Discussion
Discussion:
The equation involved in this experiment is 2H2O2 ( 2H2O + O2 . The reaction between manganese dioxide and hydrogen peroxide also will produce water and oxygen as the equation 2H2O2 ( 2H2O + O2 . Manganese dioxide did not write in the equation it is because it act as catalyst in this reaction.
The animal and plant organelle involved in this experiment is cell wall (plant), cell membrane (both animal and plant) and cytoplasm (animal and plant).
Pulping the liver causes the total surface area of the liver increase. As the total surface of liver increases, the rate of reaction between liver and hydrogen peroxide also increases. It has more total surface area compared to original liver therefore it will release more bubble when it react with hydrogen peroxide.
The liver in test tube 2 was boiled at temperature 95°C by placing inside the water bath. The high temperature will cause the denaturation of the enzyme (catalase) and the organelle inside the liver. Enzyme has an optimum temperature 37°C, when the temperature is much more higher than the optimum temperature it will cause the structure of the enzyme change. Change of the enzyme structure will cause the substrate bind to the active site. Hence, the enzyme lost it function. The denaturation of the enzyme will cause the negative result in test tube 2.
Liver and potato having the same type of enzyme which call catalase. The fresh liver give more effective result compared to potato cube, it is because the fresh liver have much more catalase activity than potato cube.
Boiling the manganese oxide will give positive result it is because it is an inorganic catalyst. But for boiled liver, it gives negative result because the enzyme in the liver has been denatured and lose it function. Furthermore, enzyme is a sensitive organic catalyst. Therefore, the result of manganese oxide is more effective than boiled liver.
The equation involved in this experiment is 2H2O2 ( 2H2O + O2 . The reaction between manganese dioxide and hydrogen peroxide also will produce water and oxygen as the equation 2H2O2 ( 2H2O + O2 . Manganese dioxide did not write in the equation it is because it act as catalyst in this reaction.
The animal and plant organelle involved in this experiment is cell wall (plant), cell membrane (both animal and plant) and cytoplasm (animal and plant).
Pulping the liver causes the total surface area of the liver increase. As the total surface of liver increases, the rate of reaction between liver and hydrogen peroxide also increases. It has more total surface area compared to original liver therefore it will release more bubble when it react with hydrogen peroxide.
The liver in test tube 2 was boiled at temperature 95°C by placing inside the water bath. The high temperature will cause the denaturation of the enzyme (catalase) and the organelle inside the liver. Enzyme has an optimum temperature 37°C, when the temperature is much more higher than the optimum temperature it will cause the structure of the enzyme change. Change of the enzyme structure will cause the substrate bind to the active site. Hence, the enzyme lost it function. The denaturation of the enzyme will cause the negative result in test tube 2.
Liver and potato having the same type of enzyme which call catalase. The fresh liver give more effective result compared to potato cube, it is because the fresh liver have much more catalase activity than potato cube.
Boiling the manganese oxide will give positive result it is because it is an inorganic catalyst. But for boiled liver, it gives negative result because the enzyme in the liver has been denatured and lose it function. Furthermore, enzyme is a sensitive organic catalyst. Therefore, the result of manganese oxide is more effective than boiled liver.