Find out what effect substrate concentrations have on the dissociation of hydrogen peroxide within the presence of the enzyme catalase.

PLAN

2a The aim of the investigation is to find out what effect substrate concentrations have on the dissociation of hydrogen peroxide within the presence of the enzyme catalase.

Catalase is the enzyme found in potato cells which specifically catalyses the decomposition of hydrogen peroxides. Hydrogen peroxide is a toxic BI product of metabolism in certain plant and animal cells. It is used in hair salons as a bleaching agent. The enzyme Catalase effectively removes it. Catalase is one of the fastest acting enzymes known. A single molecule of catalase can break 5.6 million molecules of hydrogen peroxide each minute.

This catabolic reaction breaks down hydrogen peroxide into water and releases oxygen gas, which is liberated, via effervescence.

Equation. 2H2O2(l) Catalase 2H2O(l) + O2 (g)

2b The enzyme catalase is required so that oxygen gas can be evolved. This can be achieved by placing the specimens of potato into various concentrations of hydrogen peroxide solution. Allowing the enzyme catalase to bind with the hydrogen peroxide substrate molecule to form an enzyme substrate complex. Allowing oxygen gas (product of the reaction) to evolve, which will be collected using a measuring cylinder by oxygen gas displace water out of the measuring cylinder filling it with the gaseous product of the reaction. Which will be timed in seconds in order to calculate the rate of dissociation.

4a Enzymes are extremely sensitive to environmental conditions. In order for this experiment to produce valid data the following factors are going to be taken into account. This is because any factor which alters the conformation (dependent on the tertiary and secondary structure) of the enzyme catalase will alter the shape of the active site; therefore it will affect the frequency of the enzyme substrate complex formation and thus influence the rate of the enzyme-catalysed reaction. These factors include temperature, (pH and pressure) which are consistent, enzyme concentration and substrate concentration, also the surface area of the enzyme.

4b8a In order to produce valid and reliable data from the experiment all other factors such, as temperature must be kept constant and at an optimum level. This is primarily due to the fact that if one factors then other factors can alter and can alter the result. Therefore the system and surroundings of the experiment will be kept under controlled conditions.

In order to control all extraneous variables, the following precautions will be applied to the experiment.

The experiment will be conducted in a water bath, which will maintain a constant temperature and will remain the same for all the experiments. The same size and shape and mass of potato in all the experiments, so that there will be an equal amount of surface area and enzyme catalase in each experiment in order to produce valid and reliable results.

Everyone has to do the experiment in the same place and same time. This is because the temperature will be the same for everyone as far as possible. The independent variable in the experiment is the substrate concentration. Which will be an increase in concentration of hydrogen peroxide from 4%-20% in 2% increments. This will help make the results valid and reliable. It will make it easier to rule out any anomalous results. The dependent variable is the measuring the amount of oxygen gas evolved, which will be achieved by timing in seconds how long it takes for oxygen to produce and collect by downward displacement of water in an inverted measuring cylinder to fill 2 cm 3 of oxygen gas.

Each experiment will take place in a water bath in order to maintain a constant temperature. This is due to the fact that an increase in temperature to the system will increase the amount of molecular motion (a gain in kinetic energy). Therefore the molecules of substrates and the enzymes will move more quickly, which will increase the amount of collisions. Which will increase the amount of substrate molecules colliding with the active site making the reaction faster. Which is the optimum temperature. If the temperature is reduced to below freezing then the enzyme is inactivated. Also the water in the bowl is used to keep a constant pressure on all the experiments in order to collect more accurate results

For each experiment, the same potato shall be rinsed in de-ionised water to ensure fairness in the results and reused as with using a clean needle to hold the cylinders of potato. In theory this should work because before binding, the enzyme catalase active site is relaxed. When the hydrogen peroxide substrate molecule binds to the enzyme active site it forms and enzyme substrate complex, which is known as the lock and key hypothesis. The substrate complex only lasts for a fraction of a second and the substrate molecule is released. Therefore the catalase enzyme is not used up. Potato ...

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2cEquipment list

The equipment required for the experiment.

Borer
Scalper
Ruler
Two Large bowls water.
1 Boiling tube with bung
10 cm3 measuring cylinder
Deliver tube
Stop clock
Potatoes
Needle
Pipette
Test tube wrack

Other materials for conducting this investigation is de ionised water, hydrogen peroxide for the dilutions

Diagram of apparatus

Diagram of discs of potatoes

6abcd Method

6a. The method below measures the rate of an enzyme controlled reaction varying with different amounts of substrate concentrations ranging from (4%-20%) in 2% increments. I will measure the time (using a stop clock) in seconds (cm3/sec) taken for oxygen to evolve and displace 2cm3 of water in a 10 cm3 measuring cylinder. I will repeat each concentration 3-5 times. The range of measurements should be collected in a table and analysed by drawing a graph. After each experiment the potato will be rinsed in de ionised water.

The method below outlines the procedure for conducting the experiment.

PROCEDURE

Prepare 250cm3 of required concentrations of hydrogen peroxide, using a mechanical pipette from (4%-20%) in 2% increments, by diluting the 20% concentration of hydrogen peroxide down by adding the required amount of de ionised water. Place into test tubes in a test tube wrack.
Cut Potato using a borer (1 cm3in diameter) a cylindrical piece of potatoes.
Slice the potato into 5 2-mm thick discs.
Clean needle so there is no contamination and mount disc onto needle 5-mm apart from each other.
Set up apparatus as shown in the diagram above. Filling the bowl with water.
Fill measuring cylinder with water ensuring no air bubbles are in it and hold the measuring cylinder up, ensuring delivery tube is placed inside it.
Place the 250 cm 3 of the required substrate concentration solution in the test tube.
Place the needle with potato discs into the test tube and quickly place the bung in
Start the stop clock and record the length of time it had taken for the oxygen gas to empty the measuring cylinder
Record how long it takes using a stop clock in (cm3/sec) to displace 2 cm3 of water and comment on your results.
Repeat the experiment from 0-20 % hydrogen peroxide solutions with a 2% increment interval between each recording 3-5 times.

8b/c The method, which is going to be used for the experiment will, is by collecting all the quantitative data (which is the time in seconds for the oxygen gas to displace 2 cm3 of water in the measuring cylinder) and placing it into a result table. Each experiment will be repeated 3-5 times in order to ensure reliability. The average timing can be found for each concentration and also the rate. Which will then be analysed by drawing graphs and plotting the line of best fit and draw conclusions. And then find the initial rate of the reaction after the results will be found we are then able to conduct statistical tests on them. Also the average rate of oxygen evolved can be found.

Hypothesis

For a given enzyme concentration, the rate of an enzyme controlled reaction increases with increasing enzyme substrate.

FAIR TEST.

In order to ensure the experiment was fair the following procedures were done. Firstly the different substrate concentrations were made in one batch to ensure everyone had

An equal independent variable and to increase the chance of an accurate result. The temperature was constant due to placing each test tube in a water bath with the same temperature from the tap. This is due to the fact that an increase in temperature to the system will increase the amount of molecular motion (a gain in kinetic energy). That will increase the rate of the reaction because there will be an increase probability of enzyme and substrate collisions to form complexes. The pressure was maintained at a constant level due to the water bath. All the apparatus were cleaned before the experiment was conducted. This is because it will improve the validity of results each potato was cleaned with de-ionised water to maintain a constant net movement via osmosis. Each potato was cut to the same4 mass and size and shape. This is to improve the fairness of the test. Also to make the enzyme concentration the same. The same pH level was maintained as the same potato was used. The surface area of the potato for each experiment was the same. The hydrogen peroxide can be an irritant at high concentration. Therefore goggles have to be worn. Also spillages have to be cleaned with a damp cloth.

C analysing evidence and drawing conclusions

Results

The table below shows the quantitative data obtained during the experiment with the number of repeats done for each concentration.

4a The rate of the reaction has been worked out by 2 cm3 divided by the number of seconds.

Adding up every result found for each concentration and dividing it up by how many times it was repeated have worked out the mean time calculation.

The table also shows the number of times the same concentration has been repeated.

I predict that the graph will showed that as there is an increase in concentration, there would be an increase in the rate of oxygen evolved. Then after a very high concentration, the graph will levels off because all the enzymes are saturated.

Summary table commentary

2b The table shows the quantitative data produced by the experiment. There is a general trend in the results that shows that as the concentration of hydrogen peroxide increases, the rate of oxygen dissociated increases.

The graph commentary

4b According to the graph there is a positively linear relationship, with a constantly rising slope. This shows that as variable x (concentration of hydrogen peroxide) increases as the variable y (the rate) increases. Which then levels off eventually. This is because the enzyme concentration acts as a limiting factor. The second graph (B) shows that there is another line of bets fit, which goes through more points and, shows the general trend that as the substrate concentration increases the rate increases, but the problem is that it doesn’t level off and continually increases proportionally.

6a Enzymes are biological catalysts. Their function is determined by secondary and tertiary structure. The reaction takes place in a small part of the enzyme called the active site, while the rest of the protein acts as scaffolding.

Enzymes bind temporarily to one or more of the reactants of the reaction they catalyse. In doing so, they lower the amount of activation energy needed and thus speed up the reaction. Enzymes are proteins, which generally work rapidly and are not destroyed by the reactions they catalyse. In any chemical reaction, a substrate (S) is converted into a product (P):

S P

In an enzyme-catalysed reaction, the substrate first binds to the active site of the enzyme to form an enzyme-substrate (ES) complex, then the substrate is converted into product while attached to the enzyme, and finally the product is released. This mechanism can be shown as:

E + S ES EP E + P

The enzyme is then free to start again. The end result is the same (S P), but a different route is taken, so that the S P reaction as such never takes place.

The rate of an enzyme-catalysed reaction shows a curved dependence on substrate concentration. As the substrate concentration increases the rate increases because more substrate molecules can collide with enzyme molecules, so more reactions will take place. At higher concentrations the enzyme molecules become saturated with substrate, so there are few free enzyme molecules, so adding more substrate doesn't make much difference (which increases the rate of E-S collisions).

The maximum rate at infinite substrate concentration is called vmax, and the substrate concentration that gives a rate of half vmax is called KM

6b/c The graph shows two phases at lower concentrations the rate increases directly proportional with the substrate concentration e.g. at 6% an 8% concentration. This is due to the fact that there are few substrate molecules available; therefore there are many active sites available. Therefore increasing the substrate concentration will increase the rate.

At an 18% concentration the graph starts to slope and decrease in rate this is because there is an increase in substrate molecules and all the active sites available in the enzymes are engaged in catalysis (maximum rate of reaction). The theoretical maximum hasn’t been obtained, but at the highest concentration of 20% there is no significant change in rate. The graph becomes Plato and begins to level off. This is because at high concentrations the active sites of the enzyme molecules are saturated with substrate particles. Which means that the extra substrate has to wait until the enzyme substrate complex has released the products before it enters the active site of the enzyme. Therefore the enzyme concentration acts as the limiting factor.

8a/b The graph shows some anomalous results, which are excluded in the line of best fit. The quantitative data produced shows some anomalous results.

The rate (8 and 10%) and (16 and 12%) concentration shows the same rate of dissociation. Which doesn’t fit into the first graph. This is due to the wide range of timings produced for each experiment. The 16% concentration on the first attempt gives a figure of 217 seconds whereas on the third attempt it gives a figure138 sec. that almost gives a third less then the first attempt. This can explain why the rate is the same. Another result, which is of the 12% on the fourth attempt, shows a 214-second rate. The rate proves a general trend. But some have the same rate with an increase in concentration for example the 6 and 8% show a rate of 0.008 cm3/sec. As well as the 12 and 16% that show a rate of 0.011 cm3/sec. The wide spread of reading may be due to some error in the set up of the apparatus. Firstly the seal between the glass and the bung where not connected properly and the bung kept on coming off the glass rod which then allows oxygen gas to escape gibing more time for the gas to displace 2 cm3 of water in the measuring cylinder. Therefore this will have its effect on the rate making it slower and more inaccurate, the rubber tube became very loose as well. . It may have been because the water was in the delivery tube preventing oxygen gas through the pathway. Or because the bung came out of the boiling tube, allowing the oxygen gas to escape. Hence take longer.

8c Firstly it is important to keep a constant temperature Up to the optimum temperature the rate increases geometrically with temperature (i.e. a curve shape). The rate increases because the enzyme and substrate molecules both have more kinetic energy so collide more often, and also because more molecules have sufficient energy to overcome the (greatly reduced) activation energy. Above the optimum temperature the rate decreases as more and more of the enzyme molecules denature. The thermal energy breaks the hydrogen bonds holding the secondary and tertiary structure of the enzyme together, so the enzyme (and especially the active site) loses its shape to become a random coil. The substrate can no longer bind, and the reaction is no longer catalysed. At very high temperatures this is irreversible and the enzyme is denatured.

Enzymes have an optimum pH at which they work fastest. For most enzymes this is about pH 7-8 (physiological pH of most cells), but a few enzymes can work at extreme pH, such as protease enzymes in animal stomachs, which have an optimum of pH 1. The pH level can alter the ionic charge of the acidic and basic and basic groups, which disrupts the ionic bonding which maintains the specific shape of the enzyme. Which then alters the whole shape of the enzyme

As the enzyme concentration increases the rate of the reaction increases linearly, because there are more enzyme molecules available to catalyse the reaction. At very high enzyme concentration the substrate concentration may become limiting factor, so the rate stops increasing. Also the substrate concentration is important

The amount of surface area of the potato is an important factor to consider because if there is an increased amount of surface area then the rate of reaction will be increased, as there will be more respiratory surface available. Hence the thinner and wider surface area the greater the rate of diffusion of the oxygen gas. Also the pressure is important. Also the mass is an important factor. This is because the bigger the mass the more enzymes there are. Therefore in order to maintain constant concentration of enzymes the mass has to be the same as far as possible. This will increase the validity of the results.

. The independent variable in the experiment is the substrate concentration. Which will be an increase in concentration of hydrogen peroxide from 4%-20% in 2% increments. The dependent variable is the measuring the amount of oxygen gas evolved, which will be achieved by timing in seconds how long it takes for oxygen to produce and collect by downward displacement of water in an inverted measuring cylinder to fill 2 cm 3 of oxygen gas.

Evaluation

Firstly the apparatus used in the experiment gave limitations, which effectively altered some of the result.

One of the limitations on the apparatus was estimating how much oxygen the measuring cylinder took in. Also using a measuring cylinder was really inaccurate this is because judgement had to be made. Instead a pipette would have been better, because it’s a more accurate measure of solutions. The stop clock gave limitations because there had to be another judgement of when to stop the clock whichn is makes the data become inaccurate because everyone has different reaction speeds. This could have also made an effect on the graph causing the anomalous results. This is because the times would be inaccurate. Water in the delivery tube sometimes pushed water out, making it suck back. The bung kept on coming out which allowed oxygen to escape. Also the actually mass and surface area of the potato would have an effect on the results. If the potatoes were large then the concentration of enzymes would have increase therefore more substrate molecules would collide with the active sites of the enzymes, which would increase in number. This would have increased the rate of reaction causing anomalous readings making the data invalid. Also if the potatoes had a decrease in mass due to inaccurate cutting then there would be fewer enzymes for the substrate molecules to bind with, which will decrease the rate of reaction, again causing anomalous readings.