Catalyse Investigation

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Catalyse is an enzyme that breaks down hydrogen peroxide into water and oxygen.

Hydrogen Peroxide Water + Oxygen
2H2O2 2H2O +O2

The factors that affect enzyme activity are:

· Enzyme concentration (Gareth Williams, Biology For You page 31)
· Substrate concentration (Mary + Geoff Jones, Biology page )
· Temperature (Gareth Williams, Biology For You page 31)
· pH (Gareth Williams, Biology For You page 31)

Background Information

Enzymes such as catalyse are protein molecules, which are found in living cells. They are used to speed up specific reactions in the cells. They are all very specific as each enzyme just performs one particular reaction.

Catalyse is an enzyme found in food such as potato and liver. It is used for removing Hydrogen Peroxide from the cells. Catalyse speeds up the decomposition of Hydrogen Peroxide into water and oxygen. It is able to speed up the decomposition of Hydrogen Peroxide because of the shape of the Hydrogen Peroxide molecule. This type of reaction where a molecule is broken down into smaller pieces is called an anabolic reaction.

In my investigation I will study the effect of enzyme concentration on the rate of catalyse activity.


Prediction

I predict that if I increase the enzyme concentration this will increase the rate of catalyse activity because there will be more active sites for reactions, resulting in more chemical reactions caused by successful collisions between the active site of the enzyme (catalyse) and the substrate (hydrogen peroxide)
(Gareth Williams, Biology For You pages 29-31)

Equipment
10cm3 measuring cylinder
Hydrogen Peroxide
Water bath
Pureed potato
Test tube
Delivery tube
Bung

Diagram

Plan
I will use the equipment as shown in the diagram to measure the amount of oxygen that gathers into the measuring cylinder. The amount of oxygen is directly related to the rate of catalyse activity. The more oxygen, the more catalyse activity and the less oxygen, the less catalyse activity. I will be able to measure the oxygen using the measuring cylinder.

I will be using pureed potato to provide the enzymes for this experiment. This is the best kind of potato to use because it has more surface area than whole potatoes and this means more active sites for the substrate to react with.I will be change the enzyme concentration(amount of pureed potato) in order to measure its effects on the rate of catalyse activity. I will perform the experiment using 0cm3, 1cm3, 2cm3, 3cm3, 4cm3, 5cm3 of pureed potato. In order to make it a fair test I must keep the volume the same for each test. The set volume will be 5cm3 so I will use water to keep the volume at 5cm3, this means that for example when I do the experiment with 3cm3 of pureed potato I will add 2cm3 of water. I will add 5cm3 of hydrogen peroxide each time in order to keep the test fair. I will use a measuring cylinder to measure out all these values. I will repeat the test at each value three times and then take an average to make it a fair test and I will also keep the temperature of the room the same throughout the experiment using the thermostatic controls in the laboratory. The temperature will remain at 20ºC.

These are the measurements I will be using.

TEST TUBE Enzyme(POTATO)(Cm3) WATER(Cm3) Substrate (Hydrogen peroxide)(Cm3) TOTAL VOLUME(Cm3)
1 0 5 5 10
2 1 4 5 10
3 2 3 5 10
4 3 2 5 10
5 4 1 5 10
6 5 0 4 10


Results

Test Tube Experiment No.1 Experiment No.2 Experiment No.3 Total Average
1 0 0 0 0 0
2 2.3 2.1 2.4 6.8 2.3
3 4.3 4.5 4.4 13.2 4.4
4 6.8 6.5 7.1 20.4 6.8
5 7.4 7.4 7.1 21.9 7.3
6 7.4 7.5 7.2 22.7 7.4

Conclusion
In conclusion I can see that my prediction was correct. The higher the enzyme concentration was, the more oxygen was produced. This is because the with higher enzyme concentration, there are more active sites where more successful collisions will take place between the active site of the enzyme and the substrate. The graph shows that as the enzyme concentration was increasing more and more, the rate of catalyse activity was increasing less an less which indicates that it was nearing the optimum rate of catalyse activity.

Evaluation
My experiment worked well although if I were to do it again I could do certain things differently in order to gain more accurate results. I would repeat the experiment more times as this would give me a more accurate average and I would also uses more different values in order to get a more detailed outlook on how enzyme concentration affects the rate of catalyse activity. I would also use a measuring syringe to measure the amount of oxygen because this is a much more accurate way of doing it.

Evaluation
My experiment worked well although if I were to do it again I could do certain things differently in order to gain more accurate results. I would repeat the experiment more times as this would give me a more accurate average and I would also uses more different values in order to get a more detailed outlook on how enzyme concentration affects the rate of catalyse activity. I would also use a measuring syringe to measure the amount of oxygen because this is a much more accurate way of doing it.

For this investigation I have been asked to investigate (by experimentation) the effect of substrate concentrations on the rate of the decomposition of hydrogen peroxide when catalyzed by the enzyme catalase. This is part of our work on the function of enzymes, how they work and the effects of conditions on how they work. We have learnt about the formation of enzyme-substrate complexes, the lock and key model, induced fit model, activation energies of normal reactions and enzyme-catalyzed reactions, equilibrium, specificity and denaturation.

Let me write specifically about the enzyme, catalase and the substrate, hydrogen peroxide. In organisms, hydrogen peroxide is a toxic by-product of metabolism, of certain cell oxidations to be more specific. Hydrogen peroxide on its own is relatively stable and each molecule can stay in this state for a good few years. Its decomposition therefore needs to be speeded up greatly in order to prevent it from intoxicating the cell. This is where catalase comes in.

Catalase has to be very fast acting to keep the hydrogen peroxide levels low, and it is one of the fastest acting enzymes known. It catalyses the decomposition of hydrogen peroxide, liberating oxygen gas as effervescence, each molecule of the globular protein decomposing 40,000 molecules of hydrogen peroxide per second at­ zero degrees Celsius and capable of producing an amazing 1012 molecules of oxygen per second. The equation is:

2H2O2+Free enzyme«E-S complex®2H2O+O2+Free enzyme

The E-S complex is an intermediary stage where the substrate forms temporary and reversible interactions with the enzyme. The reason that this is so much faster than the decomposition rate in the absence of a catalyst is to do with the activation energy for this route being lower than the energy it takes to simply break the bonds within the molecules because it forms an intermediary stage, but the mechanism for this is not yet fully understood.

[See diagrams]

Catalase is found in microbodies, or peroxisomes, in euchariotic cells. Peroxisomes are spherical, 0.3-1.5 mm in diameter (smaller on average than mitochondria) and bounded by a single membrane. These are derived from the endoplasmic reticulum. Peroxisome gets it name from peroxide, hydrogen. The three types of plant peroxisomes are:

Glyoxisomes, so called because they metabolise a compound called glyoxylate, are concerned with the conversion of lipids to sucrose in lipid-rich seeds.

Leaf peroxisomes are important in the process of photorespiration in which they are part of the photorespiratory pathway involving (obviously) chloroplasts and mitochondria, the three organelles often being in close proximity within the cell. The photorespiratory pathway is shown on the right. This shows how hydrogen peroxide is produced and the functions of the peroxisome as part of the photorespiratory pathway.

Non-specialized peroxisomes are a third group, which are found in other tissues.

Potato tubers contain peroxisomes, although I don’t know why. I suppose if new plants have to grow from them, they have to have all the parts of the plant with the store of food. If a potato is exposed to light it goes green, so I suppose this is proof because could not go green without chloroplasts.


Because of the increased chance of a successful collision caused by random thermal motion when there are more molecules present, I predict an increase in rate with higher substrate concentration. For low concentrations I think that the rate of the reaction will be directly proportional to the concentration of hydrogen peroxide in the solution. This is because if double the amounts of substrate molecules are in the solution, double the amount will find an enzyme molecule at the same time, if all the substrate molecules are moving at similar speeds (the average speed being directly proportional to the temperature). Therefore if there are double the amount of substrate molecules in a solution, double the amount of reactions will take place at once and the rate will be doubled.

The problem is, because of the time taken for the reaction and dissociation of the enzyme-product complex, as the concentrations of substrate increase; not all the collisions of the substrate will be successful because some active sites will be saturated (occupied by substrate/products). The frequency of this occurrence increases with the substrate concentration, and eventually the terms cancel out, leading to no rate increase with substrate concentration increase at high concentrations. This is because as the rate increases this must mean that more enzyme molecules are reacting with the substrate at one time, seeing that the reaction and dislocation time is constant at constant temperature, causing more substrate-enzyme collisions to be unsuccessful due to saturation.

This effect can be explained mathematically. The mathematical expression of the hyperbola caused by the effect explained above was developed in 1913 by two German biochemists, L. Michaelis and M. L. Menten. In the equation, VM is the theoretical maximum velocity of the reaction and KM is called the Michaelis constant.

Velocity =VM (S)

KM+(S)

The shape of the curve is a logical sequence of the active site concept; i.e., the curve flattens out at the maximum velocity (VM), which occurs when all the active sites of the enzyme are filled with a substrate. The fact that the velocity approaches a maximum at high substrate concentrations provides support for the assumption that an intermediate enzymes-substrate complex forms. At the point of half the maximum velocity, VM/2 in the diagram, the substrate concentration in moles per litre (S) is equal to the Michaelis constant, which is a rough measure of the affinity of the substrate molecule for the surface of the enzyme. The VM value for hydrogen peroxide is 1012 molecules of oxygen per molecule of catalase per second. The KM value in this case is about 5E-8.

Velocity =VM (S)

KM+(S)

I found the value of (S) by calculating the amount of moles per litre using the relative molecular mass of water and hydrogen peroxide and the Avogadro constant. This was between 0 and 6 moles per litre for the concentrations between 0% and 20%.

Therefore V=1012*5

5E-8+5

V=1E12 reactions per second per molecule of enzyme.

I estimate that there are about 50000 molecules of enzyme per square centimetre. If each cylinder has a surface area of 4 square centimetres the total amount of molecules of enzyme is 1.2E6.

1E12*1.2E6=1.2E18 reactions per second.

Each mole (6.02*1023 molecules) of oxygen takes up 24 litres.

(1.2E18/6.02E23)*24,000=0.05ml/s.

Safety: during this experiment, gloves and goggles must be worn has hydrogen peroxide is corrosive and irritant. Any spillages must be wiped up as soon as possible to avoid accidents and mishaps that could be caused by leaving them.

Apparatus: Diagram of apparatus:

· · ½ litre container

· · Boiling tubes

· · Five litre ice-cream

tub

· · Retort stand

· · Three clamps

· · Buckner flask (with

tube and bung)

· · Stop clock

· · Pipette

· · Burette

Instructions:

1. 1. Prepare boiling tubes, each with 25 ml of hydrogen peroxide in them (only prepare as many as will be used on that day). Prepare solutions of 20%, 17.5 %, 15%, 12.5%, 10%, 7.5%, 5%, 2.5% and 0%. There should be three test tubes of each concentration for the repeats.

This should give a decent range and adequate repeats to come to a conclusion. I decided that boiling tubes are the easiest way to keep the solutions until in use because they can easily be labeled and kept in a rack. Although only slowly, hydrogen peroxide still decays in the absence of a catalyst (even in a fridge). This could affect the results.

2. 2. Set the apparatus up in the way shown above, making sure that no (or as little as possible) air leaks into the burette when it is inverted with the open end underwater.

Clamping the apparatus in place enables me to concentrate on running the experiment and not have to hold anything.

3. 3. The tube must be fixed under the burette, with a funnel directing the air bubbles into the burette.

It is also possible to do it without a funnel. The funnel is to prevent the pressure of the air being pushed into the burette from causing the pipe to pop off or oxygen to be leaked from the system (squeezed out). Any pressure build up could also lead to inaccuracies.

4. 4. The burette must only have the part of it with no scale underwater (so that a maximum amount of gas can be collected to reduce the percentage error).

The gas could have also been collected in a gas syringe or over water in a measuring cylinder, but the latter is not as accurate as a burette because of the smaller graduations and larger distance between the outside and the centre of the meniscus and the former is not as reliable as a burette because of the friction in the system causing pressure and possible leaks to affect the results.

5. 5. The tube that carries oxygen to the burette must be put into the water in such a way that no air is trapped underwater that could rise into the burette to spoil the set up or the results.

This is worth doing because if a bubble goes into the burette before the experiment starts it is very annoying to have to get the waterline back to 50 ml.

6. 6. When this is all ready, prepare five 1-cm-long cylinders of potato, using borer size 4. Make sure that there are no skins on the cylinders.

1. There is no point in having fast reactions if the percentage inaccuracies in timing, caused by the short times and the percentage inaccuracies in surface area caused by cutting the potatoes into very small pieces by hand (a variation of about 0.5 mm either side), are so large that the results are inconsistent. However, too slow a rate, however low the percentage inaccuracies may be, is impractical, as I do not have all the year to do this in. I therefore decided on cylinders of 1 cm length made with a fairly thin borer1. (size 4) so that the circumference was constant (and small enough to give the cylinders a large enough surface area) and the inaccuracy was reduced (1 mm in 10=10%). Trying this out, I realised that the inaccuracy was still much greater than I would have liked it to be. I therefore weighed them too, to insure that all the pieces of potato had the same mass. This gives them the same volumes (presuming that the variations in density, caused by a gradual water loss by osmosis and the retention of the shape of the cells by the cell wall, are negligible) and therefore, if the pieces are flat ended cylinders of equal radius, the same surface area. The scales/balance weighed to an accuracy of one hundredth of a gram. My pieces were 0.5g. The inaccuracy was therefore 0.01/0.5 =2%, 1/5th of what it would be without the scales. Skin on the cylinders must be avoided because this would have a large effect on the surface area.

7. 7. Then pour the first 20 ml of hydrogen peroxide into the conical flask and start the stop clock as the bung goes into the flask.

I decided to use 20 ml because I thought that this seemed about the right amount and 20 would divide easily to dilute the hydrogen peroxide. A stop clock is most suitable because it is accurate and can easily be stopped and started without looking at it.

8. 8. After five minutes enter the amount of gas collected in the burette on the table, making sure that this point is at eye level to eliminate parallax error.

I decided to collect as much gas as possible if I had time, because this would reduce the percentage errors. If the point on the burette is not at eye level, the thickness of the burette and distance to the centre of the meniscus could throw the measurements out.

9. 9. Stop the stop clock as soon as the burette is full of gas (again, make sure that this point is at eye level to eliminate parallax error). The stop clock should be stopped the moment the bubble reaches the surface.

The moment that the bubble reaches the surface of the column of water in the Burette is better than their moment at which it comes into the Burette because one the can see it rising in that the Burette and get ready to stop the stop clock.

10. 10. Enter this time and the temperature of the solution in the Buckner flask, together with the corresponding concentration and in the correct column for which repetition of the experiment it is on the table.

I decided to take the temperature of the reactants, because this is the temperature at which the reaction occurs, and the temperature affects the rate of the reaction. I could then use the Q10 formula to “temperature balance” the results.

11. 11. Repeat the experiment three times for each concentration of hydrogen peroxide.

As I have said earlier, the more repetitions the better, and if they are not done on the same day as each other, all the better to prevent factors like which potatoes I was using and draught in the lab from affecting my average results.

For one of the reactions, take readings of the amount of gas in the burette every 30 seconds. This is to show the activity decrease curve throughout the individual reactions.

This is a summary of the steps that and I will take to reduce the amounts of errors affecting the results.

· · Keep the level of water in the ice cream tub the same and the retort stand with all the clamps and the funnel on it to make sure that the oxygen has the same distance to travel underwater each time.

· · Use a burette (reasons already explained in plan).

· · Use a compromise between factors (e.g. large volumes/longer times) to reduce percentage inaccuracies where possible.

· · Record the temperature of the hydrogen peroxide, so that I can use the Q10 formula to temperature balance the results.

· · Repeat each experiment 3 times to get an average.

· · Keep the time that I collect for (five minutes) and amount which I collect (20 ml) there same so that the reactions are at the same stage in their natural exponential decay curve when the measurements are taken.

· · Use potatoes from the same batch for all of the reactions.

This is what I will do with my results and how I will record and process them.

· · This is the table in which I will record my observations.

Join now!

I will then:

· · Plot a graph of rate against (average) substrate concentration and compare it with the prediction.

· · Calculate the amount of enzyme present and compare this with the prediction.

· · Calculate the Michaelis constant (KM) for catalase.

· · Decide whether the Q10 Formula is accurate for catalase.

· · Plot a graph of rate against concentration with all the repeats and the averages on and determine from the line of best fit which results were anomalous.

· · Decide what factor might have caused the anomalies


There are many variables that affect the results ...

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