How the concentration of enzymes controls the breakdown of Hydrogen Peroxide? Aim- We are investigating the effect of changing the concentration of the enzyme catalase has on the rate of breaking down of Hydrogen Peroxide. Background Info- This experiment will involve some soft potato. Potato has a very high enzyme count, that! |s why we use it in this experiment. This enzyme is called catalase. It is used in our body to break down food and other things that we might digest.
A catalase speeds up the rate of reactions. The enzymes are all protean molecules. We will also use Hydrogen Peroxide and washing up liquid. The catalase breaks down the Hydrogen Peroxide into oxygen and water. The more vigorous the reaction the more froth that it creates. The reaction in the experiment: Hydrogen peroxide “^3 oxygen + water 2 H 2 O 2 “^3 O 2 + H 2 O The enzyme can be used around the body to help chemical reactions, it speeds them up! The enzyme breaks up larger molecules into smaller pieces.
This is how the enzymes work below I have described some of the things that can affect how well it works. Fair test/ Factors-Temp- This can affect the movement of the particles. The higher the temp the faster the particles will move and the more collisions. If the temp of the Hydrogen Peroxide is too low the experiment will fail! High Temp- It gives them more energy so they move faster this means that they are likely to connect the substrate and the active site. The high temp also gives the particles kinetic energy.
The Term Paper on Potato An Hydrogen Peroxide And Liver And Hydrogen Peroxide
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 ...
This means that the Hydrogen Peroxide will be broken down quicker. Best Temp- This is the hottest the enzyme can be heated without the active site changing (denaturing).
If the active site changes then the reaction will no longer work. From our preliminary test we found out that the optimum temp is 25 oc this is around room temp. PH- All enzymes have an optimum PH, the wrong PH can cause denaturing.
The Ph can! |t be too acid or alkaline so keep it as near to 7 as possible. If the enzyme denatures then it is no use for breaking down the Hydrogen Peroxide. Concentration of H 2 O 2- The higher the concentration the more collisions with the active site. This means there is an increased rate of reaction. Creating more froth so the concentration will have to be the same for each of the test tubes.
The max we are going to use is 20% H 2 O 2. Volume of H 2 O 2- The volume must cover the whole of the chip, otherwise the test top/ unexposed layer of the potato wouldn! |t react with the Hydrogen Peroxide. Also the more Hydrogen Peroxide added the more likely gas given off will increase. This means that the reaction would make more froth given an unfair test.
Source of enzyme- Different potatoes will have different levels of enzymes in them. If one had more enzymes then another it may be able to breakdown the Hydrogen Peroxide quicker. We are just going to use the same potato (tes co value) for every one! Concentration of the enzyme- VARIABLE! we can change the size of the potato chip showing more or less surface area. This is changing the concentration of the enzyme. The potato shapes will all be cuboids from 1 cm x 1 cm x 1 cm – 1 cm x 1 cm x 5 cm. 1 cm 3, 2 cm 3, 3 cm 3, 4 cm 3, 5 cm 3.
The cuboids with the largest surface area has the largest concentration and vice versa. Equipment-Potato chips from 1 cm x 1 cm x 1 cm – 1 cm x 1 cm x 5 cm 2 Test tubes Cutting blade (for the potatoes) Ruler test tube rack Boiling tub Hydrogen Peroxide Measuring cylinder Stop clock Washing up liquid Stands!” Waterproof tray!” Test tube plug with thin plastic tube!” !” only in the real test only in preliminary test Preliminary test-An experiment was carried out to find the optimum temperature for working enzymes. 1. Get the moistened potato and chop to 1 cm x 1 cm x 3 cm 2. Start the water bath at 20 oc and let it warm up 3. Pour 10 cm 3 Hydrogen Peroxide into a test tube 4.
The Essay on Test Tube Starch Solution Water
Detection of Biological Molecules Introduction: Without carbon, nitrogen, hydrogen, sulfur, oxygen and phosphorus, life wouldn't exist. These are the most abundant elements in living organisms. These elements are held together by covalent bonds, ionic bonds, hydrogen bonds, and disulfide bonds. Covalent bonds are especially strong, thus, are present in monomers, the building blocks of life. These ...
Get the potato and put it in an other test tube with some water 5. Put both of these test tubes in the heated boiling tank 6. Wait for about 3-5 minutes then take out the test tubes 7. Tip the potato! |s water in the sink and add the Hydrogen Peroxide 8. Add washing up liquid 9.
Check the froth on this for every minute for 3 minutes 10. Do for all the other temperatures measured Results-Temp – 20 oc 30 oc 40 oc 50 oc 60 oc times Height of froth In mm (below) 60 s 30 12 10 20 14120 s 35 19 15 30 17180 s 45 24 19 35 20 Optimum temp! I have declared that a temperature between 20 oc and 30 oc is the optimum temperature. 30 oc was too high and 20 oc was too low so I have gone for around 25 oc. This means the reaction will have the greatest amount of heat it can have with out denaturing the enzyme. Method- Preparation: 1. Collect the equipment from the equipment list.
2. Set up the experiment as shown in the diagram. 3. Cut up 2 pieces of potato 1 cm x 1 cm x 1 cm. 4. Fill a test tube with 15 cm 3 of H 2 O 2.
5. Put the test tube in the rack and fill the waterproof tray half way with water. 6. Dip the large measuring cylinder in the water and let it fill up. 7. Use the stand to hold the cylinder upside down but so it is still in the water.
8. Push the tube from the test tube plug into the upturned measuring cylinder. Experiment: 1. One person get ready to start the stop clock and the other get a 1 cm 3 Potato. 2. Quickly drop this into the H 2 O 2, put the plug on in the test tube and start the stop clock.
3. Check that the measuring cylinder was at 0 on the side scale and watch as gas (oxygen) fills up the cylinder pushing out the water. 4. Record the amount of gas in this cylinder every 30 secs.
5. Stop after 5 minutes. 6. Detach the plug from the test tube and tip its condense away. 7. Refill the measuring tube full of water.
The Term Paper on Reaction (rate) Between Magnesium Ribbon And Hydrochloric Acid
Introduction In the experiment the magnesium reacts with the hydrochloric acid to create magnesium chloride and hydrogen. The balanced formula for this is: Mg(s) + 2HCL(aq) MgCl2(aq) + H2(g) Magnesium + hydrochloric acid Magnesium Chloride + Hydrogen Magnesium will react with hydrochloric acid, because it is higher in the reactivity series than hydrogen. The magnesium displaces the hydrogen in the ...
8. Repeat the experiment two times more using the same size potato. 9. Use these three results and find an average.
10. Change the potato! |s size (1 cm 3, 2 cm 3, 3 cm 3, 4 cm 3, 5 cm 3) after the two measurements and repeat all again. Prediction-I predict that the bigger the chip the quicker and easier that the reaction can take place. This means that the largest chip will produce the most oxygen whereas the smallest chip will produce the least.
I think that there will be a pattern between the rate of reaction and size of the chip. The more chip the faster rate of reaction, the rate of reaction should stay steady though. Results-Please Turn Over! ! ! s Analysis-Looking at the table I can clearly see that the larger the potato the more oxygen it creates. Using the table we can also see that the speed of the breakdown increases when the potato is larger. The larger potato has a larger surface to be broken down by the acid.
This is because of the fact that the quicker the reaction takes place the more oxygen that will be produced. The only exception was the 18 cm 2 and the 22 cm 2 as they were both wrong! If these values were swapped then it would of made sense. The 18 cm 2 was bigger n volumes of gas than the 22 cm 2 and it should have been the other way round. Towards the end of the results you can clearly see that the 1 st and 2 nd repetitions have a lot of difference in value. This has changed the results and made a good average.
I think that maybe on of the columns was made by a wrong way of doing the experiment. Conclusion-From this experiment you can see that raising the concentration of enzyme increases the rate of which the Hydrogen Peroxide is broken-down. This is because the more surface of the enzyme that can! SS attack!” the Hydrogen Peroxide the quicker it is going to break it down. If the enzymes can! |t touch the Hydrogen Peroxide then it will be unable to break it down.
After finding the rate of reaction (this was the average of the potato length after 300 seconds and then divide by 300, the seconds. E. g. 15 divide 300) we can say that the more potato we used (each cm it gained) the faster the rate of reaction was according to the graph.
The Essay on Ortho Quinone Enzyme Rate Reaction
Introduction In this experiment, we intended to investigate the effect of the concentration of catechol oxidase on the reaction rate of catechol to ortho-quinone. Living things, to increase the speed of chemical reactions, use enzymes. Enzymes are a type of protein. Because of the fact that enzymes are proteins, they have certain shapes. These shapes create areas in which the substrate, the ...
The graph shows a steady strait line for the first two potato sizes. After this the rate of reaction decreases heavily. This also proves my prediction is half correct. Evaluation-I think that the experiment went well.
My prediction was right and the reasons behind my chose were also correct. I am a little concerned about my results how there was such a big difference between repetition 1 and 2. I did also have trouble understanding some parts of the experiment as it got confusing. This could have been why my results were so varied, I may have set up the experiment wrong in an area and given faulty results.
If I was to do the experiment again I would understand it more and get more accurate results. I would also use something other than a potato as my source of enzyme. This could be very inaccurate as one part of the potato could contain more enzymes than another thus meaning that the larger potato chip could have less enzymes than a smaller piece. You could use a syringe to record the gas let out instead of the cylinder in water.
The syringe would be pushed out as this gas is created. My rate of reduction graph also showed great weakness in accuracy. I must of made a great mistake in my measuring to have got such strange results. The weird thing is that if I would of switched the 22 cm 3 and the 18 cm 3 chip results around we would of got perfect results.
