Specificity Constant - Online Tutor, Practice Problems & Exam Prep

The specificity constant, defined as the ratio of the turnover number (k_cat) to the Michaelis constant (K_m), measures enzyme efficiency under non-saturating substrate concentrations. This constant reflects an enzyme's substrate preference, with higher values indicating greater efficiency. The maximum value of the specificity constant is diffusion-controlled, approximately 10⁹ M^-1s^-1, characterizing catalytically perfect enzymes that rapidly convert substrates to products. Understanding these concepts is crucial for analyzing enzyme kinetics and their physiological relevance.

Specificity Constant

In this video, we're going to begin our discussions on the specificity constant of an enzyme. Now before we talk directly about the specificity constant of an enzyme, it's first important to recognize a few key points. The first key point to recognize is that when biochemists are studying or characterizing enzymes in a laboratory setting, it definitely is useful for them to study those enzymes in the lab under saturating substrate concentrations efficiency under saturating substrate concentrations. However, what's also really important to know is that the substrate concentrations within a cell or within a biological system are not always saturating. This means that saturating substrate concentrations is not always the best way to study enzymes. This has to do with the fact that under physiological conditions or conditions within a cell, the substrate concentrations are not always saturating, and instead, the tendency is for the cell to maintain substrate concentrations approximately equal to the km of the enzyme.

Additionally, another reason for why it's not always the best to study enzymes at saturating substrate concentrations is that under these conditions, it does not allow the biochemist to account for the binding affinity that an enzyme has for its substrate, or the enzyme-substrate binding affinity cannot be taken into account under saturating substrate concentrations. This is because, regardless of the binding affinity that an enzyme has for its substrate, even if the binding affinity is really low, under saturating substrate concentrations, all of the enzyme is going to be associated with the substrate to form the enzyme-substrate complex. Again, that will not allow the biochemist to account for the binding affinity that an enzyme has for its substrate under saturating substrate concentrations.

What this really means is that the maximal catalytic efficiency or the kcat that we talked about in some of our previous lesson videos, which only occurs under saturating substrate concentrations, is not always going to be the most relevant measure of the catalytic efficiency of an enzyme. Instead, a better overall measure for the catalytic efficiency of an enzyme is the specificity constant. We're going to define and talk more about the specificity constant in our next lesson video. So, I'll see you guys there.

Specificity Constant

So in our last lesson video, we said that the maximal catalytic efficiency of an enzyme indicated by the catalytic constant k_cat or the turnover number is not always the best measure of catalytic efficiency, especially since it only occurs under saturating substrate concentrations and saturating substrate concentrations are not always relevant, especially inside of a cell. And so in this lesson video, we're going to talk about another better overall measure of catalytic efficiency, specifically under non-saturating substrate concentrations or low substrate concentrations, and that's exactly where the specificity constant comes into play. And that's because, as you guys may have already determined, the specificity constant is this ratio of the k_cat to the k_M, and that's exactly what we're saying down below, is that the specificity constant is this ratio of the k_cat over the kilometers. And so this specificity constant ratio of the k_cat over the kilometers is really what determines an enzyme's preference for a substrate, specifically at non-saturating substrate concentrations or low substrate concentrations. And so it turns out that substrate preference or the preference that an enzyme has for a particular substrate, is determined by the catalytic efficiency. And so the higher the catalytic efficiency, the greater the preference an enzyme has for that substrate.

Now, it also turns out that catalytic efficiency actually depends on the concentration of substrate and whether or not those substrate concentrations saturating levels or non-saturating levels. Now, if the substrate concentrations are at saturating levels, then that means that the catalytic efficiency is going to be determined by the maximal catalytic efficiency indicated by the k_cat alone. And so under saturating substrate concentrations, the k_cat alone is what dictates the preference that an enzyme has for the substrate. However, when the substrate concentrations are at non-saturating levels or low substrate concentrations, then the catalytic efficiency is actually determined by not the k_cat alone, but instead the specificity constant, which is the ratio of the k_cat over the kilometers.

And that means that the specificity constant ratio is what determines the preference that an enzyme has for its substrate at non-saturating or low substrate concentrations. Now, what I want you guys to recall from way back in our previous lesson videos is that we talked about this enzyme called chymotrypsin, which is a peptidase that recognizes specific amino acids to cleave peptide bonds between them. And so what I want you guys to recall is that we said that chymotrypsin has a preference for the aromatic amino acids Phenylalanine, Tyrosine, and Tryptophan, and it has less of a preference for Leucine and Methionine. And so in our next video, we're going to talk about exactly how chymotrypsin's preference is influenced by these factors.

So, again, we're going to retouch on this idea here in our next lesson video. So hang on tight for this bullet point here. Now, what I want you guys to note is that this ratio, the specificity constant ratio of k_cat over the kilometers, is just another measure of catalytic efficiency. And specifically, the specificity constant ratio is the catalytic efficiency when an enzyme is not saturated with substrate or essentially at low substrate concentrations. Now this specificity constant ratio accounts for both the maximal catalytic efficiency k_cat as well as an enzyme's affinity for its substrate, which is indicated by the k_M.

Now larger ratios of this specificity constant represent more efficient enzymes, and, of course, more efficient enzymes, as we said earlier, is going to indicate a higher preference for the substrate at lower substrate concentrations. I also want to note that this specificity constant ratio of k_cat over the kilometers actually does have a maximum value that's approximately equal to 109 with units of inverse molarity inverse seconds. And we're going to touch more on this idea later in our course, so also hang on for this bullet point as well until later in our course.

So this here concludes our introduction to the specificity constant ratio of k_cat over kilometers, and in our next video, we're going to be able to talk even more about the specificity constant ratio and how it applies to chymotrypsin's preference. And so I'll see you guys in that video.

Specificity Constant

Alright. So here we have a table that looks a lot more complicated than what it actually is, and all it's really giving us is information about 3 different enzymes: urease, penicillinase, and chymotrypsin, which we're already familiar with from our previous lesson videos. Notice that in this first column right here, we're given the catalytic constants or the k_cat or the turnover number in units of inverse seconds for all of these enzymes and their substrates. The k_cat or the turnover number only occurs under saturating substrate concentrations, making it the maximal or the max catalytic efficiency that only occurs under these conditions. Looking at this value of 10,000 here, just to refresh our memories on what exactly that means. A value of 10,000 here means that just one molecule of the enzyme urease, specifically under saturating substrate concentrations, can convert 10,000 molecules of substrate into product per second. This is incredibly fast, especially in comparison to all of these other values that we have here. For that reason, we can indicate the catalytic constant speed in this column, by putting up arrows in red as fast and down arrows in blue as slow. 10,000 here is pretty fast, especially in comparison to these other values. For that reason, we can go ahead and give it 5 up arrows right here.

Moving on to this next column right here, notice what we have is the Michaelis constant, or the K_M, in units of molarity for all of these substrates. Ultimately, the K_M is just a measure of an enzyme's binding affinity for its substrate. The larger the K_M value is, the weaker the affinity will be. Looking at this value here for urease, it has a value of 10 raised to the negative 2. This makes this number, in comparison to all of these other numbers, the greatest or the largest number. Recall that the larger the value of the K_M, the weaker the affinity will be. Because this is the largest value, we can say that it has a weak affinity and we can give it 2 down arrows here. In this final column, we have the specificity constant ratio, which we introduced in our last lesson video. It is the ratio of the k_cat over the K_M, essentially the ratio of this column over this column right here. Ultimately, we're going to take the ratio of the arrows of the first and second columns so that we can also get arrows for the specificity constant. Down arrows are going to cancel out up arrows, and these 2 down arrows will cancel out with these 2 up arrows, leaving us with 3 up arrows. We can put 3 up arrows next to the specificity constant for the enzyme, urease.

Moving on to penicillinase here, notice that its catalytic constant of 2,000 is much smaller than the catalytic constant of 10,000. But still, penicillinase with a value of 2,000 can convert 1,000 molecules of substrate into product per second. So, it's still pretty fast, not quite as fast as urease, but we can go ahead and give it 3 up arrows here. Taking a look at its K_M, notice that it's raised to 10 to the negative 5th here. Compared to all of these other numbers, this value is the smallest number. The smaller the K_M value is, the stronger the affinity is. We can say that penicillinase has the strongest affinity for its substrate. In this column right here, we can indicate the strength of the affinity by giving it 2 up arrows. Ultimately, to get the arrows for the specificity constant of penicillinase, all we need to do is sum up these arrows. The 3 up arrows plus the 2 up arrows give us 5 up arrows. We can go ahead and indicate 5 up arrows here.

Then moving on to chymotrypsin, with substrates of phenylalanine, tyrosine, and tryptophan. We already know from our previous lesson videos that chymotrypsin has a preference for these enzymes it recognizes for cleavage. The mnemonic that helps us memorize chymotrypsin's preference is "free your worries like me", where we know that it has a preference for the aromatic amino acids phenylalanine, tyrosine, and tryptophan, and it has less of a preference for the amino acids leucine and methionine. Working with just chymotrypsin's substrates of phenylalanine, tyrosine, and tryptophan, if we were to convert this 100 here into a catalytic speed, it's less than penicillinase, needing fewer up arrows. We'll go ahead and give it 2 up arrows here. The K_M for this group of substrates, raised to 10 to the negative 4th along with these other three K_M values. Even though they vary a little because they're all to 10 to the negative 4th, we'll assume that they all pretty much have the same value, and we'll give them all one up arrow here for the strength of the affinity. Ultimately, what we can get over here for this set of substrates is 2 up arrows plus 1 up arrow, giving us 3 up arrows here.

Comparing chymotrypsin for leucine and methionine, the 0.63 catalytic constant compared to 100 is significantly less. So, this pair is much slower when working with leucine and methionine but still faster than with lysine which is not preferred at all. For leucine and methionine, we'll put one up arrow here, and for lysine, a value of 0.02 gets a down arrow, very slow. Summing these arrows, one up arrow plus one up arrow gives 2 up arrows here for the specificity constant. One down arrow will cancel out one up arrow, leaving us with 0 up arrows over here. Ultimately, this gives us a visual representation of the value of the specificity constant ratio.

The specificity constant was introduced as a measure of the catalytic efficiency, specifically at non-saturating or low substrate concentrations. It also determines the enzyme's preference for the substrate but specifically at low or non-saturating substrate concentrations. Notice that the catalytic constant also determines an enzyme's preference for substrate just like the specificity constant ratio does. The difference between these two is that the catalytic constant only determines an enzyme's preference for substrate under saturating substrate concentrations, whereas the specificity constant determines an enzyme's preference for its substrate at non-saturating or low substrate concentrations. Therefore, an enzyme's preference for its substrate will depend specifically on the levels of substrate concentration and whether or not the substrate concentrations are saturating or low. Moving to this next category, phenylalanine, tyrosine, and tryptophan are more preferred at saturating substrate concentrations because they have a higher maximal catalytic efficiency and are also more preferred under low substrate concentrations due to a higher specificity constant. You can see how leucine and methionine are less preferred at saturating substrate concentrations, having one up arrow here. They are less preferred than phenylalanine, tyrosine, and tryptophan at low substrate concentrations but still more preferred than amino acids that are not preferred at all. The specificity constant ratio is a balance of the k_cat and the K_M. We'll be able to apply a lot of these concepts that we've learned here moving forward in our practice problems. So, I'll see you guys there.

Use the data in the chart below to provide answers to the following problems:

A) List the substrates from most preferred to least preferred under physiological conditions. 

            a) B, A, C.         b) C, B, A.         c) B, C, A.         d) A, C, B.

B) List the substrates from most preferred to least preferred under saturating [S].

            a) B, A, C.         b) C, B, A.         c) B, C, A.         d) A, C, B.

Alright. So here we have a 2-part practice problem broken up into part a and part b. And it says to use the data in the chart below, shown right here, in order to provide the answers to the following problems. And so, if we take a look at this chart, notice that in the first column it has the substrate, and we have these 3 substrates, substrate a, substrate b, and substrate c. And then in the second column, what we have is the catalytic constant, or the 𝐫͂, or the turnover number, which we know is the maximal catalytic efficiency under saturating substrate concentrations, and it has units of inverse seconds, and so we have this 𝐫͂ value for each of these 3 substrates. And then, of course, in the final column, what we have is the Michaelis constant or the Kilometers in units of molarity for each of these 3 substrates. And so, notice that part a wants us to list the substrates from the most preferred to the least preferred, specifically under physiological conditions. And so, what we need to recall from our previous lesson videos is that by physiological conditions, it means conditions within the cell. And within a cell, we know that they typically maintain their substrate concentrations approximately equal to the value of the ☕𝒴. And so what this means is that the concentrations are not going to be at saturating levels. And if the substrate concentrations are not saturating levels, that means that an enzyme's preference is actually going to be dictated by the specificity constant. And so recall from our previous lesson videos that the specificity constant is just this ratio of the 𝐫͂ over the ☕𝒴, and it has units of inverse molarity, inverse seconds. And so notice that we have all of the data that we need to determine the specificity constant, right here. And so we have the 𝐫͂ and we have the ☕𝒴. So all we need to do to get the specificity constant ratio is take the 𝐫͂ value and divide it by the ☕𝒴 value. And so we can do that for each of our 3 substrates here. So if we do that for this first substrate, if we take 0.36 and divide it by 0.07 1, you'll get an answer of about 5.07, and of course the units are inverse molarity, inverse seconds. Now if we do the same for substrate b and take 2.8 divided by 0.025, we'll get an answer of 112 for the specificity constant with units of inverse molarity, inverse seconds. And then, of course, if we do it the same for substrate c and take 0.1 4 and divide that by 0.015, we'll get the specificity constant of about 9.33 inverse molarity inverse seconds. And so now that we have our specificity constant, we know that this is what's going to determine the preference that the enzyme has for the substrate at these non-saturating levels of substrate concentration. And so the greater the specificity constant, the higher the preference. And we want to list these from the most preferred to the least preferred, so we want the highest preference to come first. So the, notice that substrate b is the one that has the highest value for the specificity constant. So we expect b to come first, and be the most preferred, which means that we can already eliminate option b here, because it starts with substrate c. And we can also eliminate option d, since it starts with substrate a. Now what we can see is that substrate c has the 2nd highest specificity constant at 9.33, which means that it will be the next preferred, and so we have to have b followed by c, which clearly means that it's option c here that's going to be the correct answer. We have substrate b as the most preferred, substrate C as the next preferred, and substrate A as the least preferred under physiological conditions based on the specificity constant values. And so, of course, we can say that it is option c here that is the correct answer for part a. Now moving on to part b, it says to do essentially the same thing, list the substrates from the most preferred to the least preferred again. But this time, instead of being under physiological conditions, it's actually under conditions of saturating substrate concentrations. Now recall from our previous lesson videos that when an enzyme is completely saturated with substrate, then its initial reaction velocity can approach the Vmax, and the enzyme will be able to operate at its maximal catalytic efficiency. And so when it comes to the preference, specifically under saturating substrate concentrations, It is actually the maximal catalytic efficiency, 𝐫͂, alone that will dictate the preference. And so what that means is that instead of using the specificity constant ratio, like we did under physiological conditions, this time we're going to use the 𝐫͂ alone to determine the preference at saturating substrate concentrations. And so notice that the highest, 𝐫͂ is substrate B, followed by substrate a, and then substrate c has the lowest catalytic constant. And so that means that that is the exact order that we want our answer to be. Substrate b, substrate a, and then substrate c. So that means that it is option a that best matches that order, and all of these other options are incorrect. So, basically, the main takeaway that you can get from this problem is that when the concentrations of the substrate are not saturating, specifically under physiological conditions, we expect that the substrate concentrations are not saturating, then the preference is dictated by the specificity constant. However, if the substrate concentrations are at saturating levels, then instead of the specificity constant determining preference, it's actually the maximal catalytic efficiency determined by the 𝐫͂ that determines the preference. And so this here concludes our 2-part practice problem, and I'll see you guys in our next video.

Specificity Constant

So in our previous lesson videos on specificity constant, we briefly mentioned that the specificity constant ratio of k_cat over kilometers actually has a maximum value, and it cannot be higher than that maximum value. And so it turns out that this limit to the maximum value of the specificity constant is actually diffusion controlled. And so, the maximum value of this specificity constant of k_cat over Kilometers, not only is it diffusion controlled, limited, but it's also limited by the rate constant K₁, which we know is the free enzyme and the free substrate association rate constant. And so, this is because the free enzyme and the free substrate association will actually occur via diffusion. And so what this means is that the free enzyme and the free substrate can only associate with each other to form the enzyme substrate complex as fast as the maximum rate of diffusion in the solvent. And of course, in biological systems, the solvent is going to be water, and so, therefore, we can say that the maximum values of both the rate constant k₁ and the specificity constant of k_cat over Kilometers are going to be equal to the maximum rate of diffusion in water, which is going to be 10⁹ inverse molarity inverse seconds.

And so down below on the left-hand side over here, what we have is our typical enzyme catalyzed reaction. And so all I want you guys to know is that the free enzyme and the free substrate can only associate with each other via diffusion. And so what this means is that this rate constant here of k₁ is going to be limited by the maximum rate of diffusion. And so the k₁ value cannot be any greater than the maximum rate of diffusion. And, also because we know from our previous lesson videos that the Michaelis constant kilometers is expressed as the ratio of the sum of the enzyme-substrate complex dissociation rate constants, or k_-1 plus k₂, over the association rate constant k₁. Because k₁ is included in the Kilometers, what this means is that the Kilometers is also going to be limited by the diffusion, the max rate of diffusion. And, because the kilometers is included in the specificity constant, which is the ratio of the k_cat over the k_m, that means that this specificity constant is also going to be limited by diffusion. So up above, notice what we said is that, the rate constant k₁, the kilometers, and the specificity constant of k_cat over kilometers are all directly limited by the maximum rate of diffusion in water, which again is 10⁹ inverse molarity inverse seconds, just as we said up above.

And so what's interesting to note here is that an enzyme whose specificity constant ratio is actually equal to this diffusion controlled max value that we indicated up above, 10⁹ inverse molarity inverse seconds, is actually referred to as a catalytically perfect enzyme. And so a catalytically perfect enzyme is essentially an enzyme who, upon the substrate binding to the enzyme, the enzyme will immediately and almost instantaneously convert the substrate into the product. And so, that's why we're, we refer to these enzymes as catalytically perfect. And so, it's important to note here that the specificity constant ratio does have a maximum value, and it's going to be 10⁹ inverse molarity inverse seconds, and this maximum value is a diffusion-controlled limit. And so this here concludes our lesson, and, we'll be able to get some practice utilizing these concepts moving forward in our practice problem. So I'll see you guys there.

Which of the following options is correct concerning the turnover number (k _cat) and the specificity constant?

• A. k_cat reveals how well an enzyme works & its preference for S.
• B. Specificity constant is defined as (k_cat)(K_m).
• C. A large k_cat indicates a less efficient enzyme.
• D. k_cat = V_max/[ES].
• E. Specificity constant is defined as K_m / k_cat.
• F. A small K_m indicates a more efficient enzyme.

Use the Lineweaver-Burk plot to help you calculate the V_max, k_cat, K_m and specificity constant for the enzyme.

Assume the [E]_T = 2.9 nM. Hint: Pay close attention to units. 

V_max = ___________. 

k_cat = ___________. 

K_m = ___________.

k_cat / K_m = ___________.

Alright. So this practice problem wants us to use the Lineweaver Burk plot to help you calculate the Vmax, the kcat, the Km, and the specificity constant for the enzyme. And so it wants us to assume that the total enzyme concentration is nanomolar, and it gives us this hint to pay very close attention to the units. And so we can provide the answers to this problem in these spaces that are provided. And so what we need to recall from our previous lesson videos is that a Lineweaver Burk plot is also known as a double reciprocal plot because it plots the reciprocal of the initial reaction velocity on the y-axis, and it also plots the reciprocal of the substrate concentration on the x-axis. And of course, a Lineweaver Burk plot will allow us to form a line from our data, and the intercepts of this line, specifically the x and the y intercepts, are going to reveal critical information about the enzyme. And so recall that the y intercept is going to be exactly where the line of our data crosses the y axis, so that's going to be right at this particular point here. And recall that the y intercept on a Lineweaver Burk plot is also going to be a reciprocal. It's going to be the reciprocal of a velocity since the velocity is on the y-axis. So, it's actually going to be the reciprocal of the maximum velocity Vmax. And so we can see here that the numbers on the y axis are increasing incrementally by 5 here, so we know that this must be associated with 5, so the y intercept is equal to 5. So all we need to do to calculate the Vmax is to solve for the Vmax here. So we can move the Vmax up to this side by multiplying both sides of this equation here by Vmax. And so when we do that, what we get is, we get that 5, times the Vmax here on the right, and that's gonna be equal to 1. And then, essentially we can continue to solve for the Vmax, so we can divide both sides of the equation by 5. So when we do that, what we get is Vmax is equal to 1 over 5, which is of course equal to 0.2 and, the Vmax, the maximum reaction velocity is gonna have the same units as the initial reaction velocity, which is in units of micromolar per second here. And so, we can say here that the units are in units of micromolar per second. And so, okay now that we've got the Vmax, we can go ahead and plug that in up here. So it's going to be 0.2 micromolar per second. Alright, so now we can go on and move on to the x intercept here, and so recall that the x intercept is going to be where our line here crosses the x-axis, so that is going to be right here at this particular point. And recall that the x intercept is also going to be a reciprocal, except it's going to be a reciprocal of something that's associated with a substrate concentration, which would be the Km. It's gonna be the reciprocal of the Km, 1 over Km. However, we need to recall that because the x intercept always falls into this negative region here of our Lineweaver Burk plot, that our reciprocal here is actually going to be negative, and so the x intercept is actually equal to the negative reciprocal of the kilometers. And so, here we can see that the x intercept is equal to a value of negative 2, so we can set this equal to negative 2. And then of course, all we need to do is solve for the Km, so we can move it up to this side by multiplying both sides of the equation by Km. And so what we end up getting is that, negative 2 times Km on the right is equal to negative 1. And then of course, we're gonna continue to isolate for the kilometers, so, we can get rid of the negative 2 here by dividing both sides of the equation by negative 2, which is of course equal to just 0.5 and, because Km is an exact substrate concentration that allows us to get, the initial reaction velocity to half Vmax. We can say that the Km is gonna have a unit of a concentration. So it's gonna have the same units of concentration here. And so because the units of concentration are micromolar, then we're gonna also have micromolar for our Km. And so we can say that our Km up here is going to be 0.5 micromolar. Alright. So now what we can do is calculate the kcat. So recall that the kcat, which we'll do over here, is equal to the ratio of the Vmax over the total enzyme concentration or the Et here. And so we just solved for the Vmax, so we know that we can plug that in. And here we're given the Et, so, that is in 2.9 nanomolars. But what we need to do is make sure that all of the units match each other and correspond with each other concentration here has units of nanomolar. So let's convert these units of nanomolar into micromolar, so that we can plug it into our expression. And so, if we take the do that down here, so if we take the total enzyme concentration, which is equal to 2.9 nanomolars, and essentially we convert this into micromolar, we can say that there are 1,000 essentially, what we need to do is we can see that these units cancel here, and so if you take your calculator and you do 2.9 divided by 1,000, you'll get the answer of 0.0029 and of course the only units that are left are units of micromolar. And so, now that we have our total enzyme concentration, matching the units in our

Explain the steps you could take to accurately find the K _m, V_max, and specificity constant for an enzyme from the following kinetic data, assuming the experiments were all done with [E]_T = 0.1 mM.

Step #1: ___________________________________________

Step #2: ___________________________________________

Step #3: ___________________________________________

Step #4: ___________________________________________

Step #5: ___________________________________________

Alright. So this practice problem wants us to explain the steps that we could take to accurately find the \( K_m \), the \( V_{\text{max}} \), and the specificity constant for an enzyme from the following kinetic data in this table right here. It instructs us to assume that the experiments were conducted with a total enzyme concentration equal to 0.1 millimolar. Note that this practice problem is not actually asking us to calculate the Michaelis constant, the \( V_{\text{max}} \), or the specificity constant. Instead, it's just asking us to explain the steps that we could take to accurately determine them, without actually performing those steps. Notice below the chart where we have the substrate concentration in units of molarity in the first column, and the initial reaction velocity of the enzyme-catalyzed reaction in units of per second in the second column. We know from our previous lesson videos that to accurately determine, especially the \( K_m \) and the \( V_{\text{max}} \), we should plot our enzyme kinetics data on a Lineweaver-Burk plot over a Michaelis-Menten plot because a Lineweaver-Burk plot allows us to determine the \( K_m \) and \( V_{\text{max}} \) more accurately than a Michaelis-Menten plot. Recall that a Lineweaver-Burk plot, also known as a double reciprocal plot, plots the reciprocal of the initial reaction velocity on the y-axis and the reciprocal of the substrate concentration on the x-axis. The first step we need to take with this data is to take the reciprocals of the substrate concentration and the initial velocity, \( V_0 \), to plot them on a Lineweaver-Burk plot. So, the first step is to take the reciprocals of the substrate concentration and the \( V_0 \), and plot them on a Lineweaver-Burk plot. We're not going to take these steps, but merely explain them here. Imagine we have 6 data points plotted onto our curve like this, showing different data points on our Lineweaver-Burk plot. The second step is to determine a line of best fit and then get the equation for this line of best fit. Recall that the equation for a line is \( y = mx + b \), and we know that the Lineweaver-Burk equation resembles the equation of a line, substituting enzyme kinetics variables. Once we determine the line of best fit and the equation for this line, we can use the equation to determine the intercepts, both the x and y intercepts. The x-intercept of a Lineweaver-Burk plot is going to be where our line crosses the x-axis, revealing the reciprocal of the Michaelis constant, \( K_m \), but note that because the x-intercept falls into the negative region of the plot, it's the negative reciprocal of \( -1/K_m \). The y-intercept is where the line crosses the y-axis and is the reciprocal of the maximum reaction velocity, \( V_{\text{max}} \). Using the x and y intercepts, we can determine \( K_m \) and \( V_{\text{max}} \), having set the values of these intercepts into our respective ratios and solving for the kinetics variables. This allows us to accurately determine the \( K_m \) and the \( V_{\text{max}} \). Discussing the specificity constant, which is the \( k_{\text{cat}} \) over \( K_m \), to get this constant, we need to determine \( k_{\text{cat}} \), equal to \( V_{\text{max}} \) over the total enzyme concentration, given as 0.1 millimolar. With \( k_{\text{cat}} \) calculated, dividing it by \( K_m \) gives us our specificity constant. These steps summarize how to accurately determine the Michaelis constant, the \( V_{\text{max}} \), and the specificity constant of an enzyme from enzyme kinetics data. This concludes this practice problem, and I'll see you guys in our next video.

The specificity constant is obtained at low [S] via variable substitution into the Michaelis-Menten equation (V_max = k_cat[E]_T). Considering this about the MM-equation, what is the relationship between changes in [S] & V₀ when the [S] is super small and well below the K_m?

• A. The [S] term cancels out completely in this equation, so there is no effect of changing substrate concentration.
• B. The [S] term in the numerator is negligible, so there is no impact of changing substrate concentration.
• C. Because the enzyme has reached V_max, there is no effect of changing substrate concentrations on enzyme velocity.
• D. [S] term in the denominator is negligible compared to K_m, so the relationship between [S] & V₀ is directly proportional.

Here’s what students ask on this topic:

What is the specificity constant in enzyme kinetics?

The specificity constant, often denoted as k_cat/K_m, is a measure of an enzyme's catalytic efficiency under non-saturating substrate concentrations. It combines the turnover number (k_cat), which indicates how many substrate molecules one enzyme molecule can convert to product per second, with the Michaelis constant (K_m), which reflects the substrate concentration at which the reaction rate is half of its maximum. A higher specificity constant indicates a more efficient enzyme, as it suggests a high turnover rate and strong substrate binding affinity.

Why is the specificity constant important in enzyme kinetics?

The specificity constant is crucial because it provides a more accurate measure of an enzyme's efficiency under physiological conditions, where substrate concentrations are typically low and not saturating. Unlike k_cat alone, which only applies under saturating conditions, the specificity constant (k_cat/K_m) accounts for both the catalytic turnover and the enzyme's affinity for its substrate. This makes it a better indicator of how well an enzyme performs in a biological context.

How is the specificity constant calculated?

The specificity constant is calculated using the ratio of the turnover number (k_cat) to the Michaelis constant (K_m). The formula is:

$\frac{k}{}$_cat/K_m

Here, k_cat represents the number of substrate molecules converted to product per enzyme molecule per second, and K_m is the substrate concentration at which the reaction rate is half of its maximum. This ratio provides a measure of the enzyme's efficiency at low substrate concentrations.

What is the maximum value of the specificity constant?

The maximum value of the specificity constant is diffusion-controlled and is approximately 10⁹ M^-1s^-1. This limit is due to the rate at which the enzyme and substrate can diffuse together in solution, typically water in biological systems. Enzymes that reach this maximum value are considered catalytically perfect, meaning they convert substrate to product almost instantaneously upon binding.

How does the specificity constant relate to enzyme efficiency?

The specificity constant (k_cat/K_m) directly relates to enzyme efficiency by combining the enzyme's turnover rate with its substrate binding affinity. A higher specificity constant indicates a more efficient enzyme, as it suggests the enzyme can rapidly convert substrate to product even at low substrate concentrations. This makes it a valuable metric for comparing the performance of different enzymes under physiological conditions.