Effect Of Substrate Concentration On The Activity Of Catalase

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Effect Of Substrate Concentration On The Activity Of Catalase

Effect Of Substrate Concentration On The Activity Of Catalase

AIM

This is an experiment to examine how the concentration of the substrate hydrogen peroxide affects the rate of reaction of the enzyme catalase.

PREDICTION

I predict that 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. When this saturation point is reached, then adding 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. 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.

BACKGROUND INFORMATION

Enzymes such as Catalase 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.

Catalase is an enzyme found in food such as potato and liver. It is used for removing Hydrogen Peroxide from the cells. Hydrogen Peroxide is the poisonous by-product of metabolism. Catalase speeds up the decomposition of Hydrogen Peroxide into water and oxygen as shown in the equations below.

Formula:

Catalase

Hydrogen Peroxide---------------------->Water + Oxygen

Catalase

2H2O2------------------->2H2O+O2

It is able to speed up the decomposition of Hydrogen Peroxide because the shape of it's active site matches 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.

APPARATUS

1. Gas Syringe

2. Metal Stand

3. Yeast Catalase

4. Hydrogen Peroxide

5. Test Tubes

6. Beakers

7. Test Tube Rack

8. Stop Watch

9. Pipette

10. Pipette Filler

11. Tap Water

METHOD

To test out how the concentration of hydrogen peroxide affects the rate of reaction first set up the apparatus below.

[Aparatus picture not reproduced]

1. Add 2cm3 of yeast to one test tube. Add 4cm3 of hydrogen peroxide solution at a concentration of 20% to the other test tube. Use a pipette to measure out the volumes. It is very important to accurately measure the amounts of Hydrogen Peroxide, Yeast and water to ensure a fair test.

2. Pour the hydrogen peroxide solution into the test tube containing the yeast and immediately put the gas syringe bung on the end of the test tube, at the same time start the stopwatch.

3. Bubbles should start to rise up the tube and the gas syringe will move outwards, as soon as the gas syringe passes the 30cm3 mark stop the stopwatch and note the elapsed time down to the nearest 1/10th of a second.

4. Repeat the experiment with hydrogen peroxide concentrations of 16%, 12%, 10%, 8%, 4% and 0%. The 0% concentration of hydrogen peroxide solution is done as a control solution to show that at 0% concentration no reaction occurs. The different concentrations of Hydrogen Peroxide are made by adding tap water to the 20% Hydrogen Peroxide in the correct amounts. The table below shows what amounts of Hydrogen Peroxide and water are needed to make the solutions.

5. Repeat all the tests at least three times so that an average can be obtained. Repeating the experiments several times will help to produce better and more accurate results as any inaccuracies in one experiment should be compensated for by the other experiments. Note all the results in a table such as the one below.

The rate can then be worked out by

Rate=30/Average Time

This gives the rate in cm3 of oxygen produced per second, this is because I am timing how long it takes to produce 30cm3 of oxygen. From these results a graph can be plotted with concentration on the x-axis and time taken on the y-axis.

I am using yeast catalase as opposed to catalase from apples, potatoes or liver because it is easier to get the desired amount of yeast catalase by simply measuring it off. To obtain catalase from a substance such as potato would involve crushing it and with that method you would never be sure of the concentration of the catalase. If the catalase was used up then another potato would have to be crushed and this could produce catalase of a totally differen from studentcentral.co.uk t concentration which would lead to inaccuracies in the experiment making this an unfair test.

To ensure this is a fair test all the variables except for the concentration of Hydrogen Peroxide must be kept the same for all the experiments. Variables that must not be altered include:-

Temperature, yeast concentration, type of yeast, batch of yeast, volume of yeast, volume of hydrogen peroxide, air pressure and humidity.

When measuring the volumes of Hydrogen Peroxide, Yeast and Water the measurement should be taken by looking at the scale at an angle of 90 degrees to it to avoid any parallax error.

RESULTS

I carried out the above experiment and these results were obtained.

All the times are in seconds. The average results are all written down to one decimal place because although the stopwatch gives results to two decimal places it is impossible to get accurate times to two decimal places due to the fact that our reaction times are not fast enough to stop the stopwatch precisely. I then worked out the rates of the reactions with the equation

Rate=30/Average Time

From these rates I was able to plot a graph of the rate of reaction against concentration of Hydrogen Peroxide.


CONCLUSION

When the concentration of Hydrogen Peroxide is increased, the rate of reaction increases at a directly proportional rate until the concentration of Hydrogen Peroxide reaches about 16%. If you double the concentration of Hydrogen Peroxide then the rate of reaction doubles as well. When the concentration is doubled from 8-16% the rate goes up from 1.65-2.97 Cm3 Oxygen produced per second, which is an increase of 1.8 times. I would expect the rate to increase two times if the Hydrogen Peroxide concentration is increased two times because there are twice as many substrate molecules which can join onto the enzymes active sites. The reason that the number is less than two times could be put down to the fact that at 16% the Enzyme's active sites may already be close to being saturated with Hydrogen Peroxide. There may also be some experimental error which causes the inaccuracies.

After 16% the increase in the rate of reaction slows down. This is shown by the gradient of the graph going down. At this point virtually all the active sites are occupied so the active sites are said to be saturated with Hydrogen Peroxide. Increasing the Hydrogen Peroxide Concentration after the point of saturation has been reached will not cause the rate of reaction to go up any more. All the active sites are being used so any extra Hydrogen Peroxide molecules will have to wait until an active site becomes available.

The theoretical maximum rate of reaction is when all the sites are being used but in reality this theoretical maximum is never reached due to the fact that not all the active sites are being used all the time. The substrate molecules need time to join onto the enzyme and to leave it so the maximum rate achieved is always slightly below the theoretical maximum. The time taken to fit into and leave the active site is the limiting factor in the rate of reaction.

The diagram below shows what happens.

EVALUATION

To help make this experiment more accurate, I repeated it three times and then used the average of all the results to plot a graph with a line of best fit. I tried to keep all the variables except for the concentration of Hydrogen Peroxide the same for all the experiments. However, in reality it is impossible to keep all the variables precisely the same. For example:

a) There is a slight delay between pouring the Hydrogen Peroxide into the yeast, putting the bung on and starting the stopwatch. This will slightly affect all the results but as I carried out all the three steps in the same way for all the experiments it should not make any difference to the overall result.

b) It is also impossible to precisely measure out the amounts of Hydrogen Peroxide, Yeast and Water each time. As the scale on the pipettes shows the volume to the nearest mm3 the volume of the solutions that I used should be correct to the nearest mm3. The volume of gas in the test tube to start with is slightly affected by the amount which the bung is pushed down each time, if the bung is pushed down further then the volume in the tube will be less so the 30cm3 of gas is reached faster.

c) Due to the fairly slow speed of our reactions it is only possible to measure the time of the reaction to the nearest 0.1 second even though the stopwatch shows the measurements to the nearest 0.01 second.

ANOMOLIES

The plotted results on the graph produce a straight line of best fit to begin with which then goes into a curve of steadily decreasing gradient. The only anomalies are the results at 8% and 10%. The result at 8% is slightly above the line of best fit and the 10% result is slightly below it. This is probably due to an experimental error involving one of the factors mentioned above.

EXTENSION

This experiment could be improved in a number of ways. It could be repeated more times to help get rid of any anomalies. A better overall result would be obtained by repeating the experiment more times because any errors in one experiment should be compensated for by the other experiments.

Using more concentrations of Hydrogen Peroxide would have produced a better looking graph and I would have liked to use concentrations higher than 20% to extend the graph so that the maximum possible rate of reaction could be reached.

The problem of the delay between pouring in the Hydrogen Peroxide, bunging the test tube and starting the stopwatch could have been limited by getting another person to start the stopwatch when the hydrogen peroxide was poured into the tube.

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Investigate the effect of temperature on the activity of catalase (from potato)

Arefin Khan

Aim: Investigate the effect of temperature on the activity of catalase (from potato).

Introduction: Enzymes are biological catalysts. They speed up metabolic reactions in the body but remain chemically unchanged themselves. Enzymes contain an active site. This is a region, normally a depression or cleft, to which another molecule may bind. This molecule is known as the substrate, and is usually specific to the active site of the particular enzyme, which breaks it down. Substrates will not usually fit into any other active sites other than that of the enzyme it is specified to. This can be explained as a lock and key model, where the lock and key are specific to each other, only, that there are many of the same kinds of lock and key when it come to the enzymes.

Just as lock and keys have three-dimensional shapes, proteins are also three-dimensional. Usually, there is only one active site on an enzyme; however there can be more. 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:

2H2O2 2H2O + O2

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.

catalase

2H2O2 2H2O + O2

This type of reaction where a molecule is broken down into smaller pieces is known as a catabolic reaction.

In order to investigate the effect of temperature on the activity of catalase, I will record the amount of oxygen released when hydrogen peroxide is broken down.

Variables: There are quite a few variables which can alter the rate of reaction, and need to be kept constant. They are as follows:

a) PH: at too high PH, the enzyme is denatured due to the loss of H+ ions. The same applies for too low a PH level, where too many H+ ions would attach to the negative regions of the enzyme, changes its shape and causing it to denature.

b) Concentration of enzyme: The higher the concentration, the higher the rate of reaction will be. With a larger number of catalase molecules, the chance of successful collisions between enzyme and substrate will be increased. In order to keep this constant, I will make sure I use the same volume of tissue (potato) containing catalase each time I conduct the experiment.

c) Surface area: The previous also applies to this.

d) Mass of tissue: Here it needs to be taken into account that different potatoes will not give the same mass, even if equally sized pieces are cut. Different potatoes will not have exactly the same water and catalase content.

Arefin Khan

The mass will be kept constant in the same way as surface area and concentration of enzyme.

Prediction: 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.

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Enzymes have a very specific three-dimensional shape, held together by ionic and hydrogen bonds. If the amino acids are too vigorous in their motion, then, these bonds will brake. Once the bonds have been broken, the enzyme is said to have become denatured. As a result of becoming denatured, the enzymes’ rate of activity becomes less because the enzyme loses its specific three-dimensional shape. The enzyme will start to become denatured after around 40ºC as enzyme activity is usually at its optimum at this temperature. After this, the rate of reaction will probably deteriorate. After 60ºC, there is likely to ...

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