Chemistry
Experiment 9 :
The effect of substrate concentration on enzyme activity
Objective
To find out the relationship between the substrate concentration and the rate of reaction by determining the number of bubbles produce
Hydrogen peroxide (H2O2) is a by-product of biochemical metabolism. An accumulation of hydrogen peroxide can be deadly, so it has to be decomposed. One of the decomposing factors is an enzyme called Catalase. Catalase breaks hydrogen peroxide into water and oxygen.
The chemical formula for the reaction is,
2H2O2 2H2O + O2
Since this is a decomposition reaction, it is exothermic. Although hydrogen peroxide can gradually degenerate itself, it decomposes much faster with the help of Catalase, because Catalase lowers the activation energy, the minimum energy barrier that hydrogen peroxide molecules have to overcome to decompose.
Since the enzyme lowers the activation energy, the rate of reaction increases without consuming enzyme. In this experiment, the substrate is hydrogen peroxide.
The gas pressure sensor is used because the reaction produces oxygen gas. The faster the reaction is, the faster the pressure will increase. Thus, by examining the change of pressure overtime, the rate will be calculated and analyzed. For this investigation, the initial rate of the reaction is examined.
Safety precautions
All enzymes are potential allergens therefore eye and skin should be avoided
Wear eye protection throughout the experiment and avoid rubbing eyes if the enzyme solution on hands in contact.
Chemical Reagants
Hydrogen peroxide solution (5-9 moldm-3)
Materials / Equipment:
Potato
Cock
Borer
Test tube
Knife
Pipette
Boiling tube
Delivery tube
Stopwatch
Method:
1. 5.0cm3 of 2.0moldm-3 hydrogen peroxide solution is poured into tube 1. The bug is placed and followed by water to tube 2 to the depth indicated in figure below.
2. Using a sharp knife and ruler, 10 pieces of 1mm thick disc of potato are cut. 3.
Objective
To find out the relationship between the substrate concentration and the rate of reaction by determining the number of bubbles produce
Hydrogen peroxide (H2O2) is a by-product of biochemical metabolism. An accumulation of hydrogen peroxide can be deadly, so it has to be decomposed. One of the decomposing factors is an enzyme called Catalase. Catalase breaks hydrogen peroxide into water and oxygen.
The chemical formula for the reaction is,
2H2O2 2H2O + O2
Since this is a decomposition reaction, it is exothermic. Although hydrogen peroxide can gradually degenerate itself, it decomposes much faster with the help of Catalase, because Catalase lowers the activation energy, the minimum energy barrier that hydrogen peroxide molecules have to overcome to decompose.
Since the enzyme lowers the activation energy, the rate of reaction increases without consuming enzyme. In this experiment, the substrate is hydrogen peroxide.
The gas pressure sensor is used because the reaction produces oxygen gas. The faster the reaction is, the faster the pressure will increase. Thus, by examining the change of pressure overtime, the rate will be calculated and analyzed. For this investigation, the initial rate of the reaction is examined.
Safety precautions
All enzymes are potential allergens therefore eye and skin should be avoided
Wear eye protection throughout the experiment and avoid rubbing eyes if the enzyme solution on hands in contact.
Chemical Reagants
Hydrogen peroxide solution (5-9 moldm-3)
Materials / Equipment:
Potato
Cock
Borer
Test tube
Knife
Pipette
Boiling tube
Delivery tube
Stopwatch
Method:
1. 5.0cm3 of 2.0moldm-3 hydrogen peroxide solution is poured into tube 1. The bug is placed and followed by water to tube 2 to the depth indicated in figure below.
2. Using a sharp knife and ruler, 10 pieces of 1mm thick disc of potato are cut. 3.