Biochemistry Enzymes IV: Lineweaver-Burk Plots

Hello and welcome back to Educator.com; welcome back to Biochemistry.0000

Today, we are going to continue our discussion using this thing called the Michaelis-Menten equation that we introduced in the last lesson.0004

We introduced this idea of the velocity max and this thing called the K_m, the Michaelis-Menten constant- very, very important.0012

In this lesson, what I am going to do is give you an alternate version of that, where instead of estimating V_max and estimating the K_m like we did in the previous lesson, we are going to develop an actual analytical method, where we actually come up with some really, really exact values for this.0022

Let's just jump right on in; it is actually quite simple.0042

OK, let's go ahead and start off with our Michaelis-Menten equation again.0046

That is fine; I guess I can stick with black.0052

It is not a problem; we said that rate of the reaction, the speed at which it is going, is going to equal some maximum velocity times this substrate concentration over this Michaelis-Menten constant plus the substrate concentration.0054

This is our initial equation; well, let's do something to this equation.0070

Let's actually reciprocate their left side and the right side.0074

Let's just flip it and then, manipulate it algebraically to see what we might get.0078

When you do that - excuse me - you end up with the following.0083

You end up with 1/v₀ = Km + S/V_max - excuse me - x S.0088

Now, we have everything under 1 denominator; there is only a single term of the denominator.0100

Let's separate those out and see what happens; you end up with 1/v₀ = Km/V_max x S + S/V_max x S.0104

These Ss cancel; let me rearrange this.0132

This term right here is the same as Km/V_max x 1/S.0136

I end up with 1/v₀ = Km/V_max x - let me put that in parentheses - 1/S + 1/V_max.0142

This equation, it is called the double reciprocal equation.0160

Double reciprocal because I took the reciprocal of the left and the right, also called the Lineweaver-Burk equation.0165

That is what is important; notice the form of this.0180

This is Y = mx + B, and this is the equation for line.0184

Y is 1/v₀; M the slope, is Km/V_max.0191

1/s is the independent variable, and 1/V_max, that is the Y intercept.0199

Here is what you do; when you are given rate-concentration data like the previous lesson.0207

What you do is you create 2 new columns; you take the reciprocal of the substrate concentrations that you used.0214

That is going to be another column; those are going to be your X axis, and you take the reciprocal of the velocity values that you have got, and that is going to be your other column.0220

That is going to be your Y axis; you are going to plot this out.0232

When you plot this out, here is what you are going to get.0237

On this axis, we have our 1/S.0248

On this axis, we have 1/v₀.0252

When you plot this out, the 1/v₀ versus the 1/S, you are going to end up with a line, some straight line like that.0256

Here is what is great; the Y intercept, well, that is equal to...let me actually rewrite the equation here again, so we see it.0267

We had 1/v₀ = Km/V_max x 1/S + 1/V_max.0278

Well, this Y intercept because it is 1/v₀, it is not v₀ versus S anymore.0290

It is 1/v₀ versus 1/S.0296

You get a straight line; where it hits the Y axis, that is equal to 1/V_max.0300

You literally read this number; you said it equalled to V_max.0309

You switched the 2; you solved for V_max.0313

It is an analytical way of finding V_max.0314

Where it hits the X axis here, this is equal to -1/Km.0318

The slope of this equation is equal to Km/V_max.0329

That is the slope; that is the Y intercept.0338

We have an analytical method based on rate-concentration data.0341

Create 2 new columns of data: 1 over concentration, 1 over velocity.0346

Graph that; take the 2 intercepts.0350

Solve for V_max, and solve for Km; now, we have a precise way.0354

We do not have to, anymore, estimate where the V_max is going to be and then, take half of that and then, find the Km.0359

Now, we have a nice analytical procedure for finding V_max and Km- that is it.0368

That is all that is happening here; again, when given data that includes substrate concentration and various initial rates, in other words, when you are given substrate initial rate data, when you are given data that includes substrate concentrations and various initial rates, we form new data, and the new data we form is 1 over those substrate concentration and 1 over the initial rates and then, plot this.0374

From the plot, we calculate V_max and K_m- pretty standard when dealing with a new enzyme.0455

You have purified a new enzyme; one of the first things you do is you run these initial rate experiments, and you find its K_m, and you find its maximum velocity under a certain set of conditions and for that particular substrate- that is it.0460

That is it; it is literally that simple.0478

Let's go ahead and do an example; let's see what we have got.0481

Carboxypeptidase is a digestive enzyme of the pancreas, and its activity was monitored under several initial substrate concentrations.0488

The data is as follows.0495

In this particular case, the substrate concentration happens to be in millimoles per liter, 0.1, 2, 10, 20, 40, 60, 120, 1000, 2000.0500

The initial velocities that were measured are here, millimoles per liter per minute.0513

And again, it is a rate; it is the rate at which the concentration is changing initial rate- 0.2, 5, 4.8, 19, all of these numbers here.0518

Well, given concentration rate data, we create 2 new columns.0528

We take 1/S, 1 over these; that is this column here, and then, we take 1/V, 1 over all of these.0532

That is this column here; notice what is different, though.0540

Excuse me; 1 over these numbers, these numbers are going to start to get very, very small.0544

What I have done is instead of writing them as, let's say for example 10.5, 0.05, 0.025, I have went ahead, and I have multiplied them by 10^-3.0548

Notice here, that is why I have this x 10^-3; this is actually 100 x 10^-3, 25 x 10^-3.0562

I did this in order to make my graph a little easier to deal with.0570

I do not want to graph 0.1, 0.5, 0.025, 0.010, 0.005.0575

I am just scaling the graph; that is all I am doing.0582

When you read the information off the graphs, be very, very careful.0586

Make sure you actually include that number, and you will see what we deal with in just a minute.0590

Again, for the 1/S, and this, I have gone ahead and expressed these as 10^-3.0595

For example, this value is actually 50 x 10^-3.0600

This one is 12.8 x 10^-3 or 0.0128.0604

12.8 is easier to graph than 0.0128.0610

That is the only reason we are doing this to make it convenient for ourselves.0614

What we are going to be doing is we are going to be graphing this now on the X axis; this on the Y axis, we are going to draw our best line, and we are going to read off the Y intercept and the X intercept.0618

We are going to solve for V_max and K_m- very, very nice, the beautiful, beautiful thing.0627

OK, let's see what we have got here.0634

OK, let's go ahead and draw our graph; let me go ahead and actually do this in black.0638

What you are going to end up with is something that looks like this.0644

I am not going to make it 2; OK, we have got 10, 20 and 30.0648

OK, we have got 10, 20, 30, 40 and 50.0660

When we go ahead and plot all of this, we are going to end up with some line that looks like this.0668

Let me go ahead and mark this point, and let me mark this point.0674

I am going to have something like here and then, a 10 maybe here and then, a 20 like there and there and then there.0681

We are going to end up with some lines; what we do is we end up doing it best fit.0697

You do not connect the dots; you are doing the best fit line.0700

It is not going to be exactly a straight line, but it is actually going to be pretty close surprisingly.0704

You are going to end up with something that looks like...I know it does not look like a straight line, but well, you know what, let me try to make this a little bit better.0708

You are going to end up with something that is like...let me not put a point there.0724

Let me not put a point there just yet; let me go ahead and do some values like there and there.0730

What you are going to end up with is the following; erase these...maybe something that looks like that.0739

You graphed it; you have a best fit line, and then, you have that point; and you have that point.0752

Now, let's go ahead and just read them off.0758

This is 10; this is 20.0762

This is 30; this is 10, 20, 30, 40 and 50.0765

Recall, this is 1/S, and recall, this is x 10^-3.0771

This is the 1/v₀ axis, and recall, this is x 10^-3.0776

So, when I read this one off right here...I will do this one down here, and I will actually do it in blue.0786

The Y intercept, that was the V_max.0796

We have 1/V_max is equal to...and when you read it off the graph, it ends up being 6.1 x 10^-3.0800

You just read it right off the graph: 6.1 x 10^-3.0803

When you solve this, you get a V_max is equal to, of course, 1/6.1 x 10^-3.0818

It is going to equal 164 mmol - I will do the molarity that way - per minute- that is it.0827

This is our V_max, an analytical way of coming, 164 mmol/min.0842

That is the fastest that this enzyme is going to go under these conditions.0850

Now, we will go ahead and do this one; this is -1/K_m, and remember, again, it is x 10^-3.0854

When you read this one, you are going to get...when you read it right off the graph, it is going to be -15.5 x 10^-3.0864

When you solve this, you end up with the following; you end up with a K_m equal to 64.5mmol/L- that is it.0875

The K_m has units of concentration because remember what we said: K_m is the concentration of substrate at which the speed of enzyme turnover is 1/2 the maximum velocity.0891

So, when the substrate concentration is 64.5mmol/L under these conditions, the speed of the reaction, that is going to be somewhere in the neighborhood of 82mm/L/min- that is it.0906

This is an analytical way of finding the K_m and the V_max- that is it.0925

That is what this Lineweaver-Burk plots do; you can use the rate concentration data if you want.0932

Estimate V_max, read it off the graph, or you can do something a little bit more analytical and make a Lineweaver-Burk plot.0937

These Lineweaver-Burk plots are actually going to play a very, very important role when we talk about enzyme inhibition in the next lesson- very, very important analytical tool.0945

One caveat, as we said, we have a purified enzyme; one of the first things we do is find the K_m and find the V_max for the particular enzyme.0957

Now, it is very, very important - I will just leave you with this - to remember that K_m and V_max give very little information about the actual individual steps in a reaction, in other words, what is happening at the molecular level.0965

I will say not in a reaction, and we know this, but in an enzyme catalyzed reaction, but it is a place to start.1024

And again, sometimes, all you need is a place to start.1041

When we talk about kinetics, the Michaelis-Menten kinetics, steady state kinetics where all of this is coming from, this K_m and this V_max, understand that these are kinetic data.1045

These are kinetic parameters; what that means is they talk.1054

They relate to rate; that is what kinetics is.1058

It is how fast something is going, or in the case of K_m, how much substrate will allow me half the maximum velocity.1059

Again, velocity still plays a role in this.1069

This does not really tell me anything much more than that.1073

It is more historical for all practical purposes, I mean, still one of the things that you are going to do.1076

I am not saying that it does not give you information; of course, it gives you information.1081

And again, it is a place to start, but ultimately what we are concerned with is mechanism.1085

Mechanism is the individual steps- what is happening at the atomic/molecular level between an enzyme and a substrate that converts a substrate to a product.1090

The K_m and the V_max do not give you much information; they are just a place to start.1101

Do not think that you are missing something, that if you have a certain K_m or if you have a certain V_max, that is necessarily going to tell you something about the mechanism.1106

It is not; there are lots of experimental tools and lots of other things that need to be done that allow us to find out what is happening individually in a given active site of a given enzyme with a given amino acid sequence.1114

There are other things that we do; this information does not give us information about those steps, and ultimately, it is those steps that we want to consider.1131

But again, this is part of biochemistry; it is part of the normal routine that people go through on a daily basis in laboratories.1141

You are going to have to know something about it.1148

That is all, but do not feel that you are missing anything; do not try to extrapolate any other information other than what is given.1151

This is all that it is, no more and no less, and it is all based on those 2 assumptions- the steady state assumption, that the breakdown of the enzyme substrate complex and the formation of the enzyme substrate complex happen at the same rate; and, of course, ultimately, this is based on Michaelis and Menten simplifying assumption that this thing happens in 2 steps, and the second step, the breakdown of the enzyme substrate is the slowest step.1159

We need to know where these things come from.1186

We want to see the forest from the trees; I cannot reiterate this enough.1190

The K_m and the V_max are just kinetic parameters; they do not tell you anything about how the substrate turns into product in the active site.1195

Other things are used to extract that information.1205

We will be looking at mechanisms, of course, individually in a couple of lessons, and, of course, when we talk about metabolic pathways, we are definitely going to be talking about enzyme mechanisms - which proton is being transferred, which amino acid is involved in the catalytic mechanism, things like that - but these do not give you that information directly.1210

OK, thank you for joining us here at Educator.com.1230

We will see you next time for discussion of enzyme inhibition; take care, bye-bye.1234