Kcat - Online Tutor, Practice Problems & Exam Prep

The catalytic constant, or k_cat, also known as the turnover number, measures the maximum substrate conversion rate per enzyme molecule under saturating substrate concentrations. It is calculated as the ratio of V_max to total enzyme concentration. In contrast, the Michaelis constant (K_m) indicates the binding affinity of an enzyme for its substrate. Understanding both k_cat and K_m is crucial, as high binding affinity does not guarantee high catalytic efficiency. The reciprocal of k_cat represents the time for a single catalytic event, highlighting the speed of enzyme action.

Kcat

In this video, we're going to finally introduce our last enzyme kinetics variable, the catalytic constant or the k_cat. Now, as tempting as it might seem, the k_cat does not actually stand for a kitty cat. Instead, it stands for the catalytic constant. And so the catalytic constant or the k_cat is really just the rate constant for the rate-limiting step or the slowest step of an enzyme-catalyzed reaction. And so what we need to note is that the rate-limiting or the slowest step of an enzyme-catalyzed reaction ultimately is going to dictate the potential for the theoretical maximal reaction velocity, or the V_max. And so in other words, what we're saying is that a reaction cannot go any faster than its slowest step, and it’s the slowest step of the reaction that dictates the V_max. And because the catalytic constant is the rate constant for the rate-limiting slowest step, that means that the catalytic constant is going to dictate the potential for the V_max. And recall that the V_max can only occur at saturating substrate concentrations. So for that reason, we want to associate the catalytic constant k_cat with saturating substrate concentrations. Now, what I want you guys to note is that for simple enzyme-catalyzed reactions, which pretty much incorporate all of the enzyme-catalyzed reactions that we're going to cover in this course, they can be expressed by this image shown down below on the left, where we have 3 relevant rate constants, k₁, k_-1, and k₂, initially at the very very beginning of the enzyme-catalyzed reaction. Now, also recall from our previous lesson videos that of these three reactions that we see here in this expression, it is k₂ that's actually going to be the rate-limiting or the slowest step of the reaction for simple enzyme-catalyzed reactions. And so because k₂, we're going to assume, is the rate-limiting, the rate constant for the rate-limiting step. That means that we can say that k₂ is going to be equal to k_cat when it comes to simple enzyme-catalyzed reactions, and that's exactly what we're saying up above, is that the k_cat, we're going to always assume, is equal to the value of the k₂. And so recall that k₂ is just the rate constant for the product formation. And we can see that since k₂ here is directly involved with product formation. Now if we were to specifically write the rate law for this reaction here that utilizes k₂, what we would get is this rate law right here. So recall that the rate law is just another way to write or express the reaction velocity V, and this includes the initial reaction velocity. And so, essentially, the reaction velocity is going to be equal to the rate constant k₂ times the concentration of the reactant or the substrate here, which is the enzyme-substrate complex. Now if we were to determine the rate law that uses the catalytic constant, then notice that we can determine that over here on the right. So what we need to recall is that the catalytic constant dictates the potential for the V_max and is associated with saturating substrate concentrations. And when the enzyme is completely saturated with substrate, then we can say that this initial reaction velocity shown here, which can also be expressed as the change in product concentration over the change in time, approximately equals the V_max, and it's totally okay to substitute the initial reaction velocity with the V_max when the enzyme is saturated with substrate. Now, also note that the concentration of the enzyme substrate complex will approximately equal the total concentration of enzyme concentration of enzyme-substrate complex with the total concentration of enzyme when it's saturated with substrate. And then, of course, because we know that it's the catalytic constant, this rate constant for the rate-limiting slowest step that dictates the V_max, whenever the V_max is incorporated like this, we know that it's okay to substitute in the k₂ that we see here with the k_cat. And so we can say that k_cat can go into the space here. And so really we can see that this expression here is a way to express the theoretical maximal reaction velocity V_max. Now if we wanted to define the k_cat, all we need to do is take this total enzyme's concentration and move it below the V_max, and that's what we get right here. So we can say that the k_cat is also equal to this ratio of the theoretical maximal velocity V_max, over the total enzyme concentration. And so, even though we said that the k_cat is not a kitty cat, if it were a kitty cat, then you would imagine that we would have to take it to the vet every now and then. And so hopefully, by remembering that, that'll help you remember how the k_cat is the ratio of the V_max over the total enzyme concentration. And so this here concludes our introduction to the catalytic constant or the k_cat, and we're going to continue to define and express how to calculate the k_cat as we move forward in our next video. So I'll see you guys there.

Kcat

So in our last lesson video, we introduced the catalytic constant or the \(k_{\text{cat}}\) of an enzyme. In this lesson video, we're going to continue to talk about and define the catalytic constant \(k_{\text{cat}}\), specifically how to calculate and interpret the \(k_{\text{cat}}\). Recall from our last lesson video that in order to calculate the catalytic constant or the \(k_{\text{cat}}\), we're going to need both the total enzyme concentration and the theoretical maximal reaction velocity or the \(V_{\text{max}}\) of the enzyme. We defined the \(k_{\text{cat}}\) as the ratio of the \(V_{\text{max}}\) over the total enzyme concentration.

To further define and interpret the meaning of the catalytic constant or the \(k_{\text{cat}}\), here we're defining the value of the \(k_{\text{cat}}\) as the maximum amount of substrate that is being converted into product per second by just one single enzyme molecule specifically under saturating substrate concentration. We'll be able to further explain exactly what this means when we get to our image. It's also important to know that the catalytic constant or the \(k_{\text{cat}}\) is commonly referred to as the turnover number. The unit of \(k_{\text{cat}}\) or the turnover number is inverse seconds.

In our image, notice on the left that we have an image to help you remember that the \(k_{\text{cat}}\) is also known as the turnover number. I wouldn't recommend doing this, but if you were to invert a cat and then just drop it, it would probably turn over and land on its feet. By remembering this image, hopefully that will help you remember that the \(k_{\text{cat}}\) and the turnover number are the same thing. It's important to recognize for your practice problems moving forward that the turnover number is exactly the same thing; it's equal to the value of the \(k_{\text{cat}}\) as we defined already in our previous lesson video.

The table below shows four different enzymes in the first column, the function of each enzyme in the second column, and in the third column, the catalytic constant or the \(k_{\text{cat}}\), which is also equal to the turnover number, and we know from above in our previous lesson video that it has units of inverse seconds. For this specific number, 40,000,000, it means that one single molecule of the enzyme catalase, specifically under saturating substrate concentrations, can convert 40,000,000 molecules of substrate into product per second. On the opposite end, the enzyme DNA Polymerase 1 has a \(k_{\text{cat}}\) of 15, much smaller in comparison.

This number of 15 means that one single molecule of the enzyme DNA Polymerase 1, specifically under saturating substrate concentrations, is capable of converting 15 molecules of substrate into product per second. DNA Polymerase 1 is considered to have a fairly slow maximal catalytic efficiency. Recall from previous biology courses that DNA Polymerase 1 is involved with DNA replication, so this enzyme wants to work at a slower rate to minimize the amount of errors during DNA replication. Thus, by looking at the \(k_{\text{cat}}\), we also get a sense for the functionality and the importance of either catalyzing very quickly lots of substrates into product per second or catalyzing slowly a small amount of substrate being converted into product per second.

In the column with the reciprocal of the catalytic constant, we have \(1/k_{\mathrm{cat}}\) with units of seconds. It represents the time for just one single catalytic event to occur by one molecule of the enzyme. Notice that it says \(2.5 \times 10^{-8}\) seconds for catalase, an incredibly small number, reflecting rapid catalytic events. In comparison, the reciprocal of \(k_{\mathrm{cat}}\) for DNA Polymerase is \(0.07\) seconds, highlighting a slower catalytic rate.

Ultimately, the catalytic constant or the \(k_{\mathrm{cat}}\) or the turnover number alone is used by biochemists as a measure of the maximal catalytic efficiency of an enzyme specifically under saturating substrate concentration. This is one of the more important takeaways: that the \(k_{\mathrm{cat}}\) operates this quickly only when it is saturated with substrate. This concludes our discussion on catalytic efficiency and the \(k_{\mathrm{cat}}\) as well as the turnover number. We'll be able to get some practice applying these concepts as we move forward, and I'll show you an example of using the \(k_{\mathrm{cat}}\) in an example problem in our next video. I'll see you there.

Kcat Example 1

Alright. So here we have an example problem that's asking us what the turnover number for carbonic anhydrase is if the V_max is equal to 60,000 molarity per second, and the total enzyme concentration is equal to 0.1 molar. Recall from our previous lesson videos that the turnover number is the same thing as the k_cat. We need to recall that the k_cat can be defined by the ratio of the V_max for the enzyme over the total enzyme concentration. The k_cat is just a measure of the maximal catalytic efficiency of an enzyme. Notice that we are given the V_max as 60,000 molarity per second so we can plug that in. This will be over our total enzyme concentration, which is given as 0.1 molar. When we do this, notice the unit of molarity cancels and we are left with one over seconds here, so our answer will be in inverse seconds.

60000 0.1 = 600000

This answer matches with answer option B. So we can indicate here that B is the correct answer for this example. If the k_cat is equal to 600,000 inverse seconds, it means that a single molecule of the carbonic anhydrase enzyme is capable of converting 600,000 molecules of substrate into product per second under saturating substrate concentrations. That is a lot of molecules converted into product per second. This concludes our example and we will be able to get some more practice utilizing these concepts in our practice problem. I'll see you guys there.

To calculate the turnover number of an enzyme, you need to know:

• A. Total enzyme concentration [E]_T.
• B. V₀ of the catalyzed reaction when [S] >> [K_m].
• C. V₀ of the catalyzed reaction at low [S].
• D. K_m of the enzyme.
• E. Both a & b.

If 10 μg of an enzyme (MW = 50,000 g/mol) is added to a solution containing a [substrate] 100 times greater than the K_m, it catalyzes the conversion of 75 μmol of substrate into product in 3 min. What is the enzyme's turnover #?

a) 1.25 x 10⁵ min^-1

b) 2.5 x 10⁴ min^-1

c) 1.5 x 10² min^-1

d) 3.5 x 10⁶ min^-1

• A. 1.25 x 10⁵min^-1
• B. 2.5 x 10⁴min^-1
• C. 1.5 x 10²min^-1
• D. 3.5 x 10⁶min^-1

Kcat

In this video, we're going to compare and contrast the catalytic constant or the kcat or the turnover number to the Michaelis constant or the KM of an enzyme. First, I want you guys to recall from our previous lesson videos that the catalytic constant or the kcat or the turnover number is a measure of the maximal catalytic efficiency of an enzyme, specifically under saturating substrate concentrations. In comparison, the Michaelis constant or the KM is instead a measure of the binding affinity that an enzyme has for its substrate. We already knew these two pieces of information from our previous lesson video.

Really, the only thing that I want you guys to learn in this video is that the binding affinity, that an enzyme has for its substrate or the enzyme's binding to its substrate is a completely separate event to enzyme catalysis. Essentially, what this means is that moving forward in our course, we want to think of enzyme binding to its substrate and enzyme catalysis as 2 completely separate events. What this means is that an enzyme could have a really high binding affinity to its substrate and bind really well, but that does not necessarily mean that the enzyme has a high capacity to convert the substrate into product efficiently.

Essentially, what we're saying here is that the KM, or the binding affinity, that an enzyme has for its substrate is going to be completely separate and it's not necessarily going to indicate the kcat that an enzyme has or the maximal catalytic efficiency that an enzyme has. Even though the kcat and the KM both indicate 2 completely separate events, it's important that when a biochemist is studying an enzyme-catalyzed reaction that they consider both the kcat and the KM of the enzyme.

Moving forward in our course, we're going to talk a little more about how the kcat and the KM, can be studied by biochemists to reveal important information about an enzyme, specifically when we talk about the specificity constant later in our course. But for now, this is the conclusion of our comparison of the kcat and the KM, and really this is the only learning objective that I want you guys to know. That here concludes our lesson and I'll see you guys in our practice videos.

Studies with mutated forms of an enzyme show that changing some active-site amino acids decrease the enzyme's turnover number (k_cat) but do not affect the K _m of the reaction. What is the best interpretation of these results?

• A. The K_m of the enzyme for the substrate does not depend on amino acid side chains found in the active site.
• B. The two terms, K_m and turnover number, are inversely proportional.
• C. The transition state for this reaction is formed prior to the formation of the ES complex.
• D. Amino acids involved in stabilizing transition-state complexes can be different than those affecting the ES-complex.

The turnover number for an enzyme is known to be 5000 min ^-1. From the following set of data, determine both the K_m and the total amount of enzyme E_T.

A) What is the K_m of the enzyme?

a) 1 mM. 

b) 2 mM. 

c) 4 mM. 

d) 1000 mM.

B) What is the total amount of enzyme?

• A. 0.001 μM.
• B. 0.01 μM.
• C. 0.1 μM.
• D. 10 μM.

Here’s what students ask on this topic:

What is kcat in enzyme kinetics?

The catalytic constant, or kcat, also known as the turnover number, is a measure of the maximum number of substrate molecules converted to product per enzyme molecule per second under saturating substrate conditions. It is calculated using the formula:

$k\mathrm{cat}=\frac{V\max }{Et}$

where Vmax is the maximum reaction velocity and Et is the total enzyme concentration. kcat provides insight into the catalytic efficiency of an enzyme.

How do you calculate kcat from Vmax and enzyme concentration?

To calculate kcat, you need the maximum reaction velocity (Vmax) and the total enzyme concentration (Et). The formula is:

$k\mathrm{cat}=\frac{V\max }{Et}$

For example, if Vmax is 100 µmol/min and the total enzyme concentration is 0.1 µM, then:

$k\mathrm{cat}=\frac{100}{0.1}=1000 s⁻1$

This means each enzyme molecule converts 1000 substrate molecules to product per second under saturating conditions.

What is the difference between kcat and Km in enzyme kinetics?

kcat and Km are both important parameters in enzyme kinetics but measure different aspects. kcat, or the turnover number, indicates the maximum number of substrate molecules converted to product per enzyme molecule per second under saturating substrate conditions. It reflects the catalytic efficiency of the enzyme. Km, or the Michaelis constant, measures the substrate concentration at which the reaction velocity is half of Vmax. It indicates the binding affinity of the enzyme for its substrate. A low Km means high affinity, while a high Km indicates low affinity. Both parameters are crucial for understanding enzyme behavior.

Why is kcat important in enzyme kinetics?

kcat is important because it provides a measure of the catalytic efficiency of an enzyme. It tells us how many substrate molecules an enzyme can convert into product per second when the enzyme is fully saturated with substrate. This information is crucial for understanding the enzyme's performance under optimal conditions and comparing the efficiencies of different enzymes. Additionally, kcat helps in designing and optimizing industrial processes and in drug development by identifying highly efficient enzymes for specific reactions.

What are the units of kcat?

The units of kcat are inverse seconds (s^-1). This unit indicates the number of substrate molecules converted to product per enzyme molecule per second under saturating substrate conditions. For example, a kcat of 100 s^-1 means that each enzyme molecule converts 100 substrate molecules into product every second when the substrate is in excess.