To Investigate (By Experimentaion) the Effect of Substrate Concentrations On the Rate of the Decomposition of Hydrogen Peroxide When Catalyzed By the Enzyme Catalase.

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For this investigation I have been asked to investigate (by experimentaion) the effect of substrate concentrations on the rate of the decomposition of hydrogen peroxide when catalyzed by the enzyme catalase. This is part of our work on the function of enzymes, how they work and the effects of conditions on how they work. We have learnt about the formation of enzyme-substrate complexes, the lock and key model, induced fit model, activation energies of normal reactions and enzyme-catalyzed reactions, equilibrium, specificity and denaturation.

Let me write specifically about the enzyme, catalase and the substrate, hydrogen peroxide. In organisms, hydrogen peroxide is a toxic by-product of metabolism, of certain cell oxidations to be more specific. Hydrogen peroxide on its own is relatively stable and each molecule can stay in this state for a good few years. Its decomposition therefore needs to be speeded up greatly in order to prevent it from intoxicating the cell. This is where catalase comes in.

Catalase has to be very fast acting to keep the hydrogen peroxide levels low, and it is one of the fastest acting enzymes known. It catalyses the decomposition of hydrogen peroxide, liberating oxygen gas as effervescence, each molecule of the globular protein decomposing 40,000 molecules of hydrogen peroxide per second at­ zero degrees Celsius and capable of producing an amazing 1012 molecules of oxygen per second. The equation is:

2H2O2+Free enzyme«E-S complex®2H2O+O2+Free enzyme

The E-S complex is an intermediary stage where the substrate forms temporary and reversible interactions with the enzyme. The reason that this is so much faster than the decomposition rate in the absence of a catalyst is to do with the activation energy for this route being lower than the energy it takes to simply break the bonds within the molecules because it forms an intermediary stage, but the mechanism for this is not yet fully understood.

[See diagrams]

Catalase is found in microbodies, or peroxisomes, in euchariotic cells. Peroxisomes are spherical, 0.3-1.5 mm in diameter (smaller on average than mitochondria) and bounded by a single membrane. These are derived from the endoplasmic reticulum. Peroxisome gets it name from peroxide, hydrogen. The three types of plant peroxisomes are:

Glyoxisomes, so called because they metabolise a compound called glyoxylate, are concerned with the conversion of lipids to sucrose in lipid-rich seeds.

Leaf peroxisomes are important in the process of photorespiration in which they are part of the photorespiratory pathway involving (obviously) chloroplasts and mitochondria, the three organelles often being in close proximity within the cell. The photorespiratory pathway is shown on the right. This shows how hydrogen peroxide is produced and the functions of the peroxisome as part of the photorespiratory pathway.

Non-specialized peroxisomes are a third group, which are found in other tissues.

Potato tubers contain peroxisomes, although I don’t know why. I suppose if new plants have to grow from them, they have to have all the parts of the plant with the store of food. If a potato is exposed to light it goes green, so I suppose this is proof because could not go green without chloroplasts.


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.

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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 ...

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