Proteins- Enzymes

Last lecture on proteins! w00t w00t! I can finally see the light!

You should know the difference between a 1st order and a 2nd order reaction and be able to write a rate equation for a simple reaction.

You should understand that in a reversible reaction at equilibrium there is no net reaction in either direction, even though the reaction is occurring in both directions, the rates of these reactions are equal. You should follow how this leads to the definition of the equilibrium constant for a reversible reaction and know what this is in terms of rate constants and equilibrium reactant and product concentrations.

Uhhhh I'm already stuck. (Clearly I stopped paying attention in these lectures after the first few. Admittedly, I had a very hard time staying awake during these lectures on proteins.) My notes weren't very helpful either, and neither were the slides, so a big thank you to the UC Davis ChemWiki, where I pinched the following from (paraphrased, of course- I'm not that dishonest).

A first-order reaction is a reaction in which the rate is directly proportional to the concentration of one of the reactants. Most of these take the form A -> B. In these reactions, the rate can be found by calculating -(dA/dt), or dB/dt, or k[A], where [A] is the concentration and k is some constant.

A second-order reaction is a reaction in which the rate is directly proportional to the product of the concentration of the reactants. These can take the form A + B --> C, or 2A --> B. In the first case, the rate is k[A][B], while in the second the rate is k[A][A], or simply k[A]^2.

Note that the above were all for irreversible reactions. In reversible reactions, you also need to take the products into account, as I'm going to show you now.

In a reversible reaction, you can calculate the forward and reverse reactions separately, as shown above for the first- and second-order reactions. The overall forward rate is equal to the forward rate minus the reverse rate. For example, in the reaction A + B <--> C + D, the overall forward rate is equal to kf[A][B] - kr [C][D] (f and r are meant to be subscripts).

Equilibrium is defined as the state in which there is no net reaction in either direction. In this case, kf[A][B] - kr [C][D] = 0. This can be rearranged to give kf[A][B] = kr[C][D] and then kf/kr = ([C][D])/([A][B]) (note that all of the concentrations in this case have to be the concentrations at equilibrium). kf/kr can also be called Keq (with "eq" as a subscript... when will Blogger put a subscript option? Maybe I should ask them to...), or the equilibrium constant.

You should understand the difference between ΔG and ΔG° and the relationship between ΔG° and the equilibrium constant for a reaction.

Delta G is the change in free energy for a reaction, while Delta G degrees is the change in free energy under specific conditions, namely the conversion of 1 mole of each of the products to 1 mole of each of the reactants. The relationship between these two can be summarised as follows (for a reaction A + B <--> C + D):

ΔG = ΔG° + RT ln(([C][D])/([A][B]))

where R is the gas constant (8.314J/mol/degrees Kelvin) and T is the absolute temperature in Kelvin.

When the reaction is at equilibrium, there is no net formation of the products (or the reactants, for that matter) and thus the free energy change is 0:

ΔG° + RT ln(([C][D])/([A][B])) = 0

This can be rearranged to:

ΔG° = -RT ln(([C][D])/([A][B]))

which, since the reaction is at equilibrium, can be simplified even further to

ΔG° = -RT ln K(eq)

You should know the difference between ΔG and ΔG‡ and know that it is ΔG‡ that determines the rate constant for the reaction and hence the rate of the reaction. You should understand that ΔG‡ is always positive, but ΔG can be positive or negative and that if it is negative the reaction will proceed spontaneously. 

Delta G double dagger (sorry, not sure what to call it, but I think that's what the lecturer called it so I'm just going to copy him) is the activation energy of a reaction. I explained activation energy in one of my earlier Chemistry 3AB posts. Essentially, it's the energy required to reach the high-energy transition state. This is a considerable barrier to cross and thus the rate of reaction is limited by the proportion of molecules that have enough energy to cross this barrier. As the transition state is always of high energy, Delta G double dagger is always positive.

As I mentioned before, the free energy change is represented by Delta G. It can be positive, if the products are of a higher energy state than the reactants, or negative, if the inverse is true. If Delta G is positive, energy needs to be added to allow the reaction to occur; if Delta G is negative, the reaction will proceed spontaneously (though it may be very slow if the activation energy is high).

You should be able to sketch a free energy diagram illustrating the difference between a catalysed reaction and the same reaction that is uncatalysed. You should be able to estimate values of ΔG and ΔG‡ from a free energy diagram of a reaction where values are indicated on the free energy axis.

See my earlier post on reaction rates because I really cbf drawing diagrams right now.

You should know the definition of catalytic power.

Catalytic power is defined by the rate constant of the catalysed reaction divided by the rate constant of the uncatalysed reaction. If the catalysed reaction proceeds much faster, this ratio will be large, indicating a high catalytic power. If the catalysed reaction only proceeds a little bit faster, this ratio will be relatively small, indicating a low catalytic power.

You should understand that a catalytic site on a enzyme is part of its tertiary structure where substrates bind and catalysis occurs. You should also understand that the positioning of amino acid residues in the active site determines the position and orientation of bound substrates and also provides for the specificity of enzymes for particular substrates.

As I've probably mentioned 50 billion trillion times before, structure is related to function. The catalytic site, or active site, requires that a substrate fit there before it can carry out its work. The ability of an enzyme to accommodate a substrate properly depends on the amino acid residues surrounding the active site, as well as the shape of the substrate itself.

You should be able to explain why the induced fit model of enzyme action better explains enzyme catalysis than the lock and key model and understand how an enzyme stabilises the transition state conformation(s) of a substrates(s) by forming extra bonds between this conformation and the active site of the enzyme.

The "lock and key" model is the one that you've probably heard all throughout high school (and which I've used a bit in this blog as well). Basically it goes like this: substrates have a particular shape, designed to fit in the binding sites of specific enzymes, just like how a key fits into a lock.

Even though the lock and key model is pretty elegant and easy to understand, there's an even better model available. You see, the lock and key explains how a substrate and an enzyme can fit together, but it doesn't explain why the substrate-enzyme complex remains stable even as the reaction proceeds. A better model is the induced fit model, which compares the enzyme as a glove and the substrate as a hand. Just as a glove may have to stretch to accommodate the hand, the enzyme may have to accommodate the substrate.

An important concept behind the induced fit model is that the enzyme can exist in a lower-energy state and a higher-energy state. When the enzyme is in a higher-energy state, it can accommodate the substrate. As the reaction proceeds, the enzyme can change shape slightly to accommodate the transition state, allowing for stability at this state as well. Eventually the products are released from the enzyme.

You should be aware of the requirement of some enzymes for cofactors.

Cofactors are other molecules etc. that some enzymes require for them to work. I've already mentioned one of these- pyridoxal phosphate (PLP). Aside from PLP, there are many other cofactors that can be found in enzymes, such as ions and vitamins. The activity of specific cofactors depends on the reaction in question.

You should understand why most enzymes lose activity above about 45°C and be aware of the existence of thermostable enzymes from thermophilic organisms.

At a high temperature, denaturation of enzymes may occur. Denaturation is the unfolding of a protein so that it no longer adopts its required configuration. This occurs at around 45 degrees Celsius due to the weak hydrogen bonding that is responsible for most of the folding of a protein. Some thermophilic organisms (organisms that are adapted to high temperatures) have more thermostable enzymes that allow them to survive in the heat. These may contain more covalent disulfide bonds which are more resistant to changes in temperature.

You should understand that in many cases the activity of an enzyme is dependent on pH because of the presence in the active site of amino acid side chains that undergo pH-dependent protonation or deprotonation, but which need to be in a particular protonation state in order to allow substrate binding or for catalysis to occur. You should also understand that a bell-shaped pH activity curve results from having 2 such ionisable residues in the active site, one of which needs to be protonated and one of which needs to be deprotonated for substrate binding/catalysis to occur.

As mentioned in my post about the acid base chemistry of amino acids, ionisable groups of amino acids can exist in a protonated or deprotonated form, depending on the pH. This affects charge, which in turn can affect substrate binding or catalysis. Sometimes there are multiple amino acids in the active site that are affected by pH. Hence, pH is often important for enzyme activity, with activity peaking at the optimal pH and decreasing as the pH increases or decreases.

YUSSSSSSSSSSSS NO MORE AMINO ACIDS OR PROTEIN LECTURES TO GO THROUGH FOR THIS UNIT!!!!!!!!!!!

Now it's time to go to sleep...