Aim In this investigation I will investigate the effect of the enzyme Catalase on Hydrogen Peroxide by measuring the volume of oxygen gas produced in a certain time. All living cells contain Catalase in them and so instead of using a Catalase solution I will use a yeast solution. The reaction of this experiment will be as follows: Hydrogen Peroxide Water + Oxygen Hypothesis My hypothesis for this experiment is that I think as I increase the concentration of Catalase, the rate of oxygen produced will increase as long as there is always excess substrate (H2O2).
I am going to now explain what an enzyme is and this will help me to justify my hypothesis. An enzyme is a protein which acts like a biological catalyst, and the definition of a catalyst is a substance that speeds up a reaction, but is not chemically changed at the end of the reaction and therefore is never used up in a reaction. From preliminary work I have done on enzymes I know this to be true because I have done an experiment in which I measured the mass of Manganese (IV) oxide (MnO2) (also a catalyst of H2O2) before adding it to Hydrogen Peroxide.
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Then after the reaction had finished I measured the mass again. And as I suspected the mass remained exactly the same and so this proves the fact that the catalyst is unchanged. Every enzyme has an active site. Active sites are where the reactions happen. The reactions occur when a substrate molecule (hydrogen peroxide) fits into one of the specialised active sites of a Catalase enzyme. This is called the lock and key method. I will know explain this in more depth. Every enzyme is specialised to only one type of reaction because of the shape of its active site.
Enzymes can only fit one type of molecule into their active sites; this is shown by the diagram below The diagram above shows one type of enzyme and two types of substrate and that only one of the substrates fit into the enzyme. This fit means there would be a successful reaction. Enzymes can have two different types of reactions, a breaking down reaction, Catabolic and a building up reaction, Anabolic. The reaction that I will be doing is a catabolic reaction and it will go as follows:
Catalase also has a special feature in its make up; it has four active sites instead of one so this means that more than one reaction can happen at the same time on the same molecule of Catalase. I will now try to prove my hypothesis in two ways, both of Collision Theory. Collision theory is a theory that states particles are always moving and colliding with each other and that reactions happen when molecules collide with each other. But every single reaction needs a certain amount of energy when collisions happen in order to react.
This energy is called the activation energy. A catalyst speeds up a reaction by providing another route for the reaction which requires less activation energy. The following diagrams help me to show this by using the Maxwell-Boltzmann Distribution (this shows the number of particles with a certain amount of energy): The first diagram shows that only a certain amount of particles have enough energy to react, but when the catalyst is added, the new route allows more particles to react because of the fact that the activation energy is lower.
This therefore shows that if there is no catalyst present in the reaction (zero concentration of yeast (control of my experiment)) the rate of oxygen produced will be lower than if there was a catalyst present. In my investigation I will be changing the concentration of yeast (Catalase) in the reaction and my hypothesis states clearly that as concentration of yeast increases so must the rate of oxygen produced. I am now going to try and prove this again by Collision Theory, but this time not of catalysts but of concentration.
In Collision Theory it is well known that if you increase the concentration of particles in a certain area, the number of collision will increase. This can be shown by two very simple diagrams: These two diagrams show that if the concentration is increased the number of collisions will increase, in the low concentration diagram there are two collisions and in the high concentration diagram there are four collisions. Since there are more collisions there will probably be more successful reaction and therefore a higher rate of reaction which will mean a faster rate of oxygen produced.
Another way of putting this would be to say that if the concentration of the enzyme is increased there is more Catalase and so more active sites that can catalyse the substrate at one time, which means that the reaction rate would be higher leading to a faster rate of oxygen being produced. Also I expect to get a straight line graph of rate of oxygen produced against concentration of the enzyme because the higher the concentration of the enzyme, the higher the amount of active sites which can catalyse the substrate molecules to produce oxygen.
However this would only happen if there was always excess substrate because if all the substrate was catalysed then the rate of oxygen produced would level off. Method In this experiment I will measure the volume of oxygen produced in cm3, within a certain time, when the Catalase breaks down the H2O2 by using a gas syringe. The gas syringe will be able to measure the volume of gas produced because the equipment will be set up so it is air tight when the reaction start (set up as above). I did some preliminary work on this experiment in order to find the best concentrations and best time range I could use for my own reactions.
All the H2O2 that I used and I will use was and will be 100% solution 20 volume. First I tried to find the right concentration by just seeing how much gas the reaction was giving off in 10 seconds. I used for these test 5 cm3 of H2O2 with 10 cm3 of yeast, 10 cm3 of H2O2 with 10 cm3 of yeast and 5 cm3 of H2O2 with 5 cm3 of yeast. I found from these tests that the reaction with 10 cm3 of yeast was too vigorous because the froth it made from the gas rising in the liquid rose all the way into the gas syringe, and so therefore the best concentration was 5 cm3 of H2O2 with 5 cm3 of yeast.
After having done this I began to find the best possible time range. I did this by reacting 5 cm3 H2O2 with 5 cm3 of Catalase and measuring the volume of gas given off at different times. I found that 15 seconds after the start of the reaction was a good time range because it gave the reaction plenty of time and the volume of gas given off was about 75 cm3 which means that for my own reactions if I have anything that gives off more gas within this time will not go off the scale of the gas syringe.
In the reactions there will permanently be 5 cm3 of H2O2 and then I will use 5 cm3 (100% yeast), 4 cm3 (80% yeast), 3 cm3 (60% yeast), 2 cm3 (40% yeast) and 1 cm3 (20% yeast) of the yeast solution, but since I have to have a total volume of 10 cm3 I will then add the corresponding amount of water to the total volume in order to get 10 cm3. (For example 5 cm3 of H2O2 + 3 cm3 of Yeast + 2 cm3 of water = 10 cm3 total volume). Also, as a control I will have 0 cm3 (0% yeast) of yeast with 5 cm3 of water and 5 cm3 of H2O2. This control experiment will show me that pure water will not react or catalyse the hydrogen peroxide.