Catalytic Power and Biological Efficiency

Back to the love saga of enzymes and substrates! Unfortunately there probably won't be anywhere near as much steamy action in this instalment :(

Firstly, a reminder about some of the basics to do with free energy diagrams. (I've also covered this in a post that was actually written in year 11, but I felt like that diagram wasn't the best as it used different symbols to those used in these BIOC2001 lectures.)

Anyway, on one side you have your reactants, and on the other side you have your products. Joining the two is a curve thing. As the y-axis suggests, the higher up the reactants, products or transition state are, the higher the energy of those particular substances.

The difference between the free energy of the products and reactants, or ΔG, is the free energy change in the overall reaction. If this is negative, the reaction will proceed spontaneously. This does not mean that the reaction will proceed quickly. It could take 2 milliseconds or 2 googol years for all you know. All "the reaction will proceed spontaneously" means is that the reaction can take place without any help from anybody.

The main factor determining the speed of the reaction (out of the stuff on the diagram, at least), is the activation energy. From the diagram, the activation energy of the forward reaction is ΔG^‡_f (the diagram doesn't have the double dagger because I couldn't be bothered drawing one in) while the activation energy of the reverse reaction is ΔG^‡_r. The lower the activation energy, the higher the rate constant, and vice versa.

In fact, there's a nice fancy equation for this, because as you know, nerds would be nothing without their equations:

k_f = (k/h)Te^-ΔG‡_f/RT
where k and h are constants (Boltzmann's constant and Planck's constant, respectively), T is temperature (presumably in Kelvin?) and R is the gas constant. So basically, there's a shitload of constants. Also, same goes for the reverse reaction, but swap the fs for rs.

As you can see from the equation, k_f increases as T increases (since the whole thing is multiplied by T), but decreases as ΔG^‡_f increases (as e is raised to a negative power).

Now let's look at what happens when catalysts are involved! Catalysts are like matchmakers- they help the reactants get together and do wonderful stuff! Here is a diagram for a general catalysed reaction. The black curve shows the normal pathway, whereas the red line shows the catalysed pathway.

From the graph, you should be able to see that ΔG^‡_f and ΔG^‡_r have changed, but ΔG is still the same. Hence a catalyst cannot change whether or not a reaction proceeds spontaneously, but it can change the activation energies and hence the rate constant. It has these effects on both forward and reverse reactions.

Now let's go back to looking at how this applies to our good friends Enzymes and Substrates! Enzymes and substrates are a bit wacky, as even the most basic enzyme-catalysed reaction has at least two intermediates that the reaction has to go through. The first intermediate is ES, when the enzyme and substrate are together, and the second is EP, in which the pair are still together after the substrate has been transformed into a product. Here's the graph:

When calculating the activation energy, you need to use the highest of the three "transition state" peaks (or however many peaks there are if you're looking at a more complex reaction). From there, you can work out the effect on reaction rate and so forth.

Sometimes there may be even more intermediates in the reaction. For example, enolase catalyses the transformation of 2-phosphoglycerate (2-PG) into phosphoenolpyruvate (PEP). It does this by first converting 2-phosphoglycerate into an enolic intermediate. Hence there are even more transition state "peaks" in this reaction:

1. Binding of enzyme to 2-PG
2. Conversion of 2-PG into enolate (enolate is still bound to the enzyme)
3. Conversion of enolate into PEP (PEP is still bound to the enzyme)
4. Release of PEP from the enzyme

The highest peak (conversion of enolate into PEP in this case) is used to calculate overall activation energy for the reaction.

Now, you might have wondered why the activation energy is always positive. Or maybe you haven't wondered that, but I don't care, I'm telling you anyway. It all comes down to a simple equation:

ΔG^‡ = ΔH^‡ - TΔS^‡
where ΔG^‡ is the activation energy, ΔH^‡ is enthalpy (internal energy of a reaction), T is temperature and ΔS^‡ is entropy (which is kinda like the "degree of randomness" of a reaction, but I'm sure shitloads of physicists would slap me for such a simplistic explanation).

Anyway, what you have to know is this: ΔH^‡ is positive during a reaction as some bonds are broken and that leads to greater enthalpy and ΔS^‡ is negative as molecules are being positioned precisely for the reaction to occur so there is a decrease in random motion. As a positive subtracting a negative is positive, ΔG^‡ is always positive.

Arrhenius Plots

I'm sorry, I just felt like I'd gone way too long without breaking things up with a heading. So there's a heading. Hope you appreciate it.

Anyway, let's go back to that equation I mentioned earlier, the k_f = (k/h)Te^-ΔG‡_f/RT one. Yeah, that one. It looks a lot less scary when you simplify all those constants at the beginning into one letter, I promise!

k_f = Ae^-ΔG‡_f/RT

See?! I told you so! You can also tidy it up by making k_f just a generic k, and ΔG^‡_f into E_a (which I'm guessing is short for "energy of activation" or something):

k = Ae^-E_a/RT

Now, nerds like graphs. So let's turn this into a graph! Exponentials suck though... hmm... let's make this linear by using logs!

ln k = ln(Ae^-E_a/RT)
ln k = ln A + ln(e^-E_a/RT)
ln k = ln A - (E_a/RT)

Rearrange this very slightly and you get:

ln k = ln A + (1/T)(-E_a/R)

You can then use this to make a nice, neat linear graph of ln k against 1/T. The slope of this plot is (-E_a/R) and the intercept is ln A. Aren't linear equations great? (And to think that I used to hate them so much when I was in year 8 or 9...)

Let's see what we can do with our knowledge of this plot!

First we'll start with another equation, because equations are great! (Yup, I'm a nerd. I freely admit it.)

E_a = ΔH^‡ + RT

This can be rearranged to give ΔH^‡ = E_a - RT, which is useful if you already know what E_a is. (And you can work out E_a by finding out the slope of the Arrhenius plot, which is equal to (-E_a/R), and multiplying that by the negative of the gas constant.)

You can also get an expression for ΔG^‡ as well, though it's not quite as nice looking. Near the beginning, I mentioned the equation k_f = (k/h)Te^-ΔG‡_f/RT which can eventually be rearranged to give ΔG^‡ = -RT ln (hk_f/kT). (Yes, there's two ks in there. I kept the first one as k_f to separate them out. In the equation I gave, k_f is the rate constant whereas the other k is Boltzmann's constant. Why they couldn't have picked a different letter is beyond me. Some places, like the lecture slides, may italicise one of the ks to separate them out.)

Anyway, now you have ΔH^‡ and ΔG^‡, you can substitute those into the ΔG^‡ = ΔH^‡ - TΔS^‡ equation to find ΔS^‡. Because I'm sure you were dying to know that.

So overall there are a fair few things you can find out from the Arrhenius plot and a bit of manipulation: activation energy, change in enthalpy, change in free energy and change in entropy. Before you go too crazy though, only calculate these parameters at the temperatures that are on the Arrhenius plot. Calculating at temperatures before and after is extrapolating, which might not be appropriate. Oh, and just so you know, all of these parameters are for the transition state with the largest activation energy.

Catalytic Power

Yay, last bit! Catalytic power is an easy way to compare how good enzymes are. It's very easy to calculate- you simply divide the rate constant of the catalysed reaction by the rate constant of the uncatalysed reaction. For example, the hydrolysis of urea is pretty slow when uncatalysed, with a rate constant of 3*(10^(-10)) s^-1. When catalysed by jack bean urease, however, the rate constant is 3*(10^4) s^-1. This gives a catalytic power of (3*(10^4))/(3*(10^(-10))) = 3*(10^14).