An investigation to measure the rate of reaction of hydrogen peroxide decomposition by catalase in yeast

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Nicholas Thorburn                             Block I                                                                NPTL

An investigation to measure the rate of reaction of hydrogen peroxide decomposition by catalase in yeast

For this investigation I have been asked to investigate (by experimentation) the effect of substrate concentrations on the rate of the decomposition of hydrogen peroxide when catalysed 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. This experiment will look at various concentrations of substrate and enzyme, and will also look at how this will affect the rate of decomposition, using graphs and tables.

In organisms, hydrogen peroxide is a toxic waste product of metabolism.  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 is used. 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, losing oxygen gas as effervescence.

Due to an enzyme's functional shape its substrate binds only at a unique location on the enzyme's surface, called the active site, where the chemical reaction occurs. The analogy used to explain this specific interaction is that of a lock & key. Just as uniquely shaped keys will only fit into, and open, a matching lock, so it is with enzymes and their substrates. The wrong key may fit into the lock, but nothing can happen because the match of shapes is not correct. This fit is so specific that the change in a single hydrogen atom in a molecule usually means that its specificity for a certain enzyme is lost; that is, it may not even bind in the active site or even if it does the enzyme will be unable to do anything chemically to it.

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The equation is: 

2H2O2 (aq) ⇒ 2H2O(l)+O2 (g)

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.  Therefore if there is 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, as long as there are enough catalase enzymes to cope. When there are not enough catalase enzymes to cope, the reaction will tail off, as described in the graph below:

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