To determine how pH affects catalase activity in potato
Hydrogen peroxide is a byproduct of several metabolic reactions in living cells. If allowed to accumulate it can cause oxidative damage. Therefore, it is constantly removed by catalase which decomposes it to water and oxygen. Catalase is found in nearly all living organisms exposed to oxygen. The optimum pH of catalase in potatoes is about 7. In this lab, you will observe the effect of changing pH conditions on catalase enzyme activity by reacting hydrogen peroxide with catalase present in potato pulp (pH 3.5 – 7.5).
- 3% hydrogen peroxide
- pH meter
- Potato pulp
- LabQuest oxygen detector apparatus
- Stir bar
- Stir plate
- Pipette tips
- Collect 50 ml of potato pulp in two separate 150 ml beakers
- Using 0.1 M NaOH and 0.5 M HCl, adjust the pulp to the following pH
- Student Group # 1: 7 and 3.5
- Student Group # 2: 7 and 4.5
- Student Group # 3: 7 and 5.5
- Student Group # 4: 7 and 6.5
- Student Group # 5: 7 and 7.5
- Transfer the sample with pH = 7 to the oxygen detection sample bottle and add a stir bar
- Add 50 ml of the 3% hydrogen peroxide to the sample bottle and begin oxygen detection using LabQuest. Collect data (oxygen in ppm) for 10 minutes
- Transfer your data to the whiteboard for the class to record
- Repeat the procedure with the second adjusted sample
- Create a graph showing oxygen concentration in ppm (y-axis) versus time in minutes (x-axis) for each pH level. Note: Since everyone will collect readings at pH 7, the pH 7 oxygen readings should be averaged. Plot all 6 curves on the same graph (i.e. pH = 3.5, 4.5, 5.5, 6.5, 7.0 and 7.5)
- Based on the graph, was 7.0 the optimum pH for catalase in potatoes? If not, which pH was optimum based on your findings? Is this supported by literature?
- Where does hydrogen peroxide come from in living systems? What is the role of catalase in living systems?
- How does pH affect the structure, shape, and biological activity of enzymes