Introduction Enzymes are catalytic proteins. The purpose of a catalyst is to speed up metabolic reactions by lowering the free energy of activation or activation energy. Activation energy is known as the amount of energy needed to push the reactants over an energy barrier, so that the downhill part of the reaction can begin (Campbell 151). In an enzyme catalyzed reaction, the enzyme binds to its substrate, which is the reactant an enzyme acts on. In the reactions, the enzymes are very specific, where only a restricted region of the enzyme molecule binds to the substrate.
This region is known as the active site (Campbell 152). The specificity of an enzyme results from its shape; the shape is form by the amino acid sequence since enzymes are proteins. If the shape of the substrate fits the shape of the active site, the enzyme will alters its shape so the active site embraces the substrate and maintains a firm grip, known as induced fit. This allows for great variations of enzymes. Abstract In this experiment, my lab partners and I tested how time effects a catalase reaction. The amount of hydrogen peroxide was recorded after the reaction for the certain time given has taken place.
We used sulfuric acid to stop the reaction with the catalase from occurring. This process is known as denaturing (Campbell 152). The potassium permanganate in this experiment was used as hydrogen peroxide indicator. It determined the amount of hydrogen peroxide remaining after the reaction occurred. Based on our experiment we observed that the time does play a crucial role on the catalase reaction because as we increased the time of the reaction, the amount of hydrogen peroxide decreased. Problem: How does time affect the amount of hydrogen peroxide remaining after the catalase and H2O2 reaction?
Hypothesis: I believe if you shorten the time of the catalase and H2O2 reaction, then there will be more H2O2 will be present. Materials and Methods Materials: Beakers, 50mL of catalase, 60 mL of H2O2, 60 mL sulfuric acid, 10 mL of KMnO4, distilled water, 20mL potassium permanganate, syringes, 10 beakers, paper, pen, stopwatch, gloves, goggles Procedure: 1) Set of your lab table and gather all materials needed. 2) Label 5 beakers that will receive the catalase with a time of 10,20,30,40 and 50 seconds. 3) Repeat step two for the baseline reaction that will not be receiving the catalase.
4) Gather the syringes and separate them with the appropriate chemicals. Get one syringe for the hydrogen peroxide, sulfuric acid, distilled water, and the catalase. 5) In the first beaker labeled 10 seconds, add 10 ml of H2O2 and 10ml of the catalase then wait 10 seconds and add 10ml the sulfuric acid to stop the reaction. 6) Add the potassium permanganate to the beaker and mix well until a brown or pink color is present. This measures the amount of hydrogen peroxide present after the reaction has taken place.
7) Repeat steps 5 and 6 for the rest of the other time-labeled beakers reciveing the catalase and record the amount of hydrogen peroxide present after the reaction has occurred with the potassium permanganate. 8) Now test the baseline reaction without the catalase by repeating steps 5 and 6. 9) Record and organize your data. 10) Clean up your lab table and materials.
Variables: Independent Variable: The amount of time of the H2O2 catalase reaction. Dependent Variable: The amount of time of H2O2 left after you stop the reaction with the sulfuric acid.
Control: The beaker with distilled water, H2O2 (hydrogen peroxide), H2SO4 (sulfuric acid) and KMn04 without the catalyse. Constant: H2O2, sulfruric acid, 10 mL catalase. Data: Time (seconds) 10 seconds20 seconds30 seconds40 seconds50 seconds Amount of H2O2 remaining with the presents of the catalase 1. 7 mL1. 5 mL1. 0 mL. 8 mL. 6 mL Conclusion In conclusion, my hypothesis was correct, the shorter the reaction time of the catalase the more H2O2 was present. So, therefore the amount of H2O2 decreased as the time of the reaction increased.