How Concentration Effects the Rate of Reaction free essay sample

As the concentration of enzyme increases it will have an effect on the rate of reaction. The suggested optimum will be 10g as the higher the mass of catalase the more enzymes, meaning more space for reactions to occur. In the space given there will be more enzymes reacting with the substrate, which in this experiment is hydrogen peroxide (H2O2), until it has reached its optimum, where the volume of oxygen produced will not increase due to the fact that there will not be enough H2O2 left for the enzymes Quantitative Hypothesis By doubling the mass of potato, the rate of reaction will double also. 2n=2r Scientific Explanation Enzymes are made from amino acids, and which are proteins. When an enzyme is formed; it is made by stringing together between 100 and 1,000 amino acids in a very unique order, the chain of amino acids then folds into a unique shape. That shape allows the enzyme to carry out specific chemical reaction â€Å"Enzymes are biological catalysts catalysts are substances that increase the rate of chemical reactions without being used up†. The place where these substrate molecules fit is called the active site. Enzymes are proteins produced by micro-organisms, in turn speeding up the chemical reactions. These can be found in the cell’s gene. The lock and key model is a key way to express the way in which the chemical reactions that will be taking place during our investigation on the concentration of enzymes. Every enzyme molecule has an active site, the part of which the substrate joins on to. Enzymes are really specific as they only speed up one reaction. This is because, for an enzyme to work, a substrate has to be the correct shape to fit into the active site. These actions are called the ‘lock and key’ hypothesis, because the substrate fits the enzyme just like a key fits into a lock. Impacts of Temperature and pH The activity of enzymes is strongly affected by changes in pH and temperature. Each enzyme has its pH and temperature optimum, its rate can only decrease at values below that point. The ascending part of the temperature curve, in the graph below, reflects the general effect of increasing temperature on the rate of chemical reactions. The descending portion of the curve reflects the loss of catalytic activity as the enzyme molecules become denatured as the temperature increases By changing the temperature, changes in the rate of an enzyme-controlled reaction occur. A higher temperature increases the rate at first; however if it gets too hot bonds that hold the enzyme (e. g. hydrogen bonds) break. This changes the shape of the enzyme’s active site and so the designated substrate will no longer fit and the enzyme will become inactive. This process is called denaturing. All enzymes have an optimum pH that they work best at. If the pH is too high or too low, it interferes with the bonds holding the enzyme together. This again begins the process of denaturing, as the active site changes shape. Our Catalase Our enzyme can be found in raw cooking potatoes and will be reacting with the catalase, hydrogen peroxide. Hydrogen peroxide (H2O2) is a by-product of respiration and is made in all living cells. The optimum temperature for our enzyme, is about 37 °C (36 °C -39 °C), this is the same body temperature of human. Its optimum pH is around 7 (6. 8-7. 5). The reaction for the catalase can be shown in this formula: Variables There are three different types of variables that I will be using in this investigation these are Independent, Dependent and Controlled. The independent variable is the thing that you choose to change at set intervals; for this investigation my independent variable will be the mass of the potato. I will use a range starting from 2g and from there, increase in steps of 2. This will give me a general trend and positive correlation, when I come to drawing my graph. The dependent variable is something that you measure; for this investigation I will be measuring the volume of oxygen. This will be my dependant variable. This will be measured through the use of a cylinder and bung and delivery tube. The controlled variables are the things that are kept the same throughout the investigation. The controlled variables in my investigation are the volume of , pH buffer 7 and the temperature. I will always use 4ml of and 4ml of pH buffer 7, to ensure my results are accurate and have more precise results. By keeping the temperature the same, at 37 °, we allow the enzymes to work at their optimum temperature amp; pH level for each repeat and potato mass, in turn creating a fairer test. Strategy B Justification of range For the concentration of enzyme we have investigated grams of 2g, 4g, 6g, 8g and 10g. We have investigated these as we want to get a trend to support our hypothesis that suggests that the more catalase you have the higher the rate of reaction. If we were to get a graph to support our hypothesis we could also include positive correlations to our hypothesis and scientific evidence to support our results. Justification of Interval We decided to have 3 intervals per set mass. By having these 3 repeats we were able to increase the validity of the results, increasing the fairness of the test through making sure that all the variables, with an exception of the independent – mass of each potato, are kept the same e. . pH 7 buffer, temperature, volume of , same measuring cylinder etc. This also in turn gives us a chance to spot ant anomalies that may have occurred during the experiment, making it simple to recognise the anomalies and trends among thee results and graphs produced. Equipment Justification Water Bath| Sometimes direct heat, from a flame, can damage instrum ents or impair an experiment. Hot water would heat the chemicals more slowly, and therefore may keep them more stable. There is also no chance of an open flame, from a bunsen burner for example, catching on to the contents you are trying to heat and causing a fire. Knife| To cut each disk of potato. By cutting it up into disks we maintain the same amount of mass for each repeat and by dicing them up into further disks it encourages the cells to beak open, releasing more enzyme and speeds up the rate of reaction in the process| Potato| This is our source of catalase; by using potatoes, which are abundant in catalase to prevent photorespiration damage it will make separating the catalase enzyme simple and easy. | Hydrogen peroxide| H2O2 is a by-product of respiration and is made in all living cells. Hydrogen peroxide is harmful and must be removed as soon as it is produced in the cell. Cells make the enzyme catalase to remove hydrogen peroxide. | pH buffer| pH 7)-Enzyme activity often changes with pH, and some enzyme reactions will affe pH. Buffers stabalize pH so results between ntubes are likely to emphaisie differences between samples, not variations in pH. | Glass beakers 250cm3| these will hold heated water and a testube containing cataalase. It having a volume of 250cm3 will allow a suffient amount of water. | Craft Knife| to be able to cut the potato pieces effienctily and precisly| Whit Tile| this allows the pupil to be able to clearly distinguish the different groups of potato without getting confuesed as to the different masses each group of potato has. | Weighing Scales| increases the accuracy to 0. 01g for each group of potato. This alows the test to be fairier. | Test Tubes| conatains the pH buffer, potatoes and hydrogen preoxide in it. | Pipette| is marked up to 3mls and goes up in 1mls. Through the markings you are able to determine exactly how much hydrogen peroxide/pH buffer you have and need. Thermometer 0-100 °C| this measured the temperature oof the water. With it being marked in individual degrees the accuracy, when reading the temperature, is increased greatly. | Stop Watch| to make sure you don’t run over time, it is also consistemt as it goes up in both aseconds and miliseconds, increasing accuracy| C: Conclusion Of Results Prelimanry Experiments Preliminary Resu lts | Volume of Oxygen ( ) Produced in 3 Mins (cm )| Mass(g)| Repeat 1| Repeat 2| Repeat 3| Average| 2| 24| 18| 29| 23. 7| 4| | | | | | Exploded| | | | 8| Exploded| Exploded| | | 10| Exploded| | | | Key Orange writing = With disks As you can see from the table, our results were in no way relatable to each other. We could out this down to the fact that as we increased the mass there was a more likely chance of it exploding. When we identified this problem after the 2nd repeat of mass 8g as an obvious reoccurrence and not a simple outlier, we decided to change our amounts so that we could attempt to get results that we will be able to get an average from. As visible in the table, we decided that it might have been either the mass or the amount of pH that caused the excess amount of oxygen to be produced. T We decided at that time that by cutting the pieces of potato into disks would decrease the chance of left over oxygen being produced. However after testing it out we found out that it only decreased the amount of time it took for the test tube containing the ph7 buffer, and potato disks, to explode due to the excess amount of oxygen produced. We also tried lowering the amount of pH buffer and to 4ml instead of the original 7mls, to reduce the amount of substrate available, whilst still using mashed potatoes. This also did not work as we were still using mashed potatoes meaning that the cells were broken easily, releasing more enzyme, which in turn sped up the reaction. Our preliminary method therefore had to be changed. Preliminary Method 1. Weigh your potato and dice into small pieces 2. Use a pestle and mortar to mush up the potato 3. Measure out 7mls of pH buffer and 4. Set up the water trough, beakers, delivery tube and 10 ml measuring cylinder. 5. Set the correct temperature of water using a thermometer, kettle and tap water regulate the temperature. 6. Place the pH buffer and into the test tube and put the same test tube into the beaker. 7. Place the potato in the test tube and close it off with the bung and delivery tube 8. As soon as you’ve done that press start on the timer 9. After 3 minutes record the amount of oxygen has been transferred, this can be done by calculating the water displacement 10. Repeat this and change the mass of potato each time after each interval. By going through the preliminary experiment we were able to identify that 7mls of and pH buffer was too much so we decreased it to 4mls and by dicing the potato instead of mashing it up we were able to reduce the probability of the test tube exploding. By identifying these problems I was able to alter the method so that it will exclude any unnecessary problems from occurring in the actual experiment. Final Method 1. Weigh your potato and dice into small pieces 2. Measure out 4mls of pH buffer and 3. Set up the water trough, beakers, delivery tube and 10 ml measuring cylinder. 4. Set the correct temperature of water using a thermometer, kettle and tap water regulate the temperature. 5. Place the pH buffer and into the test tube and put the same test tube into the beaker full of heated water. 6. Place the potato in the test tube and close it off with the bung and delivery tube 7. As soon as you’ve done that press start on the timer 8. After 3 minutes record the amount of oxygen has been transferred, this can be done by calculating the water displacement Main| | Volume of Oxygen ( ) Produced in 3 Mins (cm )| Mass(g)| Repeat 1| Repeat 2| Repeat 3| Outlier Replacement(s)| Average| 2| 8. 0| 8. 0| 9. 0| | 8. 3| 4| 8. 0| 17. 0| 17. 0| | 17. 0| 6| 19. 0| 23. 0| 25. 0| | 22. 3| 8| 26. 0| 21. 0| 31. 0| | 26. 0| 10| | | | | | 9. Repeat this and change the mass of potato each time after each interval. Key Purple numbers = Outliers Orange numbers = Average excluding the outlier The blue highlighted numbers = highest The yellow highlighted numbers= lowest