Effect of Concentration of Catalase in Decomposition of Hydrogen Peroxide

Effect of Concentration of Catalase in Decomposition of Hydrogen Peroxide
Aim: To show an increase in the concentration of catalase affects the volume of oxygen produced from hydrogen peroxide. Hypothesis Hypothetically, rate of the decomposition of hydrogen peroxide will increase greatly in the beginning by increasing the concentration of catalase, therefore the volume of oxygen released will be greater (increased). Background and Introduction Enzymes, natural catalysts, are special proteins that function to lower the activation energy required for a reaction to occur, thus increasing the rate of the reaction. Chemical reactions can and do occur without the use of enzymes, but the rates are usually considerably lower. Enzymes have many features, which make them suitable for their highly specific jobs. First of all, they have unique active binding sites. These binding sites allow enzymes to bind only specific substrates, while excluding those not involved in the reaction. There are many factors that affect the enzyme and so the rate of reactions. Catalase is an enzyme which decomposes hydrogen peroxide (H2O2) to form water (H2O) and oxygen (O2). Hydrogen peroxide is a toxic by-product of metabolism, of fatty acid oxidation to be more specific.
Hydrogen peroxide on its own 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 intoxication the cell. This is where catalase comes in. It should be very fast acting to maintain low levels of hydrogen peroxide. Therefore, it is one of fastest acting enzymes known. Equation of this decomposition reaction is: 2 H2O2 à 2 H2O + o2 Catalase is located predominately in peroxisomes within the cytoplasm of cells (specially liver cells). Peroxisomes, (which gets the name from peroxide, hydrogen) are spherical, 0.3 ? 1.5 mm in diameter and bounded by a single membrane. They contain a number of specialized enzymes. Catalase is found in the peroxisomes of potato cells. In this experiment I will use potato tubers as a source of catalase. There are many factors which can affect the rate of decomposition, and so the volume of the oxygen released, these include: a. Temperature b. Concentration of Catalase c. pH level and changes d. Concentration of Hydrogen peroxide I will investigate the affect of an increase in the concentration of catalase, on the the volume of oxygen released from the decomposition of hydrogen peroxide. Materials Needed
Cork borer
Beakers for water baths
Beehive shelf
Measuring cylinder
Rubber bungs
15 Test tubes
Scalpel
Plastic ruler
Syringe
Buffer tablets
Stop Watch
The Procedure of experiment 1. Set up a beehive shelf, which will hold an upside down water-filled measuring cylinder, inside a filled water bath. 2. Position one end of a piece of plastic tubing under the water, into the beehive shelf and inside the measuring cylinder. Connect the other end to the rubber bung. 3. Using the cork borer, take a sample of potato, cut this piece in cm intervals using the scalpel and plastic rule. Cut a piece the length of 2 cm. Put the potato into a test tube, use buffer solution to keep pH level constant and then fit in the rubber bung. 4. Draw 10 cm3 hydrogen peroxide solution into a syringe. Connect to the other hole of the rubber bung with a short length of tubing. 5. Slowly inject the hydrogen peroxide solution into the test tube. Bubbles of oxygen must be produced. The amount of oxygen released from the hydrogen peroxide should be recorded from the level shown in the measuring cylinder after 5 minutes. 6. Using 2 cm, 3 cm, 4 cm, and 5 cm lengths of potato repeat the steps 1 to 5. Repeat the experiment for three times for each length (size) of potato. 7. Take average of the two results of each condition and work out the amount of oxygen given off per minute. A graph must be plotted from these values, to show the rate of reaction for each condition. Variables which affect the rate of reaction of the enzyme should be taken into consideration in this experiment and kept constant. Such as:
Temperature; temperature should be kept constant, if there is any
change of temperature between experiments, the rate of reaction
and so the result (volume of oxygen produced) will be affected
greatly. For example a rise in temperature would cause the rate of
reaction to be great, and so more oxygen is given off in a short
period of time.
Substrate concentration; concentration of hydrogen peroxide used
must be constant, since it will also affect the rate of reaction
and so the volume of oxygen produced. So I used same amount of
hydrogen peroxide all through the experiments.
pH; pH level should be kept constant. Each enzyme work best in a
specific pH level, so changing the pH level would affect the rate
of reaction. Therefore this value must be kept constant. I used
buffer tablets to keep pH constant.
The only variable that is not kept constant is the size of potato (so the amount of enzyme catalase) which should be the only variable to affect the dependant variable ? the amount of oxygen produced from the 10 cm3 of hydrogen peroxide. Observations Mass of potato (g) Volume of oxygen (cm3 per 5 minutes) Reading 1 Reading 2 Reading 3 Average Volume of oxygen ( cm3 min -1) 1 6 5 5.5 5.5 1.1 2 9 10 9.5 9.5 1.9 3 15 14 16 15 3 4 20 20 20 20 4 5 24.5 23 24.5 24 4.8 You can observe from the table of results that as the size of potato increased the volume of oxygen produced per minute also increased. Rate of reaction can be determined and studied by plotting a graph of these results. As you can also see from the graph, there is a clear increase in the volume of oxygen produced when a larger sample of potato is reacted with a fixed amount of hydrogen peroxide. These results support the hypothesis. Conclusion: When hydrogen peroxide is added to a piece of fresh potato (enzyme catalase) it begins to fizz violently. This is a sign that two substances are reacting together. This shows that the catalase is decomposing the hydrogen peroxide so fast that it can be seen. The enzyme catalase and the substrate, hydrogen peroxide, join together to make the products oxygen and water. This is called the lock and key method. The enzyme catalase is specific to this reaction. Because the reaction is being heated in the water bath it goes much faster and is easier to see the increase in oxygen evolved. Catalase is found in potato tissue, so as the size of potato tissue increased, the concentration of the enzyme catalase reacted increased. The results are due to the increased enzyme catalase concentration used to hydrolyse hydrogen peroxide into water and oxygen. Therefore, the greater concentration of enzyme, the greater rate of reaction for a fixed concentration of substrate. Graph and results clearly show that the rate of reaction drops suddenly as mass of potato is increased from 4 to 5 g. This is most probably because of that the maximum volume of oxygen is already produced within the set time, and also the number of occupied enzyme active sites. At a certain point, all of the active sites of the enzymes will be occupied so any extra Hydrogen Peroxide molecules will have to wait until an active site becomes available. After a certain mass of potato, the increase in the rate of reaction will come to a halt since all enzymes will be working on the amount of substrate, hydrogen peroxide, present, a practical maximum rate of reaction (slightly below the theoretical maximum) will be reached. At that point increasing substrate concentration would cause the increase in the rate of reaction to continue. But in this experiment we studied the affect of concentration of catalase, so amount of substrate, hydrogen peroxide was kept constant. Certainly, increased enzyme (catalase) concentration results in an increased rate of reaction in the production of oxygen. Limitations: To enable this experiment to be completed as accurate as possible, I repeated it three times for each size of potato sample, and then used the average of all the results to best plot a graph. I tried my best to keep all the variables apart from the one I was testing (catalase concentration) the same. However and unfortunately in practice it is impossible with the basic apparatus I had to keep all measurements precisely the same. For example: 1- Water bath was difficult to keep at the constant temperature due to heat loss to the surroundings more hot water would have to be added to heat it up again but this is imprecise. 2- Oxygen bubbles may have been trapped in the tube connecting the test tube and the measuring cylinder for some time before actually entering the measuring cylinder, when the timing actually started. 3- One of the most inaccurate pieces of apparatus used was the measuring cylinder this is extremely inexact because if the meniscus isn?t exactly on the line the measurement is wrong. 4- There are slight delays between the start of timing and the start of reactions & appearance of bubbles. This will slightly affect all the results for each individual experiment but as I carried out all the steps in the same way, it should not make any negotiable difference to the overall result. Evaluation: 1. To gain a more accurate result more replicates could be taken to get a more actual average. 2. The catalase can?t be measured in potato, a much better catalase substitute would be yeast this can be weighed and measured. Also once the potato is used another would have to be used and this could produce catalase of a totally different concentration. 3. An electrically monitored water bath would be useful this could keep the water in the correct temperature range accurately. 4. Starting timing from the moment the enzyme and substrate were mixed would eliminate some inaccuracies

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