The concentration of enzyme (the one I am investigating) also affects the rate of reaction. A higher concentration of enzyme increases the reaction rate because the numbers of active sites of the enzyme are increased. There are more available active sites for the substrate to get in contact with and for the induced fit hypothesis to take place. The substrate has a higher chance of inducing an active site because there are more available sites. They are more spread out and have an increased surface area.
Lowering the enzyme concentration will decrease the number of active sites available therefore there will be more substrate molecules than there will be available active sites. The reaction will be slower. It will take longer because there are less available sites for the H2O2 molecules to induce.
Eventually, increasing the concentration of enzyme will have no effect on the rate of reaction (see point x) because the number of substrate molecules has become the limiting factor. What this means is that there are now only a fixed number of H2O2 molecules to induce the active sites so when all the H2O2 molecules have been broken up, there will be no more substrate molecules left to induce the further active sites.
Inhibitors slow down the rate of reaction. They are an essential form of cellular control, allowing reaction rates to slow down when needed. Competitive inhibitors have the same structure to the substrate, and they fight for a place to induce the enzyme. Non-competitive inhibitors change the shape of the enzyme, including the active site so substrate molecules cannot induce them.
Prediction:
As a result of my hypothesis, I predict that if I increase the concentration of the potato enzyme, the faster the reaction rate of the enzyme catalase with H2O2 will be. This is because there are more available active sites for the H2O2 molecules to get into contact with and induce with. To increase the concentration of the potato enzyme I will cut a certain sized piece of potato into smaller chunks. For example I will begin with one piece of a 4cm potato. Then I will cut another 4cm piece of potato into two 2cm pieces. This increases the surface area of the potato, resulting into more areas of the active sites, left exposed to the H2O2 to react and induce with. The active sites of the potato enzyme will be increased and they will have a larger surface area on the potato. Therefore the hydrogen peroxide will react more rapidly with the catalase because the numbers of active site of the enzyme has been made larger.
Method:
Equipment: Side arm test tube, cork borer, scalpel, syringe with cork and pipette, 800ml beaker, stopwatch, potato, hydrogen peroxide, Burette, tile and burette stand.
Firstly fill the beaker up with about 600ml of water. Next get the burette. Make sure the tap end is closed. To check it is closed pour a small amount of water into the burette and make sure no water comes out of the end. When it is air tight, fill the burette with water. Put your thumb over the opening of the burette. Turn the burette around so that your thumb is now at the bottom. Gently lower the burette into the beaker of water, keeping your thumb over the hole so that no water comes out. When the burette is in the water, release your thumb. Clamp the burette to the burette stand to make it steady. The burette should be about 3cm from the bottom of the beaker so that you can thread the side arm test tube up the burette.
Next get the side arm test tube. Using the side arm, thread this up the bottom end of the burette. Get a potato and obtain some potato using the cork borer. Cut the potato to the desired size for the experiment using the scalpel and tile. Put this in into the side arm test tube. Then get 4ml/4cm3 of hydrogen peroxide using the syringe with cork and pipette.
Plug the pipette and cork into the side arm test tube. Make sure the cork is tight in the test tube. This prevents oxygen from escaping. As soon as you press the syringe start the stopwatch. When the catalase is reacting with the hydrogen peroxide, oxygen will be produced and will travel up the burette through/as bubbles. The water level should decrease as the volume of oxygen increases. Measure how much oxygen is produced every 30 seconds by reading the water level. I will measure how much oxygen is produced every 30 seconds for 10 minutes.
In this experiment I will be varying the concentration of catalase, i.e. the size of the potato. I will be start with 4cm of potato. I will then increase the concentration of potato catalase by cutting another 4cm potato in half so that there will be two 2cm pieces of potato. Then I will increase the enzyme catalase by cutting another 4cm piece of potato into 4 quarters so there will be four 1cm pieces of potato. I will increase the catalase concentration by cutting another 4cm potato into eight 0.5cm pieces. My maximum size of the potato will be 4cm; my minimum will be 0.5cm.
To keep the test fair and reliable I need to keep the volume of hydrogen peroxide I use the same and the temperature it is at, which will be room temperature. I should keep the time of 10 minutes the same throughout the experiment. The potato should always be in contact with the liquid substrate to maintain a reliable result, so when I will be using a 4cm potato with 4ml of H2O2, I will need to keep the side arm test tube vertically straight so the liquid substrate is in contact with the whole potato. I will repeat the experiment for all the different concentrations at least twice, so that I mat be able to take a reliable average.
Results:
Experiment 1:
Experiment 2:
Averages:
The plan managed to obtain some reliable results shown in the averages. They show a clear pattern. As the concentration of potato is increased, the volume of oxygen produced also increased. There are a couple of anomalous results with the amount of oxygen produced at different concentrations. The reasons for this will be explained in my evaluation.
Analysis:
I can make a mathematical prediction to the relationship between the enzyme concentration and the rate of reaction. I will need to calculate the surface area of the potato by finding the area of a cylinder. From there I will calculate the time it takes for the reaction and link this to the surface area of the potato.
The surface area of a cylinder is 2πrh + 2πr2. Therefore if I use a 4cm piece of potato the surface area of this potato will be:
2π x 0.3(radius of potato) x 4(the height) + 2π x 0.32 = 7.536 + 0.5652 = 8.1012
So the surface area of a 4cm piece is 8.1012cm
When I cut a 4cm in half to make two 2cm pieces we have a larger surface area:
2π x 0.3 x 2 + 2π x 0.32 = 3.768 + 0.5652 = 4.3332
As there are two pieces we multiply this answer by 2. : 4.3332 x 2 = 8.666
Here is a table to show the surface area for each concentration.
The table above shows how much hydrogen peroxide gets in contact with the surface of the potato. As the surface area gets bigger, the more oxygen is produced and the faster the reaction becomes.
Conclusion:
The results from my graph show quite a clear trend. In the graph with the comparison of the average rate of oxygen produced at different concentrations of potato, I can conclude that as the concentration of potato catalase increases, the amount/volume of oxygen produced increases as well as its speed of reaction.
For example in the seventh minute of the reaction, a concentration of one 4cm piece of potato had produced 9.2 cm3 of oxygen; a concentration of two 2cm pieces of potato had produced about 10cm; a concentration of four 1cm pieces of potato produced 10.7 cm3 of oxygen and a concentration of right 0.5 cm pieces of potato produced 11.5cm3 of oxygen.
To calculate the rate of reaction I noticed that as the concentration of potato catalase increased, the gradient of the line that represented the speed of reaction became steeper which indicated an increase in speed in the reaction. I checked the steepness of the lines by using a ruler. I also read off the graph to see how long it took for a certain amount of oxygen to be produced at each concentration. For example, to produce 7.5cm3 of oxygen, it took 3minutes for a concentration of eight 0.5cm pieces, 3minutes 30 seconds for four 1cm pieces, about 4mintes for two 2cm pieces, and 5mintes for one 4cm piece. These are the quantitative patterns that I have noticed.
When I increase the concentration of the potato, I am also increasing the surface area of the potato. By increasing the surface area, I am increasing the number of available active sites for the substrate to induce. This shows that when I increase the surface area, I increase the rate of reaction.
This reinforces my prediction of the relationship between the concentration of potato and the speed of reaction that I made earlier: an increase of concentration of potato enzyme increases the rate of reaction because the number of available sites for the substrate to react and induce with, has been increased, giving a higher chance for the collision and reaction between the substrate H2O2 and enzyme potato catalase.
Evaluation:
Overall the results obtained from the experiment were quite pleasing because they prove that as you increase the concentration of potato enzyme, the rate of reaction increases. However, there were a few anomalous results. As the concentration of potato catalase increased, the line representing the speed of reaction became slightly irregular. For example, in the graph with a concentration of four 1cm pieces of potato, in the 6th minute of experiment 2, the amount of oxygen produced seems out of place. In the graph with a concentration of eight 0.5 cm pieces of potato, in the 7th minute of experiment 1, the amount of oxygen produced is also anomalous. In the same graph, in the 5th minute of experiment 2, the amount of oxygen produced is also anomalous. However in the comparison of the averages, the results have a regular straight line. Judging by the graphs produced, I think that the experiment was quite accurate.
The anomalies can be cause by various factors. There are two main factors that I have noticed. The first factor could be that as we increase the concentration of potato, the number of pieces increases. When the potato begins to react with the H2O2, the reaction produces effervescence, which causes movement between the potatoes. Sometimes they stick together or the surface of the test tube or sometimes they float to the surface. Therefore not all the surfaces on/of the potato are exposed to the H2O2 - not all active sites have been exposed to the H2O2 molecules, so the reaction slows down and less oxygen is produced.
Another factor is that an air bubble could surface earlier or later than usual. What I mean by this is that for example in the graph with a concentration of eight 0.5cm pieces of potato, between the 6th and 7th minute there was a sudden increase of oxygen. This could be because the oxygen rose to the surface before I read the reading of the water level. So a bubble of air could have rose to the surface in the 28th second and I read every 30 seconds so, the bubble that rose earlier in the 28th second, should have rose in the 33rd second for example. This is perhaps what made some of the results anomalous. Air bubbles being trapped in the side arm of the test tube could affect this.
Other factors, which could have affected the precision of the results, could be that the cork in the test tube was not tight enough and some of the oxygen in the reaction could have escaped. The burette tap opening may have allowed small amounts of oxygen into the burette, making the results less accurate.
Another major factor was the syringe with cork and pipette. This was probably the least accurate instrument used in the experiment. When the pull force on the piston was sucking in the H2O2 liquid, it was hard to keep the pulling force for the suction the same; so different amounts of H2O2 would have been drawn in even if it read that 4ml of H2O2 had been withdrawn. The pull force/suction was also affected by the tightness of the cork. Sometimes the cork was loose, so oxygen escaped, making the suction less powerful. Also, the pipette kept some of the liquid H2O2, and we did not measure how much H2O2 was inside the pipette. We assumed that all the liquid H2O2 was inside the syringe.
The small variant amounts of H2O2 would have affected the experiment because it would have increased/decreased the amount of substrate molecules reacting and inducing with the active sites. Therefore the chances of inducing and reacting were more varied and the amount of oxygen produced and rate of reaction would also be affected. Another factor could be the inhibitors slowing down the reaction at different times.
Generally I was quite pleased with my results and I think the procedure to obtain these results was suitable. I would however make a couple of changes to the experiment to make my results more reliable and easier to follow.
Firstly I would change the way I recorded how much oxygen was produced. Rather than record ten minutes worth of oxygen produced, I could have recorded the time taken to produce a certain amount of oxygen. However I am still able to see how long it took for a certain amount of oxygen produced at a certain time as shown in my conclusion. To make my results more reliable I would have repeated the experiment at least one more time, but time was one of the factors that prevented me from doing this.