Page 2 Enzyme concentration and substrate concentration also play a role in catalysis (Campbell 2005). The more enzymes available, the quicker the reaction will occur until the substrate is all used up. More substrates will also mean a faster reaction rate until the enzyme is fully saturated so that it cannot continue increasing its activity. Inhibitors interact with an enzyme so that its activity is altered. If a molecule decreases or stops the rate of the reaction, it is an inhibitor. These substances regulate how fast an enzyme acts to something foreign. Inhibitors work by unfolding or destabilizing bonds that denature the enzyme. Some inhibitors can block or even change the shape of the active site. The purpose of this lab is to determine how the rate of enzyme activity changes due to several different environmental conditions and how long the reaction will take. This lab focuses on one particular enzyme, catalase. Its purpose is to prevent buildup of hydrogen peroxide, a waste product of cellular activity, by deconstructing it down into two water molecules and oxygen gas. 2H
O + O
+ catalase Most enzymes are adapted to the environments that they react in. For example, pepsin is found in the stomach and is suitable to acidic pH levels. However, some enzymes have optimal levels that are sustainable for them. In order to find out we will put catalase in specific environments that will certainly affect its normal reaction rate. We will be using pH, temperature, inhibitors, enzyme concentration, and substrate concentration to see what results will come due to different quantities of each of the variables that will be used. Using different variables on catalase will inform students how enzymes work in unstable environments exposed to different substances that will denature the enzyme, speed the reaction rate, or decelerate the reaction.
My hypothesis for this lab is that the when catalase concentration has been increased, the reaction rate will increase at an exponential growth, because the substrates will produce no significant change in
Enzymes: Natural Catalysts
Enzymes are catalytic proteins, meaning they speed up chemical reactions without beingused up or altered permanently in the process. Although various enzymes use different methods,all accomplish catalysis by lowering the activation energy for the reaction, thus allowing it tooccur more easily. Enzymes have very specific shapes (conformations). Part of the conformationis the active site of the enzyme, where the actual catalysis occurs. The specific molecule or closely related molecules on which an enzyme functions is known as its substrate. Shape playssuch an important role in enzymatic catalysis that often even isomers of a substrate will berejected. Once the substrate enters the active site, it may begin a process known as induced fit inwhich the enzyme perfectly conforms to the molecule to allow for more efficient catalysis.Changes in environment can severely impact enzyme catalysis in both negative and positiveways. Each enzyme has specific ranges at which it optimally functions; in general, increasing thetemperature will help the reaction along, until the point at which the protein degrades anddenatures. Denatured proteins will often return to their original state, after the removal of thedenaturing agent, except when they are degraded multiple levels.
1.Peel a fresh potato tuber and cut the tissue into small cubes.2.Weigh out 50 grams of tissue.3.Place the tissue, 50 mL of cold distilled water, and a small amount of crushed ice in a prechilled blender.4.Homogenize for 30 seconds at high speed.5.Filter the potato extract using cheesecloth.6.Pour the filtrate into a 100 mL graduated bylinder and add cold distilled water to bring upthe final volume to 100 mL.
Label eight 50 mL beakers as follows: 100 units/mL, 80 units/mL, 75 units/mL, 60units/mL, 50 units/mL, 25 units/mL, 10 units/mL, 0 units/mL.
Prepare 40 mL of enzyme for each of the above concentrations in the following ratio of enzyme:distilled water – 40:0, 32:8, 30:10, 24:16, 20:10, 10:30, 4:36, and 0:40.9.Using forceps, immerse a 2.1cm filter paper disc into the prepared catalase solution for 5seconds.10.Remove the disc and drain for 10 seconds on a paper towel.11.Place the disc at the bottom of the first substrate solution. The oxygen produced from the breakdown of the hydrogen peroxide by catalase becomes trapped in the fibers of thedisc, thereby causing the disc to float to the surface of the solution12.Measure (using a stopwatch) the reaction time for the amount of time from when the discwas placed at the bottom of the beaker until the disc floats on top of the solution. (ratewill be measured in seconds). The rate (R) of the reaction is calculated as R = 1/t.13.Repeat this procedure twice for each enzyme concentration and average the results.