When varying the concentration of either amylase or starch, I do not require such a large volume. Using a 250cm3 volumetric flask for just making up 30cm3 of a solution would mean a lot of solution is wasted. Therefore the solution will just be made in the beaker in which the solid is weighed out. However, to ensure that the total volume of water added to the solution is accurate, I will use a burette to add the water. This procedure is shown below.
- As above
- As above
-
Using a burette I will add the required volume of distilled water to the beaker- for all concentrations other than the standard concentrations I am using- i.e. 4% amylase and 1% starch- the required volume of distilled water is 30cm3.
- Using a glass rod I will then dissolve the solid in the distilled water, and thoroughly stir the solution to allow the solid to be evenly dissolved in all the solution.
In this way I can make up the concentration I require accurately without wasting a lot of solution.
Making the Concentrations of the Chemicals
Variable 1- Temperature
In order to take samples of the solution it is essential that I have enough starting solution to carry out the whole experiment.
- The table above is a rough table showing the results I intend to collect. As you can see there are, in total, 15 readings for each temperature and therefore 75 readings in total.
-
The cuvette volume, which will measure the light transmission, is approximately 3cm3. I will therefore allow for spillage etc and assume that for each sample I require 4cm3.
In order for the reaction of the amylase enzyme and the starch substrate to be fair, the two substances will be in the same ratio throughout this variable. From previous data research I have found out that 1cm3 Starch 1%: 1cm3 Amylase 5% results in a fairly quick reaction. I will therefore use 1cm3 starch 1% to 1cm3 amylase 4% as the two standard concentrations.
I can now begin to plan how I will make these concentrations.
For this variable I require 60cm3 of amylase/starch solution for each temperature. Therefore in total I need 420cm3. As the enzyme and the substrate are going to be added 1cm3 starch to 1cm3 amylase, I will require 210cm3 amylase 4% and 210cm3 of starch 1%.
4% amylase
4% of 210 = 8.4
Therefore I need to mix 8.4g of amylase powder with 210cm3 of distilled water to make a 4% amylase solution. The specific details of making up the solutions are shown in my method later on.
As well as this I require 210cm3 starch 1%.
1% Starch
1% of 210 = 2.1
Therefore I need to mix 2.1g of starch powder with 210cm3 of distilled water to make up a 1% solution.
Variable 2- Amylase Concentration
- Again the table above is a rough one showing the kind of data I hope to record. Again there are 15 readings for each concentration. However because the amylase concentrations are different I will have to make up each concentration separately.
-
15 readings x 4cm3 (maximum cuvette volume) means I will need to make a total of 60cm3 solution for each concentration.
-
As I am having 1cm3 amylase: 1cm3 starch, I will only require 30cm3 of each to make the combined total.
- Below is how I shall make each concentration:
Amylase 5%
5% of 30 = 1.5
Therefore I need to mix 1.5g of amylase powder with 30cm3 of distilled water to make a 5% amylase concentration.
Amylase 4%
4% of 30 = 1.2
Therefore I must mix 1.2g of amylase powder with 30cm3 of distilled water to make a 4% amylase concentration
Amylase 3%
3% of 30 = 0.9
Therefore I must mix 0.9g of amylase powder with 30cm3 of distilled water to make a 3% amylase solution
Amylase 2%
2% of 30 = 0.6g
Therefore I must mix 0.6g of amylase powder with 30cm3 of distilled water to make a 2% amylase solution
Amylase 1%
1% of 30 = 0.3g
Therefore I must mix 0.3g of amylase powder with 30cm3 of distilled water to make a 1% amylase solution
As the reactants are being put in as 1cm3 amylase: 1cm3 starch I will require 180cm3 of starch solution.
1% starch
1% of 180 = 1.8g
Therefore I must mix 1.8g of starch powder with 180cm3 of distilled water to make a 1% starch solution
Variable 3- Starch Concentration
My 3rd variable will be to see how the action of amylase is affected by substrate concentration. Below is a rough table showing some of the main details I will try and obtain.
The total solution required for each concentration is:
15 x 4cm3 (max cuvette volume) = 60cm3
As the enzyme and substrate and being added in equal quantities I will require 30cm3 of starch solution and 30cm3 of amylase solution for each concentration
Starch 5%
5% of 30 = 1.5
Therefore I need to mix 1.5g of starch powder with 30cm3 of distilled water to make a 5% starch solution.
Starch 4%
4% of 30 = 1.2
Therefore I must mix 1.2g of starch powder with 30cm3 of distilled water to make a 4% starch solution
Starch 3%
3% of 30 = 0.9
Therefore I must mix 0.9g of starch powder with 30cm3 of distilled water to make a 3% starch solution
Starch 2%
2% of 30 = 0.6g
Therefore I must mix 0.6g of starch powder with 30cm3 of distilled water to make a 2% starch solution
Starch 1%
1% of 30 = 0.3g
Therefore I must mix 0.3g of starch powder with 30cm3 of distilled water to make a 1% starch solution
As the reactants are being put in as 1cm3 amylase: 1cm3 starch I will require 180cm3 of starch solution.
Amylase 4%
4% of 180 = 7.2g
Therefore I required 7.2g of amylase powder with 180cm3 of distilled water to make a 4% amylase solution
Variable 4- Copper Sulphate Inhibitor
As you can see in the above table, the copper sulphate concentrations are very small. I have decided these volumes based on previous research, in which I investigated the action of copper sulphate on catalase. I therefore expect the reaction to occur somewhat the same, with small concentrations having large effects on the output of the enzyme.
Once again my standard solutions for this variable will be amylase 3% and starch 1%. As there are 90 readings in total and a cuvette volume of 4cm3, I will require about 360cm3 total solution. I therefore require approximately 180cm3 of amylase and 180cm3 of starch.
Amylase 4%
4% of 180 = 7.2
Therefore I must add 7.2g of amylase powder to 180cm3 of distilled water to make a 4% amylase solution
As the reactants are being put in as 1cm3 amylase: 1cm3 starch I will require 180cm3 of starch solution.
1% starch
1% of 180 = 1.8
Therefore I must add 1.8g of starch powder to 180cm3 of distilled water to make a 1% starch solution
To make the copper sulphate concentrations I will make a total of 10cm3 copper sulphate/water solution.
0.1% Copper Sulphate
0.1% of 10 = 0.01g
Therefore I must add 0.01g of copper sulphate to 10cm3 of distilled water to make a 0.1% copper sulphate solution
0.2% Copper Sulphate
0.2% of 10 = 0.02
Therefore I must add 0.02g of copper sulphate to 10cm3 of distilled water to make a 0.2% copper sulphate solution
0.3% Copper Sulphate
0.3% of 10 = 0.03
Therefore I must add 0.03g of copper sulphate to 10cm3 of distilled water to make a 0.3% copper sulphate solution
0.4% Copper Sulphate
0.4% of 10 = 0.04
Therefore I must add 0.04g of copper sulphate to 10cm3 of distilled water to make a 0.4% copper sulphate solution
0.5% Copper Sulphate
0.5% of 10 = 0.05
Therefore I must add 0.05g of copper sulphate to 10cm3 of distilled water to make a 0.5% copper sulphate solution
Scientific Theory
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 a certain enzyme can break down only substrates with a certain shape. For this reason amylase can only break down starch, due to the substrates fitting into the active site. It cannot however break down lipids due the lipid substrates having a structure that does not allow it to fit in the active site of the amylase enzyme.
Amylase is a complex 3 dimensional globular protein. The active site, which is usually a cleft in structure, contains some amino acids that 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. The long amino acid chains coiling up on themselves bring about this structure. 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 R groups determine the shape of the active site in an enzyme. 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 speeds up the breakdown of a substrate
The above diagram relates to how the amylase enzyme in my reaction works. In the left diagram we can see that the enzyme and substrate are in a mixture. The substrate, which in my investigation is starch, 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 that form between the R groups of the enzyme’s amino acid and the substrate. These bonds are weak and thus are not covalent.
The lock and key diagram can understand the specific shape of the enzyme. The substrate above fits the shape of the active site, so can bind with it. Any other shape will not fit this active site. A lock and a key theory can resemble this 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 amylase enzyme cause the starch to break down. The temporary bonds, which form during this process, cause a higher tendency for the breakdown of a substance, which in turn reduce the activation energy.
Now that we know what an enzyme is we can see how altering its surroundings would cause a change in the rate of the reaction.
Temperature
For a reaction to occur the particles must collide with a certain minimum kinetic energy. The size of this kinetic energy needed varies between reactions due to different bond enthalpies. An enzyme works by reducing the activation energy as shown in the diagram below.
Energy
Progress of Reaction
As you can see, the enzyme lowers the activation energy, which gives a greater number of particles the minimum energy required for the reaction to occur. The energy profile diagram shows the smaller peaks, which arise as a result of the enzyme. The first is from the formation of the enzyme-substrate complex. After this stage the enzyme-product complex forms, which also requires energy but less so than the enzyme-substrate complex. Finally the enzyme and products move away, and the enzyme can then be used again to convert a starch molecule to maltose.
In a solution like mine, which will contain amylase enzymes and starch, the particles have a range of different kinetic energy. Most of the particles will be moving at moderate speeds, others will have slightly greater kinetic energy and some will have slightly less. When the temperature of the reactants rises, they move around faster and have a greater amount of kinetic energy.
This means that of those substrates that collide with the enzymes, the amount of energy of the impact is more likely to exceed the activation energy. The amylase enzyme works by lowering the activation energy further, so that a larger number of molecules have the required energy and can cause a reaction. This is illustrated in the diagram below.
Number of collisions
With Kinetic Energy,
E
Kinetic Energy (E)
The above diagram shows that only a small proportion of the molecules have the energy E to overcome the activation energy (which in this case is 50kg mol-1), and to cause a reaction to occur. If we now however raise the temperature the graph will look like the one shown below.
Distribution curves showing effect of a temperature rise of 10K on the proportion of reactions with greater than 50kg mol-1
From the graph you can see that by increasing the temperature by 10 Kelvins the graph has shifted to the right- i.e. there is a higher average kinetic energy of each particle. There is a much higher proportion of molecules with greater than 50kg mol-1, which means more collisions will be successful enough for a reaction to occur.
Therefore increasing the temperature will raise the average kinetic energy of the reactants, which will enable a larger number of reactions to occur. This is why I have chosen to vary temperature, as I believe it will alter the rate of the reaction. However raising the temperature too far will cause the enzyme to become denatured. This means the bonds holding the tertiary structure of the enzyme together will be overcome, and the active site will have altered in shape. As a result the temporary bonds that occur between the substrate and the enzyme cannot form. This would therefore cause the reaction rate to decrease.
Rate of Reaction
Temperature
The diagram above summarises how the overall reaction rate varies with temperature. The rate begins to decrease as more and more enzymes become denatured. The diagram below shows a rough representation of how the structure of the enzyme changes when it is denatured.
As you can see the active site, which was once a cleft in the structure has basically disappeared. Bonds cannot form between the starch and the amylase, so starch is not converted to maltose.
Concentration
Increasing the concentration of either the enzyme or substrate will cause a rise in the rate of the reaction. This is because there will be more collisions occurring between the enzymes and the substrates.
Number of collisions
With Kinetic Energy,
E
Activation energy
Kinetic Energy (E)
The diagram above shows that when the concentration of the solution is 3%, there are few molecules that have the kinetic energy needed for a reaction to occur. The area between the curve and the x-axis represents the number of molecules in total.
When the concentration is raised to 5%, however, the curve shifts upwards. This means there are more molecules in total, and therefore there are a larger number of molecules with the energy needed to react. For this reason, more reactions occur at a time when the concentration is raised.
During an enzyme-catalysed reaction there are several stages that occur between the start product and the end product. This can be summarised in the flow chart shown below.
Figure 1
Enzyme + Substrate Enzyme-Substrate Complex
(E) (S) (ES)
Enzyme-Product Complex
(EP)
Enzyme + Product
- (P)
The flow chart shows that by increasing the enzyme concentration, more substrates will be able to be form enzyme-substrate complexes at a time. The rate will therefore increase. If, however, the substrate becomes a limiting factor- i.e. there is insufficient substrate concentrations, then the rate will stop rising and level off. If, however, there is too much substrate, the reaction rate can begin to slow down. This will only occur with extremely high concentrations of substrate. It occurs because there are so many substrates that they cannot move around freely and lack the kinetic energy required for the reaction to occur.
Inhibitor
I have now explained how an enzyme works without an inhibitor. However when there is an inhibitor present the reaction is very different. When copper sulphate is added to the reaction, instead of enzyme-substrate complexes forming, enzyme-inhibitor complexes form, assuming the inhibitor is competitive. However the copper sulphate could be a non-reversible inhibitor. This could be seen in that even when more substrate is added, the rate doesn’t increase.
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, and 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. Due to this change in structure the enzyme cannot function. As well as this, copper sulphate breaks disulphide bonds, which are present in enzymes. It is these bonds and others that hold the specific shape of the enzyme. Due to the bonds 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.
Rate Equations and Order of Reactions
The rate equation gives information about the slowest step in a reaction and can therefore be used to determine the mechanism of a reaction. Figure 1 on page 18 shows the reaction mechanism for an enzyme- catalysed reaction. The slowest step is referred to as the rate-determining step. When there is a low substrate concentration, the slowest step will be the enzyme and substrate coming together before forming the enzyme-substrate complex. However when there is a high substrate concentration, clearly the rate is dependant on the enzyme concentration, as there are insufficient enzymes to substrates.
For a reaction
A + B Products
Rate = k [A]m[B]n R
M and N are referred to as the order of the reaction and gives information about how the concentration of A or B affects the reaction rate. To find the rate constant the temperature must be kept constant and the concentrations also.
In my investigation I shall be using starch and amylase. After doing practical work I will then be able to find out the order of reactions. This will then allow me to judge how the concentration of each reactant affects the rate of reaction.
For a reaction with first order with respect to a reactant, the concentration of that reactant is proportional to the rate of the reaction. This can be seen when the initial rate of reaction is plotted against the concentration of that reactant, and a straight line is produced.
For a reaction with second order with respect to a reactant, the concentration of that reactant is proportional to the square of the rate of reaction. This means at low concentrations it has little effect on the reaction rate, but as the concentration progressively increases, so does its effect on the rate of the reaction.
For a reaction with zero order with respect to a reactant, the concentration of that has no effect on the reaction rate. This can be explained my looking at the reaction mechanism- not by just looking at a balanced overall chemical equation. This time, when the initial rate is plotted against the concentration of the reactant, a horizontal line comes off from the y-axis. The example below adapted from Chemical Ideas, shows how a reaction can occur as second order in respect to a reactant.
Example reaction showing how a reactant can be zero order
CH3 CH3
CH3 – C – Br + OH- CH3 – C – OH + Br-
CH3 CH3
1 CH3 CH3
CH3 – C – Br CH3 – C+ + Br-
CH3 CH3
2 CH3 CH3
CH3 – C+ + OH- CH3 – C – OH
CH3 CH3
The overall reaction is found to be first order in respect to (CH3)3CBr and zero order with respect to OH-. The order of reaction is therefore 2. This means that the rate is directly proportional to the concentration of (CH3)3CBr, but the concentration of OH- has no effect on the rate of the reaction. Why is this so?
The effect of OH- has been found to not affect the rate of reaction because it occurs in two stages. In the first stage, which is shown in step 1, the C-Br bond breaks heterolyticaly. This therefore means the OH- concentration doesn’t affect on the rate of this bond breaking.
The second part of the reaction, shown in step 2, is the reaction of the carbocation with the OH-. This stage of the reaction occurs very quickly, like most nucleophilic reactions. Stage 1, however, is slower than this and so the rate depends only on the concentration of (CH3)3CBr. Step 1 is therefore called the rate-determining step. The rate equation for this reaction is therefore
Rate = k [(CH3)3CBr]
This shows that the rate equation, as stated earlier, cannot be found just by looking at an overall balanced equation. Instead the mechanism of the reaction must be found.
In an enzyme-catalysed reaction, the enzyme uses up the substrate. Therefore, to find the order or reaction of that substrate, its concentration must be kept the same. This can be done by just working out the initial rate of the reaction- i.e. when the substrate concentration exceeds the enzyme concentration.
Therefore, after doing my practical, I will be able to analyse the results and see what order of reaction the reactants are.
Diagram taken from Biology 1 Advanced Sciences page 26 by Mary Jones, Richard Fosbery and Dennis Taylor
Diagram taken from Biology 1 Advanced Sciences page 42 by Mary Jones, Richard Fosbery and Dennis Taylor
Diagram taken from Salters Advanced Chemistry: Chemical Ideas 10.2 page 225
Diagram taken from Biology 1 Advanced Sciences page 48 by Mary Jones, Richard Fosbery and Dennis Taylor
Example taken from Salters Advanced Chemistry: Chemical Ideas 10.3 page 235