To investigate and compare the effect of temperature on the activity of catalase in plant and animal tissues.

Amreen Khadri 12Be Investigating the Activity of Catalase

Investigating the Activity of Catalase

Aim: To investigate and compare the effect of temperature on the activity of catalase in plant and animal tissues.

Hypothesis: The rate of reaction will increase as the temperature of the hydrogen peroxide increases up to the optimum point.

Introduction: Catalase is an enzyme involved in the breakdown of hydrogen peroxide (H2O2) in cells. Hydrogen peroxide is toxic to most organisms and it is an oxidizing agent, which contributes to the deteriorating changes that occur with aging in animals. Hydrogen peroxide has been implicated in the degenerative middle ear disease, otitis media. Some energy releasing reactions in cells produce hydrogen peroxide. This is acidic, and can thus, kill cells. Normally, hydrogen peroxide decomposes to form hydrogen and oxygen. However, this process is very lengthy. There is an enzyme known as catalase in cells, which dramatically increases the rate of decomposition of hydrogen peroxide. This type of reaction where a molecule is broken down into smaller pieces is known as a catabolic reaction.

H202 → H20 + 02

Enzymes are biological catalysts, which increase the rate of reactions by lowering the activation energy needed for the reaction to take place. The activation energy is the amount of energy needed for molecules to react when they collide. Molecules need to collide in order to react, this is known as the collision theory. When they collide they may not react, as a certain amount of energy is required to break bonds, this energy is the activation energy. Enzymes are made of a long amino acid chain; within this some molecules are attracted to each other, so the chain folds in on itself to form a 3D shape.

An area on the surface of the enzyme is known as the active site. This is where reactions take place to form or break down substances. Enzymes are specific which means a particular enzyme only works on one substance known as its substrate. For example, the substrate of amylase is starch and the substrate of lipase is fats. They only have one substrate because the active site is formed in a different shape for each enzyme, where only one substance can fit. The ‘lock and key’ hypothesis states that the enzyme is like a lock, which will only have one key. The substrate shown is the only substance that fits the enzyme. An enzyme substrate complex is the compound formed when the substrate is attached to the active site, it is only in this form for a short time while the substrate is being broken down. Enzymes can break own substances, known as catabolism, or can join substances together, known as anabolism. Together they form metabolism, which is every chemical reaction in the body.

Factors affecting the rate of reaction:

The rate of an enzyme-controlled reaction is measured by the amount of substrate changed, or the amount of product formed in a given time. In this particular experiment, the rate of reaction will be measured by the quickness of the evolution of oxygen gas. When investigating the effect of one factor on the rate of reaction of a catalyzed reaction, all other factors must be kept constant and at optimum levels. Only initial rates should be taken, as the reaction will slow down as it proceeds due to concentrations of substrate reducing.

Concentration

Substrate and enzyme concentrations are both factors that affect the rate of an enzyme-controlled reaction. If the substrate concentration is kept high and all other factors are kept constant, the rate of reaction is proportional to the enzyme concentration

BELOW RELATIONSHIP BETWEEN ENZYME CONCENTRATION AND THE RATE OF AN ENZYME–CONTROLLED REACTION.

For a fixed enzyme concentration, the rate of an enzyme reaction increases when substrate concentration increases. The theoretically possible maximum rate (Vmax) is never achieved, but the reaction eventually reaches a point when increasing substrate concentration only gives negligible change in rate. The explanation for this is that almost all the active sites in the enzyme molecules are filled because of the high concentration of substrate. Substrate molecules that cannot find a vacant active site have to wait until the enzyme/substrate complex has released the products so the active site is once again free to take more substrate.

BELOW: EFFECT OF SUBSTRATE CONCENTRATION ON THE RATE OF AN ENZYME-CONTROLLED REACTION.

Temperature

Increasing the temperature and therefore heat energy makes the molecules of the enzymes and the substrate move about faster, thus increasing the chance of the enzyme molecules colliding with the substrate molecules. The optimum temperature is the one that gives maximum movement of molecules. Increasing the temperature beyond the optimum will cause a drop in the rate of reaction because the enzyme will be DENATURED. The shape of the active site, which is vital for the enzyme to function, is lost. Most mammalian enzymes have an optimum temperature of about 37-40°C. Some enzymes have higher optimums e.g. the enzymes of bacteria living in hot springs may have an optimum temperature of 70°C. In this experiment, tissues of both plant and animal are going to be compared.

BELOW: EFFECT OF TEMPERATURE ON THE RATE OF AN ENZYME CONTROLLED REACTION.

At a constant temperature every enzyme works most efficiently over a narrow pH range. The optimum pH is that at which the maximum rate of reaction is achieved. If the pH is changed to go above or below this range then the rate of reaction will decrease. A pH change will disrupt the ionic charge of the acidic and basic groups, the ionic bonding that helps the protein retain its folded structure will break down. The shape of the active site will change, so the ...

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BELOW: EFFECT OF TEMPERATURE ON THE RATE OF AN ENZYME CONTROLLED REACTION.

Inhibitors

Inhibitors are small molecules that can reduce the rate of enzyme-controlled reactions. Competitive inhibitions occur when a compound has a structure that is similar to that of a substrate, so the inhibitor is able to fit into the active site. The inhibitor molecule doesn’t take part in any reaction, but its presence in the active site means the substrate cannot enter the active site. Non-completive irreversible inhibition is when chemical reagents combine permanently with the enzyme and this change the structure of the enzyme, causing it to lose its specificity.

Enzyme cofactors

Cofactors are none protein substances which influence the functioning of enzymes. They include activators that are essential for the activation of some enzymes. Coenzymes also influence the functioning of enzymes although are not bonded to the enzyme. Unless enzyme cofactors were present in the potato tissue containing the Catalase, they were not included in this investigation and therefore would not have affected the rate of reaction and the results of this experiment.

Prediction:

I predict that the higher the temperature, the higher the rate of reaction up to a certain point. This is due to the fact that the particles gain kinetic energy and subsequently move around more vigorously. Thus, the chance of there being a successful collision between the enzyme and substrate molecule increases as reacting particles with collide more frequently with increased kinetic energy. The point at which the rate of reaction stops increasing and starts to decrease is the optimum temperature and I predict that the optimum temperature for the animal tissue will be higher than that of the plant tissue because the plant tissue that I am using is potato which is usually grown in cooler climatic conditions under ground. The animal tissue will have an optimum temperature similar to that of humans (i.e. around 40°C). After the optimum temperature is reached, the rate of reaction will start to decrease because extremely high molecular activity will occur causing the secondary and tertiary structure of the enzyme to deteriorate. The hydrogen bonds will break causing the specific shape (active site) of the enzyme to be lost (denaturation).

Methodology:

Put on Lab coat and goggles and then set up 40°C water bath (using a Bunsen burner°) and set up apparatus as shown in the diagram.

Clamp Measuring Cylinder Rubber Tube Rubber Bung Clamp

Water Basin Beehive Shell Beaker Stand Boiling Tube Delivery Tube

Label the boiling tubes as 10°C, 20°C, 30°C, 40°C and 50°C so there are a total of 5 labelled boiling tubes.
Blend 50g of liver in a blender. Measure out 3g of liver.
Fill all 5 test tubes with 5cm3 of Hydrogen Peroxide and put into a test tube holder.
Set up the 40°C water bath using a kettle, tap water and beaker. Cover beaker with bubble wrap to ensure the temperature is maintained.
Place delivery tube in water bath beaker and then clamp in place. Then place the boiling tube containing H2O2 labelled as 40°C in the 40°C water bath. Wait a few minutes until the boiling tube and delivery tube reach the temperature of the water bath and check this using a thermometer.
When the correct temperature is reached add the hydrogen peroxide into the delivery tube containing the liver.
Replace bung immediately and start timing. Every 30 seconds measure the amount of oxygen released using the graduated measuring cylinder and record the results in a table.
Rinse out delivery tube and clean using a bottlebrush.
Measure another 3g of liver.
Now set up the water bath at 30°C and place the boiling tube labelled as 30°C and the clean delivery tube in the water bath. Make sure delivery tube is clamped in.
Leave it in the water bath for approximately 4 minutes. Once the temperatures of the potato and hydrogen peroxide are equal, add the hydrogen peroxide to the delivery tube and place bung on top of the delivery tube immediately. Start the stopwatch and at intervals of 30 seconds for 5 minutes, measure the amount of oxygen evolved (measure the reduction in water). Record results in the table.
Repeat steps 9-12 for 20°C, 10°C and 60°C.
Repeat steps 1-14 one more time and take an average of the results
Label the boiling tubes as 10°C, 20°C, 30°C, 40°C and 60°C so there are a total of 5 labelled boiling tubes.
Blend 50g of potato in a blender. Measure out 3g of potato.
Fill all 5 test tubes with 5cm3 of Hydrogen Peroxide and put into a test tube holder.
Set up the 40°C water bath using a kettle, tap water and beaker. Cover beaker with bubble wrap to ensure the temperature is maintained.
Place delivery tube in water bath beaker and then clamp in place. Then place the boiling tube containing H2O2 labelled as 40°C in the 40°C water bath. Wait a few minutes until the boiling tube and delivery tube reach the temperature of the water bath and check this using a thermometer.
When the correct temperature is reached add the hydrogen peroxide into the delivery tube containing the potato.
Replace bung immediately and start timing. Every 30 seconds measure the amount of oxygen released using the graduated measuring cylinder and record the results in a table.
Rinse out delivery tube and clean using a bottlebrush.
Measure another 3g of potato.
Now set up the water bath at 30°C and place the boiling tube labelled as 30°C and the clean delivery tube in the water bath. Make sure delivery tube is clamped in.
Leave it in the water bath for approximately 4 minutes. Once the temperatures of the potato and hydrogen peroxide are equal, add the hydrogen peroxide to the delivery tube and place bung on top of the delivery tube immediately. Start the stopwatch and at intervals of 30 seconds for 5 minutes, measure the amount of oxygen evolved (measure the reduction in water). Record results in the table.
Repeat steps 22-25 for 20°C, 10°C and 60°C.
Repeat steps 15-25 one more time and take an average of the results

Risk assessment:

During this investigation, I will use catalase from liver and pea cells to speed up the breakdown of hydrogen peroxide. Hydrogen peroxide is an oxidizing agent, as oxygen will be given off during the reaction. This implies that it could aid in the burning of fires. Due to this, reasonable care must be taken to ensure that the oxygen is not directly exposed to the flame of the Bunsen burner while water is being heated. As well as this, hydrogen peroxide can be very harmful if it enters the eyes. Thus, goggles must be worn. Also lab coats will also be worn to protect skin. The blade, which will be used to cut the potato, is also another hazard. It must be handled with appropriate care, as it is sharp. Glass apparatus must be handled with care so that they do not break. If they do, that is also another hazard, and should be cleared up and disposed of in an appropriate fashion.

Fair testing:

While conducting the experiment, it has to be taken into account that a few variables will have to be controlled so as to ensure that the test is fair.

1. The pH of the reactants should be kept constant. This will not be hard to ensure, as the PH of the reactants does not vary significantly during the course of the reaction.

2. The temperature of the reactants will have to be kept constant during the reaction. This is a variable, which will be harder to control. There will be used, a beaker of 150cm3 of water at the appropriate temperature into which will be immersed, the boiling tube containing the reactants once their temperatures have equalised and stabilised. This will ensure that the temperature of the reactants is stable for longer as the larger volume of water in the beaker will maintain its temperature for longer. Also, the same volume of water will be added to the beaker each time a new test is to be done, so as not have any differences in how long the water maintains its heat for.

3. It must be ensured that when taking readings off the measuring cylinders and thermometers etc, the readings are taken with the graduated markings at eye level, so as to minimize the risk of inaccurate readings.

4. It must be ensured that the same amount of potato is used for each temperature by using an electronic scale.

Results for Liver

The figures in the tables show the amount of oxygen evolved in cm3

Q10 calculations:
A temperature increase gives ore energy to gives more energy to the substrate and the enzyme so they are more likely to collide and react. The frequency of the collisions with the right activation energy will increase so the rate of reaction will increase. The rate of increase is shown by a mathematical coefficient known as Q10, which states that a ten-degree rise in temperature will cause the rate of reaction to approximately double so the calculations below should give a value which is near to 2. However at high temperatures enzymes will begin to denature so this coefficient can only be used until 40°C.

Q10 = Rate of reaction at x°C+10°C --------Rate of reaction at x°C

Rate of reaction at 20°C 9.647 = 1.42
Rate of reaction at 10°C 6.8

Rate of reaction at 30°C 11.5 = 1.19
Rate of reaction at 20°C 9.647

Rate of reaction at 40°C 13.6 = 1.18
Rate of reaction at 30°C 11.5

Results for Peas:

Q10 = Rate of reaction at x°C+10°C --------Rate of reaction at x°C

Rate of reaction at 20°C 0.113 = 1.13
Rate of reaction at 10°C 0

Rate of reaction at 30°C 0.211 = 1.87
Rate of reaction at 20°C 0.113

Rate of reaction at 40°C 0.415 = 1.97
Rate of reaction at 30°C 0.211

Conclusion:
The amount of gas produced is proportional to the rate of reaction because if the rate of reaction doubles then twice as many reactions are occurring per second so the amount a gas produced is doubled. The rate of reaction can be found by dividing the gas produced by the time. This has been done in the results.

From the graph of temperature against rate of reaction we can see the highest rate of reaction, of 0.013 /s, occurs at 40°C. The rate is slow, 0.00125 /s, at 10°C it then rises with temperature until it reaches its maximum at 40°C. This happens because at low temperatures the hydrogen peroxide has less energy and moves more slowly. It will collide with the catalase less often, meaning the frequency of collisions is low. They are less likely to have the right activation energy so there are less collisions resulting in reactions. This will mean the rate of reaction will be low. At higher temperatures the hydrogen peroxide has enough energy to reach its activation energy and it is also colliding more often, so the rate of reaction will be higher.

After 40°C the rate of reaction falls, this is because the catalase begins to denature. When enzymes denature the attractions between amino acids in the enzyme break and the enzyme begins to return to its original shape. The shape of the active site also changes so it cannot break down the hydrogen peroxide.

The rise in rate of reaction between 10°C and 40°C complies with the Q10 coefficient, as a 10°C rise causes the rate of reaction to approximately double. This can be shown by dividing the higher rate of reaction by the lower one. For example dividing the rate of reaction at 20°C by the rate of reaction at 10°C should give a figure close to 2.

These figures are all close to 2, they will not be exact because Q10 is only an approximate and the results are not perfect.

From the table of results showing the average volume of gas produced every 30 seconds we can see that at 50°C the enzymes denature within 2 minutes, as gas is no longer produced. This happens because the enzyme takes time to heat up, while it is still reacting with the substrate. Once it is heated to the temperature of the hydrogen peroxide not all the enzymes are denatured. At 60°C the potato is heated faster and it takes 90 seconds for the enzymes to denature, at 70°C the potato is heated even faster and it takes 30 seconds to denature.

The results agree with my hypothesis because I have predicted that the rate of reaction will rise between 10°C and 40°C, and the rate will fall after 40°C. The graph obtained for the results is also similar to the one predicted, and the results seem to follow as predicted.

From conducting the experiment and gathering data I can conclude that the rate of reaction between catalase and hydrogen peroxide rises as the temperature of the mixture rises. This happens until 40°C, after which the rate of reaction falls because the catalase begins to denature. When enzyme denature attractions between the amino acids break so the enzymes loses its shape. The active site will no longer have its unique shape and the enzyme will be unable to react with its substrate. This is because only the right active site shape can break down hydrogen peroxide, according to the ‘lock and key’ hypothesis, which suggests the substrate, like a key, will only have one lock, enzyme, it fits into. The reaction follows Q10 until 40°C, because a 10°C rise will give enough energy to the substrate to increase the number of collisions and give more molecules the right activation energy to react when the collide. This will double the rate of reaction.

Two anomalous results occurred during the experiment.

1. During the repeat reading of the experiment at 10°C the reading at 90 seconds is higher than that of 120 seconds. This does not affect the analysis as the reading was ignored when taking averages.

2. During the 50°C experiment the first time the volume of gas produced stopped increasing between 120 and 180 seconds. Gas was then released, it may have been trapped in the delivery tube. When drawing the best fit line this was taken into account, so it shouldn’t affect the analysis.

Evaluation

The experiment was conducted successfully, the results obtained indicate a clear pattern which can be used to draw and support a valid conclusion. The experiment could not be conducted as planned because there was not enough time to repeat each experiment twice. However, one repetition was conducted which did make the results more reliable.

The results are reliable because the experiment was a fair test. This was done by keeping all variables constant. The concentration of hydrogen peroxide was not changed, however, it did vary as it naturally decomposed into water and oxygen. It also decomposed more during the higher temperature experiments because the hydrogen peroxide had more energy. This was a slight change and could not vary the volume of the gas produced significantly. The same volume oh hydrogen peroxide was used. The surface area of the potato was kept constant by using the same size cork borer and cutting it to the same size.

The results are accurate because a narrow measuring cylinder was used, so the volume measured is more accurate. Hot and cold water were mixed to achieve accurate temperatures.

An anomalous results occurred during the repeat reading of the 10°C experiment, the reading at 90 seconds is higher than that at 120 seconds. This reading was ignored when taking the average so It does not affect the analysis. It occurred because the measurement was misread, it may have been 0.1 cm3, instead of 0.2 cm3. Another anomalous result occurred during the first taking of the 50°C experiment, gas was released at 210 seconds when the experiment seemed to have stopped. The gas may have been trapped in the delivery tube and should have been released earlier in the experiment. This was taken into account when drawing the best fit line on the graph, so does not affect the analysis.

The method used was good enough to achieve reliable readings , but it can be improved by measuring the mass of the gas lost, this would be more accurate as digital readings would be taken. Using a smaller frequency and a larger range of temperatures would give more evidence for the conclusion. However would require more time and equipment.

Other improvements are, using thinner measuring cylinders, to measure out the hydrogen peroxide and the gas produced. Using a thermocouple thermometer to accurately measure the temperature. The hydrogen peroxide took time to heat up as the test tube is glass and is insulated. Using a better material would save time and would have allowed the plan to be completed. Using electronic equipment to take readings a exactly 30 seconds, would eliminate human error.

Further work that would extend the investigation and give more evidence to the conclusion would be, to use different concentrations of hydrogen peroxide and lengths of potato, to see how these affect the rate of reaction. Using other substances with catalase, like liver, to see how enzyme concentration affects the rate.