Catalase Lab

Assessment of Catalase Function Lab
Introduction
The purpose of this lab report was to test and measure the rate of substrate destruction by an enzyme, we tested the destruction of hydrogen peroxide by the enzyme catalase. Hydrogen peroxide is a poisonous by product of metabolism that can damage cells if it is not removed. Catalase is an enzyme that speeds up the breakdown of hydrogen peroxide into water and oxygen gas.
H2O2 + catalase → H2O + O2 A catalyst is a substance that lowers the activation energy required for a chemical reaction, and therefore increases the rate of the reaction without being used up in the process. Catalase is an enzyme, a biological catalyst. Hydrogen peroxide is the substrate for catalase.
Notes: 1.) One
unit of pure catalase decomposes 1 ɥmole H2O2 per minute at 25°C at pH7. 2.) Assume that 1 mL of KMnO4 reacts with 1 mL of H2O2. When the time is increased I believe that the amount of KMnO4 used will increase.

Part A: A visible change will occur when the catalase solution is added, another change will occur when it is boiled, then cooled and added to the solution.
Part B: I predict that the amount of KMnO4 used will be approximately 5 mL, or less.
Part C: I predict that for each time interval, it will use slightly more KMnO4, and therefore slightly less H2O2
Materials and Methods The materials and methods we used were supplied by the teacher.
Results
Part A: As the 1 mL of fresh catalase solution was added to the 10 mL of the 1.5% H2O2 solution bubbles formed. When we transferred the 1 mL of boiled and then cooled catalase extract to 10 mL of 1.5% H2O2 it became cloudy.
Part B: It took 4.5 mL of KMnO4 to titrate the H2O2 in the solution. The readings for the amount of KMnO4 used in titration in are found in Table
1
Part C: Readings for the digestions and titrations for the samples at 10, 30, 60, 120, 180, and 360 seconds from are displayed in Table 1.
The amount of peroxide used against time is displayed in Figure 1.

Table 1: Qualitative Catalyzed Decomposition | KMnO4 (mL) | Time (sec) | | 10 | 30 | 60 | 120 | 180 | 360 | Baseline | 4.5 | Final Reading | 23.5 | 25.3 | 27.6 | 32.5 | 35.5 | 44.9 | Initial Reading | 21 | 23.5 | 25.5 | 28.5 | 33 | 40.6 | KMnO4 Used | 2.5 | 2.3 | 2.1 | .5 | 0.4 | 0.3 | H2O2 Used | 2 | 2.2 | 2.4 | 4.0 | 4.1 | 4.2 |

Figure 1
Slope (10 to 30 sec) = Δy Slope (30 to 60 sec) = Δy Δx Δx = 2.2 – 2 = 2.4 – 2.2 30 – 10 60 - 30 = 0.01 = 0.0067
Slope (60 to 120 sec) = Δy Slope (120 to 180 sec) = Δy Δx
Δx = 4 – 2.4 = 4.1 - 4 120 – 60 180 - 120 = 0.0267 = 0.00167
Slope (180 to 360 sec) = Δy Δx = 4.2 – 4.1 360 - 180 = 5.5 x 10-4
Discussion
Part A: In adding catalase to H2O2, a chemical change was observed in the production of bubbles (O2), which displays that the enzyme catalyst was speeding up the decomposition of the H2O2. The enzyme was the catalase, while the substrate was the H2O2. My predictions for this section were correct a visible change did occur when the catalase solution was added, O2 gas was formed, and then became cloudy when boiled.
Part B: The baseline was determined as 4.5 mL, it was determined to show the extent of the reaction and the amount of H2O2 that was in the solution after the catalyzed reaction. The enzyme concentration determined the baseline. My predictions were correct, the baseline was determined as less than 5 mL. This number should have been lower; this could be due to incorrect measurements
Part C: Firstly the graphed data displays an almost constant rate of change. The rate of reaction was lowest from 180 to 360 seconds, and the highest from 60 to 120 seconds. It was lowest during the end because there was a decreasing amount of substrate available due to denaturation, it would not have surpassed the amount of baseline used (4.5 mL). The longer the solution was swirled for, the more H2O2 was used. This is why the shape of the graph (Fig. 1.) increased steadily, and then began to even out. To discuss the effect of boiling on enzyme activity, the boiling subjected the enzyme to a high temperature and a non-ideal condition for its activity, which denatured its structure. The reaction was not catalyzed because of the inability to form the enzyme-substrate. When sulphuric acid was added to the catalase it lowered the pH below the range where catalase was functional. Lowering the pH range took away the H+ ions, therefore denaturing it. This change in pH caused the side chains of the amino acid to change its charge which resulted in a change of protein. My predictions were correct for this section. I predicted that the lower the temperature of the enzyme activity the lower the kinetic energy which causes fewer collisions. I predicted this because I know that the higher the kinetic energy the higher the temperature and the more collisions, lowering the temperature would have an opposite effect. Therefore lowering the temperature would create a slower reaction rate. To test the effects of varying pH, temperature, or enzyme concentration on rate of reaction I would design an experiment. I would gather beakers of hydrogen peroxide in controlled amounts, and add varying amounts of catalase to each. Each would react for 30 seconds, and then would be stopped with 5 mL of H2SO4. Lastly it would be titrated to see the effects. .