alpha as 2. So we know alpha is going to be equal to 2, which is equal to this expression right here. So it's gonna be equal to 1 plus again, the concentration of inhibitor is given to us as 10 nanomolar, so we can put 10 nanomolar here, over the inhibition constant Ki. And then, again, we're going to algebraically solve for this Ki here. So we can start by subtracting 1 from both sides of the equation. And when we do that, we get 2 minus 1, and 2 minus 1 is, of course, 1. And that's going to be equal to this ratio right here, 10 nanomolar over the inhibition constant, k I. And again, we're going to algebraically solve for this k I, so we can move it up to this side of the equation by multiplying both sides of the equation by k I, and then we can divide, both sides of the equation by 1. And so when we do that, what we get is the k I is just going to be equal to 10 nanomolar over 1, which is, of course, the same thing as just 10 nanomolar. And so what we're saying is that the inhibition constant is going to be 10 nanomolar if the concentration of inhibitor is 10 nanomolar. And so, here, what we can say is that the KI is going to be equal to 10 nanomolar, and this this here concludes our, 3 part practice problem. And so I'll see you guys in our next video.

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