Enzymes are biological catalysts. They carry out thousands of chemical reactions that occur in living cells. They are large proteins made up of amino acids.
In an enzyme catalysed reaction, the substrate (in this case H2O2) binds to the active site of the enzyme (catalase). The catalase and H2O2 is held together by hydrophilic and hydrophobic bonds, ionic bonds, disulphide bonds and hydrogen bonds. This is known as the enzyme substrate complex.
Each enzyme is specific for a certain reaction (lock and key mechanism) because its amino acid sequence is unique and causes it to have a unique 3D structure consisting of alpha helixes, beta pleated sheets and the four structures (primary, secondary, tertiary and quaternary structures).
Catalase has four sub-units represented by the four different colours. Each sub-unit contains a heme group. The heme group is responsible for carrying out catalase activity. Catalase is one of the fastest enzymes known. Its turnover rate (number of substrate molecules an enzyme can act upon) is about 40,000,000 molecules per second. Catalase is very efficient. Its high rate shows its capability for detoxifying hydrogen peroxide and preventing the formation of carbon dioxide bubbles in the blood.
Catalase is found in the liver, potatoes, kidney beans, chickpeas and many other living cells. Catalase speeds up the decomposition of hydrogen peroxide into water (H2O) and oxygen (O2). H2O2 is a by-product of metabolism and it is highly toxic. Its rapid conversion is therefore very important.
This is the reaction when catalase breaks down hydrogen peroxide:
2H2O2 =====> 2H2O + O2
Hydrogen =====> Water + Oxygen
Peroxide
The volume of oxygen released will indicate the rate of reaction. In my experiment I will measure the volume of oxygen produced. This is the dependent variable. I will find out the results by depending on the volume of oxygen consumed. The independent variable is the time allowed for soaking. This should produce different volumes of oxygen depending on how long the beans are soaked.
There are certain factors that affect the rate of enzyme activity.
As the pH is lowered the enzyme will tend to gain positive H ions. Eventually enough side chains will be effected so that it causes the disruption of the enzymes shape. When the pH is increased the enzyme will loose positive H ions, and in turn, will eventually loose its shape. In both cases the substrate will not fit the active site of the enzyme. The enzyme will be denatured and it will not carry out its job.
An increase in temperature effects the rate of reaction in two ways.
- Collision theory - When heat energy is increased, the kinetic energy also increases. This causes the substrate and enzyme molecules to move faster causing more successful collisions. The more often they collide the greater the rate of reaction.
- Increasing the temperature can, on the other hand, denature the enzyme (enzyme looses its catalytic features). When enough heat is added the molecules start to vibrate. As the vibrations increase it starts to break hydrogen bonds, ionic bonds, hydrophilic bonds, hydrophobic bonds, etc. this causes the 3D shape to change. The substrate can no longer fit into the active site. The catalytic properties have now been lost.
The actual affect of temperature combines these two properties to reach an optimum temperature. This allows the enzyme to work at its best without denaturing the enzyme.
The Active site of an enzyme can be used again and again as long as it is not permanently denatured. This property allows enzymes to work efficiently at a low concentration. Provided that all conditions are ideal for an enzyme catalysed reaction, and there are excess substrate molecules, the rate of reaction is directly proportional to the enzyme concentration. But if the substrate molecules are limited the graph shows its outcome. (See graph below).
An enzyme catalysed reaction increases as the substrate concentration increases but only up to a point. Where the substrate is at a low point, the active sites of the molecules are not all used up. As the substrate increases it gets to the point where all the active sites are full (i.e. occupied) by the substrate. Therefore a further increase will make no further difference. (See graph below).
Method
- I took a handful red kidney beans and ground them until I was satisfied. I did not have to measure the time it took to grind because all the beans that I needed were ground at once.
- After grinding them I measured accurately 2g of beans on the pan balance in a weighing boat. I did this 15 time. I decided to use 2g of beans for each experiment. I had to conduct 5 experiments. I decided to soak the beans for 15, 30, 45, and 60 minutes. One of the experiments was done without any soaking at all. I had 5 experiments and needed to repeat each experiment 3 times.
- After measuring the beans I put each 2g into a separate test tube. Three of the fifteen experiments were not to be soaked. So, I did them first.
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I added 10cm cubed of 2H2O2 and 2cm cubed of pH 7 buffer into a test tube using a syringe. I set up the water bath (as shown in the diagram below).
- I added the kidney beans to the solution (pH 7 buffer and 2H2O) and put the lid of the delivery tube straight on and immediately started to time the reaction.
- After 2 minutes I removed the delivery tube and measured the volume of O2 produced. I recorded the results and repeated the experiment a further two times (total of 3 times).
- The other 12 experiments were conducted the same but they were soaked as follows.
- After weighing and separating the beans into test tubes I used a 10cm-cubed syringe to add 10cm cubed of water to each test tube, containing 2g of ground kidney beans. I left the beans to soak for their appointed times.
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Once the soaking was complete the contents was filtered so that the water could be extracted.
- Once the beans were ready I opened up the filter paper and scooped up the contents, carefully, with a spatula. I put the beans into the test tube containing the pH 7 buffer and 2H2O2, and followed the same steps as stated earlier.
Variables kept constant:
- 2g of kidney beans
- 10cm cubed of water for soaking
- Temperature of 2H2O2
- 10cm cubed of H2O2
- 2cm cubed of pH 7 buffer solution
- timing kept at 2 minutes for each experiment
Variables not kept constant:
Development of Prediction
My prediction states that water is needed for catalase activity to take place. The reaction is dependent on water. A mature seed contains 10% water. This explains the decomposition of 2H2O2 when it goes through no soaking at all. The seed at this point (no soaking) is very stable and activity will surely increase as more water is added.
The seed structure determines what goes in and out of the kidney bean has a protective layer called a testa.
The catalase exists within the testa. The catalase can only be activated when the testa breaks apart.
Metabolic activity occurs in the embryo