1. Enzymes- biological catalysts Enzymes are proteins, eg. amylase, lipase, protease Activity depends on tertiary and quaternary structure and the specificity of enzyme action

2. Tertiary structure- intermolecular forces (strong and weak) van der Waals forces Ionic bonds Hydrogen bonds Disulphide bridges

3. Amylase- degrades starch Protease- degrades protein (HIV) Lipase- degrades lipids

4. Efficiency Very effective: Can speed up a reaction 10 8 to 10 20 times An enzyme can catalyse a specific reaction, e.g. degrade starch to maltose Enzymes operates under mild conditions (temp, pH etc)- or they get denatured Inorganic catalysts often works under harder condition, often on solid surfaces etc.

5. Catalyst Speed up the rate of a thermodynamically unfavourable reaction Doesn’t change or isn’t consumed in the process Doesn’t alter the  H or  G in the reaction Doesn’t change the position of an equilibrium or the equilibrium constant The reaction takes another path

6. A catalyst lowers the activation energy (E a )- the energy needed to start a reaction. If the E a is reduced the reaction will go faster

7. Enzymes lower the E a Reactant + Enzyme  ES  Product + Enzyme Substrate, S Enzyme-Substrate complex The enzyme can now catalyse a new substrate molecule

8. Induced-fit: the tertiary structure of the enzyme changes

9. Inhibition of enzymes The presence of other molecules, inhibitors, decrease the enzymes activity Irreversible competition- the inhibitor binds covalently to the enzyme- it is POISONED Reversible competition- the inhibitor binds loseley and can let go again

10. Competitive inhibition A competitive inhibitor binds to the active site, but can’t change it into products Often chemically similar to the substate

11. Non-competitive inhibition A non-competitive inhibitor does not bind to the active site, but changes the active site Often heavy metal ions

13. Enzyme kinetics- Michaelis-Menten equations Substrate + Enzyme  SE-complex  Product + Enzyme If we have no substrate present, then there will be no reaction, so at [S] = 0, the rate will be 0. Adding substrate [S] to the vessel will increase the amount of substrate binding to the enzyme and the reaction rate will increase. If we keep adding substrate however, there will come a point when all the enzymes are working as fast as they can and are essentially saturated, therefore, the rate will not be able to increase regardless of how much extra substrate is added. Thus, the plot of increasing substrate will be a curve that eventually flattens out reaching a maximum reaction rate (known as Vmax).

14. www.mymcat.com/wiki/Enzyme_Inhibition Substrate + Enzyme  SE-complex  Product + Enzyme substrate From the Michaelis-Menten equations, a constant (K m ) is used which relates the equilibrium constants of the forward and reverse reactions all together. This point, where [S] = K m, has a rate exactly equal to V max /2. A high Km value implies the enzyme is not very good at holding onto its substrate whereas a low Km implies the enzyme has a very high affinity for its substrate.

15. Competetive inhibitor Increasing the concentration of the inhibitor will decrease the reaction rate, however, V max can still be achieved if enough substrate is added to out compete the inhibitor. The right graph will shift where the x-intercept (1/K m ) is, but the y-intercept (1/V max ) will stay the same.

16. Non-competetive inhibitor When noncompetitive inhibitors are added to a reaction, V max reduces as the enzyme can no longer function as well. The K m for the reaction does not change because while the enzyme isn't catalyzing the reaction as well, it can still bind the substrate just as well as it did before. K m is a measure of the affinity of the substrate to the enzyme. In the plot to the right the y-intercept increases (because V max in 1/V max is decreasing) but the 1/K m position stays the same.