Find out how a certain factor that is changed will affect the reaction rate of enzymes

Investigating Enzymes

By Hannah Robinson

Aim: The aim of this experiment is to find out how a certain factor that is changed will affect the reaction rate of enzymes.

Enzymes are globular proteins which act as biological catalysts. A catalyst is a molecule which speeds up reactions without being used up itself. All globular proteins have a tertiary structure, which makes them very easily altered by pH and temperature. Enzymes are different to other globular proteins as they have an active site to which other molecules can bind. The molecule is called the substrate for that enzyme. The shape of the active site allows the substrate to fit perfectly and be held in place by temporary bonds between the substrate and the R groups of the enzymes amino acids.

An amino acid

The combined structure is called the enzyme-substrate complex.

The active site for all molecules of one enzyme is made up of the same arrangement of amino acids, and therefore has a highly specific shape. There is usually only one active site on each enzyme molecule and only one type of substrate molecule that will fit into it. The enzyme is said to be specific for that substrate. This is known as the lock and key hypothesis.

The enzyme may catalyse a reaction in which the substrate molecule is split into two or more molecules. It may also catalyse the joining of two or more molecules together as when making a dipeptide.

Interaction of the R groups of the enzyme and the atoms of the substrate can break or encourage the formation of bonds in the substrate molecule. When the reaction is complete, the product or products leave the active site. The enzyme is unchanged by this process, so it is then available to receive another substrate molecule.

Some research on enzymes has suggested that the active sites of enzymes may not be exactly the right shape for the substrate to begin with. It is thought that when the substrate combines with the enzyme it causes a small change to occur in the shape of the enzyme molecule, therefore enabling the substrate to fit more snugly into the active site. This is called the induced fit hypothesis.

Reactions proceed because the products have less energy than the substrates. However most substrates require an input of energy to get the reaction going. This energy required to initiate the reaction is called activation energy.

In mammals, body temperature is often higher than the temperature of the air around them. In humans, it is 37ºC, but even this temperature is often not high enough to give most of the substrates the activation energy which they need to change into products.

Enzymes help to decrease the amount of activation energy needed to start the reaction by holding the substrate or substrates in such a way that their molecules can react more easily. This helps reactions to take place rapidly at low temperatures.

Variables

There are many variables that could be changed in this experiment to investigate rates of reaction.

Temperature has a major effect on enzymes. At low temperatures, the reaction takes place very slowly. This is because the molecules are moving slowly and so substrate molecules will not often collide with the active site, and so binding between substrate and enzyme is a rare event. As the temperature rises, the enzyme and substrate molecules move faster, causing collisions to happen more frequently so the substrate molecules enter the active sites more often. They also collide with more energy and so the bonds are broken more easily and the reaction can occur. As the temperature continues to increase the rate of reaction will continue to increase but after a certain temperature the enzyme molecules vibrate so energetically that some of the bonds holding the enzyme molecule in its precise shape begin to break, and the enzyme, especially the active site, loses its shape and so the enzyme can no longer work. We call this denaturation. The enzymes in our body work best at around 40ºC. we call this the optimum temperature for enzymes. When the temperature rises above 40ºC the enzyme will begin to become denatured and will carry on losing its shape until it is fully denatured. This process cannot be reversed. This is a graph to show the effects of temperature on an enzyme.

pH can also affect the rate of reaction of enzymes. pH is a measure of the concentration of hydrogen ions in a solution. The lower the pH, the higher the hydrogen ions in a concentration. Hydrogen ions can interact with the R groups of amino acids, affecting the way in which they bond with each other and therefore affecting their tertiary structure. The optimum pH for most enzymes is seven as it is fairly neutral conditions. If the pH varies too much from the optimum, the enzyme will become denatured. This is a graph to show the effect of pH on enzymes.

Substrate concentration also affects the rate of reaction. As the substrate concentration increases the rate of reaction also increases. This is because as the substrate concentration increases collisions happen more frequently with the enzymes active site. This will continue to increase the rate of reaction as long as there is a plentiful supply of enzyme. However if it continues to increase, then at a point all the active sites of the enzymes will be continuously in use and so increasing the substrate concentration will have no further effect. The enzyme cannot work any faster as it is working at its Vmax rate. This graph shows this effect.

Enzyme concentration will affect the rate of reaction. At low enzyme concentrations there is great competition by the substrate for active sites and so the rate of reaction is low. As the enzyme concentration increases there are more active sites available and so the reaction can proceed at a faster rate. Eventually, increasing the enzyme concentration will have no further effect because the substrate concentration becomes the limiting factor as it has all been used up. This graph shows the effects of increasing the enzyme concentration.

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Inhibitors are regulatory chemicals which slow down or stop enzyme catalysed reactions. There are two types of inhibitor, competitive and non-competitive. Competitive inhibitors are of a very similar shape to the substrate so can fit into the enzymes active site. Therefore they prevent the substrate from entering the active site and binding there. This can be reversed by increasing the substrate concentration, and so it is a competitive reversible inhibitor. Sometimes the inhibitor remains permanently bonded to the active site and therefore causes a permanent block for the substrate. This is called non-competitive irreversible inhibition.

Some inhibitors can bind to a different part of the enzyme, other than the active site. This can disrupt the normal arrangement of the hydrogen bonds and hydrophobic interactions holding the enzyme molecule in its tertiary structure. This makes the active site unsuitable for the substrate to fit into. This is another type of non-competitive inhibitor. It can be reversible or irreversible depending on whether the inhibitor binds briefly or permanently with the enzyme.

Aim: I am going to investigate how different types of potatoes affect the rate of reaction of enzymes.

I have found some information in the website crystal.uah.edu about catalase. It has told me very useful information about where catalase is found and what job it performs in living organisms.

Catalase is an enzyme which is found in all plant and animal cells. It works very rapidly. It is found in tissues, where it speeds up the decomposition of hydrogen peroxide into water and oxygen. Catalase is composed of four subunits. Each subunit contains a heme group, which is responsible for carrying out catalase's activity. The reaction of breaking down hydrogen peroxide is performed by two types of reactions, called oxidation and reduction. Each of the subunits in catalase uses the energy from electrons to decompose the hydrogen peroxide. Catalase functions by the oxidation of Iron within its heme group. It removes an electron from two molecules of hydrogen peroxide to form two water molecules and one oxygen molecule. Catalase is located in a cell organelle called the peroxisome. Peroxisomes in animal cells are involved in the oxidation of fatty acids, and the synthesis of cholesterol and bile acids. Peroxisomes in plant cells are involved in respiration and symbiotic nitrogen fixation.

One molecule of catalase can turn six million substrate molecules into products every minute. This high rate shows an importance for the enzymes capability for detoxifying hydrogen peroxide and preventing the formation of carbon dioxide bubbles in the blood.

Potatoes are a good source of this enzyme.

Potatoes

In my experiment I will use six types of potatoes. They are:

Maris Piper
New
Organic
Baking
Sweet
Boiled Maris Piper

Potatoes can be chopped or grated to release their enzymes.

I have found the nitritional information for two types of potato that I am investigating on the internet site www.wholehealthmd.com. However, I could not find information about the other types of potato I am using.

The nutritional information for one medium potato is:

Sweet potato

The nutritional information for one sweet potato is:

Organic potatoes

Organic means the potatoes have been grown in a totally organic environment. They are 100% natural. They do not contain any pesticides or chemicals.

Buffer: I will use buffer in my experiment to make sure that the pH stays constant throughout the experiment. A buffer is a chemical or combination of chemicals that can resist change in pH by collecting the excess acid or base. The main buffers important in living organisms are blood proteins, such as haemoglobin and albumin, and inorganic buffers systems in body fluids, such as the hydrogen carbonate system. This system acts as a buffer because an extra acid tends to shift the equilibrium to the left, so the extra hydrogen ions are neutralised by the hydrogen carbonate ions.

Prediction: It is very difficult to predict which of the potatoes will work fastest as I do not have information about how much catalase the different types of potatoes contain and so I cannot compare them. However I do know that the boiled maris piper potato should not produce a reaction. This is because when they are boiled, their temperature will rise to 100ºC and the enzymes in the potato will become denatured. This means that their active site will be damaged and so when added to hydrogen peroxide, no reaction should take place. I also think that the organic potato may react better than the others as it has not been exposed to any chemicals which could act as inhibitors and prevent the hydrogen peroxide from entering the active sites of the potatoes enzymes.

Preliminary Tests: During my preliminary tests I need to find out which size potato chips are best to use for the experiment, how much hydrogen peroxide should be used, and how long I should conduct each experiment for.

I am going to test three lengths of potato cylinders, 3cm, 4cm and 5cm, cut into 1cm pieces. I will test them in 5cm³ of hydrogen peroxide and 10cm³ of hydrogen peroxide. I will test them for 30, 90 and 120 seconds to see how much gas is produced.

Method: I will measure out the correct amount of hydrogen peroxide, using a measuring cylinder. I will cut my potato strips using a corer and a knife to the correct length, making sure the diameter is the same each time by ensuring the same corer is used. I will place these in a conical flask. I will fill up a burette with water and place it, using a clamp stand to support it, in a bowl of water. I will have a bung to place over the conical flask with a delivery tube attached to it. The other end will be placed up the burette. A stop clock will be used to time the experiment. I will also use 5cm³ of buffer solution to keep the pH of the solution constant. I will pour the 10% hydrogen peroxide into the conical flask and quickly put the bung in to it. I will then start the stop clock immediately and record the amount of gas produced every 30 seconds.

Results:

Using 5cm³ of 20% hydrogen peroxide – How much oxygen is released in 120 seconds

The 4cm potato chip seemed to give an odd result on 30 and 60 seconds as the number should be higher than that of the 3cm chip but lower than the 5cm chip because in the difference in surface area for the reaction to take place. This may have been because of a blockage in the delivery tube or not cleaning the equipment thoroughly.

Using 10cm³ of 20% hydrogen peroxide – How much oxygen is release in 120 seconds

The results give a wider range with 10cm³ of hydrogen peroxide and with 5cm chips of potato.

Using 10cm³ of 20% hydrogen peroxide and 5cm potato chips I will repeat the experiment recording the gas produced every 30 seconds for 5 minutes to find out if this would be a suitable time span for my experiment.

This seems a suitable time span for my experiment as I get a good increase in gas released.

Conclusion: From these tests I have decided to use 10cm³ of 20% hydrogen peroxide, 5cm potato chips cut into 1cm chips, and conduct my experiment for a total of 5 minutes, recording the gas released every 30 seconds. I will use 5cm³ of buffer in my experiment, as it is a suitable amount for the amount of hydrogen peroxide I am using.

I will use the results of my preliminary tests to finalise the workings of my experiment.

I need to make sure that all my variables remain constant and there are no other factors that will affect my experiment.

I will try to keep the temperature as constant as possible. Each time I conduct the experiment I will do so in the same place in the room, which is away from windows so the temperature outside the room should not affect my results.

I will keep the pH the same in all of my trials by using a pH buffer solution. This will ensure that the pH remains constant throughout.

I will keep the substrate, hydrogen peroxide, concentration constant by using the same strength hydrogen peroxide each time (20%) and ensuring that the amount used stays the same and is accurately measured.

No inhibitors should be present in my experiment but I will ensure the equipment is washed thoroughly before I begin my experiment so no substances used in previous experiments will affect my results.

The enzyme concentration may vary from potato to potato but the surface area of the potato chips will be kept the same and so there will be no difference in the amount of the enzyme that could be available to the substrate.

I am using the burette method for this experiment as I find it an accurate and simple way to produce as precise and reliable results as possible. I could also use the method of counting bubbles produced or use a gas syringe. However, I think the method of counting bubbles could be inaccurate as it would be hard to keep track of the bubbles produced if the reaction happened quickly. I also chose the burette method as it measures more accurately than the gas syringe. The gas syringe measures to one ml where as the burette measures to 0.1 cm3. It is easy to set up accurately for each experiment and can be read accurately at the relevant time intervals.

The equipment needed for this experiment is:

hydrogen peroxide
six varieties of potato (maris piper, new, organic, baking, sweet, boiled maris pipers)
pH buffer solution
water
a measuring cylinder
a syringe
a burette
a delivery tube attached to a bung
a bowl
a clamp stand
a corer
a knife
a stop clock
goggles

Method: I will measure out 10cm³ of 10% hydrogen peroxide using a measuring cylinder and wearing goggles for safety. I will cut cylinders of potato from the first potato using a corer and cut the cylinders into 1cm chips using a knife. I will place 5 of these in the conical flask for the first trial. I will measure out 5cm³ of pH buffer solution, using the syringe, and add this to the conical flask. I will fill up the burette and the bowl with water and place the burette in the bowl, and hold it using a clamp stand. I will put the delivery tube up the burette and have the bung ready to put in the conical flask. I will pour the hydrogen peroxide into the conical flask and quickly place the bung over the mouth of it. I will start the stop clock immediately and record the oxygen given off every 30 seconds for 5 minutes.

During this I have to make sure that a number of factors do not change other than variables to ensure accurate results.

I will keep these factors the same.

I will make sure that I use the same corer each time I repeat the experiment so the potato chips are always the same size and have the same surface area.
I will use the same burette each time.
I will use the same bung and delivery tube each time as different delivery tubes may be different lengths and so this may make my results inaccurate.
I will ensure I use the same conical flask each time so that the area in which the reaction takes place does not change.
I will wash the equipment between experiments so that the equipment is not contaminated.
I will make sure I always measure the quantities accurately.
I will make sure I record the amount of oxygen given off accurately and at the right time intervals.

There are a number of safety aspects of this experiment.

I will wear goggles when handling hydrogen peroxide so it is not able to enter my eyes.

I will make sure my hair is not hanging down at any point during the experiment.
I will ensure that all spillages are wiped up as hydrogen peroxide is dangerous and has the appearance of water on a work surface.

Results:

Initial rate of Reaction: I have worked out the initial rates of reaction at 120 seconds from my graph. I have rounded them to three decimal places.

Maris Piper: 1.15 = 0.01 cm³/sec

120

New: 4.2 = 0.035 cm³/sec

120

Organic: 3.6 = 0.03 cm³/sec

120

Baking: 5.0 = 0.042 cm³/sec

120

Sweet: 5.6 = 0.047 cm³/sec

120

Maris Piper Boiled: 0 = 0 cm³/sec

120

I have plotted my results on to a graph to show which potatoes had the highest reaction rate. (See graph 2)

Conclusion: As you can see from my graph, the potato that produced the most oxygen was the sweet potato. This also had the highest reaction rate. The baking potato also produced oxygen at a fast rate, closely followed by the new and organic potatoes. The maris piper did not produce as much oxygen in comparison with the other potatoes and also took over 60 seconds to produce any oxygen at all. As I predicted the boiled maris piper potato did not produce any oxygen. This is because when I boiled the potato the enzymes were denatured as the temperature reached 100ºC, which is a much higher temperature than the enzyme can stand before its tertiary structure is altered. Because of this, the enzymes active site is damaged and it becomes denatured. Therefore in that experiment, no gas was released as the hydrogen peroxide had no enzymes to help the reaction.

I do not know why the sweet potato reacted best as I could not find enough information about enzymes in potatoes. Sweet potatoes are quite different in appearance to the other potatoes as they are orange inside and longer that the others. This may also mean that their enzymes act differently than the other potatoes or that they have a higher concentration of them. That would account for the high reaction rate. If there was a higher concentration of enzymes, then more collisions would happen with the substrate and so binding between enzyme and substrate would happen more frequently, therefore the reaction would happen quicker. I predicted that the organic potato may react better than the other potatoes as it had not been in contact with any chemicals which could have acted as inhibitors. However my results show that this was not the case. The organic potato did not react very quickly and did not produce a lot of oxygen. My prediction may have been inaccurate because the other potatoes could have been sprayed with chemicals to help their growth that would not affect the enzymes or act as inhibitors. We also only used the inside of the potato, and removed any skin and so it is possible that any chemicals that were sprayed on the potatoes did not get to the inside. As I could not find information on the enzymes in potatoes I cannot explain fully why the potatoes reacted as they did but it is likely that some potatoes have a higher concentration of catalase than other potatoes.

Evaluation: My experiment mostly went to plan and was suitable for the results I wished to produce. However there were some problems that occurred during it. There were three main sources of error. Operator errors such as cutting the potato chips to exactly the same size which was very hard as I could not ensure that when I cut the chips into 1cm pieces, I was cutting a straight line down. I may have been cutting at a slight angle, which would give some chips a larger surface area. I tried to measure the substances used as precisely as possible and so I think that the amounts used should have been accurate. An equipment error was that the buffer solution changed throughout the experiment. When it ran out a new buffer solution had to be made through dissolving a tablet in water. However, the tablet did not always dissolve completely before it was used, as there was not much time in the lesson to conduct the experiment. This may have caused slight fluctuation in the pH when the reaction was taking place. Therefore the control of pH levels may have not been accurate. One design error with the equipment was that the delivery tube leading to the burette needed enough pressure to build up from the reaction taking place before it released the gas into the burette. This meant that often there may have been be a gap where none, or not much gas was given off and then a sudden rush of gas would cause the amount of gas released to rise greatly.

I think my results are mostly quite accurate. There are a few anomalous results, as shown in my table in red. They do not fit in with the pattern of the rest of the results. I have left these results out of my averages.

Some whole columns of results do not fit in with the other two trials. This seems to suggest that there may have been a fault in the equipment. It is possible that on that trial the delivery tube was not connected properly or the bung was not pushed right into the conical flask. It may also have been because of possible faults in measuring out the quantities of hydrogen peroxide or another substance. It may also have been because the equipment was not washed properly and previous experiments were influencing the results.

In the baking potato experiment, in the second trial the reaction took a little while to produce results similar to others. This may have been because of the method used, as it often took a while for the gas pressure to rise up enough in the delivery tube, and for gas bubbles to be released. This may have been a reason why the reaction took a while to begin. In the boiled maris piper experiment, the results should have all shown nought as the enzymes should have been denatured, but in the first trial there was some gas released. This suggests that the equipment had some remains of other enzymes in, or possibly that the delivery tube still had gas in it from the last experiment. Once I have removed the anomalous results my other trials are more accurate. They are similar in number indicating that the results are reasonably accurate.

If I repeated this experiment I would take much greater care in washing the equipment between each trial and also ensuring that the delivery tube did not contain any gas before each experiment was begun. This would ensure more reliable results as no factors from previous experiments would effect the experiment. It also may be more accurate to use a gas syringe to measure the oxygen produced. This would conquer the problem of creating pressure in the delivery tube and should therefore produce more reliable results, although this method may not be as accurate, as the measuring units on a gas syringe are larger than on a burette. If I had had more time I would have taken more care to ensure everything was exactly the same in each experiment. I would have checked the temperature with a thermometer before each trial, to ensure the same conditions, and I would have ensured the buffer was properly dissolved. This would make my results more reliable as the conditions would have been exactly the same for each trial.

My conclusion may not be entirely accurate as I did not have the resources available to me to test the potatoes in other ways to try and find out about their individual properties. I also could not easily find information on this subject. It is also difficult to know if all the potatoes of a certain species have been treated in the same way and so if I used a different make of the same types of potatoes, it may produce different results. If I had had more time I could have tried to investigate some of these other factors.

Hannah Robinson