Fundamental Enzyme Kinetics part 2: Kinetic Parameters

And now it's time to go back into the labs of the Relationships Institute of EnzymeLand!

I promised you pretty graphs in the last post, and now you get your pretty graphs! Here's the first one, which if I remember correctly is called the Michaelis-Menten graph:

Here we have the rate of reaction (v) on the y-axis and the substrate concentration [S] on the x-axis. V_max is the maximum rate of reaction. As you can see, V_max isn't represented anywhere on that red line at all. Instead, it is asymptotic to that line (i.e. approaches but never touches). This is because you technically need infinite substrate concentrations to reach V_max, which isn't really very practical. V_max/2 and subsequently K_m can also be approximated from this graph (I say "approximated" because it can be very hard to tell where the V_max asymptote is). Note that K_m is a substrate concentration and not a rate.

As I've just alluded to, it can be hard to work out where that V_max asymptote is on a curvy graph like that. Also, curvy graphs are harder to work with than linear graphs. That's why the Lineweaver-Burk plot was invented. The Lineweaver-Burk plot plots the reciprocal of v against the reciprocal of [S].

The equation for this line is essentially just the Michaelis-Menten equation turned upside down (i.e. the reciprocal of the Michaelis-Menten equation). I can't be bothered showing you the rearranging step-by-step, so here's the end product:

1/v = 1/V_max + (K_m/V_max)(1/[S])

Anyway, from this equation and the basic rules of working with linear equations, (K_m/V_max) is the slope of the curve and 1/V_max is the y-intercept. The x-intercept is equal to -(1/K_m).

Although reciprocals might seem a bit fiddly to work with, the fact is that the Lineweaver-Burk plot is actually easier to obtain estimates of V_max and K_m from. You see, the Michaelis-Menten plot requires that you figure out where the asymptote is, which isn't always easy. Meanwhile, here 1/V_max is bang smack on the y-axis. You do have to do some simple rearranging to find out what V_max is from there, but it's a helluva lot easier to do that than guess at an asymptote.

Now for some other cool tricks that you can use with these enzyme parameters!

Turnover Number

Firstly, let's play around with k_cat and V_max. From the previous post, V_max = k_cat[E]_total. This can be rearranged to give you k_cat = V_max/[E]_total. kcat, a.k.a. the rate constant of the [ES] -> E + P reaction, can also be called the "Turnover Number." It gives a guide as to how many substrates can be transformed into products in a unit of time (usually seconds, so normally turnover number is given by s^-1).

Affinity and Catalytic Efficiency

Now let's play around with K_m! From my previous post, K_m = (k_-1 + k_cat)/k₁. Now let's have a look at what happens if k_cat is substantially larger or smaller than the other values!

If k_cat is much smaller than k_-1, then K_m ~ k_-1/k₁. Incidentally, k_-1/k₁ is also equal to the dissociation constant for [ES] (since it is the rate constant for the dissociation of ES into E and S, divided by the rate constant of the association of E and S to form ES). Hence, when k_cat is much smaller than k_-1, K_m can also be seen as a measure of the affinity of an enzyme for its substrate. The lower the dissociation constant (and hence the K_m), the greater the affinity that the enzyme has for its substrate. (Awwww. Get that K_m low enough, and I bet they'll be giving each other roses on a daily basis.)

If k_cat is much larger than k_-1, then K_m ~ k_cat/k₁. This can be rearranged to give k₁ ~ k_cat/K_m. That might not mean much to you now, so let's look at another equation first. In particular, I want to take you back to the Michaelis-Menten Equation, which was this:

v = (V_max[S])/(K_m + [S])

Now, remember how V_max = k_cat[E]_total? Substitute that in:

v = ( k_cat[E]_total[S])/(K_m + [S])

Now, when [S] is much smaller than K_m, the denominator becomes pretty much equal to K_m so that you get this:

v = (k_cat/K_m)[E]_total[S]

Now, remember how rate constants are the concentrations of the reactants multiplied by a rate constant? Well, here we have two reactants, [E] and [S], so (k_cat/K_m) essentially becomes the rate constant of [E] and [S] banging each other. (What? I meant that the enzyme and substrate are colliding with each other! What did you think I meant?!)

Anyway, since (k_cat/K_m) is the rate constant here, it's also a measure of catalytic efficiency, which to my understanding is how well the enzyme transforms a substrate into a product.

Now, back to k₁ ~ k_cat/K_m. This equation almost seems to suggest that when k_cat is much larger than k_-1, the catalytic efficiency is roughly equal to the rate constant of E and S getting together to form ES. From the lecture slides, it says that the significance of this is that the catalytic efficiency can be no higher than the rate of diffusion controlled binding of E to S, which makes sense: product can't be formed if the substrate and enzyme aren't already an item, so to speak.

Alright, that's enough equations for this post. The next post is going to include betrayal! Competition! Steamy bed action! You won't want to miss it!