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Aim: to investigate the effect of Ho concentration on the activity of the enzymes catalase

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Aim: to investigate the effect of H²o² concentration on the activity of the enzymes catalase.

Background info


Enzyme is a protein molecule, which can be defined as biological catalyst.  It is used to speed up the rate of reaction, but remain unchanged at the end of the reaction. Each enzyme has a specific shape of active site that allow certain substrate to bind into it. In the reaction, substrate binds into the active site of enzyme. This is a lock and key theory. The enzyme is a lock and the substrate is a key. However, It is not enough for the substrate just to fit into an active site, it need an attractive force to keep it in place. The attractive forces are:

* Electrostatic attraction between oppositely charged groups
* Hydrogen bonding
* Permanent dipole-permanent dipole forces
* Instantaneous dipole-induced dipole forces

Also there is a minimum amount of energy which colliding particles need in order to react with each other. If the colliding particles have less then this minimum energy, then they just bounce off each other and no reaction occurs. This minimum is called the activation energy.

When the substrate into it, an enzyme-substrate complex is formed.

In this investigation enzyme Catalase behaves as a catalyst for the conversion of hydrogen peroxide into water and oxygen and is able to speed up the reaction because the shape of its active site matches the shape of the hydrogen peroxide molecule.

 Below is a diagram showing the lock and key mechanism through which catalase works: As the concentration of hydrogen peroxide increases, there is more substrate to fit into the active sites of the enzyme, catalase.


The chemical equation for the breaking down of hydrogen peroxide into water and oxygen is as follows: This reaction is performed by two types of reactions called oxidation (losing electrons) and reduction (gaining electrons). Catalase functions by removing an electron from a molecule of hydrogen peroxide (H²O²) to form a water molecule (H²O) and an oxygen molecule (O²).

Catalase + 2H²O² Catalase + 2H²O² + O²

Dependent and independent variables to be tested

Independent variable

Concentration of H²O²

Dependent variable

Rate of reaction of catalyse- volume of O² production

Fair Test

To make it a fair test, I will keep the temperature remain constant. This is because temperature could affect my result due to the fact that high temperature causing the reactant to gain more kinetic energy. Therefore increase the rate of reaction. To do this, I will keep my sample in a water bath, so that the temperature will constant.

Preliminary work


  • G-clamp
  • Boiling tubes
  • Syringes
  • Measuring cylinder
  • Water
  • Clamp stand
  • Delivery tube
  • Stopwatch
  • Rubber bungs


  1. Get all the apparatus set up
  2. Inject 5ml of hydrogen peroxide and catalase into the test tube which is connected with the delivery tube.
  3. Place the bung and cover the hole with plasticine
  4. Start the timing by using stopwatch
  5. Record the change in oxygen every 30 second for 4 minutes.
  6. Do the same for the next concentration.


No of bubbles

Time (s)






























In this experiment, I measured the volume of oxygen produced by counting the number of bubbles produced. The higher the number of bubbles produced in 30 seconds meant a faster reaction rate.


Through this experiment, I found out that the method I used is not accurate as my courting of oxygen might not be 100% correct and also the measuring cylinder didn’t give me an accurate result as it has small graduation and big scale. To improve it, I will use a burette instead of measuring cylinder in my actual experiment. Also I will do more concentrations so that I can compare my results. Moreover, during the experiment I will measure the volume instead of counting the number of bubbles because counting the number of bubbles given out during the reaction is not accurate due to the fact that some bubbles are bigger and some are smaller.


I predict that the higher the substrate concentration, the faster the rate of reaction. This is because collision will occur more frequently due to there will be more hydrogen peroxide (substrate) molecules available to collide with the enzyme’s active sites. This means at a lower substrate concentration the rate of reaction will be lower because there will be less substrate so it will take them longer to collide with the enzymes active sites and bind with them.image02.png

You can clearly see in the graph, which shows that when the concentration of enzyme is maintained constant, the reaction rate will increase as the amount of substrate is increased. However, at some point, the graph shows that increasing the amount of substrate does not increase the reaction rate. The line begins to level off and stay level at B on the graph. This is called Vmax. image03.png

At point A there is very little substrate and a lot of enzyme. An increase in the concentration of substrate means that more of the enzyme molecules can be utilized. As more enzymes become involved in reactions, the rate of reaction increases. At some point near B, all the enzymes are being involved in reactions. When this happens, some of the substrate must "wait" for enzymes to clear their active sites before the enzyme can fit with them (like a "lock and key"). After point B, the reaction rate remains flat because increasing amounts of substrate must wait before they can fit with their enzyme.

Also, the chances of the substrate entering an active site will increase. If the amount of enzyme remains constant, the rate of reaction will increase up to a point. The rate of reaction increases until all of the active sites of the enzyme molecule are being used. If excess substrate is added, the rate of reaction will not increase any further because the enzyme is working as fast as it can. At this point, Vmax is constant. Vmax is the maximum rate at which chemical reactions can occur (V = velocity).

The rate of an enzyme controlled reaction increase in proportion to the concentration of the enzyme (assuming there is excess substrate). If substrate is not in excess (e.g. constant) the graph tales off as it comes to a point where the reaction reaches maximum velocity, Vmax.

I predict that the higher the concentration of hydrogen peroxide there is, more oxygen will be produced in the allotted time. When there is a 100% concentration of hydrogen peroxide, in my case 5ml, I think the most oxygen will be produced. From then on, as the concentration gets weaker, I think less oxygen will be produced and at a slower rate. I would expect 100% to react quickest because it has the most hydrogen peroxide molecules in it. With more of these molecules inside the solution, it is more likely that reactions will take place. This means that a reaction is more likely to take place in a shorter time, making the rate of reaction quicker. After 100% producing the most oxygen, I then expect it to be 60% (3ml), then 40% (2ml), and then 20%(1ml). An increase in the substrate concentration will produce a corresponding increase in the rate of reaction. Therefore, when the concentration is doubled, as in from 1ml to 2ml, I also expect the amount of oxygen produced in the reactions to double. I think this because if the solution of hydrogen peroxide is stronger, the catalyse will have more hydrogen peroxide molecules to break down. If there is a weaker concentration of hydrogen peroxide solution, then there will be less hydrogen peroxide molecules for the enzyme to break down, and less oxygen will be produced in the allotted time.

Before any change takes place on collision, the colliding molecules must have a minimum kinetic energy called the Activation Energy shown on the energy level diagrams below (sometimes called reaction profile/progress diagrams). image04.png

The purple arrow up represents this minimum energy needed to break bonds to initiate the reaction, which is the activation energy.



Why it need to kept constant

How it will be regulated


If we let one test tube have more time than the other then it
would give it more time to react. So if the time is different in each concentration, this could make the result not accurate and it is hard to compare the results.

Use a stopwatch to time the experiment for each concentration. Therefore this makes sure that this is a fair test.

Volume of solution

If the volume of solution were different in each sample, this would make the experiment not a fair test due to the fact that if the volume of solution is big, it will take a longer time to heat it up.

I will use 10ml syringe to measure the volume of solution instead of 20ml syringe. This can increase the accuracy of my result and also it could reduce the percentage error of syringe.


As the temperature increases, the speed of the reaction will increase. This is because the particles that have been heated will gain kinetic energy and move faster which increase the speed of the reaction. Therefore there is more chance of collision between reactant molecules and the rate of reaction will increase. However, if the temperature were too high, the enzyme would become denatured. This is due to more vibration in the enzyme, which cause the bonds inside the enzyme’s structure to break and this change the shape of the active site of the enzyme. Therefore, the substrate cannot fit into it and no reaction will happen. So this will lower the rate of reaction. In the other hand, if the temperature is too low, this will slow down the rate of reaction too because enzyme will has less kinetic energy which there is less chance of collision between enzyme and the substrate.

I will do my experiment in a water bath so that the temperature will remain constant at all the time. I will check the temperature by using a thermometer.


Each enzyme has an optimum PH, if the PH is too high or too low, this will lead to the breaking of the ionic bonds that hold the tertiary structure of the enzyme in place. The enzyme begins to lose its functional shape, particularly the shape of the active site, such that the substrate will no longer fit into it, the enzyme is said to be denatured. Also changes in pH affect the charges on the amino acids within the active site such that the enzyme will not be able to form an enzyme-substrate complex.

I will use distilled water during the experiment which mean that the enzyme and substrate will be working in a neutral condition.

Concentration of enzyme

The concentration of the enzyme solution also affects the rate of an
enzyme reaction. If there were more enzyme molecules than substrate
molecules then the substrate would be a limiting factor, as there
would not be enough substrate to continue forming products. When all
the active sites are in use, the optimum rate will have been achieved.
But the reaction will take place very quickly and then finish if no
more substrate is added because the substrate will run out.

I will use a set amount of concentration of enzyme. So that we can easily compare the result.

Competitive inhibitor

It is really affective to the reaction. The more the competitive inhibitors, the slower the rate of reaction. This is because Competitive inhibitor has the specific shape as the active site of the enzyme; therefore it can bind in and block the substrate to bind in which would slow down the reaction. But this is only for temporary, unlike the non- Competitive inhibitor; although it does not has the specific shape as the active site, but it can bind into the other parts of enzyme, which change the shape of the active site. So the substrate cannot fit into it and no reaction will happen.





Reason for choice



Hold the burette

Prevent the burette fall down

Boiling tubes


It is used to insert the solution in

It is suitable for small amount of solution to react, as there is less space, so the solution will mix together easily.



Measure the quantity of the solution

Takes an accurate amount of volume making the amount of volumes taken accurate.



Measure the volume of oxygen

It gives an accurate result as it contains big graduation. So it is easy to take readings from it.

Distilled water


Dilute the concentration of hydrogen peroxide

As this is pure, it will have no impurities of which may affect this experiment by causing the concentration to be slightly different to what it should be.

Clamp stand


Hole the G-clamp

Prevent the G-clamp fall down

Delivery tube


Let oxygen to pass through it

Easy to transferred oxygen to burette



To time the experiment

Give a accurate result

Water bath


To hold water

It is big enough to hold both water and the delivery tube.

Rubber bungs


To stop air from passing into the test tubes, and evaporation to take place.

Prevents air and dirt to interfere with osmosis and prevents any water from being evaporated.


  1. Take all the safety precautions before the experiment and set up equipment as shown in the diagram.
  2. Use the syringe to get 5ml of calalase and inject this into the 6 boiling tubes.
  3. Use the thermometer to check if the temperature on the water bath is at 37 degree.
  4. Then remove the bung and place the with the delivery tube. Inject the peroxide into the boiling tube via the bung. Cover the hole with plastercine and start the stopwatch.
  5. Record the change in oxygen every 30 second for 2 minutes 30 seconds.
  6. Do the same for all concentrations on the dilution table and repeat each dilution twice.

Dilution table

Concentration of solution (mol dm-³)

Concentration of hydrogen peroxide (ml)

Distilled water (ml)



















Percentage error – showing us how accurate the experiment was

Percentage error = (error / reading) x 100

Syringe (10ml) – 0.1/10 x 100 = 1%

Burette (25ml)-  0.05/25 x 100=0.2%

Total Percentage error = 1 + 0.2%=1.2%

Overall the percentage error is quiet low, which suggest that my result is really accurate.


To ensure safety during the experiment, safety precautions must be met which include:

  • Wearing safety goggles or glasses to protect chemicals such as iodine in this case from going in the eyes.
  • Standing up and making sure nothing is in the way such as bags and coats/jackets.
  • Make sure the equipment require is only on the work surface to prevent accidents.
  • Keeping breakable equipment such as boiling tubes away from the edge of the table or work surface.

Empty result table

Analysis image05.png

The results show that as the concentration of hydrogen peroxide decreases, the amount of oxygen being released into the measuring cylinder also decreased. This is because the reaction was working at a slower rate due to the fact that the water had diluted the solution. The result of dilution means that there would be fewer enzymes to react with and therefore less oxygen would have been produced.

 This confirms what I said in my prediction, because as the concentration of hydrogen peroxide was weakened, there were fewer hydrogen peroxide molecules for the catalyse to break down, therefore less hydrogen peroxide was broken down and less oxygen was produced.

According to the biological theory, the rate of an enzyme controlled reaction increases in proportion to the concentration of the enzyme (assuming there is excess substrate). The 0.4M concentration is twice as strong as the 0.2M concentration so the rate of reaction should be double what it was for the 0.2M concentration. The relationship between rate of reaction and concentration should be linear i.e. by doubling the concentration of H²0² you double the rate of reaction.

My prediction that the 100%(5ml) concentration would produce the most oxygen in the shortest time was correct, and I was also correct in thinking that 60% would be next, then 40% and then 20%. There is a pattern to the results, as the rate of reaction decreases when the concentration does.

My results support my original predictions. The higher the concentration of substrate, the quicker the rate of reaction because there are more molecules for the enzyme to break down. As a result of this more oxygen can be produced, and my results, averages and graph show this. The graph also supports this, with the total average volume of oxygen increasing along with the concentration of the substrate.

When the concentration of the substrate doubles, i.e. from 20% to 40%, I expected the amount of oxygen to exactly double with it. In one of the experiments, this did happen. The amount of oxygen in the 2minutes 30 seconds was 3.16cm for the 20% and 6.62cm for the 40%, which shows a clear double in the results. As you can see from the graph, the rate of reaction does increase when the H²0² concentration increases. From looking at my graph we can see that when the concentration of the hydrogen peroxide is 0.2 Molar the amount of oxygen produced is 3.16 cm3 and when the concentration is 1.0 Molar the oxygen produced is 15.00 cm3. These results show that a strong concentration will produce more oxygen than a weak concentration.

Rate of reaction = the amount of oxygen / time

The graph below shows the rate of the reaction:


The line has the classic shape of a rate of reaction graph. It starts off steep, becoming shallower until it levels off. This can tell the rate of reaction at any particular time by the slope (gradient) of the line.
The steeper the slope of the graph, the faster the reaction at that point. At the beginning, the reaction goes faster because there are more reactants to react. Whereas at the end, the reaction is slowing down because there are not enough reactants to react.


Rank order

Minor error

Why it is an error?



The substrate will become more active as it is placed into the solution first.

This is due to the temperature in the water bath, which caused the substrate to gain more kinetic energy. This mean that it is more likely to collide with the enzyme and the rate of reaction would increase.

Put the substrate and enzyme into the solution simultaneously.

So that they will have the same amount of kinetic energy.

I used my plasticine to cover the hole in the bung.

Some of the oxygen might have escape as the chemical placed into the solution.

Use your thumb to cover it

This could reduced the amount of oxygen escaped from the test tube

Human reaction time

The reaction time of the person who timing the experiment will be different at all the time. This would decrease the accuracy of my result.

To solve this problem, I will do a reaction time test after the experiment.

So that I can find out the percentage error of the stopwatch.

Rank order

Major error

Why it is an error?



The concentrate of hydrogen peroxide

The technician of who provided this solution may have made an error into providing an accurate solution. As a result, my result may have been affect due to different concentrations being used of which were not even recorded causing a continuous error and to an extent, the whole experiment unreliable as the water potential of a potato cannot be measure accurately.

Make my own sucrose solution

This will cause a more accurate measurement of sucrose solution because I will know what has and has not been put into this sucrose solution therefore is something does go wrong in the experiment I will have a reason why.

Temperature of the water bath might be change.

Therefore the substrate will gain less kinetic energy which do not give me accurate result

Use thermometer to check the temperature before doing next concentration.

This could make sure that my results are accurate.

Parallax error

When I had used a 10mm syringe to measure the volumes of the hydrogen peroxide, due to their being a meniscus and probably bad eyesight, the readings of level of volumes may have been taken down incorrectly my results unreliable.

Put a white piece of paper behind the syringe before measurement.

This will allow me to see the volume of solution within this apparatus more clearly allowing less of a percentage error and result more reliable.

Some of the hydrogen peroxide is trap on the delivery tube

The light can break the hydrogen peroxide, which decrease the accuracy of my result.

Cover the delivery tube with black paper

So that the light cannot go through it and break the H²O² which increase the accuracy of my result.

Overall I think this experiment is really successful due to the fact that all my results matched my prediction. There are only a few anomalous results as shown on the graph, which is circled. However, most of my result is lie on the line of best fit

Reliability-This investigation could have been more reliable but it was limited to the resources of the school. This meant that only two sets of results could have been taken. This reduced the reliability of the investigation because the averages of two sets of results are not as good as three sets. If the investigation were done again a minimum of 3 set would be taken. This would increase reliability of the results. This would allow the results to be more valued so the analysis could be proven. Overall this investigation has gone well. The results would have been better if 3 sets had been taken by the schools resources limited it to only two.

Here is a table to show how reliable my results are.

Volume of oxygen produced (cm³) for each Concentration of Hydrogen peroxide (mol dm-³) in 150 seconds

Concentration of hydrogen peroxide

(mol dm-3)

Set 1

Set 2





Very reliable




Slightly reliable




Fairly reliable




Slightly reliable




Fairly reliable




Slightly reliable




Fairly reliable

For accuracy I tried to be as accurate as possible, however there were some things I didn't control. The results might not have been as accurate as they could have been when timing. I only used a stopwatch, which relied on my reaction, to stop the stopwatch. Everyone's reaction times are different so is not particularly the most accurate way of timing.

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