- 28ml hydrogen peroxide at all times
- 1ml water at all times
-
60cm2 surface area of potato pieces
- Maximum 1ml copper sulphate solution
- Total of 30ml volume of reaction mixture
Method
I will set up my apparatus as shown in the diagram earlier. Using a knife I will then cut the skin off the potato. After this I will cut the potato into long strips 1cm wide by using the potato chipper. As stated from my preliminary work, I will then cut the potato into cubes, which each have a total surface area of 6cm2. I will put 10 of these pieces into a conical flask, giving a total area of 60cm2.
Using a syringe, I will then make up the various concentrations I require. This will be done by putting 28ml of hydrogen peroxide into a small beaker and then making up the other 2ml of water/copper sulphate using another syringe. I will not pour the various liquids directly in the conical flask one at a time as a reaction would already be underway and so would make the results unreliable. This is why I will add the combined solution at the same time.
When the solution has been made and the level of water in the burette measured, I will add the solution to the conical flask, replace the bung and start the timer. Every 20 seconds I will then record the volume of water in the beaker. The difference between the initial volume and the obtained volume will indicate the amount of oxygen produced. After 2 minutes I will then stop the reading, and prepare for the next one. The volumes of copper sulphate I use will be 0ml, 0.2ml, 0.4ml, 0.6ml, 0.8ml and 1.0ml. I will wash the conical flask with water each time I finish a reading to prevent any hydrogen peroxide or copper sulphate from the last reading affecting the results.
Once I have carried this out for the 6 readings, I will do two repeats of each concentration. I will then plot the results in a table.
Scientific Prediction
As I stated in my earlier prediction, I think that the rate of the reaction will decrease as the concentration of copper sulphate inhibitor increases. I think this because there will be more copper sulphate molecules in the solution which, along with the hydrogen peroxide molecules, will also compete for the active site of the catalase enzymes. If this happens then it is clear that less hydrogen peroxide will break down during the reaction, as the enzymes will also be breaking down the copper sulphate.
An enzyme can be defined as a biological catalyst and like any other it is affected by the conditions it is in. The enzyme has a specific shape and this means only substrates with a certain shape can be broken down by a certain enzyme. E.G. protease enzyme can only break down protein molecules, due to the substrates fitting into the active site. It can not however break down lipids due the lipid substrates having a structure which does not allow it to fit in the active site of the protease enzyme.
An enzyme is a complex 3 dimensional globular protein. They are produced by all living cells and speed up a reaction, without altering themselves. The Active site, which is usually a cleft in structure, contains some amino acids which carry out the breakdown of a substance.
H R O
Diagram showing structure of a simple
N C C amino acid
H OH
The enzyme is the quaternary structure of a protein. It is held together by several bonds, which are hydrogen bonds, ionic bonds, disulphide bonds and hydrophobic interactions between non- polar side chains. This structure is brought about by the long amino acid chains coiling up on themselves. Hydrogen bonds then form between the –CO groups of one amino acid and the –NH group of another, which hold this shape in place. This is called an α- helix and is the secondary structure. This structure may coil up into a precise three- dimensional shape which is the tertiary structure.
The shape of the active site in an enzyme is determined by the R groups. The large variety of different R groups means different shape active sites can exist, explaining why the enzymes are specific to one substrate type.
The diagram below taken from ‘Biology 1’ illustrates how an enzyme works.
Diagram showing how an enzyme catalyses the breakdown of a substrate
The previous diagram relates to how the catalase enzyme in my reaction works when no copper sulphate is present. In the left diagram we can see that the enzyme and substrate are in a mixture. The substrate, which in my investigation is hydrogen peroxide, moves into the active site of the enzyme. The two then bind and form an enzyme-substrate complex. It is held in place using temporary bonds which form between the R groups of the enzyme’s amino acid and the substrate. These bonds are weak and thus are not covalent
The specific shape of the enzyme can be understood by the lock and key diagram. The substrate above fits the shape of the active site, so can bind with it. Any other shape will not fit this active site. This can be resembled by a lock and a key theory in that if the key, i.e. the substrate, is not the right shape it will not fit in the lock, which is the enzyme.
Finally the interactions between the substrate and active site of the catalase enzymes cause the hydrogen peroxide to break down into two smaller products. The temporary bonds which form during this process cause a higher tendency for the breakdown of a substance, which in turn reduced the activation energy. During this process the enzyme is unchanged. This allows the protein to break down other hydrogen peroxide molecules so the reaction, as long as the substrate doesn’t run out, occurs at the same rate. The equation below shows what is happening during this process.
Hydrogen Peroxide (l) Hydrogen Peroxide (l) + Oxygen (g)
H2O2 H2O2 O2
I have now explained how an enzyme works without an inhibitor. However when there is an inhibitor present the reaction is very different. When the inhibitor copper sulphate is added to the reaction, instead of enzyme-substrate complexes forming, enzyme-inhibitor complexes form. This is a competitive inhibitor and as a result the rate of the reaction decreases. I predict that the copper sulphate, however, is a non-reversible inhibitor, due to the reaction slowing down so rapidly. I saw this in my preliminary results, in that even a small amount of inhibitor greatly reduced the rate of the reaction.
Diagram showing how a non-competitive inhibitor works
This diagram shows how a non-competitive permanent inhibitor works. It binds with some part of the enzyme, even though it doesn’t have the same shape. In doing so the inhibitor modifies the structure of the active site. This means the substrates are no longer the correct shape to fit the active site and form an enzyme-substrate complex. As a result the reaction greatly slows down, as this enzyme is now out of use.
The reason the enzyme changes shape can be explained by the reactivity series. Iron, which is present in enzymes, is displaced by the copper which is more reactive. Due to this change in structure the enzyme can not function. As well as this copper sulphate breaks disulphide bonds which are present in enzymes. It is these bonds and others which hold the specific shape of the enzyme. Due to the bond breaking the shape and structure changes, making the enzyme not function correctly.
For these reasons I believe that the rate of the reaction will decrease when more copper sulphate is added. For these reasons I predict my graph will look something like the one shown below.
Rate of Reaction
0ml copper sulphate
0.2ml “ “
0.4ml
0.6ml
0.8ml
1.0ml
Time
Safety
Due to hydrogen peroxide being corrosive I shall wear rubber gloves to ensure I do not spill any on my hands. I will also wear safety goggles to protect my eyes from any hydrogen peroxide that could get in my eyes and cause a big hazard. As well as this I will try to work in clean surroundings, with tidy worktops to prevent spillage of any liquid or breakages. When using the knife to cut the potatoes I will take extra care. Finally I shall prevent dropping potato bits on the floor as this causes a large hazard in which I could slip and cause injury to myself and others.
Strategy
As you can see from the previous table I have explained why I have chosen some apparatus over others. I have done this mainly due to the fact that I will be able to obtain more accurate readings. As a result the information I obtain will be more reliable and will show clear trends and patterns. For example the potato chipper is more accurate to cut the potato with then a knife because the holes are all the same size. Therefore all the potato cubes I obtain will be the same size, which erases the chance of second variables- which for this example is surface area. Even if the holes in the chipper are equally bigger than they should be, my accuracy may decrease, yet my results will be just as reliable because that factor applied to all pieces of potato, not just one or two. Therefore a clear trend will still be seen. This can also be applied for when I place the bung on the conical flask. When it is pushed down more air is forced through the delivery tube, so more water will be displaced from the conical flask. This will undoubtedly make all the results slightly inaccurate, but due to it happening to all the readings I take, the error will be cancelled out.
I have also chosen to use precise apparatus as there is a smaller percentage error. This means I can measure more accurately and produce better results.
Analysis
From my results I can see that as the concentration of 0.5M copper sulphate increases, the volume of oxygen produced decreases. This is clearly visible on the graph. I can also see from the graph that the lines are fairly straight which means the rate of the reaction is constant.
These results agree with my prediction in that the copper sulphate is an inhibitor and will slow down the rate of the reaction. I stated in my prediction that as the volume of inhibitor increases there are less successful collisions between the enzyme and the hydrogen peroxide molecules. I can see that when no copper sulphate was present the reaction was faster which obviously implies that the hydrogen peroxide is in competition with the copper sulphate.
The diagram below shows the collision theory in action which applies to changing concentrations of copper sulphate
Hydrogen Peroxide
Copper Sulphate
Water
Catalase
Consider the probability of a collision between any molecule and the enzyme is equal. If this is true then the probability of a hydrogen peroxide molecule colliding with the catalase enzyme:
No. of hydrogen peroxide molecules / Total No. of substrates
= 8/16
= ½
The probability of a copper sulphate colliding with catalase is 1/8. Therefore it is clear that the reaction will be going fairly quickly. For every 8 collisions of hydrogen peroxide with catalase there will be 2 collisions of copper sulphate with catalase. This idea illustrates how the reaction occurred when there was little or no copper sulphate- i.e. there was a faster rate of reaction.
In this diagram the probability of hydrogen peroxide colliding with catalase is still ½. This is because the number of hydrogen peroxide molecules in the solution has not changed. However there are now more copper sulphate molecules. The probability of one of these colliding with the catalase enzyme is:
No. of copper sulphate molecules / Total No. of substrates
= 4/16
= ¼
Therefore for every 8 collisions of hydrogen peroxide with catalase there are now 4 collisions of copper sulphate with catalase. This means the hydrogen peroxide is competing more with the copper sulphate molecules and as a result the rate of the reaction decreases. This illustrates why the reaction slows down in my experiment as the concentration/ volume of copper sulphate increases.
For these reasons I can say that copper sulphate is an inhibitor and reduces the rate of the reaction. This is true for generally all of my results, and although they are not all accurate, they are reliable as they show a similar trend.
Calculating rates of the reaction
Due to the majority of air displaced in first 20 seconds being air forced out of conical flask I will calculate the gradient between 20 to 40 seconds. This will reduce the error margin and present me with more accurate results.
Initial rate of reaction = gradient of line
Gradient= vertical change/ horizontal change
= (4.3-2.4) / (40-20)
= 0.095cm3 of gas produced per second
Gradient= vertical change/ horizontal change
= (3.5-2.4) / (40-20)
= 0.055cm3 oxygen produced per second
Gradient= vertical change/ horizontal change
= (3.7-2.1) / (40-20)
= 0.08cm3 of oxygen produced per second
Gradient= vertical change/ horizontal change
= (3.1-1.9) / (40-20)
= 0.06cm3 of oxygen produced per second
Gradient= vertical change/ horizontal change
= (2.3-1.2) / (40-20)
= 0.055cm3 of oxygen produced per second
Gradient= vertical change/ horizontal change
= (2.3-1.3) / (40-20)
= 0.05cm3 of oxygen produced per second
From the graphs above I can deduce that the rate of the reaction is moderately constant and occurs at a steady rate. I can say this because the line on the graph is straight which means the rate is neither falling nor increasing. This implies that the number of collisions between the hydrogen peroxide molecules and the catalase enzymes is fairly constant throughout each reading. This also means that the enzymes are working at their fastest rate at this ‘setting.’
I believe that the copper sulphate is a non- competitive inhibitor. I think this because the effect of a very low concentration- even under 0.02% drastically reduces the rate of the reaction by up to a factor of 2. The copper sulphate inhibitor works by attaching itself to some part of the enzyme and altering the shape of the active site. This may be because copper is more reactive than the iron found inside the enzyme. As a result the iron is displaced and the enzyme can not function as it would usually do. This change in structure changes the specificity of the enzyme. Bonds holding the enzyme in place, such as the disulphide bonds and strong covalent bonds break. This in turn causes the active site to change shape.
When the altered enzyme collides with a hydrogen peroxide molecule there is not a tight fit between the two. As a result temporary hydrogen bonds do not form between the R groups of the enzyme and the substrate. As it is these interactions which break down the hydrogen peroxide into oxygen and water, the substrate is not broken down and must collide with an unchanged enzyme. This therefore means the rate of the reaction decreases sharply and can be seen by the results I have obtained. The lock and key diagram shown below illustrates how the enzyme would work but does not due to a ‘loose fit.’
1. 2. 3.
Substrate Enzyme
In diagram 1 the substrates move towards the enzyme and locks with its active site and forms an enzyme-substrate complex. During this period interactions occur between the R groups of the enzyme and the substrate, which causes the substrate to break down, as shown in diagram 3. This theory applies to the enzymes in my reaction that had not reacted with the copper sulphate. Those enzymes could therefore function properly as their structure had not been altered.
1 2
This diagram shows the effect copper sulphate is having on the enzyme. This illustrates how the copper sulphate reacts with the iron and causes the enzyme’s bonds holding it in shape to break. The active site, as you can see above, has clearly changed in shape. As a result a tight fit can not occur between the two. This means the temporary bonds can not form and due to this the substrate is not broken down. This is what occurred in my reaction when I added copper sulphate and is the reason for the reaction rate decreasing. It is therefore clear how enzymes function using the lock and key diagram. The key, which is the substrate, can not fit in the lock, i.e. the active site of the enzyme. As a result the door can not be opened, which in reaction terms means the reaction can not occur.
The reaction between catalase enzyme and hydrogen peroxide is found in the human body. Catalase is found in the liver, and it is here that poisons are broken down. This means therefore that the hydrogen peroxide is a poison. Hydrogen peroxide is a corrosive one and would cause much harm to the body if left how it is. The enzyme is able to break up the hydrogen peroxide into oxygen and water, shown in the equation below.
Hydrogen Peroxide (l) Oxygen (g) + Water (l)
2H2O2 O2 2H2O
The enzyme works by lowering the activation energy required for the reaction to occur. As a result more hydrogen peroxide is converted more quickly, which means hydrogen peroxide causes less damage to our body than it could.
Energy
Reactants
H2O2
Progress of Reaction
Activation energy required without enzyme
Activation energy required with enzyme
As you can see from the graph above the presence of an enzyme means the activation energy required for the reaction to occur is lessened. The Ea (Activation energy) can be thought in terms of pushing a boulder up a hill. Doing so on your own is difficult and uses more energy. If however you have a shortcut the energy you require is less. This is exactly how an enzyme works- by finding a route for the reaction to occur which requires less energy. In this way enzymes speed up the rate of a reaction.
Diagram taken from Biology 1 Advanced Sciences
Diagram taken from Biology 1 Advanced Sciences
