Form 4 Biology
Data Logger Experiment
Enzyme activity
Objectives
To investigate the factors affecting the rate of reaction of the enzyme catalase
Instruction
Use the Pressure Sensor to measure the change in gas pressure inside a boiling tube
containing hydrogen peroxide (H2O2) and enzyme catalase. Compare the rate of
enzyme activity for the same mixture under two different conditions: change in pH
and change in temperature. Use the Science Workshop to record and analyze the
data.
Apparatus and Materials (per group of 4)
Pressure Sensor and connector*
Rubber tubing #
5 boiling tubes and test tube rack
One-hole stopper
Water bath (at boiling temperature)
Measuring cylinder, 10 ml
Pestle and mortar
Cork borer
1 M hydrochloric acid
1M sodium hydroxide solution
3 % hydrogen peroxide solution
Ice cubes
Potato
Water
*The Pressure Sensor is calibrated and connected to the connector
#One end of the tubing is connected to the connector, the other end is connected to the
one-hole stopper
Method
1. Prepare some potato extract using the cork borer, pestle and mortar.
2. Pour 10 ml of hydrogen peroxide solution into a boiling tube.
3. Add 2 ml of potato extract to the boiling tube.
4. Put the one-hole stopper into the tube and click the ‘REC’ button.
5. Gently shake the boiling tube from side to side. Click the ‘STOP’ button after 150
seconds.
6. Remove the stopper from the boiling tube. Dispose the peroxide mixture and clean
the boiling tube thoroughly.
7. From the Graph display, find the minimum pressure and the maximum pressure.
8. In the result table on the next page, record the minimum pressure as the starting
pressure and the maximum pressure as the ending pressure.
9. Calculate and write down the difference in pressure and hence the enzyme activity.
10. Repeat the experiment with each of the following mixtures.
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Mixture (use 2 ml of potato extract and 10 ml of hydrogen peroxide for each)
A Potato cylinder (1cm length) + hydrogen peroxide
B Potato extract + 1ml sodium hydroxide + hydrogen peroxide
C Potato extract + 1ml hydrochloric acid + hydrogen peroxide
D Chilled potato extract + hydrogen peroxide
E Boiled potato extract + hydrogen peroxide
Result Table
Mixture Starting pressure
(kPa)
First
A
B
C
D
E
Ending pressure Pressure difference Enzyme activity
(kPa)
(kPa)
(kPa/min)
Discussion
Answer the following questions after the experiment.
1. What is the gas given out in the mixture? How can you test the gas?
2. What does the graph of the reaction between hydrogen peroxide and catalase (first
mixture) tell you about enzyme activity?
3. What is the effect of adding the base (sodium hydroxide) to the solution of peroxide?
What does this tell you about the range of conditions in which catalase may be
effective?
4. What is the effect of adding the acid (hydrochloric acid) to the solution of peroxide?
What does this tell you about the range of conditions in which catalase may be
effective?
5. What is the effect of cooling the catalase before adding it to the solution of
peroxide?
6. What is the effect of heating the catalase to boiling before adding it to the solution
of peroxide?
7. How did the effect of cooling compare to the effect of boiling the catalase? How
can you explain the difference between these two trials?
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8. Are the amounts of gas given out from each mixture the same? If not, which
mixture gives out the largest amount of gas? Which mixture gives out the least
amount of gas? Explain such differences.
Suggested Answers (Teacher’s reference)
1. The gas is oxygen. Put a glowing splint into the boiling tube, the splint relights in
oxygen.
2.
The graph shows that the rate of reaction is not constant, it changes as the
reaction progresses. The reaction rate is very fast in the beginning but slows
down gradually.
3.
The base raises the pH in the mixture. The rate of reaction is slower and
eventually the reaction stops. The pressure change is about half as great as when
no base is added. It means catalase can tolerate a temporary rise in pH, but when
it is a prolonged rise in pH, catalase becomes denatured.
4.
The acid lowers the pH in the mixture. The rate of reaction is very low or zero. It
means catalase cannot tolerate low pH and it is denatured.
5.
Cooling slows the rate of reaction. The reaction requires over twice as much time
to reach the maximum pressure.
6.
The reaction stops when catalase is heated to boil.
7.
Cooling does not denature the enzyme, it only makes the enzyme inactive or less
active, therefore the reaction rate is slower. Boiling denatures the enzyme and
therefore it loses the catalytic power.
8.
Each mixture gives out different amount of gas. The first mixture gives out the
largest amount of gas, the mixtures with acid added and boiled enzyme give out
the least amount of gas. This is because the enzyme is denatured in both mixtures,
no reaction takes place.
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