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Find out what effect substrate concentrations have on the dissociation of hydrogen peroxide within the presence of the enzyme catalase.

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PLAN 2a The aim of the investigation is to find out what effect substrate concentrations have on the dissociation of hydrogen peroxide within the presence of the enzyme catalase. Catalase is the enzyme found in potato cells which specifically catalyses the decomposition of hydrogen peroxides. Hydrogen peroxide is a toxic BI product of metabolism in certain plant and animal cells. It is used in hair salons as a bleaching agent. The enzyme Catalase effectively removes it. Catalase is one of the fastest acting enzymes known. A single molecule of catalase can break 5.6 million molecules of hydrogen peroxide each minute. This catabolic reaction breaks down hydrogen peroxide into water and releases oxygen gas, which is liberated, via effervescence. Equation. 2H2O2(l) Catalase 2H2O(l) + O2 (g) 2b The enzyme catalase is required so that oxygen gas can be evolved. This can be achieved by placing the specimens of potato into various concentrations of hydrogen peroxide solution. Allowing the enzyme catalase to bind with the hydrogen peroxide substrate molecule to form an enzyme substrate complex. Allowing oxygen gas (product of the reaction) to evolve, which will be collected using a measuring cylinder by oxygen gas displace water out of the measuring cylinder filling it with the gaseous product of the reaction. Which will be timed in seconds in order to calculate the rate of dissociation. . 4a Enzymes are extremely sensitive to environmental conditions. In order for this experiment to produce valid data the following factors are going to be taken into account. This is because any factor which alters the conformation (dependent on the tertiary and secondary structure) of the enzyme catalase will alter the shape of the active site; therefore it will affect the frequency of the enzyme substrate complex formation and thus influence the rate of the enzyme-catalysed reaction. These factors include temperature, (pH and pressure) which are consistent, enzyme concentration and substrate concentration, also the surface area of the enzyme. ...read more.


Each potato was cut to the same4 mass and size and shape. This is to improve the fairness of the test. Also to make the enzyme concentration the same. The same pH level was maintained as the same potato was used. The surface area of the potato for each experiment was the same. The hydrogen peroxide can be an irritant at high concentration. Therefore goggles have to be worn. Also spillages have to be cleaned with a damp cloth. C analysing evidence and drawing conclusions Results The table below shows the quantitative data obtained during the experiment with the number of repeats done for each concentration. % Conc. of H2O2 Experiment x1 in seconds Experiment x2 in seconds Experiment x3 in seconds Experiment x4 in seconds Experiment x5 in seconds Mean time sec's Rate in (cm3/sec) 20 114 195 150 109 180 150 0.013 18 163 194 170 154 152 167 0.012 16 217 174 138 194 174 180 0.011 14 181 235 184 192 175 193 0.010 12 164 183 171 214 175 181 0.011 10 264 281 274 251 252 265 0.008 8 225 268 244 285 297 264 0.008 6 374 307 352 344 0.006 4 890 4a The rate of the reaction has been worked out by 2 cm3 divided by the number of seconds. Adding up every result found for each concentration and dividing it up by how many times it was repeated have worked out the mean time calculation. The table also shows the number of times the same concentration has been repeated. I predict that the graph will showed that as there is an increase in concentration, there would be an increase in the rate of oxygen evolved. Then after a very high concentration, the graph will levels off because all the enzymes are saturated. Summary table commentary 2b The table shows the quantitative data produced by the experiment. ...read more.


Therefore in order to maintain constant concentration of enzymes the mass has to be the same as far as possible. This will increase the validity of the results. . The independent variable in the experiment is the substrate concentration. Which will be an increase in concentration of hydrogen peroxide from 4%-20% in 2% increments. The dependent variable is the measuring the amount of oxygen gas evolved, which will be achieved by timing in seconds how long it takes for oxygen to produce and collect by downward displacement of water in an inverted measuring cylinder to fill 2 cm 3 of oxygen gas. Evaluation Firstly the apparatus used in the experiment gave limitations, which effectively altered some of the result. One of the limitations on the apparatus was estimating how much oxygen the measuring cylinder took in. Also using a measuring cylinder was really inaccurate this is because judgement had to be made. Instead a pipette would have been better, because it's a more accurate measure of solutions. The stop clock gave limitations because there had to be another judgement of when to stop the clock whichn is makes the data become inaccurate because everyone has different reaction speeds. This could have also made an effect on the graph causing the anomalous results. This is because the times would be inaccurate. Water in the delivery tube sometimes pushed water out, making it suck back. The bung kept on coming out which allowed oxygen to escape. Also the actually mass and surface area of the potato would have an effect on the results. If the potatoes were large then the concentration of enzymes would have increase therefore more substrate molecules would collide with the active sites of the enzymes, which would increase in number. This would have increased the rate of reaction causing anomalous readings making the data invalid. Also if the potatoes had a decrease in mass due to inaccurate cutting then there would be fewer enzymes for the substrate molecules to bind with, which will decrease the rate of reaction, again causing anomalous readings. By Miss Reema Kohli ...read more.

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