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
Fe3+ 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. 1.11.1.6. 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:
H2O2
Catalase
H2O + 1/2 O2
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 (kcat/KM)
approaching the diffusion-controlled limit (108 to 109 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
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 Fe3+ 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, H2O2) for 5 minutes on ice. The reaction
will then be quenched (stopped) by the addition of H2SO4. The amount of
H2O2 remaining in the reaction mixture (enzyme + substrate + cofactor +
buffer) after 5 minutes of catalase action will be determined by titration
with potassium permanganate (KMnO4), 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 0oC per
mL.
Does performing the experiment at
0oC 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.
Reagent
Phosphate buffer (mL)
H2O2 (mL)
Water (mL)
5 x 10-5 M KCN (mL)
6 N H2SO4 (mL)
Catalase Soln. (mL)
Catalase previously boiled for 2 min. (mL)
Tube 1
5.0
1.0
0.5
--1.0
0.25
---
Tube 2
5.0
1.0
0.5
----0.25
---
Tube 3
5.0
1.0
0.5
------0.25
Tube 4
5.0
1.0
0.25
0.25
--0.25
---
Tube 5
5.0
1.0
--0.5
--0.25
---
Let all tubes stand 5 minutes at 0oC
6 N H2SO4 (mL)
---
1.0
1.0
1.0
3. Fill a buret with 0.005M KMnO4 and use this reagent to titrate all reaction tubes (1 –
5) to the pink endpoint. Record these data (mL KMnO4 used) in the following table.
Tube 1
Tube 2
Tube 3
Tube 4
Tube 5
mL KMnO4
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 (µmolsi) or after (µmolsf) the addition of enzyme. The conversion factor is
12.5. It was calculated using the following relationship:
103µL/mL x 0.005µmol KMnO4/µL KMnO4 x 2.5µmol H2O2/µmol KMnO4
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 (µmolsi) or
at the end of the 5-minute incubation (µmolsf).
The quantity of substrate converted (decomposed) to product in five minutes defines
µmolsdecomp.
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).
1.0
Parameter
Calculation
µmolsi H2O2
mL KMnO4
µmolsf H2O2
mL KMnO4 x 12.5
µmolsdecomp H2O2
µmolsi - µmolsf
(Tube 1)
Tube 1
x 12.5
(the µmols calculated for T1 will
be used for T2 - T5)
Catalase activity µmolsdecomp / 0.25
(µmols/min/mL)
Inhibition (%)
Activity T2 – Activity Tx
Activity T2
Record the data in your notebook:
x 100
Tube 2
Tube 3
Tube 4
Tube 5
Catalase STUDY GUIDE
1. The Enzyme Commission number for the enzyme catalase is 1.11.1.6. 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 37oC.
performed near 0oC. Why?
The catalase experiment was
3. Which of the following reactions is catalyzed by catalase?
a. H2O2 ______> HO-OH + H2O
b. H2O2 ______> H2O + 1/2 O2
c. H2O2
______> 1/ H O + O
2 2 2
2
d. H2O2 + 1/2 O2
______> H O + O
2 2
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 KMnO4 are used in all the tubes
5.
In one student's experiment, 15 mL of standard KMnO4 brought Tube 1 to the end
point, and 10.5 mL of the KMnO4 brought Tube 2 to the end point. How many
micromoles of H2O2 were decomposed in Tube 2?
a.
b.
c.
d.
e.
131.25
56.25
318.75
187.5
None of the above.