Investigate the action of the Enzyme Catalyse

Linsey Belford L5V

Biology Coursework

To Investigate the action of the Enzyme Catalyse

Aim: The this experiment is to investigate the action of the enzyme catalase using potato and hydrogen peroxide. I am trying to find out the productive rates of the hydrogen peroxide and the potato. I am changing the surface area of the potato but I am still using the same mass to make this fair.

Prediction: I predict that the larger the surface area the faster the productive rate and the more gas produced because the catalase has more active sites for the hydrogen peroxide to fit into.

Catalysts are used to speed up specific reactions in cells. They are all very unique as each enzyme only performs one particular reaction.

Catalase is the enzyme which I am using. It is in the cells of living organisms. It is found in potato and liver so I am using potato in my investigation. The catalase speeds up the decomposition of Hydrogen Peroxide (H2O2) into water and oxygen.

catalase

Hydrogen Peroxide Water + Oxygen

2H2O2 2H2O + O2

It is necessary for Hydrogen Peroxide to be broken down as it is produced by many chemical reactions in the cells and it is very dangerous. As it is such a dangerous substance it must be broken down immediately.

Enzymes are proteins. The molecules have a three dimensional shape. This shape contains a small space, dent or area of which it is exactly the correct size and shape for a molecule, of the enzymes substrate, to fit into. In my case for this experiment the catalase molecules have a space for the hydrogen peroxide to fit into. This space is called an active site. When the hydrogen peroxide molecule slots into the active site, the catalase ‘adjusts’ the hydrogen peroxide molecule and pulls it out of shape, it then splits into the product molecules, which are oxygen and water. Once the product molecules have left the active site, the active site is then ready to do the same to another hydrogen peroxide molecule. The active site and the molecule are often known as ‘lock and key’.

(Reference: Jones and Jones, )

The diagram below shows how an enzyme works:

There will be several conditions that will affect my experiment. They will affect both the results and accuracy of my experiment. These are:

The temperature: The room temperature can have an effect on my experiment. I can control this by keeping the catalase and Hydrogen Peroxide in the same room at the same temperature. An increase in temperature will speed up the production of O2 unless the temperature is over 40°C. This is because when the Hydrogen Peroxide is heated the molecules vibrate more meaning they have more energy to carry out the reaction. If the temperature is over 40°C the catalase will be damaged by the heat therefore it will not work.

Concentration: I will use the same concentration of 15 volume throughout my investigation. This is because the stronger the concentration the faster the reaction so it would be an unfair test if I used different concentration throughout. The reason a reaction usually happens faster is because if there is a higher concentration it means there are more parts (molecules) to collide and react within the same amount of ml.

E.g. In the concentration of 10 vol. there are 100 active molecules that are going to react but in the concentration 5 vol. There will be 50 active molecules that will collide and react. (This was only an example. Not actually correct)

This is a preview of the whole essay

Surface Area: This is the factor that I am going to change. I will measure it as accurately as possible. The diameter will always be the same, as I will use the same cork borer for every cylinder. The length will be cut as accurately as possible using a ruler. To measure the surface area of a cylinder (which is the shape I am using) the formulae is 2πr² + (πdxh). This is working out the net of the shape, as I do not want the volume.

Mass: The mass of potato will always be the same. It will be as accurate as possible and I will use the weighing scales to do this.

Amount of H2O2: The amount of Hydrogen Peroxide will always be the same. This is so that the same amount of Hydrogen Peroxide is “attacking” the potato at once and it is fairer.

Preliminary: During my preliminary experiment I tested the accuracy of my experiment and worked out the quantities needed for my actual experiment. Using a conical flask and a bung with a syringe attached I measured the amount of oxygen produced. I am using different surface areas of potato but they will all have the same mass. They will be cylindrical in shape and will all start off of the length of 4cm before being divided (or not in one case). The potato is put in groups and the surface area recorded of each group. I used 10cm³ of hydrogen peroxide and measured this accurately with a measuring cylinder. The potato was placed inside the conical flask with the bung on. The conical flask had a delivery tube attached to it and this was fed into a beaker of water. To work out the gas I was using the displacement of water and to do this I had to make sure that I did not let any air in. I used 15 vol (concentration) of hydrogen peroxide. I found that this was best as it was not the strongest but it was still fairly strong.

The equipment was set up as shown below:

The Hydrogen Peroxide was sucked up using the syringe and 10cm³ was sucked up. The syringe with the hydrogen peroxide in it was attached to the bung on the conical flask. Everything was ready so the Hydrogen Peroxide was released into the conical flask. The stop clock was started immediately and the gas (oxygen) started to displace the water. I think it will be more accurate in my actual experiment to use a burette because this has more accurate measurements and is easier to use.

The syringe is left attached to the bung and this is an error which I made during one of my preliminary experiments. I did this experiment again. Every thirty seconds the volume of gas was checked and recorded.

My results for my preliminary experiment for the surface area of 14.14cm² were:

I have not included a column for the concentration, as the concentration is the same for every experiment.

In my preliminary experiments I only recorded results for 2 minutes, as it would take too long otherwise to carry out the experiments and only one experiment was carried out for each surface area.

The amount of Hydrogen Peroxide will always be the same. This is to keep the experiments under all the same conditions except for the variable which is the surface area.

While doing my preliminary experiment I realised several things. I did not seem to have enough time and was not organised enough. I found it easier prepare the potato before starting any of the experiments so that is what I will do.

Actual Method

Apparatus needed:

Conical flask with delivery tube, bung and 10cm³ syringe,

Litre beaker,

15vol (concentration) Hydrogen Peroxide,

Burette,

2 Large Potatoes,

Stop Clock,

Cork Borer (1cm diameter),

Chopping Knife,

Chopping Tile,

8 weighing boats

Heatproof Mat,

Weighing Scales,

Ruler.

This is how the apparatus should be set up:

This will be the actual method for my experiment:

Cut out the potato using the cork borer and cut into the required sections and surface areas. Weigh the cylinders of potato separately in the weighing boats and measure the lengths that they are cut. Make sure the surface area has been recorded accurately. Keep the potato in the weighing boats to avoid confusion. Use a ruler with millimetres for cutting.
Measure 10cm³ of hydrogen peroxide (15 vol) and suck this up into the syringe. Always wear safety glasses, as hydrogen peroxide is very irritant to the skin. (To measure against the measurements on the side, use the bottom line of the base on the syringe)
Be careful with the sharp knife when cutting. Fill the burette with water making sure the tap is closed. Let the gas in to the level 50ml (it is a reading on the side) this is actually only a volume of approx 10 cm³. Let the air in to the top reading on the burette. The measurements should be taken using the bottom of the menisci.
Connect up the delivery tube to the burette in the tub of water ready for the displacement of air.
Place the potato into the conical flask and place the bung on.
Place the syringe onto the bung and add the hydrogen peroxide to the potato in the conical flask.
Start the stop clock and every 30 seconds record the volume of gas.
After a five minutes (300 seconds) the last measurement should be taken and then recorded and the experiment should be repeated for the same surface area. For more accurate experiments the experiment for each surface area should be done at least twice
If the results are similar they are consistent. If they differ quite a lot, another experiment should be taken for accuracy. The set of results that are most similar should be used. After the experiments are completed the next surface area should be done and two of each experiments should take place for each surface area.
The results should be collected in a results table. The conditions in each experiment should be exactly t he same except for the surface area which is the variable. The used Hydrogen Peroxide should not be used again and a fresh piece of potato should be used for each experiment.

I will use the burette to measure the amount of water displaced. When the water is replaced it is done so by a gas. The ruler will be mainly used for measuring the length of the potato. All the equipment such as the conical flask, bung, delivery tube, water and the burette are to be used for collecting the gas. It will be as accurate as possible as I will do my best not to let any of the gas escape.

I think my method is a good way of doing this experiment as it is easy to perform and can collect results accurately and efficiently.

My experiment went well and I collected the results in a safe way. There was only one experiment that did not have results that fitted in so I did this experiment again and used the two sets of results that were most similar.

The results are shown on the attached page.

I worked out my surface areas using these measurements:

My range of readings will be the smallest surface area to the largest. That is from 14.14cm² to 43.98cm. I used the formulae: 2πr² + (πdxh) to work out the surface areas for my potato cylinders. e.g. for surface area 14.14cm² : 2 x 3.142 x 0.5² + (3.142 x 1 x 4cm). This took a bit of working out with a calculator and all of the surface areas are rounded to two decimal places as some had around eight decimal places.I chose these surface area as I mostly doubled the circular faces which was easier than working out each surface area exact. This would be very time consuming.

My results cover a good range as they have small and large surface areas.

My results were then placed in a table. (See table ‘results collected’) On this table it shows that two experiments took place for each surface area. The average of these two results is also shown in a column to the side of the results. This table was too complicated to work from so the average volume of gas was taken and placed in a separate table. (See table ‘average gas produced next to each other’). Comparing these results in a table was still not very useful but it was clear to see that the average volume of gas produced did rise when the surface area was increased.

A line graph shows all the surface areas against each other. It shows fairly clearly that the lines all go up. Showing that as the time goes by more gas is produced. (See ‘graph 1’) It also shows that as the surface area increases the amount of gas produced also increases.

My prediction was about the rate of reaction so once I had collected all the results for the volume of gas produced and the average I set about finding the rate of reaction. The rate of reaction is the volume of gas divided by time. I worked this out using my average gas produced results and the seconds it took for that gas to be produced. I put all the new data into a table. (See table ‘rates of reaction’) I needed to compare these results so I put them into the table called ‘rates of reaction next to each other’. This displayed the results easily so I could make a graph. The graphs had to be generated on the computer, as the results were extremely close. These six graphs did not compare against each other. They were to show how during the experiment the rate of reaction changed. (See ‘rate of reaction graph 2,3,4,5,6+7’) There is only one problem with these graphs and that is that they are different scales that mean they cannot be easily compared.

I needed to work out the rate of reaction average to make a main graph to prove my prediction. I used the averages I had worked out earlier to put them on a graph. (See graph 8).

Graph 8 shows that as the surface area increases the productive rate increases this means that they are directly proportional.

In my evidence I have found out that as the surface area of the catalase is increased the amount of gas produced and the rate of reaction at which the reaction occurs increases.

In all my evidence there is a general trend of that the surface area and rate of reaction are proportional. The graphs 2,3,4,5,6, and 7 are not included in this as they relate time and rate of reaction which are different.

In all of my evidence I do not really have any results that stand out i.e. extreme results not making sense. A line of best fit is not appropriate for graph 1 as the points change over each other in some points and lines of best fit would confuse the matter. A line of best for graph 8 (rate of reaction against surface area) may be appropriate to show how the graph is going up generally.

Conclusion

Using my evidence I have concluded that the as the surface area increases the rate of reaction increases also. This proves that my prediction is correct. This is because if there is a larger surface area there is likely to be more area for the active sites for the enzyme substrate to fit into and be broken down. The larger the surface area the more active site there are so they can work faster. This means that more Hydrogen Peroxide will be broken down into water and oxygen. The H2O2 molecule fits into the active site like ‘lock and key’ and the catalase then distorts the H2O2 and releases it. If there are more active sites doing this then more oxygen is going to be released and so the rate of reaction will speed up. My results have ‘agreed’ with my prediction and the scientific knowledge which is earlier in my plan. On my graph the reason that the smaller surface areas have a slower rate of reaction is because the concentration is the same the active site number is different. This is because although here are the same number of H2O2 molecules some may have to wait to be broken down whereas in the larger surface areas there are enough active sites to satisfy the number of H2O2 so fewer have to wait to be broken down. The maximum rate of reaction is when all the sites are being used but in reality this is never reached due to the fact that not all the active sites are being used at the same time.