An experiment to determine the effect of substrate concentration on catalase PDF

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Lab Report on determining the effect of substrate concentration on catalase...

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An Experiment to Determine the Effect of Substrate Concentration on Catalase

Introduction Enzymes are biological catalysts that speed up metabolic reactions. The active site of an enzyme, which has a specific shape, will bind to a complementary substrate forming an enzyme-substrate complex speeding up the rate of reaction. The bonds in the substrate break and form products, leaving the enzyme chemically unchanged in the reaction. Due to the specificity of the active site’s shape, each enzyme will catalyse usually only one particular substrate through the ‘Induced Fit’ Theory. This theory entails the active site changing shape slightly and moulding itself to fit around a certain substrate (Parsons, 2008). Hydrogen peroxide (H2O2) is a by-product of various metabolic reactions and a dangerously powerful oxidising agent thus making it very toxic to the body (Clegg, 2014). The enzyme, catalase is produced in the liver and breaks down hydrogen peroxide to water and oxygen therefore stopping the accumulation of the toxin (Hocking et al., 2008).

2H2O2



catalase

2H2O + O2

Hydrogen peroxide will gradually decompose naturally however catalase increases the rate of reaction of the decomposition as shown in the equation above. This particular reaction is catabolic as hydrogen peroxide breaks down into two smaller products and gives out energy (exergonic). As the enzyme speeds up the rate of reaction, the activation energy needed for the reaction is reduced. This means that the reactants are more likely to collide. In this investigation, the hydrogen peroxide acts as the substrate. The purpose of this experiment is to examine the relationship between the substrate concentration on enzyme activity and its rate of reaction. The rate of reaction can be examined by calculating the speed it takes for the potato disc (enzyme) to reach the surface of H 2O2 (substrate) solution as the potato disc becomes buoyant due to the oxygen gas produced (Scientific American, 2012).

Hypothesis I predict that the rate of reaction will increase as the substrate concentration (H2O2) increases. This is because there will be a rise in collision frequency between the substrate and active site of the enzyme (catalase), assuming the factors affecting enzyme activity remain constant such as the temperature, pH and enzyme concentration. However, once the saturation point is reached, there will be no further effect on the rate of reaction. At this point there will be an excess of substrate molecules and not enough enzymes to bind to and so the enzyme concentration becomes the limiting factor.

Method 6 labelled test tubes were numbered to ensure that there was no confusion when setting up the dilutions. The dilutions were set up as shown in the table below: Table 1: Dilutions made up for H 2O2 Final concentration Test Tube of H2O2 (%) 1 1.00 2 1.00 3 0.10 4 0.25 5 0.50 6 0.00

Dilution Factor 0 0 1 in 10 1 in 4 1 in 2 1

Volume of distilled water to add (cm3) 0 0 18 15 10 20

Volume of H2O2 to add (cm3) 20 20 2 5 10 0

The dilutions were calculated using the DF (Dilution Factor) = C 1 / C2 = V1 / V2 formula to make a final volume of 20 cm3. For instance, in test tube 3 the final concentration is 0.10% and the dilution factor is 1 in 10 20 cm3 / Solute Volume = 10 solute volume = 20 cm 3 / 10 = 2 cm3 Therefore 2 cm3 of H2O2 should be added and 18 cm 3 of water to make up a total volume of 20 cm3. It was essential to do the dilution in the test tube individually as hydrogen peroxide decomposes naturally into water and oxygen. Therefore, by performing the experiment one test tube at a time, the reaction was delayed. This ensured that when the potato discs were submerged into the solution, they were able to produce reliable and accurate results to investigate the rate of reaction in different substrate concentrations. In each experiment, the distance was measured from the bottom of the boiling tube to the meniscus of the liquid. This information was used to calculate the speed of the potato homogenate floating to the surface of the solution. The disc-shaped filter paper was dipped into the potato homogenate and blotted on a paper towel to get rid of the excess potato on the disc. This was to ensure that each disc had similar amounts of potato and therefore reacted with the hydrogen peroxide and yielded consistent results. Once the disc was dropped to the bottom of the test tube, the timer was immediately turned on as the catalase (enzyme) in the potato would react with the hydrogen peroxide (substrate) and reached the top of the test tube. This was because one of the products of hydrogen peroxide decomposition is oxygen and caused the potato disc to become more buoyant causing it to float upwards. The timer was also immediately stopped once the potato disc reached the surface of the solution. The submersion of the potato disc was repeated three times in total to provide trustworthy results and for an average to be calculated. Test tube 1 was left to do last as this experiment was used as a negative control. The disc being dropped to the test tube had no potato homogenate dabbed onto it to show that the substrate concentration had no effect without the enzyme catalyst.

Results Table 2: Time taken for potato disc to reach the top of the tube against concentration of H2O2 Final concentration of H2O2 (%) 0.00 0.10 0.25 0.50 1.00 1.00 (control)

Time taken for potato disc to reach the top of the tube (secs) 1 2 3 Mean 90 91 116 99 60 63 62 62 30 31 37 33 15 18 16 16 -

The effect of substrate concentration (H 2O2) on catalase in the potato disc can be analysed by calculating the speed using the following formula: Speed (mm s-1) = distance (mm) time (s) For example, In 0.10% of H2O2  Speed = 57/99 Speed = 0.58 mm s-1 (2 d.p.)

Table 3: Distance potato disc travelled in tube and speed against the concentration of H2O2 Final concentration of H 2O2 (%) Distance disc travelled (mm) Speed (mm s-1) 0.00 60 0 0.10 57 0.58 0.25 59 0.95 0.50 56 1.70 1.00 58 3.63 1.00 (control) 59 0

Conclusion Gathering from the results of the experiment, the speed of the potato disc increases as the concentration of H 2O2 rises. This supports my hypothesis in that the rate of reaction speeds up as the concentration of H 2O2 increases. This is due to the fact that more oxygen gas is produced at higher substrate concentration as the potato disc becomes buoyant a lot faster meaning more substrate molecules are able to bind to the catalase enzymes in the potato homogenate. The more substrate particles there are, the higher the collision frequency which implies a higher rate of reaction thus supporting my graph. Graph 1 indicates that there is a positive correlation between the concentration of H 2O2 and catalase activity as there is an increase in speed of the potato disc reaching the surface of the solution in the test tube. This is due to the increase in enzyme-substrate collision frequency and rate of reaction. However, my results do not fully support my hypothesis in that graph 1 does not reach a ‘saturation point’. I predicted that after a certain point in substrate concentration, the rate of reaction would stay the same and would have no further effect due to limiting factors such as enzyme concentration. This suggests that the graph would plateau at the saturation point and the speed would remain constant due to the lack of effect on the rate of reaction as the active sites would have all been used up due to the excess substrate molecules. The investigation was conducted within the ranges of 0 to 1.00% of H2O2 concentration therefore my hypothesis could not have been fully supported within such a small range. Evaluation The results reinforce the first part of my hypothesis in the sense that the rate of reaction increases as the substrate concentration increases. However, the final statement in the hypothesis about the rate of reaction remaining constant after the saturation point is not portrayed in the results. This is due to the maximum substrate concentration (H 2O2) of the experiment reaching only 1.00%. Therefore, this is limiting the investigation of the full effect of substrate concentration on catalase activity as there is no data above 1.00% of H 2O2 concentration collected. This implies that at 1.00% H 2O2 concentration, there are still active sites available for substrates to bind to as the speed of the potato discs in the H 2O2 solution is still increasing meaning the rate of reaction is also increasing. In future, the investigation could include a higher percentage of the substrate concentration in order for there to be an excess of substrate molecules and produce a graph with a plateau to show the saturation point and the limiting factor. An example of a reliable aspect of the investigation includes the filter disc papers being cut to the same diameter. This controlled variable meant each potato disc had the same surface area to be pushed upwards by the bubbles of oxygen gas produced. Hypothetically, the bubbles should push the disc on both sides evenly and so will rise to the surface quickly with ease. However, in each trial of submerging the potato discs in the H 2O2 solution, it is impossible to control the way the potato disc floats. The formation of the bubbles of oxygen gas causes the disc to be pushed upwards to the surface of the solution in different ways. For instance, the bubbles could form only on one side so they get trapped underneath the disc until it flips and floats upwards eventually. Therefore, this took more time to reach the surface of the test tube in comparison to other potato discs and produced inaccurate results. This is true in the 0.10% H2O2 concentration in the third trial of submerging the

potato disc as it produced an anomalous result. Despite this, a majority of the data remained consistent. When the disc would float upwards, some would end up sticking to the walls of the test tube. So, the catalase enzyme concentration would be compromised on some of the potato disc making the results less reliable. To avoid this, the trial could be repeated again to avoid anomalous data. Also, when blotting the potato discs, some were coated with the potato homogenate unevenly. Therefore, the uneven distribution of the potato homogenate of each disc would have impacted the results and given less accurate data. This was due to lack of practical accuracy and could have easily been avoided by being more focused on the experiment and being more consistent with the handling of the potato discs. There were controls set up in the investigation which included the filter disc paper without the potato homogenate being submerged in 1.00% concentration of H 2O2. The result of this negative control suggested that the decomposition of H 2O2 was sped up by the catalase in the potato discs as it was visible in the other trials that oxygen gas was produced in the form of bubbles which caused the potato discs to float to the surface. Also, the potato disc submerged in 0.00% concentration of H 2O2 showed that H2O2 was necessary for the potato disc to float. This is because there would be no substrate for the catalase enzymes in the potato disc to bind to (Parsons, 2008). The temperature was also controlled to an extent as the experiment was carried out under room temperature. Although there may have been some fluctuation in the temperature, the data would not have been impacted greatly. Moreover, the potato homogenate was kept in ice to prevent it from denaturing, especially when exposed to environments with temperature fluctuations. Keeping the enzyme in ice maintains its tertiary structure by avoiding intermolecular forces from breaking (Clegg, 2014). Various compounds can influence enzyme activity. For instance, inhibitors prevent enzymes from binding to it substrate and so affect the rate of reaction. There are two different types of inhibition – competitive and non-competitive. Inhibitors that resemble the substrate shape in order to compete for the active site are called competitive inhibitors (Clegg, 2014). Therefore, if reversible competitive inhibitors are added at the same concentration to hydrogen peroxide in each tube, the rate of reaction will decrease because the active sites of the catalase are occupied by the inhibitor instead of the H2O2. Non-competitive inhibitors do not attach to the active site but binds to a different part of the enzyme causing the active site to change shape. Therefore, the substrates can no longer fit into the active site (Kent, 2000). This would decrease the rate of reaction if reversible non-competitive inhibitors were added into each tube and the overall velocity of the experiment for both inhibitors would be reduced. Overall, the results given follow the trend as hypothesised however due to the limitations of the experiment having a low range of substrate concentration. The investigation could have been carried out further to prove the rate of reaction remains constant when enzyme concentration becomes the limiting factor as the substrate concentration would have reached its saturation point.

References Clegg, C.J. (2014) Cambridge International AS and A Level Biology. London, United Kingdom: Hodder Education. Hocking, S., Kennedy, P., Sochacki, F. and Winterbottom, M. (2008) OCR A2 Biology Student Book. Edited by Sue Hocking. Oxford, United Kingdom: Gardners Books. Kent, M. (2000) Advanced Biology. Oxford: Oxford University Press. Parsons, R.D. (2008) AS-Level Biology OCR Complete Revision & Practice. Edited by Amy Boutal, Ellen Bowness, Joe Brazier, Charlotte Burrows, Tom Cain, Katherine Craig, Laurence Stamford, and Jane Towle. United Kingdom: Coordination Group Publications Ltd (CGP). Scientific American (2012) The Liver: Helping Enzymes Help You! Available at: https://www.scientificamerican.com/article/bring-science-home-liver-helping-enzymes/ (Accessed: 1 December 2016)....