Hydrogen peroxide decomposition by potato catalase
Prediction
Because of the increased chance of a successful collision caused by random thermal motion when there are more molecules present, I predict an increase in rate with higher substrate concentration. For low concentrations I think that the rate of the reaction will be directly proportional to the concentration of hydrogen peroxide in the solution. This is because if double the amounts of substrate molecules are in the solution, double the amount will find an enzyme molecule at the same time, if all the substrate molecules are moving at similar speeds (the average speed being directly proportional to the temperature). Therefore if there are double the amount of substrate molecules in a solution, double the amount of reactions will take place at once and the rate will be doubled.
The problem is, because of the time taken for the reaction and dissociation of the enzyme-product complex, as the concentrations of substrate increase; not all the collisions of the substrate will be successful because some active sites will be saturated (occupied by substrate/products). The frequency of this occurrence increases with the substrate concentration, and eventually the terms cancel out, leading to no rate increase with substrate concentration increase at high concentrations. This is because as the rate increases this must mean that more enzyme molecules are reacting with the substrate at one time, seeing that the reaction and dislocation time is constant at constant temperature, causing more substrate-enzyme collisions to be unsuccessful due to saturation.
This effect can be explained mathematically. The mathematical expression of the hyperbola caused by the effect explained above was developed in 1913 by two German biochemists, L. Michaelis and M. L. Menten. In the equation, VM is the theoretical maximum velocity of the reaction and KM is called the Michaelis constant.
Velocity =VM (S)
KM+(S)
The shape of the curve is a logical sequence of the active site concept; i.e., the curve flattens out at the maximum velocity (VM), which occurs when all the active sites of the enzyme are filled with a substrate. The fact that the velocity approaches a maximum at high substrate concentrations provides support for the assumption that an intermediate enzymes-substrate complex forms. At the point of half the maximum velocity, VM/2 in the diagram, the substrate concentration in moles per litre (S) is equal to the Michaelis constant, which is a rough measure of the affinity of the substrate molecule for the surface of the enzyme. The VM value for hydrogen peroxide is 1012 molecules of oxygen per molecule of catalase per second. The KM value in this case is about 5E-8.
Velocity =VM (S)
KM+(S)
I found the value of (S) by calculating the amount of moles per litre using the relative molecular mass of water and hydrogen peroxide and the Avogadro constant. This was between 0 and 6 moles per litre for the concentrations between 0% and 20%.
Therefore V=1012*5
5E-8+5
V=1E12 reactions per second per molecule of enzyme.
I estimate that there are about 50000 molecules of enzyme per square centimetre. If each cylinder has a surface area of 4 square centimetres the total amount of molecules of enzyme is 1.2E6.
1E12*1.2E6=1.2E18 reactions per second.
Each mole (6.02*1023 molecules) of oxygen takes up 24 litres.
(1.2E18/6.02E23)*24,000=0.05ml/s.
That means that I would be expecting to collect about 15 ml of the gas in five minutes (ignoring the decrease in substrate concentration). It would therefore be unrealistic to plan to fill the burette, which holds 50 ml, and a burette is therefore sufficient for the reaction.
