Enzyme Catalysis

Enzymes are proteins that act as catalysts to regulate metabolism by selectively speeding up chemical reactions in the cell without being consumed during the process. During the catalytic action, the enzyme binds to the substrate – the reactant enzyme acts on – and forms an enzyme-substrate complex to convert the substrate into the product. Each type of enzyme combines with its specific substrate, which is recognized by the shape.

There's a specialist from your university waiting to help you with that essay.
Tell us what you need to have done now!

order now

In the enzymatic reaction, the initial rate of activity is constant regardless of concentration because the number of substrate molecules is so large compared to the number of enzyme molecules working on them. When graphed, the constant rate would be shown as a line, and the slope of this linear portion is the rate of reaction. As time passes, the rate of reaction slowly levels with less concentration of the substrate. This point where the rate starts to level is called the Kmax, in which the peak efficiency of enzymes is reached.

In order to start the reaction, reactants require an initial supply of energy called activation energy. The enzymes work by reducing the amount of free energy that must be absorbed so that less required energy leads to faster rate of reaction. The rate of catalytic reactions is affected by the changes in temperature, pH, enzyme concentration, and substrate concentration. Each enzyme has an optimal temperature at which it is most active; the rate of reaction increases with increasing temperature up to the optimal level, but drops sharply above that temperature.

Most enzymes have their optimal pH value that range from 6 to 8 with exceptions, and they may denature in unfavorable pH levels. An increase in enzyme concentration will increase 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 not completely full. The enzyme used in this lab is catalase, a common catalyst found in nearly all living organisms. Catalse is a tetramer of 4 polypeptide chains, each consisting of more than 500 amino acids. Its optimum pH is approximately 7, and optimum temperature is about 37 °C.

The primary catalytic reaction of catalase decomposes hydrogen peroxide to form water and oxygen as shown by the equation: 2 H2O2 > 2 H2O + O2 . Within cells, the function of catalase is to prevent 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 potato extract) affects the rate of reaction by using titration to measure and calculating the decomposition rate of hydrogen peroxide (H2O2) to water and oxygen gas with enzyme catalysis. Part II – Material and Methods

In Part 2A, I tested for catalase activity by using the seriological pipette to transfer 10mL of H2O2 into a beaker. The serological pipette was utilized in all transfer of substances in this lab because of its high quality and accuracy in measurement, especially with delicate control of volume and graduations that extend all the way to the top. Then, I used another serological pipette to add 1mL of catalase in the beaker. After observation, I analyzed and recorded the results. The above procedure was repeated with the boiled catalase solution using another beaker and serological pipette.

I analyzed and recorded the results after examination. In Part 2B, I established the baseline to determine the amount of H2O2 present in the nominal solution without adding the enzyme. I used serological pipettes (for the same reason mentioned above) to transfer 10mL of H2O2 in a beaker previously 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 sulfuric acid was used to lower the pH and thereby stopping the catalytic activity.

Using the serological pipette, I removed 5mL of the mixture into a different beaker to assay for the H2O2 amount through titration. This was done particularly through the titration technique because it can determine the concentration of a reactant – in this case, remaining amount of H2O2 – with volume measurements. After recording the initial burette reading, I placed the assay beaker underneath a burette containing KMnO4 and gradually added the titrant with controlled drops while gently swirling the beaker until the color of the mixture turned permanently pink or brown.

Then, I recorded the final burette reading. The potassium permanganate was specifically used 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 measured the rate of H2O2 decomposition with enzyme catalysis in 5 different time intervals of 10, 30, 60, 120, and 180 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 – Discussion

In Part 2A, the enzyme activities of catalase and boiled catalase were observed. According to the data, the bubbles began to form in the mixture when the catalase was poured into H2O2. The bubbles are the O2 that results from the breakdown of H2O2 as the catalase takes effect. In the case of boiled 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 data 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 significantly 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 collected data of initial reading and final reading was used to calculate the baseline of 4. 7mL KMnO4, which is proportional to the amount of H2O2. The 4 groups combined data as a class and took the average of the 4 baselines by liminating the highest and lowest number and taking the average of remaining 2 numbers. The established 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 separated by the type of catalase (pure and potato extract) with the lines of best fit drawn. The trend that should have shown in all 4 graphs was a steady increase from zero in the beginning and a gradual leveling off into a horizontal line towards the end. However, the actual results did not exactly come out as expected. In Graph A1, the data of Group 1 did steadily increase in the beginning, but the amount in 120 seconds was off and the data of Group 3 started with a negative amount, which went up and down throughout the time intervals.

In Graph B1, the data of Group 2 started with a steady increase and slightly declined towards the end although the graph started at a negative 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. Out 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 gradually smoothed out into a curve 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 both graph’s line of best fit, which relatively supports the background information.

The rate of enzyme activity on the reaction would decrease with lowered temperature since the lowered average kinetic energy of the molecules decrease the chances of the enzyme colliding and binding with the substrate. Also, the enzyme may be denatured with low enough temperature. The function of catalase is inhibited by sulfuric acid. The sulfuric acid removes the enzyme’s 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 dependent on the structure.

Part V – Error Analysis The data from Part 2D did not completely support the background information, which could be explained by errors that was made 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 trouble 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 fluctuation in the graph drawn from the data of Group 3 and Group 4. This could be due to any measurement error made during the lab, such 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 conclude 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 amount of H2O2) is 4. 4mL, 4. 2mL, 3. 9mL, 4. 2mL, and 3. 9mL over 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 gradually 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.

Add a Comment

Your email address will not be published. Required fields are marked *


I'm Garry

Would you like to get such a paper? How about receiving a customized one?

Check it out