Investagating the Action of the Enzyme Catalase On the Surface Area of a Potato.

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INVESTAGATING HOW INCREASING THE SURFACE AREA OF A POTATO AFEECTS THE AMOUNT OF OXYGEN PRODUCED

AIM:  My aim in this investigation is to examine how the amount of oxygen produced is affected by increasing the surface area of a potato (increasing the enzyme catalase in the experiment). I will be changing the surface area of a potato and observe how long it takes the oxygen to produce as I increase the surface area of the potato to produce oxygen. The substrate I will be using will be hydrogen peroxide and the biological catalyst will be the potato.

VARAIBLES: In this investigation, I will increase the enzyme catalase in the experiment by increasing the surface area of the potato, and observe how it affects the amount of hydrogen peroxide decomposed, producing oxygen at the end of the reaction.  The main variables in this investigation that affect the rate at which hydrogen peroxide is broken down are, the concentration of the hydrogen peroxide, the quantity of the hydrogen peroxide solution used, the temperature of the solution, the time at which the experiment is carried out for, the pH value of the catalyst and the surface area of the potato used.

        If the concentration of the solution is high, more hydrogen peroxide will be decomposed faster producing more oxygen. This is because if the concentration of the reactant is high and the reactants are more concentrated, then there are more particles for each unit of volume in the hydrogen peroxide.

Since there are more particles, there is a higher possibility that the particles will attack the surface area of the potato causing a reaction and forming the product oxygen and water. Since there are more particles in the solution causing more collisions, the rate of which the hydrogen peroxide is decomposed, (the rate of reaction), will also be fast allowing more of the products to be formed in a short period of time.

        Another variable that affects this experiment is the amount of temperature applied to the experiment. If a high temperature is applied to the reaction then this will give the particles inside the hydrogen peroxide more kinetic energy. As the molecules gain the kinetic energy, they tend to move around more faster, which also gives the particles a higher chance of colliding and causing a reaction to take place. Since the molecules have more energy to move around, they are more likely to overcome their activation energy barrier, which is the minimum amount of energy needed to over come the barrier. Now that the molecules have more energy they are likely to attack the surface area of the potato, causing a faster rate at which the substrate is decomposed.

Also, if the temperature of the experiment is too high, the catalase will not be able to decompose the substrate because the enzyme will be damaged at 500C. However, if the temperature is to low, the rate at which the substrate will be decomposed may be slower because enzymes work at there fastest at 370C.  

        The surface area of the catalyst (potato in this experiment) used in the experiment can also affect the rate of reaction. If two cubes of the catalyst potato are taken which are the same mass and one is cut into smaller pieces and the other is not, the smaller pieces would form more products at the end of the reaction than the large cube. This is because the small pieces provide more surface area for the hydrogen peroxide to attack and be broken down.

                

As more pieces of the catalyst are added, the more oxygen is produced. This is because as there are more potatoes this will provide the hydrogen peroxide to attack more of the potato. This will cause more of the catalase enzyme (in the potato) to decompose the hydrogen peroxide to form more oxygen at a faster rate. Therefore it will be faster to break down the hydrogen peroxide and also faster to form oxygen at the end of the reaction if the surface area of the substrate is increased.

        Time also affects the reaction. This is because if the catalyst is left in the solution for a longer period of time, more oxygen will be produced. This is because the hydrogen peroxide particles have more time to attack the surface area of a potato, which will cause the catalase to decompose the hydrogen peroxide. As the catalyst is left in the hydrogen peroxide for longer, this gives more time for the amount of oxygen produced to build up.

        The volume of hydrogen peroxide used in the experiment will also affect the experiment. This is because if there is more hydrogen peroxide solution surrounding the piece of potato, there will be more particles in the solution. Therefore, when the investigation takes place the particles overcome their activation energy barrier and then attack the surface area of the potato. As there is more hydrogen peroxide molecules, their will be more collisions of the solution which will attack the surface area of the potato, which will then decompose the potato quickly producing oxygen at a faster rate.  

        The pH of the hydrogen peroxide solution also affects the rate at which the oxygen is produced. This is because the pH of a potato is naturally 7; so therefore, it will decompose the potato fastest if the pH of the solution is 7. In this investigation, the input variable is the surface area of the potato and the output variable is the amount of oxygen collected.

        The amount of sunlight also affects the experiment because if there is a lot of sunlight, it will decompose the hydrogen peroxide solution. This is because sunlight can break the substrate so I have got to make sure that the substrate is kept away from sunlight.

        As well as the variables above I also have to control some key factors to allow me to gain reliable and accurate results. I will always make sure that the measuring cylinder used to measure the amount of oxygen is the same so that each experiment gets a fair advantage. I will also make sure that the measuring cylinder is clear so that when the oxygen is being collected, it can be seen clearly and therefore, I can record reliable results.

        I will also keep the same conical flask so that the width of the conical flask is the same for each experiment, which makes it fair for potatoes in the solution. This is because, if the bottom of the flask is wider, the potato will not be surrounded by the solution as much as when the flask is narrower. The diagram below shows this.

        

I will always judge the amount of oxygen collected at eye level so that an accurate result is obtained. The thickness of the tube should always be same at each time so that it is fair on every other experiment. This is because a thicker tube can allow more oxygen to be collected when compared to a thinner tube so therefore, the same tube should be used. When judging the amount of oxygen collected the same distance should be used to give a fair advantage to each experiment. Also, the same person should standardise the judgement so that each time the oxygen is collected, the same person knows how to standardise the experiment.  

BACKGROUND INFORMATION: Hydrogen Peroxide is chemical compound of hydrogen and oxygen with the formula H2O2. Hydrogen peroxide is a harmful and toxic substance found in the cells of living organisms. It needs to be broken down to a harmless product so that it can be used. Therefore, it is decomposed in all living things to make oxygen and water as shown on the next page.

 Hydrogen peroxide is also a pure, anhydrous, colourless, sweet liquid with a precise gravity of 1.44. It damages the skin and has a metallic taste. Hydrogen Proxide solidifies at -0.41° C (31.4° F). To delay the break down of the Hydrogen Peroxide into water and oxygen, organic substances, such as acetanilide, are added to the solutions.

 Also, hydrogen peroxide is kept in dark bottles at low temperature so that it does not decompose because sunlight and high temperature cause the hydrogen peroxide to decompose. The decomposition of hydrogen peroxide is very slow unless a catalyst is present. Collecting the oxygen formed and measuring the volume of the hydrogen peroxide at certain times after the start of the reaction can allow me to find the rate at which the reaction takes place.  

Organic substances, such as potatoes or acetanilide, decompose hydrogen peroxide. If a potato were used to decompose the hydrogen peroxide, the following would happen. Firstly the potato contains an enzyme called catalase. When a cube of potato is dipped into a solution of hydrogen peroxide (which is the substrate), the enzyme catalase on the surface area of the potato will be attacked by the hydrogen peroxide. The molecule of the substrate hydrogen peroxide has to fit in to the contours of a region in the catalase enzyme molecule called the active site. The shape of the catalase enzyme and the active site of the enzyme catalase let the molecule of the substrate fit in and lock as a key.

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 When the catalase enzyme loosens slightly the substrate molecule (the hydrogen peroxide molecule) is pulled apart producing two new substances. The substances formed from the decomposition of the hydrogen peroxide are oxygen and water. The two new substances are separated and released. This process may be repeated many times until the hydrogen peroxide solution cannot be broken down any further.  The harmful hydrogen peroxide is decomposed to form harmless product such as water and oxygen.  The diagram below shows the process of the decomposition of the substrate.

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