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 …show more content…
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 …show more content…
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