Peroxide Experiment

Topics: Catalase, Oxygen, Poison Pages: 3 (745 words) Published: February 20, 2013
Hydrogen Peroxide Experiment
To determine whether living tissue contains catalase enzyme (which is liver) and investigate in which circumstances the enzyme works efficiently. Background
Catalase in an enzyme that the breakdown of hydrogen peroxide into oxygen and water, since hydrogen peroxide is a poisonous chemical that produced continually by all reaction in living cells. I will be using animal tissues to determine if the enzyme catalase will create a reaction. Tools

* Chicken Liver (Fresh)
* Test Tube
* Hydrogen Peroxide
* Small knife
* Weight Scale
* Stopwatch (more than 1)
1. Gather all needed materials
2. Start cutting the liver, and measure into 5 different mass (1-5 gr) 3. Put Hydrogen Peroxide to the test tube
4. Insert all cut livers to the tubes at the same time
5. Start record the time of 3 minutes
6. With another stopwatch, record the time of each reactions 7. Insert data to the data table
The more liver in grams are put in the test tube filled with hydrogen peroxide, the larger the reaction would be due to its mass. Variables:
Independent: The amount of liver in grams
Dependent: The height and time of hydrogen peroxide reaction Controlled: Time, the amount of hydrogen peroxide, test tubes. RESULT DATA TABLE
Test Tube | Liver (Grams)| Time (Minutes)| Hydrogen Peroxide (Ml) | Result (Height)| Time of Reaction| A| 1| 3| 2| 0.3 cm| 7.4 s|
B| 2| 3| 2| 3.1 cm| 24 s|
C| 3| 3| 2| 6.4 cm| 50 s|
D| 4| 3| 2| 10.5 cm| 84 s|
E| 5| 3| 2| 13 cm| 168 s|

So from the data below, the more liver in grams were inserted in tube filled with hydrogen peroxide, the reaction of the liver to hydrogen peroxide will be bigger. It is shown in the column of Height result that Test Tube A, with the smallest amount of liver is only producing 0.3 cm, while Test Tube E, with the largest amount of liver produces 13 cm of reaction on...
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