Enzyme Mechanics

The post I wrote on hemoglobin makes me feel as though I understand it on some quantum level.

Like, you have no idea how long that took. I texted Finn shortly afterward and told him, “I aM HEMOGLOBIN.”

[ahem] But today, I’m going to write the post I meant to write in the first place.

That’s right! We’re going to talk about enzyme mechanics! Aren’t you excited?

I sure am!


Enzymes are, as emphasized before, absolutely essential to metabolism. However, what I didn’t mention before was that the enzymes that catalyze metabolic processes come in two varieties. The majority of reactions are catalyzed by a kind of enzymes called Michaelis-Menten (MM) enzymes. My professor calls them the workhorses of metabolism, and they’re so efficient that they usually aren’t the slowest part of a metabolic pathway. A significantly smaller number of metabolic reactions are catalyzed by allosteric enzymes, which control metabolism extremely well by reacting to allosteric inhibitors and allosteric activators. If that sounds like Icelandic to you, congrats! You’re about where I was during this lecture.

Before we get into distinguishing the two, let’s talk about some general characteristics. First of all, as probably makes intuitive sense to you, the reaction velocity of any given reaction is dependent on the concentration of the substrate (reactants). If you’ve had a touch of Gen Chem II, this will make more sense as an equation:

v = kf[S]x

There’s a lot of information in this equation, so let’s break it down. First, you can see that the rate of the reaction is dependent on two factors: the rate constant, kf, and the concentration of substrate, [S]. The rate constant is a function of the conditions of the experiment (pressure, temp, etc.), which tells us that, under a constant set of conditions, the velocity of a reaction is dependent on the concentration of the substrate. You’ll also notice that [S] has an exponent attached to it; that means that there can be different orders to the dependence of the reaction on [S]. If x = 1, velocity increases linearly with substrate concentration (2x [S]? 2x v). If x = 2, velocity increases quadratically (2x [S]? 4x v). There’s also the case of x = 0, wherein velocity is independent of substrate concentration. In all cases, the order of the reaction must match the mechanism that the substrate goes through to become a product.

Now we can bring in the enzymes. There’s a general mechanism that all enzymes follow when producing their product. First, an enzyme, E, binds its substrate, S, forming an enzyme-substrate (ES) complex. This can easily be reversed. Second, the enzyme does chemistry on its bound substrate (this is really a number of steps), producing a product still bound to it as an EP complex. Finally, it releases the product, liberating product and regenerating the unbound enzyme. (Since enzymes work in both directions, you can reverse this, too!) This can be written as a nifty equilibrium:

E + S   ES EP E + P

An important thing to note about this equation is the slowest step is the equilibrium between ES and EP, since this is the chemistry step. The rate constant of this step is kcat, which is the rate constant of the slowest step of the chemistry that converts substrate to product. Said another way, the rate of the whole process can only be as fast as the rate of the slowest step in converting substrate to product. Provided that we assume that we’re putting enzyme into a solution with only substrate (no product), we can write this as another equation:

v = kcat[E][S]

where [E] is the concentration of enzyme and [S] is the concentration of substrate.

This is similar to the rate equation above, except now, you’ll notice, the rate of the conversion of S to P by an enzyme is dependent on three factors: the rate constant of the slowest chemistry step, the concentration of substrate, and the concentration of enzyme.

Overall, this is a second order reaction. If you double enzyme concentration or substrate concentration, you’ll double the velocity. If you double enzyme concentration and substrate concentration, you’ll quadruple velocity.

Again, that kind of makes intuitive sense, doesn’t it? Your enzyme-catalyzed reaction should get faster when you have lots of stuff to make into products and when you have lots of enzyme to make the product. Easy peasy lemon-squeezy.

Ahaha, now we need to amp it up a notch. Let’s look at the specific kinetics of Michaelis-Menten enzymes.

Michaelis-Menten enzymes follow their own kind of kinetics that can be described using rate constants. Remember the equation I gave you up there about the mechanism of action of an enzyme? Let’s talk about a version of it where we’re primarily concerned with the enzyme-substrate complex, ES:

E + S ES —> —> E + P

Here, we’re concerned with two primary things: the rate of formation of the ES complex, and the rate of loss of the ES complex. The rate of the formation of the ES complex from E and S we designate k1, and the rate of dissociation of the ES complex into E and S we designate k2. The rate of conversion of ES to EP is the rate of the slowest step in the chemistry that converts substrate to product, which we previously designated kcat.

Now, let’s assume that we wanted to keep our concentration of ES constant. That means that the rate of formation of ES (described by k1) would have to be equal to the rate of its loss (described by k2 + kcat). When we divide k2 + kcat by k1, we get what is called the Michaelis constant, Km. This constant tells us how well the enzyme binds its substrate. (We’ll get to that in a bit, but for now, trust that, the bigger the Michaelis constant, the better the enzyme lets go of substrate.)

We can also get a maximum velocity, Vmax, by multiplying the rate of the chemical step (kcat) by the amount of enzyme present. That makes sense, if you think about it, since, like we said before, the maximum rate of production of a product with an enzyme should be limited by the amount of enzyme and the speed at which it can do chemistry.

Putting all of this together, we can relate the velocity of a MM enzyme to its maximum velocity and its ability to bind substrate like this:

v = Vmax[S]/ Km + [S]

When graphed, this gives you a hyperbolic curve. If you remember our study of myoglobin, you’ll understand why I felt the need to get a grip on that before I really honed in on this: it follows the same basic pattern. When substrate concentrations are high, the velocity of a MM enzyme barely changes with increases in [S]—v approaches Vmax. When substrate concentrations are low, velocity increases linearly with increased substrate concentration—v is essentially Vmax[S]/Km. Finally, when the concentration of substrate equals the Km, the velocity of the reaction is half of the maximum velocity.

All right. Now that we’ve done all that, let’s backpedal for a sec and talk about what the constants in the velocity equation actually mean. First, we can talk about kcat, which is pretty important, turns out. You see, under saturating conditions (when all enzyme is bound to substrate), the kcat tells us the efficiency of the enzyme independent of the enzyme concentration. Practically speaking, kcat tells you the number of molecules of product the enzyme produces in a unit of time. If you divide 1 by kcat, you get the lifetime of product formation, or the amount of time it takes to form one product molecule. So, for example, if your kcat was one million, you could make a million product molecules per second, or make one product molecule in a millionth of a second.

Yeah, that’s pretty darn fast.

The other constant, Km, tells you, as I said above, how well the enzyme binds its substrate. It makes sense if you break it down: with a small kcat relative to k1, and k2, Km essentially becomes the ratio of the rate constant of the dissociation of ES to make E and S with the rate constant of association of E and S to make ES. If Km is small, it means that the association of E and S is faster than the dissociation of ES, and the enzyme binds substrate tightly. The opposite is also true. That means that, if you compare Km for two substrates of an enzyme, the one with the smaller Km is the one that the enzyme binds better.

Finally, we can put all of this together and talk about something called catalytic efficiency. As it turns out, the faster an enzyme does the chemistry that converts substrate to product (high kcat) and the better it binds to its substrate (low Km), the more efficient it is. Put in an equation, catalytic efficiency is defined as follows:

CE = kcat/Km

There is a limit to how big catalytic efficiency can be, as you might expect. That number is 108/M(s), which is the diffusion limit of molecules in a buffer at standard temperatures. However, because enzymes are awesome (and fast as heck), there are a handful that get pretty close to this limit. In fact, my professor said that there are even enzymes that exceed it. Tell me this stuff ain’t cool. I dare you.

Okay, so enzymes are fast. That’s the whole point, right? (Well, unless you’re looking for accuracy, like if you happen to be a DNA polymerase…) But what if you want to slow them down? Can you even slow them down?

Of course! That’s where inhibitors come in, dearest reader! There are three kinds of inhibitors for MM enzymes: competitive, uncompetitive, and noncompetitive.

Competitive inhibitors, as you’d expect from their name, compete with substrate to bind with the enzyme. This introduces another equilibrium into the equation I gave you above:

E + I EI

Because Km is a function of the rate of formation and disassociation of ES, I binding to E increases the apparent Km of the enzyme (decreases its apparent ability to bind substrate). However, it doesn’t affect Vmax, since, if you put enough substrate in with the enzyme, it will eventually compete off the inhibitor. This means that competitive inhibitors lower the velocity of product formation best when the concentration of substrate is near Km.

In contrast, uncompetitive inhibitors bind to the ES complex following an equilibrium like this:

ES + I ESI

This decreases both the Km and the Vmax. It decreases the Km because it reduces the amount of ES present in the reaction, and it decreases the Vmax because no amount of substrate will be able to out-compete the inhibitor. What this means practically is that the most noticeable reduction in enzyme velocity with uncompetitive inhibitors shows up at high substrate concentrations.

Finally, noncompetitive inhibitors work sort of as a fusion of the two previous inhibitors; they bind both with unbound enzyme and the ES complex. This doesn’t change the Km, since the inhibitor is affecting both E and ES equally. However, it does decrease Vmax, because substrate concentration can’t compete off the inhibitor. Noncompetitive inhibitors, as you might expect, work like a mix of competitive and uncompetitive inhibitors, reducing reaction velocity equally well at all substrate concentrations.

Phew! All right, let’s take a breather. I’m going to go eat a bit of a coffee protein bar. We’re only halfway there, remember?

(What the frick-frack diddly dack patty wack snick snack…)

All right! Now that we’ve talked at length about Michaelis-Menten enzymes, let’s shake things up and look at a smaller but more finely tuned category of enzymes called allosteric enzymes. Allosteric enzymes, like hemoglobin, have subunits in a tense (T) state at low substrate concentration. However, as substrate concentration increases, so does catalytic efficiency, until the catalytic efficiency is pretty high at high substrate concentration. Halfway between T- and R-states, the velocity of the reaction is half its maximum.

As you probably already had deduced, when this is graphed, it produces a sigmoidal curve. If you read my post on hemoglobin (or if you know a lot about hemoglobin), you know that the sigmoidal curve of hemoglobin affords it some finesse in its binding of oxygen that myoglobin can’t hack. Does it work the same way here, with MM and allosteric enzymes?

Yeah, actually. Things are actually easy, for once.

You see, there are a whole host of allosteric activators and allosteric inhibitors that just love to mess with these enzymes, affecting their rates at different substrate concentrations. An allosteric activator will bind to a site other than the active site of the enzyme and stabilize the R-state of the enzyme, making it faster at every substrate concentration than an unmodified enzyme. The halfway point in velocity moves to the left, where it occurs at a lower substrate concentration.

In the same way, an allosteric inhibitor will bind to an allosteric enzyme and do two things—compete off the substrate, and stabilize the T-state. When it does this, it decreases the velocity of the reaction at every substrate concentration. The halfway point to maximum velocity moves to the right, where it occurs at a higher substrate concentration.

Okay, that’s nice, but why the heck is it important? I just said these enzymes are relatively rare, right?

Yes, but they’re extremely important in the regulation of metabolism. Think about it: your body doesn’t want to just constantly be building things and breaking things, willy-nilly. Your cells freak out if they have too much of x, too little of y, etc., etc. It would drive them nuts to not be able to control the amounts of things that are made and when they’re made. Luckily, allosteric enzymes are used to regulate when metabolic pathways switch on and how fast they go.

The example my professor gives is the example of pyrimidine biosynthesis. In the first step, the enzyme that will begin the synthesis is an allosteric enzyme that is activated by ATP. That means that the synthetic pathway won’t move quickly until the cell has sufficient energy to fuel the synthesis. Once that switch is flipped, the allosteric enzyme makes the first molecule in the pathway, and the MM enzymes take over, pushing that first piece through the cycle until UTP is finally made. Once UTP (the product) builds up in sufficient quantities, it acts as an allosteric inhibitor on the first enzyme, competing off the ATP and switching off the enzyme. After the allosteric enzyme stops making the first molecule in the chain, the whole synthesis slows to a trickle until the cell again needs to make UTP.

So, you’ve got your Michaelis-Menten enzymes doing all of the hard stuff, and you’ve got your allosteric enzymes acting as switches for the reactions. All in all, it’s a pretty nifty setup, and it’s part of a bigger picture—a big, beautiful picture of metabolism. There are a lot of catabolic and anabolic metabolisms that we’re going to look at, but first, we need to better understand what drives those pathways in the first place. That’s where metabolic flux comes in.


Questions? Comments? Salt? Tears? Put ’em below!

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