8 Measuring the Activity of the Enzyme Catalase

Transcription

1 8 Measuring the Activity of the Enzyme Catalase Enzymes perform work! Objectives: A) To determine the activity of the enzyme catalase in a solution and B) to observe the effects of heat and cyanide inhibitor upon this activity. Introduction: As you know, enzymes represent the class of proteins that catalyze reactions. Catalase is an enzyme found in most living cells. It is abundant in blood, liver, and kidney. It has been isolated in crystalline form and it has a molecular weight of about 250,000 daltons. Catalase is a conjugated enzyme in that it has a heme group. This heme-containing enzyme has Fe 3+ at the center of the heme group. Beef liver catalase has 4 heme groups per molecule. The International Union of Biochemists has assigned numbers to all characterized enzymes as an aid in classifying their functions. The first digit in this number identifies the class to which an enzyme belongs. The enzyme commission number for catalase is E.C This number indicates that catalase is an oxidoreductase meaning that it catalyzes an oxidation-reduction reaction. Catalase is very specific in its reaction, which is the conversion of hydrogen peroxide to water and oxygen: Catalase H 2 O 2 H 2 O + 1/ 2 O 2 Hydrogen peroxide is a product of a number of oxidation reactions in cells, and it can be very damaging to them. One of the functions of catalase is to degrade hydrogen peroxide and thus prevent the accumulation of toxic levels of the oxidant. Because hydrogen peroxide build up would be extremely dangerous to cells, it needs to be removed as quickly as it forms. Consequently, catalase has evolved to be one of the most efficient enzymes created having a turnover number (k cat /K M ) approaching the diffusion-controlled limit (10 8 to 10 9 M -1 s -1 ) under physiological conditions. There are many factors that can affect an enzyme s activity. These factors are generally referred to as effectors. Examples of effectors are temperature, ph, inhibitors, and activators. Effectors such as temperature and ph generally affect the noncovalent interactions that occur in an enzyme s structure. Inhibitors and activators can affect the enzyme s

2 structure and/or its interaction with a substrate, prosthetic group, or other ligand. In today s experiment, the activity of catalase will be measured (assayed) under the following conditions: 1) with substrate alone; 2) in the presence of substrate and a denaturant; and, 3) in the presence of substrate and an inhibitor. The inhibitor we will use is cyanide (CN - ). CN - is a poison to all respiring organisms. It effectively shuts down electron transport and, eventually, oxidative phosphorylation by bonding to the Fe 3+ of heme groups. Such being the case, extreme caution must be exercised while performing today s experiment. Under all assay conditions, the enzyme will be incubated with excess substrate (hydrogen peroxide, H 2 O 2 ) for 5 minutes on ice. The reaction will then be quenched (stopped) by the addition of H 2 SO 4. The amount of H 2 O 2 remaining in the reaction mixture (enzyme + substrate + cofactor + buffer) after 5 minutes of catalase action will be determined by titration with potassium permanganate (KMnO 4 ), a very strong oxidizing reagent. The amount of substrate remaining in the mixture is inversely proportional to the activity of the enzyme. Catalase activity is defined as the number of micromoles of hydrogen peroxide decomposed in 5 minutes at 0 o C per ml. Does performing the experiment at 0 o C adversely affect the enzyme s structure? Procedure: 1. Transfer 0.25mL of a catalase solution to a single eppendorf tube. Place this tube in a boiling water bath and boil for two minutes. Note: The sample must be kept hot until you are ready to add it to Tube 3 (below) at the appropriate time. 2. Set up five test tubes as indicated in the following table. Make additions in the precise order given in the following table.

3 Reagent Tube 1 Tube 2 Tube 3 Tube 4 Tube 5 Phosphate buffer (ml) H 2 O 2 (ml) Water (ml) x 10-5 M KCN (ml) N H 2 SO 4 (ml) Catalase Soln. (ml) Catalase previously boiled for 2 min. (ml) Let all tubes stand 5 minutes at 0 o C 6 N H 2 SO 4 (ml) Fill a buret with 0.005M KMnO 4 and use this reagent to titrate all reaction tubes (1 5) to the pink endpoint. Record these data (ml KMnO 4 used) in the following table. ml KMnO 4 Tube 1 Tube 2 Tube 3 Tube 4 Tube 5 Calculations - The following information may be useful in completing the subsequent table. There is a conversion factor used to calculate the quantity of substrate in the reaction tube either before (µmols i ) or after (µmols f ) the addition of enzyme. The conversion factor is It was calculated using the following relationship: 10 3 µl/ml x 0.005µmol KMnO 4 /µl KMnO 4 x 2.5µmol H 2 O 2 /µmol KMnO 4 When this relationship is multiplied by the volume of KMnO4 (ml) used, the resulting number will indicate the µmols of substrate present in the tube either initially (µmols i ) or at the end of the 5-minute incubation (µmols f ). The quantity of substrate converted (decomposed) to product in five minutes defines µmols decomp. The catalase activity will be calculated by dividing the quantity of substrate converted to product in five minutes by the volume of enzyme used (0.25mL).

4 Parameter Calculation Tube 1 Tube 2 Tube 3 Tube 4 Tube 5 µmols i H 2 O 2 ml KMnO 4 (Tube 1) x 12.5 (the µmols calculated for T 1 will be used for T 2 - T 5 ) µmols f H 2 O 2 ml KMnO 4 x 12.5 µmols decomp H 2 O 2 µmols i - µmols f Catalase activity (µmols/min/ml) µmols decomp / 0.25 Inhibition (%) Activity T 2 Activity T x x 100 Activity T 2 Record the data in your notebook:

5 Catalase STUDY GUIDE 1. The Enzyme Commission number for the enzyme catalase is What type of reaction does catalase catalyze? a. hydrolysis b. bond cleavage with an agent other than water c. oxidation-reduction d. isomerization 2. Most enzymes are optimally active near 37 o C. The catalase experiment was performed near 0 o C. Why? 3. Which of the following reactions is catalyzed by catalase? a. H 2 O 2 > HO-OH + H 2 O b. H 2 O 2 > H 2 O + 1 / 2 O 2 c. H 2 O 2 > 1 / 2 H 2 O 2 + O 2 d. H 2 O / 2 O 2 > H 2 O 2 + O 2 4. In the "Catalase" experiment, titration of the solution in Tube 1 (the Control) tells you. a. how many micromoles of hydrogen peroxide are decomposed in 5 minutes b. how many micromoles of hydrogen peroxide are decomposed in 5 minutes in all tubes (Tubes 1 through 5) c. how many micromoles of hydrogen peroxide are initially present in all tubes d. how many milliliters of KMnO 4 are used in all the tubes 5. In one student's experiment, 15 ml of standard KMnO 4 brought Tube 1 to the end point, and 10.5 ml of the KMnO 4 brought Tube 2 to the end point. How many micromoles of H 2 O 2 were decomposed in Tube 2? a b c d e. None of the above.