On day 45, pods can be removed from the dried plants and seeds can be harvested. This is
the end of a complete life cycle.
For your experiment, you will grow two pots of Brassica rapa plants. After they have been
allowed to grow for a few days, you will trim them back and then apply an experimental
setup to one of the pots. You will then observe and measure them over the rest of their life
cycle.
Microbial Ecology
In the second section of the module, you will collect microbial samples from the
environment which you will then grow and study in lab. You will use microscopic and
biochemical techniques to characterize the broad classes of microbes that you isolated. You
will learn how to make slides and use an optical, 1000x microscope, and how to perform a
Gram stain on a microbial sample.
Classical biological
taxonomy relied heavily on phenotypic variation, i.e. observable characteristics and traits.
The taxonomic system that we use today owes a heavy debt to Carl Linnaeus and his
binomial naming system (Genus followed by species). Linnaeus also formalized the
taxonomic hierarchy system, which we use today in its modern form – Domain Kingdom
Phylum Class Order Family Genus Species. In this way, every species can be slotted into
the hierarchy of life on earth, with each subsequent step down the hierarchy being a subset
of the previous branch.
Today, the highest taxonomic rank in standard use is the Domain, which is separated into
three branches – Archaea, Bacteria, and Eukarya. The first two are composed of prokaryotic
microorganisms, representing the overwhelming majority of species diversity on Earth, but
barely understood (or not known at all) to those who originally set up the taxonomic
classification system. Animals and plants are (relatively) easy to classify into species by
simple observation, but prokaryotes are much more difficult, so once it became understood
58
that they represented their own distinct branch on the tree of life, there then needed to be
methods to
identify and classify them.
The simplest way to classify bacteria is to simply observe them, particularly their shape and
color. Some of the nomenclature we still use reflects this classification – coccus (as in
Streptococcus or Staphylococcus) is Latin for “grain” or “seed”, because bacteria that have a
coccus shape are spherical like a seed. Similarly, bacillus (Latin for “stick”) refers to a
bacteria that is rod-shaped. Spirillum and Spirochaeta are spiral-shaped bacteria.
Cyanobacteria are also known as blue-green algae, for their color; these are the only
photosynthetic bacteria capable of producing oxygen.
Simple morphological taxonomy
only gets you so far, so other methods were developed to help differentiate organisms.
One of the most powerful was developed as a simple method to stain cells for observation,
but was later discovered to divide bacterial species into two fairly large groups; that method
is known as the Gram stain. In Gram staining, cells are treated with a dye (crystal violet)
that reacts with the peptidoglycan in bacterial cell walls; bacteria that have a thick layer of
peptidoglycan are stained purple when they react with the dye, and these are known as
59
“Gram positive”, while bacteria with a thin layer of peptidoglycan are not permanently
stained by the dye; these cells are known as “Gram negative”. In Gram staining a counterstain is applied to the cells (usually safranin) which stains cells pink, but any cells that were
stained with the crystal violet will retain their purple color, and therefore Gram positive and
Gram negative cells can be distinguished from one another under a microscope.
There are a number of other biochemical tests that can be applied to bacteria to
differentiate them. These tests are often, but not always, dependent on the presence or
absence of certain enzymes that mark species differences. Two tests that you will carry out
in this lab are the oxidase test, which identifies if a species produces certain cytochrome c
oxidases, and the catalase test, which tests for the presence of catalase, an enzyme that
catalyzes the decomposition of hydrogen peroxide. These tests will help you narrow down
the microbe that you are looking at, but likely will not be enough to pinpoint it to a single
species.
In modern microbial taxonomy, genetic methods are often used to distinguish between
species, commonly by sequencing the 16S ribosomal rRNA, a component of the 30S small
subunit of prokaryotic ribosomes. These are highly conserved between different species,
because they evolve slowly over time, meaning that species that have evolved from a
common ancestor can be identified by the degree to which their 16S rRNA has diverged. In
future modules, you will learn the process by which species can be identified and
catalogued by PCR; for this module, we will be using morphological and biochemical tests;
these methods are not mutually exclusive, but often are used to reinforce one another.
Pre-Lab Questions:
Day 1:
Design an experiment to conduct on your plants. See the “Experimental design” section,
below, for possible ideas. Describe the setup in paragraph form. What will you test? How
will you test it?
Day 3:
Define the following terms:
⦁
Cotyledon
⦁
True leaf
⦁
Hypocotyl
⦁
Nodes
⦁
Silique
Day 4:
⦁
What is peptidoglycan
⦁
How does it cause the difference in the Gram stain between Gram negative cells
60
and Gram positive cells
⦁
What is the difference between a simple optical microscope and a compound
optical microscope?
⦁
If the objective lens in the microscope you will be using magnifies at 100x, and the
total magnification of the image is 1000x, what is the magnification of the eyepiece
lens?
⦁
Why do we put oil on the slide before viewing the image at 1000x magnification?
Day 6:
For this experiment, we will do several biochemical tests. These tests are not remotely
exhaustive, and only scratch the surface of the possible tests one can do on an unknown
bacteria. Name and describe three other biochemical tests, and whether E. coli and M.
luteus would be positive or negative for those tests.
Part 1 – Plant ecology
Procedure
Day 1: Setup, planting and experimental design
You will grow your plants in the microscope room, which has been outfitted with a coolwhite fluorescent lighting system. Each table will work as a single unit during this part of the
module.
⦁
Using a pair of scissors, cut a small X in the bottom of the two smaller plastic
containers on your bench.
⦁
Wet a cotton wick thoroughly, and then insert it through the hole you cut.
⦁
Add about 60 mL of potting mix into each container.
⦁
Spread 18 fertilizer pellets over the top of each.
⦁
Add another 120 mL of potting mix, covering the pellets.
⦁
Sprinkle water onto the soil until you see water begin to drip from the wick.
⦁
Place one seed disk on top of each of your containers. Then lightly cover the seed
disk with no more than about 25 mL of potting mix.
⦁
Pour water into the larger containers, to serve as a reservoir. Place the smaller
containers into the larger ones so that the wick reaches into the water. Gently
water the potting mix from above using a Pasteur pipette to wet the seed disk.
61
⦁
Place the containers under the fluorescent light, making sure to label your
containers. The potting mix surface should be about 5-8 cm below the light source.
Day 2: Observation and Experimental Design
Observation:
Observe your containers. Take note if anything has changed.
Experimental design
With your group, come up with an experiment to test on your plants - one container will be
your untreated control, and the other will be your experimental group. Some ideas to get
you started:
⦁
Growth hormones.
⦁
Plant spacing and growth
⦁
Water relations: excesses and deficiencies
⦁
Photoresponses: light intensity and photoperiod
⦁
Salt injury
⦁
Herbicide resistance
⦁
Acid rain impact
⦁
Environmental pollutants
⦁
Microbe-plant interactions
⦁
Pollen and pollination (bee sticks)
Day 3: Thinning and experimental setup
⦁
Choose four to six healthy plants that are spaced far apart in each of your
containers. With scissors, carefully clip off the other seedlings at the soil surface,
leaving only the ones you have selected (if you have chosen an experimental design
that involves plant spacing or the affect of the number of plants on growth, then
trim your plants accordingly)
⦁
Carry out your design. If you need chemicals, talk to your TA or instructor about
making solutions. The simplest way to apply chemicals is to mix them into the soil
in water.
Days 4 – 8: Watering, measurements and observation
62
⦁
The Brassica rapa will need to be watered regularly. In order to do this, you will
need to fill the reservoir (the large container) so that the water is touching the wick.
Do this during every class period
⦁
Rinse and clean out the water reservoirs once per week. This will help to prevent
the buildup of algae in the reservoir.
Measurements and observation:
⦁
Observe your plants during each class period. Note things like color changes,
the emergence of true leaves/flowers/seeds, and any other noticeable
differences you see. Pictures are always helpful.
⦁
For quantitative data, you can measure:
⦁
Plant height
⦁
Number of leaves
⦁
Number of open flowers
⦁
Leaf size (length/width)
⦁
Number of nodes
⦁
Number of hairs on first true leaf (requires the use of the microscope)
⦁
Plant weight (can only be done at the end of the experiment)
And anything else you can think of.
Part 2 – Microbial Ecology
Procedure
In this experiment, you will begin by collecting environmental samples. This is a groupbased, not table-based, exercise; work in groups of 3.
Day 1: Sample collection
⦁
Your group will need two plates for this experiment, one LB and one Nutrient Agar.
Label each on the bottom with your group identification.
⦁
Each group will need two sterile swabs. Choose two locations to collect samples
from, this can be tabletops, desktops, floors, or any other surface of your choosing.
Rub the end of your swab over the surface and then transfer to your plates.
⦁
You will want to do zig-zag streaks across your plates, trying to cover the widest
63
area possible.
⦁
Incubate your plates at 37C. They will be allowed to incubate for 24-48 hours.
Day 2: Isolation of colonies
⦁
Remove your plates from the incubator. Examine them and write down your
observations in your lab notebook.
⦁
You are going to pick a
single colony to study. Pick one that has relatively sharp borders and a uniform
color, and that can be picked in isolation from its neighbors. Avoid colonies with a
powdery texture (molds)
⦁
Use a sterile loop to pick that colony and then streak on a fresh plate of the same
type of medium that your initial colony grew on, using the T-streak method:
⦁
Incubate your plates at 37C.
Day 3: Liquid cultures
You are now going to grow your cells up in liquid culture, in order to propagate the cells for
further study.
⦁
You will need one 15 mL Falcon tube containing either LB or Nutrient Medium. This
should be the same growth medium that is in your plates.
⦁
Pick a single, isolated colony from your Day 2 plate using a sterile loop. Insert the
loop into your culture tube and spin it to dislodge any cell debris.
⦁
Place the cap on top of your tube. Do not screw the cap on! It’s important that you
grow your cells aerobically, and screwing the cap on will not allow for oxygen
replenishment. Instead, tape the cap in place using labeling tape.
⦁
Give your tube to your TA. They will be placed on a 37C incubator overnight.
64
Day 4
and 5: Gram staining and microscopy
Today, you will use the microscopes to observe your cells, and learn how to carry out a
Gram stain. You will compare your cells to two known cultures, one Gram negative (E. coli)
and one Gram positive (Micrococcus Luteus)
⦁
Remove your liquid culture from the refrigerator. You will also need cultures of the
two known organisms.
⦁
Using a Pasteur pipet, place a drop of each culture on a slide. You will want to label
the slide first, to make sure you know which sample came from which culture.
⦁
Air dry the slide(s)
⦁
Once they have sufficiently dried, heat fix the bacteria to the slide by passing it
carefully through the flame of your alcohol burner. Pass quickly through the flame
several times – do not hold the slide over the flame for a prolonged period, since it
will get hot. You can use slide clamps to hold the slide. This creates a smear.
⦁
Cover the smear with crystal violet and let it sit for 30 seconds. This is the primary
stain and it will stain Gram positive cells purple.
⦁
Carefully rinse away the crystal violet with water from a squeeze bottle. Allow the
water to trickle over your smear, do not hit the smear directly, since you don’t want
to dislodge the cells.
⦁
Cover the smear with Gram’s iodine and allow it to sit for 60 seconds. This traps the
crystal violet stain in the cells that took it up.
⦁
Carefully rinse with water.
65
⦁
Blot, do not rub, the smear using a Kimwipe until dry.
⦁
Cover the smear with ethanol and immediately, gently rinse with water. This
decolorizes the Gram negative cells, so it is important that you do not let it sit in this
for long, because eventually all of the bacteria will begin to degrade.
⦁
Cover the smear with safranin and let stand for 45 seconds. This counter-stains cells
that did not take up the crystal violet to a pink color.
⦁
Carefully rise with water and again blot dry with a Kimwipe.
At this
point, you will observe the bacteria under the microscope using the 100x objective. You will
use immersion oil to do this, and you can put the oil directly on the Gram stain without a
cover slip.
⦁
Important: if you need to move your microscope to get it into position, lift it up by
66
holding the portion of the frame in the back that curves around and carrying it to its
new destination. Never drag the microscope across a surface, and never lift it by
holding the stage or the objectives.
⦁
Plug your microscope in and turn on the LED illumination via the switch on the
bottom (1). The dial next to the light switch (2) adjusts the brightness.
⦁
Make sure the light path selection knob (3) is set to eyepiece, not camera.
⦁
Before you place your slide on the specimen holder, rotate the course focusing knob
(7, the big knob not the small one) so that the stage is fully lowered.
⦁
Press the holding lever (4) forwards and place your slide on the stage so that it is
held by the specimen holder.
⦁
Move the X and Y axis knobs (5) so that the smear is centered under the objective.
⦁
Turn the revolving nosepiece (6) so that the 4x objective (the short red one) is
pointed at your slide. Rotate the nosepiece by turning the wheel, not by holding
the objective.
⦁
Adjust the interpupillary distance (8) so that you can look through the eyepieces
comfortably. If you are not wearing glasses, move the folded eye shades (9) into
their unfolded position.
⦁
Now you can use the focusing knob to bring the specimen into focus. Use the
course focusing knob to make large adjustments, and the fine knob to make small
adjustments.
⦁
Switch to the 10x objective (yellow) and bring the specimen into focus again. From
this point forward, you need to be very careful about adjusting the course focusing
knob. The 40x and 100x objectives will be situated very close to the slide. If you
carelessly adjust the course focusing knob when either of those objectives are
engaged, you could damage the slide or the objective. You should be able to
proceed without adjusting it further.
⦁
Engage the 40x objective. Use the fine focusing knob to focus the specimen.
⦁
Move the revolving nosepiece so that it is between the 40x and 100x objectives.
You will now apply immersion oil to your slide, directly on top of the smear.
⦁
Rotate the nosepiece so that the 100x objective is engaged. It should be touching
the oil on the slide; if it isn’t, ask your TA for help.
⦁
You should now be able to see your Gram stained cells. Make notes of the
appearance of the cells; their shape, their size, and their color. You can take a
67
picture by placing your phone’s camera to the eyepiece. You want to note whether
the cells are Gram positive (purple) or Gram negative (pink) and whether they are
rod-shaped, spherical shaped or something else.
⦁
Repeat the process for each of your smears.
⦁
When you are done: gently wipe the objective lens with lens paper in order to
remove the oil. It is important that you do not leave residue on the lens. Dispose of
your slides in the sharps container provided.
⦁
Turn off the microscope and unplug it. Return the stage to its lowest position, and
place the 4x objective into the viewing position.
Day 5: Plates and Peptone Water
⦁
You will need a fresh LB or Nutrient plate, as well as one tube containing Peptone
Water
⦁
Pick a single, isolated colony from your Day 2 plate using a sterile loop. T-streak on
your fresh plate and incubate at 37C
⦁
Pick another colony from your Day 2 plate and inoculate the Peptone Water using a
sterile loop.
⦁
Insert one Kovac’s Reagent Strip (Indole test) right beneath the cap, so that the strip
is suspended in the head space of the tube and will not fall into the liquid
⦁
Place the cap on top of your tube. Do not screw the cap on tightly! It’s important
that you grow your cells aerobically, and screwing the cap on will not allow for
oxygen replenishment. Instead, tape the cap in place using labeling tape, making
sure your reagent strip stays suspended.
⦁
Give your tube to your TA. They will be placed on a 37C incubator overnight.
Day 6: Biochemical tests
68
You have determined whether your cells
are Gram negative or Gram positive. You will now do three other biochemical tests to help
narrow down its identification.
Test 1 – Catalase
⦁
Place a drop of 3% H2O2 on a slide.
⦁
Using a sterile inoculating loop, scrape
cells from your plate and mix them in the hydrogen peroxide.
⦁
If catalase is present, you will see a rapid evolution of oxygen gas within 5-10
seconds, as evidence by bubbling.
⦁
If you see no bubbles, or only a few scattered bubbles, then catalase is negative.
https://i2.wp.com/microbe
online.com/wpcontent/uploads/2013/10/c
atase-test.jpg
Test 2 – Oxidase
⦁
Take one strip of oxidase paper for each of the cultures in question.
⦁
Pick a colony from your plate using an inoculating loop and smear on the diagnostic
69
strip of the paper.
⦁
Observe the inoculated area for a color change to deep blue or purple within 10-30
seconds. Purple indicates a positive result for oxidase, otherwise the color should
not change.
Test 3 – Indole
⦁
Read the Indole test strip in your Peptone Water tube. If it is pink, the test is
positive. No change to the color of the strip means a negative test.
You now know what shape your cells are, their Gram staining status, and whether they are
catalase and oxidase positive or negative. Using these pieces of information, as well as the
appearance of their colonies on a plate, you should be able to narrow down the possibilities
as to what species your microbe belongs to.
Observing Chemical Kinetics with the
Iodine Clock Reaction
ISC Lab I – Kinetics Module
Written by Adam Smith and Jose Rodriguez Corrales
Safety: Standard laboratory personal protection, goggles, lab coat, gloves.
Skills: Basic reaction set-up, data collection, and interpretation.
Introduction
70
Rate of reaction
Rate of reaction is a measurement of how fast reagents are consumed and/or products are
formed in a chemical reaction. For a general equation of the form:
A + 2B → 3 C
the average rate of reaction over a period of time can be defined as:
where the square brackets refer to the molar (mol/L) concentration of each chemical and
lowercase t stand for time (usually expressed in seconds). Since two moles of substance B
are consumed by every mole of substance A, the rate of reaction has to be “standardized”
by dividing by the corresponding stoichiometric coefficients as displayed above. Note that
the concentration of reagents decreases while that of products increases when the reaction
takes place. Thus, a negative sign needs to be used when calculating rate of reaction using
the reagents.
Chemical reactions require bond breaking and/or making. Consequently, it is intuitive to
think that a higher concentration of reagents speeds up a reaction, but this is not always
true. The rate of reaction at any particular moment (instantaneous rate) can be expressed
as the multiplication of a constant and the concentration of each reagent elevated to a
power. This equation is called the rate law of the reaction. For the model reaction
presented above:
rate= k [A]x[B]y
where k is the rate constant (units will depend on the equation), while x and y correspond
to the partial orders of reaction with respect to A and B. Exponents x and y are NOT
necessarily the stoichiometric coefficients and must be determined empirically in the lab.
The partial order of reaction is usually 0, 1, or 2, and in rare cases, ½. The overall order of
reaction is the sum of all partial orders of reaction.
Collision theory
The collision theory establishes that a reaction requires molecules (or atoms) to collide with
other with the right orientation and energy to break and form chemical bonds. Given the
small size of molecules and atoms, collision of three or more of them with the right
orientation is not highly probable. However, some reactions consume several equivalents of
each reagent (i.e. one of A and two of B in the aforementioned equation). Thus, more
“complex” reactions (those that require more than two molecules to collide) might undergo
several “simpler” reactions, denominated elementary steps, in what constitutes the
mechanism of reaction. For example, the formation of three equivalents of C from one of A
and two of B could happen through the following mechanism:
A + B → D (step 1)
71
D + B → 3 C (step 2)
In the proposed mechanism, D is an intermediate: a molecule that is produced in one of the
elementary steps and consumed in a latter one. Note that the “sum” of steps 1 and 2
produces the overall reaction described above.
Furthermore, step 1 and step 2 might happen at different rates. If step 2 is faster than 1,
intermediate D will have a short lifetime in solution and will be consumed almost as soon as
it is generated. Consequently the rate of reaction will only depend on the reagents needed
in step 1, and the order of reaction with respect to A and B will be 1. The second equivalent
of B is not needed until Step 2, which is fast and does not impact the overall rate of
reaction. In other words, the order of reaction for an “overall” reaction represents how
many molecules of each reagent are consumed in the slowest elementary step of the
reaction mechanism, also known as limiting step. Thus, the rate law is not impacted by the
stoichiometry of the overall reaction, but it depends only on that of the limiting step.
As mentioned above, product formation depends on the collision of molecules with the
right orientation and energy. Unlike molecular orientation, the energy of the molecules can
easily be controlled and monitored through an extensive property: temperature.
Temperature is a measurement of the average kinetic energy of the molecules that make up
a piece of matter. Higher temperatures translate into faster molecular motion, which leads
to more frequent collisions between molecules and higher energy (momentum) upon
collision. Thus, most reactions occur faster at higher temperatures.
The collision theory establishes that collision converts the kinetic energy of the molecules
into internal energy. More specifically, the bonds of those molecules that have collided will
vibrate and can break, while this energy can also be used to overcome electronic repulsion
between the molecules and allow formation of new bonds. The molecular arrangement
where some bonds are being broken and/or formed is known as a transition state. The
minimum energy required for the reactants to collide and form the transition state is called
the activation energy.
Iodine clock experiment
In today’s lab, you will measure the rate of reaction and determine the rate law, rate
constant, and activation energy of a simple chemical reaction. In this experiment, you will
investigate the rate of the reaction for:
The rate law for this reaction is given by:
72
(1)
Where the partial and overall orders of reaction and the rate constant are to be
determined. Note that we leave water out of the rate equation.
None of the reagents or products of this reaction absorbs visible light, which prohibits use of
Vis-spectrometry. However, starch reacts with I3- (one of the products) to produce a dark
blue colored complex. If starch alone was added to the reaction mixture, the solution would
turn dark blue as soon as all of the reagents were mixed. Thus, the reaction rate will be
monitored by a “clock reaction”. A known amount of thiosulfate ion will be added to each
reaction. Thiosulfate reacts rapidly with I3- as shown in the following reaction.
Since I3- is immediately consumed, the solution remains colorless as long as there is any
thiosulfate left in solution. When all of the thiosulfate is consumed, the I3- formed in the
initial reaction is free to react with the added starch indicator, and the solution will turn
dark blue. From the above reaction, two moles of S2O32- are consumed by every mol of I3-.
Therefore, the reaction rate can be written as:
And given that the color change is only observed once all thiosulfate has been consumed,
the reaction rate can be rewritten as:
(2)
You will use the dilution formula M1V1 = M2V2 (M1 is the stock concentration, V1 is the
volume of stock solution added, M2 is the concentration of the reactant in the reaction
solution, and V2 is the total volume of the reaction solution) to determine the initial
concentration of the reactant (in this case, M2). Once you have calculated the initial S2O32for each experiment and timed the formation of the blue color, you can determine the rate
of reaction for each experiment by using equation 2.
Order of reaction
The partial order of reaction for each reactant can be found by comparing two experiments
in which the concentrations of only that reactant changes. As an example, we can change
the initial concentration of I- to determine its partial order of reaction.
(3)
(4)
73
(5)
Equations 3-5 explain show how to determine the partial order of the reaction with respect
to I-, but they can be used to determine the partial rate of reaction of any reactant, simply
by changing which rates are compared. Remember that you will use equation 2 to
determine the rate of each reaction.
The overall reaction order can be determined by summing x, y, and z. Once the partial
orders are known, the rate constant can be determined for each experiment using equation
1, and then the rate constants can be averaged to obtain a mean rate constant.
Temperature dependence and the Arrhenius equation
Activation energy is the energy required for a chemical reaction to occur. For instance, if we
look at an exothermic reaction A+B→C+D, the energy diagram would look similar to the
energy diagram shown in Figure 1.
Figure 1. Energy diagram for the exothermic reaction A+B→C+D.
Note that this reaction is exothermic, because it releases more energy than is required to
initiate the reaction (net change in free energy). However, in order to initiate the reaction,
an input of energy is required (activation energy, Ea).
The Arrhenius equation relates the reaction rate constant, k, to the activation energy and
the temperature. This equation is given in Equation 6, where k is the rate constant, A is the
frequency factor, R is the gas law constant (8.314 J/mol*K), and T is the temperature in K.
(6)
If we take the natural log (ln) of both sides, the equation can be reduced to variables that
can be calculated:
74
(7)
If k is known at two different temperatures, we can determine Ea by subtracting the two
equations from one another and rearranging to get Equation 8.
(8)
Pre-lab questions
⦁
⦁
Define the following:
⦁
Intermediate
⦁
Overall order of reaction
⦁
Activation energy
⦁
Transition state
Consider the following chemical equation:
3A + 2B 3C
The overall rate of the reaction = k[A]x[B]Y. Determine x and y from the following
data:
Trial
[A]
[B]
Initial rate
1
0.1 M
0.1 M
0.5 M/min
2
0.1 M
0.2 M
1.0 M/min
3
0.2 M
0.1 M
2.0 M/min
Hint: use equation 5.
What is the overall order of this reaction?
Procedure
75
Determination of the Rate Law:
⦁
Create table 1 in your lab notebook.
⦁
Label four clean beakers 1 through 4.
⦁
Using Table 1, add the specified amount of H2O, KI, Na2S2O3, starch solution, pH 4.7
buffer, and acetic acid to beaker 1. At this point, do not yet add the hydrogen
peroxide (H2O2).
⦁
Measure out the specified amount of hydrogen peroxide into a separate beaker.
⦁
Record the temperature in beaker 1. Record this in your lab notebook.
⦁
Record the pH in each beaker, and record the values.
⦁
With your stop watch ready, quickly add the peroxide and start timing.
⦁
Stop the timer when the color of the solution changes (this can take up to 5
minutes, although it will usually happen within 2-3 minutes. Hit stop when the bulk
of the reaction mixture has turned blue; the most important thing is to be
consistent with your timing.
⦁
Record the time required for the color change.
⦁
Repeat steps 2-6 for beakers 2, 3, and 4.
Table 1. Initial volumes of reactants for determining the rate laws.
Experiment
Volume of H2O
(mL)
Volume
of
Volume
of
0.050
M KI
(mL)
0.050 M
Na2S2O3
(mL)
Volume
of starch
solution
(mL)
Volume
of pH
4.7
buffer
(mL)
Volume
of 0.30 M
acetic
acid (mL)
Volume
of 0.80
M H2O2
(mL)
1
125
25
5
5
30
0
10
2
100
50
5
5
30
0
10
3
115
25
5
5
30
0
20
4
100
25
5
5
30
25
10
76
Reaction
time (s)
Determination of the activation energy:
⦁
Create table 2 in your lab notebook.
⦁
Using the values in Table 2, add the specified amount of H2O, KI, Na2S2O3, starch
solution, pH 4.7 buffer, and acetic acid to beaker 1. Again, do not add the peroxide
yet.
⦁
Measure the pH of the solution and record it in your lab notebook.
⦁
Using one of the hot plates, heat a water bath to 40-45°C.
⦁
Remove the water bath from the hot plate and place the reaction beaker in the
water bath.
⦁
After the temperature of the reaction solution reaches ~40°C, with your stop watch
ready, quickly add the peroxide and start timing.
⦁
Stop the timer when the color of the solution changes.
⦁
Record the time required for the color change.
⦁
You will compare the reaction rate in this reaction to the reaction rate and
temperature in beaker 1 in the previous section.
Table 2. Volumes of reactants for the determination of activation energy
Experiment
5
Volume
of H2O
(mL)
Volume
of
Volume
of
0.050 M
KI (mL)
0.050 M
Na2S2O3
(mL)
25
5
125
Volume
of starch
solution
(mL)
Volume
of pH 4.7
buffer
(mL)
Volume
of 0.30 M
acetic
acid (mL)
Volume
of 0.80
M H2O2
(mL)
5
30
0
10
For Your Lab Report
⦁
Determine the initial reactant concentrations for each experiment using the dilution
formula M1V1 = M2V2 (M1 is the stock concentration, V1 is the volume of stock
solution added, M2 is the initial concentration in the reaction solution, and V2 is the
total volume of the reaction solution). Remember that [H+]=10-pH. Record the values
in the calculations table.
⦁
Determine the rate of each reaction from the time required for the formulation of
77
Reaction
time (s)
the blue starch complex using Eq. 2. Record these values in a calculations table.
⦁
Determine the order of the reaction using experiments 1 through 4 and Equation 5.
Record for each the order of the reaction for each reactant to 2 significant figures,
and then rounded to the nearest integer value. Record the overall order of the
reaction.
⦁
Determine the rate constant for experiments 1 through 4 and record these values in
the calculations table. Calculate the average rate for reactions 1 through 4.
⦁
Determine the activation energy from Equation 8 using data from experiments 1
and 5.
Table 3. Report final concentrations of reagents in the reaction mixture in a table like the one shown below.
Experiment
[KI]
[Na2S2O3]
[H2O2]
[H+]
rate
k
1
2
3
4
Solve for the rate first, then solve for x, y, and z using the equation 5. Next, the constant k
can be solved for by using equation 1.The activation energy, Ea, can be solved for using
equation 8.
78
Discussion questions:
⦁ What units will the activation energy be in?
⦁ What was the purpose of timing the reaction at different reagent concentrations
and temperatures?
⦁ Suppose a catalyst was added to this reaction? What could be a possible scenario
for the effect on the time for the formation of the product and consequently the
activation energy, Ea?
Enzyme Kinetics
ISC Lab I – Kinetics Module
Purpose: To study the kinetics of an enzymatic reaction, and calculate the kinetic constants
KM and Vmax
Apparatus: A spectrophotometer
Chemicals:
⦁
Lactase (in the form of Lactaid)
⦁
Phosphate buffer, pH 7.0
⦁
ONPG (ortho-Nitrophenyl-β-galactoside)
79
⦁
Sodium carbonate
Skills: Dilutions, data analysis and interpretation.
From U Mass Amherst, http://bcrc.bio.umass.edu/intro/content/enzyme-kinetics-labprotocol
Introduction
Lactose metabolism in E. coli is dependent on the control of expression of a set of genes
known as the lac operon. When E. coli cells are grown in the presence of the sugar lactose, a
disaccharide composed of monomers of glucose and galactose connected by a β(1-->4)
glycosidic bond, the lac operon turns on and triggers the expression of the gene that
encodes β-galactosidase (β-gal). β -gal is an enzyme that breaks down lactose into D-glucose
and D-galactose. Because it’s very difficult to directly and accurately measure lactose or
glucose in solution, we can use a false substrate (known as an analog) for the β -gal enzyme
known as ONPG (ortho-nitro-phenyl-galactoside). ONPG has a structure similar to lactose,
so it can bind to the enzyme and be cleaved. The resulting product of that cleavage is
galactose and ortho-nitrophenolate; ortho-nitrophenolate is yellow in color and can be
measured by spectrophotometry.
The amount of yellow product in the tube at any given time depends on two things:
⦁ The amount of β -gal in the solution (how much β -gal is made by the gene being
transcribed into mRNA, and mRNA being translated into protein)
⦁ How fast β -gal converts ONPG to galactose and ortho-nitrophenolate (enzyme
kinetics).
In today's lab we are going to focus specifically on the enzyme kinetics part of the equation.
In order to do this, we are going to be measuring the enzymatic activity of lactase (also
known as lactose-phlorizin hydrolase, or LPH), which is a member of the β -galactosidase
family of enzymes. Like β -gal, lactase breaks lactose down into galactose and glucose;
lactase is found in the small intestine of humans and other mammals.
Figure 1. The normal substrate for lactase in the body, lactose, broken down into galactose and
glucose
Lac operon
Although every cell in the body contains a full copy of the organism’s genetic code, how
cells behave and what they do depends on which genes are expressed at any given time.
80
Expression of a gene means that the gene is actively translated into mRNA, which is then
typically translated into a protein, oftentimes an enzyme which then catalyzes a specific
chemical reaction. It is these reactions that carry out the function of a cell on a minute to
minute basis, and therefore understanding which genes are turned on (and off), and how
genes are turned on and off, is one of the most important pieces of information in
understanding how a cell functions.
Genes are turned on or off by way of a promoter, a sequence of DNA upstream of the gene
that controls the expression of the gene by RNA polymerase (the enzyme responsible for
synthesizing mRNA). The promoter contains specific DNA sequences that allow RNA
polymerase to bind; a promoter can be turned on by the attachment of an activator
protein, or turned off by being bound by a repressor protein, and other regulatory
mechanisms are responsible for the formation of activator and repressor proteins. These
complex cascades of actions and reactions of genes turning on and off and affecting other
the expression of other genes in turn form the network of cell signaling that tells a cell what
to do at any given time.
An operon is a set of genes under the control of a single promoter, generally related to one
another, such that the cell functions with the entire set turned on or off as a group. In E.
coli, the lac operon contains several genes (lacZ, lacY, and lacA) that, when activated by the
presence of lactose in the medium, allow for the transport and hydrolysis of lactose. The
lacZ gene encodes β-galactosidase, but the other genes are also important for the
processing of lactose by the cell. This operon has been extensively studied, and acts as a
typical model of the operon structure, including up- and down-regulation of its expression.
Lactase
Lactase in humans is encoded by the LCT gene found on chromosome 2, and the gene is
expressed exclusively by mammalian small intestine enterocytes (absorptive cells that line
the inner surface of the intestines). Human infants typically express high levels of lactase,
but transcription of the gene generally declines with age, which results in a condition in
adults called lactose intolerance. Some populations exhibit a continued lactase expression
into adulthood (called lactase persistence or lactose tolerance), due to a mutation in the LCT
gene promoter. This mutation is believed to have occurred between 5 and 10 thousand
years ago, and involves a single base change in the promoter region which prevents binding
of the transcription factors that normally down-regulate expression in adulthood.
Because of its similarity to β-gal, lactase is also able to convert ONPG to galactose and
ortho-nitrophenolate, which can be used to measure enzyme activity in vitro. In this case,
we will be dealing with the lactase enzyme in a cell free assay. This will allow us to measure
the enzyme kinetics without the added complication of gene regulation/gene expression. An
easily obtained source of lactase is Lactaid pills, which are sold over the counter at
pharmacies and grocery stores to treat lactose intolerance.
81
Figure 2. Lactase enzyme breaking down ONPG to galactose and ortho-nitrophenolate, which appears
yellow.
All cells carry out a variety of chemical reactions in order to survive and function. While
many of these reactions would eventually occur without enzymatic help, enzymes act to
make reactions happen more rapidly, and also act to order reactions into coordinated
building or degrading processes. In this way, individual reactions can be linked into specific
pathways. The specificity of enzymatic reactions is determined by the structure of the
enzyme and its targets, known as substrates. Enzymes are proteins, and their ability to bind
to other molecules and assist in chemical reactions is determined by their structure. The
structure of a protein, in turn, is determined by the sequence of amino acids that are linked
together. Once the amino acids have been linked together, the chemistry of the individual
amino acids and their order in the chain will cause the folding of the chain into a three
dimensional structure. As a result of this folding, amino acids that were far apart in the
string may now be neighbors, connected by hydrogen bonds or other intermolecular forces.
Specific amino acids will make up what is known as the active site, where substrates bind
and are converted into products.
82
Figure 3. Sequence showing substrate binding to enzyme to form the enzyme-substrate complex and
then generation of product.1
Changes in the three dimensional configuration of an enzyme (and hence the position of
specific amino acids) will affect the ability of the enzyme to catalyze its specific reaction. If
the entire shape of the enzyme is disrupted (such as by high heat or strong acids,) all
catalytic activity will be lost. However, subtle changes in shape, such as those caused by
gene mutations or those caused by drugs binding to the enzyme, or those caused by
changes in environmental temperature, pH or ionic balance, may either increase or
decrease catalytic activity. It is important to remember that the enzyme itself is not changed
over the course of the reaction, as it is neither a substrate nor a product, and at the end of
the reaction is released to bind a new substrate. Enzymes simply lower the amount of
energy required to get the reaction going (activation energy) by providing a chemical
environment that encourages conversion of substrate to product. As more substrate
molecules are now able to clear the lowered bar of activation energy, the rate of reaction
will increase.
It is possible to measure the rate of a catalyzed reaction, which is the concentration of
product formed over a period of time, either by measuring the concentration of substrate
(how much has been used up) or product (how much has formed). There are a variety of
variables that influence the rate of reaction: amount of substrate, amount of enzyme,
83
amount of product, shape of the enzyme, temperature, and pH, among others.
Making an enzyme stock solution and measuring its activity
To measure an unknown molecule, one first needs to measure a sample of it at all
wavelengths to determine the optimum wavelength.
Ortho-nitrophenolate has an absorbance peak at 420 nm.
84
Next, one needs a
way to relate the absorbance reading from the spectrophotometer to the amount of orthonitrophenolate product in the tube. In order to make that correlation, one would need to
make a series of known dilutions of ortho-nitrophenolate and then measure their
absorbance, plotting the concentration of the ortho-nitrophenolate vs. the absorbance at
420 nm. This is known as a standard curve. This has already been done for you, and the
standard curve is shown below.
The important piece of information to take from this graph is that, for any given absorbance
(y), one can calculate the concentration of ortho-nitrophenolate (in nM) by using the
equation:
Y = 0.0009x
(equation 1)
Pre-lab Questions
1. Where on the enzyme does the substrate bind?
2. What does it mean when one says an enzyme is denatured?
3. What is a catalyst?
4. What is a coenzyme? What is a cofactor? What is the difference(s) between the two?
5. What is meant by optimum temperature? Optimum pH?
6. Do all enzymes in the human system have the same optimum temperature? Optimum pH? Explain
your answer.
Procedure
85
Part 1 – Making and testing the lactase stock solution
⦁
Take one pill of Lactaid and crush it into a powder with a mortar and pestle.
⦁
To the powder, add 5 mL of 0.1 M phosphate buffer (pH 7.0) and dissolve.
⦁
Centrifuge for 5 minutes at 3000 rpm
⦁
Pour just the supernatant into a fresh tube. This is your enzyme stock.
At this point the amount of lactase activity in your stock is unknown, so it would be
good to determine a proper concentration of lactase to use for the rest of your
experiments. Usually a 1:1000 dilution is a good concentration of enzyme to carry out
the experiments. To do this you'll make a series of dilutions of this first stock and test
them.
⦁
Make a set of 4 serial 1:10 dilutions, as shown in the table below. Begin by adding 1
mL of your stock enzyme solution from step 4 to 9 mL of pH 7.0 buffer, as shown in
solution 1. Mix thoroughly. This is your 1:10 dilution.
⦁
Next, add 1 mL of the solution you made in step 5 to 9 mL of pH 7.0 buffer and mix.
This is your 1:100 dilution.
⦁
Follow the table for solutions 3 and 4, each time using the dilution you made in the
previous step. In this way you will have made 4 dilutions (1:10, 1:100, 1:1000, and
1:10,000).
Solution
Amount
Transferred
Source
pH 7.0 buffer
(diluent)
Step dilution
Final
concentration
1
1 mL
Stock
9 mL
1/10
1:10
9 mL
1/10
9 mL
1/10
9 mL
1/10
2
3
4
1 mL
1 mL
1 mL
#1
#2
#3
When measuring the kinetics of an enzyme, we try to work with relatively low
86
1:100
1:1000
1:10,000
concentrations. However, it’s important to choose a concentration that is neither too dilute
(hard to measure activity if there isn't much enzyme) nor too concentrated (the
spectrophotometer has a ceiling limit for absorbance, after which it can't differentiate
similar solutions from one another). To begin, we will test the 1:1000 dilution (solution 3).
⦁
Connect your spectrophotometer to your laptop, and go to “Experiment”, and then
“Calibrate”. Allow the lamp to warm up.
⦁
Add 1 mL of pH 7.0 buffer to a cuvette, and blank the spectrophotometer with it.
⦁
To test the activity of the 1:1000 dilution, make a new reaction tube and add to it
the following, making sure to add the enzyme last, and then immediately starting a
timer:
⦁
⦁
⦁
⦁
⦁
⦁
3.5 mL of pH 7.0 buffer
0.5 mL of 1% ONPG (substrate)
0.5 mL of 1:1000 enzyme stock (solution 3)
Watch the tube carefully, and when a pale yellow color appears in the tube, add 0.5
mL sodium carbonate to stop the reaction and stop your timer. Write down the
total time of your reaction in your notebook.
Add 1 mL from the reaction tube to a cuvette and measure the absorbance at 420
nm. The absorbance should be between 0.1 and 1; if it was < 0.1, you stopped the
reaction too early, and if it was > 1, you let the reaction proceed too long. If it was
not between 0.1 and 1, repeat step 10.
Use equation 1 from the introduction to calculate the nM of product in the tube
from the absorbance, and then divide by the time in seconds to calculate nM of
product formed/time.
If step 11 lasted for anywhere from about 30 seconds to 2 minutes, you can proceed to step
14. If step 11 happened in less than 30 seconds, that means your enzyme solution is too
concentrated. Repeat step 10 with your 1:10,000 dilution. If step 11 lasted more than 2
minutes, that means your enzyme solution is too dilute; repeat step 10 with your 1:100
dilution.
The goal here is to have an enzyme dilution (1:100, 1:1000, or 1:10,000) that reacts in the
30 second to 2 minute range. Once you have that, you can proceed to step 14; you will want
to use that dilution for all steps going forward.
Part 2 – Calculating Vmax and Km for lactase and ONPG
Now that you have a good working concentration of lactase, you will collect data that will
allow you to calculate the two main measurements of enzyme kinetics: Vmax (maximum rate
of reaction for this enzyme and this substrate under these conditions) and Km (the Michaelis
87
constant, a measure of binding affinity).
Vmax is the maximum rate possible (high substrate concentration)
Km is the substrate concentration at ½ Vmax.
In order to determine Vmax, and Km, you will graph Rate of reaction (o-nitrophenolate/time,
as calculated in step 13) versus Substrate (ONPG) concentration.
In order to do this, you will react different concentrations of ONPG with your enzyme
solution. Set up the following reactions:
Reaction
pH 7.0 buffer
Enzyme
solution
ONPG
ONPG
concentration
1
3.5 mL
0.5 mL
0.1 mL
0.2%
2
3.5 mL
0.2 mL
0.4%
3
3.5 mL
0.3 mL
0.6%
4
3.5 mL
0.4 mL
0.8%
5
3.5 mL
0.5 mL
1%
⦁
⦁
⦁
⦁
0.5 mL
0.5 mL
0.5 mL
0.5 mL
To begin, add the solutions in reaction 1 to a test tube. Add the enzyme solution
last! Start a timer.
When the reaction has turned pale yellow, add 0.5 mL sodium carbonate and note
the time of the reaction. Measure the absorbance at 420 nm.
Use equation 1 to calculate the approximate nM of ONP product.
To determine the rate, divide the nM of product made by the amount of time you
88
⦁
⦁
ran the reaction in seconds.
Repeat steps 14-17 for the rest of the reactions.
Run all 5 reactions again in order to get duplicated data (more data is better!) Note
that you can run reactions at the same time (you don’t have to wait for one to stop
in order to start the next one,) but you’ll probably want to stagger your reactions.
For your lab report
⦁
Graph the reaction rate for your 5 reactions vs. the substrate concentration (in %
ONPG). (If you were able to do duplicates of your data, take the average rate at
each concentration of ONPG.) This is known as a Michaelis-Menten plot. Here is a
sample of what your data should look like:
⦁
Convert your data to the Lineweaver-Burk form. Do this by graphing 1/reaction rate
(Vi) vs. 1/product concentration [S]:
89
⦁
⦁
Calculate Vmax and Km for lactase. To do this, create a best-fit line with your
Lineweaver-Burk data and determine the y=mx + b equation. The point at which the
line intersects the Y-axis is 1/Vmax, which can be determined by setting x = 0 in the
equation and solving for Y. Then, by inverting this value, you have calculated Vmax.
The point at which the line intersects the X-axis is -1/ Km, which can be determined
by setting y = 0 in the equation and solving for X. Convert this to Km by inverting the
value (1/x) and reversing its sign (- to +). \
Include both your Km and Vmax values in your lab report, as well as your MichaelisMenten and Lineweaver-Burke plots.
References
⦁
Voet, D.; Voet, J. G. Biochemistry, 3rd Ed.; John Wiley & Sons, Inc., 2004 Hoboken, NJ.
⦁
Lipson, K. L.; Fonseca, S. G.; Ishigaki, S.; Nguyen, L. X.; Foss, E.; Bortell, R. Rossini, A. A.;
Urano, F. Regulation of insulin biosynthesis in pancreatic beta cells by an endoplasmic
reticulum-resident protein kinase IRE1; Cell. Metab. 2006, 4, 245–254.
⦁
Aldridge, W. N.; Reiner, E. Enzyme Inhibitors as Substrates, Frontiers of Biology, Volume 26;
Neuberger, A. (Ed.); Tatum, E. L. (Ed.); North-Holland Pub. Co., 1972 London.
⦁
Copeland, R. A. Evaluation of Enzyme Inhibitors in Drug Discovery: A Guide for Medicinal
Chemists and Pharmacologists, Methods of Biochemical analysis, Volume 46; John Wiley &
Sons, Inc. 2005, Hoboken, NJ.
⦁
Hill, J.W.; McCreary, T.W.; Kolb, D.K. Chemistry for Changing Times, 12th Ed.; Pearson, 2010
San Francisco.
See also
Michael Blaber Biochemistry website at Florida State University College of Medicine:
http://www.mikeblaber.org/oldwine/BCH4053/Lecture25/Lecture25.htm
90
Biochemistry Powerpoint lecture at Ohio State:
http://class.fst.ohio-state.edu/fst605/605p/Enzymekinetics.pdf
91
Enzyme Kinetics - Inhibition
ISC Lab I – Kinetics Module
Purpose: To study the kinetics of an enzymatic reaction under the influence of a potential
inhibitor.
Apparatus: A spectrophotometer
Chemicals:
⦁
Lactase (in the form of Lactaid)
⦁
Phosphate buffer, pH 7.0
⦁
ONPG (ortho-Nitrophenyl-β-galactoside)
⦁
Sodium carbonate
⦁
Sugars (lactose, galactose, glucose)
Skills: Dilutions, experimental design, data analysis and interpretation.
From U Mass Amherst, http://bcrc.bio.umass.edu/intro/content/enzyme-kinetics-labprotocol
Inhibition
There is another important factor in the control of enzyme activity, above and beyond those
factors that were discussed in the previous lab, and that is the presence/absence of
inhibitors. Inhibitors may be endogenous - part of the natural regulation of enzyme activity
within the cell, or they may be exogenous - drugs that are designed to regulate abnormal
enzyme activity. Inhibitors can either be reversible (meaning that, when the inhibitor is
removed from the system, the enzyme will return to normal function) and irreversible
(meaning that the inhibitor alters the enzyme in such a way as to make it non-functional
even when the inhibitor is removed.) This lab will focus on reversible inhibition, which can
be further sub-divided into three broad categories – competitive, non-competitive, and
uncompetitive inhibition.
In competitive inhibition, the inhibitor resembles the chemical structure of the true
92
substrate, and it competes with the substrate for binding to the active site. The inhibitor is
only able to bind the free enzyme, and cannot bind the substrate-enzyme complex. Since
whether the chemical reaction occurs or not is based solely on whether substrate or
inhibitor binds (because they compete, Km increases), adding substrate in large amounts
until there is more substrate than inhibitor will eventually return the rate of reaction to
normal (Vmax is unchanged).
Figure 4. General
scenario for when a competitive inhibitor binds to the active site of an enzyme. The active site of the enzyme is
blocked and therefore the enzyme cannot convert substrates to products.
In non-competitive inhibition, the inhibitor binds to the enzyme at a site different than the
active site. Binding to this site by the inhibitor changes the overall shape of the enzyme,
slowing the processing of substrate into product once it is bound to the enzyme. This form
of inhibition does not prevent the binding of the substrate (Km is unchanged) but the
reaction rate at high levels of substrate is decreased (Vmax is lower).
93
Figure 5. General scheme for an enzyme capable of non-competitive inhibition. An inhibitor can bind to the allosteric site
and change the conformation of the binding site.
In uncompetitive inhibition, the inhibitor will only bind to the enzyme when the enzyme has already
bound to substrate (the enzyme-substrate, or ES, complex). Binding of the inhibitor to the ES
complex prevents the substrate from being converted into product, which causes a decrease of both
Vmax and Km
94
Figure 6. General scheme for an enzyme capable of uncompetitive inhibition. Inhibitor binds to the ES complex,
stopping the formation of product.
In today’s experiment, you will test one sugar to determine whether it is an inhibitor of
lactase, and if so what type of inhibition it engages in. Each group will test either lactose,
galactose, or sucrose.
Procedure
The procedure for today’s lab will be similar to the procedure for the previous lab. To begin
with, you will need to make an enzyme solution, and then you will need to repeat the
previous experiment by measuring the enzyme activity with different concentrations of
ONPG substrate. You will then do the same experiment again, but add inhibitor to the
mixture.
⦁
Take one pill of Lactaid and crush it into a powder with a mortar and pestle.
⦁
To the powder, add 5 mL of 0.1 M phosphate buffer (pH 7.0) and dissolve.
⦁
Centrifuge for 5 minutes at 3000 rpm
⦁
Pour just the supernatant into a fresh tube. This is your enzyme stock.
⦁
Make a set of 3 or 4 serial 1:10 dilutions, as shown in the table below. Begin by
adding 1 mL of your stock enzyme solution from step 4 to 9 mL of pH 7.0 buffer, as
shown in solution 1. Mix thoroughly. This is your 1:10 dilution.
⦁
Next, add 1 mL of the solution you made in step 5 to 9 mL of pH 7.0 buffer and mix.
95
This is your 1:100 dilution.
⦁
Follow the table for solution 3. If you used your 1:1000 dilution in the previous lab,
you can stop here. If you used your 1:10,000 dilution in the previous lab, make the
4th serial dilution.
Solution
Amount
Transferred
Source
pH 7.0 buffer
(diluent)
Step dilution
Final
concentration
1
1 mL
Stock
9 mL
1/10
1:10
9 mL
1/10
9 mL
1/10
9 mL
1/10
1 mL
2
1 mL
3
4 (optional)
1 mL
#1
#2
#3
1:100
1:1000
1:10,000
⦁
Connect your spectrophotometer to your laptop, and go to “Experiment”, and then
“Calibrate”. Allow the lamp to warm up.
⦁
Add 1 mL of pH 7.0 buffer to a cuvette, and blank the spectrophotometer with it.
⦁
You can choose which enzyme dilution to work with. Best practice is to use the
same dilution you used in the previous lab. At this point, you will repeat yesterday’s
experiment. It is important for time management purposes that you run multiple
reactions at the same time, rather than wait for one reaction to complete before
starting the next.
Reaction
pH 7.0 buffer
Enzyme
solution
96
ONPG
ONPG
concentration
1
3.5 mL
2
3.5 mL
3
3.5 mL
4
3.5 mL
5
3.5 mL
⦁
0.5 mL
0.5 mL
0.5 mL
0.5 mL
0.5 mL
0.1 mL
0.2%
0.2 mL
0.4%
0.3 mL
0.6%
0.4 mL
0.8%
0.5 mL
1%
Next, you will set up a new set of reactions, and include your inhibitor. We will call
the concentration of inhibitor in this set of reactions “I”
Reaction
pH 7.0 buffer
Enzyme
solution
Inhibitor
solution
ONPG
ONPG
concentration
1
3.0 mL
0.5 mL
0.5 mL
0.1 mL
0.2%
2
3.0 mL
0.5 mL
0.2 mL
0.4%
3
3.0 mL
0.5 mL
0.3 mL
0.6%
4
3.0 mL
0.5 mL
0.4 mL
0.8%
5
3.0 mL
0.5 mL
0.5 mL
1%
⦁
0.5 mL
0.5 mL
0.5 mL
0.5 mL
Set up a third set of reactions, with twice as much inhibitor. We will call this
concentration “2I”
Reaction
pH 7.0 buffer
Enzyme
solution
Inhibitor
solution
ONPG
ONPG
concentration
1
2.5 mL
0.5 mL
1.0 mL
0.1 mL
0.2%
97