Introduction: Your background info should focus on the role of enzymes as. a catalyst not general info on how to speed up the rate of a chemical

Transcription

1 Introduction: Your background info should focus on the role of enzymes as a catalyst not general info on how to speed up the rate of a chemical reaction. You also need to keep clear the difference between a catalyst and the enzyme catalase. Begin by discussing enzymes- what they are, how they work and why they are specific. Then discuss all factors that affect the rate of enzyme activity. Finally discuss catalase. A ccatalase is a very crucial catalyst that is known to catalyze the conversion of hydrogen peroxide to water and oxygen (Crook, January 15, 2010). How it does this is it binds temporarily to the reactant which lowers the activation energy needed and therefore increasing the speed of the reaction (Massengale, January 6, 2010). What the rate of reaction is, intuitively defined as how fast a reaction takes place (Wikipedia, January 15, 2010). (Don t use Wikipedia and write in your own words don t use quotes) Some reactions occur slowly, such as the oxidation of iron, yet others, such as the combustion of butane in a fire occurs rapidly (Wikipedia, January 15, 2010). The speed depends on how much catalase is present. Along with the amount of catalase present, there are other factors that affect the rate of reaction. (There is a difference between catalase and a catalyst; catalase is a specific enzyme that catalyzes the breakdown of H2O2) A major factor that deals with the rate of reaction is the amount, or concentration, of the catalase (catalase is not a substrate) used. Increasing the concentration allows for more collisions to occur which in effect creates more products at a quicker pace. Now increasing the concentration will only work until the point of saturation is reached. When

2 this situation occurs, it will not matter how much substrate is added the reaction will level off and not increase due to all the enzymes in the reaction being taken up with substrate. Stirring and surface area also have an effect on rate of reaction. Without stirring, a reaction will occur prematurely and thus slow down the rate of reaction. With surface area, the rate increases because smaller pieces of matter have greater surface area, thus giving more of a chance for a particle to hit the surface and react. Temperature also affects the rate of reaction, as well as an enzyme which aids in the proceedings of that reaction. When particles are heated, they gain kinetic energy and ultimately move faster, creating more collisions and thus increasing the rate of reaction (Doc Brown s Chemistry, November 0, 2009). Yet temperature does affect the enzymes in a negative way if their optimal condition is exceeded. An enzyme s optimal temperature is 7 C, and once this temperature is surpassed, it starts to denature. An enzyme denatures because the high temperature disrupts the weak bonds making up an enzyme, such as hydrogen and ionic bonds, thus altering its shape and consequently destroying its function (Bitesized Bio, October 11, 2007). Contrastingly, if the temperature is too low, then there is not enough kinetic energy for the enzymes to move around with and cause collisions, thus producing no reaction ( Enzymes, April 26, 2002). Light works similarly in increasing the rate of reaction, in that it uses beams of photons to break chemical reactions occurring in the solution, and with this they overcome the particle s activation energy, increasing the rate of reaction. Catalysts work a lot like light does, as they also break down chemical bonds and lower the activation

3 energy, providing a pathway for the reaction to take place (Doc Brown s Chemistry, November 0, 2009). ph has a major effect on the way in which an enzyme functions. Certain enzymes work best in either acidic or basic environments, yet when they become too acidic or too basic according to their preference, the enzyme starts to change shape. The globular conformation of the protein is altered and thus changes the shape of the active site, not allowing the substrate to bind to it and hence obliterating its ability to function (Pearson, January 2010). Salinity also has a similar affect, in that it alters the shape of the enzyme, thus making it useless as the substrate cannot bind to its active site ( Enzymes, April 26, 2002). Inhibitors also directly and indirectly affect the way in which an enzyme functions. The direct way an inhibitor impedes the function of an enzyme is through competitive inhibition. The inhibitor mimics the substrate and binds to the active site, leaving the actual substrate to find another enzyme. Even though the inhibitor binds to the active site, no reaction occurs because it lacks the chemical components which bind to the active site and allow for products to be formed (Worthington, January 18, 2010). A way competitive inhibition can be overcome is to add more substrate, which would exceed the amount of inhibitors and thus be able to bind to the active sites. The other form of inhibition is noncompetitively. This is where the inhibitor binds to an area other than the active site. This binding alters the shape of the protein and changes the shape of the active site, not allowing the substrate to bind correctly.

4 There were two research questions that guided this experiment. The first was, How fast does catalase work? and Which variables increase a catalase s rate of reaction? The purpose of this lab was to determine find how differing variables would affect an enzyme s solution s rate of reaction. The variables used included temperature, concentration of catalase, and light/dark environments. There were six hypotheses tested in this lab. They included, If the concentration of hydrogen peroxide wasis increased, then the rate of reaction will increase., If the potato solution with hydrogen peroxide wasis exposed to darkness, then the rate of reaction will decrease., and If the potato solution wasis exposed to heat, then the rate of reaction will increase and more gas will be produced. Furthermore, the hypotheses stated that, If the potato solution wasis exposed to ice cold temperatures, then the rate of reaction will decrease., If the potato solution with hydrogen peroxide wasis exposed to sunlight, then the rate oif reaction will increase., and If a one exposes potato solution was exposed to certain variables and measures how much gas was emitted from the reaction was measured, then one can use that toto the rate of reaction can be determined. determine the rate of reaction. The way this lab was conducted was differing concentrations of substrate were tested to see how much gas was produced by allowing the enzyme, being the catalase, to catalyze the breakdown of hydrogen peroxide, which was the substrate in this experiment. The potato solution, along with 10 ml of hydrogen peroxide, was also tested in a hot water bath, an ice bucket, outside in the sunlight, and the inside of a dark microwave. All variables were set for five minutes and after the time was up the amount of gas produced was measured.

5 Materials/Methods: The first step in the experiment was to gather all materials necessary to conduct the lab. They included potatoes and a blender to make the potato solution, safety goggles, a knife, a funnel, water, and a graduated cylinder. Other materials used were hydrogen peroxide, test tubes (radius measuring 1.5 mm), a funnel, hot water bath, bucket, ice, large beaker, microwave, and a ruler. (Don t list materials) The first variable tested was differing concentrations of substrate. Ten 10 ml of hydrogen peroxide wereas measured with a graduated cylinder and then poured into a clean test tube. Then 10 ml of potato solution wereas measured out with a graduated cylinder and added poured into the same test tube. The solution was then allowed to react for five minutes and when the time was up, the amount of gas emitted in the form of foam was measured with a ruler. This procedure was then repeated for 20 ml, 0 ml, and 5 ml of hydrogen peroxide. All test tubes and graduated cylinders were cleaned after use. The next variable tested was temperature. Ten10 ml of hydrogen peroxide wereas measured into a graduated cylinder and then poured into a test tube. Then 10 ml of potato solution wereas measured and addedpoured into the same test tube. The test tube was then submerged into a hot water bath with water measuring 40 C for five minutes. When the time was up, a ruler was used to measure how much gas was emitted from the reaction (see Figure 1). All test tubes and graduated cylinders were cleaned after use. Figure 1: Set-up of test tube in hot water bath (Describe what picture shows)

6 After the hot water bath was used, the solution was tested in a cold temperature. Ten10 ml of hydrogen peroxide wereas measured with a graduated cylinder and poured into a test tube. Then 10 ml of potato solution wereas measured with the graduated cylinder and added poured into the same test tube. The test tube was then submerged into a bucket of ice measuring at 0 C for five minutes. When the time was up, a ruler was used to measure how much gas was emitted from the reaction (see Figure 2). All test tubes and graduated cylinders were cleaned after use. Figure 2: Set-up of Solution in a Bucket of Ice f(describe in detail what picture shows) Finally, different environmental conditions were tested. Ten10 ml of hydrogen peroxide wereas measured into a graduated cylinder and poured into a test tube. Then 10 ml of potato solution wereas measured into a graduated cylinder and poured into the same test tube. The tube was then taken outside into the sunlight for five minutes and

7 when the time was up, a ruler was used to measure how much gas was emitted from the reaction. Similarly, the affect darkness had on rate of reaction was measured. Ten10 ml of hydrogen peroxide wereas measured into a graduated cylinder and poured into a test tube. Then 10 ml of potato solution wereas measured into a graduated cylinder and poured into the same test tube. The test tube was then put into a large beaker, which was then placed inside of a dark microwave for five minutes. When the time was up a ruler was used to measure the amount of gas emitted. All test tubes and graduated cylinders were cleaned after use. After all data was collected, how much gas produced (in mm ) was calculated and then used to find the rate of reaction. Data/Results: Figure : Calculations for how much gas was produced - Concentration o Constant: 10 ml = π(radius)(height) = π(1.5 mm) 2 (25mm) = mm o Triple: 0 ml = π(radius)(height) = π(1.5 mm) 2 (42 mm) = mm o Double: 20 ml = π(radius)(height) = π(1.5 mm) 2 (85 mm) = mm o Reduced: 5 ml = π(radius)(height) = π(1.5 mm) 2 (65 mm) = mm - Hot Water Bath = π(radius)(height) = π(1.5 mm) 2 (120 mm) = mm - Bucket of Ice = π(radius)(height) = π(1.5 mm) 2 (20 mm) = mm - Light = π(radius)(height) = π(1.5 mm) 2 (9 mm) = mm

8 - Dark = π(radius)(height) Figure 4: Calculations to find out rate of reaction - Concentration o Constant: 10 ml amount of gas produced time mm 5 minutes mm / min = π(1.5 mm) 2 (40 mm) = mm o Triple: 0 ml amount of gas produced time mm 5 minutes mm / min o Double: 20 ml amount of gas produced time mm 5 minutes mm / min - Hot Water Bath amount of gas produced time mm 5 minutes mm / min - Bucket of Ice amount of gas produced time mm 5 minutes mm / min o Reduced: 5 ml amount of gas produced time mm 5 minutes mm / min - Light amount of gas produced time mm 5 minutes mm / min - Dark amount of gas produced time mm 5 minutes mm / min Table 1: Table showing all of the variables used and how much gas was produced with

9 Volume (mm) each and the rate of reaction of each Concentration of what for both variable and 2 nd column Formatted: Superscript Variable Concentration Rate of (mm) Reaction Concentration Constant (10 ml) mm/min Concentration Double (20 ml) mm/min Concentration Triple (0 ml) mm/min Concentration Reduced (5 ml) mm/min Hot Water Bath mm/min Bucket of Ice mm/min Light mm/min Dark mm/min Figure 5: Graph of all different variables and the amount of gas each produced Volume of Gas Produced Through Different Variables Variables Table 2: Table explaining how much gas was produced and rate of reaction produced with differing concentrations of hydrogen peroxide Variable Concentration - Constant (10 ml) Concentration - Double (20 ml) Volume (mm) Rate of Reaction mm/min mm/min

10 Volume (mm) Concentration - Triple (0 ml) Concentration - Reduced (5 ml) mm/min mm/min Figure 6: Graph showing volume of gas produced from differing concentrations of hydrogen peroxide Volume Produced from Differing Concentrations of Hydrogen Peroxide Variable Table : Table showing differing temperatures and amount of gas and rate of reaction produced Variable Volume (mm) Rate of Reaction Hot Water Bath (40 C) Bucket of Ice (0 C) mm/min mm/min Figure 7: Graph explaining temperature differences and the amount of gas produced

11 Volume (mm) Volume (mm) Volume Produced From Differing Temperatures Variable Table 4: Table showing gas produced and rate of reaction for dark and light variables Variable Volume (mm) Rate of Reaction Light Dark mm/min mm/min Figure 8: Graph describing how much gas was produced from differing environmental conditions Volume Produced With Differing Environmental Conditions Variable Once all the data was collected, it was further analyzed and used to find the amount of gas each variable produced, as well as the rate of reaction of each variable.

12 When altering concentrations of substrate, the constant (10 mlof what ) produced 14,1.88 mm of gas and reacted at a rate of 2, mm /min. Doubling the amount of hydrogen peroxide to 20 ml created 48, mm of gas and had a reaction rate of 9,7.42 mm /min. When the hydrogen peroxide was raised to 0 ml, the gas produced was 24,047 mm and had a rate of reaction of 4, mm /min. Then the concentration of hydrogen peroxide was reduced to 5 ml and it produced 7, mm of gas and reacted at a rate of 7,44.22 mm /min (See Table 2). As one can see by looking at Figure 6shows that, doubling the concentration of substrate to 20 ml worked the best in emitting the most gas out of all the differing concentrations of substrate tested, thus showing that it increased the rate of reaction the most (See Figure 6). When temperature was tested, it produced a huge range that was to be analyzed. Putting the solution into a hot water bath produced 68,706.6 mm of gas and reacted at a rate of 1,741. mm /min (See Table ). The bucket of ice, on the other hand, only produced 11, mm of gas and had a reaction rate of 2, mm /min (See Table ). When comparing the two differing temperatures, one can see that the hot water bath produced the most gas through the reaction, thus showing that it increased the rate of reaction more than the ice did (See Figure 7). (Remember rd person only do not use one) Formatted: Superscript Then differing environmental conditions were measured. When the solution was exposed to sunlight, it produced 22,29.66 mm of gas and had a reaction rate of 4,465.9 mm /min (See Table 4). The dark environment produced 22, mm of gas and reacted at a rate of 4, mm /min (See Table 4). As one can see by looking

13 ataccording to Figure 8, the dark environment produced more gas through the reaction, thus increasing its rate slightly more than the sunlight did (See Figure 8). Conclusions: When reviewing all of the analyzed data, it has been shown that increasing a solution s temperature has the most affect on a solution s rate of reaction, yielding the most gas produced out of all variables by reacting at a rate of 1,741. mm /min. This is due to the increased kinetic energy the catalase hads within the potato solution: with this increase in kinetic energy, more collisions occurred and thus increased the rate of reaction the most. According to Michael J. Gregory, Higher temperature generally causes more collisions among the molecules and therefore increases the rate of a reaction. More collisions increase the likelihood that substrate will collide with the active site of the enzyme, thus increasing the rate of an enzyme-catalyzed reaction (Michael J. Gregory, April 18, 2006). (Put in your own words, don t quote. The ice had a very insignificant impact on the rate of reaction, producing a rate of only 2, mm /min. This wasis due to the fact that there was a lack of energy due to the cold temperature. This caused a very small change and generated little gas production, yielding a slow rate of reaction. A similar lab was done in order to see how the rate of reaction was affected with varying catalysts and increasing the temperature. One (you can t use one only can use rd person, you will need to reword sentence to eliminate) timed how long it would Formatted: Superscript take the solution to change colors with the addition of a catalyst at differing temperatures. The data showed that at 8 C, the lowest temperature the solution was faced with, it took

14 the solution 117 seconds to change color, when at 45 C, it only took it 26 seconds to change color ( Reaction Kinetics, January 18, 2010). (Was this an enzyme catalyzed reaction? If not don t use) This showeds that with colder temperatures, the rate of reaction is stalled due to lack of collisions between the substrates and the enzymes. What do published works on catalase show about its rate of reaction and temperature? When dealing with concentration, doubling the amount of substrate worked best by reacting at a rate of 9,7.42 mm /min. This doubling of concentration of the substrate worked the mostre efficiently because it allowed for all more of the substrate molecules in the potato solution to collide with all the active sites available on the enzymes, thus increasing the rate of reaction. This wasis similar to the 5 ml of hydrogen peroxide, which had a rate of 7,44.22 mm /min. Because there was less substrate, there were lessmore collisions and the chance for the solution to become saturated was highly unlikely. This explains why the 0 ml did not work as well, only reacting at a rate of 4, mm /min. Due to the excess of substrate when compared to enzyme, all the active sites were saturated with substrate and an increase in concentration would have no affect on the rate of reaction because there were no more available active sites present to bind that would allow the creation of an enzyme-substrate complex. If saturation was causing the rate difference then 0 would have had a similar rate as 20ml not less. Doc Brown (date) also advocates(not a good choice of word) substrate concentration increasing the rate of reaction because there is a chance of a fruitful collision forming products being proportional to the concentration (Doc Brown, November 0, 2009). (Again your own words, don t quote)

15 It was strange that that tthe solution had awith the higher rate of reaction wasin the dark rather than in the light was unexpected. The dark, reacting at a rate of 4, mm/min, exceeded light s rate, as it was only 4,465.9 mm /min. Light, being a form of energy, should have had more of an impact than darkness, which has no energy involved with it. This could be explained due to a personal error of accidentally shaking the test tube before measuring the amount of gas produced which would have mixed the reactants together causing a reaction. There is a lack of data on how light affects an enzyme s ability to increase rate of reaction, but according to Doc Brown, light breaks bonds and thus lowers activation energy needed for a reaction to occur. An example of this would be photosynthesis, in that sunlight is required for chlorophyll to absorb it and use it to initiate the process of making glucose (Doc Brown, November 0, 2009). But do most chemical reactions in organisms occur in the light? In the dark? A limitation to this lab was there was not an applicable source of darkness available, so the microwave was used as a last resort. This was not a true dark setting and could have had an impact on the results. Testing the potato solution in sunlight also was not very consistent and the results could have been altered by wind, a dry spell, or drops/raises in temperature. An error that occurred was during the first few trials, the wrong sized test tubes were used, not allowing for adequate measuring of gas produced. This was fixed by using a test tube with a larger radius which allowed for the correct amount of gas to be contained and measured. Another error was there was a lack of a control throughout the whole experiment, (which gave nothing for the results that were obtained to be compared to) What does this mean. This could be fixed by measuring the

16 amount of gas produced and rate of reaction of hydrogen peroxide alone, and then comparing that to the results acquired. Why would this be done what variable(s) are you eliminating? A final error that occurred was the maximum rate oif reaction was not measured due to the fact that only 10 ml of substrate was used throughout the lab, yet 20 ml of substrate proved to work best when increasing the rate of reaction. This could be corrected by obtaining the maximum substrate needed first and then proceeding with the experiment. Another limitation to this lab consisteds of when the hot water bath was used, it was previously being occupied by another experimenter who was heating the water to a different temperature than this lab called for. This could have affected the results in that as the water was being heated to 40 C, it took time for this to occur, yet the test tubes were put in anyway due to a lack of time. The thermometer read 40 C by the time the test tube was submerged for five minutes, but this does not mean the temperature was consistent throughout the time interval, resulting in an inconsistency in this lab. (Time and resources was another limitation, in that the experiments had to be done in a certain amount of days, as well as the fact that all resources had to be shared with everyone throughout the class.) Not an error. Omit. Another problem is using bubbles since large bubbles burst so the height may not be a way to get an accurate measurement of gas produced. Further Experiments:

17 There are many directions this lab could go from here in order to conduct further experiments. Instead of pouring the hydrogen peroxide and potato solution together first and then putting them in the hot water bath, one could put them in separately and after five minutes was up, they could pour them into a test tube together and measure the amount of gas produced through bubbles from there. The same concept can be used for the bucket of ice. Instead of pouring the hydrogen peroxide and potato solution into a test tube first and then putting them into a bucket of ice, one could put these different solutions into separate tubes and ice them separately, then after five minutes pour them into the same tube and measure the gas produced. (This should have been an error or limitation discussed in conclusion.)these further experiments would reveal accurate results and eliminate the possibility of mixing the two solutions beforehand, thus increasing the rate of reaction on accident. Another experiment that could be done through this lab is to first measure the gas produced and rate of reaction of hydrogen peroxide alone, as this would provide a constant for the lab to be based on. These needs to be done at the same time to eliminate any variables that would affect results not done first. The time in which the test tubes were placed in their differing variables could also be altered as a further experiment. Five minutes could have been too long for the solutions to sit in their conditionsvariables, so as an extension to this lab, one(don t use one, reword sentence to eliminate) could measure the rates at time intervals as to determine if prove that the rate of reaction was constant throughout the whole experiment. The way in which the gas was measured could also be changed. Instead of measuring the bubbles produced with a ruler, which

18 could be inaccurate because larger bubbles pop, one could use a gas sensor in order to measure how much gas was produced, this being way more accurate and precise (Pasco, January 18, 2010). One could also measure the gas produced through volume displacement, which would yield a more exact number of the amount of gas emitted. Measuring the maximum rate of reaction could also prove as an extension to this lab. Instead of using 10 ml of substrate as a constant, one could first measure differing concentrations of hydrogen peroxide to see which one would create the most rate of reaction. Once this is obtained, this amount can then be used as the constant to provide the most accurate and highest results. Testing other conditions? References: Bitesize Bio. Why Do Enzymes Have Optimal Temperatures? October 11, January 18, < Crook, James. Catalase An Extraordinary Enzyme. July 5, 200. January 6, < Doc Brown s Chemistry. Factors Increasing the Speed-Rates of Chemical Reactions. November 0, November 2, < Enzymes. April 26, January 18, < Lizarrago, Melissa. Lab partner, helped conduct experiment. November 2 December 11, Massengale, C. Enzyme Rate of Catalase. Janaury 6, Michael J. Gregory. The Biology Web. Enzymes. April 18, January 18, < %20101%20Laboratory/Enzymes/Enzymes.htm>

19 Pasco. Chemistry. All That Fizz: CO 2 Gas. January 18, < Pearson. ph and Enzyme Function. January January 18, < Reaction Kinetics. January 18, < on_kinetics.htm> Wikipedia. Reaction Rate. January 15, January 18, < Worthington. Effects of Inhibitors on Enzyme Activity (Introduction to Enzymes). January 18, <