Catalase Design Experiment Essay

To determine what effect pH has on the rate of reaction in the presence of the enzyme catalyse. Observing the enzyme activity on potato rods will discover this. The amount of carbon dioxide produced indicates the rate of reaction.

Hypothesis

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The enzyme will have an optimum pH of around neutral. Either side of the optimum pH will have relatively high rates of reaction and far away from the optimum pH will have a low rate of reaction.

The enzyme catalase is found in many tissues including potato and liver. It is important because it breaks down the hydrogen peroxide formed in metabolism. Hydrogen peroxide is toxic and if it were not broken down it would kill the cells. As it does this, it liberates carbon dioxide that can be collected.

2H2O2 2H2O + O2

Materials and Apparatus

* Small test tube

* Potato

* Cork Borer

* Scalpel

* Boiling Tube

* Test tube

* 2 x 3 cm� prepared buffered solutions of pH 4, 6, 7, 8, 9

* 10 x 2cm� Hydrogen Peroxide

* Rubber Bung

* Glass Delivery Tube

* Rubber tubing

* Stopclock

* Beaker-250 cm�

* Water (approximately 250 cm�)

Control Sample

* Small test tube

* 5 x Boiled potato slices

* Boiling Tube

* Test tube

* 3 cm� prepared buffered solution of pH 4

* 2cm� Hydrogen Peroxide

* Rubber Bung

* Glass Delivery Tube

* Rubber tubing

* Stopclock

* Beaker-250 cm�

* Water (approximately 250 cm�)

* Heating apparatus – bunsen burner, gauze, tripod, heatproof mat, water bath, boiling tube

Variables

Constants

* Volume of buffer solution (3cm�) – controlled because different values will affect the reaction by acting on the catalase in a different way and possibly produce different volumes of gas – which is not an accurate test.

* Volume of hydrogen peroxide (2cm�) – this is what the catalase reacts with and in turn liberates gas – would produce different volumes off gas if not constant.

* Temperature (room temperature) – if this is not kept constant the rate of reaction will be altered, and different volumes of gas will be given off within the same time constraint.

* Concentration of hydrogen peroxide (2M) – if this is not kept constant the rate of reaction will vary for each concentration used – not producing accurate results.

* Amount of time for reaction (10 minutes) – if this is not maintained at the same value then the amount of gas collected will be difficult to calculate.

* Surface area of potato slices (width of cork borer, and 2 mm in length) – a larger surface area would expose more particles for the hydrogen peroxide to react with, therefore speeding up the reaction, a smaller surface area would reduce the rate of reaction, and therefore change the amount of gas liberated.

* Same volume of water in the beaker – if not kept constant; this would change the pressure, which would make the amount of gas collected vary within the given time.

Independent Variable

* pH of buffer solution

Dependent variable

* The amount of gas collected within 10 minutes determines the rate of the reaction

Procedure

Diagram

Original Method

1. Cut a length of potato tissue using a cork borer

2. Using a scalpel cut it into slices 2 mm thick

3. The cut slices should be kept under water to prevent drying out of any enzymes or tissue fluids

4. The apparatus should be set up as shown in the diagram

5. The slices should be placed in the tube with 3 cm� buffer solution

6. Then the 2 cm� of hydrogen peroxide put into the small test tube and lowered into the large boiling tube using an elastic band attached to the test tube and a piece of string joined onto the test tube

7. The bung should be inserted and the connecting tube led into the measuring cylinder

8. Then tip the large tube so that hydrogen peroxide spills over the potato and starts the reaction

9. Start the stopclock

10. Any gas displaced is collected in the measuring cylinder

11. After 10 minutes, the move the measuring cylinder from its position over the connecting tube and alter its vertical position so that the level of water in the cylinder is the same as that in the beaker

12. Note the level of water and calculate the amount of gas displaced

13. Keep the gas in the measuring cylinder

14. Calculate the amount of gas produced per slice per minute by measuring how much gas is in the cylinder

15. Repeat the process using 5 slices of boiled potato and pH4 buffer solution for the control sample. These conditions are used because as an enzyme is denatured, the substrate is acted on by the enzyme in a usual experiment – but the control shows that no other factors are influencing the experiment.

Modifications

Step 6 was changed because the tipping of the hydrogen peroxide was making inaccuracies within the experiment, as it was extremely difficult to ensure all hydrogen peroxide was emptied from the test tube. This technically meant that different amounts of gas would be liberated. There was also another motive to this modification – the bung would act as a piston and drives any air left into the glass delivery tube out into the upturned test tube.

Results

The amount of gas liberated when catalase enzyme reacts with H2O2 with varied pH environments

pH

4

6

7

8

9

CONTROL

Replicate 1 (Cm�)

1.25

5.80

4.01

3.50

3.34

0

Replicate 2 (Cm�)

1.32

2.80

4.42

3.72

3.65

0

Mean Values (Cm�)

1.285

2.80

4.215

3.61

3.495

0

Mean volume of O2 per slice per minute (Cm�)

0.0257

0.056

0.0843

0.0722

0.0699

0

Anomalies

The result for the first replicate of pH6 is obviously anomalous when in comparison to the surrounding results. I am putting this down to the sources of error within the experiment.

Sources Of Error

* The hydrogen peroxide could not be fully emptied from its container as each time the boiling tube was tipped, hydrogen peroxide would leak out-but re-enter when the boiling tube was returned to its upright position. A possible modification is to use a larger boiling tube where it would be possible to tip the smaller test tube fully.

* There was no temperature control – a change in temperature may affect the rate of reaction.

* The potatoes used may have been different, or cut from a different part of the potato – this may have an effect no the reaction. This also produces questions such as do different potato cells contain different amounts of catalyse? Alternatively, different potatoes contain more or less enzyme? A way to eliminate this error is to use slices cut from the same rod of potato.

* It is very hard to cut the potato rods into slices exactly 2mm thick – therefore some variety in the size of the rods will occur, meaning difference in surface areas. A modification to reduce errors of this kind could be to use larger disks – making it easier to cut the slices. Then this would entail its own problems – usage of a larger container and disks in contact with each other.

* The potato disks touched each other in the hydrogen peroxide solution, and therefore did not get full surface area exposure with the hydrogen peroxide. If separated, this could produce a change in the reaction speed – presumably faster. A modification for this limitation would be to use a larger boiling tube – or other carrier apparatus, and use separators to prevent the disks from touching.

Discussion

The results obtained showed that the optimum pH that the enzyme catalyse within potatoes works at is pH 7. This is because the maximum oxygen was given off as my results show.

The pH of a solution can have several effects of the structure and activity of enzymes.

For example, pH can have an effect of the state of ionisation of acidic or basic amino acids. Acidic amino acids have carboxyl functional groups in their side chains. Basic amino acids have amine functional groups in their side chains. If the state of ionisation of amino acids in a protein is altered then the ionic bonds that help to determine the 3-D shape of the protein can be altered. This can lead to altered protein recognition or an enzyme might become inactive.

Changes in pH may not only affect the shape of an enzyme but it may also change the shape or charge properties of the substrate so that either the substrate cannot bind to the active site or it cannot undergo catalysis.

The reactant in a biological reaction is known as a substrate. Enzymes will only catalyse specific reactions of certain substrates. Some enzymes will catalyse more substrates than others, but in general enzymes are so specific that they are only able to catalyse a single reaction. The reason that enzymes are so specific is that the joining of the enzyme and the substrate must have a perfect fit. The three-dimensional structure of an enzyme has many different crevices, which will only allow certain substrates to join. Just like a lock and key, there must be a perfect fit between the enzyme and substrate for the reaction to occur. Once this combination occurs, the biological reaction will be able to take place at an extremely higher rate.

The forces that maintain the shape of the enzyme include salt bridges, hydrogen bonds, disulphide bonds, and hydrophobic interactions. Any chemicals or temperature changes can alter these forces within the enzyme, and limit the ability of the enzyme to catalyse a reaction. pH denatures catalase enzyme as it alters the forces stated above.

In this experiment, catalysis of the decomposition of hydrogen is studied with the use of a potato extract. Hydrogen peroxide is a chemical that is produced in certain human cells, and would be deadly if consumed. A catalyst, which is found in red blood cells, prevents the accumulation of this chemical. With the use of a potato extract, the biological reaction that occurs in every human body can be observed.

Conclusion

My hypothesis was proved correct as the results show – the graph shows a peak over the optimum pH7 and a smooth decreasing slope either side away from the optimum -which was expected.

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