For more information, please see full course syllabus of Biochemistry

For more information, please see full course syllabus of Biochemistry

### Enzymes VII: Km & Kcat

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Km
- Recall the Michaelis–Menten Equation
- Km & Enzyme's Affinity
- Rate Forward, Rate Backward, and Equilibrium Constant
- When an Enzyme's Affinity for Its Substrate is High
- More on Km & Enzyme Affinity
- The Measure of Km Under Michaelis–Menten kinetic
- Kcat (First-order Rate Constant or Catalytic Rate Constant)
- Kcat: Definition
- Kcat & The Michaelis–Menten Postulate
- Finding Vmax and [Et}
- Units for Vmax and Kcat
- Kcat: Turnover Number
- Michaelis–Menten Equation
- Km & Kcat

- Intro 0:00
- Km 1:48
- Recall the Michaelis–Menten Equation
- Km & Enzyme's Affinity
- Rate Forward, Rate Backward, and Equilibrium Constant
- When an Enzyme's Affinity for Its Substrate is High
- More on Km & Enzyme Affinity
- The Measure of Km Under Michaelis–Menten kinetic
- Kcat (First-order Rate Constant or Catalytic Rate Constant) 24:10
- Kcat: Definition
- Kcat & The Michaelis–Menten Postulate
- Finding Vmax and [Et}
- Units for Vmax and Kcat
- Kcat: Turnover Number
- Michaelis–Menten Equation
- Km & Kcat 36:37
- Second Order Rate Equation
- (Kcat)/(Km): Overview
- High (Kcat)/(Km)
- Low (Kcat)/(Km)
- Practical Big Picture
- Upper Limit to (Kcat)/(Km)
- More On Kcat and Km

### Biochemistry Online Course

### Transcription: Enzymes VII: Km & Kcat

*Hello and welcome back to Educator.com, and welcome back to Biochemistry.*0000

*Today's lesson, we are going to be talking about the Michaelis-Menten constant and this new constant that we are going to introduce called K _{cat}.*0004

*This particular lesson is really, really, very, very important.*0016

*I am not suggesting that the other lessons are not important, but in the biochemistry course, this quantitative study of kinetics is usually, sort of, a daunting thing for the kids.*0020

*There is just a lot of symbols floating around; there is a lot of mechanisms, a lot of reactions.*0035

*You have enzyme substrate product, enzyme substrate, enzyme substrate pro...all kinds of things.*0040

*So, it tends to get really, really confusing.*0047

*What I wanted to do is I wanted to come back; we have already discussed the Michaelis-Menten kinetics.*0051

*We have the Michaelis-Menten equation; I wanted to go back and talk about what K _{m} actually means.*0056

*I want to talk about how it is used in biochemistry, whether that usage is actually justified or not.*0062

*I just wanted to spend some more time with it- that is it.*0070

*I just want you to be as comfortable as possible because these things do come up a lot - well, of course, they come up a lot - in the literature, and it is the one area that I think kids are the most...that it alienates them the most easily.*0074

*It is really not that bad; a lot of it is actually quite intuitive.*0089

*You just need to, sort of, get past a lot of the symbolism.*0093

*In any case, let's just jump right on in, and hopefully, we can help you to get a better understanding of what all of these different things mean.*0097

*Let's see what we can do; OK, let's go ahead and start off with the Michaelis-Menten equation.*0104

*Again, let's recall the MM - you know what, I am just going to...the Michaelis-Menten equation.*0109

*It said v _{0} is equal to V_{max} x the substrate concentration over something called K_{m} plus the substrate concentration.*0125

*Basically, what this equation does is it relates the speed, the velocity, the rate of the reaction, how fast it is going, in either moles per second or molarity per second as a function of substrate concentration, which is really, really, very, very convenient.*0138

*It is a beautiful, beautiful equation; now, this V _{max}, we saw already that if you are given a particular enzyme, you can keep adding substrate, keep adding substrate.*0154

*You are going to increase the rate, but at some point, all of the enzyme molecules are going to be completely saturated with substrate.*0163

*All of the binding sites are taken over; now, enzyme can only go so fast.*0171

*So, no matter how much more substrate you add, you are not going to make it go any faster.*0176

*From your perspective, what you see - or actually I will draw it on here - is this thing that looks like this.*0181

*It ends up maxing out at some speed; there is some upper limit to how fast an enzyme can go.*0188

*You cannot push it any further than it will go if you have a certain amount of enzyme.*0193

*Now, it is true; you can add more enzyme, but that is experimental conditions.*0198

*The body does not just add more enzyme; the body needs to keep its enzyme concentration reasonably low because enzymes are not used up in a reaction.*0202

*It needs to keep them low and constant; it does not just add enzyme whenever it wants to speed it up.*0213

*It is substrate concentration that changes; V _{max} is just an upper limit on how fast the reaction will go.*0217

*This K _{m}, as it turns out, is remember, it is the concentration at which the speed it 1/2 of the maximum speed.*0224

*I did not draw it very well here; let me raise this one a little bit, so something like that.*0237

*That will be V _{max}; that is some upper limit that the speed is going to approach.*0242

*K _{m} is a constant, but it has units of concentration because you remember, it is concentration of substrate that is on the X axis and its rate, its speed that is on the Y axis.*0247

*So, this K _{m} is the concentration of substrate that I need that will make the enzyme operate at half of its maximum velocity.*0262

*Half is just a good number- not too low, not too high, right in the middle.*0275

*That is what K _{m} is, so this is the Michaelis-Menten equation.*0280

*Now, recall also from the derivation because we did do the derivation, recall that this was derived from the following postulated 2-step process.*0284

*Michaelis and Menten, they postulated this particular process- a 2-step process.*0312

*They said that you had enzyme plus substrate.*0321

*There is an equilibrium that exists between that and the complex enzyme substrate once it is attached, and then, that breaks down into enzyme plus product.*0325

*Now, there is some k _{1}, which is the rate constant for the forward reaction.*0334

*There is some k _{-1} for the breakdown of enzyme substrate back to enzyme and substrate, and then, there is k_{2}, which is the rate constant for the breakdown of the enzyme substrate complex into enzyme and product.*0339

*They postulated that this was the slow step; this was very important.*0353

*They postulated that this was the slow step- the rate-determining step.*0358

*So, when we say slow, we mean rate-determining; I will go ahead and write that down here: the slow step, the rate-determining step.*0363

*Because again, your reaction is only going to go as fast as your slowest.*0372

*OK, now, OK, let's see if we can do this.*0378

*Now, K _{m} is the amount of substrate concentration that will put your enzyme at half velocity.*0384

*Well, K _{m} is often interpreted as a numerical measure of an enzyme's affinity for its substrate.*0392

*You will often see K _{m} as interpreted as some numerical measure of an enzyme's affinity for its substrate.*0438

*I am going to show you why it is interpreted that way, and then, I am going to discuss why you have to be careful with that particular interpretation.*0448

*For the most part, it is not going to be a problem, as far as the problems that you do in your book or the things that you discuss in class, but it is very, very important not to just automatically assume that some K _{m} value that you are given or that you read off in some reference book for an enzyme is a measure of the affinity of that enzyme for its substrate.*0455

*I am going to discuss why it is interpreted that way; let's start off there.*0478

*Here is why.*0483

*OK, now, in the derivation of the Michaelis-Menten equation, if you want to go back a couple of lessons, the K _{m} was this.*0488

*It was a combination of the rate constants k _{1}, k_{2} and k_{-1}.*0502

*It was k _{2} + k_{-1}/k_{1}.*0506

*That was the K _{m}; the actual definition of K_{m} is this.*0514

*Our interpretation this way makes more sense intuitively because we have a graph that represents it, but this is the actual definition.*0518

*It is actually a combination of the rate constants of a given process.*0527

*In this case, you have a 1, 2, 3; you have a 3-step.*0531

*There is one this way, one this way, one that way; it is a combination of those rate constants.*0535

*Now, if k _{2} - let me write this a little bit better - is, in fact, rate-limiting, in other words, if that k_{2} step is really the slow step as what was postulated, then, k_{2} is a lot less than k_{-2}.*0540

*It is just a slow step; the rate constant is very, very low, very, very small compared to k _{1}, this thing.*0574

*This is small compared to this.*0583

*Well, because k _{2} is so much less than k_{1}, we can actually ignore it in the numerator.*0592

*So, K _{m} becomes K_{m} = k_{-1}/k_{1}.*0598

*This is what the K _{m} actually becomes under these conditions.*0610

*OK, now, let's investigate this; what does this mean?*0615

*When we have the rate constant in the reverse direction, divided by the rate constant in the forward direction, that is the K _{m} under conditions that we have postulated where k_{2} really is the slow step and can be ignored in the actual definition of K_{m}.*0622

*Well, here is what it means.*0634

*Let's go back to our equilibrium.*0638

*We have enzyme plus substrate in equilibrium with enzyme substrate.*0642

*This is k _{1}; this is k_{-1}.*0648

*Now, the forward rate, and you remember, anytime we write a rate law, it is going to be some constant times the concentration of the reactants.*0652

*In the forward direction, the enzyme and the substrate separately are the reactants.*0664

*The rate forward is equal to k _{1} times the enzyme concentration times the substrate concentration.*0671

*Now, the rate backward, well the rate backward, the reaction is this way.*0683

*Enzyme substrate complex, that is the reactant.*0687

*So, the rate backward - excuse me - equals k _{-1} times the enzyme substrate complex.*0691

*Well, since, we have an equilibrium, these rates are equal to each other, so let's set them equal to each other.*0702

*I will do it over here; I will write k _{1} ES = k_{-1} ES complex - ES together.*0708

*OK, now, let me go ahead and rearrange this and actually put k _{-1} / k_{1}.*0722

*What I end up with is the following; I will write it on this side.*0727

*k _{-1}/k_{1} = enzyme concentration x substrate concentration divided by enzyme substrate complex concentration.*0731

*So far, so good, OK, notice what this is.*0745

*This is the equilibrium constant.*0753

*It happens to be the equilibrium constant for the reverse reaction for...I will write it in the forward direction, though.*0764

*It is the equilibrium constant for the breakdown of the enzyme substrate complex, ES as reactant, E + S - right - reactants/products.*0779

*I am sorry, right, no, yes, right.*0790

*Equilibrium constant is product over reactant; thanks- product over reactant, product over reactant.*0792

*For this reaction, the breakdown of the enzyme substrate complex, this k _{-1}/k_{1} is the equilibrium constant.*0799

*Now, that is what the K _{m} is; the K_{m} actually ends up being an equilibrium constant.*0810

*Now, when an enzyme's affinity for its substrate is high, that means it is found mostly in the enzyme substrate complex form.*0816

*It is found in this form; because the enzyme has a high affinity for its substrate, it is going to bind to its substrate, and it wants to be in that form.*0840

*It does not want to be free enzyme, free substrate.*0848

*High affinity means high concentration of ES.*0852

*Let me go back to black here; when an enzyme's affinity for its substrate is high, then, it is found mostly in the ES state.*0856

*The concentration of ES is very high.*0892

*OK, well, look what we have; we have K _{m} is equal to, we said, k_{-1}/k_{1}.*0899

*Well, k _{-1}/k_{1} is equal to the enzyme concentration times the substrate concentration over the concentration of enzyme substrate.*0907

*This is just the equilibrium constant, again, for the breakdown of the enzyme substrate complex.*0916

*Well, if ES is high, that means if the denominator is high, that means the K _{m} is low...is very high, then, K_{m} is going to be low- that is it.*0921

*Now, this is why we actually interpret or why many biochemists interpret this K _{m} as a measure of affinity for its enzyme.*0952

*The lower the value of K _{m}, because of this equation that we have right here, that means that the concentration of enzyme substrate complex is very, very high because the enzyme has a high affinity for its substrate.*0965

*It wants to be bound to it; it does not want to be free enzymes.*0981

*When this is high, because of how the postulated process works, the K _{m} is going to be very very low.*0985

*In this particular case, if a mechanism, if the particular enzyme actually follows this Michaelis-Menten postulated process, then, it is true.*0993

*The K _{m} is a numerical measure of the affinity for an enzyme, for its substrate.*1004

*The lower the K _{m}, the higher the affinity; again, the K_{m} is just a reflection.*1009

*It is the amount of substrate you have to add in order to get to half velocity.*1015

*Well, if all you have to do is add just a little bit of substrate, that means that every little bit of substrate that you add - boom - immediately attaches to the enzyme, and the enzyme is working very, very, very, very fast or doing what it does.*1020

*You do not have to add a lot of substrate; this is why we interpret K _{m} as the affinity of an enzyme for its substrate.*1032

*This equation, right here, it is an equilibrium that is existing between the enzyme and the substrate and the enzyme substrate complex.*1042

*Now, let's go ahead and say a few words about this.*1050

*Now, many enzymes demonstrate Michaelis-Menten kinetic behavior - OK - but they do not follow the postulated 2-step process that we used to derive the equation.*1057

*They follow that kinetic behavior; they follow that hyperbolic path.*1087

*Many enzymes demonstrate Michaelis-Menten kinetic behavior, but they do not follow the postulated 2-step process with step 2 being rate-limiting.*1092

*Often, K _{m} is a much more complex function of all of the rate constants involved in a given enzyme’s mechanism.*1129

*You might have 5 different steps.*1140

*If each one of those steps is involved in some kind of an equilibrium, you have got 10 different rate constants.*1144

*K _{m} is a lot more complicated than this; often, K_{m} is a much more complex function of the various rate constants for several steps.*1148

*OK, in these cases, K _{m} is not a simple measure of enzyme affinity for substrate.*1180

*OK, let's recap what we did; we went through this process, Michaelis-Menten kinetics, using the postulated 2-step process.*1220

*Again, what we just said, there are many enzymes that actually demonstrate a Michaelis-Menten kinetic behavior.*1229

*We can find a K _{m} for it; it is not a problem.*1236

*However, just because it follows Michaelis-Menten kinetic behavior, it does not mean that the enzyme mechanism is that 2-step process that Michaelis and Menten postulated.*1240

*If it is that simple 2-step process, then yes, we can go ahead and interpret K _{m} as some numerical measure of the affinity for an enzyme for its substrate.*1260

*The lower the K _{m}, the higher the affinity, but if you are dealing with an enzyme that demonstrates Michaelis-Menten behavior, and if you know something else about that enzymes, let's say you happen to know that the enzyme actually operates with 7 steps or 5 steps or 10 steps or whatever it is, and let's say you are not really sure which one is rate-limiting or let's say even if you do know which one is rate-limiting, the K_{m} at that point, even though you can find it from either the graph or doing a Lineweaver-Burk plot and getting the K_{m}, that is not necessarily going to reflect the enzyme's affinity for its substrate.*1268

*This idea of interpreting it, this is very, very important.*1306

*Forgive me if I keep repeating myself; the interpretation of K _{m} as enzyme affinity is based only on this 2-step process.*1310

*If the 2-step process is satisfied, then yes, you can say that 1 enzyme has a K _{m} here.*1317

*One enzyme has a K _{m} of 0.5; another enzyme has a K_{m} of 0.2.*1323

*The enzyme that has a K _{m} of 0.2, I can say, it has a greater affinity for its substrate than the other one, but if it is not the case, just because it demonstrates Michaelis-Menten behavior, is no guarantee that the mechanism is a 2-step process.*1328

*If that is the case, then, you cannot use K _{m} as a measure of affinity- absolutely not.*1343

*I am not sure about the extent to which your professor is going to concentrate on this.*1350

*For all practical purposes, the problems that you do, you are going to go under the assumption that this a simple 2-step process, and we can use K _{m} to decide on affinity.*1357

*However, understand that when you get into the lab, whether it is undergraduate research that you are doing or whether you just run across this in the future, do not lock it into your head that that is what K _{m} means- it does not.*1368

*Only for a specific case can you say that this is actually a measure of affinity.*1378

*OK, now, let's go ahead and talk a little bit more and introduce a new constant called K _{cat}, which is going to give us a lot of great information, and then, we will connect it to K_{m}; and we will use it as a measure of the efficiency of an enzyme- really, really fantastic.*1383

*Now, I will do this in blue.*1400

*Under Michaelis-Menten kinetics, the K _{m}, again, is a measure of how much substrate - this is how we really want to think about it - will bring the rate, will bring the v_{0} to 1/2 V_{max}.*1406

*It will bring the rate to half maximum velocity.*1440

*OK, now, I am going to introduce a new constant, something called K _{cat}, so K_{cat}.*1444

*Now, for a reaction with several steps, K _{cat} is very easy.*1455

*It is the rate constant for the slowest step, if you happen to know what it is- that simple.*1468

*For a reaction with several steps, K _{cat} is the rate constant for the step, which is clearly rate-limiting - OK - for what is obviously the slow step.*1472

*If you have 2 steps that you are not quite sure, you have got to be careful.*1501

*If you clearly know which step is actually slowing the reaction down, then, that is going to be your K _{cat}.*1506

*OK, for the step, which is clearly rate-limiting...OK, now, from Michaelis-Menten postulate, where we had enzyme + substrate going to enzyme substrate going to enzyme + product, it was k _{2}.*1512

*That was the postulated slow step; in this particular case, your K _{cat} = k_{2}.*1538

*That is easy; now, recall that when we did the derivation for the Michaelis-Menten equation, V _{max}, it was defined as k_{2} times the total enzyme concentration.*1545

*We know what the total enzyme concentration is; we control that.*1564

*Now, because K _{cat} = k_{2}, V_{max} = K_{cat} times...no, I have just substituted the K_{cat} in for k_{2}.*1568

*Let me rearrange, and now, I have got K _{cat} is equal to V_{max} over the total enzyme concentration.*1588

*This is really, really fantastic; this is very, very important equation here.*1600

*We introduced this thing called the K _{cat}, which happens to be the same...it is the rate constant for the slowest step in a reaction if I happen to know what that slow step is.*1604

*In the case of Michaelis-Menten kinetics, I actually have a way of calculating it directly.*1614

*V _{max}, I can get either from the graph or from a Lineweaver-Burk plot, so I have that number.*1619

*Well, the total enzyme concentration, well, I picked that.*1623

*Whatever the enzyme concentration is in my particular experiment, I know that.*1627

*If I take one divided by the other, I have calculated that.*1631

*I have come up with a way of actually calculating this thing called K _{cat}; the rate constant for the slowest step in that reaction.*1635

*This is absolutely fantastic; this is a really, really, really important constant.*1642

*Well, that is fine; I will go ahead and write it down.*1653

*V _{max} is easily obtainable from a Lineweaver-Burk plot or directly from rate concentration data and ET.*1657

*Well, it is just the total enzyme concentration, which I control.*1682

*Now, there is a way of finding K _{cat} really, really easily.*1694

*Once you find V _{max}, divide by the enzyme total concentration.*1699

*You have got yourself this K _{cat}, and we are going to talk a little bit about what this means in, well, right not.*1703

*Let's do a little bit of unit analysis.*1708

*V _{max} is a velocity; OK, its V_{max} units are molarity per second.*1712

*OK, now, total enzyme concentration is in molarity.*1726

*This K _{cat} is equal to molarity per second over molarity.*1739

*You cancel those, and you get a unit of 1 over second, which you can also do it that way.*1747

*This is a unit of frequency, the Hertz.*1754

*This is really, really great; now, K _{cat} is also called the turnover number, and here is where it is really, really important, why it is a great number, the turnover number.*1758

*This per second, it means cycles per second.*1775

*When I get some value of K _{cat}, that is telling me in 1 second or in 1 minute or in 1 hour, whatever my time unit happens to be, that that is how many cycles the enzyme runs through.*1779

*One catalytic cycle is one cycle; if I have a K _{cat} of 600, that means in 1 second, the enzyme goes through 600 cycles.*1790

*Well, 1 cycle of the enzyme turns over 1 substrate molecule.*1800

*If the K _{cat} happens to be 600 that I have calculated for a particular enzyme, that means in 1 second, 600 substrate molecules are being produced, are being turned over- that is it.*1805

*It is a direct measure of how fast an enzyme is actually working.*1816

*OK, K _{cat} is also called the turnover number.*1823

*Now, OK, it is the number of cycles per second.*1828

*Again, it is a unit of frequency, of Hertz, and 1 catalytic cycle turns over 1 substrate molecule.*1842

*K _{cat} is a direct measure.*1866

*That is what is great about it; we love direct measures of how many substrate molecules are converted per second- that is it.*1873

*It is that simple, and again, the time unit does not have to be seconds.*1896

*It can be minutes; it can be hours.*1900

*Often, it is going to be seconds or minutes; those are the primary.*1903

*K _{cat} is a direct measure of how many substrate molecules are converted per second: 600, 6000, 60000, 5 million- whatever.*1905

*Now, let's do a little bit more; we said that V _{max}, the maximum velocity is equal to K_{cat} times the total enzyme concentration.*1918

*The Michaelis-Menten equation, which is v _{0} = V_{max} x substrate concentration over K_{m} + substrate concentration, when I substitute this V_{max}, this K_{cat} ET into V_{max}, it becomes v_{0} = K_{cat} x total enzyme concentration x substrate concentration over K_{m} + substrate concentration.*1934

*Now, K _{cat}, we found a way to take the Michaelis-Menten equation, this new constant that we defined, and we came up with this thing.*1977

*Now, we have K _{cat} and K_{m} in the same expression, but it also involves the total enzyme concentration.*1987

*This is actually really, really great; now, when the substrate concentration is a lot less than K _{m} - OK - when you have low substrate concentration, which is actually going to be the case under a lot of physiological conditions, you are not going to have a lot of substrate molecule floating around all the time.*1995

*I mean, under experimental conditions that we control, we can control the substrate concentration, but it is not always like that.*2017

*You have billions and billions of different types of molecules floating around in any given moment in its cell, outside of the cell.*2024

*The substrate concentration for a given enzyme can actually be a lot lower than the actual concentration that we need to get it to half velocity.*2030

*So, how does an enzyme actually do what it does?*2040

*Well, under these conditions, which are pretty standard conditions, watch what happens if S is a lot less than K _{m}.*2042

*We can actually drop it from this denominator; because it is a lot less than this, it becomes insignificant.*2048

*Under these conditions, v _{0} becomes K_{cat} times the total enzyme concentration times the total substrate concentration over K_{m}, or it becomes this constant, K_{cat}/K_{m} x enzyme concentration and substrate concentration.*2053

*Notice, now, our velocity, our rate, is actually has now become a second order reaction.*2080

*It depends on the concentration of 2 things; it depends on the concentration of enzyme, and it depends on the concentration of substrate.*2086

*Well, that makes sense because you remember, the reaction is actually this.*2095

*You have substrate, and you have enzyme.*2102

*When we write a rate law for this, it should be second order because we have...where both of them are actually controlling the rate.*2106

*When we write the rate law for this, remember earlier, we wrote it as any rate law is some constant times the concentration of 1 reactant, the concentration of the other reactant.*2115

*Well, here we go; we have a version of the Michaelis-Menten equation with K _{cat}, K_{m} that actually shows the dependence on the concentration of both.*2125

*OK, let me just actually write this here- initial rate.*2135

*Again, we have the E + S, the ES.*2145

*This represents the rate when under initial conditions, before actual saturation takes place because that is what we are doing.*2151

*We are looking at initial rates; that is what we are measuring.*2164

*This particular version, well, yes, the way we are looking at it actually makes a lot of sense when we are using this K _{cat}/K_{m} x ET x S because now, it is dependent on both of those things, which is more natural.*2168

*OK, now, as we approach sat...you know what, actually I am going to write this out.*2185

*I am sorry; I apologize.*2191

*I think if I write it out it will make a little bit more sense here.*2194

*Notice how this rate equation, this v _{0} equation is, now, second order, which makes sense because again, E + S, you have 2 reactants going to ES.*2198

*The initial rate before saturation - capital here - depends on both the concentration of the enzyme and the concentration of substrate.*2237

*Now, as we approach saturation, the concentration of S, now, becomes a lot bigger than K _{m}, so you can drop the K_{m} from the equation.*2265

*Now, v _{0} approaches V_{max} as expected.*2284

*Now, K _{cat}, this K_{cat}/K_{m}... let me rewrite the equation on this page just so we have it as reference.*2304

*Actually, you know what, let me do this over because I do not want things to show from previous pages.*2316

*v _{0}, we said we have K_{cat}, ET and S/K_{m} + S.*2321

*Now, as we approach saturation, the concentration of substrate becomes huge compared to the K _{m}.*2333

*So, we can actually drop the K _{m} from the denominator; when I do that, I end up with just S in the denominator.*2340

*The Ss cancel my velocity, approaches K _{cat} ET, which is equal to V_{max}.*2347

*It makes perfect sense; it is in perfect alignment with what is happening under the normal version of the Michaelis-Menten equation.*2354

*Nothing is changed; now, K _{cat}/K_{m} - this is the important number - is a number which tells us about the efficiency of the enzyme.*2362

*Again, we have a way of finding K _{m} from the Lineweaver-Burk plot or from the rate concentration data/graph- whichever we want.*2394

*We have a way of finding K _{cat} because it is just a function of the V_{max}.*2403

*We can easily find K _{cat} and K_{m}; when we take K_{cat}/K_{m}, that is what is going to give us a nice number.*2408

*This ratio is what is important.*2416

*Now, do this one in red; OK, a high K _{cat}/K_{m} mathematically means a high turnover number.*2421

*It means a high K _{cat} and a low K_{m}.*2443

*In other words, it means high turnover number and a low amount of substrate concentration needed for half velocity, needed for 1/2 of maximum velocity.*2453

*This is good; this very, very good.*2486

*This is an efficient enzyme; this is what we want.*2490

*This is an efficient enzyme.*2495

*OK, let's talk about what this means; if you have a high K _{cat}, we said that K_{cat} is a direct measure of how many substrate molecules are turned over in a given unit of time, let's say per second.*2502

*If that number is really, really high, that means the enzyme is working very, very fast.*2515

*Now, a K _{m}, if the K_{cat}/K_{m} number is high, that means the numerator is high, and the denominator is low.*2519

*So, K _{m} is low; it means that very, very little, small concentration of substrate is enough to get that enzyme to high velocity, that you can actually get it to the enzyme substrate complex quickly, or in terms of affinity, it has very high affinity for its substrate.*2529

*That means, for every little substrate molecule that comes around, the enzyme grabs it quickly and turns it over quickly.*2550

*That is the idea; high affinity is related to K _{m}.*2558

*High turnover number is related to K _{cat}; I am sorry.*2564

*Yes, high affinity, low K _{m}; high turnover number, high K_{cat}.*2568

*When those conditions are satisfied, your ratio is very, very high.*2573

*That is a very efficient enzyme; it turns over quickly, and it does not need a lot of substrate.*2577

*As the substrate comes in, boom, it is attached to the enzyme, and it is turned over.*2585

*This is a measure of efficiency; this is a very, very important number: K _{cat}/K_{m}.*2589

*OK, now, the other version, if you a have a low K _{cat}/K_{m}, well, let's just do it one at a time.*2595

*OK, a low K _{cat} means low turnover number.*2610

*In other words, that enzyme does not really turn over a lot of substrate molecules, and a high K _{m} means a high amount of substrate is needed for half maximum velocity.*2620

*This is not good.*2654

*This is not good.*2659

*This is not very efficient.*2665

*You have to put heavy concentration of substrate in order to get the rate of the reaction to actually get to half velocity.*2670

*You do not often have a lot of substrate, and if the turnover number, if the K _{cat} happens to be low anyway on that enzyme, you are going to end up really...it is just not very efficient.*2678

*It is not turning the enzyme very, very quickly; that is the low K _{cat}.*2690

*The K _{m} is high; that means you have to add a lot of substrate just to get it to a point where the reaction actually moves at a reasonable rate.*2694

*That is not very efficient; when you have a low K _{cat} and a high K_{m}, your K_{cat}/K_{m} ratio is going to be small.*2700

*Here is where K _{cat}/K_{m} is small because now the denominator is really, really high.*2714

*The numerator is really, really low, so the ratio is going to be low.*2724

*This is not efficient; we do not want this.*2728

*Well, it is not only that we do not want this, under physiological/physio conditions, the concentration of substrate is relatively low.*2733

*We need an efficient enzyme to catalyze the reactions that are important, and an efficient enzyme is where you have a higher K _{cat} to K_{m} ratio.*2760

*High turnover: K _{cat}; low affinity: K_{m}.*2780

*Here is the practical big picture.*2792

*Our practical big picture, if you are given rate concentration data, here is what you do.*2803

*This is the process- very, very simple.*2812

*The first thing you want to do is you want to find K _{m} and V_{max} either from the graph or from the Lineweaver-Burk plot.*2816

*I will just say "find K _{m} and V_{max} from graph" or better yet, from a Lineweaver-Burk plot.*2822

*Once you have the K _{m} and the V_{max}, find K_{cat}.*2837

*I should say "calculate K _{cat}".*2841

*Calculate K _{cat}, which is equal to V_{max} divided by the total enzyme concentration.*2851

*This we get from the graph; this we know.*2858

*The third step form the ratio K _{cat}/K_{m}.*2862

*They sometimes call this thing the specificity constant.*2878

*You do not necessarily have to know that; what is important is that you are actually able to get K _{m}, V_{max} and K_{cat}, and then, you form K_{cat}/K_{m}.*2882

*This K _{cat}/K_{m} is a measure of how efficient that enzyme is.*2897

*You want to have a high K _{cat}, higher efficiency, lower K_{cat}, low...I am sorry.*2902

*High K _{cat}/K_{m} ratio is high efficiency; low K_{cat}/K_{m} ratio is low efficiency.*2908

*The K _{cat}, if that is high, that is a high turnover number.*2916

*That means lots of substrate molecules per unit time.*2920

*K _{m}, you want it to be low.*2924

*That means higher affinity in general under most conditions.*2928

*That is what is going on there; now, there is an upper limit.*2934

*There is an upper limit to this K _{cat}/K_{m}.*2940

*Again, it just depends on how fast 2 things can actually come together to react, and that upper limit is about 10 ^{9} per molarity per second.*2948

*Do not worry about the unit; it is not altogether that important.*2962

*Now, we will close it off with this, and I will go back to blue.*2967

*OK, the nice thing about this K _{cat}/K_{m} is that it can be used to compare efficiencies between different enzymes or the efficiencies of a single enzyme that can have more than 1 substrate, of a single enzyme that takes, that can take - I should be more precise - more than 1 substrate.*2973

*OK, and here is why; you see, different enzymes...I should say different reactions, the different reactions that the different enzymes catalyze.*3050

*Different reactions for different enzymes, they have different uncatalyzed rates, the basic rate without catalysis, as well as different mechanisms.*3073

*K _{cat} alone or K_{m} alone, they do not provide a reference point for comparison.*3110

*The only time that you can ever use K _{m} or K_{cat} alone to compare 2 enzymes or to compare the enzyme with one substrate and the enzyme of another substrate is if the mechanisms is the same and if the uncatalyzed rate of the reaction is the same, if basically all situations are the same.*3133

*That is what we do in science; we measure things, and we try our best to assign numbers to things that we can measure, but in order to actually have value when we are comparing one measurement of another measurement for different things is if we have a point of reference.*3153

*Well, because different enzymes have different mechanisms, there is no way to actually...and they have different uncatalyzed reaction rates, well, one reaction rate might speed it up this much, and another reaction rate, which is uncatalyzed might speed it up this much, well they might end up in the same place as far as K _{cat} or K_{m} is concerned, but the actual difference that it made is going to be different.*3166

*We cannot really tell when a particular enzyme, how much of a rate enhancement that enzyme has provided for the uncatalyzed reaction.*3195

*Because we do not have a point of reference, the K _{m} for a given enzyme and the K_{cat} for a given enzyme alone, I cannot compare the K_{m}s of 2 enzymes or the K_{cat} of 2 enzymes, but if I take K_{cat}/K_{m}, that gives me a measure of efficiency; and I can compare that efficiency with the K_{cat} divided by the K_{m}, the efficiency of another enzyme.*3206

*Those 2 I cannot compare because now, I brought them to the same reference point.*3230

*That is why both numbers are important, but together, in my personal opinion, they are more important because you can actually compare different enzymes- really, really great.*3235

*OK, now, K _{cat} alone of K_{m} alone does not provide a reference point for comparison, but this K_{cat} divided by the K_{m} does.*3248

*It provides a single reference point; now, I can compare 2 enzymes, or I can compare an enzyme with 2 different substrates.*3261

*It does not matter if the uncatalyzed rates are different.*3268

*It does not matter if their mechanisms are different; I am concerned with the relative efficiencies of those 2 enzymes.*3272

*I hope that has helped clear some things up regarding this whole kinetics issue with K _{m} and V_{max} and K_{cat}.*3279

*Thank you for joining us here at Educator.com; we will see you next time, bye-bye.*3286

1 answer

Last reply by: tiffany yang

Thu Dec 5, 2013 1:51 PM

Post by tiffany yang on November 14, 2013

If Ki (dissociation constant for inhibitor) is basically just ligand-protein binding relationship, then Ki will be the ligand concentration when 50% of the enzyme active sites are occupied. Can we apply the same logic for Km?

Can Km be calculated by the concentration of ligand when 50% of enzyme active sites are occupied. From what i remember from lecture, we can't, but i'm not sure why...

0 answers

Post by tiffany yang on November 14, 2013

this is amazing! many many thanks Raffi! everything makes sense now...

1 answer

Last reply by: Professor Hovasapian

Fri Oct 4, 2013 4:04 PM

Post by Tom Beynon on October 4, 2013

Dr Raffi, from down here in Australia, i would like to say thankyou, you are a genuine legend.