This course is part of a series taught by Kevin Ahern at Oregon State University on General Biochemistry. For more information about online courses go to http://ecampus.oregonstate.edu/ http://www.youtube.com/playlist?list=PL850269AA28EF394A
Also check out the free textbook "Biochemistry Free and Easy" at: http://biochem.science.oregonstate.edu
1. The Gibbs free energy is the energy available to do useful work in reactions. The change in the Gibbs free energy for a reaction is important because it determines whether a reaction is favored, unfavored, or at equilibrium.
2. Thus, when the Gibbs free energy change is negative, the reaction in question goes forward as written, but when the Gibbs free energy change is positive, the reaction goes in reverse.
3. A related term to G is G°', which is the Standard Gibbs Free Energy change. This refers to the Gibbs Free Energy change for a reaction under standard conditions. Since most reactions occur at non-standard conditions, G is much more useful than G°'. In fact, the sign of G°' does NOT tell the direction of a reaction, except under standard conditions.
4. Chemical reactions require activation energy (I'll call it G+ here) in order to get started. Catalysts (both enzymes and non-biological catalysts) act by lowering G+. Catalysts DO NOT CHANGE G. All they do is lower the energy required to activate the reaction. While enzymes speed reactions immensely, they therefore DO NOT CHANGE THE OVERALL REACTION CONCENTRATION AT EQUILIBRIUM. They simply allow the reaction to get to equilibrium faster.
5. G is affected by the concentration of reactants and products of a reaction by the following equation
G = G°' + RTln[Products/Reactants]
Thus, as product concentrations increase, the G will become more positive.
6. If one performs an experiment in which a fixed amount of enzyme is added to 20 different tubes, each containing a different amount of substrate (molecule that the enzyme catalyzes the reaction on) and then lets the reaction in each tube go for a fixed amount of time, one will create varying amounts of product when the tubes are analyzed. The greatest amount of product will be found in the tube which had the greatest amount of substrate. If one measures the concentrations of each product and divides by the time the reaction occurred, one obtains a velocity for each reaction. A plot of the velocity versus the substrate concentration (V versus S) from the experiment looks like the binding curve of myoglobin for oxygen - hyperbolic.
7. Velocity of an enzymatic reaction is measured as the concentration of product formed per time.
8. If one lets a reaction go for a long time, it will reach equilibrium. At equilibrium, the relative concentration of products and reactants do not change. Initial velocities of reactions are therefore measured so as to avoid allow the product to accumulate and favor the reverse reaction.
9. The catalytic action of enzymes appears to be related to their ability to be at least slightly flexible. Originally, Fischer proposed a model of catalysis called the Lock and Key model. It described enzymes as inflexible and the substrate as like a key fitting into a lock. While substrates do, in fact, fit into enzymes somewhat like a key, the enzyme is NOT inflexible.
10. Koshland's model of enzyme action, called the Induced Fit model says that not only does the enzyme change the substrate (via catalysis), but the substrate also changes the enzyme shape upon binding. This transient change of enzyme shape is important for catalysis because it may bring together molecular groups (such as a phosphate and a sugar) that may not be close together in the enzyme prior to the change in enzyme shape. Remember that enzymes also work by orienting substrates together in the proper way to maximize their likelihood of bouncing together in a way that leads to making a bond.
11. Chemical changes brought about by catalysis facilitate a last change in enzyme shape to allow for the release of the products. When this happens, the enzyme returns to its original shape and remains unchanged by acting to catalyze the reaction.
12. At very high concentrations of substrate, the enzyme gets overwhelmed (saturated) and a maximum velocity (Vmax) is approached. If one adds more enzyme, a higher maximum velocity will be reached. Vmax thus doesn't tell very much about an enzyme because its value depends on the amount of enzyme we use.
13. Km is the substrate concentration that gives an enzyme half its Vmax. Note that Km is a CONCENTRATION.
14. We define Km as the substrate concentration that gives Vmax/2. Whereas the Vmax varies, depending on the amount of enzyme that one uses, the Km is a constant for a given enzyme for its substrate.