Download Enzyme Lab

Biology of Cell Lab (BIOL1021)
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EXPERIMENT OBJECTIVE:
The purpose of this experiment is to understand enzyme
catalysis. Students will perform an enzyme assay and
determine the rate of a biochemical reaction.
BACKGROUND INFORMATION
A catalyst is a substance that accelerates the rate of
a chemical reaction without being consumed or
transformed by the reaction. The catalyst does not alter
the equilibrium constant of the reaction. Only the rate of
approach to equilibrium is changed. A catalyst is not
required in stoichiometric quantities and is often used in
trace amounts. Platinum, palladium, strong acids, and
bases are frequently used catalysts in organic chemistry
and can accelerate reactions thousands of times.
The biological catalysts known as enzymes catalyze
the great majority of chemical reactions in the cell.
Enzymes can accelerate reactions 1014 to 1020 times,
amounts that are far greater than any artificial catalyst.
Enzymes generally produce these accelerations under
the comparatively mild physiological conditions of neutral
pH, atmospheric pressure and temperatures of 37°C.
Unlike most catalysts; enzymes are generally very
specific for the reactions they catalyze. The activity
of certain enzymes can also be regulated by intracellular
concentrations of key metabolites not directly involved
with the reaction. This regulation can increase or
decrease the activity of the enzyme in a manner
adaptive to the cell’s physiological requirements at a
given time. Enzymes that are regulated in this way are
termed allosteric.
Certain proteins also contain, as integral parts of
their structure, chemical groups that are not part of the
amino acid residues but are absolutely required for
biological activity. These groups include small organic
molecules, such as certain vitamin derivatives, and
certain metal ions. Groups such as these are called
prosthetic groups. A well-known prosthetic group is
heme. Heme consists of an iron atom coordinated to the
nitrogens of a set of organic rings called porphyrin,
giving a red color in highly concentrated solutions. The
reactant molecule in an enzyme-catalyzed reaction is
called the substrate. The substrate (S) is transformed to
product (P). Before the enzyme can transform the
substrate it must first bind to it. Initial binding is noncovalent and can be in rapid equilibrium. After productive
binding has been achieved, the enzyme-substrate
complex can now generate product that is subsequently
released. The free enzyme (E) can now react with more
molecules of substrate (the enzyme has turned over).
This can be summarized using a single substrate, single
product, and non-reversible reaction:
E + S  ES  EP  E + P
If the reaction has 2 substrates, then there needs to be a
S1 and S2 in the beginning part of the equation, also if
Lab 5 – Enzyme kinetics
there are 2 products this must be reflected at the end of
the equation.
The disappearance of S or the appearance of P (or
both) can be measured as a function of time. This
relationship is the rate of the reaction. The method of
measurement is called the assay. At a fixed enzyme
concentration and fixed reaction conditions, increasing
substrate concentrations can increase the reaction rate.
The probability of forming more ES complex increases
when there are more substrate molecules present.
Generally, the substrate concentration is thousands of
times greater than the enzyme concentration in kinetic
studies performed in vitro. At the early stages of the
reaction, if the substrate concentration is in great
excess, the rate is approximately linear with time and is
termed the initial velocity (v).
[S]1 - [S]2
T1 - T2
where [S]1 is the molar concentration of substrate at
some initial time, T1, and [S]2 is the substrate
concentration at a later time, T 2. The reaction rate can
also be expressed in terms of the appearance of
product:
[P]2 - [P]1
T2 - T1
Note that the concentration of substrate decreases
with time and the concentration of product increases with
time. Graphically, this can be represented with the
substrate concentration on the y-axis and time on the xaxis. The decrease in the substrate concentration with
time will generate a curve. The rate of decrease is
fastest at the earliest times of the reaction since the
substrate concentration is comparatively high. The rate
of decrease diminishes at later times because the
substrate concentration is lower and the reaction is
slower. Within small time intervals there will be sections
of the curve that are approximately linear and the rate
can be estimated. The rate of an enzyme reaction
cannot be increased indefinitely by continuously
increasing the substrate concentration. At some
substrate concentration, all the enzyme molecules are
bound to substrate and are involved in some stage of the
catalytic cycle. Under these conditions the enzyme is
saturated with substrate and no further increase in
reaction velocity is observed. This does not mean that
the enzyme is not functioning, it only means that the rate
is at its maximum
Only a relatively small portion of the enzyme
molecule is involved with substrate binding and
catalysis. This region is called the active site. The active
site contains the critical amino acid residues and, if
applicable, the prosthetic groups required for activity.
Hydrogen peroxide is a toxic by-product of aerobic
oxidation and certain processes in intermediary
metabolism. All aerobic life forms have evolved methods
Biology of Cell Lab (BIOL1021)
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of enzymatic peroxide detoxification. Almost all the cell
types in mammals contain catalase, with liver, kidney
and erythrocytes being particularly rich sources. The
enzyme catalase (H2O2:H2O2 oxidoreductase) catalyzes
the rapid decomposition of hydrogen peroxide by the
following reaction:
2 H2O2 ———> 2H2O + O2 (gas)
Bovine liver catalase has a molecular weight of
approximately 250,000 and is a tetramer of 4 identical
subunits. The enzyme contains 4 heme prosthetic
groups per molecule, one per subunit. The heme forms
part of the active site and there are 4 active sites per
molecule of enzyme. The peroxide oxygens are
believed to be coordinated to the heme iron during one
phase of the catalytic cycle. Another molecule of
peroxide is then used to complete the reaction. Catalase
has one of the highest turn over rates known. Over 3.6 x
107 molecules of hydrogen peroxide are decomposed
per minute per molecule of enzyme.
We will use a colorimetric assay to measure the
concentration of hydrogen peroxide based on the
oxidation of iodide by peroxide:
2 I- + 2 H+ + H2O2 ———> 2 H2O + I2
The generation of iodine imparts a brown-red color
to the solution. The color intensity increases with the
Lab 5 – Enzyme kinetics
peroxide concentration. Consequently, catalase
mediated decomposition of hydrogen peroxide will
decrease the color intensity of the assay with time. The
stoichiometry of this reaction creates 1 mole of iodine for
every mole of hydrogen peroxide. The assay has been
designed to use a minimum of acid (less than 10 mM
HCl). For a typical catalase assay the iodine
concentration generated is less than 0.2 mM.
Student Experimental Procedures
The enzyme catalase catalyzes the decomposition of
hydrogen peroxide (substrate) to water and oxygen gas
(products). In this experiment:
• Catalase will be added to a buffered solution of
hydrogen peroxide. A time course of the reaction will be
obtained by removing aliquots from the reaction tube
every 30 seconds.
• These aliquots will be added to separate tubes of
assay solution. The assay solution denatures the
enzyme, catalase, which destroys its activity.
• The iodide (I-) in the assay solution is oxidized by any
remaining peroxide, producing a red-brown iodine (I2)
solution. The color intensity can be quantitated in the
spectrophotometer and the rate of the reaction
determined.
• The concentrations of peroxide and enzyme in the
reaction are approximately 1.8 milliMolar and 5
nanoMolar respectively.
The class will perform this lab in groups of three. Each group will perform the reactions at either 4, RT or 37.
All the results from each groups assay will be placed on the board. It is your responsibility to get the data written
down. You will calculate the average absorbance obtained for each time point at each temperature. Your
homework will be to perform the analysis described at the end of the lab sheets and answer the study questions.
This will be turned in at the start of the next lab. Please answer questions on a separate sheet of paper.
PREPARATION OF ASSAY TUBES
1. With a water resistant pen, label 6 empty test tubes (at the top) to indicate blank and the various reaction times (0 to 2
minutes): B (blank)
0
0.5
1.0
1.5
2.0
2. With a 5 ml pipet, transfer 3 ml of assay solution to each of the tubes. Set pipet aside for disposal.
3. With a 1 ml pipetor with a tip, transfer 0.3 ml (300 l) of diluted buffer to the blank. Mix by tapping. Discard tip.
PREPARATION OF CONTROL AND REACTION TUBES
4. Label one of two remaining test tubes "Con" (for control) and the other "Rxn" (for reaction).
5. With a 5 ml pipet, dispense 1.8 ml of Enzyme Reaction Cocktail to each of the two tubes.
6. With a 1 ml pipetor with tip, add 0.3 ml (300l) of dilute phosphate buffer to the control tube (Con). Discard pipet tip.
7. Remove 0.3 ml of liquid from the control tube (Con) and add to the assay tube labeled 0. Set the "Con" tube aside as
this is your negative control. Discard pipet tip.
Biology of Cell Lab (BIOL1021)
Lab 5 – Enzyme kinetics
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8. The remaining steps will be conducted using the Rxn tube. Make sure you are prepared for the following steps as once
you add the enzyme, you must start collecting samples every 30 seconds.
Enzyme reaction cocktail is added. Hydrogen
peroxide that is not catalyzed by the enzyme
catalase will oxidize iodide to give a brown-red
color. The color intensity increases with the
peroxide concentration. Upon addition of
enzyme catalase, the reaction will begin. Start
timing the reaction immediately and aliquot 0.3
ml of this mixture to tubes labeled 0.5, 1.0, 1.5
and 2.0 at 30 second intervals. Do NOT start
this step until you are really ready. Make sure
you understand each of the following steps.
Preparation and Monitoring of Timed Reactions
9. With a 1 ml pipetor with tip, add 0.3 ml of diluted catalase (Enzyme) to the tube you labeled Rxn. Mix. Start timer, or
note the second hand on the clock or watch. Discard pipet tip and immediately get a new one to take samples.
10. With the 1 ml pipet, remove 0.3 ml from the Rxn tube and at 0.5 min. (30 seconds), add it to tube 0.5. Mix.
11. With the 1 ml pipet, remove 0.3 ml from the Rxn tube and at 1 minute, add it to tube 1. Mix.
12. With 1 ml pipet, remove 0.3 ml from the Rxn tube and at 1 minute 30 seconds, add it to tube 1.5. Mix.
13. With 1 ml pipet, remove 0.3 ml from the Rxn tube and at 2 minutes, add it to tube 2.0. Mix. Discard pipet tip.
14. Wait 4 minutes after your last time point to allow full color development.
DATA COLLECTION
Spectral readings can now be taken. You may insert your test tubes directly into the instrument. Wipe the outside of each
tube to remove any dust or fingerprints that may interfere with the reading.
15. Zero the instrument with the tube B solution (Blank) according to your instructor’s directions. Be sure the instrument is
set at 500 nm wavelength. The instrument should read 0 absorbance with the blank solution (no color).
16. Remove the blank and record the absorbance values for each solution in tubes 0 to 2.0. Record the results in the
table. If you are using an instrument such as a Spec 20, it should take approximately two to three minutes to complete
your readings.
Time (Min)
Assay
Solution
Diluted Buffer
Volume Con
Volume Rxn
Blank
3 ml
0.3 ml
---
---
0
3 ml
---
0.3 ml
---
0.5
3 ml
---
---
0.3 ml
1.0
3 ml
---
---
0.3 ml
1.5
3 ml
---
---
0.3 ml
2.0
3 ml
---
---
0.3 ml
A500
Biology of Cell Lab (BIOL1021)
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Lab 5 – Enzyme kinetics
Analysis
The reaction rate can be obtained by graphing the absorbance data versus time. However, the rate can also be expressed
in terms of substrate consumed.
1. To express your data in terms of molar concentration of peroxide:
Absorbance x 11 = Molarity of hydrogen peroxide in Rxn tube.
e
e is the extinction coefficient for this assay system and has been determined by your instructor and is 3.7 x 103.
Multiplication by 11 (dilution factor) gives the peroxide concentration in the reaction tube. Scientific notation will make the
calculations more convenient.
2. Get all the data from the board that was done by all of the groups in the lab. For each temperature, calculate the
average absorbance for each time point. In case you forgot: Average is the sum of all observations at 1 time point divided
by the number of observations. You should end up with 3 sets of numbers to graph on the graph paper.
3. Graph the peroxide concentration on the y-axis versus time on the x-axis.
4. Draw the best straight line through the data points. You may notice some curvature to the data points. This is normal,
especially between 0 and the first time point, and between later time points. You are making a linear approximation.
5. Determine the rate of change in the molarity of hydrogen peroxide with time. The rate is equivalent to the slope of the
line. Pick a time, go vertically up to the line, then horizontally to the y-axis. Determine the concentration in this way for the
next time point.
rate = [peroxide]1 - [peroxide]2
| time1 - time2 |
Express the rate as molarity change per minute.
Study Questions
1. Did you observe bubbles in Step 8 of the experiment? What gas do you expect the bubbles to contain? Assume you
had boiled the enzyme solution before adding it to the peroxide(boiling would cause what to happen to the protein?).
Would you expect to see bubbles?
2. Why did the color intensity of your peroxide assays decrease with time?
3. What makes the rate of a reaction of an enzymatic reaction decrease?
4. Assuming optimal reaction conditions (pH, temperature, etc.) how could you increase the rate of the reaction other
than increasing the substrate concentration? HINT: Something besides temperature.
5. An active preparation of catalase was exposed to the proteolytic enzyme, trypsin that breaks proteins into small
fragments. The catalase preparation was found to be inactive when it was re-assayed. Why?
6. Concentrated solutions of catalase have a red color. Why?
7. What was the overall effect on the catalase reaction when performed at 4? At 37? Hint: what happened to the slope
of the lines for these 2 temperatures?
8. Which of the following generalized enzyme- catalyzed reaction schemes best describes the catalase reaction? Hint:
Remember the chemical reaction for catalase.
a. E + S  ES  EP  E + P
b. E + S1 + S2  ES1S2  EP  E + P
c. E + S  ES  EP1P2  E + P1 + P2
d. E + S1 + S2  ES1S2  EP1P2  E + P1 + P2