For more information, please see full course syllabus of Biochemistry

For more information, please see full course syllabus of Biochemistry

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### Enzymes III: Kinetics

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
- Enzymes III: Kinetics
- Rate of an Enzyme-Catalyzed Reaction & Substrate Concentration
- Graph: Substrate Concentration vs. Reaction Rate
- Rate At Low and High Substrate Concentration
- Michaelis & Menten Kinetics
- More On Rate & Concentration of Substrate
- Steady-State Assumption
- Rate is Determined by How Fast ES Breaks Down to Product
- Total Enzyme Concentration: [Et] = [E] + [ES]
- Rate of ES Formation
- Rate of ES Breakdown
- Measuring Concentration of Enzyme-Substrate Complex
- Measuring Initial & Maximum Velocity
- Michaelis & Menten Equation
- What Happens When V₀ = (1/2) Vmax?
- When [S] << Km
- When [S] >> Km

- Intro 0:00
- Enzymes III: Kinetics 1:40
- Rate of an Enzyme-Catalyzed Reaction & Substrate Concentration
- Graph: Substrate Concentration vs. Reaction Rate
- Rate At Low and High Substrate Concentration
- Michaelis & Menten Kinetics
- More On Rate & Concentration of Substrate
- Steady-State Assumption
- Rate is Determined by How Fast ES Breaks Down to Product
- Total Enzyme Concentration: [Et] = [E] + [ES]
- Rate of ES Formation
- Rate of ES Breakdown
- Measuring Concentration of Enzyme-Substrate Complex
- Measuring Initial & Maximum Velocity
- Michaelis & Menten Equation
- What Happens When V₀ = (1/2) Vmax?
- When [S] << Km
- When [S] >> Km

### Biochemistry Online Course

### Transcription: Enzymes III: Kinetics

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

*Today, we are going to start talking about enzyme kinetics.*0004

*Kinetics is just a fancy word for rate of a reaction, how fast does this reaction go.*0009

*Kinetics is, sort of, the beginning of our understanding of mechanism.*0016

*With biochemistry, with all of chemistry, our ultimate goal is to understand the mechanism, how something goes from one thing to another, in the case of an enzyme, how an enzyme takes a particular substrate and how it converts it to product.*0022

*Each individual little step, every single electron, where it goes, every single atom, every single hydrogen ion that is transferred from here to there- that is our ultimate goal because if we understand the mechanism, then we know how to control it.*0038

*We know how to fiddle with it, how to do whatever we want to it because we understand how it works.*0050

*Kinetics is, sort of, the first step of understanding that mechanism.*0057

*It is the oldest approach to understanding mechanism, and it is still very, very, very important.*0062

*Let's jump in and see what we can do; now, we are going to deal with the quantitative aspects of kinetics, and I am going to go through a derivation of the Michaelis-Menten equation.*0069

*Hopefully, it will not be too bad; ultimately, we are not going to be concerned with the derivation.*0082

*Your teacher will let you know about the extent to which you need to know the derivation or not the derivation, but we will go through it.*0086

*It is the final form of the equation and how the enzyme behaves; that is what is important.*0093

*It is more of a global understanding; let's see what we can do.*0098

*OK, now, as we said, the fundamental approach - let's see here - to studying a mechanism is to determine the rate of an enzyme-catalyzed reaction/rxn and more importantly and how this rate actually changes with changes in substrate concentration.*0102

*In other words, if I start with a certain amount of enzyme that will hold constant and if I add a certain amount of substrate to this enzyme solution and if I measure how fast these substrate molecules are being turned over into product, that gives me a rate.*0175

*It will often be expressed in terms of, let's say, millimoles or micromoles per minute or perhaps in molarity, millimolar or micromolar per minute or per second.*0195

*A rate is some change in some value per unit time.*0205

*That is what a rate is, how fast something is happening; well, if I start with a different initial concentration, does that change the rate?*0209

*As it turns out, the answer is yes.*0216

*Ultimately, substrate concentration controls how fast or how slow an enzyme-catalyzed reaction is going to go.*0220

*Now, it is never always this simple.*0227

*With normal chemical reaction, it is not even that simple; you can just imagine how complex it can become in the case of an enzyme-catalyzed reaction where you have, maybe, more than 1 substrate, where you have multiple steps in a mechanism, not just 1 or 2 but perhaps 7 or 8 before the enzyme releases the product.*0231

*Again, we are going to be making some simplifying assumptions to make it tractable, so that we can actually deal with it, but ultimately, it is about how the substrate concentration affects how fast an enzyme-catalyzed reaction can go.*0250

*OK, let's see what we have.*0266

*Now, since the substrate concentration will denote it like that, S with a bracket.*0270

*Concentration - you remember from chemistry - always has brackets, and those brackets are normally in moles per liter.*0277

*In the case of biochemistry, it will usually be millimolarity, occasionally micromolarity because you are talking about pretty small amount, at least as far as the enzyme is concerned, and often the substrate is well.*0283

*Since S changes, as the reaction proceeds, we simplify matters by measuring initial rates.*0296

*And you remember or you should remember, at least, or I hope you remember initial rates from general chemistry when you were doing the kinetics portion, although I do understand that the kinetics portion of general chemistry is generally that portion of chemistry that one would like to forget.*0326

*That is the one that seems to leave the mind the quickest because it was so very odd compared to the other concepts because it was heavily mathematical.*0339

*In any case, as you know, you put a substrate in with an enzyme, the substrate concentration actually changes.*0347

*So, as an experimenter, how can one...and since the substrate concentration is changing, how can I actually try different substrate concentrations to get some, sort of, a relationship between substrate concentration and speed of the reaction?*0355

*Well, if we measure how fast the reaction is going in the beginning, right when we put the substrate and the enzyme together, let's say, within the first 30 seconds or, at most, the first minute, we never want to let it go past that.*0370

*Well, during that timeframe, because the substrate concentration in general is usually going to be so much more than the enzyme concentration - right - enzyme, well, is not used up in a reaction, so the enzyme concentration is very small.*0383

*One enzyme molecule can handle millions of substrate molecules.*0398

*Because that is the case, in the first part of a reaction, the first 30 seconds, the first minute, maybe the first minute and a half, the substrate concentration does not really change all that much, especially since the substrate concentration is so much bigger than the enzyme concentration.*0401

*So, for all practical purposes, the substrate concentration that I start the experiment with, it is going to be held constant.*0417

*If I start with 10 millimoles or 20 millimoles or 30 millimoles, even though substrate is being used up, compared to the 20 millimoles, 30 millimoles, 40 millimoles, that I start off with, it is going to be completely negligible.*0425

*Now, we can actually do that; even though substrate concentration changes with an enzyme-catalyzed reaction because of the difference, because there is so much substrate, I can pretend that it does not really change.*0440

*Now, I can actually run my experiment; I can actually collect some data.*0450

*Since S changes to the reaction when we simplify matters by measuring initial rates of reaction, how fast the reaction is going at the beginning before other complications start to slow the reaction down.*0456

*Now, as we said, since the concentration of S is so much greater than the concentration of enzyme during the first minute or 2, the substrate concentration is essentially constant.*0470

*OK, now, what this does is this allows us to choose various initial concentrations - like I said, 10, 20, 40, 80, 160, however - then, plot the rate of reaction that we measure as a function of these different initial concentrations.*0510

*In other words, what we are saying is that the rate - I will call it initial velocity, so we will use the symbolism that is pretty common, well, actually, you know what, let me go ahead and just write it as rate first, and I will go ahead and use the symbolism - is going to be some function of S.*0568

*In other words, you are going to have S on the X axis.*0590

*OK, that is going to be your independent variable, and the rate is going to be your dependent variable.*0594

*That is going to be your Y axis.*0598

*Rate or speed of the reaction, we will often symbolize with v _{0}, initial velocity.*0602

*Sort of, this is 0 making reference to that fact that we are dealing with an initial rate.*0608

*v _{0} is equal to some function of the substrate concentration.*0613

*Different substrate concentrations, different values, and you end up getting this curve.*0620

*Well, when we do this, when we actually run the experiment, we get the following.*0625

*We take some enzyme; we start with different substrate concentrations.*0631

*Let's say we run 10 different initial concentrations, we get 10 data points.*0636

*We measure the speed at the beginning of the reaction.*0640

*We get the following graph.*0645

*We get something that looks like this, so let's take a look at this- very, very, very important graphical representation.*0650

*This is a plot of substrate concentration on the X axis and reaction rate on the Y axis.*0657

*Do not worry about what the V _{max}, half V_{max} and K_{m} mean right now.*0664

*We will get to that in a minute when we talk about this Michaelis-Menten equation that we want to deal with.*0668

*OK, if I start with an initial concentration of a substrate concentration of 0, well, nothing is going to happen.*0673

*As I increase my initial substrate concentration, I am going to measure a rate.*0682

*I try another initial concentration, I am going to measure an initial rate.*0688

*Another experiment, another concentration, I measure another rate; I do this several times.*0693

*In this case, it looks like 1, 2, 3, 4, 5, 6, 7, 8, 9, 9 experiments.*0698

*You can do as many as you like; obviously, the more that you do, the better data you get.*0703

*When I connect all of those dots, I get this thing.*0707

*This is a very, very, very characteristic enzyme behavior- reaction rate versus substrate concentration.*0712

*Let's just talk a little bit about what is happening here just qualitatively; let's get a sense of what is happening.*0722

*You notice in the beginning almost this...well, that is OK.*0728

*At low concentrations of substrate, what happens is you have basically almost a straight line here - right - of 2 about right here?*0735

*For very, very low concentrations, what you have is, sort of, a linear behavior, in other words, just something that is like the rate equals some constant times the concentration of S.*0745

*That is just some line - that is it - a linear relationship, but as the substrate concentration increases, notice that the speed actually starts to slow down.*0760

*You might ask yourself, well why is that, why does it slow down?*0772

*Well, think about it; it actually makes sense.*0775

*You only have so much enzyme; you keep adding substrate.*0778

*You keep adding substrate; you keep adding substrate.*0781

*At some point, you have added so much substrate; around this point, you have added so much substrate that, now, the enzyme is completely tied up with substrate.*0784

*In other words, it is bound to the substrate; it is converting it to product.*0795

*The minute it releases that product, yes, it is a free enzyme again, but immediately, another substrate binds to it.*0798

*So, for all practical purposes, at some point, you have tied up all of the enzymes that can turn molecules over, that can turn substrates over.*0804

*It does not matter how much more substrate you add; you can keep adding substrate.*0813

*You are not going to make the reaction go any faster because there is only so much enzyme there that can react, that can actually do the job.*0816

*That is why you see this tapering off behavior, and you know intuitively that at some point, you are just going to keep adding so much substrate- that is it.*0824

*There is only so fast that the enzyme is going to turn these substrates over.*0835

*That is what this is here; this is an upper limit on the speed of the enzyme.*0840

*Enzymes achieve this, what we call, maximum velocity, maximum speed.*0844

*There is some maximum speed faster than which the enzyme...it is not just going to go.*0850

*No matter how much more substrate or no matter what else you do to it, it is just not going to go any faster than it can go, than it can turn over the molecules.*0855

*Let's write all this out.*0864

*At low substrate concentrations, the rate increases.*0870

*So, there is an increase; as you increase concentration, the rate, it is faster.*0878

*The rate increase linearly, or I will just say "almost linearly", which seems to make sense.*0882

*I mean it is reasonably intuitive; now, as the substrate concentration increases, the rate slows down with each increasing concentration, and notice, you are having a big increase in the initial concentration; but you are getting a very, very little increase in rate.*0890

*It is definitely slowing down here; the rate increase slows until a point is reached where increasing the substrate concentration has no effect on the rate.*0912

*OK, this V _{max} is exactly what you think it is, this maximum velocity, V_{max}, the fastest this enzyme can go.*0950

*OK, it is here that the enzyme is saturated.*0976

*In other words, it is completely tied up with substrate; it is completely saturated, and no matter how much more S is added, it will only go as fast as it can turn substrate over or convert substrate to product.*0989

*Again, what we would like here is to have some, sort of, quality, be able to explain this behavior.*1031

*We explain this behavior by saying that at some point, your initial substrate concentration, the minute you put it into this enzyme solution, it just binds up all of the enzyme.*1037

*Now, there is no free enzyme floating around; here, there was free enzyme.*1048

*The more substrate you added, the more free enzyme actually started to operate.*1052

*So, you have got a faster turnover overall, but at some point, your substrate is going to completely saturate the enzyme.*1059

*Now, all of the free enzyme is bound up, and since there is no more enzyme that is free, you can add as much substrate as you want.*1066

*It is not going to make it go any faster than it is already going; that is its maximum velocity.*1074

*Now, when you do these experiments, when you do reaction rate and substrate concentration experiments, you are probably wondering where does this V _{max} come from.*1079

*OK, if your data is reaction rate versus substrate concentration, this maximum velocity is something that you are going to have to estimate.*1087

*You are going to have to take a look at your data, and you are going to have to extrapolate some maximum velocity.*1095

*Usually, it is not a problem; you just go a couple of units above something like that, and you draw a line.*1100

*Now, in the next lesson, we will talk about a way to actually get an analytical method for coming up with the V _{max}, the maximum velocity, which is a really, really fantastic method.*1106

*Do not worry about that; we will be talking about that, but when you have just reaction rate and concentration data, this is something that is just extrapolated.*1116

*It is estimated; you look at your graph, and you just draw a line based on experience or whatever other information you have at your disposal.*1122

*OK, now, what we have here is empirical behavior.*1132

*You took some enzymes; you took some initial concentrations, measure the rates; this is how an enzyme behaves.*1138

*What Michaelis and Menten did is they postulated a few things based on what is happening, and they decided to derive an equation to see if they can come up with some equation that explains this behavior- OK, that is it.*1143

*They wanted to have some mathematical formula that matches this.*1156

*That is usually how it works in science; you come up with some data.*1162

*You graph it, and then, you try your best to go back and theoretically see if you can find an equation which fits the data.*1166

*That is usually how it works; we, sort of, learn it in reverse especially when we are learning things like math and physics.*1174

*We learn the formulas, and then, simultaneously we will talk about the data, but this is how it really happens.*1178

*You have the data first, the graph first, and then, you try to fit the equation to the graph by making some assumptions and fiddling around with a bunch of equations until something works.*1186

*Alright, now, this is empirical behavior.*1198

*OK, Michaelis and Menten postulated the following; let me actually move here.*1210

*Again, any theoretical discussion begins with a series of postulates, and then, you, sort of, take it from there; and you see where your math leads you, or you see where your chemistry leads you theoretically, and then, you compare it with the data.*1227

*That is how it works; Michaelis and Menten postulated the following.*1240

*They postulated that the enzyme and the substrate, there is some k _{1} - rate constant - and k_{-1}, and they form something called an enzyme substrate complex.*1247

*Enzyme substrate bind, you have this enzyme substrate complex.*1262

*They assumed that this was the fast step, that this happens very quickly, and then, of course, the enzyme substrate complex, that is the one that actually breaks down into enzyme and product.*1267

*So, there is another equilibrium here.*1279

*They call this k _{2}, and they call this -2 into enzyme and product.*1283

*Their postulate was that this was the slow step, and if you remember from general chemistry, the slow step in a mechanism, in a series of steps, that is the step that controls the rate of the reaction.*1288

*As it turns out, now, because step 2 is slower, the overall reaction rate depends on this concentration of enzyme substrate rather than just substrate- that is it.*1301

*That is what they postulated, that this was the slow step, and this behavior actually depends on the concentration of enzyme substrate, not necessarily just the concentration of substrate.*1341

*They decided to take it from there.*1357

*Let's see; let's go ahead and take a look at this one more time.*1364

*Again, under low concentrations of sub...well, let's write all this out.*1374

*When the concentration of substrate is low, it is still going to be mostly free enzyme.*1382

*In other words, the enzyme substrate complex has not formed.*1399

*It is not saturated; the enzyme is not saturated with substrate.*1403

*It is still mostly free enzyme.*1406

*The rate still depends just on substrate concentration.*1411

*Remember what we had?*1417

*Let me draw it over here; we said E + S goes to E _{S}, and then, from here, goes to enzyme + product.*1420

*At low concentrations of substrate, we are still mostly free substrate and free enzyme.*1430

*The speed of the reaction is still just contingent on the substrate concentration.*1436

*The rate is still just a direct function of substrate concentration/S- the substrate concentration.*1442

*That is what explains the linearity; when it is just depending on one thing, you are generally going to get some, sort of, a linear behavior.*1460

*OK, now, but as the concentration of substrate/S rises, the enzyme, now, is mostly in its bound state, right?*1468

*The enzyme becomes saturated; the enzyme becomes bound up with the substrate.*1498

*Now, it is no longer here; now, it is mostly this, thus, the second step of this Michaelis-Menten postulate takes over, and that is where we experience deviation from linear behavior.*1502

*We have deviation from a linear behavior, and this deviation comes from the fact that it is no longer contingent just on substrate concentration; but it depends on enzyme substrate concentration.*1535

*Clearly, from here to here, things get rather complicated because now, it is the enzyme substrate that is controlling how fast the reaction is going.*1552

*OK, now, a second assumption here or postulate, if you will, I will say in these proceedings, is the steady state postulate, steady state assumption.*1562

*OK, I will just call it assumption.*1590

*The steady state assumption...again, we have to, sort of, make things as easy as possible for us that we can deal with the mathematics in reasonably decent way.*1602

*The steady state says this; the formation of the enzyme substrate complex is quick - right, a fast step, this is the fast step - and once it is formed, the concentration of this enzyme substrate complex remains constant.*1611

*That does not fluctuate; it achieves a steady state.*1644

*It does not mean things stop; it just means it has achieved a steady state.*1648

*This concentration is not going wildly up and down.*1654

*If that were the case, there is no way we can handle this mathematically, at least not in any simple fashion.*1658

*A second assumption of proceedings is the steady state assumption; this is a very important assumption.*1664

*The formation of the enzyme substrate complex is quick, but once it is formed, the concentration of enzyme substrate, it stays reasonably constant.*1669

*OK, in other words, that is the rate of enzyme substrate breakdown is the same as the rate of enzyme substrate formation- that is it.*1691

*In other words, you saturated your enzyme; now, you have enzyme substrate complex.*1720

*As quickly as the enzyme turns over the substrate and kicks out a product, another substrate actually binds.*1724

*You are never going to end up with some enzyme substrate concentration that goes from 10 to 50 to 3 to 5 to 47 to 8.*1731

*Once it gets there, that is it; it usually just stay there.*1740

*That is the steady state assumption; it allows us - in a minute, as you will see - to set rates equal to each other so that we can make the mathematics a little bit more tractable.*1743

*OK, now, let's go ahead and rewrite what we have here.*1753

*We have our enzyme plus our substrate.*1758

*It is going to form our enzyme substrate complex, and that is going to break down into enzyme plus product.*1762

*This is the Michaelis-Menten 2-step mechanism; now, we say mechanism here.*1768

*Mechanism means individual steps in a reaction- what atoms move, which electrons move.*1773

*We talk about mechanism a little bit more broadly in terms of steps.*1782

*I will call it a mechanism; this is not a mechanism formally.*1786

*This is more just reaction steps because lots of things might happen between here and here.*1789

*Even though this is one step at the molecular level, at the atomic level, several things might be going on.*1795

*Again, there is a rate constant in this direction; there is a rate constant in this direction- the breakdown of the enzyme substrate complex.*1800

*Well, the breakdown of the enzyme substrate complex to product has its own rate constant, K, and the formation of enzyme product.*1811

*Well, enzyme and product might come back together to form enzyme substrate, right?*1822

*This is always in equilibrium; things go back and forth.*1825

*This we will call it k _{-2}; now, in this case, the k_{-2}, this step, it can be ignored.*1828

*It can be ignored because in the beginning of a reaction, which is really all we are concerned with, the concentration of product is negligible.*1842

*The first minute or 2 of a reaction, the concentration of product does not matter.*1858

*Because the concentration of product does not matter, the reformation of enzyme substrate complex from product and enzyme, we can pretend that it does not really happen.*1862

*We can just, sort of, knock this out of consideration, so when we do our mathematics, we are only going to be concerned with this step forward, this step backward and this step forward.*1872

*Enzyme substrate forming enzyme substrate complex, enzyme substrate complex breaking down into enzyme substrate free or breaking down into enzyme product- that is it.*1884

*That is all we are going to be concerned with when we deal with mathematics.*1894

*Rate is determined by how fast this enzyme substrate complex breaks down- that is it.*1901

*That is how fast it is; it is controlled by that thing.*1917

*It breaks down to product; we know that the rate, which we are calling v _{0}, initial velocity.*1923

*Now, because the Michaelis-Menten postulate was that this second step is the slow step, that is the rate-determining step.*1934

*That means that the rate law for this, the rate equation that equates concentration of this to how fast a reaction is going, is going to be this: the rate constant of this step times the concentration of the reactant.*1942

*That is how we do it if you remember from chemistry and kinetics.*1959

*The rate of a reaction is equal to some rate constant times the concentration of the particular reactant in that reaction.*1963

*In this particular case, this is the fast step; that does not control the rate.*1971

*The slow step, this step, from E _{S} to E + P, that is what controls the rate.*1976

*Therefore, this is the reaction we are concerned with- this way, right?*1980

*We said we can ignore this; we are concerned this way; the rate of our ultimate overall reaction is going to be the rate at which this breaks down.*1987

*That rate is given mathematically by k _{2} x S; k_{2} is the rate constant.*1994

*E _{S} is the concentration of the enzyme substrate complex- basic, first order kinetics.*1999

*This exponent here is 1; we are not going to get into what the order of the reaction is right now.*2005

*We are just going to stay nice and simple, so let me go back to black.*2010

*OK, now, we have this; this is our starting point right here.*2015

*Now, the thing is, the concentration of enzyme substrate complex, this is not easily measured experimentally.*2024

*We have to find different expressions for E _{S} so that we can substitute into this expression.*2047

*In other words, we need to express E _{S} in terms of things that are easily measured, and we can do that.*2055

*Let's find some alternatives for substituting - I hope I spelled that correct - into this E _{S} thing.*2062

*OK, excuse me.*2086

*Now, we are going to get into the mathematical derivation, and again, please do not worry if you do not necessarily follow this derivation.*2090

*The only reason I am actually doing the derivation is so you have some sense of what is actually happening, that these things do not just fall out of the sky.*2097

*Again, it is good to see a derivation; it is good to listen, to think about what is happening just to, sort of, see it even if it is only passively.*2107

*At least your mind is starting to wrap itself around what is happening.*2116

*The ultimate form of the Michaelis-Menten equation is what we are going to be concerned with, but I think the derivation is just important to see.*2120

*We are going to introduce some thing; actually, let me go ahead and do this derivation in blue.*2129

*Let's introduce some notation here, introduce this thing called E _{T}.*2135

*This is the total enzyme concentration; OK, well, the total concentration of enzyme comes from 2 sources.*2145

*You are going to have free enzyme, and you are going to have the enzyme that is actually bound up as the enzyme substrate complex, right?*2157

*That is it; those are your sources that comes up to the total enzyme concentration.*2164

*Our total enzyme concentration is equal - oops, let me make this a little bit better - to free enzyme concentration plus the enzyme that is tied up as enzyme substrate.*2172

*These things we can measure; this we can measure.*2189

*That is easy; we know how much enzyme we are actually using.*2193

*This is very, very easy; notice, we have already started to express E _{S} in terms of things that we can measure.*2198

*OK, now, rate formation, now the rate of E _{S} formation...OK, actually let me do these a little bit separately.*2205

*Now, we are going to write 2 rate equations.*2229

*And again, a rate equation is some rate constant times the concentration of the reactants.*2233

*Now, the rate of enzyme substrate formation is equal to enzyme substrate formation comes from here.*2239

*It is this step right here, this step.*2254

*It is equal to k _{1} times the concentration of free enzyme times the concentration of S- that is it.*2258

*Again, the rate of a given reaction is the rate constant in the direction of that reaction times the concentration of the reactants.*2267

*In this case, you have that reactant and that reactant, so it is this, this.*2276

*Well, it is equal to k _{1}.*2281

*Let us rearrange E; E is equal to the total enzyme concentration minus the enzyme that is tied up as substrate.*2288

*It is equal to E total - E _{S}, and that times S- that is it.*2296

*I just substituted; I have just moved this over here.*2309

*That is E; I have substituted to here, and I end up with this.*2312

*That is my rate of formation of enzyme substrate complex.*2316

*Now, my rate of enzyme substrate breakdown because that is what is happening here.*2321

*Enzyme substrate is forming; enzyme substrate is breaking down.*2331

*Breakdown of the enzyme substrate complex can happen in 2 directions.*2336

*It can happen in this direction to form enzyme and product; it can happen in this direction to form enzyme and substrate.*2340

*We account for both; in this direction, it is equal to k _{-1} because it is now this direction, so we use that rate constant, and the reactant is if our reaction is going in this direction, that is the reactant.*2345

*It is k _{2} times the enzyme substrate complex.*2364

*Now, the reaction in this direction, again, E _{S} is the reaction, but k_{2} - that is it, they are additive - k_{2} times enzyme substrate concentration.*2368

*I hope this makes sense; again, anytime you have some reaction, AA + BB going to, let's say, CC, and if there is some equilibrium, if the reaction that we are looking at in that direction, well, it is going to be the rate constant - k _{1} - times the concentration of this and times the concentration of that because we always deal in terms of the concentration of reactants.*2385

*If we are looking at the reaction in this direction, well, the rate constant, we call it k _{-1} - that is the symbol - but now, in this direction, this is the reactant.*2413

*Because that is the reactant, it is going to be k _{-1} times the concentration of that because now, the reaction is going in this direction.*2422

*We are just not writing it left to right; the forward reaction formation, that is this one.*2430

*The breakdown happens in 2 ways; it can break down in that direction.*2438

*It can break down in that direction; each of those is represented here.*2442

*OK, now is when we are going to apply the steady state assumption.*2447

*The steady state assumption says that the rate of breakdown of E _{S} is the same as the rate of formation of E_{S}.*2452

*I am going to set this equation equal to that equation.*2458

*That is the steady state assumption; here is where the math comes in.*2465

*We need to simplify the math; if I do not have the steady state assumption, I cannot do anything here.*2470

*Let's go ahead and do that one.*2475

*OK, now, I have got k _{1} times the total enzyme concentration minus enzyme substrate concentration times substrate concentration is equal to k_{-1} enzyme substrate concentration plus k_{2} enzyme substrate concentration.*2479

*Now, after some algebra, I will just write algebra…, and you can check the algebra for yourself if you want.*2515

*You are going to get something that looks like this; you are going to get the enzyme substrate.*2523

*We are going to solve for this E _{S} thing; it is equal to the total enzyme concentration times the substrate concentration all divided by k_{2} + k_{1} / k_{-1} + S.*2527

*I will go ahead and put this, something like that.*2552

*Now, this thing, this is defined - it is a definition right now, we will give a better definition in a minute - as the K _{m}.*2556

*We call it a capital K _{m}; this is the Michaelis-Menten constant.*2570

*OK, now, I am going to rewrite this.*2588

*Because we call it K _{m}, I am going to put K_{m} in there, so I have, now, E_{S} is equal to the total enzyme concentration times the concentration of S over K_{m} + S.*2592

*Now, I have an expression for this E _{S}; that is what I wanted.*2607

*Well, all of these things are actually pretty easily measured.*2612

*That is hard to measure, but E _{T} is easy to measure; that is easy to measure, and this one we will see in just a minute.*2616

*Now, we said - let me go back to blue here - the initial velocity is equal to k _{2} x enzyme substrate concentration.*2624

*Well, now, I am going to take this, which is this, this expression, and I am going to substitute into here.*2638

*Therefore, I have v _{0} is equal to k_{2} times total enzyme concentration times substrate concentration over this thing called K_{m} + S.*2644

*Notice, this K _{m} constant, it is just a combination of the 3 constants that are involved in the reaction.*2656

*You have reaction 1, reaction 2, reaction 3- that is it.*2663

*That is all you are doing; we are just calling it 1 constant because we do not want a bunch of Ks floating around.*2667

*Alright, now, let's take a look.*2674

*Now, V _{max}, the maximum velocity occurs when the enzyme is saturated, right?*2679

*When the enzyme is fully saturated, we have that curve, that line up at the top when the enzyme is saturated.*2694

*That is when the total enzyme concentration happens to equal the enzyme substrate concentration.*2710

*In other words, when the enzyme is saturated, that means it is all enzyme substrate complex.*2720

*Well, that means that is the same as all of the enzyme concentration that I started with.*2727

*So, E _{S} happens to equal E_{T}; the total enzyme concentration, all of the enzyme is now enzyme substrate, so I can set those equal to each other.*2734

*The maximum velocity equals k _{2} E_{T}, right?*2751

*We said that v _{0} is equal to k_{2} E_{S}.*2762

*Well, E _{S} is E_{T}, so V_{max}, now, because at maximum velocity, the enzyme substrate concentration equals E_{T}, so v_{0}, which happens to be the maximum velocity now - not just some initial rate - is equal to k_{2} E_{T}, right?*2766

*That just comes from this; this, at maximum velocity, E _{S} is equal to E_{T}.*2785

*I put E _{T} in here, and instead of calling it initial velocity because the velocity is now maximum when these 2 are equal, I just call it V_{max}.*2791

*Now, I can write it.*2801

*Now, v _{0} is equal to V_{max} x S/K_{m} + S.*2807

*This is what we wanted; OK, we had k _{2} E_{T} up at the top.*2823

*We call that the V _{max}, maximum velocity; this is the Michaelis-Menten equation.*2832

*This is what we were hoping for.*2837

*This equation is the equation for that thing, that line that we say, which we are actually going to see in a second again.*2843

*This equation expresses that; now, each of these - this is what is nice - quantities, in other words, the initial rate that we measure the velocity, the maximum velocity, the substrate concentration and this Michaelis-Menten, this thing, are all easily experimentally determined and/or measurable.*2853

*We took that expression that involved the enzyme substrate complex.*2897

*We fiddled with it because we did not want to deal with the concentration of enzyme substrate complex, and we expressed it in terms of something that we can measure.*2903

*We know the substrate concentration; that is what we are controlling.*2912

*These other things are all easily measurable- that is it.*2916

*This is the Michaelis-Menten equation; this equation represents, under the postulates, the steady state postulate and the Michaelis-Menten postulate of the second step being the rate-limiting step, this describes that curve that we saw.*2922

*This says that the initial rate is contingent on...it depends on a maximum velocity.*2937

*It depends on substrate concentration, and it also depends on something called the Michaelis-Menten constant, which we will talk about a little bit more in just a second.*2945

*Now, let's see what happens when the v _{0} is equal to 1/2 of V_{max}.*2954

*Again, special cases, we like to just, sort of, see what happens mathematically.*2970

*Well, when you set v _{0} to 1/2 V_{max}, you end up with V_{max}/2 - you just plug it into that equation - equals V_{max} x the concentration of S over K_{m} plus the concentration of S.*2974

*The V _{max}s go away; you can cross multiply.*2996

*You end up with K _{m} plus the concentration of S equals 2 times the concentration of S.*3000

*Go ahead and subtract, and you get K _{m} equals the concentration of S.*3008

*This is beautiful; this says that when we measure an initial rate, which is half as fast as the enzyme is ever going to get half the maximum velocity.*3014

*It is at that, when that is the case, the K _{m} is actually equal to the substrate concentration at that point.*3028

*Whatever substrate concentration I happen to pick, that is my Michaelis-Menten constant.*3040

*That is this thing; this is the definition we want.*3044

*That is beautiful; this K _{m} is equal to the substrate concentration, which gives us a velocity equal to 1/2 V_{max}.*3048

*Again, let's go ahead and take a look at our...alright.*3080

*Yes, alright, now, we can talk about V _{max}, 1/2 V_{max}, K_{m}.*3095

*We go ahead and we run our reaction data with different initial concentrations.*3101

*We end up getting this thing; we go ahead and we take a guess, an estimate, of this maximum velocity.*3105

*For this particular enzyme, whatever it happens to be, I go ahead and once I actually estimate a maximum velocity, I take half of that maximum velocity.*3111

*I come down; I draw a horizontal line.*3120

*I see where it touches the graph, and then, I go down, then, whatever this is, it is that substrate concentration, which will allow me to be at half my maximum velocity for this enzyme- that is it.*3123

*That is the definition of K _{m}; that is the best definition of K_{m}.*3139

*It is a nice experimental definition of K _{m}; again, K_{m} is the substrate concentration - moles per liter, millimoles per liter, micromoles per liter - at which the enzyme is at half maximum velocity.*3142

*OK, you run the experiment; you can find the maximum velocity from that, then, from maximum velocity, you can take half of it, and then you can find the K _{m}.*3157

*Now, you have got yourself a perfectly valid equation: v _{0} = V_{max} x the concentration of S/K_{m} + S concentration.*3167

*Now, you can use whatever you want.*3180

*Let's say you want to slow it down a little bit for your experiment, whatever you are doing, or let's say you want to speed it up faster than half the velocity max, let's say you want the velocity to be up here, you can control the substrate concentration; and you have this equation.*3185

*In other words, you found K _{m}, and you found V_{max}- that is it.*3200

*You have an equation for that particular enzyme; that is what we want.*3205

*We want some sort of a quantitative description; now, let's take a look at this equation.*3209

*When the substrate concentration is a lot less than K _{m} on the denominator, we can ignore the S.*3215

*OK, because it does not make that big of a contribution to the denominator, we can ignore S in the denominator.*3230

*The equation becomes v _{0} = V_{max}/K_{m} x the S concentration.*3250

*As we said before, this is a linear equation, thus, the almost linearity for small concentrations of S.*3266

*In other words, under small concentrations of S, the Michaelis-Menten equation reduces to a linear equation.*3276

*It supports the data; now, when the concentration of S is a lot greater than K _{m}, now, K_{m} can be ignored.*3282

*We ignore K _{m} in the denominator, and the equation becomes v_{0} = V_{max} x S/S.*3298

*These cancel implying that v _{0} = V_{max}, which is exactly what this graph says.*3320

*As your concentration rises really, really high, so much higher than what the K _{m} is, the K_{m} becomes negligible.*3326

*In this equation, the velocity becomes V _{max}- asymptotic behavior.*3332

*As you go in this direction, you actually approach your maximum velocity.*3340

*This equation confirms low-end behavior, high-end behavior, and it also matches everything else in between.*3346

*Now, the question is this: “Do enzymes actually behave this way?".*3355

*The answer is yes; enzymes actually behave this way even for situations that are not necessarily postulated by Michaelis and Menten, where you have not this simple 2-step process, where the 2nd step - the breakdown of the enzyme substrate complex - is the slow step.*3359

*It is very, very interesting; this is characteristic enzyme behavior- steady state kinetics.*3377

*It is called Michaelis-Menten kinetics; the names in and of themselves do not really matter all that much.*3385

*What matters is the actual behavior.*3390

*Thank you for joining us here at Educator.com; we will see you next time for a further discussion of enzyme kinetics, bye-bye.*3394

2 answers

Last reply by: Gabriel Au

Fri Oct 10, 2014 7:59 PM

Post by Gabriel Au on October 9, 2014

Hey professor Hovasapian,

I have been watching many of your videos for many classes and I think you do a great job! I was wondering If you will ever do a series of videos specifically for complex variables?

1 answer

Last reply by: Professor Hovasapian

Wed Dec 4, 2013 3:40 AM

Post by Assaf Tolkowsky on December 4, 2013

Just a pointer - you have a mistake with the algebra at 42:46 - Km is (k-1+k2)/k1 and not (k1+k2)/k-1 like you wrote down

0 answers

Post by tiffany yang on November 13, 2013

Dear professor,

Can you please explain why when [E]total increases, why Km stays the same?

and what does Kcat depends on?

is catalystic efficiency independent from enzyme concentration? what does it depend on?

Thank you so much. I truly truly appreciate it.

1 answer

Last reply by: Professor Hovasapian

Sat May 4, 2013 3:37 AM

Post by Brian Phung on May 1, 2013

Hi I was wondering how do I determine the inhibition mechanism meaning if its competitive or noncompetitive inhibition by the shapes of plots of (kcat)ATP versus [I] and (kcat/Km)ATP versus [I] were used to determine the inhibition mechanism?