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Potato an hydrogen peroxide and liver and hydrogen peroxide

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  • Pages: 9
  • Word count: 2029
  • Category: Enzyme

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An investigation to compare the reaction rates between potato and hydrogen peroxide against liver and hydrogen peroxide through loss in mass.

Background information:

Catalase is an enzyme that is found in all cells. This means that it is an intracellular enzyme. And enzyme is a biological catalyst. A catalyst is some thing that speeds up a reaction without being changed itself. Because of this enzymes and catalysts can be used again and again. Enzymes are protein chains that have a primary, secondary and tertiary structure. Their primary structure shows the order and types of amino acids used to form the protein chain. The secondary structure shows the basic folding of the protein and is held in place by hydrogen bonds. The tertiary structure shows a more complex folding which gives it its globular shape. The tertiary folding of the enzyme also gives it its active site. The active sit of an enzyme is the part of the enzyme that determines what the enzyme will react with. If this active site is destroyed in any way the enzyme is said to be denatured and will no longer work.

Certain things affect the reaction rate of a substance where an enzyme is used. The concentration of the reactant will affect the reaction rate. If there is a strong concentration of reactant or catalase then there will be a faster reaction rate than if the concentration was weak. Also, the higher the temperature of the substance the faster the reaction rate will be. This also applies to the surface area of the cells containing the catalase. If the cells have a large area the reaction rate will be very fast. This is because more catalase will be exposed to the reactant and this will mean that more products are being made in a short amount of time. However if the cells have a small surface area then the reaction rate will be slow because less of the enzyme will be accessible to the product and therefore fewer products will be made in a certain amount of time. Catalase reacts with hydrogen peroxide to form water and oxygen in the cells. Hydrogen peroxide is a toxic chemical that is made naturally in the cells.

Because it is toxic it has to be removed. This is why there is catalase in the cells. Some types of cells contain more catalase than other types of cells. For instance the liver contains a lot of catalase. This is because the liver is a very active part of the body and therefore produces a lot of hydrogen peroxide. One of the liver’s major functions is the manufacture and secretion of bile, which is stored in the gall bladder and released in the small intestine. To combat this the liver cells need to contain a lot of catalase. The liver also controls glucose levels in the blood and breaks down amino acids. Plant cells on the other hand do not contain a lot of catalase. This is because plants are very inactive and do not produce a lot of hydrogen peroxide and therefore do not need to have a lot of catalase to get rid of it. The potato is a tubur, which means it is a store of food for plants. It is an extension of the root of the potato plant. It does not do any thing and for this reason should not contain a lot of catalase


My aim is to compare the reaction rates between crushed liver and crushed potato when placed in hydrogen peroxide through loss in mass and see which tissue cells contain the most catalase.


I think that the reaction rate for the liver will be faster than the reaction rate of the potato. This is because there is more catalase in the liver. This means that a piece of crushed liver will contain more catalase than a piece of crushed potato of the same mass. From my research I know that the higher concentration of catalase there is then the faster the reaction rate will be. Because liver has a higher concentration of catalase than potato it means that when it is placed in hydrogen peroxide it will produce more oxygen in a set time limit.

Pilot study:

Before I could do my experiment I had to do a pilot study to find out which measurements I should use to get the best results. First of all I tried to measure the reaction rate of 5g of crushed potato in 20 mls of hydrogen peroxide diluted with 20 mls of water. This gave a measurable result so I decided to continue with these masses and volumes of reactants except this time I tried it with crushed liver instead of potato. The reaction this time however was too volatile so I had to rethink my quantities. I tried the experiment with only 2g of crushed liver and I used a bigger beaker to contain the reaction better. This time the result wasn’t so volatile and the reactants and products stayed inside the beaker. To make sure that this would give a measurable result I repeated the experiment again but using 2g of crushed potato. The reaction was quite slow but measurable so I decided to continue with these amounts for my real investigation. I also found out that the potato reaction starts to stop after three minutes so I decided to record the mass every 20 seconds for three minutes.


First of all I weighed out 2g of the internal tissue of a potato and 2g of liver on the top pan scales. Then using the pestle and mortar I separately crushed up the two reactants. To help crush the liver a pinch of sand was placed in the mortar. Then I measured out 20 ml of hydrogen peroxide and diluted it with 20 ml of water making a 50% concentration solution of hydrogen peroxide. Next I placed my 500 ml beaker on the top pan scales and returned the weight to zero. After adding my potato to the beaker I added the hydrogen peroxide and recorded the weight in my table. Every 20 seconds after this I recorded the weight on the top pan scale. I continued to take readings for three minutes aggravating the solution after every reading.

After three minutes was over I repeated the experiment keeping everything the same except this time I used 2g of crushed liver instead of potato. Once this experiment was over I repeated the potato experiment and then I repeated the liver experiment. This was to give a more accurate set of results.


Instead weighing the difference during the reaction I could’ve measured the reaction rate through displacement. This would mean that I set up my experiment as shown below and measured the amount of water that is displaced from the measuring cylinder.

To make this a fair test I used the same concentration of hydrogen peroxide each time I did the experiment. This is because the concentration of the hydrogen peroxide could affect my results. I also made sure that I used the same mass of potato and liver, as I wanted to compare them. If they had different masses it may possibly have affected my results. The experiment was also made a fair test because I used the same top pan scales each time I did the experiment. Using different equipment may have meant I got a different accuracy in my results. I also used the same potato and the same liver for when I repeated my results. I repeated my results so that I could get an average and get more accurate results. If I had used different potatoes and liver there would have been a difference in the amount of catalase and this would’ve changed my results or made then inaccurate.

To make sure that I was making my experiment as safe as possible, I had to do a risk assessment of my experiment. First I checked out how hazardous the hydrogen peroxide was. I found out that it was corrosive, however it was safe to use in 2 mole amounts. This was why the hydrogen peroxide I used had a molarity of 2 moles. I also used gloves, safety goggles and a lab coat to make sure that I was protected.


During my reaction I noticed that the hydrogen peroxide solution fizzed and gave of a gas. My research indicates that this gas was oxygen because the formula for the reaction between hydrogen peroxide and catalase is:

Hydrogen peroxide + catalase water + oxygen

(See also graphs and tables)


From my graph I can see that my prediction was correct. I predicted that the liver would have a faster reaction rate than the potato when mixed with the hydrogen peroxide. I can tell that this is true form my graph because the line for the liver reaction rate is steeper that the line for the potato reaction rate. This means that the liver has a faster reaction rate than potato does. I can also tell by my calculations of the average reaction rates that the liver has an average reaction rate that is faster than the potatoes average reaction rate. This indicates that liver contains more catalase that potato does. I know this because my research shows that when there is a higher concentration of catalase then there is a faster reaction rate. When there is a larger concentration of an enzyme the reaction rate of the solution will speed up. This is because there are more enzymes that are able to make successful collision with the particles and cause a reaction.

This principle is based on something called the collision theory. This theory states that for two particles to react they must collide with each other at a certain speed and have a certain amount of energy and if they collide correctly they react. This is called a successful collision. If there are more enzymes then there will be a larger chance of a successful collision. Though doing my research I thought that there would be more catalase in the liver than the potato because the liver is a more active organ. It produces bile and helps control the glucose levels in the blood where as the potato is just a store for plant minerals and not much else.


Even though I get the results I was hoping for I still got some anomalous results in my graph. This could’ve been for a number of reasons. For instance because I was using very delicate measuring scales (top pan balance) a little disturbance in the air could’ve changed my results. Also because after every 20 seconds I was agitating the reactants and placing the beaker on the scales I could’ve placed it in a different place each time and this would’ve changed my results. Or when I was agitating the reactants I could agate it more sometimes and this would’ve caused a faster reaction.

I think that my experiment was very valid. This is because it gave the expected results. If my results had not shown what was expected then, I would’ve redone my experiment to find out where I went wrong.

If I were to do the experiment again I would try to do it in a place with less people. This is because whilst I was doing my experiment there were a lot of people walking around. This caused a draft around my experiment and it changed my results. I would also repeat my experiment more so that I could get a more accurate result.

I would take this experiment further by finding out if other organisms held more catalase. For instance I could see if the potato skin had more catalase in it then potato flesh does. I could also find out if kidneys had a faster reaction rate than liver. I would do the same experiment using the same measurements but I would use a wider variation of reactants.

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