Internal Assessment Biology (HL) - Comparing the Inhibition of Calatase Enzyme by Metal Ion Inhibitors

Comparing the Inhibition of Catalase Enzyme by Metal Ion InhibitorsPersonal Engagement I have been consuming honey regularly every day since I could remember until right now. I believe honey is a better sweetener than the commonly consumed sugar in Indonesia. Honey has lower sugar content despite its sweet taste because it is rich in fructose. Therefore, I often add honey to the beverages I drink, such as on tea, milk, coffee or even hot water. I also add honey on bread. I always feel very refreshed and healthier after consuming honey. Besides being a good sweetener, honey is also good for health because of its antibacterial property. Diseases are spreading around the world and a natural antibacterial like honey is a great benefit. Meanwhile, in studying biology, H2O2 (hydrogen peroxide) is well known as a harmful substance that could be broken down by catalase enzyme. However, H2O2 could actually be useful to treat infection1. It is also responsible for the antibacterial activity in honey2. If honey is contaminated with catalase enzyme, it may lose its antibacterial activity because the H2O2 is broken down. However, metal ions are well known to inhibit catalase enzyme, maintaining the antibacterial activity of honey. The interaction of this knowledge in Biology has given me an interest in investigating the inhibition of catalase enzyme by several metal ions, as it can maintain and maximize the benefits from the honey that I consume regularly. Fortunately, a study of enzyme and its inhibition are covered in Biology Chapter 2: Molecular Biology and Chapter 8: Metabolism, Cell Respiration and Photosynthesis, which help me in this investigation and lead me to choosing this topic. Background Knowledge Honey contains several enzymes (biological catalyst made of protein) such as diastase, invertase and glucose oxidase3. Glucose oxidase catalyzes the oxidation of glucose producing gluconic acid and H2O2 (hydrogen peroxide). Fig 1. Hydrogen peroxide formation in a reaction catalyzed by glucose oxidase4 1 Evolution, C., Martino, J., Walia, A., Martino, J., Walia, A., DeNicola, M., & Erickson, A. (2014). 25 Amazing Benefits & Uses For Hydrogen Peroxide. Collective-Evolution. Retrieved 22 September 2015, from http://www.collective-evolution.com/2014/02/16/25-amazing-benefits-uses-for-hydrogen-peroxide/ 2 French, V. (2005). The antibacterial activity of honey against coagulase-negative staphylococci. Journal Of Antimicrobial Chemotherapy, 56(1), 228-231. http://dx.doi.org/10.1093/jac/dki193 3 Airborne Honey,. Honey Enzymes. Retrieved 13 November 2015, from http://www.airborne.co.nz/enzymes.shtml 4 Santos, A., Soares, D., & Queiroz, A. (2010). Hydrogen Peroxide Formation in a Reaction Catalyzed by Glucose Oxidase. Retrieved from http://www.scielo.br/scielo.php?script=sci_arttext&pid=S1516-14392010000100003 1  H2O2 is a strong oxidizing agent that could kill several pathogenic microorganisms. However, H2O2 could be decomposed into H2O (water) and O2 (oxygen gas) upon contact with catalase enzyme. Fortunately, catalase enzyme could be inhibited by several metal ions. Fig 2. Decomposition of hydrogen peroxide catalyzed by catalase enzyme 5 Metal ions such as Cu2+ (copper (II) ion), Zn2+ (zinc ion) and Pb2+ (lead (II) ion) could inhibit catalase enzyme and they can respectively come from CuSO4 (copper (II) sulfate), Zn(NO3)2 (zinc nitrate) and Pb(NO3)2 (lead (II) nitrate), which are commonly found in our lives. Being a different metal ion, they can have different inhibition activity. Their amount was suspected to affect the inhibition of catalase enzyme. Therefore, an experiment was conducted to investigate how the amount of inhibitor affects the inhibition of catalase enzyme and to compare the inhibition of each substance. The catalase enzyme was obtained from chicken liver as it is known to be rich in catalase enzyme. To get the inhibition level, the reaction would be conducted in the presence of an inhibitor and in the absence of it. The amount of O2 produced would be obtained from the difference in the O2 level before and after adding catalase. The inhibition level would be calculated from the ratio of the difference in the amount of O2 produced to the amount of O2 produced when inhibitor is absent. Research Question  How does the amount (mmol) of each inhibitor (CuSO4, Zn(NO3)2 and Pb(NO3)2) affect the inhibition level (%) of catalase enzyme and how do they compare? Hypothesis  Greater amount of inhibitor will cause greater inhibition level and the greatest inhibition level will be shown by Pb(NO3)2 that uses greatest amount which could reach inhibition level around 90% since lead compound is well known to be toxic. Variables Independent Variable:  Inhibitor used: CuSO4, Zn(NO3)2 and Pb(NO3)2.  Amount of inhibitor: 0.100, 0.200, 0.300, 0.400 and 0.500 mmol. Dependent Variable:  Increase in O2 level after 40 seconds of pouring catalase enzyme. 5 Strong, S. Decomposition of Hydrogen Peroxide. Retrieved from http://stronglabnotebook.weebly.com/ 2 Controlled Variable: Table 1. Controlled variables in this investigation Factors How Why Temperature The experiment would be conducted Higher temperature will provide of reaction in a laboratory with air conditioner greater kinetic energy on the enzyme, (23oC ±0.5). set at 23oC. causing reaction rate from enzyme to produce O2 to be different. While having too high temperature may denature the enzyme causing no reaction to occur. Concentration It was controlled by using 15 g of Different concentration of enzyme of catalase liquid from squeezed chicken liver will cause different enzymatic activity enzyme measured using digital balance and and O2 will have its production rate to (15.00 g liver diluted with 300 ml of distilled be different. Higher concentration in 300 ml water to provide the catalase tends to give out higher O2 production distilled enzyme. The same solution was rate. water). used throughout all trials. Amount of It was controlled by using 2.0 ml of Different volume of enzyme will enzyme prepared chicken liver solution cause O2 to be produced at different (2.0 ml measured using 5.0 measuring rate where more enzyme used will ±0.05). cylinder. cause more O2 produced. Amount of The volume of H2O2 for each trial Amount of substrate should be kept substrate was kept constant which was 5.0 ml constant for a fair test in the amount (5.0 ml ±0.05 measured using 5.0 ml measuring of O2 gas produced. The H2O2 – 5%). cylinder while the concentration was concentration should be constant to kept constant which was 5% taken allow same rate of O2 gas production, from same container. and therefore the amount of H2O2 should be kept constant with same volume of it. Volume of The volume of container where the The O2 sensor measures the container reaction occurred was kept constant concentration of O2. The volume (Conical flask by using 250 ml conical flask for should be controlled to be able to get 250 ml). each trial. constant data of amount of O2. Total volume The total volume of inhibitor added Using different volume of inhibitor of inhibitor to the enzyme and H2O2 was kept will cause different volume of total (5.0 ml constant by adding distilled water solution in the conical flask 250 ml, ±0.05). until total volume become 5.0 ml therefore may cause the H2O2 or measured using a 5.0 measuring enzyme concentration to be different. cylinder. Period of The time was kept constant which Different period time will cause the time for the was 40 seconds recorded in the O2 increase in amount of O2 produced to increase of O2 sensor measured when increase in be different. level O2 level was detected. (40 s ±1). 3  Methodology Table 2. Apparatus used in this investigation Table 3. Materials used in this investigation No. Apparatus Quantity No. Materials Quantity 1. Digital balance 500.00 (±0.01) g 1 1. Chicken liver 100 g 2. Knife 1 2. Hydrogen peroxide 5% 400 ml 3. Cutting board 1 3. CuSO4·5H2O* 6.24 g 4. Mortar and pestle 1 4. Zn(NO3)2 4.73 g 5. Strainer 1 5. Pb(NO3)2 8.28 g 6. Stirring rod 4 6. Distilled water 1,290 ml 7. Measuring cylinder 100 (±0.5) ml 1 * CuSO4·5H2O was used instead of CuSO4 8. Measuring cylinder 5.0 (±0.05) ml 3 because CuSO4 was not available. Once 9. Beaker 100 ml 3 dissolved, CuSO4·5H2O would make a CuSO4 10. Beaker 500 ml 1 solution which is the same if made from CuSO4. 11. Volumetric flask 250 (±0.15) ml 3 solid. 12. Conical flask 250 ml 80 13. O2 sensor data logger (±102 ppm) 1 Safety Precaution and Disposal: Irritation may occur upon contact on H2O2, CuSO4, Zn(NO3)2 or Pb(NO3)2 by skin or eyes. Safety gloves and googles are recommended. If contact occurs, wash the exposed area with clean water for 15 minutes. The decomposed solution in conical flask is disposed into sink only after no more bubble of gas is produced. Glove is used when handling liver for hygiene purpose. The excess liver is discarded into trash bin. Procedures: 1. Turn on and set the laboratory’s air conditioner at temperature 23 oC while closing all the doors and windows. Put all the materials and apparatuses in the laboratory to allow the whole experiment to proceed in temperature 23oC. 2. 250 ml of solution 0.1 M inhibitor was prepared by dissolving the weighed solid (using digital balance to weigh the mass. Refer to Table 4 for its mass) using 40-60 ml of distilled water in a 100 ml beaker then stirred using stirring rod, and poured into a 250 ml volumetric flask. The beaker was filled with distilled water again and poured into the volumetric flask until the 250 ml line was reached. Then the volumetric flask was shaken. Table 4. Mass of inhibitor No Inhibitor Mass (g) ±0.01 1. CuSO4·5H2O 6.24 2. Zn(NO3)2 4.73 3. Pb(NO3)2 8.28 o The value of the required mass in Table 4 is found through the following calculation: ???? (g) = ????????????? (M) × ?????? (L) × ?? o For example, the mass required for CuSO4·5H2O (Mr = 249.685) is: ???? = 0.100 M × 0.250 L × 249.685 = 6.2421 g ≈ 6.24 g o The molecular weight of Zn(NO3)2 is 189.36 and Pb(NO3)2 is 331.2. 4 o The uncertainty in the solution concentration is: ???? ??????????? ?????? ??????????? ????????????? ??????????? = ( + ) × ????????????? ???? ?????? o The uncertainty of 0.1 M CuSO4 solution is: 0.01 0.15 ????????????? ??????????? = ( + ) × 0.1 = 0.00022 M 6.24 250 o The concentration uncertainty is very small for all solutions and therefore will be negligible. 3. Chicken liver was cut using knife on a cutting board into small pieces and grinded using mortar and pestle until the liquid was squeezed out. The liquid was separated from the residue by using strainer. 15.00 g of the liquid was taken (measured using digital balance) and mixed with 300 ml of distilled water (measured using a 100 ml measuring cylinder, three times) in a 500 ml beaker, then stirred using stirring rod. 4. 2.0 ml of the chicken liver (prepared in step 3) was measured using a 5.0 ml measuring cylinder and poured into a 250 ml conical flask. 5. The inhibitor was taken into the 5.0 ml measuring cylinder until it reached the desired volume and then added by distilled water until 5.0 ml according to Table 3. This was done to get the different amount of inhibitor, but same volume of total solution. Table 5. Different amount of inhibitor Desired Amount Volume of inhibitor 0.1 M Volume of distilled water No (mmol) ±0.005 (ml) ±0.05 (ml ±0.05) 1. 0.500 5.0 0 2. 0.400 4.0 1.0 3. 0.300 3.0 2.0 4. 0.200 2.0 3.0 5. 0.100 1.0 4.0 o The value of inhibitor volume in Table 5 is found through the following calculation: ??????? ?????? (mmol) ?????? (ml) = ????????????? (M) o For example, the volume of inhibitor solution required to get 0.500 mmol of inhibitor is: 0.500 mmol ?????? = = 5.00 ml 0.100 M o The uncertainty in the inhibitor amount is found through this formula: ?????? ??????????? ?????? ??????????? = × ?????? ?????? o For example, the uncertainty of inhibitor amount of 0.500 mmol is: 0.05 ?????? ??????????? = × 0.500 = 0.005 mmol 5.0 o The uncertainty absolute value in each desired amount of inhibitor is found to have same value. 5 6. The inhibitor was added to the 250 ml conical flask containing the liver solution, and then the flask was shaken for a while. 7. The initial O2 level in the flask was recorded by placing the O2 sensor on the opening of the flask and O2 Sensor the data collection was started. 8. 5.0 ml of H2O2 5% was taken using a 5.0 ml measuring cylinder. O2 sensor was removed from the flask opening; the prepared H2O2 was poured into the flask. The O2 sensor was immediately returned to the flask opening. After the increase in O2 level was detected, the final O2 level after 40 seconds was recorded. Liver, inhibition 9. Step 4-8 were repeated until 5 trials for each substance and H2O2 concentration, and then repeated by using other concentrations, and repeated again with other Fig 3. Laboratory set-up to inhibitors and repeated without using any inhibitor measure the increase of O2 level but using 5.0 ml of distilled water. Raw Data After pouring H2O2 to the solution of liver and inhibitor was present, the increase in O2 level value was not immediately detected. When there was increase in O2 level, bubble of gas in conical flask was observed. The data of O2 level will also be rounded to four significant figures to make the calculation simpler while maintaining the accuracy. The example of the graph for other samples will be shown in Appendix. Fig 4. Example of the obtained result in graphical display – no inhibitor trial 1 6  Below is an example of the data for no inhibitor and CuSO4 0.100 mmol. Table 6. Example of raw data – change in O2 level in no inhibitor and CuSO4 0.100 mmol Initial (104 ppm) Final (104 ppm) Increase (104 ppm) Sample Trial ±0.01 ±0.01 ±0.02 1 19.69 28.91 9.22 2 19.65 32.73 13.08 No inhibitor 3 19.65 32.07 12.42 4 19.71 32.20 12.49 5 19.60 30.41 10.81 1 19.44 26.64 7.20 CuSO4 2 19.60 26.88 7.28 0.100 (±0.005) 3 19.65 26.15 6.50 mmol 4 19.38 26.17 6.79 5 19.60 27.88 8.28 *Note: The complete raw data will be shown in Appendix. The increase in O2 level (later referred as?) is calculated through this formula: ?2 ???????? (?) = ????? ?2 ????? − ??????? ?2 ????? In sample with no inhibitor in trial 1, ?2 ???????? (?) = 28.91 × 104 − 19.69 × 104 = 9.22 × 104 ppm Data Processing The data processing and the formula will also be applied for other samples, but only CuSO4 is shown for the example of calculation. The average increase of O2 level in no inhibitor is calculated with average formula: ∑??=1 ?? 9.22 + 13.08 + 12.42 + 12.49 + 10.81 ??????? (?) = ??????? (?) = = 11.60 ? 5 The inhibition of each substance is calculated with this formula: ? ?? ?? ??ℎ?????? − ? ?? ??ℎ?????? 11.60 − ? ?? ??ℎ?????? ??ℎ??????? = × 100% = × 100% ? ?? ?? ??ℎ?????? 11.60 11.60−7.20 In CuSO4 0.100 mmol trial 1, ??ℎ??????? = × 100% = 37.9% 11.60 Table 7. The inhibition in CuSO4 0.100 mmol O2 increase Sum of Standard Sample Trial (104 ppm) Inhibition (%) Average (%) Square (SS) Deviation (%) ±0.02 1 7.20 37.9 2 7.28 37.2 CuSO4 3 6.50 44.0 37.8 135.9 5.2 0.100 mmol 4 6.79 41.5 5 8.28 28.6 7  ? ??? ?? ?????? (??) = ∑(? − ?? )2 ??? ?? ?????? ???????? ????????? (?) = √ ?=1 ? ??? ?? ?????? = (37.8 − 37.9)2 + (37.8 − 37.2)2 + (37.8 − 44.0)2 + (37.8 − 41.5)2 + (37.8 − 28.6)2 = 135.9 135.9 ???????? ????????? = √ = 5.2% 5 To compare the difference between each concentration on an inhibitor, statistical test Analysis of Variance F-Ratio Test6 will be done using CuSO4 as the example of calculation. Table 8. F ratio of CuSO4 Amount Sum of SS SS Average Total Average Total Sample (mmol) Square Within Between Inhibition (%) Inhibition (%) SS ±0.005 (SS) Group Group 0.100 37.8 135.9 0.200 39.4 51.1 CuSO4 0.300 49.3 174.0 588.1 52.1 5,251.8 4,663.8 0.400 59.6 87.0 0.500 74.6 140.0 o ?? ???ℎ?? ????? = ∑??=1 ??? = ??1 + ??2 + ??3 + ??4 + ??5 = 135.9 + 51.1 + 174.0 + 87.0 + 140.0 = 588.1 o Total Sum of Square is found using same formula as Sum of Square, but it uses all data of CuSO4 0.100 – 0.500 mmol and using the total average. o ?? ??????? ????? = ????? ?? − ?? ???ℎ?? ????? = 5,251.8 − 588.1 = 4,663.8 Table 9. F-Ratio of the inhibitors SS SS Degree of F- Confidence Critical Sample Between ??? Within ??? Freedom Ratio Level Value Group Group (??) CuSO4 4,663.8 588.1 39.80 Zn(NO3)2 49.0 4 11.7 20 20.89 (4,20) 95% 2.87 Pb(NO3)2 14.1 8.1 8.74 o ??? = ?????? ?? ????? − 1 = 5 − 1 = 4 o ??? = ?????? ?? ????? ????? − ?????? ?? ????? = 25 − 5 = 20 ?? ??????? ?????×??? 4,663.8×20 o ? − ????? = In CuSO4, ? − ????? = = 39.80 ??? ×?? ???ℎ?? ????? 4×588.1 6 Longstreet, D. (2012). How To Calculate and Understand Analysis of Variance (ANOVA) F Test.. Retrieved from https://www.youtube.com/watch?v=-yQb_ZJnFXw o 8 A chart of the inhibition activity will be made to compare the difference in the inhibition activity among different concentrations and among different samples. The previously obtained standard deviation will be used to represent the error bar on the chart. Chart 1. The comparison of average inhibition of several substances on catalase enzyme 100 97.3 97.5 97.6 98.6 96.3 90 94.1 97.3 97.8 97.8 80 95.2 70 Inhibition (%) 74.6 0.100 mmol 60 50 59.6 0.200 mmol 40 49.3 0.300 mmol 30 39.4 37.8 0.400 mmol 20 0.500 mmol 10 0 Copper (II) Sulfate Zinc Nitrate Lead (II) Nitrate Inhibitor Samples Discussion and Review The small error bar in Chart 1 showed the high precision of the data, indicating the data to be reliable to get a valid analysis and conclusion. Before H2O2 came into contact with catalase, no bubble of gas was observed because its decomposition required huge activation energy. After coming into contact with catalase, bubble of gas was observed, indicating a reaction was happening and producing O2 (oxygen gas) justified from the increase in O2 level in the O2 sensor. Catalase enzyme catalyzes the reaction by lowering the activation energy7. In catalase, there is a site in enzyme whose shape is specific to the shape of the substrate (H2O2), called active site8. The substrate binds to the active site allowing the enzyme to break down H2O2 into H2O and O2 (Fig. 2). The products (H2O and O2) will then leave the enzyme without changing the active site, allowing the used enzyme to catalyze another same reaction9, making the enzyme to be re-usable. The re-usability of enzyme was justified from the production of O2 until all H2O2 was broken down whether inhibitor was present or absent. The increase in O2 level within the presence of the solution of CuSO4, Zn(NO3)2 or Pb(NO3)2 was lower than when they are absent. This justified that those solutions were inhibitors that inhibited the activity of catalase enzyme. Metal ions are non-competitive inhibitor of enzymes10. They bind to a place in enzyme which is not the active site but altering the shape of the active site, causing molecule to no longer fit to the active site, therefore reducing the enzymatic activity11. In the structure of enzyme, there are several side- 7 Allott, A. (2014). Biology for the IB Diploma (p. 99). Oxford University Press. 8 Allot, A., & Mindorff, D. (2014). Biology Course Companion (pp. 96-97). Oxford University Press. 9 Jones, M., & Jones, G. (2010). IGCSE biology (p. 40). Cambridge University Press. 10 Chemguide.co.uk,. (2009). Enzyme Inhibitors. Retrieved 16 September 2015, from http://www.chemguide.co.uk/organicprops/aminoacids/enzymes3.html 11 Weem, M., Talbot, C., & Mayrhofer, A. (2014). International Baccalaureate Biology (4th ed., p. 229). Melton: IBID Press. 9 chains and polypeptide backbone containing oxygen, nitrogen and sulfur atoms with the potential to act as ligand to form chelate complex with various metal ions, where metal ions tend to bind to –SH (sulfhydryl functional group) in side chain of cysteine residues, disturbing the formation of disulfide bridges or replacing them with sulfur-metal-sulfur fragments12. This causes the enzyme to be denatured; the shape of the active site is changed, making the substrate to be unable to fit anymore, resulting in the inability of enzyme to catalyze the reaction. The enzyme bound to inhibitor is referred as enzyme-inhibitor complex. Sometimes in the presence of the inhibitors, the O2 level did not increase right away because of the lag phase of enzymes13 where the enzymes needed some time to separate themselves from the inhibitors after the enzyme-inhibitor complex was formed. This separation was possible because some non-competitive inhibitors could have a reversible effect14 because they did not bind very tightly with the enzyme (low enzyme affinity), and then the enzyme can catalyze the reaction after it is detached from the inhibitor. The presence of H2O2 as the substrate was the factor that triggered the separation between the enzyme and inhibitor. Although there was a delay in the O2 increase, the O2 level was measured only when O2 level started to increase to allow the comparison to be possible or else the O2 level when inhibitor is present may show no increase and cannot be compared among each other. From the results of the experiment and tested using F-test analysis of variance, the difference in the inhibition of a certain sample but using different amounts of inhibitor was statistically significant because the F ratios of the three samples are considerably greater than the critical value of confidence level 95% with degrees of freedom at (4,20). This implies that amount of inhibitor truly played role in the inhibition of enzymatic activity. The more CuSO4 was present, there was greater inhibition occurring on the enzyme (positive correlation) although the relationship was not clearly shown from the obtained results due to the large inhibition level. It was because more inhibitor caused more inhibition on the enzymes, therefore fewer enzymes could catalyze the reaction. However, this was slightly different for Zn(NO3)2 and Pb(NO3)2. In Zn(NO3)2, the inhibition of enzyme was the same for 0.300 mmol and 0.400 mmol of inhibitor, while the inhibition increased again when there was 0.500 mmol of Zn(NO3)2. However from the trend of results in Zn(NO3)2, the inhibition level observed was the same while actually it might increase slightly, but it was undetected from the O2 sensor. This deduction was made because there had already been huge inhibition when only 0.100 mmol of Zn(NO3)2 was present, and the increase in inhibition was not significant when amount of inhibitor was raised. For Pb(NO3)2, it was similar to Zn(NO3)2 that the inhibition was very large when only 0.100 mmol of it was present. Peculiarly, there was a decline in the inhibition level when more inhibitor was added after 0.300 mmol of inhibitor. This could happen because the inhibition had reached maximum level, or the binding of Pb(NO3)2 on the enzyme had been fully saturated and unable to bind to the enzyme, causing no increase in the inhibition level. Furthermore, Pb 12 Bylikin, S. (2014). Chemistry 2014. [S.l.]: Oxford University Press. 13 Worthington-biochem.com,. (2015). Enzyme Concentration (Introduction to Enzymes). Retrieved 16 September 2015, from http://www.worthington-biochem.com/introbiochem/enzymeconc.html 14 Chemguide.co.uk,. (2009). Enzyme Inhibitors. Retrieved 16 September 2015, from http://www.chemguide.co.uk/organicprops/aminoacids/enzymes3.html 10 (lead) can have several oxidation states15, causing lead compound could also behave as a chemical catalyst to decompose H2O216 by reducing H2O2 into H2O and O2 in a reduction- oxidation reaction, where Pb2+ as the reducing agent from Pb(NO3)2 was oxidized into Pb4+. Since it promoted the O2 production, the calculated inhibition level would decrease. Among the three tested inhibitors, the inhibition activity from high to low deduced from Chart 1 was respectively Pb(NO3)2, Zn(NO3)2 and CuSO4 although there was slight difference between Pb(NO3)2 and Zn(NO3)2, while there was clear difference between CuSO4 and other inhibitors. This showed that Pb(NO3)2 and Zn(NO3)2 had greater enzyme affinity compared to CuSO4; they could bind stronger to catalase, reducing its enzymatic activity more significantly. However, the enzyme affinity of Pb(NO3)2 might not differ much from Zn(NO3)2 since they had a quite similar inhibition level. However, there was a possibility that the anions, such as sulfate and nitrate ion, might get involved in enzyme inhibition, where nitrate ion inhibited catalase more than sulfate ion. So, it was possible that the difference in inhibition was not purely caused by the different metal ions, causing an unfair comparison. The investigated ions can be taken into the body through several kinds of food to improve the antibacterial activity of honey. Food rich in copper content are cooked oyster (5.71mg/100g), raw kale (1.5mg/100g) and cooked shiitake mushroom (0.9mg/100g)17. Food rich in zinc content are cooked oyster (78.6mg/100g), cooked lean beef short ribs (12.3mg/100g), and toasted wheat germ (16.7mg/100g)18. There is rarely any food containing lead because lead itself is a highly toxic metal and a very strong poison commonly found in art supplies or contaminated dust, capable of causing high blood pressure, kidney disorder and behavior problem19, therefore it needs to be avoided. Overall, food such as cooked oyster is the best source of both copper and zinc ion to improve the antibacterial activity of honey. Conclusion From the experimental results, it had been verified that CuSO4, Zn(NO3)2 and Pb(NO3)2 are the inhibitors of catalase enzyme in liver. The research question –“How does the amount (mmol) of each inhibitor (CuSO4, Zn(NO3)2 and Pb(NO3)2) affect the inhibition level (%) of catalase enzyme and how do they compare?”– can be answered. The hypothesis –“Greater amount of inhibitor will cause greater inhibition level and the greatest inhibition level will be shown by Pb(NO3)2 that uses greatest amount which could reach inhibition level around 90% since lead compound is well known to be toxic”– can also be evaluated. The amount of inhibitor played role on catalase activity, justified using F-test analysis of variance. The hypothesis is accepted in CuSO4, where more inhibitors caused greater inhibition. The similar condition also occurred when using Zn(NO3)2 although the inhibition was once same for different amount, but it increased again, therefore accepting the hypothesis. However, in 15 Clark, J. (2004). Oxidation state trends in Group 4. Chemguide.co.uk. Retrieved 15 March 2016, from http://www.chemguide.co.uk/inorganic/group4/oxstates.html 16 Catalytic Decomposition of hydrogen peroxide. Oxygen.atomistry.com. Retrieved 15 March 2016, from http://oxygen.atomistry.com/catalytic_decomposition_hydrogen_peroxide.html 17 Whitbread, D., & House, P. Top 10 Foods Highest in Copper. HealthAliciousNess. Retrieved 20 January 2016, from http://www.healthaliciousness.com/articles/high-copper-foods.php 18 Whitbread, D., & House, P. Top 10 Foods Highest in Zinc. HealthAliciousNess. Retrieved 20 January 2016, from http://www.healthaliciousness.com/articles/zinc.php 19 Healthline,. Lead Poisoning. Retrieved 20 January 2016, from http://www.healthline.com/health/lead- poisoning#Symptoms4 11 Pb(NO3)2, the inhibition level initially increased, but it decreased slightly beyond 0.300 mmol which could be caused by the catalytic activity of Pb(NO3)2 to decompose H2O2, so this partially fit with the hypothesis. It gave a possibility that each substance might have different trend in how they inhibit catalase. The inhibition level from low to high was respectively CuSO4, Zn(NO3)2 and Pb(NO3)2 where Zn(NO3)2 and Pb(NO3)2 had similar inhibition level, while there was significant difference between CuSO4 and other inhibitors. This accepts the hypothesis where lead compound has the greatest inhibition level, which is 98.6% when in 0.300 mmol. Overall, the hypothesis on the research question can be accepted. Further Research This investigation itself could not determine the type of the inhibitor; therefore it is suggested to determine the type of the inhibitor by finding the Vmax (maximum enzymatic rate) and Km (substrate concentration when rate of reaction is half of Vmax). In addition, doing a direct investigation on the effect of the ratio of catalase enzyme to its inhibitors against the antibacterial activity of honey would also be recommended. Evaluation and Improvement I am proud in doing this investigation as I could obtain sufficient data to answer my research question while reducing the experimental error. The low standard deviation indicates the high precision during experiment and sufficient amount of trials were conducted to support its precision. I learnt how to use F-test analysis of variance as a statistical tool to justify whether the difference in the inhibition level among different amount of inhibitor was significant or not. The opening of the flask fit with O2 sensor, preventing the escape of O2 gas and minimalizing the experimental error. I also used O2 sensor to monitor the O2 gas level which also gave out accurate control in the variable time. However, some experimental errors might still occur. Those errors and the possible improvements would be shown in Table 10. Table 10. Experimental errors and their improvements Limitation Improvement Zn(NO3)2 and Pb(NO3)2 powders did not fully Using smaller concentration so powder can dissolve in the solution despite being stirred and fully dissolve or using the commercially shaken, so the amount used can slightly deviate. sold Zn(NO3)2 and Pb(NO3)2 solution. pH of the solution was not controlled and might Using buffer to maintain the pH of the differ in each trial, affecting the enzymatic solution. Buffer pH 7 is suggested because activity, resulting in unfair comparison. the optimum pH of catalase is around 7. This experiment took long time and may decrease Catalase enzyme from liver can be the freshness of liver solution, causing catalase to replaced by commercially sold or standard perform poorly and unfair for all trials. catalase solution of known concentration. Slight amount of O2 gas may escape during the Use flask with longer neck or with two interval after adding catalase from liver solution necks for O2 sensor and adding the liver and before covering the flask with the O2 sensor. solution. The anions (NO3-, SO42-), could contribute to the Using substances containing same type of enzyme inhibition, and each anion could inhibit anions such as CuSO4, ZnSO4 and PbSO4 differently, resulting in an unfair comparison. to allow fair comparison. This method of investigation could be ineffective The inhibition will be investigated through because the increase in O2 level after 40 seconds how long the time taken for the O2 level to could be very small or insignificant. increase to a certain level. 12 References Airborne Honey,. Honey Enzymes. Retrieved 13 November 2015, from http://www.airborne.co.nz/enzymes.shtml Allott, A. (2014). Biology for the IB Diploma (p. 99). Oxford University Press. Allot, A., & Mindorff, D. (2014). Biology Course Companion (pp. 96-97). Oxford University Press. Bylikin, S. (2014). Chemistry 2014. [S.l.]: Oxford University Press. Catalytic Decomposition of hydrogen peroxide. Oxygen.atomistry.com. Retrieved 15 March 2016, from http://oxygen.atomistry.com/catalytic_decomposition_hydrogen_peroxide.htm Chemguide.co.uk,. (2009). Enzyme Inhibitors. Retrieved 16 September 2015, from http://www.chemguide.co.uk/organicprops/aminoacids/enzymes3.html Clark, J. (2004). Oxidation state trends in Group 4. Chemguide.co.uk. Retrieved 15 March 2016, from http://www.chemguide.co.uk/inorganic/group4/oxstates.html Dinov, I. (2015). F-Distribution Tables. Socr.ucla.edu. Retrieved 22 September 2015, from http://www.socr.ucla.edu/applets.dir/f_table.html Evolution, C., Martino, J., Walia, A., Martino, J., Walia, A., DeNicola, M., & Erickson, A. (2014). 25 Amazing Benefits & Uses For Hydrogen Peroxide. Collective-Evolution. 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Retrieved from http://stronglabnotebook.weebly.com/ Weem, M., Talbot, C., & Mayrhofer, A. (2014). International Baccalaureate Biology (4th ed., p. 229). Melton: IBID Press. Whitbread, D., & House, P. Top 10 Foods Highest in Copper. HealthAliciousNess. Retrieved 20 January 2016, from http://www.healthaliciousness.com/articles/high-copper- foods.php Whitbread, D., & House, P. Top 10 Foods Highest in Zinc. HealthAliciousNess. Retrieved 20 January 2016, from http://www.healthaliciousness.com/articles/zinc.php Worthington-biochem.com,. (2015). Enzyme Concentration (Introduction to Enzymes). Retrieved 16 September 2015, from http://www.worthington- biochem.com/introbiochem/enzymeconc.html 14 Appendix Here are some documentations of the experiment. Image 1. Liver solution Image 2. Laboratory set-up to measure O2 level Below is the complete raw data of the increase in O2 level. Table 11. Raw Data – Increase in O2 level Amount Initial (104 ppm) Final (104 ppm) Increase (104 ppm) Sample (mmol) Trial ±0.01 ±0.01 ±0.02 ±0.005 1 19.69 28.91 9.22 2 19.65 32.73 13.08 No inhibitor 3 19.65 32.07 12.42 4 19.71 32.20 12.49 5 19.60 30.41 10.81 1 19.44 26.64 7.20 2 19.60 26.88 7.28 0.100 3 19.65 26.15 6.50 4 19.38 26.17 6.79 5 19.60 27.88 8.28 1 19.57 26.39 6.82 CuSO4 2 19.52 26.72 7.20 0.200 3 19.51 26.85 7.34 4 19.55 25.95 6.40 5 19.38 26.76 7.38 1 19.83 26.81 6.98 0.300 2 19.59 25.79 6.20 15  3 19.51 24.74 5.23 4 19.37 24.47 5.10 5 19.53 25.40 5.87 1 19.69 24.43 4.74 2 19.44 23.49 4.05 0.400 3 19.43 24.58 5.15 4 19.44 24.71 5.27 5 19.41 23.64 4.23 1 19.72 22.48 2.76 2 19.88 21.72 1.84 0.500 3 19.48 22.66 3.18 4 19.62 23.07 3.45 5 19.55 23.06 3.51 1 19.26 19.89 0.63 2 19.32 20.00 0.68 0.100 3 19.38 20.09 0.71 4 19.10 19.74 0.64 5 19.00 19.75 0.75 1 19.34 19.94 0.60 2 19.34 19.87 0.53 0.200 3 19.17 19.69 0.52 4 19.27 19.89 0.62 5 19.26 19.77 0.51 1 19.10 19.27 0.17 2 19.08 19.37 0.29 Zn(NO3)2 0.300 3 19.09 19.50 0.41 4 19.08 19.39 0.31 5 19.08 19.43 0.35 1 19.08 19.31 0.23 2 19.02 19.22 0.20 0.400 3 19.04 19.35 0.31 4 19.36 19.93 0.57 5 19.09 19.31 0.22 1 19.07 19.42 0.35 2 19.18 19.49 0.31 0.500 3 19.16 19.44 0.28 4 19.18 19.45 0.27 5 17.90 18.10 0.20 1 19.12 19.49 0.37 2 19.17 19.43 0.26 0.100 3 19.25 19.52 0.27 4 19.21 19.46 0.25 5 19.21 19.45 0.24 Pb(NO3)2 1 19.30 19.62 0.32 2 19.37 19.74 0.37 0.200 3 19.25 19.41 0.16 4 19.37 19.57 0.20 5 19.61 19.83 0.22 16  1 19.34 19.51 0.17 2 19.44 19.64 0.20 0.300 3 19.38 19.52 0.14 4 19.28 19.41 0.13 5 19.50 19.65 0.15 1 19.83 20.05 0.22 2 19.62 19.82 0.20 0.400 3 19.84 20.09 0.25 4 19.17 19.54 0.37 5 19.29 19.53 0.24 1 19.99 20.47 0.48 2 19.98 20.43 0.45 0.500 3 19.39 19.76 0.37 4 20.27 20.83 0.56 5 19.64 19.92 0.28 Here is the average inhibition of the investigated inhibitors that has been calculated. Table 12. Average inhibition of several inhibitors Amount Average Sum of Square Sample Standard Deviation (%) (mmol) ±0.005 Inhibition (%) (SS) (%) 0.100 37.8 136 5.21 0.200 39.4 51.1 3.20 CuSO4 0.300 49.3 174 5.90 0.400 59.6 87.0 4.17 0.500 74.6 140 5.29 0.100 94.1 0.734 0.383 0.200 95.2 0.752 0.388 Zn(NO3)2 0.300 97.3 2.34 0.684 0.400 97.3 6.99 1.18 0.500 97.5 0.912 0.427 0.100 97.6 0.823 0.406 0.200 97.8 2.28 0.675 Pb(NO3)2 0.300 98.6 0.229 0.214 0.400 97.8 1.32 0.513 0.500 96.3 3.41 0.826 17 Here are the examples of the increase of O2 level in graphical display. Fig 5. Example of the obtained result in graphical display – no inhibitor trial 1 Fig 6. Example of the obtained result in graphical display – CuSO4 0.100 mol trial 1 18 Fig 7. Example of the obtained result in graphical display – Zn(NO3)2 0.100 mol trial 1 Fig 8. Example of the obtained result in graphical display – Pb(NO3)2 0.100 mol trial 1 19