Fundamental Enzyme Kinetics part 1: Enzyme-Substrate Couples

I am going to pursue my impossible and try and make enzyme kinetics fun, or at least interesting. Here goes!

Once upon a time, there were enzymes. And there were substrates. (Please excuse the crappy picture, the brush automatically anti-aliased which made using the fill bucket tool hard to use. When I contacted Microsoft about it, they responded with “Since your issue is about turning off only the anti-aliasing in MS paint, I will provide a complimentary support for you Mr. Customer wherein we will try to perform 2-3 basic TS if that didn’t work we can talk about our Service Plan for more advanced TS. We have our self-help options as well.  But before that let’s proceed to our basic TS.” Lol. Well, I've just found out that the pencil doesn't anti-aliase, so maybe my next pictures will be a bit better.)

Anyway, back to the story. Sometimes, these enzymes and substrates would hit it off and become enzyme-substrate complexes. (Yes, their faces moved. They're shapeshifters now. Get over it.)

Now, some nerd at the Relationships Institute of EnzymeLand decided to write a mathematical equation to show how quickly these enzyme-substrate relationships formed. These rate equations were similar to rate equations for just about any other reaction, as detailed in an earlier post about enzymes. Essentially, it took the concentration of enzyme and substrate into account: the more enzymes and/or substrates there are, the more couples that can be formed. It also took into account a mysterious rate constant called k, which probably has to do with the temperature or weather or some other factors. Anyway, the main point is that the rate at which the enzyme and substrate couple up is equal to k[E][S]. Sometimes the k is written as k₁ to distinguish it from the rate constants of other reactions.

Unfortunately, not all relationships last. Just as enzymes and substrates could form relationships, they could also break up. The ES complex could dissociate into E and S, with a rate equal to k_-1[ES]. (The -1 implies that it's the reverse reaction of the reaction with the rate constant k₁.)

Now, sometimes these relationships (enzyme-substrate complexes) were very good. Sometimes the enzyme could turn the substrate into a better (person? protein? chemical?) than before. This was known as "product formation," because apparently it's okay to refer to the inhabitants of EnzymeLand using such dehumanising terms. (Tsk tsk.)

The formation of product also had a nice little rate equation as well, equal to k_cat[ES]. The "cat" refers to catalysis, and the ES refers to the fact that these products could only come out of a healthy, happy enzyme-substrate complex.

Now, the EnzymeLand Census Bureau became very interested in the formation of these "products" (ugh, such a dehumanising term). In particular, they often wanted to know how rapidly these products could form. Unfortunately, it was very difficult to determine using the equation, because the number of enzyme-substrate couples could sometimes be hard to tell. So they had to make a compromise.

Firstly, they did some experiments on some enzymes and substrates. Under controlled conditions, they observed that, while they rapidly form couples at the start, after a short while the number of couples remained more or less constant. They called this period the "steady state." (I would make a joke about "going steady" here or something but I'm not actually 100% sure what that phrase actually means.)

Now, just a quick refresher on the equations from before. We've talked about the formation of enzyme-substrate complexes (E + S -> ES), the break-ups (ES -> E + S) and the transformations (ES -> E + P). The rate equations from these three equations (k₁[E][S], k_-1[ES] and k_cat[ES]) can be put together to find the overall rate of formation of ES. Now, you might wonder why I've deviated like this, but don't worry- I'll get to the point later.

From this, the overall rate of formation of [ES] was found to be equal to k₁[E][S] - k_-1[ES] - k_cat[ES]. (The latter two are subtracted because they are showing the rate of [ES] breaking up.)

Now, at steady state, since [ES] did not change, k₁[E][S] - k_-1[ES] - k_cat[ES] = 0. But that wasn't particularly helpful either, because [ES] was in the equation, which as I just said wasn't always that easy to determine.

Then some equation rearranging was done:

• k₁[E][S] - [ES](k_-1 + k_cat) = 0
• k₁[E][S] = [ES](k_-1 + k_cat)
• [E] = ([ES](k_-1 + k_cat))/(k₁[S]) 

... which still left the equation in terms of [ES].

(And yes, I recognise that they could've just divided both sides by (k_-1 + k_cat) after the second step to eliminate that problem. But shhh. The EnzymeLand Census Bureau works in mysterious ways. Also, I suppose it would've been just as hard for them to find the concentration of E on its own as it was for them to find the concentration of E bound to S, or [ES].)

Anyway, another thing that they worked out is that the total amount of enyzme, or [E]_total, was equal to [E] + [ES]. They then got that whopping big equation from before and substituted it in:

[E]_total = ([ES](k_-1 + k_cat))/(k₁[S])  + [ES]

which they then rearranged a bit more:

[E]_total = [ES](((k_-1 + k_cat)/(k₁[S]))  + 1)
[ES] = [E]_total /(1 + (k_-1 + k_cat)/(k₁[S]))

Now, remember how earlier on I said that the rate of product formation could be written as k_cat[ES]? Well, now all you need to do is substitute in ES:

Rate of product formation = (k_cat[E]_total)/(1 + (k_-1 + k_cat)/(k₁[S]))

Looks kinda complicated, doesn't it? Well, it can be simplified, but first we need to talk about some special parameters.

The first parameter is V_max. This is the maximum velocity of the reaction, or the fastest rate at which ES can form a product. Usually, the velocity increases as S increases (more substrates to go around), so theoretically the fastest rate should occur when S = infinity. (As for why increasing E doesn't do anything... well, enzymes tend to go out with lots of substrates, those unfaithful buggers.)

Let's go back to the equation that we had before: (k_cat[E]_total)/(1 + (k_-1 + k_cat)/(k₁[S])). When S = infinity, then k₁[S] becomes infinitely large. Since it's the denominator, when k₁[S] becomes infinitely large, (k_-1 + k_cat)/(k₁[S]) approaches 0. Therefore, when S = infinity, the rate of product formation = (k_cat[E]_total)/1, or simply (k_cat[E]_total).

And thus we have our first parameter: V_max = k_cat[E]_total.

Another important parameter is K_m. This is the substrate concentration when the velocity is half maximum, or V_max/2. Now, when the rate of product formation = V_max/2, you get this equation:

V_max/2 = (k_cat[E]_total)/(1 + (k_-1 + k_cat)/(k₁[S]))

Substitute in k_cat[E]_total/2 for V_max/2 :

k_cat[E]_total/2 = (k_cat[E]_total)/(1 + (k_-1 + k_cat)/(k₁[S]))
1/2 = 1/ (1 + (k_-1 + k_cat)/(k₁[S]))
2 =  (1 + (k_-1 + k_cat)/(k₁[S]))
1 =  (k_-1 + k_cat)/(k₁[S])
 (k₁[S]) = (k_-1 + k_cat)
[S] =  (k_-1 + k_cat)/k₁

Therefore, the substrate concentration when the velocity is half maximum is equal to (k_-1 + k_cat)/k₁. This value is also known as the K_m.

Anyway, the useful thing about this is that now we've got some shiny new parameters that we can substitute into that complex equation from before (y'know, the  (k_cat[E]_total)/(1 + (k_-1 + k_cat)/(k₁[S])) one). Since V_max = k_cat[E]_total and K_m = (k_-1 + k_cat)/k₁, we can get the following:

Rate of product formation = (V_max)/(1 + K_m/[S])

which can also be rearranged to give

Rate of product formation = (V_max[S])/(K_m + [S])

The above two shiny new equations were collectively called the Michaelis-Menten equation, after the guy who figured them out and subsequently made everyone else's lives a whole lot easier.

Anyway, I feel like this post has gone on for a long time. I'm going to end this post here, but in the next post I will describe the kinetic parameters in a bit more detail. There will be pretty graphs, I promise!