Action of the enzyme catalase.

• H2O2                                                        • thermometer                        

• potato                                                • beaker

• liver                                                        • mortar

• sand                                                        • wooden splint

                                                        • ruler                                

Procedure:

1. 2 cm3 of H2O2 is put in 8 test tubes.
2. 5 cubes (1cm3) of liver and potato is cut.
3. The followings are added to test tubes 1 to 8.

1. a sprinkle of sand
2. 2 cm3 of MnO2
3. one cube of liver
4. tube 3. is repeated, the temperature of the H2O2 is measure before and after the addition of the liver
5. tube 3. is repeated, a bung is put in the test tube, firmly hold and tested for oxygen (a glowing wooden splint is used)
6. one cube of boiled liver (boiled in a beaker of water for 15 min)
7. one cube of liver ground up (ground in a mortar with some sand)
8. one potato cube

Observations:

1. the sand sank to the bottom of the tube, the solution was yellowish
2. the resulting solution was black, a bit of foam was formed
3. yellowish solution formed with foam on the top
4. temperature of H2O2 was 27 oC, after the addition of the liver it was 32 oC
5. the wooden splint caught fire
6. yellowish solution formed, no foam
7. lots of bubbles were formed, yellowish solution
8. yellowish solution formed with foam on the top

Conclusion and evaluation:

1. Sand is the negative control because it does not contain the enzyme catalase. If any other materials act as the sand did (not producing bubbles, foam on the top of the solution), those materials do not contain catalase.

2.        MnO2 does not contain the enzyme catalase because it is an inorganic compound.

3.        Liver contains the enzyme catalase. Foam appeared as hydrogen peroxide was broken down to water and oxygen.

4.        The temperature of H2O2 increased after the addition of liver because the breaking down of H2O2 is an exergonic reaction, that means that heat is produced during the reaction.

5. The H2O2 was broken down to H2O and O2, the glowing wooden splint caught fire due to the presence of the oxygen.
6. The enzyme has disappeared due to the boiling.
7. The concentration of the catalase is higher by grounding up the liver.
8. Potato also contains the enzyme catalase.

Part II.

Materials:                                                Equipments:

• H2O                                                        • pH test strips        

• HCl                                                        • test tubes

• NaOH                                                        

• H2O2                                                        

• liver                                                        

Procedure:

1. Solutions of different pH-s are prepaired.
2. Test tubes are filled with water and drops of HCl or NaOH solutions are added
3. The pH is measured with pH test strips
4. Liver is put into the prepaired solutions
5. The cubes are removed after 15 min and are put into 2 cm3 of H2O2.

Observations:

The loss of enzyme activity when catalyse was exposed to acidic or basic solutions:

HCl (acidic)

pH 6 – a lot of foam was formed

pH 4 – few foam was formed

pH 2 – no foam was formed

NaOH (basic)

pH 8 – few foam was formed

pH 10 – more foam was formed

pH 12 – a lot of foam was formed

Conclusion:

The enzyme catalase is more active as the pH grows.

Suggestion for Part I. and II.

The test tubes should be washed befor the experiment because some chemicals might remain in them from previous experiments, disturbing the actual experiment and providing different results.

Other catalase-containing materials might be used, e.g. yeast.

All the variables must be kept the same for all the experiments. Variables that must not be altered: Temperature, concentration of liver, potato and hydrogen peroxide, type of liver and potato, volume of hydrogen peroxide, air pressure, humidity.

Repeat the experiment many times to get an average result.

The amount of substrate experimented should be in proportion with the amount of enzyme because as the substrate concentration increases, the rate of reaction will go up at a directly proportional rate until the solution becomes saturated with the substrate hydrogen peroxide (Fig.1.). When this saturation point is reached, then addision of extra substrate will make no difference. The rate steadily increases when more substrate is added because more of the active sites of the enzyme are being used which results in more reactions so the required amount of oxygen is made more quickly. Once the amount of substrate molecules added exceeds the number of active sites available then the rate of reaction will no longer go up, it will level off. This is because the maximum number of reactions are being done at once so any extra substrate molecules have to wait until some of the active sites become available.

The enzyme activity exposed to different pH was compared only within the acidic and within the basic pH, however all the pH should have been compared. Fig. 2. Shows the effect of pH on enzyme activity. The optimum rate is between pH 3 and 12.

Sources:

http://neiljohan.com/projects/biology/enzymes.htm