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Lecture Comments (5)

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.

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 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

Transcription: Enzymes VII: Km & Kcat

Hello and welcome back to, 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 Kcat.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 Km 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 v0 is equal to Vmax x the substrate concentration over something called Km 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 Vmax, 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; Vmax is just an upper limit on how fast the reaction will go.0217

This Km, 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 Vmax; that is some upper limit that the speed is going to approach.0242

Km 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 Km 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 Km 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 k1, 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 k2, 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, Km is the amount of substrate concentration that will put your enzyme at half velocity.0384

Well, Km is often interpreted as a numerical measure of an enzyme's affinity for its substrate.0392

You will often see Km 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 Km 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 Km was this.0488

It was a combination of the rate constants k1, k2 and k-1.0502

It was k2 + k-1/k1.0506

That was the Km; the actual definition of Km 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 k2 - let me write this a little bit better - is, in fact, rate-limiting, in other words, if that k2 step is really the slow step as what was postulated, then, k2 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 k1, this thing.0574

This is small compared to this.0583

Well, because k2 is so much less than k1, we can actually ignore it in the numerator.0592

So, Km becomes Km = k-1/k1.0598

This is what the Km 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 Km under conditions that we have postulated where k2 really is the slow step and can be ignored in the actual definition of Km.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 k1; 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 k1 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 k1 ES = k-1 ES complex - ES together.0708

OK, now, let me go ahead and rearrange this and actually put k-1 / k1.0722

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

k-1/k1 = 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/k1 is the equilibrium constant.0799

Now, that is what the Km is; the Km 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 Km is equal to, we said, k-1/k1.0899

Well, k-1/k1 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 Km is very high, then, Km is going to be low- that is it.0921

Now, this is why we actually interpret or why many biochemists interpret this Km as a measure of affinity for its enzyme.0952

The lower the value of Km, 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 Km 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 Km is a numerical measure of the affinity for an enzyme, for its substrate.1004

The lower the Km, the higher the affinity; again, the Km 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 Km 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, Km 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

Km is a lot more complicated than this; often, Km is a much more complex function of the various rate constants for several steps.1148

OK, in these cases, Km 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 Km 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 Km as some numerical measure of the affinity for an enzyme for its substrate.1260

The lower the Km, 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 Km at that point, even though you can find it from either the graph or doing a Lineweaver-Burk plot and getting the Km, 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 Km 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 Km here.1317

One enzyme has a Km of 0.5; another enzyme has a Km of 0.2.1323

The enzyme that has a Km 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 Km 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 Km 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 Km 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 Kcat, which is going to give us a lot of great information, and then, we will connect it to Km; 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 Km, 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 v0 to 1/2 Vmax.1406

It will bring the rate to half maximum velocity.1440

OK, now, I am going to introduce a new constant, something called Kcat, so Kcat.1444

Now, for a reaction with several steps, Kcat 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, Kcat 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 Kcat.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 k2.1512

That was the postulated slow step; in this particular case, your Kcat = k2.1538

That is easy; now, recall that when we did the derivation for the Michaelis-Menten equation, Vmax, it was defined as k2 times the total enzyme concentration.1545

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

Now, because Kcat = k2, Vmax = Kcat, I have just substituted the Kcat in for k2.1568

Let me rearrange, and now, I have got Kcat is equal to Vmax 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 Kcat, which happens to be the 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

Vmax, 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 Kcat; 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

Vmax 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 Kcat really, really easily.1694

Once you find Vmax, divide by the enzyme total concentration.1699

You have got yourself this Kcat, 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

Vmax is a velocity; OK, its Vmax units are molarity per second.1712

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

This Kcat 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, Kcat 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 Kcat, 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 Kcat 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 Kcat 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, Kcat 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

Kcat 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

Kcat 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 Vmax, the maximum velocity is equal to Kcat times the total enzyme concentration.1918

The Michaelis-Menten equation, which is v0 = Vmax x substrate concentration over Km + substrate concentration, when I substitute this Vmax, this Kcat ET into Vmax, it becomes v0 = Kcat x total enzyme concentration x substrate concentration over Km + substrate concentration.1934

Now, Kcat, 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 Kcat and Km 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 Km - 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 Km.2042

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

Under these conditions, v0 becomes Kcat times the total enzyme concentration times the total substrate concentration over Km, or it becomes this constant, Kcat/Km 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 Kcat, Km 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 Kcat/Km x ET x S because now, it is dependent on both of those things, which is more natural.2168

OK, now, as we approach 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 v0 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 Km, so you can drop the Km from the equation.2265

Now, v0 approaches Vmax as expected.2284

Now, Kcat, this Kcat/Km... 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

v0, we said we have Kcat, ET and S/Km + S.2321

Now, as we approach saturation, the concentration of substrate becomes huge compared to the Km.2333

So, we can actually drop the Km from the denominator; when I do that, I end up with just S in the denominator.2340

The Ss cancel my velocity, approaches Kcat ET, which is equal to Vmax.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, Kcat/Km - 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 Km from the Lineweaver-Burk plot or from the rate concentration data/graph- whichever we want.2394

We have a way of finding Kcat because it is just a function of the Vmax.2403

We can easily find Kcat and Km; when we take Kcat/Km, 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 Kcat/Km mathematically means a high turnover number.2421

It means a high Kcat and a low Km.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 Kcat, we said that Kcat 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 Km, if the Kcat/Km number is high, that means the numerator is high, and the denominator is low.2519

So, Km 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 Km.2558

High turnover number is related to Kcat; I am sorry.2564

Yes, high affinity, low Km; high turnover number, high Kcat.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: Kcat/Km.2589

OK, now, the other version, if you a have a low Kcat/Km, well, let's just do it one at a time.2595

OK, a low Kcat means low turnover number.2610

In other words, that enzyme does not really turn over a lot of substrate molecules, and a high Km 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 Kcat happens to be low anyway on that enzyme, you are going to end up is just not very efficient.2678

It is not turning the enzyme very, very quickly; that is the low Kcat.2690

The Km 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 Kcat and a high Km, your Kcat/Km ratio is going to be small.2700

Here is where Kcat/Km 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 Kcat to Km ratio.2760

High turnover: Kcat; low affinity: Km.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 Km and Vmax either from the graph or from the Lineweaver-Burk plot.2816

I will just say "find Km and Vmax from graph" or better yet, from a Lineweaver-Burk plot.2822

Once you have the Km and the Vmax, find Kcat.2837

I should say "calculate Kcat".2841

Calculate Kcat, which is equal to Vmax divided by the total enzyme concentration.2851

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

The third step form the ratio Kcat/Km.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 Km, Vmax and Kcat, and then, you form Kcat/Km.2882

This Kcat/Km is a measure of how efficient that enzyme is.2897

You want to have a high Kcat, higher efficiency, lower Kcat, low...I am sorry.2902

High Kcat/Km ratio is high efficiency; low Kcat/Km ratio is low efficiency.2908

The Kcat, if that is high, that is a high turnover number.2916

That means lots of substrate molecules per unit time.2920

Km, 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 Kcat/Km.2940

Again, it just depends on how fast 2 things can actually come together to react, and that upper limit is about 109 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 Kcat/Km 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

Kcat alone or Km alone, they do not provide a reference point for comparison.3110

The only time that you can ever use Km or Kcat 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 Kcat or Km 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 Km for a given enzyme and the Kcat for a given enzyme alone, I cannot compare the Kms of 2 enzymes or the Kcat of 2 enzymes, but if I take Kcat/Km, that gives me a measure of efficiency; and I can compare that efficiency with the Kcat divided by the Km, 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, Kcat alone of Km alone does not provide a reference point for comparison, but this Kcat divided by the Km 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 Km and Vmax and Kcat.3279

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