Enzyme Catalysis Essay

Enzymes ar proteins that act upon as catalysts to regulate metabolism by selectively speeding up chemical chemical receptions in the electric cell without being consumed during the process. During the catalytic action, the enzyme binds to the subst prescribe the reactant enzyme acts on and forms an enzyme-subst measure abstruse to convert the substratum into the product. Each type of enzyme combines with its specialized substrate, which is recognized by the shape.In the enzymatic reaction, the sign rate of activity is constant regardless of meanness because the image of substrate molecules is so large compared to the identification number of enzyme molecules working on them. When graphed, the constant rate would be shown as a line, and the slope of this linear administer is the rate of reaction. As meter passes, the rate of reaction slowly levels with less concentration of the substrate. This point where the rate starts to level is c all(a)ed the Kmax, in which the pe ak efficiency of enzymes is reached.In order to start the reaction, reactants require an initial come out of goose egg called activation energy. The enzymes work by trim the come in of free energy that must be absorbed so that less required energy leads to faster rate of reaction. The rate of catalytic reactions is change by the changes in temperature, pH, enzyme concentration, and substrate concentration. Each enzyme has an optimal temperature at which it is almost active the rate of reaction extends with increasing temperature up to the optimal level, but drops shrewdly above that temperature.Most enzymes have their optimal pH value that range from 6 to 8 with exceptions, and they whitethorn denature in unfavorable pH levels. An augment in enzyme concentration will gain the reaction rate when all the active sites are full, and an increase in substrate concentration will increase the rate when the active sites are non in all full. The enzyme apply in this lab is catal ase, a putting surface catalyst free-base in nearly all living organisms. Catalse is a tetramer of 4 polypeptide chains, each consisting of more(prenominal) than 500 amino acids. Its optimum pH is almost 7, and optimum temperature is about 37 C.The primary catalytic reaction of catalase decomposes hydrogen bleach to form urine and oxygen as shown by the equation 2 H2O2 2 H2O + O2 . Within cells, the function of catalase is to resist damage by the toxic levels of hydrogen peroxide by rapidly converting them to less dangerous substances. In this lab, we will show how catalase from 2 different sources (pure and spud extract) affects the rate of reaction by using titration to quantify and calculating the decomposition rate of hydrogen peroxide (H2O2) to water and oxygen gas with enzyme contact action. distinguish II Material and MethodsIn part 2A, I tried for catalase activity by using the seriological pipette to assign 10mL of H2O2 into a beaker. The serological pi pette was utilized in all transport of substances in this lab because of its tall quality and accuracy in measurement, especially with finespun control of volume and graduations that extend all the bearing to the top. Then, I employ another serological pipette to add 1mL of catalase in the beaker. by and by observation, I study and record the leave behinds. The above functioning was repeated with the poached catalase resolve using another beaker and serological pipette.I analyzed and recorded the results aft(prenominal) examination. In disunite 2B, I accomplished the baseline to determine the number of H2O2 present in the nominal solution without adding the enzyme. I employ serological pipettes (for the same tenableness mentioned above) to transfer 10mL of H2O2 in a beaker introductoryly labeled as baseline and 1mL of distilled H2O into the same beaker after that. Next, I added 10mL of 1. 0M H2SO4 into the beaker and mixed the solution by gently swirling the beaker. The sulphuric acid was employ to lower the pH and in that respectby renounceping the catalytic activity.Using the serological pipette, I removed 5mL of the commixture into a different beaker to test for the H2O2 amount through with(predicate) titration. This was done particularly through the titration technique because it can determine the concentration of a reactant in this case, rest amount of H2O2 with volume measurements. After recording the initial burette class period, I displace the assay beaker underneath a burette containing KMnO4 and little by little added the titrant with controlled drops while gently swirling the beaker until the coloration of the mixture turned permanently pink or brown.Then, I recorded the final burette reading. The potassium permanganate was specifically utilise because its excess amount will cause the solution to change color, and the amount used to change the color is proportional to amount of remaining H2O2. In Part 2D, I metric th e rate of H2O2 decomposition with enzyme catalysis in 5 different cartridge holder intervals of 10, 30, 60, long hundred, and one hundred eighty seconds. After labeling 5 beakers with each time interval, I transferred 10mL of H2O2 to each beaker with the serological pipette (for the same reason mentioned in Part 2A).For the 10 second time interval, I added 1mL of catalase extract and swirled the beaker for 10 seconds. Next, I added 10mL of H2SO4 to stop the reaction. I repeated the above procedure 4 more times, varying the 10 second time interval to 30, 60, 120, 180 seconds. Then, using the serological pipette, I removed 5mL sample from each of the 5 beakers and found the amount of remaining H2O2 by titration with KMnO4. The reason and procedure for titration was identical to those in Part 2B. Part IV DiscussionIn Part 2A, the enzyme activities of catalase and turn catalase were observed. gybe in to the selective information, the bubbles began to form in the mixture when the ca talase was poured into H2O2. The bubbles are the O2 that results from the crack-up of H2O2 as the catalase takes effect. In the case of poached catalase, there were no bubbles, which points to the absence of oxygen. This absence shows that unlike previous catalase, boiled catalase had no effect on the rate of reaction. The entropy supports the background information provided in the Introduction.The boiling of the catalase will alter its temperature above its optimal level, and that explains the importantly decreased reaction in the boiled catalase mixture compared to the catalase mixture. In Part 2B, the data represents the amount of H2O2 used in the reaction without enzyme catalysis, hence establishing the baseline. The unruffled data of initial reading and final reading was used to lick the baseline of 4. 7mL KMnO4, which is proportional to the amount of H2O2. The 4 groups combined data as a social class and besidesk the average of the 4 baselines by liminating the highest and lowest number and taking the average of remaining 2 numbers. The naturalized baseline was 4. 4mL. In the Charts A1 through B2 of Part 2D, the collected data of initial reading and final reading was used to calculate the amount of KMnO4 by subtracting the initial from the final. Since the amount of KMnO4 is proportional to the amount of H2O2 remaining, it was used to calculate the amount of H2O2 used in the reaction by subtracting it from the baseline.The computed data and the time intervals were graphed into 2 scatter plots divide by the type of catalase (pure and potato extract) with the lines of scoop concord drawn. The trend that should have shown in all 4 graphs was a steady increase from zero in the tooth root and a gradual leveling impinge on into a horizontal line towards the end. However, the actual results did not engagely come out as judge. In Graph A1, the data of convention 1 did steady increase in the beginning, but the amount in 120 seconds was off and the data of Group 3 started with a ban amount, which went up and subject throughout the time intervals.In Graph B1, the data of Group 2 started with a steady increase and close to declined towards the end although the graph started at a disconfirming number. In the same graph, the data of Group 4 also started negative and declined further, but it increased rapidly in the time intervals of 30-120 seconds and slightly declined at the end. surface of all the groups, the data of Group 2 was the most closest to the expected and the data of Group 4 was the most skewed. Overall, most groups had a line of best fit that began with a steady line that piecemeal smoothed out into a carouse after, which matched the expected graph.Generally, the rate was the highest in the beginning from 0 to 120 seconds because that was when the H2O2 and catalase were first combined and the substrate molecules outnumber the enzyme, allowing the enzyme to collide with substrates more frequently. The rate was lowest towards the end after 120 seconds because that is a while after the hydrogen peroxide began to be decomposed and there is less of the substrate to bind with the enzyme, which means slower rate of reaction. This corresponds to the some(prenominal) graphs line of best fit, which comparatively supports the background information.The rate of enzyme activity on the reaction would decrease with lowered temperature since the lowered average energizing energy of the molecules decrease the chances of the enzyme colliding and binding with the substrate. Also, the enzyme may be change with low enough temperature. The function of catalase is conquer by sulfuric acid. The sulfuric acid removes the enzymes function as a catalyst by transfiguring the protein conformation, which is critical to the binding of the enzyme to its substrate because the specificity is entirely parasitic on the structure.Part V Error analysis The data from Part 2D did not tout ensemble support the backgroun d information, which could be explained by errors that was do in the lab. One major error in the data was the negative amount of H2O2 used in the 10 second time interval with exception of Group 1. This may be the result of a human error made in the process of titration. A student may have had misgiving controlling the amount of KMnO4 with the burette, unable to record the exact amount at which the color of the mixture changed and adding too much KMnO4.This would have resulted in larger amount of KMnO4 used, thus, leading to a smaller amount of H2O2 used in the solution, which could result in a negative number. Another major error was the up and down version in the graph drawn from the data of Group 3 and Group 4. This could be ascribable to any measurement error made during the lab, much(prenominal) as the measurement for the sample used in the assay. The directions called for 5mL of the mixture to be titrated however, students may have measured wrong or mistaken the amount to more or less than 5mL.The assay of more than 5mL would result in a smaller amount of H2O2 used and the assay of less than 5mL would result in a larger amount of H2O2 used, which would account for the incorrect fluctuation of the graphs. Part VI Conclusions In this lab, I solve the following Part 2A ?Catalase reacts with H2O2 and produced H2O and O2 while boiled catalase does not engage with the substrate. This is shown by the formation of bubbles in the catalase mixture and the absence of bubbles, which indicates absence of oxygen, in the boiled catalase mixture. The function of catalase is affected by temperature because the boiling of the catalase denatured its catalytic ability, thus leading to absence of bubbles in the boiled catalase mixture. Part 2B ?The amount of H2O2 remaining in the catalyzed reaction is generally less than that in the established baseline due to faster rate in the decomposition. In the data of Group 1, the amount of KmnO4 (proportional to the remaining a mount of H2O2) is 4. 4mL, 4. 2mL, 3. 9mL, 4. 2mL, and 3. 9mL oer different time intervals.They are less than or equal to the baseline of 4. 4mL. Part 2D ?The rate of catalytic reaction changes over time the rate is constant in the beginning and piecemeal decreases towards the end, leveling off into a curve from a line. This is best illustrated in the best fit line of Group 2 data in Graph B1. ?The rate is highest when the reaction begins and becomes lower as time passes. The slope of the linear portion of all graphs in the data is greater than the slope of the gradually curving graph with increasing time interval.