Show them in graphs please - Enzyme Kinetics!

Last time, we learned about enzymes and their functions.

Now, we are here to talk about how to use some mathematics to understand more about their interactions! Shall we start?

                                          

Enzyme kinetics is the study of chemical reactions catalyzed by enzymes.

It concentrates a lot on the speed - Or I should say "Rate".

By doing this, we can study how we can alter the rate of reaction or it's functionality.

In enzyme kinetics, we have some terms to learn about before we do anything else. The first one is this: E + S ↔ ES → E + P

Where E responds to the enzyme, S as the substrate, ES as the ES complex, and P as the product. The enzyme - substrate complex can rewind back to an enzyme and its substrate. The Enzyme and it's product can do the same too.

However, because products are usually very stable, they hardly rewind back.

Assuming that you know kinetics, let's call the forward reaction of E + S → ES "Reaction 1". And "Reaction 2" for the forward reaction of ES → E + P.

Then, the rate of our reactions will be like this:

Rate1 = K1[E][S] and Rate2 = K2[ES]

Where Rate1 is the rate of reaction 1 and vice versa.

Same for K1 and K2. Got it? Ok.

Next is something called "Vmax". And... What is it?

Say that we have an enzyme that can catalyze 10 reactions per second.

If we had four of them, what would be the maximum rate?

40 reactions per second, right? That's what we call "Vmax". 

The max rate that the enzyme can catalyze per second. That's Vmax.

By the way, Vmax will always occur in every situation. Keep that in mind please.

                                          

Next is a diagram where you plot the reaction rate(V) and the substrate concentration. 

We start with no reaction rate and no substrate.

Then as we add more substrate, our rate goes up, and it becomes more flatter in the end, approaching a certain value. The point where it flattens out is when the enzyme is saturated with substrate.

Which means it's reached it's Vmax.

Now, if you look at the y - axis, you can see 1/2 of the Vmax. 

It's corresponding X axis is also shown as well too.

This X axis - Or substrate concentration is called Km.

Km, is also called the Michaelis Constant. We will talk about why it's called like that later. So this Km, is the substrate concentration when the rate is 1/2 Vmax. When people talk about this, some say that Km is 1/2 Vmax.

That is Wrong! I mean like, Really WRONG!

Km is a Substrate Concentration. Remember this and you'll be fine! 

                                          

Next is the Michaelis Menten Equation!

E + S ↔ ES → E + P

Since we have reaction 1 as the forward for E + S ↔ ES, let's get ourselves a reverse reaction and call it "Reaction -1". We can make a "Reaction -2" for the same reason, but as I said - "Products hardly reverse back". Let's sign that out.

Now, we are going to have something called the Steady - State Assumption.

At steady state, [ES] is constant. Meaning that the rate of formation of [ES] is equal to the rate of dissociation of [ES].

K1[E][S] = K-1[ES] + K2[ES]

Let's group the right side with [ES].

K1[E][S] = (K-1+ K2)[ES]

Now let's move the rate constants on the right and the others on the left!

We will get [E][S]/[ES] = (K-1 + K2)/K1

Turns out, is that this Km is actually (K-1 + K2)/K1!

Then we will get Km =[E][S]/[ES]

Let's move the [ES] from the right to the left side.

[ES]Km = [E][S]

Now let's define a new term called [E]T or "total enzyme" as [E]T = [E] + [ES]

Then we can substitute the [E] from the right to [E]T - [ES]. 

That gives us [ES]Km = ([E]T - [ES])[S].

If we distribute the right side, it's: [ES]Km = [E]T[S] - [ES][S].

By moving the [ES][S] to the left side and grouping the [ES] together, we get

[ES](Km + [S]) = [E]T[S].

If we leave [ES] on the left side only, then we get [ES] = [E]T[S]/(Km + [S]).

Our rate is equal to the rate of formation of product, Or K2[ES].

So Vo = K2[ES]. At high [S], The enzyme will be saturated with substrate. So the Vo = Vmax. And since all the substrate are bonded with the enzyme, there are no free enzymes in the solution. Meaning that [E]T = [ES] as well.

So it also means that Vmax = K2[E]T as well.

Let's multiply K2 on both sides of [ES] = [E]T[S]/(Km + [S]).

Then we get K2[ES] = K2[E]T[S]/(Km + [S]).

By using our terms above, we get Vo = Vmax[S]/(Km + [S]).

This, is the Michaelis Menten Equation!

                                          

In our usual graph, which is in a hyperbolic curve, is not that quite accurate.

For example, if you got the Vmax wrong because of where the curve was headed to, then you would get the Km wrong and everything!

This is where the Lineweaver - Burk plot comes in.

It's derivation is quite simple. 

First, get our Michaelis Menten equation Vo = Vmax[S]/(Km + [S]).

Second, take the inverse of it, making 1/Vo = (Km + [S])/Vmax[S].

Third, rewrite the equation to 1/Vo = Km/Vmax[S] + [S]/Vmax[S].

Simplifying the equation gives us 1/Vo = Km/Vmax * 1/[S] + 1/Vmax.

And that... Is the same way of saying y = mx + b!

Which gives us a straight line on the graph shown below.

With this, we can get a more accurate value of Km and Vmax!

The X - Intercept is -1/Km, and the Y - Intercept is 1/Vmax.

And there are the other values shown in the picture above.

                                         

In enzyme kinetics, we also learn about Inhibitors as well.

Inhibitors are molecules that decreases the enzyme's activity.

The first type of inhibitor is called a Competitive Inhibitor.

Competitive inhibitors - Like the name says, "Competes" with the substrate for the enzyme. And whoever comes first gets the enzyme.

Normally it's for the active site, but it can also be the Allosteric site as well too.

If we write this out, it will be: E + I ↔ EI, E + S ↔ ES ↔ E + P.

At steady state, our Km will still be [E][S]/[ES]. 

However, now we will have a new term called KI. Which is [E][I]/[EI].

Now let's remove some terms that aren't necessary.

Let's divide Km by KI. Then we get Km/KI = [S][EI]/[ES][I].

Now it looks like we have nothing else to simplify... Or is it?

Let's look at our toolbox of equations! 

Since[EI] is also a part of [E]T, [E]T = [E] + [ES] + [EI].

If we substitute [EI], then Km/KI = ([E]T - [E] - [ES]) [S]/[ES][I].

Simplifying it gives us Km/KI = [E]T[S]/[ES][I] - [E][S]/[ES][I] - [ES][S]/[ES][I]. 

Km = [E][S]/[ES], so [E]/[ES] = Km/[S].

And Vmax = K2[E]T, Vo = K2[ES]. Therefore, Vmax/Vo = [E]T/[ES].

Applying it gives us Km/KI = Vmax/Vo *[S]/[I] - Km/[S] * [S]/[I] - [S]/[I].

Grouping the [S]/[I] and moving it to the left side equals...

Km/KI * [I]/[S] = (Vmax/Vo - Km/[S] - 1).

After this, let's put the Vmax/Vo in one side and everything else in the other.

Vmax/Vo = Km/KI *[I]/[S] + Km/[S] + 1.

By grouping, we get Vmax/Vo = Km(1/KI * [I]/[S] + 1/[S]) + 1.

And then Vmax/Vo = Km([I]/KI + 1) 1/[S] + 1.

Dividing both sides by Vmax is: 1/Vo = Km([I]/KI + 1)/Vmax * 1/[S] + 1/Vmax. 

Lastly, we substitute [I]/KI + 1 into a constant called α.

So it becomes 1/Vo = Km*α/Vmax *1/[S] + 1/Vmax.

This is the Lineweaver burk equation of a competitive inhibitor.

If we graph this, we get the fallowing picture below:

The 1/Vmax stays the same. So there is no difference to the Y - Intercept.

For the X - Intercept however, is different.

If we solve what the X intercept is for the Competitive inhibitor, we get -1/Kmα.

And since α is larger then 1, our X intercept is going to be larger then the original X intercept.

In short, Vmax does not change. However, the apparent Km increases.

How does that happen? The Substrate and the Inhibitor are competing for the enzyme. So if we have a LOT of substrate then the inhibitor, we will be able to reach the original Vmax by beating the inhibitor on the way to the enzyme.

So there is no change to the apparent Vmax.

In order to reach the same Vmax, more substrate is needed to reach the same 1/2Vmax. So the enzyme has a low substrate affinity. Therefore the apparent Km increases.

The reason why I used the word "Apparent" is because the Vmax and the Km itself doesn't change. The inhibitors are making it look like the Vmax and Km changed when it's actually decreasing it's activity.

Vmax and Km are Constants. And Constants stay Constant.

                                          

Next are Uncompetitive Inhibitors.

These guys don't "Compete" with the substrate.

Instead, they bind to the Allosteric site when the Enzyme is associated with the substrate. Making an ESI complex. 

Writing it out will show this:

E + S ↔ ES ↔ E + P, ES + I ↔ ESI.

I won't do the LBP derivation all over again. Instead, I'll show it to you directly -

1/Vo = Km/Vmax *1/[S] + α`/Vmax.

Whereas α` = 1 + [I]/K`I. What is this α` and K`I?

K`I is the KI(inhibitor constant) when the inhibitor is uncompetitive,

And α` is α with [I]/K`I in it instead of [I]/KI.

They're just here to distribute which inhibitor they are in.

If we graph it, it will look like this:

It's Y intercept is α`/Vmax. And since α` is also larger then 1, The Y intercept is higher then the original one.

For the X intercept, we get -α`/Km. That makes the X intercept more lower then the original one.

So it means that the apparent Vmax and Km both decreases in an uncompetitive inhibitor. How does that happen?

The uncompetitive inhibitor binds to an ES complex only. So no matter how many substrate you add, the inhibitor is going to stick anyways. So the apparent Vmax decreases.

And as the ESI complexes gets formed, ES is decreasing over time.

So the system tries to balance the equilibrium by accepting more substrate.

This leads to a higher affinity - and thus, low Km.

                                          

The third are mixed inhibitors.

These guys are basically a "mix" of the two inhibitors above.

So if we write them out, it's like this:

E + S ↔ ES ↔ E + P, E + I ↔ EI, ES + I ↔ ESI.

It's Vmax will be decreased due to the effects of the ESI.

However for the Km, it can increase, decrease, or stay the same as before. Because it's a mixture of the competitive inhibitor's behavior(increase) and the uncompetitive inhibitor's behavior(decrease).

It's lineweaver burk equation is 1/Vo = Km*α/Vmax * 1/[S] + α`/Vmax.

In a graph, here's what it looks like:

The X intercept is -α`/Km*α. Which can be smaller or bigger then the original X intercept. It depends which is bigger - α or α`.

If α is bigger, then the X intercept is bigger then the original.

If α` is bigger though, then the X intercept is smaller then the original.

In this situation, α is bigger then α`.

For the Y intercept however, is always α`/Vmax. So the Y intercept is always bigger then the original.

                                          

The last are Noncompetitive inhibitors.

They are a type of mixed inhibitors. However we call it a noncompetitive inhibitor when α = α`!

It's graph looks like this:

The X intercept is the same as the original one since α = α`.

The Y intercept, of course is α`/Vmax. Which is bigger then the original.

                                          

And that is it! The end of Enzyme Kinetics!

I might post some more stuff about things like cooperativity.

Until then, stay tooned!

                                  2016. 5/29

                                  <해인이가>

Image credits section:

Khan academy MCAT videos.

Wikipedia: Main page, Enzyme kinetics section.

Wikibooks: Main page.

EDX: Principles of Biochemistry.