BIBC
103
Biochemical Techniques
LABORATORY MANUAL
2022–2023
Aaron Coleman
School of Biological Sciences
University of California, San Diego
Copyright © 2022 by Aaron Coleman, Meredith Gould, and Jose Luis Stephano
Copyright © 2022 by Hayden-McNeil, LLC on illustrations provided
Photos provided by Hayden-McNeil, LLC are owned or used under license
Cover Image (background): Puszaya/Shutterstock.com
Cover image by Mohamud Aden and Steven Qammouh from fall 2021 BIBC 103 section A01. The
image shows a monomer the SARS-CoV-2 virus spike protein, which facilitates the pathogen’s
entry into a host organism. The spike protein is a homotrimer, with each monomer containing an
S1 subunit (red and green top portion), which interacts with Angiotensin Converting Enzyme 2
(ACE2) on the host cell surface, and an S2 subunit (lower portion), which promotes viral fusion
with the host cell membrane. The S2 subunit contains a Fusion Peptide (purple) and two heptad
repeat regions which form alpha helices (cyan). The Fusion Peptide is a conserved region of 18
amino acids which helps anchor the virus onto the plasma membrane. Key amino acid residues
in the receptor binding domain of the S1 subunit (shown in red) interact with ACE2 to make
binding more favorable. Inset: Phe486 (yellow) of the receptor binding domain interacts with
Tyr83 (yellow) of ACE2 to form an aromatic ring stacking interaction (Huang et al., 2020. Acta
Pharmacol. Sin. doi: 10.1038/s41401-020-0485-4). This image was produced from Protein Data
Bank Structure 6VYB (Walls et al., 2020. Cell. doi: 10.1016/j.cell.2020.02.058) using PyMOL.
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Table of Contents
Table of Contents
Safety and General Lab Protocols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v
Formatting Your Lab Notebook . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix
Making Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
The Spectrophotometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Amino Acid Structures and R Group pKa Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
LAB 1
Working Efficiently in the Biochemistry Laboratory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
LAB 2
Electrophoresis and SDS-PAGE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
Project 1: Purification and Analysis of Lactate Dehydrogenase
LAB 3
Initial Purification of Lactate Dehydrogenase: Centrifugation and
Ammonium Sulfate Precipitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
LAB 4
Affinity Chromatography Purification of LDH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
LAB 5
Size Exclusion Chromatography Purification of LDH . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
LAB 6
Determination of LDH Activity Units and Specific Activity . . . . . . . . . . . . . . . . . . . . . . 71
LAB 7
SDS-PAGE of LDH Purification Intermediates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
LAB 8
Determination of LDH Isozymes by Electrophoresis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
Project 2A: Role of the Cytoplasmic Calcium Influx in
Sea Urchin Fertilization
LAB 9A Sea Urchin Fertilization: Preparation of Samples and
Examination of Egg Activation and Cell Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
LAB 10 SDS-PAGE and Western Blotting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
LAB 11 Immunodetection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
LAB 12 ELISA Determination of IP1 Levels. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
iii
Table of Contents
Project 2B: Examining Signal Transduction from the Fibroblast Growth Factor
Receptor in Cultures of NIH 3T3 Fibroblasts
LAB 9B FGF Signaling: Experimental Design and Preparation of Cell Lysates . . . . . . . . . . . . 107
LAB 10 SDS-PAGE and Western Blotting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
LAB 11 Immunodetection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
LAB 12 ELISA Determination of IP1 Levels. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
Project 3: Expression and Purification of Fluorescent Proteins
LAB 13 Isolation of Fluorescent Protein Plasmid DNA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
LAB 14 Restriction Digest and Agarose Gel Electrophoresis. . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
LAB 15 Preparation of Competent Cells and Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . 189
LAB 16 Purification and Analysis of Fluorescent Proteins. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
LAB 17 SDS-PAGE of Fluorescent Proteins. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
LAB 18 Protein Crystallization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
LAB 19 Computational Tools for Biochemistry: Sequence Analysis, Bioinformatics, and
Examination of Protein Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
iv
Safety and General Lab Protocols
Safety and General Lab Protocols
Safety and General Lab Protocols
General
•
•
•
Labs are pre-set to 68°F in order to conserve energy. The outside doors must
be kept closed for proper temperature control. Please do not prop them open!
Due to the hazardous materials in these labs, safe laboratory conduct is
expected at all times. Improper conduct will result in a loss of points and/or
expulsion from the class.
To protect the hearing of all lab users, playing any song by the bands Journey
or Coldplay is strictly prohibited.
Fire
•
•
•
•
Note locations of fire extinguishers in labs and on the verandas. These should
only be used on small fires. Note also the location of alarm pull boxes (outside, on the west, north, and south sides of York). See the “Emergency Evacuation Plan” on or near the lab door for more information.
Each lab, with doors shut, will contain fire for one hour. (The Fire Dept. is
only 5 minutes away.)
If it is safely possible, grab your personal items on the way out—there is no
way to predict how long the building will be off-limits, and emergency personnel will not allow anyone back in for any reason.
Assemble at the Revelle Plaza fountain for roll call. Do not leave until your
instructor has officially ended the lab period.
Earthquake
•
•
•
•
The Rose Canyon Fault runs by campus, so there could be a seismic event at
any time.
Take cover under lab benches during shaking, then evacuate the lab quickly.
If it is safely possible, grab your personal items on the way out.
Assemble at the Revelle Plaza fountain for roll call. Do not leave until your
instructor has officially ended the lab period.
v
Safety and General Lab Protocols
PPE
•
•
•
•
Wear proper lab attire. Open-toed shoes, shorts, etc. are unacceptable. Lab
coats and safety glasses are mandatory when anyone is working in the lab.
You must provide these yourself. Confine loose hair and clothing. Students
wearing improper attire will be asked to leave the lab.
Gloves should be worn at all times in the lab when handling reagents and
equipment.
As stated above, safety glasses are required for lab. Be aware that wearing
contact lenses in a lab may be hazardous, as they trap chemicals and are hard
to remove in a splash accident. Wear contacts only if medically necessary.
UV protection will be provided when needed (if your glasses aren’t UV-safe)
and MUST be used. Students who refuse to wear glasses will be asked to leave
the lab.
All PPE must be removed before exiting the lab for any reason.
Chemicals
•
•
•
•
•
Familiarize yourself with the chemicals you will handle. Safety Data Sheets
(SDS) are available on-line: https://blink.ucsd.edu/safety/resources/SDS/
sources/index.html
Emergency eye wash/shower stations located in the labs or just outside on the
veranda. Make note of the nearest EW/ES to your work station. Sinks may
also be used to flush chemicals from your hands or face.
Report ALL accidents, however minor, to your IA. They will provide help and/
or first aid and fill out an accident report.
First Aid kits are located at the safety area in each lab.
Clean up all spills immediately. Non-hazardous liquids can go in the sink.
Non-hazardous solids must go in the “lab trash” container—NOT the regular trash. Consult your IA before doing anything with a spill of hazardous
chemicals. Chemical spill kits are in a large white bucket at the front of the
lab for handling hazardous spills.
Chemical Waste Disposal
•
•
Small amounts of most chemicals can be washed down the sink (with plenty
of water). Always rinse glassware before placing in washtubs or back in your
locker.
Hazardous chemicals will be collected in various waste streams for eventual
pick up by EH&S. Your IA and UGL staff will inform you which chemicals are
to be handled this way.
Sharps Boxes
•
•
•
vi
All broken glassware (including used Pasteur pipettes) and other sharps are to
be placed in green “sharps” boxes. Dustpans and brushes are available in each
lab. It is not the job of your IA or staff to clean up your broken glassware! If
the broken glass is too big to fit in the sharps box, alert your IA or UGL staff.
Micropipette tips must be placed in the green puncture-resistant containers
on the bench.
Do not place any sharps in the regular trash cans—custodians are at risk of
getting poked with them!
Safety and General Lab Protocols
Equipment Breakdowns
•
•
•
•
Report all equipment malfunctions to your IA or UGL staff immediately. If
they don’t know about it, it won’t get fixed!
Each group will have a locker to store equipment, but remember that other
sections may be sharing these items with you. Leave items as you would like
to find them. Each group should check their equipment (esp. pipetters!) at the
start of lab to avoid being charged for breakage or loss. Once lab has begun,
your group assumes all responsibility for equipment.
Micropipettes, if fouled with a reagent, must be cleaned/recalibrated by staff
with special tools. You risk contaminating your samples if you continue to use
a fouled pipette.
DO NOT TAKE MICROPIPETTES APART FOR ANY REASON!! If a pipette
is working improperly, bring it to your IA, who can loan you their pipette
while yours is being fixed.
Food and Drink
•
Food and drink (including water) are not to be consumed in the lab at any
time, but may be consumed out on the veranda. Wrappers and bottles should
be thrown in trash cans located outside labs. EH&S is adamant about this
rule, so we have to be strict about it as well. Food items must be kept out of
sight in your cubby.
•
•
Remove all PPE and wash hands before leaving lab for any reason.
Remove all tape from glassware and rinse with water before placing in washtub or back in your locker. If you fail to do this, your IA will have to do it.
Leave empty reagent bottles on front bench or in refrigerator—not in washtubs. If you need more reagent, consult with your IA or UGL staff.
Use good lab practices at all times, and remember that other people will use
the lab after you.
Cleanup
•
•
vii
Safety and General Lab Protocols
viii
Formatting Your Lab Notebook
Formatting Your Lab Notebook
A fair percentage of the grade assigned in this course will be based on the quality of
your laboratory procedure, including your notebook. Your notebook should be a log
of your experimental work. It should be complete and thorough to the extent that
your peers should be able to understand your experimental findings.
Allowable Types of Notebooks and Basic Formatting
1. Refer to your instructor’s syllabus for what types of notebooks are and are not
allowable. You may not need to have a notebook with carbon-copy pages.
2. Refer to your instructor’s syllabus for information on the table of contents and
pagination.
3. Use pen only!
4. Do not rip out pages! If mistakes are made, they should be struck through with
a single line so that the error is still legible. (What you consider wrong at one
time might turn out to be right later.)
What to Enter
1. Each page should have the experiment date, the project title (if that experiment
is part of a larger project), and the title for that day’s experiment. Most of the
individual labs you will do in this class will be part of larger projects that continue
for a number of weeks, so it is important to identify both the project and the
experiment. For example, “LDH Purification” would the project title and “Size
Exclusion Chromatography” would be the experiment title. Be sure to include
these on every page where you enter material into your notebook, not just the
first page for the experiment. If you are working on two different experiments
on the same day, be sure to space them out in your notebook so there are some
blank pages between them.
2. It is not necessary to copy the details of an experimental procedure, provided
you reference the source of the procedure (e.g., the pages in the lab manual).
However, any variations from the original procedure must be documented
in the notebook. (Making a flowchart of the procedures can be helpful, if the
experimental protocol is difficult to follow in the manual.)
ix
Formatting Your Lab Notebook
3. You must enter all data you collect directly into your notebook in a timely
manner, that is, while you are conducting the actual work. When possible, data
should be recorded in tables. You will collect some spectrophotometry data that
will be printed by the instrument. These printer tapes are your raw data, and
the tapes or copies of the tapes should be neatly taped into your notebook and
labeled. Be sure these are permanently fixed into your notebook—tucked into
the pages does not count!
4. Remember to write down any relevant observations you make as well, especially
if something seems unexpected or unusual.
5. Include any calculations for dilutions, etc., that you do during the experiments.
That way, you can backtrack and troubleshoot if something goes wrong.
6. You will be required to include some data analysis in your notebook that may
include calculations and statistical analysis, sketches of graphs and/or printouts
of graphs prepared on a computer, and printouts of electrophoresis gel images.
Any printouts should be permanently fixed into your lab notebook.
7.
Include a brief summary of any results you obtained. Typically this only requires
one or two sentences. You should also briefly describe anything that went wrong
with the experiment.
Always refer to your instructor’s specific instructions for what to include in your
lab notebook.
x
Making Solutions
Making Solutions
I. Molar Solutions
Molecular Weight
The smallest atom (hydrogen) is assigned a mass of one. One H atom weighs
1.66 × 10–24 g. One gram of H contains 6.0221 × 1023 H atoms. This number of atoms
(“Avogadro’s number”) is called a mole. The mass (molecular weight) of NaCl is 58.44
times the mass of the H atom; therefore, 58.44 grams of NaCl contains 6.0221 × 1023
NaCl molecules, or one mole of NaCl. A one molar solution is one mole dissolved in
one liter. Therefore a 1 M NaCl solution contains 58.44 g/l of NaCl.
How would you make 300 ml of a 0.25 M KCl solution? The MW of KCl is 74.56.
Multiply the molecular weight × the molarity × the volume:
(74.56 grams/mole) × (0.25 moles/liter) × 0.3 liters = 5.592 grams
(moles and liters cancel out)
How would you make 800 ml of a 10 mM solution of KCl?
74.56 × 0.01 M × 0.8 l = 0.596 grams (596 mg)
Practical Consideration
Suppose you want to make 1 liter of 1 M NaCl. If you add 58.44 g of NaCl to one liter
of water, the final volume will be > 1 l. (Why?) Therefore you should first dissolve
the salt in less than a liter, then adjust the final volume to 1 l.
II. Percent Solutions
Solids
Unless stated otherwise, % means grams/100 ml. A 2% solution of NaCl would have
2 grams in 100 ml. A 0.01% solution would have 0.01 grams (10 mg) in 100 ml.
1
Making Solutions
How would you make 25 ml of a 0.02% solution of bromophenol blue?
0.02 g
xg
=
100 ml
25 ml
100x = (0.02)(25)
x = 0.005 g (5 mg)
Liquids
Substitute ml for grams. 95% ethanol has 95 ml ethanol/100 ml solution.
Sometimes % solutions have the notation w/v (weight/volume) or v/v (volume/volume).
III. Dilutions
Often it is convenient to make concentrated stock solutions, then dilute them to get
the desired final concentrations. Dilutions are calculated with the formula:
V1 × C1 = V2 × C2
a.
where V1 = volume of solution 1,
C1 = concentration of solution 1, etc.
Suppose you have a stock solution of 4.8 M NaCl and you want to make 250 ml
of a 0.1 M solution. How much NaCl stock and how much water should you mix?
(V1)(4.8 M) = (250 ml)(0.1 M)
V1 = 5.2 ml NaCl stock + 244.8 ml water
b. If you have a stock solution of 10 mg/ml bromophenol blue, how do you make
50 ml of a 0.02% solution?
First, make the concentration units uniform—either mg/ml or %—then do the
calculation.
10 mg/ml = 0.01 g/ml = 1 g/100 ml = 1%
V1 (1%) = (50 ml)(0.02%) V1 = 1 ml
or
0.02% = 0.02 g/100 ml = 20 mg/100 ml = 0.2 mg/ml
V1 (10 mg/ml) = 50 ml (0.2 mg/ml)
V1 = 1 ml
1 ml stock + 49 ml water
2
Making Solutions
c.
Suppose you have 10 ml of a protein solution in 150 mM NaCl and you want to
increase the NaCl concentration to 0.5 M using a 3 M stock solution of NaCl.
How much 3 M NaCl should you add?
(10 ml)(150 mM) + (x ml)(3000 mM) = (10 + x ml)(500 mM)
1500 + 3000x = 5000 + 500x
2500x = 3500
x = 1.4 ml
d. Sometimes the stock solution just says it is “10×”. This means it is 10× concentrated and it is understood that you want a 1× solution. If you have a 10× buffer
solution, how much do you use to make 80 μl of a solution containing that buffer?
V1 (10×) = 80 μl (1×)
V1 = 8 μl
IV. Serial Dilutions
These are a series of dilutions, each made by diluting the previous one. For example,
here is a serial 1:10 dilution of a 0.01 M stock solution of a hormone (H):
Final H
concentration
Tube 1
Tube 2
Tube 3
Tube 4
100 µl 0.01 M H
900 µl buffer
100 µl tube 1
900 µl buffer
100 µl tube 2
900 µl buffer
100 µl tube 3
900 µl buffer
10 –3 M
10 –4 M
10 –5 M
10 –6 M
Reasons for performing a serial dilution are:
1. The dilution is more accurate when you pipette larger volumes. For example, to
make 1 ml of 10 –5 M H from the 10 –2 M stock, you would add only 1 μl to 1 ml
buffer.
(V1 × 10 –2 M = 1 ml × 10 –5 M; V1 = 1 μl)
2. You want to make a large dilution without using large volumes because it is
inconvenient or the substances are expensive. For example, you could make the
above solutions in the following manner, but this would waste a lot of buffer.
100 μl of 10 –2 M stock + 9.9 ml buffer = 10 –4 M
100 μl of 10 –2 M stock + 99.9 ml buffer = 10 –5 M
100 μl of 10 –2 M stock + 999.9 ml buffer = 10 –6 M
3
Making Solutions
3. You don’t know the concentration that will be effective in the experiment you
want to do, so you make a range of serial dilutions and test them all to see which
one(s) will give results.
In the example, the final volume in tubes 1–3 is 900 μl since you took 100 μl out to
make the next dilution. Suppose you want to have at least 1 ml of each solution left
over after you take 1/10th of the volume out. You could increase the volumes (e.g., add
150 μl stock to 1.35 ml buffer) or you could be a Super Calculator and end up with
exactly the desired volume in each tube except the last one. In this case 1 ml will be
9/10ths of the initial stock + buffer volume. If 1 ml = 9/10ths, 1/10th = 1 ml ÷ 9 = 0.111.
So you would add 111 μl of 0.01 M H stock to 1 ml buffer to make the 10 –3 M stock,
then add 111 μl of this to 1 ml buffer to make the 10 –4 M stock, etc. This leaves 1 ml
in each tube except tube 4, which will have 1.111 ml.
This method works for any dilution. For example, suppose you want to make serial
1:5 dilutions of the stock solution and have 1 ml in each tube. 1 ml will then be 4/5ths
of the initial stock + buffer volume. If 1 ml = 4/5, 1/5 = 0.25 ml, so add 0.25 ml stock
to 1 ml buffer, then take out 0.25 ml for the next dilution, etc. The general formula
would be:
(final volume desired) ÷ (dilution factor – 1) = amount of stock that should be added.
Suppose you want 3 ml of serial 1:4 dilutions. 3 ml ÷ 3 = 1 ml. Add 1 ml stock to
3 ml buffer, mix, take out 1 ml for the next dilution and so on.
4
The Spectrophotometer
The Spectrophotometer
Spectrophotometry is one of the most common methods used in biology research.
Spectrophotometry can be used to identify substances or to determine the amount
of a substance by measuring how much light it absorbs. In order to understand spectrophotometry, you should first know something about electromagnetic radiation.
Background on Electromagnetic Radiation
Light is a form of electromagnetic radiation, which consists of discrete packets of
energy called photons. Light is also treated as a wave phenomenon characterized by
wavelength and frequency. Wavelength is defined as the distance between adjacent
peaks (or troughs), and may be designated in meters, centimeters, or nanometers (10–9
meters). The energy associated with a specific form of electromagnetic radiation is
inversely proportional to its wavelength; in other words, the longer the wavelength,
the less energy.
Wavelength
Amplitude
Direction
of travel
©Hayden-McNeil, LLC
As you can see in Figure 0.1, visible and UV light are just a small part of the total
electromagnetic spectrum (we define visible light as the range of wavelengths within
the spectrum that the eye can respond to). Although sunlight or white light appears
uniform, it is actually composed of a broad range of radiation wavelengths in the
ultraviolet (UV), visible, and infrared (IR) portions of the electromagnetic spectrum.
When white light strikes a colored substance, a subset of the mixed wavelengths is
absorbed. The wavelengths that are not absorbed are what determine the apparent
color of the object.
5
The Spectrophotometer
FIGURE 0.1 Electromagnetic spectrum.
The visible spectrum
−4
10
1 -unit
−2
10
2
4
6
10
10
10
10
???
Radio
8
10
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Red
Radar
Spark discharge
10
Heat waves
1
Yellow Orange
Infrared
Ultraviolet
−6
10
Green
X-rays
−8
10
Blue
Gamma
Indigo
Cosmic
Violet
12
10
Å nm
μ
mm
m
km
1 angstrom
1 micron
1 millimeter
1 meter
1 kilometer
14
16
10
10
Wavelength (nanometers)
FIGURE 0.2 Visible spectrum.
Red
700
600
740
©Hayden-McNeil, LLC
625
Orange
590
Yellow
565
Green
Cyan
500
400
500 520
Blue
435
Violet
380
Wavelength (nanometers)
Matter can capture electromagnetic radiation and convert the energy of a photon
to internal energy. This is the process of absorption. In absorption, the energy of
the photon is transferred to a valence electron in the absorbing molecule. When
this occurs, the electron in the molecule is raised to an “excited state,” in which it
occupies a higher energy orbital. The ground-state orbital and excited-state orbital
represent different energy levels for that molecule. Since the difference between these
two energy levels is a discrete amount, or quantum, of energy, only photons with a
matching amount of energy can be absorbed by that molecule. The amount of energy
in the photon is determined by its wavelength on the electromagnetic spectrum. In
other words, specific compounds absorb specific wavelengths of light.
6
The Spectrophotometer
The Spectrophotometer
The spectrophotometer operates by passing a beam of light through a sample and
measuring the intensity (number of photons) of light reaching a detector.
As previously mentioned, the extent to which a sample absorbs light depends strongly
upon the wavelength of light. For this reason, spectrophotometry is performed using
monochromatic light, which is light in which all photons have the same wavelength.
The components of the spectrophotometer are:
1. The light source. Sometimes, there are two different lamps present: a tungsten
incandescent lamp for visible wavelengths and a hydrogen or deuterium lamp
for UV wavelengths. However, in newer machines, a xenon lamp is usually used
since it can provide both UV and visible wavelengths of light. Light from the
lamps is directed to the dispersion device by a moveable reflecting mirror.
2. The dispersion device (also called the monochromator) separates the light into
its component wavelengths, and can be rotated to direct the light of the desired
wavelength to the exit slit (3). When you move the wavelength selector of the
spectrophotometer you are rotating the dispersion device. The dispersion device
is either a prism or a diffraction grating.
3. The exit slit helps to cut out stray light of undesired wavelengths.
4. The sample chamber where the cuvette is inserted. The path length through the
sample chamber is the distance through the solution through which the light
must pass. All standard spectrophotometers have a path length of 1 cm.
5. The detector. This is usually a light-sensitive photomultiplier tube (PMT) that
emits electrons when light hits it.
6. The “electronics” convert the electronic signal to the digital or analogue display
and you read absorbance or transmittance.
FIGURE 0.3 Spectrophotometer diagram.
Light source
Lens
Entrance slit
Lens
Dispersion
device
Photodetector
Green
Blue
Digital readout
Red
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Filter
Sample
Exit slit
Lens
7
The Spectrophotometer
Transmittance and Absorbance
In using a spectrophotometer, the intensity of light (I0 ) passing through a blank is
first measured. The blank is a solution that is identical to the sample solution except
that the blank does not contain the solute that absorbs light. Then the intensity of
light (I) passing through the experimental sample solution is measured.
The data is used to calculate two quantities: the transmittance (T) and the absorbance (A).
Transmittance is
T = I / I0
and percent transmittance is
%T = (I / I0 ) × 100
Thus, the percent transmittance is the percentage of incident light that is transmitted
through the sample (the % of incident light that is not absorbed).
Absorbance is calculated from the transmittance as follows
A = –log10 T
so that the greater the absorbance, the less the transmittance and vice versa. For
example, if 100% of the incident light is transmitted, I / I0 = 1 and A = 0. If 50% of
the light is transmitted through a sample, I / I0 = 0.5 and then A = 0.3.
We typically use absorbance rather than transmittance because absorbance is a linear
function of concentration of the solute whereas transmittance is not.
60
A
%T
80
40
20
0
0
0.2 0.4 0.6 0.8 1.0
Path length (cm)
8
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0
©Hayden-McNeil, LLC
100
0
0.2 0.4 0.6 0.8 1.0
Path length (cm)
The Spectrophotometer
Absorption Spectrum
An absorption spectrum is the absorption of light by a compound as a function of
wavelength.
The spectrum of an atom or molecule depends on its molecular structure, making
absorption spectra useful for identifying compounds.
Each compound thus has a wavelength at which it absorbs maximally and a characteristic absorption spectrum.
FIGURE 0.4 The absorption spectrum for a compound.
300
©Hayden-McNeil, LLC
Absorbance
λ max
400
500
600
Wavelength (nm) or λ
The compound in the absorption spectrum above has an absorption peak (or λmax)
between 400 and 500 nm, which are the violet and blue wavelengths of light. Therefore, this compound absorbs violet and blue light and would appear the color of the
wavelengths of light that it does not absorb. A compound may have more than one
absorption peak for UV and visible light. However, there is typically a single, mostprominent peak in any region of the spectrum that can be used to identify a λmax.
The absorbance at that wavelength can be used to measure the concentration of the
compound in solution.
9
The Spectrophotometer
Beer’s Law
Absorbance varies linearly with both the cell path length and the analyte concentration. These two relationships can be combined to yield a general equation called
Beer’s law.
A=εlc
Where l is the light path in cm and c is the concentration of the analyte in M. The
quantity ε is the extinction coefficient or molar absorptivity. This value varies with
the wavelength of light.
ε is defined as the absorbance of a 1 M solution of the compound at a given wavelength through a 1 cm path length. The units of ε are M–1cm–1.
Compounds with a high molar absorbtivity are very effective at absorbing light (of
the appropriate wavelength) and, hence, low concentrations of a compound with a
high molar absorptivity can be easily detected.
If ε is known for a compound, the concentration of that substance can be determined
from its absorbance. The path length in a standard spectrophotometer is 1 cm and
therefore cancels out.
For example:
ε280 M–1cm–1 = 2500 means that a 1 molar solution of this molecule will have an
absorbance at 280 nm of 2500 in a 1 cm cuvette. If an unknown solution of the
compound was analyzed in a spectrophotometer and the absorbance at 280 nm
was found to be 0.38, simply divide this value by ε to determine the concentration.
c = A/ε
c = 0.38/2500 = 0.000152 M
10
The Spectrophotometer
Light Scattering
Light can also interact with matter in a way that does not transfer energy to the molecules through a process called light scattering. In this case, the light is refracted or
reflected by the matter it interacts with. Although absorption does not occur, light
scattering may produce absorbance that can be measured by a spectrophotometer.
Absorbance from light scattering by particles can introduce errors when one wants
to measure the absorbance of molecules in solution. However, it can be useful for
measuring cell densities. For example, light scattering is commonly used to monitor
the growth of bacterial cultures. The greater the concentration of bacteria, the more
light is scattered and fails to reach the detector. This is registered as an increase
in absorbance. Light scattering is also used to measure densities of sperm, yeasts,
microalgae, etc. Different types of cells scatter light differently so it is first necessary to make a calibration curve with known concentrations determined by actually
counting cells (e.g., with a hemocytometer). Once the calibration curve is made, it is
quick and easy to monitor cell densities with the spectrophotometer. The wavelength
is not critical; however, if the solution is colored (e.g., bacterial culture medium),
it is better to choose a wavelength where the solution has minimum absorbance.
11
The Spectrophotometer
12
Amino Acid Structures and R Group pKa Values
Amino Acid Structures and R Group pKa
Values
The charge state of the structures is shown for pH 7. The single letter codes for each
amino acid is given in parentheses following the name.
Nonpolar, aliphatic R groups
COO–
H3N+ C H
Polar, uncharged R groups
COO–
COO–
CH3
H
Alanine (A)
Glycine (G)
COO–
H3N+ C H
CH2
CH
CH3 CH3
COO–
+
H3N C H
CH2
CH
CH3 CH3
Valine (V)
H2 N
Isoleucine (I)
CH2
CH2
CH2
+NH
3
Lysine (K)
pKa = 10.5
CH2
Cysteine (C)
COO–
+
H3N C H
CH2
C
CH2
C
H2N
O
O
Glutamine (Q)
Negatively charged R groups
COO–
COO–
H3N+ C H
CH2
CH2
NH
C +NH2
NH2
C NH
Arginine (R)
pKa = 12.5
SH
Asparagine (N)
Positively charged R groups
COO–
+
H3N C H
CH2
CH2
CH2
CH3
COO–
+
H3N C H
CH2
H3N+ C H
H3N+ C H
H C CH3
CH2
Methionine (M)
COO–
COO–
H3N+ C H
CH3
Threonine (T)
Proline (P)
COO–
S
CH3
COO–
H
C
H2N+
CH2
H2C
CH2
H3N+ C H
CH2
H C OH
Serine (S)
COO–
Leucine (L)
H3N+ C H
COO–
+
H3N C H
CH2OH
H3N+ C H
CH
C N
H
H3N+ C H
CH2
COO–
CH2
COO–
Glutamate (E)
pKa = 4.3
Aspartate (D)
pKa = 3.7
Aromatic R groups
Histidine (H)
pKa = 6.0
©Hayden-McNeil, LLC
COO–
+
H3N C H
CH2
COO–
+
H3N C H
CH2
COO–
H3N+ C H
CH2
COO–
+
H3N C H
CH2
C CH
NH
OH
Phenylalanine (F)
Tyrosine (Y)
Tryptophan (W)
13
Amino Acid Structures and R Group pKa Values
14
LAB 1 • Working Efficiently in the Biochemistry Laboratory
1
Pressmaster/Shutterstock.com
LABORATORY
Working Efficiently in the
Biochemistry Laboratory
Welcome to Biochemical Techniques! You will be conducting many engaging and
thought-provoking experiments this quarter. We want you to be successful in all of
your laboratory work for this class, so we will use this first lab session to introduce
you to the equipment you will be working with and help you perfect the skills that
are essential for any experimental work. Bear in mind, the abilities you take away
from this class, which include manual, mathematical and analytical skills, will be
a valuable asset not only in the laboratory but in any technology-oriented career.
A. Introduction to the Equipment and Instruments You Will Be Working With
In order to work efficiently in the lab, you will need to know what equipment is available to you and to understand how and when it is used. Take a look in your locker
and on the bench around your workspace and identify the following items.
1.
Micropipettes—On your bench is a rack holding six micropipettes.
What are these used for? The micropipettes are used to transfer small volumes
in a range indicated by the number on the pipette; the P-1000 can measure 100
μl to 1000 μl, the P100 can measure 10 μl to 100 μl, and the P-10 can measure
1 μl to 10 μl. Realistically, the P-10 is not very accurate below 2 μl, so try not to
pipette volumes less than 2 μl if accuracy is important. Note that the other 3
micropipettes on the rack are for the group beside you.
15
LAB 1 • Working Efficiently in the Biochemistry Laboratory
2. Micropipette tips
What are these used for? These are used to hold the liquid being transferred by
the micropipettes. The small tips go on the P-10, P-40, and P-200 micropipettes,
and the large tips go on the P-1000 micropipettes. These tips are not sterile. You
will need to refill the tip containers when they are empty. Find where the boxes
of micropipette tips are located in the cabinets. You should wear gloves when
refilling the tip boxes to prevent substances on your skin (proteases!) from getting on the tips and potentially damaging your samples.
Waste and cleaning streams? These tips are disposable. They are considered
sharps waste, so eject them into the green plastic sharps tubs.
3. Pi-Pump—This is the green plastic device with the white barrel and thumb-roller.
What is this used for? The Pi-Pump is used with the glass serological pipettes to
transfer liquids in volumes greater than 1 ml. If you need to transfer 5 ml, do not
use the P-1000 to transfer 1 ml five times. This takes longer and prevents other
students from accessing the reagent stock bottles, and increases the chance of
contaminating the stock solution.
4. Glass serological pipettes—Find where these are located in in the cabinets around
the lab.
What are these used for? These are used with the Pi-Pump to transfer larger
volumes of liquid. What sizes of pipettes are available?
Waste and cleaning streams? The glass serological pipettes are not disposable.
After use, place them into the pipette washing barrels that are located in the
sinks.
5. Beakers and graduated cylinders—These are in your locker.
What are these used for? These are used to contain and measure larger volumes
of liquids. What sizes of graduated cylinders do you have available?
6. Microfuge tubes
What are these used for? These are tubes that hold liquid volumes less than
about 1.25 ml. They can be used to centrifuge the liquids in your microfuge.
Find the container of microfuge tubes in your locker. These are not sterile. You
are expected to refill the microfuge tube container in your locker at the end
of each lab session. Find where the bags of microfuge tube are located in the
cabinets. You should wear gloves when refilling the microfuge tubes to prevent
substances on your skin (proteases!) from getting in the tubes and potentially
damaging your samples.
Waste and cleaning streams? Microfuge tubes are disposable. Remove any liquid
(using an aspirator flask) before discarding and then place them in the white lab
trash bucket.
16
LAB 1 • Working Efficiently in the Biochemistry Laboratory
7.
Microfuge
What is this used for? The microfuge is used to centrifuge samples in microfuge
tubes. You will use the microfuge in an exercise later in this lab.
8. Vortex—This is on the bench in your workspace.
What is this used for? This is used to mix liquids. Set it on touch mode and press
a microfuge tube or larger tube into the rubber cup on top to vortex the liquid.
9. Stir plate—This is on the bench in your workspace.
What is this used for? This is used to stir larger volumes contained in a beaker.
When you use this, you will check out a stir bar from your IA and place it in the
beaker.
10. Spectrophotometer—Find the spectrophotometer that is located in your
workspace.
What is this used for? When you have time, carefully read “The Spectrophotometer” on page 7 of this text. This is the single most important instrument
you will be using in this class and it is essential you know how to use it correctly.
Instructions for using the instrument are posted by each spectrophotometer.
Open the lid of the sample chamber and look inside. Find the light path. In which
direction does it cross the sample chamber? Where does the cuvette go?
11. Cuvettes—These will be supplied on the front bench. Take one now for each
member of your group. You will use it for an exercise later in this lab.
What is this used for? These hold the liquid that is measured in the spectrophotometer. Look at the cuvette. This size cuvette can hold 1–2 ml of sample. To
take an accurate absorbance reading in the spectrophotometer, it needs to be
filled with at least 1 ml of sample. The diagram indicates the required 1 ml fill
level on the cuvette. Now look at the bottom of the cuvette. One dimension is
longer than the other, so how should you orient the cuvette in the spectrophotometer? The light path in the spectrophotometer is designed to be 1 cm. Orient
the cuvette so that the long dimension is parallel with the light path.
Bottom
view
1 mL
1 cm
©Hayden-McNeil, LLC
Waste and cleaning streams? These cuvettes are disposable. Empty each cuvette
using the aspirator flask, then place the cuvettes in the white lab trash bucket.
17
LAB 1 • Working Efficiently in the Biochemistry Laboratory
12. Waste Aspirator Flasks
Liquid waste can never go in any trash receptacle, and containers must be empty
before disposal. You will use aspirator flasks attached to a vacuum line to remove
any liquids. Note that there are different flasks for hazardous and non-hazardous
waste. Non-hazardous flasks may be emptied in the sink when full. Hazardous
flasks will be emptied by UGL staff for EH&S pick-up.
B. How to Safely Open Microfuge Tubes Containing Chemical Reagents
Each group member should perform this exercise. Fill a microfuge tube with 1 ml
of water. Shake the tube. Hold a paper towel up to the tube and flick open the cap
with your thumb so that it will pop open toward the paper towel.
Did any drops of water land on the paper towel? Is this how you would want to open
the microfuge tube if it contained a chemical reagent? The most common cause of
chemical exposures in this class is the flicking open of microfuge tube caps.
Close the lid on your tube and now open it in a way that will prevent any liquid on
inside of the cap from being dispersed onto you or into the lab. Turn the tube so that
it will open away from you (the hinge is facing you), place your thumbnail under the
lip of the cap and slowly open it. Taking simple precautions such as this will help to
make a safer work environment.
C. Pipetting and Dilution Math Skills Test
Being able to accurately pipette small volumes will be essential for your experiments
this quarter. Additionally, much of the work you will be doing will involve making
dilutions of samples. These skills are core competencies that we want everyone to
establish early in the class. As everyone has already taken the BILD 4 Introductory
Biology Lab (or an equivalent biology lab), you may have already developed these skills.
This exercise will test these skills to be sure you can perform them at the required
level. When you are ready to do the exercise, you will do the following.
1. Let your IA know you are ready to do the exercise. They will call you to the front
bench when they are ready to proctor you.
2. Bring a calculator and pen or pencil to the front bench to do the exercise. Your
IA will provide everything else.
3. You will make a dilution of a dye solution in one well of a 96-well plate. A bottle
of DI H2O (the diluent) and tube of the blue dye will be present on the bench
along with micropipettes and tips.
4. Your IA will point to the well you will use, and tell you to make X μl (a volume
from 100–300 μl) of a 1:X dilution (the dilution will be specified) of the blue dye
in the phosphate buffer diluent.
5. You will have exactly 3 minutes to perform the dilution.
18
LAB 1 • Working Efficiently in the Biochemistry Laboratory
6. Your IA will compare the absorbance of your dilution well to a series of standards. Your absorbance must fall within +/– 0.05 absorbance units of the average
standard absorbance to pass.
Here is an example of a dilution you might need to perform in this exercise: Prepare
250 μl of a 1:60 dilution of the dye. Do the math first and write down the volumes
of the diluent and the dye that you will use before you start pipetting. How would
you do this calculation?
•
•
The dilution you want to perform is 1:60, which means 1 part of dye diluted
into 60 parts total. Your dilution factor is 60, so you essentially want 1/60th
of your 250 μl final volume to be the dye. Therefore, 250 μl divided by 60 =
4.16 μl (round to 4.2 μl) of the dye.
Subtract 4.2 μl from 250 μl to get the volume of diluent you will need; 250 μl –
4.2 μl = 245.8 μl. Round this to 245 μl. You can only set the P-1000 in 5 μl increments and the difference is negligible.
What is the best way to pipette the solutions?
•
•
Add the diluent to the well first. It is easier to pipette a smaller volume into a
larger volume.
Don’t forget to mix the solution once you have added the dye. Do this by gently stirring with the pipette tip.
You do not need to do this exercise during this lab session if you do not feel ready.
There is a practice station set up in the lab at which you can draw dilution flash cards
and time yourself for performing the dilution. Your IA will periodically measure the
absorbances in the plate reader so that you can anonymously see if you passed. You
do need to pass the exercise proctored by your IA in order to pass the class. However,
you can take it as many times as necessary with no point deduction.
D. Working with the Microfuge and Pelleted Materials
Be sure that you have received instructions on how to use your benchtop microcentrifuge before beginning this exercise. You will use a solution of bovine serum albumin
(BSA—a readily available protein obtained from cattle) that has been precipitated
with ammonium sulfate. This solution contains 5% BSA and ammonium sulfate at
a concentration of 80% of saturated (prepared by adding 0.561 grams of ammonium
sulfate per ml of the original BSA solution). You will learn about precipitating proteins
with ammonium sulfate in Lab 3. The BSA protein is no longer in solution, but is a
suspension of small protein aggregates that can be brought down by centrifugation
to form a “pellet.” Working with pelleted materials requires some practice, and each
group member should perform the exercise on their own.
a.
Pipette 50 μl of the BSA solution into a microfuge tube, and prepare a “balance”
tube that contains 50 μl water.
b. Place the BSA tube and balance tube directly opposite each other in your benchtop microcentrifuge, and centrifuge the tubes for five minutes at full speed. Both
group members may centrifuge together.
19
LAB 1 • Working Efficiently in the Biochemistry Laboratory
c.
Remove the tubes from the microcentrifuge and note the white pellet at the
bottom of the BSA tube.
d. Carefully remove all of the supernatant (the liquid on top of the pellet) without
disrupting the pellet. If you disrupt the pellet, place all of the supernatant back
in the tube and centrifuge again for five minutes. Show the tube to your TA so
they can confirm you have successfully removed all of the supernatant.
e.
Add 20 μl of water on top of the pellet. Resuspend (dissolve) the pellet in the
water by gently pipetting up and down with your micropipette, without causing any bubbles to form in the liquid. To avoid adding bubbles to your tube, do
not press the plunger all the way down when bringing the liquid in and out of
the pipette tip. If your resuspended solution contains any bubbles, repeat the
exercise beginning with step a. Show your resuspended pellet to your TA so they
can confirm the absence of bubbles.
E. Dilution and Spectrophotometry Exercise
In this exercise, you will pipette a small volume of a viscous solution, which is difficult to do accurately. The dye solution you will use has 0.2% brilliant blue FCF (a
common food coloring) in 50% glycerol. You will transfer 1 μl into 1 ml of water
and measure the resulting absorbance in your spectrophotometer. This will let you
measure the accuracy of your pipetting to see how close you were to 1 μl. First, however, you will need to convert your absorbance value into a percent concentration
(grams/100 ml) for the diluted dye.
You should use your best technique when pipetting the dye! The glycerol makes the
dye solution viscous. You should just touch the very end of the pipette tip to the dye
solution and then slowly raise the barrel of the micropipette. Do not submerge the tip
in the dye or it will coat the outside of the tip and you will transfer too much volume.
After bringing up the dye, look at the tip. One μl will just be visible in the very end
of the tip, but you should see it. If you don’t see anything, you haven’t brought up
any dye. Once you have dye in the tip, touch it to the water, and press the barrel to
release it. Go past the first barrel stop to purge all the dye, then raise the tip out of
the water before raising the barrel. Then place the tip back in the water and carefully
pipette in and out two or three times to rinse any residual dye from the tip (do not
go past the first stop). It is also a good idea to gently stir with the pipette tip while
you are doing this.
1. Each group member will do this exercise. Each of you should get a microfuge
tube and fill it with 1 ml of water.
2. Pipette exactly 1 μl of the dye solution into the water.
3. Use your vortex to thoroughly mix.
20
LAB 1 • Working Efficiently in the Biochemistry Laboratory
4. Get one cuvette for each group member (use the cuvette you took for part A)
and transfer the diluted dye solution into the cuvette. Get one additional cuvette
for a blank, and fill it with 1 ml of water. Be sure there are no fingerprints on the
lower part of the cuvette where the solution sits, and that there are no bubbles.
5. Set the wavelength of your spectrophotometer to 625 nm. Insert the blank and
zero the instrument.
6. Measure and record the absorbance for each of your diluted solutions.
7.
To determine the accuracy of your pipetting, you need to calculate the corresponding molar concentration of the diluted dye from each absorbance value,
and then convert the molar concentration to percent concentration (grams/100
ml). This will allow you to compare the concentration of your diluted dye to
what you know to be the “true value” for the dilution. The true value is the dye
concentration you would obtain if your pipetting was 100% accurate. In this
case the true value is stock dye concentration (0.2%) divided by 1000, which is
the dilution factor.
Calculate the molar concentration from your absorbance value first. You will use
Beer’s law to do this (refer to page 10 for more information), where A = εlc. For
brilliant blue FCF, ε625 = 97000 M−1cm−1. Next convert the molar concentration
to percent concentration. The MW of the dye is 792.8 grams/mol.
8. Once you have the percent concentration of your diluted dye solution, calculate
the Percent Error of your pipetting.
Percent error =
X–X true
# 100
X true
How accurate were you? Keep in mind that pipetting such a small volume of a
viscous liquid will inherently have a lot of error.
21
LAB 1 • Working Efficiently in the Biochemistry Laboratory
22
LAB 2 • Electrophoresis and SDS-PAGE
2
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LABORATORY
Electrophoresis
and SDS-PAGE
In this lab your will learn how to perform a type of electrophoresis called SDS-PAGE.
This is an acronym for sodium dodecyl-sulfate polyacrylamide gel electrophoresis.
This type of electrophoresis is designed to separate proteins by molecular weight,
and it is a widely used method for analyzing proteins. Because the stained SDS-PAGE
gel has a band for each protein present at its respective molecular weight, it provides
information on the number of proteins present and their relative abundance. SDSPAGE is the first step for more specific analytical techniques like Western blotting.
Because you typically want to compare the abundance of one or more proteins
between lanes, it is critical to be able to load a fixed amount of protein into each lane
of the gel. This requires some math skills and good pipetting technique, and this lab
is designed to help you learn these.
Before coming to lab, you should watch the short YouTube video on assembling
and running the polyacrylamide gel electrophoresis units you will be using.
A link to this video is available on your instructor’s Canvas site.
23
LAB 2 • Electrophoresis and SDS-PAGE
A. Introduction to Electrophoresis
Creating the Electric Field
Electrophoresis is a variety of methods for the separation of charged molecules in an
electric field. The “gel” in gel electrophoresis is the semi-solid material into which the
sample is loaded, and it is typically agarose or polyacrylamide. The gel prevents the
sample molecules from diffusing away and separates them as they migrate through
it. The gel is permeable to the buffer solution around it, so the current of the electric
field is carried through the gel. Once the current starts, the charged sample molecules
will move toward the electrode that carries the opposite charge. The sample is most
often prepared so the molecules to be analyzed will carry a net negative charge. It
is loaded at the negative electrode end and the molecules migrate toward the positive electrode. Proteins are typically analyzed on polyacrylamide gels because the
smaller pore size of polyacrylamide is more compatible with the molecular weights
of proteins. Proteins diffuse more rapidly through the larger pore size of agarose gels.
This random diffusion from kinetic energy tends to scatter the protein molecules
as they move through the electric field and results in poor resolution and diffuse
bands. That being said, you will successfully run proteins on an agarose gel in Lab
7! Nucleic acids are run on either agarose or polyacrylamide, depending on the
fragment size. Large DNA and RNA molecules, longer than about 200 base pairs or
nucleotides, are better separated on agarose, while polyacrylamide produces better
resolution for small fragments.
Regardless of the type of gel and molecules being separated, the same basic principles apply to how the electric field is generated. The electrophoresis apparatus that
contains the gel and electrodes is connected to a power supply. When the power
supply is turned on, electrons are pushed through the lead to the negative electrode,
producing an accumulation of negative charge. At the same time, electrons are leaving the positive electrode and returning to the power supply, producing a positive
charge. This creates the electrical potential for the electric field. The field must follow
the properties defined by Ohm’s law:
V = IR
Where V is the voltage with units of volts, I is the current with units of amps
or milliamps, and R is the resistance with units of Ohms. In order to produce
the electric field, charged particles (ions) must move in response to the electrical
potential; this establishes the current and corresponding voltage. So, what are the
ions that will produce the current? They are not the electrons from the negative
electrode; the electrons do not move through the gel! Rather, the electrophoresis
buffer contains the ions that will produce the current. A basic knowledge of Ohm’s
law helps in understanding what is occurring during an electrophoresis run, and in
avoiding potential pitfalls that could ruin your experiment.
24
LAB 2 • Electrophoresis and SDS-PAGE
FIGURE 2.1 Electron movement and electrolysis during electrophoresis.
Negative
electrode
e–
e–
2H2 + 4OH–
4e– + 4H2O
e–
Sample wells
e–
R–(–)
Semi-solid gel
R–(– –)
Power
source
e–
e–
e–
4e– + O2 + 4H+
e–
Positive
electrode
2H O
2
©Hayden-McNeil, LLC
The ion concentration of the buffer is critical as it affects the current through the
gel. The greater the ion concentration, the greater the current. It is important to note
that the rate at which the charged sample molecules move in the gel is determined
by the voltage, and not the current. The greater the voltage, the faster they will move
toward the electrode of opposite charge. The power supply can be set to run the
electrophoresis at constant voltage or constant current. On the constant voltage setting, you set the voltage at which you want to run the gel and the power supply will
vary the current to maintain that voltage. On the constant current setting, you set
the current and the voltage will vary to maintain it. The resistance often increases
during an electrophoresis run, and the power supply will need to vary either the
current or voltage to compensate.
Because the buffer provides the ions to produce the current, the nature and concentration of the ionizable compounds in the buffer are important considerations.
The ions in the buffer are compounds where the charge is dependent on the pH of
the solution. For DNA gels, acetate or borate provide the ions and Tris is used to
control the pH and also acts as a counter ion (TAE or TBE buffers). For protein gels,
glycine provides the ions in Tris-glycine buffers. The choice of ionizing compound
is dictated by the pH at which the electrophoresis needs to be run. By having the
correct compound at the right pH, the degree to which the compound is ionized,
and therefore the total ions in solution and amount of current, can be precisely controlled. An ion concentration that is too high will produce too much current. Current
produces heat, and as the amount of heat produced is exponentially proportional to
the current, if the current is too high, the gel can rapidly overheat.
25
LAB 2 • Electrophoresis and SDS-PAGE
H = I 2R
(H is the amount of heat with units of joules or calories)
With any type of gel, gas bubbles are produced at the positive and negative electrodes
during the electrophoresis run. What gas is being produced and why? Accumulation
of charge at the negative and positive electrodes drives a chemical reaction that splits
water molecules. This is the process of electrolysis (see Figure 2.1). At the negative
electrode, electrons are consumed to yield hydrogen gas and hydroxide ions. At the
positive electrode, electrons are produced by splitting H2O to yield oxygen gas and
hydrogen ions. Where do you expect to see more gas bubbles being produced, at
the negative or at the positive electrode? Looking for the appearance of these gas
bubbles at the beginning of an electrophoresis run is a good way to confirm that the
gel apparatus and power supply are working correctly; in other words, that you have
established an electric field that will cause the charged molecules in your sample to
move through the gel.
What Determines How Fast a Charged Molecule Will Move in the Gel?
Factors such as the voltage and the density of the gel will affect how rapidly charged
molecules will migrate toward the electrode of opposite charge. These factors can be
considered extrinsic since they are not determined by the sample molecules being
analyzed and affect all the sample molecules equivalently. But since the analysis
depends on separating the sample molecules, what factors determine this? Three
intrinsic factors, meaning the chemical nature of the sample molecules themselves,
determine how rapidly they will migrate in the gel. These are:
•
•
•
The size or molecular weight of the molecules
The net charge of the molecules (really the mass to charge ratio)
The shape or conformation of the molecules
Each of these factors will independently affect the migration rate. In most analyses,
however, you want to separate the sample molecule based on differences in only one
of these factors, which one depending on the type of information that needs to be
determined. The majority of electrophoresis on proteins and nucleic acids is done
to separate them by molecular weight to provide a snapshot of which molecules are
present in the sample. In order to separate sample molecules by only one factor, the
other two factors must be equivalent between molecules or the differences must be
neutralized in some fashion so that they don’t affect the migration rate.
Separation by molecular weight occurs because larger molecules will collide more
with the molecular lattice of the semi-solid gel, and this slows their movement
through the gel. Smaller molecules will fit more easily through pores in the gel lattice and will therefore move more quickly. For both agarose and polyacrylamide,
different concentrations can be used in preparing the gel to optimize the separation
of molecules within a specific size range. For charge, the greater the net charge of
the molecule the greater the attraction to the opposite electrode and the faster the
26
LAB 2 • Electrophoresis and SDS-PAGE
movement through the gel. However, this assumes equivalent molecular weight. It is
actually the mass to charge ratio that determines the effect of net charge on migration rate. For example, if two molecules both had a net charge of –2 but one of the
molecules was twice the molecular weight of the other, the larger molecule would
have only half the attraction to the positive electrode as the smaller molecule. All
nucleic acids have an equivalent mass to charge ratio because each nucleotide residue
in the chain carries one negative charge on the phosphoryl group. A double-stranded
DNA of 50 base pairs has a net charge of −100 (excluding the single extra charge on
the 5'-terminal phosphate) and a DNA strand of 100 base pairs has a net charge of
−200, but both have a mass to charge ratio of −2 per base pair. This is not the case
with proteins, however. The mass to charge ratio of proteins can vary widely based
on their isoelectric point, and steps must be taken to neutralize the difference in net
charge in order to separate them strictly based on molecular weight.
Differences in the folding or conformation of large biological molecules will also
affect their movement in the gel. A long, rod-like protein would have a more difficult time fitting through the gel molecular lattice than a compactly-folded globular
protein of the same molecular weight. Linear DNA molecules move more slowly on
an electrophoresis gel than supercoiled, circular DNA molecules of the same length,
as the latter fits more easily through the pores of the gel lattice. Conformational differences due to folding can be removed by denaturing the sample molecules prior
to running them on the gel. This is particularly important for proteins. All electrophoresis procedures can be classified as either denaturing or native. In denaturing
electrophoresis gels, steps are taken to denature all the molecules in the sample
to their unfolded conformation. Native gel electrophoresis, on the other hand, is
done under conditions that will preserve the natural (“native”) conformation of the
sample molecules. This preserves the biological activity of sample. For example, the
catalytic activity of enzymes or the binding affinity of antibodies would be retained
at the end of the electrophoresis run.
B. Background on SDS-PAGE
The gel in SDS-PAGE is polyacrylamide. This is a polymer of covalently cross-linked
acrylamide and bis-acrylamide subunits (see Figure 2.2). The polymer is porous, and
water, buffer ions, and the proteins being separated pass through these pores once
the electric field is established. By adjusting the concentration of acrylamide monomers, the average pore size can be controlled to produce polyacrylamide gels that
are optimized for separating proteins within a particular molecular weight range.
Percent (g/100 ml)
Acrylamide
8%
10%
12%
15%
Average Pore
Radius in nm
160
140
90
70
MW Range
Fractionated
40–300 kDa
30–200 kDa
20–150 kDa
10–80 kDa
27
LAB 2 • Electrophoresis and SDS-PAGE
The polyacrylamide gel is produced by mixing a solution of acrylamide and bisacrylamide with the polymerization initiating agents TEMED (tetramethylethylenediamine) and ammonium persulfate ((NH4)2S2O8). These initiate a free radical-based
chain reaction that forms covalent bonds between the acrylamide and bis-acrylamide.
The polymerization occurs rapidly, so once the polymerization initiating agents have
been added the solution must be quickly poured between the glass plates that will
hold the gel. Preparing polyacrylamide gels is technically challenging and requires
a lot of practice before being able to produce useable gels each time you pour one.
Additionally, acrylamide and bis-acrylamide are potent neurotoxins. A lab coat,
safety glasses, and gloves should be worn when handling unpolymerized acrylamide
solutions. Extra care should be taken when working with dry acrylamide to prepare
the monomer solutions because the light powders easily form particulate dust that
is an inhalation hazard. Dry acrylamide should be handled in a fume hood. Fortunately, we will use pre-cast polyacrylamide gels in this class so you will not have to
prepare them. PPE and gloves should always be worn when handling the pre-cast
gels, however, in case some unpolymerized acrylamide remains on the gel.
FIGURE 2.2 Formation of polyacrylamide.
O
H2C
C
H
C
NH
CH2
NH2
C
C
H
Acrylamide
NH
C
C
H
O
Bisacrylamide
H2C
CH2
CH
CH2
H2C
O
CONH2
CONH
CH
CH
CH2
CH2
CONH
CONH
CH2
CONH
CH2
CH
CH2
CONH
CONH2
CH2
CH
CH2
CH
CH2
CONH
CH2
CONH
28
Bisacrylamide
CH
CH
CONH2
CH2
CH
Acrylamide monomer
LAB 2 • Electrophoresis and SDS-PAGE
SDS-PAGE is designed to separate proteins strictly based on their molecular weight.
Since proteins vary dramatically in net charge and conformation, these factors must
be neutralized in order to accomplish this. The detergent sodium dodecyl-sulfate
(SDS) is present in the gel and the buffer solution that surrounds it, and it plays a
critical role in neutralizing the charge and conformation factors. The amphipathic
SDS molecules interact with proteins through both charge and hydrophobic interaction. The high SDS concentration (0.1%) drives maximal binding to the proteins
in the sample and they become coated with SDS. As each dodecyl-sulfate molecule
carries a negative charge, this has the effect of coating the proteins with negative
charges. This overwhelms the intrinsic charge on the proteins and gives them a strong
net negative charge. The number of SDS molecules bound depends on the size of
the protein; a 100 kDa protein will bind roughly twice as many SDS molecules as
a 50 kDa protein because there is more polypeptide for them to interact with. This
gives all the proteins in the sample an equivalent mass to charge ratio and removes
their intrinsic net charge as a factor that will affect how fast they move in the gel.
Sodium dodecyl-sulfate
O
O
S
O
CH3
–
O
Na
+
The sample proteins in SDS-PAGE are denatured before they are loaded into the gel,
and this prevents differences in conformation from affecting their migration rate.
As a detergent, the SDS also acts as a denaturing agent to disrupt the interactions
that hold the protein in its folded conformation. It is a relatively weak denaturing
agent, though, and SDS alone will not disrupt very stable protein conformations.
The sample proteins are also treated with a compound called 2-mercaptoethanol.
Mercaptoethanol is a potent reducing agent and its purpose is to break any disulfide
bonds that may be holding together the folded conformation of a protein. Disulfide
bonds are covalent bonds between the R-groups of two cysteine residues in distant
parts of the polypeptide. Disulfide bonds are not disrupted by SDS or by heat, so
breaking them with mercaptoethanol is required to completely denature these proteins. Mercaptoethanol and SDS are in the sample buffer that is added to prepare
the protein samples for running on the gel (see section E). The sample buffer also
contains other reagents that aid in the loading of the samples and in tracking the
progress of the electrophoresis run. The most important element for ensuring that
the sample proteins are completely denatured, however, is heat. After the sample
buffer is added, the samples are heated at 100°C for 2–10 minutes.
29
LAB 2 • Electrophoresis and SDS-PAGE
FIGURE 2.3 Breakage of a disulfide bond.
disulfide
bond
SH
S
S
2-mercaptoethanol
HS
©Hayden-McNeil, LLC
If you look carefully at an SDS-PAGE gel you may notice that the region of the gel
just below the wells seems to be a different consistency than the rest of the gel. This
is indeed the case. The top and bottom parts of the gel are poured separately and
have different acrylamide concentrations (3% on top and 7−15% below) and pH. The
top is the stacking gel, and the rest of the gel is referred to as the running gel (or
separating gel). The running gel is where proteins are actively separated by molecular
weight. The purpose of the stacking gel is to compress all of the proteins in the sample
into a thin band so that they all enter the running gel simultaneously. When you
load your SDS-PAGE gel you will note that the sample takes up most of the volume
of the well, and it takes some time for the electric field to pull all of the sample into
the gel. This means that without the stacking gel, proteins of the same molecular
weight would enter the running gel at different times, and this would result in poor
resolution of the protein bands. You can read about how the stacking gel works to
compress the sample proteins in section D.
FIGURE 2.4 The stacking gel and the running gel.
Sample well
−
Stacking gel
Running gel
+
©Hayden-McNeil, LLC
30
LAB 2 • Electrophoresis and SDS-PAGE
At the completion of the SDS-PAGE run, all proteins of the same molecular weight
should be present in the gel as a single, high-resolution band, distinct from other
molecular weight bands on the negative to positive electrode axis. You cannot see
the bands in the gel, however, without staining them. The most common staining
method is Coomassie staining, which uses the dye Coomassie Brilliant Blue in a solution of acetic acid and methanol. The dye binds to all of the proteins present in the
gel and the methanol and acetic acid act to fix the gel to preserve it and the protein
bands. The methanol precipitates the proteins, which helps to preserve them, and
also precipitates the SDS to remove it. The acetic acid also favors precipitation, and
confers a net positive charge to the proteins (see Lab 8) which aids in binding of the
negatively charged Coomassie Blue. Usually, the gel is left in the fixing and staining
solution overnight, and then it is rinsed in several changes of the fixing solution
without the Coomassie Blue to remove the unbound dye. Silver staining is a more
sensitive staining technique that can detect smaller quantities of protein, but it is
more expensive and time consuming. Antibodies can also be used to detect specific
proteins that may present. To do this, the gel is not stained but instead the proteins
are transferred out of the gel and onto a filter where they can interact with the
antibodies. This technique is called Western blotting and you will do this in Lab 10.
C. SDS-PAGE Procedure
The objective of this lab exercise is to give you practice in the skills required to prepare and run protein samples on an SDS-PAGE gel. You will run at least two more
SDS-PAGE gels in this class that are part of larger projects. All of the equipment and
reagents, including the polyacrylamide gel rigs, 5× sample buffer, and MW markers, will be the same for each gel you run, so take the time to familiarize yourself
with them now. You and your lab partner will each prepare a set of 4 BSA samples
that contain 0.25 μg, 0.5 μg, 1 μg, and 5 μg of protein and run them in 4 contiguous
lanes of your gel. After you run the gel and stain it, you should have a set of bands of
increasing intensity as shown in Figure 2.5. With careful work you should be able to
detect the 0.25 μg band. The banding pattern and corresponding molecular weights
of the MW markers you will use throughout the class are shown in Figure 2.6.
31
LAB 2 • Electrophoresis and SDS-PAGE
FIGURE 2.5 Relative band intensities for the indicated amounts of BSA protein following
Coomassie staining with a rapid stain protocol.
0.25 μg
0.5 μg
1.0 μg
5.0 μg
MW Markers
©Hayden-McNeil, LLC
FIGURE 2.6 MW markers (1 kDa = 1000 amu or 1000 g/mol)
kDa
~180
~130
~100
100
~75
5
~63
63
~48
48
~35
~28
~17
17
~10
©Hayden-McNeil, LLC
1. Get 15 μl of the 1 μg/μl BSA stock from the front table (you and your partner
should take 15 μl each in separate tubes). Do not take more than 15 μl; this
volume is sufficient for preparing your SDS-PAGE samples.
32
LAB 2 • Electrophoresis and SDS-PAGE
2. You want to prepare 4 samples with the indicated amount of BSA protein. Each
sample will have a final volume of 20 μl; this is the maximum amount that you
can load into the gel. You will need to add 4 μl of 5× sample buffer, therefore
you will add 16 μl of BSA solution and/or water to bring the volume up to 20 μl.
Think about how you will prepare each sample so that you will be as accurate and
efficient as possible. It is important to remember that pipetting small volumes
with the P-10 micropipette is inherently inaccurate. When accuracy matters
(and it does for this exercise), don’t pipette volumes less than 2 μl. Use the table
below to guide how you prepare your samples. Write out the table with all the
indicated volumes in your lab notebook. You should write out a similar table in
your lab notebook each time you run an SDS-PAGE gel.
Volume
5X Sample
Buffer
Final
Volume
0.25 μg BSA
4 μl
20 μl
0.5 μg BSA
4 μl
20 μl
1 μg BSA
4 μl
20 μl
5 μg BSA
4 μl
20 μl
Sample
Dilution, if
any
Volume
BSA
Volume
H2 O
Lane # on
Gel
3. Prepare 600 ml of 1× tris/glycine electrophoresis buffer. The 10× stock is on
the front table.
4. Obtain a 10% precast polyacrylamide gel. Be sure you are wearing gloves while
you handle the gel and assemble the gel apparatus. Remove the gel from its
packaging. The storage buffer in the packaging contains 0.02% sodium azide,
which is highly toxic. Absorb this onto a paper towel and carefully dispose of
it in the white lab trash bucket. (This very small amount does not need to be
collected as hazardous waste.)
5. Peel the green strip from the bottom of the gel. Now remove the comb. To do
this, brace the bottom of the gel against the bench while firmly grasping the gel
on both sides, and use your thumbs to gently push it out.
6. Assemble the gel apparatus. Fill the top chamber with buffer until the level is
above the level of the lower gel plate. Pour the remaining buffer into the lower/
outer chamber.
7.
Use a disposable glass Pasteur pipette to flush out the wells. Position the end
of the pipette over the well and squeeze and release the bulb several times to
push buffer into the well. This will remove any unpolymerized acrylamide that
33
LAB 2 • Electrophoresis and SDS-PAGE
could block your sample from entering the well. Let the gel sit while you prepare
your samples.
8. Prepare your samples, following the table you completed for step 2. Add the
water first, then the BSA solution, and the 5× sample buffer last. Stir gently
with the pipette tip to mix.
Safety Note: The 5× sample buffer contains 25% 2-mercaptoethanol. This
compound is toxic and is hazardous by inhalation and skin contact. Wear
gloves and remove the glove immediately if you come in contact with the 5×
sample buffer. Add the 5× sample buffer to your samples in the fume hood,
being sure to use a new tip for each sample. It is okay to take your tubes out
of the hood once the 5× sample buffer has been diluted into your samples.
9.
Heat your samples at 95°C for 3 minutes. Place the samples on ice for a minute
or two to remove the heat, and then spin them in your microfuge for 1 minute
to bring down any condensation. Keep the samples at room temperature until
you are ready to load them; do not place them back onto the ice as this could
cause the SDS to precipitate.
10. Before loading your samples into your gel, practice loading using the IA demo gel.
11. Wait until both you and your partner have your samples ready before you load.
Before you load, check the buffer level in the top chamber to be sure it is still
above the lower gel plate; a slow leak may cause it to drop without you noticing.
Adjust the gel assembly and/or top off the buffer if necessary. Load your samples
into 4 adjacent wells. Gently rest the pipette tip between the gel plates (do not
push the tip down!), and slow press the plunger. Change tips between samples.
12. Load 5 μl of the MW markers into one lane of the gel.
13. Place the lid on the gel assembly and plug the leads into the power supply. Set
the power supply to run constant voltage and set the voltage to 120 V.
14. Start the gel. Find the volts/current/time toggle switch and set it to display the
current. It should take about 22 mA of current for each gel unit connected to
the power supply to get to 120 V (for two gels you should be at 44 mA). It is a
good idea to check this each time you run a gel. If the current is significantly
less than 22 mA per gel at the beginning of the gel run, this could indicate a
problem. Let your IA know if this is the case.
15. Run the gel until the tracking dye reaches the bottom of the gel; this takes about
an hour. Check the gel periodically to be sure there is adequate buffer in the top
chamber and that the tracking dye continues to move. Check the current again
toward the end of the run. It will have dropped significantly; this is normal. Why
does it decrease during the run?
34
LAB 2 • Electrophoresis and SDS-PAGE
16. Stop the gel and unplug the leads from the power supply. Then remove the lid.
Take the electrode assembly out of the buffer tank and dispose of the buffer in
the sink (it is not hazardous). Remove the gel plates from the electrode assembly.
17. Transfer 15 ml of the rapid stain Coomassie blue solution into a staining box.
Use tape to label the box with your group number
18. Follow your IAs instructions for how to use the green metal key from your gel
kit to separate the gel plates. The gel will stick to one of the two plates. Gently
tease the gel away from the plate so that it neatly drops into the staining solution.
There are green plastic gel blades on the front bench to help with this. Uncurl it
with your finger (be sure you are wearing gloves) if necessary.
19. Incubate the gel in the staining solution for 30 minutes with gentle agitation on
the orbital shaker.
20. After 30 minutes, check to be sure you can see your bands. Leaving the gel in
the staining container, aspirate the staining solution into the waste flask on the
middle bench. Add distilled water from the sink to gel to rinse it, and leave the
gel in the water.
21. Remove the gel from the water and place it on a lightbox. Examine your bands.
The MW of BSA is 66.5 kDa (66,500 g/mol). Did your bands migrate to the
expected position on the gel? Are your band intensities similar to those in Figure
2.5? Your IA will take a picture of the gel and place the image file on Canvas.
You must print the gel image and add it to your lab notebook for the next lab
session. Label each lane of the gel. It is okay to write in the lane labels in pen
once the image is in your lab notebook.
22. Rinse the staining box and return it to the bin box. Place the gel in the gel waste
bucket on the middle bench (NOT the lab trash bucket).
Were you able see a band for the 0.25 μg sample? This is getting close to the detection
limit for Coomassie blue stains (with optimized, overnight staining procedures you
can actually detect close to 0.05 μg). Later in the class, you will perform a Western
blot in which you will use antibodies against a specific protein in order to detect
very small amounts of it. Western blotting analysis is about 1000 times more sensitive than Coomassie staining, so with the right antibodies you could easily detect
0.25 ng or less of BSA.
D. Supplemental Information
How the Stacking Gel Works
H
H3
N+
C
O
C
H
OH
H3
N+
C
O
C
O−
H2N
H
O
C
C
H
H
H
pH < pI
pH = pI
pH > pI
O−
35
LAB 2 • Electrophoresis and SDS-PAGE
The purpose of the stacking gel is to compress all of the proteins in the sample into
a thin band so that they enter the running gel simultaneously. This way proteins of
the same MW will be present in a discrete, well-resolved band in the gel after the
electrophoresis, regardless of the time it took for all of the sample to enter from the
well. This is accomplished by a drop in the voltage that occurs across the stacking
gel, produced by differences in the conductivity between the top and bottom of the
stacking gel. To understand how this voltage drop is produced, it is necessary to be
familiar with the composition and pH of the stacking gel, the running gel, and buffer
that fills the reservoirs of the electrophoresis unit (tank buffer).
Tank buffer:
Tris, pH 8.3; glycine; SDS. (The pI of glycine is 6.0.)
Stacking gel:
Tris, pH 6.8; 3% acrylamide; SDS; low concentration of Cl– ions.
Running gel:
Tris, pH 8.8; 5 to 15% acrylamide; SDS; low concentration of Cl– ions.
The Cl− ions in the stacking and running gels come from the HCl that was used to
adjust the pH of the gel (Cl− is also present in the 2× sample buffer). There is no Cl−
in the tank buffer. There is no glycine in either of the gels before the electrophoresis
begins, and the glycine in the tank buffer carries a strong negative charge at pH 8.3
(see diagram above). When the glycine enters the stacking gel, it finds itself at pH
6.8, very close to its pI of 6.0. Therefore, its charge and mobility decrease drastically.
Since the Cl− ions in the stacking gel are highly mobile, a large voltage drop develops
between the leading Cl− ions and the trailing glycine ions with the proteins sandwiched in between. From Ohm’s law we can define:
V (volts) = I (current, amperes) × R (resistance, ohms) (V= IR)
Conductivity, G, is the reciprocal of resistance, so
V= I
G
Since the charge on glycine decreases drastically when it moves into the stacking
gel, this creates a region just below the wells where conductivity is less. However,
the current is constant throughout the stacking gel, so V must rise locally (V = I/G).
This local increase in V accelerates the proteins migrating through this part of the
stacking gel, just below the wells. The small highly negative Cl− ions, on the other
hand, will tend to move more rapidly, and as they pull ahead they create a region of
greater conductivity. Here, V will decrease locally, so the proteins migrating through
this part of the stacking gel, closer to the running gel, will slow down. The end result
is that the proteins are compressed into a thin stack between the leading Cl− and
the trailing glycine.
As the glycine moves into the running gel (pH 8.8), it acquires a strong net negative
charge again and moves ahead of the proteins, leaving them to move through the
pores of the gel according to their size. The low concentration of acrylamide in the
stacking gel produces a pore size that is too large to separate the proteins, aiding in
compressing them together.
36
LAB 2 • Electrophoresis and SDS-PAGE
E. Supplemental Information
Recipes
10× Electrophoresis Buffer
• Trizma base 0.25 M, pH 8.3 (30.275 g/l)
• Glycine 1.92 M (144.13 g/l)
• SDS 1% (10 gm/l)
• Should be pH ~8.3 without adjusting (no Cl should be present)
5× Sample Buffer
• 375 mM tris, pH 6.8
• 45% glycerol
• 25% 2-mercaptoethanol
• 11.25% SDS
• 0.023% bromophenol blue
Fixing and Staining Solution
• 40% methanol
• 0.2% Coomassie Blue (0.2 g/100 ml)
• dissolve CB in methanol first, then add 10% acetic acid
• 50% dH2O
Destaining Solution
• 40% methanol
• 10% acetic acid
• 50% dH2O
37
LAB 2 • Electrophoresis and SDS-PAGE
38
LAB 3 • Initial Purification of Lactate Dehydrogenase: Centrifugation and Ammonium Sulfate Precipitation
3
Ron Kloberdanz/Shutterstock.com
LABORATORY
Initial Purification of Lactate
Dehydrogenase: Centrifugation and
Ammonium Sulfate Precipitation
Over the next three labs (Labs 3–5) you will purify the enzyme lactate dehydrogenase
(LDH). You will learn a variety of protein purification techniques that are frequently
used in biochemistry. You will try two different column chromatography methods,
affinity chromatography and size exclusion chromatography, and see which gives a
better purification of LDH. In mammals, different LDH isoenzymes are produced
in different body tissues (e.g., heart vs. skeletal muscle vs. kidney). Isoenzymes (or
isozymes) are different proteins that catalyze the same chemical reaction. You will
begin this project with a crude homogenate of a porcine (from a pig) tissue, either
skeletal muscle or heart. After purifying the LDH, you will analyze your final product
and samples taken at various points during the purification to determine the amount
of LDH catalytic activity and total protein that is present. With this information,
you will determine the most effective purification strategy for LDH. You will also
determine the identity of the LDH isozymes you have purified, and assess the isozyme
expression pattern in different tissues.
A. Introduction to Lactate Dehydrogenase (LDH)
Enzymes are proteins that act as catalysts to increase the rate of biochemical reactions. The reactions necessary to make life possible only occur at an adequate rate
with the help of enzymes. LDH catalyzes the conversion of pyruvate to lactate by
39
LAB 3 • Initial Purification of Lactate Dehydrogenase: Centrifugation and Ammonium Sulfate Precipitation
removing two electrons and a H+ ion from NADH (an oxidation reaction) and adding them to pyruvate (a reduction reaction). A second H+ is taken out of solution to
make lactate.
H3C
O
O
C
CO−
OH O
H+
+
NADH
H3C
C
CO−
+
NAD+
H
pyruvate
lactate
NAD+/NADH is an electron carrier coenzyme that functions to shuttle electrons
around the cell. It participates in the reactions catalyzed by a number of different
metabolic enzymes. The oxidized form is NAD+ (not carrying electrons) and the
reduced form is NADH (carrying electrons).
During low oxygen conditions in the cell, LDH functions to convert the pyruvate
produced from glycolysis into lactate, and in doing so replenishes the supply of NAD+
in the cytoplasm. In the absence of oxygen, pyruvate cannot be broken down in the
mitochondria to produce ATP. Glycolysis yields a small amount of ATP on its own
(2 ATP per glucose) and this is enough to keep the cell going until normal oxygen
levels are restored. However, glycolysis requires a constant supply of NAD+ to keep
running, and with no oxygen present to receive electrons from the end of the electron transport chain, all of the NADH/NAD+ will be stuck in the NADH form. LDH
allows glycolysis to keep running by converting some of this NADH back to NAD+.
This rescue of glycolysis by LDH is metabolically expensive, as it converts a valuable
fuel molecule (pyruvate) to a nonusable form (lactate). At some point the lactate
needs to get oxidized back to pyruvate.
In mammalian skeletal muscles, the ability to keep generating ATP in anaerobic
conditions is critical during periods of prolonged muscular activity. During heavy
exercise the blood supply cannot deliver oxygen to the muscles quickly enough, and
the muscle cells generate large amounts of lactate. The lactate is transported into the
bloodstream where it is taken up by the liver. There, LDH converts lactate back to
pyruvate by catalyzing the reaction in reverse (reducing NAD+ to NADH). The liver
then uses the pyruvate to synthesize new glucose molecules, which are distributed
out to the blood where they replenish the glucose supply back in the muscle cells.
This cycle of glucose to lactate in muscle and lactate back to glucose in the liver is
referred to as the Cori cycle.
40
LAB 3 • Initial Purification of Lactate Dehydrogenase: Centrifugation and Ammonium Sulfate Precipitation
FIGURE 3.1 The Cori cycle.
Muscle
Glucose
Pyruvate
Lactate
Glucose
Lactate
Glucose
Pyruvate
Lactate
©Hayden-McNeil, LLC
Liver
Active LDH is a tetramer, a complex composed of four separate polypeptide chain
subunits. These subunits are either H type (strongly expressed in the heart) or M type
(strongly expressed in skeletal muscle). Different combinations of these two subunits
in the tetramer produce the five different LDH isozymes. The H- and M-type subunits have different numbers of charged amino acids, giving them distinct isoelectric
points. Because of this, the different LDH isozymes can be separated and identified
by electrophoresis. You will learn more about the different isozymes in Lab 8.
The subunit polypeptides (both H and M) have a MW of 36,600 daltons (36.6 kDa),
making the active LDH tetramer 146.4 kDa in size.
One dalton is equivalent to one atomic mass unit, or one gram per mole. 1000
daltons = 1 kilodalton, abbreviated kDa. Because proteins are large molecules
with large MW values, it is more convenient to use units of kilodaltons. The
LDH tetramer has a MW of 146,400 grams/mol, but by convention we refer
to this as 146.4 kDa.
When considering a strategy for purifying a protein, it is helpful to know as many
of its biochemical properties as possible. These include its MW, isoelectric point
(pI), and the molecules that it interacts with or binds to (such as enzyme substrates).
These properties dictate the types of purification procedures that will be most useful
41
LAB 3 • Initial Purification of Lactate Dehydrogenase: Centrifugation and Ammonium Sulfate Precipitation
to purify the protein and how the protein will separate relative to other proteins.
The cellular localization of the protein is also important. LDH, along with all of the
enzymes for glycolysis, is found in the cytosol (the liquid component of the cytoplasm
with all of its soluble proteins). This information is critical for determining the initial
steps for isolating the protein. You would use a very different approach for purifying
an enzyme from the nucleus of a cell vs. the cytosol.
B. Protein Purification Strategies
The basic concept in protein purification is simple; you want to remove all of the other
proteins, lipids, nucleic acids, and any other “contaminating” molecules and end up
with only your target protein. You may utilize different techniques to accomplish
this, however, depending on the properties of the protein you wish to purify and its
source. Whether the starting material is animal tissue, parts of a plant, bacterial
cells, or cultured animal cells grown in vitro will significantly affect your purification
strategy. This being said, any protein purification strategy will have certain common
elements to it. It is useful to examine these common elements to shed some light
on what you will be doing in your LDH purification and how this might differ if you
were purifying a different protein.
1. The first step in any purification scheme is to homogenize the source material.
This breaks open the tissue and cells, and is done by grinding or blending the
source material in a buffer solution. The homogenized material is called a crude
homogenate, or lysate. For your LDH purification, a crude homogenate will be
prepared for you. While homogenization breaks open the cell membrane, most
organelles remain relatively intact. Sonication, disruption of the source material
with high-frequency sound waves, is often used to break open bacterial cells.
Detergents may also be used to help solubilize cell membranes. Source material
that lacks a cell wall, connective tissue, or a fibrous component can often be
homogenized with detergents alone. The choice of detergent is critical; it should
not denature or compromise the activity of target protein.
2. The next step typically involves separating soluble and insoluble fractions of
the crude homogenate. At one level this simply means separating fatty, fibrous,
or otherwise insoluble material from the proteins or components that remain
in solution. In terms of cellular localization, however, this means separating
different cellular compartments and organelles. This separation is typically
accomplished with one or more centrifugation steps. Filtration may also be used,
particularly for plant source material that contains a lot of fibrous cellulose.
Centrifugation and separation of cellular organelles: A centrifuge spins material
at very high speeds, applying a force that is directed outward from the center of
the rotor. This is relative centrifugal force (RCF) and it is quantified relative to
the force of gravity. An RCF of 1000 is 1000 times the force of gravity, or 1000 ×g.
RCF is dependent on two variables: the speed at which the centrifuge rotor is
42
LAB 3 • Initial Purification of Lactate Dehydrogenase: Centrifugation and Ammonium Sulfate Precipitation
spinning in revolutions per minute (rpm) and the radius of the rotor (distance
from the rotor center to where the sample is placed). A longer radius will produce
greater RCF at the same speed in rpm. The speed of many centrifuges is set in
RPM, often because different rotors with different radii can be interchanged in
the same centrifuge. To run a centrifuge at a desired RCF, the corresponding
rpm for a particular rotor radius can be looked up in reference tables.
RCF generated by the spinning rotor will push particles and large molecules to
the bottom of the centrifuge tube. Depending on the RCF applied, smaller, less
dense particles will remain in the liquid fraction (the supernatant), while larger,
denser particles will collect at the bottom of the centrifuge tube (the pellet).
The rate at which a particle moves toward the bottom of the centrifuge tube at
a given RCF is called the sedimentation rate. Larger, denser particles have a
greater sedimentation rate. Because cellular organelles have different sizes and
densities, they can be separated by centrifugation. Fractionating cells in this way
is a critical tool for isolating proteins with a particular cellular localization.
Cellular Organelle
RCF Required to Pellet That Organelle
Nuclei
500 ×g
Mitochondria
9,000–11,000 ×g
Peroxisomes, lysosomes
10,000–12,000 ×g
Microsomes, membrane fragments
100,000 ×g
Multiple centrifugation steps may be employed to isolate a particular organelle;
larger organelles can be pelleted first, followed by smaller organelles in subsequent centrifugations. The density of the solvent or medium in which the particles
are suspended also affects the sedimentation rate. Changing the density of the
medium can optimize separation of organelles of similar size. Sucrose can be
added to increase the density of the medium, either at a single concentration or
in a gradient of increasing concentration.
Your first purification step for LDH is centrifugation at 20,000 ×g. Which cellular
fraction are you isolating?
3. A precipitation step is often performed next. This serves to remove a large portion of the undesired protein, although in a relatively non-specific manner, and
to reduce the volume of the preparation. The complete purification of a protein
to the point where only trace amounts of contaminating protein remain requires
targeted purification steps that are tailored to specifically isolate the target protein based on its biochemical properties. One or more column chromatography
techniques are typically used for this. These targeted purification techniques
are not as efficient when a large quantity of contaminating proteins is present.
43
LAB 3 • Initial Purification of Lactate Dehydrogenase: Centrifugation and Ammonium Sulfate Precipitation
Furthermore, putting large volumes of a preparation though chromatography
columns is problematic, particularly for size exclusion chromatography. Therefore, precipitation steps are used to remove a bulk of the contaminating protein
and reduce the volume before chromatography steps are performed.
A precipitation step causes protein to aggregate and come out of solution. Centrifugation is then used to pellet the precipitated proteins. The precipitation
can be achieved in a variety of ways, some of which denature the proteins and
some of which do not. A few examples include the addition of organic solvents,
changing the pH, and the addition of salts at high concentration. Precipitation
with the salt ammonium sulfate is the most commonly used method because it
precipitates proteins without denaturing them.
Ammonium sulfate precipitation: This technique fractionates (separates) proteins based on differences in their relative solubility in water. All globular proteins are solvated with a layer of water molecules that participate in hydrogen
bonding with the charged and polar R-groups of amino acids on the surface of
the protein. This helps to stabilize the protein in its folded conformation—it is
more favorable for hydrophobic R-groups to be tucked away toward the center
of the protein. When ammonium sulfate is added to the solution, the NH4+ and
SO42− ions also hydrogen bond with the water molecules. Hydrogen bonding of
water with ammonium sulfate ions is higher energy and therefore favored over
hydrogen bonding of water with R-groups on proteins. At high concentrations
of ammonium sulfate, more water is used to solvate the salt ions so less is available to solvate the proteins. Once a protein is no longer solvated, it will slightly
unfold and expose its hydrophobic R-groups. Therefore, hydrophobic interactions between proteins are increasingly favored until the proteins aggregate and
come out of solution. Proteins that are less soluble in water (that have a higher
degree of hydrophobicity) will precipitate out at lower ammonium sulfate concentrations, whereas a higher concentration is required to precipitate proteins
that are more soluble (lower hydrophobicity). The precipitated proteins are not
denatured. When the pelleted proteins are resuspended in water, they refold to
their original conformation.
Ammonium sulfate is used because of its high solubility in water, and the fact
that its solubility is not very temperature dependent. A saturated solution is
3.9 M at 0°C and 4.1 M at 25°C. Because high, close to saturating concentrations
are required to precipitate proteins, the convention is to refer to the ammonium
sulfate concentration as a percentage of the saturating concentration. The term
“cut” is used because the precipitation is fractionating or separating proteins.
Therefore, a 40% cut uses an ammonium sulfate concentration that is 40% of
saturated (about 1.6 M at 0°C).
44
LAB 3 • Initial Purification of Lactate Dehydrogenase: Centrifugation and Ammonium Sulfate Precipitation
In your LDH purification you will perform two ammonium sulfate precipitations
to fractionate the proteins. You will first do a 40% cut where the LDH will
remain in the supernatant after centrifugation, followed by a 65% cut where
the LDH will be in the pellet. Following both of these, you will have separated
all the proteins that went into the first cut into how many separate fractions?
C. Exploratory LDH Purification
Rather than simply carrying out a predetermined set of purification steps with the
idea of maximizing the final yield of LDH, you will undertake the purification on more
of an exploratory basis. You will determine which combination of purification steps
would be most efficient for purifying LDH to a single protein. You will be provided
with 100 ml of crude homogenate and put all of it through an initial centrifugation
to generate a clarified homogenate (part B, number 2) and differential ammonium
sulfate precipitations (part B, number 3) to generate the 65% cut pellet. From this
partially purified intermediate, you will try two different column chromatography
steps to see which gives better results. You will assess your purification results by
measuring how much LDH vs. how much contaminating protein is present at each
step of the purification. This will be done in Labs 6 and 7. The diagram below outlines
the order of purification steps and where samples need to be taken.
LDH Purification Scheme
Crude Homogenate
20,000 ×g
Centrifugation
Clarified Homogenate
Ammonium Sulfate
Precipitations
65% Cut Pellet
Affinity
Chromatography
Size Exclusion
Chromatography
LDH
LDH
©Hayden-McNeil, LLC
45
LAB 3 • Initial Purification of Lactate Dehydrogenase: Centrifugation and Ammonium Sulfate Precipitation
D. Procedure
1. Pre-chill one 250 mL centrifuge bottle in your ice bucket. Once it is cold, get
100 ml of unknown crude homogenate from your TA. This was prepared by
homogenizing the tissue in 50 mM phosphate buffer at pH 7.5. It is very important
to keep the homogenate and all resulting purification intermediates and samples
cold. This will keep the activity of proteases to a minimum and prevent your
LDH from being degraded. Have a bucket of ice ready ahead of time. For all of
the procedures you will perform over the next four labs, always think about how
you can keep your LDH material as cold as possible.
2. Centrifuge at 20,000 ×g for 15 minutes at 4°C. When the bottles are loaded
into the centrifuge rotor, they must be balanced so that each bottle has the
same weight as the one directly across from it. For this first centrifugation, you
should balance your bottle against that of another lab group. This will help get
everyone through this first step more quickly. For all the other centrifugation
steps in today’s lab, you will balance against a water blank that you will prepare.
Safety Note: The TAs are the only ones who will start or stop the centrifuges.
Do not open the centrifuge without a TA present.
3. After the centrifuge is done, gently carry your bottle back to your bench so as
not to disturb the pellet. The supernatant is your clarified homogenate, your
first purification intermediate. Carefully decant the supernatant into a graduated cylinder to measure the volume, then pour the supernatant into a 150 ml
beaker with a stir bar (prechilled). Write the volume in your lab notebook; you
will need this number to calculate enzyme activity units in Lab 6. Also, take a
0.5 ml sample of it, place it in a well-labeled microfuge tube, and freeze it down
for analysis in Lab 6.
4. You will now do a 40% ammonium sulfate cut, precipitating proteins by adding
ammonium sulfate to a concentration of 40% of saturated. This concentration will
precipitate non-LDH proteins. Weigh out 0.242 grams of ammonium sulfate per
ml of your clarified homogenate. Place your 150 ml beaker inside a plastic dish
(located in your locker bin box) that contains a small amount of ice water, then
place the dish onto a stir plate. With rapid stirring, add the ammonium sulfate
to the beaker slowly, over the course of 2 to 3 minutes. (Why is it important to
add the ammonium sulfate slowly?) Keep stirring until all of the ammonium
sulfate is dissolved, and then stir slowly on ice for 15 minutes.
5. Centrifuge in a chilled 250 ml bottle at 20,000 ×g for 15 minutes. Carefully
decant the supernatant and measure its volume. Remove the pellet using a bottle
brush (located at sink), and discard.
46
LAB 3 • Initial Purification of Lactate Dehydrogenase: Centrifugation and Ammonium Sulfate Precipitation
6. Increase the ammonium sulfate concentration to 65% of saturated (a 65% cut)
by adding another 0.166 grams per ml, to the volume of the supernatant you
obtained in step 5. Add it as before, with rapid stirring on ice, and then stir slowly
for 15 minutes. This cut will precipitate the LDH.
7.
Centrifuge at 20,000 ×g for 15 minutes. Carefully remove the supernatant, and
rest the centrifuge bottle upside down on a Kimwipe for a minute to let the
residual supernatant drain away. Save a 0.5 ml sample of the supernatant in a
clean microfuge tube and label it “65% cut supernatant.” Keep the sample on
ice. Measure the volume of the remaining supernatant and record this in your
lab notebook, then discard the supernatant.
8. Resuspend the pellet in 4 ml of cold phosphate buffer, pH 7.5. Use a 5 ml graduated plastic pipette. The pellet will be streaked up along the side of the centrifuge
bottle. It will require some work to get the entire pellet into solution. You can
use the tip of the pipette to break up the pellet and scrape it down from the side
of the bottle while you are pipetting the buffer up and down (be careful not to
suck the solution up into the Pi-Pump). The resuspended pellet is your next
purification intermediate, the 65% cut pellet. Measure your final volume, and
record it in your lab notebook. Take a 200 μl sample, place it in a well-labeled
microfuge tube, and freeze it down for Lab 6.
9.
Place the resuspended 65% cut pellet into a new 15 mL plastic centrifuge tube.
Label the tube and store this in the refrigerator until the next lab.
10. After you have completed your initial LDH purification, prepare your size exclusion chromatography column that you will use in Lab 5. The instructions for
doing this are on page 66, and your TA will demonstrate the procedure.
End-of-Lab Checklist:
❏ Clarified homogenate, 65% cut pellet, and 65% cut supernatant
samples in freezer
❏ LDH prep in 15 mL centrifuge tube in refrigerator
❏ Return stir bar
47
LAB 3 • Initial Purification of Lactate Dehydrogenase: Centrifugation and Ammonium Sulfate Precipitation
48
LAB 4 • Affinity Chromatography Purification of LDH
4
Shutterstock.com
LABORATORY
Affinity Chromatography
Purification of LDH
In this lab, you will take part of your 65% cut pellet from Lab 3 and further purify the
LDH by affinity chromatography. Chromatography separates compounds via their
interaction with an immobilized (insoluble) substrate. In column chromatography,
the immobilized substrate is in the form of tiny beads composed of cross-linked
polysaccharides such as dextran, agarose, or cellulose. These beads are sometimes
referred to as the column resin. Other molecules can be linked to the beads to
facilitate the separation, as in ion exchange and affinity chromatography, or the
separation can occur from interaction with the beads themselves, as in size exclusion chromatography.
In column chromatography, a stream of buffer is fed into the top of the column,
and flows down through the resin by gravity until it comes out the bottom. The
sample is loaded into the top of the column with the buffer. The column output can
be collected into a series of tubes, each collecting a small volume. This way, when
the compound being purified elutes (flows from the column), it is contained away
from all of the contaminating molecules from which it has been separated. This is
referred to as collecting fractions of the column output and is done to maximize
the degree of purification achieved.
49
LAB 4 • Affinity Chromatography Purification of LDH
The three major types of column chromatography are:
1. Size exclusion chromatography—separates compounds by size or MW
2. Ion exchange chromatography—separates compounds by charge
3. Affinity chromatography—separates compounds based on their interaction
with a specific “ligand” or binding partner
You will perform size exclusion chromatography as your final LDH purification step
in Lab 5, and you will do a brief tutorial on ion exchange chromatography in Lab 8.
A. Introduction to Affinity Chromatography
Affinity chromatography is widely used in biochemistry and molecular biology, as it
provides a fast and easy way to achieve a high degree of purification for a compound
or protein. Separation in affinity chromatography is based on the specific binding of
the protein being purified to a binding partner, or ligand, that is covalently linked to
the beads. This binding is considered to be an affinity interaction because it is based
on the shape of the molecular structure of both the protein of interest and the ligand.
An affinity interaction has both high specificity and high affinity. High specificity
means the ligand will bind only to the protein of interest or to a group of proteins
with similar structure. High affinity means that there is much strength in the binding
interaction and that the two binding partners will stick together at low concentrations and not come apart easily. Binding affinity can be quantified by measuring
the dissociation constant (Kd), which is the molar concentration at which there is
half-maximal binding of a ligand to its receptor or binding partner. Some examples
of affinity interactions are the binding of an antibody to its antigen, the binding of
a substrate molecule to catalytic binding pocket of an enzyme, and the binding of a
growth factor or hormone molecule to its cell-surface receptor protein. The specificity and strength of affinity interactions are what make affinity chromatography so
efficient at separating out a target protein. You can think of it as a way of reaching
in to a crude mix of proteins and pulling out only the protein you are interested in.
Once the protein or compound of interest is bound to its ligand on the resin, it can
be removed from the column by adding a high concentration of salt (NaCl), which
disrupts charge attractions that participate in the affinity interaction. This can be
done with a single, high concentration of NaCl, or with a gradual increase in NaCl
concentration that is referred to as a gradient. A NaCl gradient produces a further
degree of purification because different proteins will elute at different NaCl concentrations. To produce a gradient, two electric pumps mix changing amounts of buffer
from two reservoirs (one with NaCl and one without NaCl) to create a steady, linear
increase in the NaCl concentration of the buffer going into the top of the column. By
collecting fractions during the NaCl gradient, proteins that were initially bound to
50
LAB 4 • Affinity Chromatography Purification of LDH
the column can be separated by differences in their affinity for the ligand. Proteins
that have a lower affinity for the ligand (lower binding strength) will be displaced
from the resin at a lower NaCl concentration, whereas proteins that bind the ligand
more tightly require a high NaCl concentration to disrupt the binding.
Another way to elute the bound protein is by adding a high concentration of a competitor molecule that can compete for binding to the ligand. The competitor usually
has a similar structure to the compound of interest, or the part of the protein that
binds the ligand. The competitor, therefore, also binds to the ligand with high specificity and high affinity, and at high enough concentrations can displace the bound
target protein. The competitor can also be added in a gradient to separate bound
proteins by differences in ligand affinity.
Group-Specific Ligands
Some ligands will bind to a family of different proteins that share a common characteristic or structural feature. Affinity resins that have common enzyme cofactors
linked to the beads are often used to purify enzymes based on their ability to bind
these cofactors.
TABLE 4.1 Group-Specific Affinity Resins.
Ligand
Group Specificity
Concanavalin A-sepharose
Glycosylated proteins
5'-AMP-agarose
Enzymes with NAD+ cofactors,
ATP-dependent kinases
Cibacron Blue-sepharose
Enzymes with nucleotide cofactors
Ni-NTA-agarose
Proteins with histidine tags
Protein-A-agarose
IgG-type antibodies
Poly(U)-agarose
Nucleic acids containing poly(A)
sequences
Affinity resins with the ligand Cibacron Blue are frequently used to purify enzymes
with nucleotide cofactors, including enzymes that utilize NAD+. Cibacron Blue is
a dye used in the textile industry, but this compound was found to have a strong
affinity for the NAD+, and ATP binding sites of enzymes. Cibacron Blue fits into
the active site cleft on LDH in a way that mimics the binding of NAD+. We will use
Cibacron Blue-agarose to further purify our LDH.
51
LAB 4 • Affinity Chromatography Purification of LDH
FIGURE 4.1 Cibacron Blue 3GA.
O
Agarose
NH2
SO3H
N
O
NH
NH
N
N
SO3H
NH
A
HO3S
Measurement of Total Protein Eluting from the Column
It is helpful to know how much protein is coming out of the bottom of the column
at different times during the chromatography run. Most of this protein will be contaminants that are being removed by the purification, but at some point this will
include the target protein. This can be monitored by measuring the absorbance at
280 nm of the buffer flowing from the column. All proteins have an absorption peak
at 280 nm due to the absorption of light at this wavelength by the aromatic side
chains of tryptophan, tyrosine, and phenylalanine amino acids. The flow from the
column can be directed into a small UV spectrophotometer to continually monitor the column output. If the absorbance is plotted vs. increasing column elution
volume (the amount of buffer that has flowed from the bottom of the column from
the time of loading the sample), the peaks in the resulting graph (or UV trace) will
show when different proteins are eluting from the column. This is referred to as a
chromatogram. Elution volume here is analogous to time.
FIGURE 4.2 A chromatogram.
A280
Relative enzyme activity
Load
sample
Washes
Elution in NaCl gradient
Elution volume
©Hayden-McNeil, LLC
In an affinity chromatography run, the UV trace can be used to indicate when the
column has been washed sufficiently to remove all of the unwanted proteins that
do not bind to the resin. After these proteins have all been removed, the protein of
interest can be eluted. Individual fractions collected during the elution can be assayed
for activity of the target enzyme, and this information can be superimposed on the
52
LAB 4 • Affinity Chromatography Purification of LDH
UV trace to correlate the elution of the enzyme to a particular protein peak. We do
not have spectrophotometers that we can connect to our column output available
in the lab, but you will use the spectrophotometers available in the lab to measure
the absorbance at 280 nm from periodic sampling of your column output at known
elution volumes. By doing this, you will be able to create a chromatogram for your
affinity column run.
B. Procedure
Important: Read through the entire procedure before you begin. You should
have a well-labeled table in your lab notebook that is ready for you to record
your absorbance readings.
Preparing the Cibacron Blue-Agarose Column
Safety Note: The stock bottles of the chromatography resins (Labs 4 and
5) contain the compound thimerosal at a concentration of 0.005%. While
this concentration is relatively low (for reference, some vaccines contain
0.001–0.01%), internalization of significant amounts through oral or dermal
routes can cause mercury poisoning. Gloves are required while handling the
resin stocks and initial column flow-through.
1. The columns you will use in this experiment are 1 cm diameter by 10 cm length.
Pour 12 ml of a 1:1 slurry into the column to obtain 6 ml of packed beads. See
page 66 of Lab 5 for instructions on pouring the column. Remember to keep
the slurry well mixed while you are pipetting it by swirling the flask. After all
12 ml of slurry have been added, let the beads settle for a few minutes before you
continue. In affinity chromatography, it is not as critical that the beads are well
packed as it is in size exclusion chromatography, so we can pour the column on
the same day we run it.
2. Drain the storage buffer from the column until the meniscus is at the level of
the packed beads. It is critical that you do not let the buffer level in the column
drop below the level of the packed beads. If the beads run dry, air bubbles may be
introduced that will impede the flow of the column. Collect the storage buffer so
that you can dispose of it in the thimerosal waste carboy on the middle bench.
3. You need to equilibrate the column before you load your sample onto it. This
will flush out the remaining thimerosal and any salts that may have been in the
storage buffer. Flow 12 ml (two “column volumes” of 6 ml each) of phosphate
buffer through the column and stop the flow once the buffer meniscus reaches
the level of the beads. Collect the entire 12 ml of column output in a beaker and
dispose of this in the thimerosal waste carboy. The column is now ready for the
sample.
53
LAB 4 • Affinity Chromatography Purification of LDH
Preparing the Sample
To save time, perform these steps while you are pouring and equilibrating
the column.
1. Retrieve your 65% cut pellet. Take 3 ml and transfer it to a 50 ml centrifuge tube.
Place the centrifuge tube on ice. Be sure that you have at least 2 ml of 65% cut
pellet remaining after transferring the 3 ml; if not, consult your IA.
2. Some material may have precipitated in your sample over the last few days in the
refrigerator. This could clog the column, so as a precaution you will centrifuge
the 65% cut pellet to remove any unwanted precipitate. Prepare a balance tube
of the same weight and centrifuge at 5,000 RPM for 5 minutes at 4°C.
3. Carefully remove the supernatant, leaving any solid material behind, and temporarily place it in a new 5 mL Falcon tube. Discard the pellet if there is one.
4. Get 20 microfuge tubes and number them sequentially 1 through 20. Mark the
1 ml level on each tube.
Running the Column
Get a 10 ml graduated cylinder and place it under the column to collect the output.
From this point on you will need to measure the volume of buffer flowing from the
column. By doing this, you will know the elution volume (Ve) that corresponds to
each absorbance (A280) reading you will take. Ve is simply the volume of buffer that
has flowed from the column at any given point during the run. Ve starts at zero and
begins accumulating as soon as your LDH sample enters the column. When the
graduated cylinder fills to 10 ml, empty it and place it back under the column, and
record a plus 10 ml tally in your lab notebook.
1. Warm your 65% cut pellet to room temperature just before loading it onto the
column.
2. Carefully load your 65% cut pellet onto the column without disrupting the beads.
3. You will now begin running the column and taking periodic A280 measurements
of your column output. Blank your spectrophotometer with phosphate buffer.
Open the stopcock valve to begin the column flow and collect the first ml in a
plastic cuvette by letting the column drip directly into the cuvette. The narrow
part of the cuvette holds exactly 1 ml. Stop the column while you measure A280.
This should be recorded in your lab notebook as A280 at 1 ml Ve. After measuring
A280, empty the contents of the cuvette into your graduated cylinder under the
column. Rinse the cuvette with dH2O so that it will be ready for your next A280
reading. This first A280 reading should be close to zero and establishes a baseline
for protein peaks.
54
LAB 4 • Affinity Chromatography Purification of LDH
Keep in mind that you can start and stop the column as often as you need to.
4. Take two more A280 readings for the second and third ml of elution (Ve = 2 ml
and Ve = 3 ml). You will see the A280 readings rise quickly. Be sure to return the
sample from the cuvette to your graduated cylinder so that you can accurately
keep track of Ve. Stop the column with the meniscus at the level of the beads
once all of the 65% cut pellet has entered the column.
5. You will now wash the column by adding phosphate buffer. This will remove all
of the residual proteins that are not binding to the beads. Continue to track Ve,
and take an A280 reading at every third ml of the wash. When all of the unbound
proteins have been removed, A280 will be close to baseline. Continue washing
until A280 is less than 0.15 or until there is an hour and a half remaining in the
lab session.
6. You will now elute the bound proteins by adding 1 M NaCl to the column.
(What other proteins might be bound to the Cibacron Blue resin besides LDH?)
Remove the graduated cylinder from under the column. From this point, you
will collect 1 ml fractions into the sequentially numbered tubes. Fill the column
with 1 M NaCl in phosphate buffer. It is okay if you still have some phosphate
buffer without NaCl in the column.
7.
Take an A 280 reading on every second fraction. You can stop the column to
catch up if you need to. To take the reading, pour the 1 ml from the tube into
the cuvette, measure the absorbance, and then return that fraction to its numbered microfuge tube. Place all the fractions on ice once you have measured the
absorbance. Be sure to rinse the cuvette well between readings.
8. Continue collecting fractions until A280 drops to baseline (less than 0.15). Add
more phosphate buffer plus 1 M NaCl as needed. Watch your time. It requires
45 minutes to complete the lab from the time you are done collecting fractions,
so you should proceed to the next step at this point even if your A 280 reading is
not yet at baseline.
9.
Your LDH should now be off of the column and in your fraction tubes. Flow
two column volumes of phosphate buffer with 2 M NaCl through the column to
strip away any remaining proteins from the resin (collect this output in a beaker
and dispose). You can begin your spot tests for LDH activity while you are doing
this. When you are done, recycle the resin into the flask on the middle bench,
using the vinyl hose and a water bottle as shown by your TA.
10. Before the next lab session, plot the chromatogram of your affinity chromatography run and fix this in your lab notebook. The y-axis is A280 nm. The x-axis
is a single, continuous elution volume (Ve) for the entire run, including loading
the sample, washing, and elution.
55
LAB 4 • Affinity Chromatography Purification of LDH
Preparing the Column
Pour 12 ml well-mixed slurry into
column, settle and drain buffer.
Flow through goes in
hazardous waste container.
Flow through 12 ml phosphate buffer.
Flow through goes in
hazardous waste container.
STOP when buffer meniscus
reaches top of beads.
Preparing the Sample
Centrifuge 65% cut pellet 5 minutes.
Discard pellet.
Decant supernatant to clean test tube.
Running the Column
Load supernatant from 65% cut pellet
onto column.
Run column and collect output in
10 ml graduated cylinder.
Measure A280 at 1 ml, 2 ml,
and 3 ml. Return to cylinder
to track elution volume.
Do not let the liquid in the column
drop below where the resin begins.
Wash the column with phosphate buffer
to remove the unbound proteins.
Measure A280 of every
third ml. Return to cylinder
to track elution volume.
Wash the column until A280 has
returned to baseline, or until
1.5 hours remain in the lab period.
56
LAB 4 • Affinity Chromatography Purification of LDH
Eluting the Sample
Flow phosphate buffer + 1M NaCl
through column. Collect 1 ml fractions
in individual microfuge tubes.
Measure A280 of every other
fraction. Return samples to
microfuge tube.
STOP when A280 < 0.1 OR
when 45 minutes remain
in lab period.
Wash column with 12 ml
phosphate buffer + 2M NaCl,
recycle resin into flask at
front of room.
Spot test on all fractions.
Spot Test for LDH Enzyme Activity
Now your job is to determine which fraction tubes contain your LDH. To do this
you will use a quick (but not very quantitative) assay for LDH catalytic activity called
the “spot test.” This assay is done in a 96-well plate, using one well per fraction tube.
The background for this assay will be covered in detail in Lab 6. Briefly, the reduction of NAD+ to NADH by LDH leads to the subsequent reduction of the compound
nitro-blue tetrazolium (NBT). Reduction of NBT causes a shift in its absorption so
that it has a peak (λmax) at 610 nm and becomes blue in color. Therefore, the appearance of blue color in a well indicates that LDH catalytic activity is present in the
corresponding fraction tube.
1. Prepare a “master mix” that contains all the reagents you will need for the assay
except for the chromatography fractions. Using a master mix is a much quicker
way to prepare multiple samples that have the same components and cuts down
on pipetting errors.
Reagent
Volume for
One Reaction
Volume for
Ten Reactions
25 mM lactate in
100 mM CAPS
buffer
90 μl
900 μl
10 mM NAD+
10 μl
100 μl
2 mM NBT*
2.5 μl
25 μl
2 mM PMS*
0.65 μl
6.5 μl
Total volume
(rounded)
100 μl
1 ml
Volume for
X Reactions
*Safety Note: The NBT and PMS should be considered toxic. Wear gloves
when handling them, and treat waste as hazardous.
57
LAB 4 • Affinity Chromatography Purification of LDH
You will need to do one reaction for each fraction tube you have to assay, as well as
a positive and negative control. Prepare enough master mix for this many reactions,
plus enough for two extra reactions to adjust for pipetting error and make certain
you have enough (number of fraction tubes + 4 = X reactions).
2. Get a 96-well plate. Well #1 will be the positive control, and well #2 will be the
negative control, followed by consecutive wells for each fraction tube. Note in
your lab notebook what sample will go in each well.
3. Pipette 100 μl of the master mix into each well that you will use.
4. Add 2 μl of the original 65% cut pellet to the positive control well. Mix by stirring briefly with the pipette tip. Nothing will be added to the negative control
well. The positive control should turn bright blue within a minute or two.
5. You will now add your fraction samples to the wells. The reaction is time dependent, so to find the fractions that have significant LDH activity you must look
for blue color appearing in the wells within a short amount of time from adding
the sample. Very small, insignificant amounts of LDH will produce blue color
after a long period of time. Be sure to change pipette tips between each fraction.
What could happen if you didn’t change tips?
Add 2 μl from each fraction tube to the appropriate well. Start timing 5 minutes
from the time you add the first fraction to the first well. Keep track of how long
it takes you to add the 2 μl to all of the wells. Score the wells for LDH activity
after 5 minutes (add some extra time for the wells that received the 2 μl last).
LDH activity is indicated by the appearance of a strong blue color.
6. Take a picture of your spot-test plate to print and paste into your lab notebook.
7.
Combine all of the fractions that have significant LDH activity (the column
fractions, not the 96-well plate wells). Measure the combined fraction volume
and then place it in a 50 ml screw-cap tube. This is your affinity-purified LDH.
Keep it on ice or in the refrigerator.
8. Aspirate all spot test wells and any leftover master mix into the waste flask on
the middle bench. Rinse the spot test plate well with DI water and return to the
plate bin box.
End-of-Lab Checklist:
❏ 65% cut pellet back in refrigerator
❏ Affinity-purified LDH in refrigerator
58
LAB 4 • Affinity Chromatography Purification of LDH
C. Supplementary Information
Quantifying Binding Affinity by the Dissociation Constant
High-specificity, high-affinity interactions between molecules are central to the function of biochemistry and molecular biology. This can be a small molecule binding to a
protein (e.g., oxygen binding to hemoglobin) or a protein binding to another protein.
One of the molecules is called the “ligand” and the other is called the “receptor.” In
the case of a small molecule–protein interaction, the small molecule is always the
ligand and the protein is the receptor (therefore, oxygen is a ligand that binds to hemoglobin). In the case of two soluble proteins that bind to each other, the designation
of ligand and receptor is arbitrary. When one of the binding pair is not in solution
it is designated the receptor. This is the case for cell membrane proteins; all protein
hormone and growth factor receptors are cell membrane proteins. The number of
receptors present on the cell surface represent a given number of “binding sites” for
the ligand protein. The concentration of the ligand in solution outside the cell can
vary, determining how many of the binding sites are occupied. For convenience, in
affinity chromatography we refer to the protein that we are trying to purify as the
“protein” and its binding partner on the beads as the “ligand.” However, it is more
accurate to consider the protein of interest as the ligand (since it is free in solution),
and binding molecule linked to the beads as the receptor.
Because of the prevalence of affinity interactions in biochemistry, it is useful to have
a way of quantifying the strength of these interactions. We can consider the binding of a ligand (L) to its receptor (R) to form the ligand-receptor complex (LR) as a
reversible chemical reaction.
L+R
k1
k –1
LR
In this reaction, k1 is the rate constant (a value inherent for any chemical reaction that
describes how fast that reaction proceeds for any given concentration of reactant)
for the ligand coming together with the receptor, and k–1 is the rate constant for the
dissociation (coming apart) of the ligand from the receptor. k1 is called a second-order
rate constant because there are two reactants (L and R) involved in the reaction, and
it has units of M–1 sec–1. k–1 is called a first order rate constant because there is only
a single reactant (the LR complex), and has units of sec–1.
It is easier to describe the binding strength, or affinity, in terms of the dissociation
of ligand and receptor, or the reverse of the reaction shown above. The dissociation
constant (Kd) is the equilibrium constant for dissociation reaction:
Kd =
[L] [R]
= k –1
[LR]
k1
59
LAB 4 • Affinity Chromatography Purification of LDH
where the concentrations of ligand, receptor, and ligand–receptor complex are the
molar concentrations when the reaction has reached equilibrium. The equilibrium
constant is merely the ratio of products to reactants once a chemical reaction has
reached equilibrium, and no longer continues in either the forward or reverse
directions. Kd has units of molar concentration. The weaker the binding of a ligand
to a receptor, the higher the concentrations of free ligand and receptor will be at
equilibrium and the higher the ratio of products to reactant, or Kd. Conversely, if a
ligand binds to a receptor very tightly, the ratio of free ligand and receptor over the
bound complex of the two will be very low, giving a low Kd. Therefore, the higher the
affinity of a ligand–receptor interaction, the smaller the value of Kd.
It is also useful to think of ligand binding in terms of the fraction of the total amount
of receptor that is bound by ligand. This fraction is dependent on the concentration
of ligand in solution and is referred to as θ (theta).
i=
binding sites occupied
[LR]
=
[LR] + [R]
total binding sites
This fraction can also be thought of as the number of binding sites for the ligand that
are occupied (bound) at that ligand concentration over the total number of binding
sites available, which is equivalent to the total amount of receptor protein present.
θ can be determined experimentally over increasing ligand concentrations:
FIGURE 4.3
1.0
©Hayden-McNeil, LLC
θ 0.5
Kd
[L] (ligand concentration)
60
LAB 4 • Affinity Chromatography Purification of LDH
There is a hyperbolic relationship between the fraction of the receptor that is bound
(θ), and the concentration of the ligand, with an asymptote at θ = 1. Starting from the
definition of Kd based on equilibrium concentrations of free ligand and receptor over
receptor–ligand complex, the equation can be rearranged and solved in terms of θ:
Kd =
[L] [R]
[LR]
[LR] =
[L] [R]
Kd
[L] [R]
[LR]
Kd
=
i=
[LR] + [R]
[L] [R]
+ [R]
Kd
Multiply each term by K d
[R]
i=
[L]
.
[L] + K d
This is the equation for the hyperbolic relationship between θ and [L], and it demonstrates that Kd is equal to the molar concentration of ligand at which there is onehalf maximal binding to the receptor (when [L] = K d, θ = 0.5).* In other words, half
of all the receptors present have been bound by the ligand, when the concentration
of the ligand reaches the value of Kd. Therefore, in a high affinity interaction, only
a very small concentration of ligand is required to bind a significant fraction of the
receptors present. If you consider this in terms of a protein hormone binding to its
cell-surface receptor, the Kd for that binding gives an idea of the concentration of
the hormone required to produce a biological response in target cells that have that
receptor. The Kd is the molar concentration of the hormone at which one half of the
binding sites on the target tissue, or one half of all the receptors, have been bound.
Indeed, the extracellular concentrations of most hormones are typically close to the
Kd for binding the receptors on their target cells.
* This assumes that for L + R <--------> LR, the total concentration of L in solution is not significantly
changed by receptor binding. In most experimental systems, the concentration of L far exceeds the fixed
number of binding sites.
61
LAB 4 • Affinity Chromatography Purification of LDH
TABLE 4.2 Dissociation Constants for Some Known Affinity Interactions.
Ligand
Receptor
Kd
Insulin
Insulin receptor
1 × 10 –10 M
Fibroblast growth factor,
FGF-2
High-affinity tyrosine kinase cell-surface
receptor
2 × 10 –11 M
Fibroblast growth factor,
FGF-2
Low-affinity proteoglycan
cell-surface receptor
2 × 10 –9 M
Anti-HIV antibody
(human)
HIV surface protein gp41
4 × 10 –10 M
Ca2+
Calmodulin
(a calcium-binding signal
transduction protein)
3 × 10 –6 to 2 × 10 –5 M
Affinity interactions occur over a range of dissociation constants, and the terms
“high-affinity” and “low-affinity” can be relative to a particular system or class of
interactions. In general, any binding that occurs with a Kd of less than 10 –8 M or so
is considered high-affinity. Interactions that have a much higher Kd (10 –7 to 10 –3 M)
can still be very specific and have a significant affinity, and thus are still considered
affinity interactions. Antibody-antigen interactions typically occur in a Kd range of
10–12 to 10–8 M. Most enzyme-substrate interactions are somewhat lower affinity, with
a Kd range of 10 –7 to 10 –3 M. The protein, avidin (a component of egg white), binds
to the small molecule, biotin, with such a high affinity (10 –15 M) that the interaction
is considered irreversible.
62
LAB 5 • Size Exclusion Chromatography Purification of LDH
5
Shutterstock.com
LABORATORY
Size Exclusion Chromatography
Purification of LDH
You will now further purify the LDH in your 65% cut pellet by size exclusion chromatography. You will compare the results from this purification to those from the
affinity chromatography.
A. Introduction to Size Exclusion Chromatography
Size exclusion chromatography is another of the three major types of column chromatography, including ion exchange and affinity chromatography. Size exclusion
chromatography separates molecules by size, or molecular weight, and is also called
gel filtration or molecular sieve chromatography. In size exclusion chromatography,
the resin consists of spherical beads made from cross-linked dextran polymers. The
trade name for this type of resin is Sephadex. The beads are not solid, but have a
sponge-like interior with pores that permit access to the interior space. Molecules
that are larger than the pore diameters cannot enter the beads and will pass quickly
through the column, whereas smaller molecules will enter the beads and be slowed
in their movement through the column.
63
LAB 5 • Size Exclusion Chromatography Purification of LDH
FIGURE 5.1 Size exclusion chromatography.
Small protein
Large protein
il,
c Ne
©Hayden-M
C
LL
There are different grades of Sephadex that have different amounts of cross-linking
in the dextran beads and, therefore, different average pore diameters. The more
cross-linking there is in the dextran polymer, the smaller the average diameter of
the pores in the bead, and vice versa. The grade of Sephadex we are using is G100.
The fractionation range of G100 is approximately 4,000–150,000 daltons. Molecules larger than 150 kDa are over what is called the exclusion limit and will not
enter the pores to any extent. All molecules larger than the exclusion limit will elute
from the column “first” and together, in what is called the void volume. Molecules
smaller than 4 kDa will enter the beads to the maximum extent. Their movement
through the column will be retarded by the same amount, so they will elute “last”
and together. Molecules between these two sizes (those within the fractionation
range of the beads) will enter the beads to different extents, based on their molecular
weight, and be slowed accordingly. These molecules will elute in different fractions
from the bottom of the column. The exclusion values are only approximate because
they don’t always reflect the actual diameter (referred to as the Stokes radius) of the
molecule. For example, a globular molecule could enter pores that exclude a linear
molecule of the same molecular weight. Different grades of Sephadex have different
fractionation ranges.
64
LAB 5 • Size Exclusion Chromatography Purification of LDH
TABLE 5.1 Fractionation Ranges of Sephadex.
Grade of Sephadex
Fractionation Range
(daltons)
Volume of Hydrated Beads
(ml/g dry beads)
G10
0–700
2–3
G15
0–1,500
2.5–3.5
G25
1,000–5,000
4–6
G50
1,500–30,000
9–11
G75
3,000–80,000
12–15
G100
4,000–150,000
15–20
G150
5,000–300,000
20–30
G200
5,000–600,000
30–40
Obviously, G10 is not useful for separating proteins, but is often used for removing
salts and other small molecules from protein solutions. It can also be used for changing the buffer composition and can be used as alternatives to dialysis.
In the column, the volume outside the beads is called the void volume, Vo. The
volume contained inside the beads is Vi , and the total volume of the column is Vt
whereas Vm is the volume occupied by the dextran polymer (matrix) itself. VT = Vo
+ Vi + Vm. (Note that this total volume represents the entire volume of the column,
not just the volume that is accessible to liquids. This other volume is more useful for
calculations, and is represented as Vt.)
To calibrate a column, the void volume is determined by loading the column with a
high molecular weight dye like Blue Dextran (MW 2 × 106) and measuring the volume
of buffer that comes off until just before the dye begins to elute. This volume is equal
to Vo and is also equal to the elution volume (Ve) for Blue Dextran. The total volume
of the column is determined by loading a low molecular weight dye like Orange-G
(MW 452.38) along with the Blue Dextran. (Both of these calibration dyes are typically loaded at the same time along with the sample to be separated.) The elution
volume for Orange-G is determined in the same way as the elution volume for Blue
Dextran was determined—by measuring the volume of the buffer that comes off of
the column until just before the dye begins to elute. The elution volume for OrangeG is equal to the total volume of the column. The inner volume can be determined
by subtracting the Ve of Blue Dex (which is the same as Vo) from the Ve of Orange-G
(which is the same as Vt), since Vt = Vi + Vo.
As you run your column, you will notice that the volume of the fractions containing
the blue color is smaller than the volume of the fractions containing the orange color.
There are two main reasons for this difference. One is that there was more time for
diffusion of Orange-G molecules to occur, since they remain in the column longer
than the Blue Dextran molecules. The second reason is that the entrance and exit
65
LAB 5 • Size Exclusion Chromatography Purification of LDH
of Orange-G from the pores in the beads is random, and some molecules will exit
from the pores sooner than others; therefore, there will be a spread of average path
lengths with some molecules spending more time inside the beads than others, even
though they are all the same size.
Although most molecules separate according to their molecular size, aromatic
substances (those with phenolic rings, of which Orange-G is an example) have
some affinity for Sephadex and therefore migrate more slowly through the column.
Another factor that can cause molecules to migrate more slowly is the ionic strength
of the elution buffer. There are a small number of carboxyl groups on the Sephadex
matrix; therefore, if the concentration of cations in the buffer is very low, molecules
with positive charges can be retarded. This problem can be eliminated by increasing
the salt concentration.
Other factors will also influence the degree of separation. These include:
1. The volume of the sample placed on the column. The larger the volume, the
longer it takes for all of the sample to enter the column, and the more spread
out all of the proteins will be. Always load your sample in as small a volume as
possible.
2. The length of the column. Longer columns give better separation.
3. The speed of elution. The faster the column is flowing, the less chance there
is for material on the column to separate. A slower flow rate will give better
separation.
B. Procedure for Size Exclusion Chromatography
Preparing the Sephadex Column
The chromatography column you will use for this experiment is a 1 cm diameter
by 30 cm length glass cylinder that holds the resin, or beads. You will need to fill
the column with 20 ml of packed Sephadex G100. This is done on a day before you
actually run the column, so that the beads have time to sufficiently settle and pack
down in the column. The dry Sephadex was previously allowed to hydrate and
swell in pH 7.5 phosphate buffer + 0.005% thimerosal (an antimicrobial agent). The
mixture of hydrated Sephadex and buffer is called a slurry. In a 1:1 slurry there are
equal volumes of swollen Sephadex and buffer. Therefore, to obtain 20 ml of packed
beads, you need to pour 40 ml of slurry into the column. Your TA will demonstrate
the different parts of the glass chromatography column and how to assemble it.
66
LAB 5 • Size Exclusion Chromatography Purification of LDH
Safety Note: The stock bottles of the chromatography resins (Labs 4 and
5) contain the compound thimerosal at a concentration of 0.005%. While
this concentration is relatively low (for reference, some vaccines contain
0.001–0.01%), internalization of significant amounts through oral or dermal
routes can cause mercury poisoning. Gloves are required while handling the
resin stocks and initial column flow-through.
1. The stopcock is already attached to the bottom of the column, and it does not
need to be further tightened onto the column—overtightening the stopcock onto
the column may crack the plastic and cause the column to leak. If the column
leaks during the week before Lab 5, it will need to be re-poured just before use,
which may affect your results. Also, it is recommended that you always hold the
stopcock with one hand while opening/closing the valve with the other hand.
This prevents the stopcock from loosening (and thus leaking). Note that the blue
cap has a small plug on top that seals the intake spout—be sure to keep track of
all the pieces.
2. Mount the column on the ring stand. Be sure it is vertical.
3. Pipette 20 ml of water into the column. Mark the top of the water level.
4. Test the column before adding the Sephadex: Open the stopcock and let the water
begin to drip from the column. Sometimes the filter in the bottom of the column
dries out and prevents water from flowing. If this is the case with your column,
gently tap the top opening of the column with your palm. The air pressure will
force the water into the filter, and once it is wet, the column should flow. If the
column still won’t run, check with your IA.
5. Now test the column for leaks: Fill the column back to the top with water. Dry
off the stopcock with a Kimwipe, and check it for leaks. Monitor the column for
5–10 minutes. If it leaks, check if the stopcock has worked loose. The stopcocks
occasionally go bad, and they can be easily replaced. It is better to take the time
now to check for leaks, than to find out a week later that your column has run
dry! If you cannot stop a leak, check with your IA. Once the column is leak-free,
dump out the water.
6. With a plastic transfer pipette, add about a ml of phosphate buffer into the column. Open the stopcock to let the liquid flow through the filter at the bottom
of the column. Check to be sure there is no air trapped at the bottom.
7.
Measure out 40 ml of 1:1 Sephadex-phosphate buffer slurry into a 50 ml graduated cylinder using the provided pipette. (Dedicated Sephadex cylinders will
be provided—do not use your group cylinder for Sephadex!) Swirl the flask of
Sephadex slurry well before pipetting. The beads settle out quickly, and care
must be taken to keep them suspended in a 1:1 slurry while transferring them.
67
LAB 5 • Size Exclusion Chromatography Purification of LDH
8. Cover the graduated cylinder with Parafilm and invert it a few times to resuspend
the beads; in a single pour add as much of the slurry as the column will hold.
Make sure there are no air bubbles.
9.
When the column is full of slurry, place a beaker under the column and slowly
open the stopcock to let excess buffer flow through. As the liquid level falls, add
more slurry to the top of the column.
Important: Do not drain all the buffer from the slurry. As the beads settle,
you want to have a layer of clear buffer on top. As soon as you have added
all of the slurry to the column, close the stopcock. Let the beads settle until
you can see a few inches of clear buffer on top.
10. Drain some more of the buffer. You want to end up with two or three inches of
clear buffer on top of the settled beads. You can add phosphate buffer (without
beads) to the top of the column if you need to.
11. Seal the column tightly for storage until Lab 5. Close the top of the column with
the cap, and make sure the intake spout in the cap is sealed with the plug. Use
tape to label both the column and the cylinder with your section and group
number.
12. The buffer you collected while pouring the column contains thimerosal. Dispose
of it in the thimerosal waste carboy on the middle bench.
Preparing the Sample
1. Retrieve your 65% cut pellet. Transfer 700 μl to a pre-chilled microfuge tube
and place on ice. This is what you will run on the size exclusion column.
2. Centrifuge the size exclusion sample in your benchtop microfuge for 3 minutes.
This will remove any material which may have precipitated and prevent it from
clogging the column. Transfer the supernatant to a clean tube and place it on
ice. You may not see a pellet.
3. Add 200 μl of Blue Dextran and 100 μl of Orange-G to the size exclusion sample.
Mix well by stirring with the pipette tip. This will give you a total sample volume
of 1 ml to load onto the size exclusion column. This is about the maximum volume you would want to load onto the column. Why wouldn’t you want to load
a larger volume?
Running the Column
1. Get 20 microfuge tubes and number them sequentially 1 to 20.
2. Run the clear buffer on top of the packed beads, plus an additional 25 ml of
phosphate buffer through the column.
68
LAB 5 • Size Exclusion Chromatography Purification of LDH
Important: Stop the buffer flow just as the meniscus of the buffer reaches
the level of the packed beads. In all column chromatography, you must
never let the level of buffer drop below the level of the packed beads or air
bubbles will be introduced into the column that will impede its flow. The key
here is to load your sample in as small a volume as possible without letting
the column run dry.
3. Because the resin contains thimerosal as a preservative, all of the column output,
up until you begin collecting fractions, should be disposed of in the thimerosal
waste carboy on the middle bench.
4. Use the narrow plastic transfer pipettes, provided on the front bench, to load your
sample on to the column. Gently pipette the sample with the Blue Dextran and
Orange-G directly onto the top of the packed beads without disrupting them.
Once your sample is on the column, open the stopcock and let the column slowly
flow and drain into a beaker (no fraction tubes yet). Stop the column when the
meniscus is once again at the level of the beads. Gently pipette 1 ml of buffer
onto the beads, and then run it through; this will get all of your sample into the
column. Now you can fill the top of the column with buffer and allow it to run
continually, at a rate of one drop every 10 to 15 seconds. The slower the flow rate
of the column, the better your separation will be. As the column runs, you will
see the Blue Dextran and Orange-G separating in the column. Refill the top of
the column with buffer as necessary.
5. When the Blue Dextran is 3 inches from the bottom of the column, start collecting 1 ml fractions, sequentially, in your numbered test tubes. Place the filled
test tubes on ice.
6. Continue collecting fractions until all of the color, including the Orange-G, has
eluted from the column. While the column is running, you can begin preparing
the spot test for LDH activity that you will need to do on each fraction tube.
7.
When you are done collecting fractions, flush the column with 20 ml of deionized water to remove any residual material from the beads. Recycle the beads
into the recycling flask on the middle bench, using the vinyl hose as you did with
the affinity agarose in Lab 4.
Spot Test for LDH Activity
Your LDH is in your fraction tubes, at an elution volume (Ve) somewhere between
that of the Blue Dextran and the Orange-G. Conduct a spot test to determine which
fractions contain LDH activity. See the instructions on pages 57–58 for how to
perform the spot test. You will need enough master mix to assay each fraction tube,
plus two controls, plus two extra for pipetting error. Use the 65% cut pellet as the
positive control again. Take a picture of your spot test results and add this to your
lab notebook before the next lab session.
69
LAB 5 • Size Exclusion Chromatography Purification of LDH
Determine which wells produce significant blue color after five minutes. Whether to
include wells that turn slightly blue is a judgement call. The more wells you combine,
the better your recovery of LDH will be, but the greater the chance of including
contaminating proteins. Conversely, if you only take strongly blue wells you may
getter better purification, but you will recover less LDH. Combine your LDH positive fractions and transfer them to a 50 ml screw-cap tube. Measure the volume and
record this in your lab notebook. Keep on ice or in the refrigerator.
End-of-Lab Checklist:
❏ 65% cut pellet back in refrigerator
❏ Size exclusion-purified LDH in refrigerator
70
LAB 6 • Determination of LDH Activity Units and Specific Activity
LABORATORY
6
Determination of LDH Activity
Units and Specific Activity
You will now have a chance to go back and carefully analyze the samples from each
of the purification steps you have performed over the last few labs. What you want
to know is, a) how much LDH enzyme is present at each purification step? and, b)
how pure (free of contaminating proteins and other cellular materials) is the LDH
at that step?
In order to quantify how much of an enzyme is present, rather than determining the
moles or mass of the enzyme in a sample, it is more desirable to find out how much
catalytic activity is present for that enzyme. This way, damaged or inactive enzyme
molecules will not throw off the determination of how much functional enzyme you
have. The catalytic activity is referred to as enzyme activity and is analogous to the
reaction rate for the chemical reaction catalyzed by that enzyme. Therefore, enzyme
activity is defined as the amount of product of the catalyzed reaction produced per
unit time, and has units of μmol/min.
To assess how pure the LDH is at each purification step, we will determine the total
amount of protein present in that step. The vast majority of the protein present at
the early purification steps is not LDH. Therefore, you expect to see the amount of
total protein decrease in the column chromatography purifications. LDH may make
up the majority of the total protein present in the column chromatography products.
You will have to assess this. To do this, you will also need the information provided
71
LAB 6 • Determination of LDH Activity Units and Specific Activity
by the SDS-PAGE gel you will run in Lab 7. After measuring the LDH activity and
protein concentration for each purification step, you will sum up your data in a
purification table.
A. Samples and Dilutions
Gather all of your LDH samples from the freezer and refrigerator and place them
on ice. Use the 65% cut pellet from the refrigerator, and leave the frozen sample in
the freezer. You will only use the freezer sample if your refrigerator sample appears
to have very low LDH activity. The table below indicates which assays will be run
on each sample, and if the sample needs to be diluted before running the assay. The
activity assay and Bradford assay are only accurate within a limited concentration
range for LDH and total protein. Using the dilutions provided in the table will give
you the best chance for having both assays work on the first try. However, because
each LDH preparation is different, you may need adjust the dilution of one or more
of your samples to obtain accurate results. If you change the dilution for one assay
(activity or Bradford), the original dilution may still be best for the other. Remember
that they are measuring very different things.
Sample
Dilution
Assays
Clarified homogenate (CH)
1:10
LDH activity and Bradford
65% cut supernatant (65S)
none
LDH activity only
65% cut pellet (65P)
1:10
LDH activity and Bradford
Affinity-purified LDH (A)
none
LDH activity and Bradford
Size exclusion-purified LDH (S)
none
LDH activity and Bradford
1. If your samples were frozen, thaw them by warming with your fingers. Gently
mix them by inverting the tubes and then spin them with a quick pulse in your
benchtop microcentrifuge to get all the liquid back to the bottom of the tube.
2. If you are preparing a dilution, make 100 μl of the diluted sample in phosphate
buffer in a clean microfuge tube. Keep the diluted sample tubes on ice.
3. If you are not preparing a dilution of that sample, you may still want to transfer
100 μl to a clean microfuge tube to use for performing the assays. This will keep
your original sample safe and decrease the chances of contaminating it.
B. Background on Measuring LDH Activity
The LDH activity assay you will be performing today is based on the conversion of
NAD+ to NADH. You will set up assay reactions with high concentrations of lactate
and NAD+ to force the reaction to proceed in this direction (see page 40). Since
NADH is a product of the reaction, LDH activity can be quantified as μmols of
NADH produced per minute. The rate of NADH formation is measured by spectrophotometry. The reduction of NAD+ to NADH changes its absorption spectrum,
giving NADH a strong peak at 340 nm that is not present for NAD+.
72
LAB 6 • Determination of LDH Activity Units and Specific Activity
To measure LDH activity, the increase in absorbance at 340 nm will be measured
over time. This is referred to as a kinetic assay, in which the reaction rate is used to
accurately determine the enzyme activity units present. Note that it is important
to catch the linear part of the curve for increasing absorbance. This linear increase
corresponds to the initial reaction rate for the production of NADH. As the reaction
approaches equilibrium, the reaction rate will slow, and the increase in absorbance
will become nonlinear. Therefore, the increase in absorbance from the nonlinear
part of the curve does not accurately reflect the amount of LDH enzyme present.
The absorbance values could be taken on an ordinary spectrophotometer by taking
readings at regular intervals (e.g., every 10 seconds) and recording the data by hand.
However, spectrophotometers are available in the lab that can be programmed to
do this for you and give you a printout with a list of the time and absorbance values.
FIGURE 6.1 Conversion of NAD+ to NADH.
H
O
H
C
NH2
O
CH2
O
P
O
O
O
P
OH
O−
O
O
HO
N
R
NADH
NH2
N
CH2
NH2
N
NAD+
HO
O
C
+ 2e− + H+
N+
O−
H
N
N
OH
1.0
NAD+
0.6
NADH
0.4
©Hayden-McNeil, LLC
Absorbance
0.8
0.2
0
220
240
260
280
300
320
340
360
380
Wavelength (nm)
73
LAB 6 • Determination of LDH Activity Units and Specific Activity
How the Spot Test for LDH Activity Works
You used the spot test as a rapid way of analyzing the amount of LDH activity in
your chromatography fractions in Labs 4 and 5. The spot test is also based on the
conversion of NAD+ to NADH, but in this case the NADH goes on to transfer 2 e– to
a second electron acceptor, phenazine methosulfate (PMS). The reduced PMS then
donates the electrons to a third and final electron acceptor, nitro-blue tetrazolium
(NBT). The reduced form of NBT absorbs strongly at 610 nm, giving it a blue color.
This type of assay is referred to as a cycling assay because the NAD+/NADH and
PMS cycle back and forth between their oxidized and reduced forms, driven by a
continuous influx of electrons from lactate. Since the appearance of blue color is not
measured over time, the assay is not quantitative and is only good for a qualitative
determination of whether LDH activity is present.
Lactate
NAD+
PMSred
NBTox
Pyruvate
NADH
PMSox
NBTred
Blue color
C. Procedure for LDH Activity Assay
The activity assays are set up and run in cuvettes, so that you can measure the increase
in absorbance at 340 nm. Each cuvette will contain 90 mM CAPS buffer, pH 10, 22.5
mM lactate, and 1 mM NAD+, to which a small amount of the LDH sample will be
added. The timing for starting and reading the assay must be exact, so read through
the entire procedure before beginning.
1. Prepare a master mix by making a 1:10 dilution of the 10 mM NAD+ into the
100 mM CAPS/25 mM lactate stock. You will need 1 ml of master mix for each
assay, and you have five samples to run, plus a blank. Some of your samples will
most likely need to be run more than once to get a good absorbance curve, so
prepare enough master mix for 10 assays. Note that the master mix should be
room temperature when you begin running the assays.
2. Get five UV transmissible cuvettes (these have a red dot), mark one for a blank,
and fill each with 1 ml of master mix. You will need to reuse the cuvettes to run
all your assays. After running an assay, empty the cuvette and thoroughly rinse
with water; it is critical to remove any residual LDH activity. The other materials
you will need are: P-10 pipetter, 200 ul tips, and small Parafilm squares.
3. Use the spectrophotometer at your group’s work station to run the activity assay.
Let your IA know when you are ready to begin; they will show you how to run
the assay. Place your blank in the instrument. Use the diluted samples you prepared in part A, and run one sample at a time. Quickly but carefully, add 2 μl of
sample to the cuvette, cover the top with Parafilm, invert twice to mix, place it
in the instrument, and start the spectrophotometer.
74
LAB 6 • Determination of LDH Activity Units and Specific Activity
4. The instrument will give you a plot of absorbance vs. time and the tabulated
numbers. Decide whether you can measure the initial reaction rate and get a valid
ΔA/min from the assay. To be sure you are catching the initial reaction rate, the
absorbance at t = 0 seconds should be less than or equal to 0.1. If the absorbance
rises too quickly, repeat the assay with a greater dilution of that sample (less
enzyme gives a slower reaction rate). If the absorbance doesn’t increase much,
repeat the assay with a smaller dilution of that sample (more enzyme gives a
faster reaction rate).
5. Perform a “good” assay for each of your five purification samples, and obtain a
printout of the data points for each. You may use the same blank each for each
sample. Be sure to rinse the cuvette very well with dH2O before using it for
another assay. Any residual LDH activity sticking to the cuvette will throw off
your data. You may need to adjust the dilution of one or more of your samples
to obtain an accurate ΔA/min. Do this by increasing or decreasing the dilution
of the sample, and continue to add 2 μl for each assay. When you are done, rinse
the cuvettes well and return them to the box on the front bench (these will be
used by other sections, unlike the other cuvettes we use).
D. Calculating LDH Activity Units
Enzyme activity is defined as the amount of product from the catalyzed reaction, in
micromoles, produced per minute. Therefore, one LDH activity unit (U) corresponds
to the production of one micromole of NADH in one minute (1 U LDH = 1 μmol
NADH/min).
1. First determine the change in absorbance that occurred over the one minute
time period for each assay. This is the ΔA/minute. It is most easily calculated
by subtracting the absorbance (A) at zero seconds from the A at 60 seconds.
A more accurate method is to plot A vs. time (in seconds) in MS Excel. Add a
linear trendline and display the equation for the line. The slope is ΔA/second.
Multiply this by 60 to get ΔA/min. Determining ΔA/min by subtracting initial
from final absorbance is sufficiently accurate under most circumstances. If any
of your final LDH unit values seem problematic, you can recalculate them using
ΔA/min from the trendline slope.
2. To determine how many LDH activity units you have purified, first convert your
ΔA/min value to the corresponding change in molar concentration of NADH,
using Beer’s law (A = εlc). The molar extinction coefficient (ε) for NADH is
6220 M–1cm–1 at 340 nm.
3. The volume in your cuvette (for the reaction that produced this change in molar
concentration) was 1 ml, so multiply by this volume in liters to determine the
number of moles of NADH produced in one minute. Then, simply convert this
value to micromoles (what factor do you multiply by?), and you have the number
of LDH units in your cuvette.
75
LAB 6 • Determination of LDH Activity Units and Specific Activity
4. What you really want to know is how many units of LDH you had at a particular
step of the purification. Next, calculate the units per ml for that purification step.
This is the relative activity (U/ml). To do this, you need to know what volume
was added to your assay. This is the small volume of the LDH intermediate that
you added to the cuvette to start the reaction. The number of units you calculated in step 2 all came from this small volume of that intermediate. Therefore,
divide by this volume in ml to get U/ml. For example, if you added 2 μl, divide
by 0.002 ml. Don’t forget to multiply by your dilution factor if you diluted your
LDH sample for the activity assay.
5. To calculate the total LDH activity units for that purification intermediate,
multiply your relative activity (U/ml) by the total volume of that intermediate
(ml).
E. Background on the Determination of Total Protein Concentration
One method for determining protein concentration is to measure the absorbance
of the protein sample at 280 nm. The ring structures in the side chains of the amino
acids tryptophan, tyrosine, and phenylalanine absorb strongly at this wavelength.
Individual proteins have set extinction coefficients at 280 nm (ε280), based on the
number of these amino acids that they contain. If the amino acid sequence of the
protein is known, the ε280 can be calculated using bioinformatics programs that are
available on the internet. Absorbance at 280 nm, however, is not useful for determining the total protein concentration in a sample of mixed proteins. There is no single
ε280 and many of the proteins in the sample may be unknown.
For protein mixtures, or for pure proteins whose extinction coefficients are not
known, several assays are available that produce colored reaction products detected
by visible light absorption. None of these are perfect, so the choice of which one to
use depends on a comparison of advantages and disadvantages. Criteria for choosing an assay include: sensitivity, accuracy, and precision, as well as how fast and
easy it is to perform. Another important criterion is whether the assay is sensitive
to interfering substances that may be present in the protein sample (e.g., detergents,
reducing agents, acids or bases, etc., used in the purification process). The most common assay used to measure total protein concentration is the Bradford assay, and
this is the assay you will use for your LDH purification. The Bradford is faster and
does not require the sample to be heated, making it preferable to some of the other
colorimetric methods for protein determination. This assay is based on the change
in absorbance of the dye Coomassie Brilliant Blue G when it binds to proteins. The
dye is reddish (absorbs 465 nm blue light) when no protein is present, and turns blue
(absorbs 595 nm orange light) when it binds to proteins.
76
LAB 6 • Determination of LDH Activity Units and Specific Activity
FIGURE 6.2 Coomassie Brilliant Blue G.
CH2CH3
N+
CH2
SO3−
CH3
CH3
C
CH3CH2O
NH
N
CH2
CH2CH3
SO2Na
FIGURE 6.3 Determination of the presence of protein using the Bradford assay.
+ Protein
©Hayden-McNeil, LLC
Absorbance
− Protein
340
455
570
685
800
Wavelength (nm)
At low pH, the dye carries a negative charge and binds to arginine side chains, and
more weakly to lysine and histidine (the other two amino acids with positively charged
side chains at low pH). The dye also binds to tryptophan, tyrosine, and phenylalanine
(amino acids with aromatic side chains) via hydrophobic interactions and Van der
Waals forces. Coomassie Blue must be used in an acidic, low pH solution because
at higher pH the dye will deprotonate and the absorption max of the dye when it is
not bound to protein will shift to 595 nm. Therefore, the Bradford assay cannot be
used when the protein is in an alkaline solution. Also, some detergents, such as SDS,
will interfere with the binding of Coomassie Blue to protein.
In order to interpret the results of an assay like the Bradford assay, we need to
include a standard protein. The readings that we get on the spectrophotometer will
be impossible to translate into protein concentrations without a known amount of
a standard protein. A commonly used standard protein is bovine serum albumin
(BSA) because it is inexpensive and readily available in pure preparations.
77
LAB 6 • Determination of LDH Activity Units and Specific Activity
F. Bradford Protein Assay and Calculation of Specific Activity
The 5× stock solution of the dye we are using (Bio-Rad 500-006) contains 25% methanol, 25% dH2O, and 50% phosphoric acid. Note: all samples with Bradford reagent
must be collected in the aspirator flask on the middle bench. It will be diluted for
you to 1× with dH2O before lab (the diluted solution is stable for at least two weeks
at room temperature). You will use BSA as the standard. It is prepared as a 1 μg/μl
solution in dH2O.
1. Get 15 clean microfuge tubes. Label them as indicated on the following tables.
Add standards, samples, and dH2O as indicated in the tables. Use the sample
dilutions you prepared in section A. Note: you will not run the Bradford assay
on the 65% cut supernatant.
TABLE 6.1 BSA Standards.
Tube Label
MQ Water
BSA Standard
BSA0
100 μl
0 μl
BSA1
99 μl
1 μl
BSA2
98 μl
2 μl
BSA4
96 μl
4 μl
BSA6
94 μl
6 μl
BSA8
92 μl
8 μl
BSA10
90 μl
10 μl
μg Protein in Tube?
TABLE 6.2 LDH Purification Samples.
78
Tube Label
MQ Water
LDH Sample
CH2
98 μl
2 μl clarified homogenate (1:10)
CH10
90 μl
10 μl clarified homogenate (1:10)
65P2
98 μl
2 μl 65% cut pellet (1:10)
65P10
90 μl
10 μl 65% cut pellet (1:10)
A2
98 μl
2 μl affinity purified LDH
A10
90 μl
10 μl affinity purified LDH
S2
98 μl
2 μl size exclusion purified LDH
S10
90 μl
10 μl size exclusion purified LDH
LAB 6 • Determination of LDH Activity Units and Specific Activity
2. Collect your aliquot of Bradford reagent from the middle bench using the 25 ml
cylinder and 50 ml glass beakers provided. DO NOT use your plastic beakers
& cylinders for Bradford—they will stain blue! Add 900 μl of Bradford reagent
to each tube. Note that the methanol in the Bradford reagent makes it very
volatile—pipette 450 μl × 2 to avoid aspirating the reagent into the barrel of the
micropipette. Vortex the tubes to mix, and let them sit for a couple of minutes.
Do not place them on ice.
3. Take four of the non-UV transmissible cuvettes. Blank the spec using the BSA0
sample, and measure the absorbance of each sample at 595 nm. After taking the
absorbance reading for a sample, transfer it back to the original tube (in case
you need to read it again). Measure your BSA standard samples first, going from
lowest protein conc./absorbance to highest. This way you will not need to wash
the cuvettes between the BSA standards. Wash the cuvettes well before starting
your LDH purification samples and between the pairs of LDH samples.
4. For each pair (2 μl and 10 μl) of purification intermediate samples, you need to
obtain at least one absorbance value that falls inside the range of absorbance
values obtained from your BSA standards. You may need to adjust the dilution
of the sample to do this (adjust the dilution of the sample and continue to add
2 μl and 10 μl per assay).
5. After you have measured your samples, aspirate all tubes that have Bradford
reagent into the Bradford aspirator flask on the middle bench. Place empty
cuvettes in the white lab trash buckets. Rinse your 50 mL Bradford beaker and
return it to the bin box.
6. Make a standard curve with μg BSA (not μg/ml; why?) on the x-axis and A595 on
the y-axis. Do not include the BSA0 values in your plot (forcing your standard
curve to go through the origin will make it less accurate). Add a best-fit straight
line through your points—this is a linear trend-line in MS Excel.
7.
Using your standard curve, determine the μg protein in each of your LDH
unknowns. This is the mass of protein in the cuvette when you took the absorbance reading. For each sample choose the one volume (i.e., CH2 or CH10) that
has an absorbance value closest to center of the curve. The reason that you used
two different volumes of each LDH sample was to increase the chance that one
of them will be at a concentration at which an accurate reading can be taken
(i.e., in or close to the absorbance range of the BSA standards).
8. Calculate the concentration of total protein in each LDH purification step, in
mg/ml. To do this, determine the μg protein in the cuvette using your standard
curve. Then divide by the number of μl of sample added to the cuvette (e.g., if A10
had 20 μg of protein, then the concentration was 2 μg/μl). Convert this to mg/
ml. (What factor do you multiply by to convert μg/μl to mg/ml?) Next, multiply
by the dilution factor, if it was diluted, to get the actual protein concentration.
79
LAB 6 • Determination of LDH Activity Units and Specific Activity
You can then determine the total amount of protein present in that step of the
purification by multiplying by the volume of that intermediate; this is the same
volume you multiplied by in section D, step 5.
9.
Calculate the specific activity for your purification steps. Specific activity is
defined as units of enzyme activity per mg total protein (U/mg), and it is the best
indicator of how pure an enzyme sample is. Divide the total amount of activity
at that step (U) by the total amount of protein present (mg). Would you expect
the specific activity of the clarified homogenate to be higher or lower than the
other purification steps?
End-of-Lab Checklist:
❏ Save all your original samples and return them to the freezer or refrigerator
❏ Dispose of Bradford assay waste in Bradford aspirator flask
G. Purification Table
Now that you have quantified the LDH activity and total protein from each purification step, you will summarize this information in a purification table. This will
allow you to critique the effectiveness of each purification technique you performed.
Effectiveness is determined by the recovery of LDH by that purification step; you want
to lose as little LDH as possible in the purification process. Effectiveness is also determined by the degree of purification achieved. The more total protein removed and
the better the subsequent increase in the specific activity, the better the technique.
In a traditional purification table, the enzyme activity, total protein, and specific
activity values are provided for each intermediate, where each purification step was
performed sequentially. Since the idea is to maximize the recovery and purity of
the target enzyme, percent yield and fold purification values are also given to track
changes through each subsequent step relative to the amount and purity of the
enzyme in the starting material.
You performed an exploratory LDH purification, however, where the goal is to determine which combination of purification techniques, and in which order, would be
the most effective if you wanted to design a strategy to purify a large quantity of very
pure LDH. Therefore, your purification table will be somewhat different and will be
read in a different way. The two column chromatography steps were not consecutive
but rather both started with the 65% cut pellet. You are also including the LDH activity in the 65% cut supernatant, not because this was the product of the ammonium
sulfate precipitations, but rather to see how much LDH might have been lost in this
process. In your purification table you will also include relative activity (LDH U/ml)
and protein concentration (mg/ml) values. Because these values are dependent on
the volume of each purification intermediate, they do not provide useful information
80
LAB 6 • Determination of LDH Activity Units and Specific Activity
for critiquing the purification techniques used, and so they are not typically given
in a purification table. You are including them so that your IA can confirm you are
multiplying by the correct volumes when they grade your work.
Since it is crucial to know how much LDH was lost by a purification step in order
to understand how effective that step was, you will also calculate the Percent LDH
Recovery for each purification procedure you performed. This value tells you how
much of the LDH that went into that purification step was recovered. To calculate
this, divide the total activity units that came out of that step (this is the Total Activity value) by the number of units that went into the purification step. The number
of activity units that went into the step is not necessarily the total activity from the
previous step. For the affinity and size exclusion chromatography, only part of the
65% cut pellet, and therefore a fraction of the 65% cut pellet’s total activity, went into
each chromatography purification. Think about how you will need to calculate this.
Purification
Intermediate
Relative
Activity
(U/ml)
Total LDH
Activity
Units (U)
LDH
Recovery
by That
Step (%)
Protein
Concentration
(mg/ml)
Total
Protein
(mg)
Specific
Activity
(U/mg)
Clarified
Homogenate
65% Cut Pellet
65% Cut
Supernatant
AffinityPurified LDH
Size ExclusionPurified LDH
Fill in all of the blanks in the table and carefully examine how the values change.
Which proved to be a more effective technique, the affinity chromatography or the
size exclusion chromatography? Once you have assessed the effectiveness of each
purification technique, design a purification strategy that will maximize the yield
and purity of LDH. You can change the parameters or method used for any individual
technique. You can also change the order of the techniques used, remove a technique,
and even add a new technique that you have not performed. Just be sure that you
can provide a rationale for each step in your new purification strategy based on your
exploratory purification table. You also need to be sure that your new purification
strategy works out logistically in terms of what you know to be the requirements for
performing a particular purification technique. For instance, it would not be feasible
to perform size exclusion chromatography on the clarified homogenate because the
81
LAB 6 • Determination of LDH Activity Units and Specific Activity
volume is far too large. Keep in mind that the initial crude homogenate you started
with in /DELVQRWOLVWHGLQ\RXUSXUL¿FDWLRQWDEOHDQGWKHFODUL¿HGKRPRJHQDWHLV
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Note: A printed copy of your purification table will be required for your lab
notebook, in addition to submitting it with the written assignment for your
purification analysis.
82
LAB 7 • SDS-PAGE of LDH Purification Intermediates
7
Shutterstock.com
LABORATORY
SDS-PAGE of LDH Purification
Intermediates
A. Qualitative Analysis of LDH Purification Intermediates
In this lab, you will perform SDS-PAGE that will allow you to resolve LDH in your
purification intermediates from contaminating proteins of different MW. In Lab 6
you used the Bradford assay to quantify the total protein content of your purification
intermediates. In your first two intermediates (clarified homogenate and 65% cut
pellet) the vast majority of the protein present is contaminating, non-LDH protein.
The contribution of LDH to the total protein mass in these steps is negligible. In
your affinity-purified and size exclusion-purified LDH, however, LDH itself accounts
for a significant portion of the total protein content. It is not possible to distinguish
between contaminating protein mass and LDH mass with the Bradford assay, so how
can you really tell the purity of your LDH? The SDS-PAGE analysis of your different purification steps will give you a qualitative assessment of the purity. While not
quantitative like the Bradford assay, it will tell you how much of the protein mass is
LDH, and how much is contaminating protein of higher or lower MW.
You will run pre-stained size standards of different MWs, and a sample of LDH
standard to help you determine which band on your stained gel corresponds to LDH.
Remembering that LDH is a tetramer of four identical subunits with a total MW
of 146 kDa, at what MW position do you expect to find LDH after running the gel?
83
LAB 7 • SDS-PAGE of LDH Purification Intermediates
B. SDS-PAGE Procedure
You will use precast 10% polyacrylamide gels for your SDS-PAGE. The set-up will be
the same as you did in Lab 2. Since you will only use 4 lanes for your samples, you
will share a gel with the group next to you.
1. Make 600 ml of 1× tris-glycine running buffer from the 10× stock. Set up the gel
in the electrophoresis unit and add buffer to both the inner and outer chambers.
The buffer in the inner chamber should come almost to the top of the highest glass
plate to ensure the top of the gel will be in contact with the buffer. Slow leakage
from the inner chamber to the outer/lower chamber will sometimes cause the
buffer level to drop, so you will need to check this periodically during the run.
2. Clean out the wells by squirting buffer into each well with a Pasteur pipette.
3. Retrieve your clarified homogenate, 65% cut pellet, affinity-purified LDH, and
size exclusion-purified LDH samples. For the frozen samples, briefly vortex them
and then spin at full speed in your microcentrifuge for 2 minutes to pellet any
precipitated material. If any significant pellet is visible, transfer the supernatant
to a clean tube and discard the pellet. Place on ice.
4. Calculate the volume of each LDH purification sample that contains 20 μg of
protein; this is the mass of total protein that you want to load into each lane of
your gel. It is difficult to accurately pipette volumes less than 2 μl, so you should
make an appropriate dilution of your sample rather than pipetting volumes less
than this. Your sample will also contain 4 μl of 5× sample buffer and you want
to load the sample in a total volume of 20 μl. Calculate the volume of water you
need to add so that your purification sample plus water comes to 16 μl. Do not
add more than 20 μl to each well; add less total protein if your samples are too
dilute. Add your sample volumes and other information to your lab notebook,
using the same table format as on page 33.
Safety Note: The 5× sample buffer contains 25% 2-mercaptoethanol. This
compound is toxic and is hazardous by inhalation and skin contact. Wear
gloves and remove the glove immediately if you come in contact with the 5×
sample buffer. Add the 5× sample buffer to your samples in the fume hood,
being sure to use a new tip for each sample. It is okay to take your tubes out
of the hood once the 5× sample buffer has been diluted into your samples.
5. Label a clean microfuge tube in which to prepare each gel sample, then add the
water first, followed by the LDH purification sample.
6. Add the 4 μl of 5× sample buffer and stir gently with the pipette tip to mix. Do
not pipette up and down or vortex because the SDS in the sample buffer will
make a foamy mess.
7.
84
One of the two lab groups sharing a gel should prepare a sample of the LDH
standard. Add 5 μl LDH standard, 11 μl water, and 4 μl of 5× sample buffer.
Mix gently as above.
LAB 7 • SDS-PAGE of LDH Purification Intermediates
8. Heat the samples for 3 minutes in the 95°C water bath,, then let them cool to
room temperature (do not place them on ice). Place in the microcentrifuge with
the hinges pointing out and spin for 1 minute. If any precipitated material has
formed a pellet after the spin, be careful not to load this into the gel. Place the
pipette tip into the tube on the side away from the hinge when it is time to load.
9.
Identify the left-most well in your gel; this will be lane 1. Load your samples in
precisely this order: clar. homog., 65% cut, affinity, then size exclusion LDH for
the first group in lanes 1–4; 5 μl MW markers in lane 5; the LDH standard in
lane 6; clar. homog., 65% cut, affinity, and size exclusion LDH for the second
group in lanes 7–10.
10. Place the lid on the electrophoresis unit and run at 120 V. Run until the tracking dye is within 1 cm of the bottom of the gel. This will take about an hour
and 15 minutes.
11. When the gel run is complete, disassemble the apparatus and carefully separate
the plates holding the gel. Add the gel to 15 ml of rapid stain Coomassie blue
solution in a staining box.
12. Incubate the gel in the staining solution for 30 minutes with gentle agitation.
After 30 minutes, aspirate the staining solution into the waste flask on the middle
bench and add distilled water from the sink. Place the gel on the light box and
take a picture to document your results. Refer to Figure 2.6 for the MW marker
key. You should print the gel image and add it to your lab notebook for the next
lab session. Label each lane. It is okay to write in the lane labels with pen once
the image is in your notebook.
13. Rinse the staining box and return it to the bin box. Place the gel in the gel waste
bucket on the middle bench (NOT the lab trash bucket).
85
LAB 7 • SDS-PAGE of LDH Purification Intermediates
86
LAB 8 • Determination of LDH Isozymes by Electrophoresis
8
Robyn W/Shutterstock.com
LABORATORY
Determination of LDH Isozymes
by Electrophoresis
In this lab, you will finally determine which LDH isozymes you have purified. The
crude homogenate you began the purification with was made by homogenizing either
porcine heart tissue or porcine skeletal muscle tissue. Once you know what isozymes
you have isolated, you will know what your starting material was. Before you begin
this experiment, you must have calculated the relative activity (U/ml) for your LDH
purification samples. You need these values to know what volume of your LDH to
load into the electrophoresis gel.
The SDS-PAGE gels you have run separate proteins by MW. The LDH isozymes you
have purified are more or less identical in MW, but they have different isoelectric
points (pI) and, thus, different net charges at any given pH. Therefore, the different
isozymes will migrate at different rates in an electrophoresis gel, allowing you to
determine which one you have isolated. This electrophoresis experiment will use an
agarose gel rather than polyacrylamide. You will visualize your LDH isozyme bands
in the gel with a stain for LDH enzymatic activity. This stain works the same way as
the spot test you used in Labs 4 and 5.
A. Background on LDH Isozymes
An intact LDH enzyme molecule is a complex of four subunits that are bound
together by a combination of charge and hydrophobic interactions (this is referred
to as the quaternary level of protein structure). Each subunit consists of a single
87
LAB 8 • Determination of LDH Isozymes by Electrophoresis
polypeptide chain of 36.3 kDa. There are two different types of LDH subunits, the
H type (for heart) and the M type (for muscle). They are encoded on separate, nonallelic genes and have slightly different amino acid sequences (see the amino acid
sequence alignment in Table 8.2). Different combinations of these two subunits give
rise to five LDH isozymes. Different tissues produce different isozymes, controlled
by the differential expression of the genes for the two subunit types.
TABLE 8.1 LDH Isozymes.
LDH Isozyme
Subunit Composition
Tissue Expression
LDH-1
HHHH
Heart and red blood cells
LDH-2
HHHM
Heart and red blood cells
LDH-3
HHMM
Brain and kidney
LDH-4
HMMM
Skeletal muscle and liver
LDH-5
MMMM
Skeletal muscle and liver
TABLE 8.2 Alignment of Human M-type (top) and H-type (bottom) LDH Amino Acid
Sequences.
LDH Amino Acid Sequences
MATLKDQLIYNLLKEEQT-PQNKITVVGVGAVGMACAISILMKDLADELA
MATLKEKLIAPVAEEEATVPNNKITVVGVGQVGMACAISILGKSLADELA
49
50
LVDVIEDKLKGEMMDLQHGSLFLRTPKIVSGKDYNVTANSKLVIITAGAR
LVDVLEDKLKGEMMDLQHGSLFLQTPKIVADKDYSVTANSKIVVVTAGVR
99
100
QQEGESRLNLVQRNVNIFKFIIPNVVKYSPNCKLLIVSNPVDILTYVAWK
QQEGESRLNLVQRNVNVFKFIIPQIVKYSPDCIIIVVSNPVDILTYVTWK
149
150
ISGFPKNRVIGSGCNLDSARFRYLMGERLGVHPLSCHGWVLGEHGDSSVP
LSGLPKHRVIGSGCNLDSARFRYLMAEKLGIHPSSCHGWILGEHGDSSVA
199
200
VWSGMNVAGVSLKTLHPDLGTDKDKEQWKEVHKQVVESAYEVIKLKGYTS
VWSGVNVAGVSLQELNPEMGTDNDSENWKEVHKMVVESAYEVIKLKGYTN
249
250
WAIGLSVADLAESIMKNLRRVHPVSTMIKGLYGIKDDVFLSVPCILGQNG
WAIGLSVADLIESMLKNLSRIHPVSTMVKGMYGIENEVFLSLPCILNARG
299
300
ISDLVKVTLTSEEEARLKKSADTLWGIQKELQF–
LTSVINQKLKDDEVAQLKKSADTLWDIQKDLKDL
332
334
(See page 13 for a key to the single letter amino acid codes.)
88
aa#
LAB 8 • Determination of LDH Isozymes by Electrophoresis
The H and M subunits have different numbers of charged amino acids, giving each
LDH isozyme unique charge characteristics. It is this difference in charge that allows
LDH muscle and heart isozymes to be separated by electrophoresis. To understand
what determines the “net” charge on a protein at any given pH, it is necessary to
understand what produces the charge on individual amino acids in its polypeptide
chain.
The carboxyl groups in the side chains of the “acidic” amino acids (aspartic acid and
glutamic acid) are capable of dissociating an H+ ion. In the process of giving off the
H+ they go from being neutral to gaining a single negative charge.
RCOOH RCOO– + H+
The “basic” amino acids (arginine, histidine, and lysine) have side chains that contain
amino groups or ring structures that contain nitrogen atoms. These nitrogen atoms
can bind to an additional H+ and acquire a positive charge.
RNH2 + H+ RN+H3
Examine the structures of the charged amino acids on page 13. Each R-group has
its own pKa value that dictates when the H+ will be “on” or “off.” As the pH changes,
the concentration of H+ changes, and these amino acid R-groups will either bind H+
or dissociate H+, depending on their pKa values. The R-group structures on page 13
are shown at pH 7. Look at the pKa value and R-group for each of the five charged
amino acids. Is the amino acid shown in its proton donor or proton acceptor form?
How will the charge on the R-group change when it converts to the other form?
As the pH increases (H+ concentration drops) the carboxyl groups of the acidic amino
acids will go from being neutral to negative in charge as they dissociate their H+.
At the same time, the amino groups in the basic amino acids will go from having
a positive charge to being neutral. The net charge on a protein is the sum of all the
charges on its individual amino acid side chains. Moreover, this net charge is not
static, but changes as the pH of the protein’s environment changes. The N-terminal
α-amino group and the C-terminal α-carboxy group carry a single charge each (+
and –, respectively) that cancel each other out at physiologic pH. The important
points here are:
a.
For all proteins, as the pH increases the net charge on the protein will decrease
(get more negative).
b. At some pH value the number of positively charged amino acid side chains will
be equal to the number of negatively charged amino acid side chains, and the net
charge on the protein will be zero. This is the isoelectric point (pI). This value
is unique for every protein and is determined by the numbers of the different
acidic and basic amino acids it contains, and their individual pKa’s.
89
LAB 8 • Determination of LDH Isozymes by Electrophoresis
c.
As the pH drops below the pI of a protein, it becomes increasingly positive in
charge. As the pH rises above the pI of a protein, it becomes increasingly negative in charge.
The different numbers of acidic and basic amino acids in the M and H subunits give
each LDH isozyme a unique pI and, therefore, a slightly different charge at any given
pH. Before you do the experiment, look up the pI values for the individual M and H
subunits. Do part B before you come to lab.
B. Determine pI Values for Porcine Heart and Muscle LDH Isozymes
To interpret the pattern of bands you will get from your electrophoresis experiment
in this lab, you need to be able to predict which isozymes, muscle or heart, will
move more quickly in the gel. The migration rates of the isozymes are determined
by their isoelectric points, and where these values fall in relation to the pH of the
electrophoresis gel, which is 9.5. Follow the steps below to find the pI values for the
different isozymes.
TABLE 8.3 Sus scrofa domestica LDH.
Subunit
NCBI Accession number
Gene
H
AAA50438
LDH-B
M
AAA50436
LDH-A
1. First you will need to look up the amino acid sequences for the H and M subunits.
Go to the National Center for Biotechnology Information homepage (https://
www.ncbi.nlm.nih.gov/) and note the search field at the top. Click the box to the
left that says All Databases and select Protein from the dropdown menu. One
at a time, use the accession numbers from the table above to find the sequences.
2. The output will display the entry linked to that accession number in the NCBI
protein database. Click FASTA at the top left, just under the title. This gives you
just the amino acid sequence with a header.
3. Open a new browser page and go to the EXPASY proteomics Web site (https://
www.expasy.org/). Click on Resources A..Z and select ProtParam from the
alphabetized list. The ProtParam program will analyze the amino acid sequence
and then list several of the biochemical properties of that protein, including the
pI.
4. Copy and paste the amino acid sequence for one of the subunits, omitting the
header (copy all the single-letter codes starting on the second line) into the
search field and hit Compute Parameters. Repeat this for the other subunit.
5. Find the pI values for the H and M subunits and write them into your lab
notebook.
90
LAB 8 • Determination of LDH Isozymes by Electrophoresis
6. From the pI values of the subunits you can easily estimate the pI for each LDH
isozyme. The pI for each isozyme is the average of the four pI values for the
subunits composing the isozyme. Therefore, if the pI value for the H subunit
equals x and for the M subunit equals y, then the pI for the H4 isozyme is 4x/4,
the pI for the H3M isozyme is (3x + y)/4, etc. Write the pI value for each of the
5 LHD isozymes in your lab notebook.
Now that you know the isoelectric points of the heart- and muscle-type isozymes,
you can determine which isozymes will migrate more quickly toward the positive
electrode at the gel pH of 9.5. You will need to do this to identify your LDH unknown.
C. Electrophoresis Procedure
1. Make 300 ml of 1× Tris-Glycine running buffer from the 10× stock.
2. Prepare 50 ml of 1% agarose in 1× Tris-Glycine running buffer in a 250 ml flask.
(A 1% solution is 1 gram in 100 ml, so how much agarose powder will you weigh
out?)
3. Heat the flask in the microwave oven to get the agarose into solution. Watch
the flask as it is heating and as it just starts to boil, remove it and swirl. (Leather
gloves are provided for handling the hot flask.) Repeat until all of the granular
agarose crystals have disappeared.
4. Let the agarose cool for a few minutes as you prepare the electrophoresis unit.
Don’t let it sit for too long or it will solidify in the flask.
5. Prepare the electrophoresis unit:
Slide the baffles into the grooves next to the gel bed, handles to the outside. The
baffles should rest on the small rectangular baffle blocks and remain extended
above the edges of the gel bed.
Before pouring the gel, place the comb/bridge assembly over the gel bed. Loosen
the white screws holding the comb to the bridge, and adjust the comb depth.
Gently retighten the adjusting screws. The teeth of the comb should be just above
the bottom of the gel casting plate and should not touch the plate—otherwise
the samples will leak out when they are loaded in the wells. Also, the comb needs
to be at the end of the gel that is next to the negative electrode—decide which
end this is before you pour the gel.
Pour all of the agarose solution onto the gel bed. Quickly position the gel comb
at the desired location. Immediately after pouring, rinse the flask with water
to remove any residual agarose before it hardens onto the flask. Let the agarose
solidify for 25 minutes. It will turn slightly opaque when it is solid.
91
LAB 8 • Determination of LDH Isozymes by Electrophoresis
6. Prepare your samples to load on the gel while the agarose is cooling. You will
run four samples:
a. 65% cut pellet
b. Affinity or size exclusion-purified LDH, whichever has a larger relative
activity
c. LDH standard-1, and
d. LDH standard-2.
Get four clean microfuge tubes and label them accordingly.
7.
Determine the volume in μl that contains 1 LDH U for both of your LDH purification samples. Prepare a tube for each purification sample that has 1 U LDH
plus water to 16 μl. Do not add less than 2 μl or more than 16 μl of the LDH (add
from 2–16 μl based on the relative activity). This will ensure there is enough
LDH in your sample and that you will be able to load all of it into the gel. Add
4 μl of 5× native gel sample buffer, and mix gently by stirring with the pipette
tip. This should give a volume of 20 μl to load onto the gel.
The 5× sample buffer contains glycerol to make the sample denser so that it will
sink into the well when loading. It also contains a tracking dye, bromophenol
blue, which will allow you to monitor the progress of your electrophoresis.
8. Your TA will give you aliquots of LDH standard-1 and standard-2. Add 2 μl of
each standard to the appropriate tube. Add 14 μl of dH 2O, and then add 4 μl of
5× sample buffer. Mix gently by stirring.
9.
When the agarose gel has solidified, gently pull out the comb/bridge assembly.
Remove the baffles.
10. Fill the gel chamber with 1× Tris-Glycine running buffer until the buffer just
covers the top surface of the agarose gel. There should be about 2 or 3 mm of
buffer on top of the gel.
11. Load your samples into the wells, and write a key in your lab notebook so you
know what sample is in each lane. The activity stain you will use to visualize
the LDH gives large, diffuse bands in the gel, so leave two or three empty lanes
between each sample.
12. Put the lid on the chamber. Plug the leads into the power supply, and then turn
it on. Set the power supply to run at a constant voltage of 80 volts, and start the
current.
Safety Note: Always make sure the current is not running when you plug in
the leads or unplug the leads from the power supply. You can start and stop
the current as many times as you need to during the run without harming the
experiment. Never run an electrophoresis unit without the lid.
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LAB 8 • Determination of LDH Isozymes by Electrophoresis
13. Run the gel until the tracking dye is almost to the bottom of the gel. This will
take from 1½ to 2 hours.
14. When the gel is finished running, carefully remove it from the electrophoresis
unit and place it in a staining container. Pour 40 ml of the activity stain over
the gel and place it in the 37°C incubator for 20 minutes, or until the bands have
developed. The activity stain is the same mix of reagents used in the spot test
for LDH activity.
15. Remove the gel and cover it in plastic wrap so that there is a single layer of plastic
over both sides of the gel, and seal the edges by making several folds of the wrap.
Safety Note: The PMS and NBT in the staining solution should be considered
toxic. You will need to remove the gel from the activity stain with your hands,
so wear two pairs of gloves [a small pair beneath and a larger pair over these]
to avoid contact with these agents. Keep the activity stain contained and do
not splash it around the lab.
Dispose of the used activity stain in the waste aspirator flask on the middle
bench.
16. Place wrapped gel on the light box or on a white sheet of paper and look at your
bands. Which isozymes are LDH standard-1 and standard-2? Which isozymes
did you purify? How many individual isozymes (bands) can you see in each lane?
Your IA will take a picture and place the image file on Canvas. Print this, add
it to your lab notebook, and label the lanes.
17. Dispose of the gel (with cling wrap) in the gel waste bucket on the middle bench
when you are done; do not throw it in the trash.
End-of-Lab Checklist:
❏ Return all LDH purification samples to refrigerator
❏ Dispose of stained gel and staining solution in separate hazardous
waste containers
D. Supplementary Information
Ion Exchange Chromatography
In looking at your gels you may have noticed that some samples contained small
amounts of other isozymes that are not the LDH isozymes expressed at high levels
by that tissue. Skeletal muscle is known to produce small amounts of heart-type
LDH isozymes. Contamination with H4 and H3M isozymes can result from erythrocytes (red blood cells) that were in the muscle tissue when it was harvested. If you
needed to remove these contaminating isozymes, how could you do this? In general,
93
LAB 8 • Determination of LDH Isozymes by Electrophoresis
electrophoresis is not an efficient way to purify large amounts of a protein, and is
used more for analytical purposes.
Ion exchange chromatography is used to separate proteins based on differences
in their isoelectric point. It can be done on a large scale and is frequently used in
protein purifications. In an ion exchange resin, a charged chemical group such as
DEAE (diethylaminoethyl) is covalently attached to the beads.
(bead)–OCH2CH2N+(C2H5)2
DEAE sepharose is an ion exchange resin that has DEAE groups attached to sepharose beads. This resin is called an anion exchanger since the DEAE amino group
has a positive charge and will bind to negatively charged anions. Typically, a column
containing this material is prepared and the protein solution is passed through it.
Proteins with a net negative charge at the pH of the column buffer will tend to bind
to the DEAE. The stronger their charge, the more tightly they bind. Tightly bound
proteins can be released from the beads by changing the pH (to change the charge
on the protein) or increasing the salt concentration of the buffer (to add competing
anions). Salt and/or pH gradients are commonly used to separate proteins with ion
exchange columns. In order to decide what pH to use for the most effective binding
and elution of your protein, you need to know the pI of the protein.
Design a purification step to separate heart and muscle LDH isozymes using ion
exchange chromatography with DEAE sepharose. What buffer pH will you use to
load your sample onto the column? How will you elute the bound proteins?
E. Recipes for Native Gel Electrophoresis
1× Tris-Glycine Runner Buffer
• 20 mM tris
• 20 mM glycine
• 2 mM EDTA
• pH 9.5
5× Native Gel Sample Buffer
• 0.2 M tris
• 0.2 M glycine
• 10 mM EDTA
• 0.06% bromophenol blue
• 25% glycerol
• pH 8.8
LDH Activity Stain
• 0.1 M tris
• 1% lactate (0.11 M)
• 0.05% NAD+ (754 μM)
• 0.005% nitro-blue tetrazolium (NBT, 61.2 μM)
• 0.0005% phenazine methosulfate (PMS, 16.3 μM)
94
LAB 9A • Sea Urchin Fertilization: Preparation of Samples and Examination of Egg Activation and Cell Division
LABORATORY
9A
Sea Urchin Fertilization:
Preparation of Samples and
Examination of Egg Activation
and Cell Division
Over the next three labs, you will be investigating some of the cellular and molecular
events that occur during fertilization of the sea urchin egg. In this first lab, you will
be able to observe the “activation” of fertilized eggs under the microscope, followed
by the first cell division as the embryo begins to divide. In the next two labs you will
be looking at how the activity of the enzyme MAP kinase changes at fertilization,
and what the triggers are that cause this change in activity. To do this, you will perform an experiment called a Western blot. This involves separating proteins by size
with SDS-PAGE, and then transferring them out of the gel and onto a filter where
you will use antibodies to specifically detect MAP kinase. In addition, you may also
perform a fourth lab where you will look at the initial signal transduction events
that occur upon fertilization. For this you will use another technique that utilizes
antibodies, called an ELISA.
A. Introduction to Sea Urchin Fertilization
Sea urchins reproduce by casting large numbers of gametes out into the environment of the ocean. To ensure successful reproduction, sea urchins have evolved
mechanisms that allow the sperm and egg to find each other, and that guard against
95
LAB 9A • Sea Urchin Fertilization: Preparation of Samples and Examination of Egg Activation and Cell Division
multiple sperm fertilizing one egg (polyspermy). We will be using gametes from the
California purple sea urchin, Strongylocentrotus purpuratus. Depending on the time
of year, we may also have gametes from the white-tan urchin, Lytechinus pictus.
Sea urchin eggs are very large cells (about 0.1 mm in diameter) and are easily seen
with low magnification on a light microscope. The plasma membrane of the sea
urchin egg is surrounded by another membrane called the vitelline envelope, which
remains tightly tethered to the plasma membrane by proteins on the unfertilized
egg. The vitelline envelope is not a phospholipid bilayer like the plasma membrane,
but rather a lattice of interconnected proteins. Just inside the egg plasma membrane
are thousands of vesicles (small, membrane-bound sacks), about 1 μm in diameter,
called cortical granules. When a sperm binds to and fuses with the egg, it releases
materials into the egg cytoplasm that signal for protein and DNA synthesis to begin
in preparation for cell division. One of the earliest signal transduction events to occur
is the activation of an enzyme called phospholipase C. The activity of this enzyme
causes one of the better understood events of fertilization, an influx of calcium ions
from the endoplasmic reticulum to the cytoplasm. Calcium is an important second
messenger in cell signaling, in that it functions to transmit signals from one part of
the cell to another. Therefore, the calcium concentration is very low in the cytoplasm
of unfertilized eggs, about 0.1 μM, and temporarily increases to 1 μM upon sperm
binding to propagate the fertilization signal throughout the egg. Calcium is a direct
trigger for exocytosis of vesicles. At fertilization it causes the cortical granules to
fuse with the plasma membrane and release their contents into the space between
the membrane and the vitelline envelope.
FIGURE 9A.1 Unfertilized vs. activated sea urchin eggs.
Unfertilized egg
Activated egg
Nucleus
Cortical
granule
Plasma
membrane
Fertilization
envelope
LL
C
Vitelline
envelope
den
©Hay
il,
Ne
Mc
The cortical granules contain a protease that cuts the proteins tethering the vitelline envelope to the egg. They also release a muco-polysaccharide into the space
beneath the vitelline envelope that produces an osmotic gradient and causes water
to rush into this space. These events cause the vitelline envelope to rapidly expand
away from the egg surface. The inflated membrane that now surrounds the egg is
called the fertilization envelope. This is one of the ways in which polyspermy is
96
LAB 9A • Sea Urchin Fertilization: Preparation of Samples and Examination of Egg Activation and Cell Division
prevented. The cortical granules also contain another enzyme called a peroxidase.
Utilizing hydrogen peroxide that gets produced in the egg cytoplasm at fertilization,
the peroxidase cross-links proteins in the fertilization envelope. This hardens the
envelope into a protective shell that keeps the early embryo safe in the unpredictable
environment of the ocean.
You will have a chance to observe expansion of fertilization envelopes at fertilization
during today’s lab. The expansion of the fertilization envelope along with the other
immediate changes to the egg that occur at fertilization are termed egg activation.
Egg activation occurs within the first 1 to 2 minutes after fertilization. By 12 minutes
after fertilization, the sperm nucleus has migrated to the egg nucleus in the center of
the cell; by 40 minutes, DNA synthesis has begun; and by 180 minutes, the zygote has
undergone its first cell division or cleavage (for S. purpuratus). In today’s experiment,
you will produce an artificial calcium influx in the cytoplasm of unfertilized eggs
(no sperm) through the use of a reagent called a calcium ionophore (see section C).
This artificial calcium influx induces exocytosis of the cortical granules and other
fertilization events and produces egg activation without fertilization. An artificial
calcium influx will produce egg activation, but will this also lead to cell division in
the absence of fertilization? This is the first of three questions that you must answer
during this project, and you will be able to answer it with today’s experiment.
1. Is a cytoplasmic calcium influx sufficient to produce cell division in an unfertilized egg?
B. Introduction to MAP Kinase
MAP kinase is an enzyme that is well-known to play an important role in keeping
oocytes (eggs) in an arrested state until fertilization. In mammalian and frog eggs,
which arrest in metaphase II of meiosis, active MAP kinase keeps the eggs from
completing meiosis until after fertilization. Sea urchin eggs complete meiosis and
are kept in G0 (mitotic arrest) by the activity of MAP kinase.
MAP kinase is an acronym for mitogen-activated protein kinase, and this enzyme
is a protein kinase. Protein kinases transfer phosphate groups from ATP to serine,
threonine, or tyrosine amino acids that are in a specific recognition sequence of amino
acids on substrate, or recipient, proteins. This phosphorylation changes the activity
or function of these target proteins. MAP kinase phosphorylates a variety of target
proteins that carry out its downstream effects, which, in unfertilized sea urchin eggs,
is preventing the egg from beginning DNA synthesis and entering mitosis. MAP
kinase is itself regulated by another protein kinase called MEK. MEK recognizes a
conserved Thr-Glu-Tyr sequence on the MAP kinase polypeptide and phosphorylates
the threonine and tyrosine. In mammals these are Thr 202 and Tyr 204. MAP kinase
is completely inactive unless it is phosphorylated on these two amino acids. Once
phosphorylated, MAP kinase can, in turn, phosphorylate a number of cytoplasmic
proteins, including other protein kinases. Importantly, phosphorylation also allows
97
LAB 9A • Sea Urchin Fertilization: Preparation of Samples and Examination of Egg Activation and Cell Division
MAP kinase to enter the nucleus where it phosphorylates and regulates the activity of a variety of transcription factors. Some of these transcription factors are key
regulators of the cell cycle.
MEK
MAPK
Inactive form
protein phosphatase
P
–MAPK– P
Active form
MAPK
Inactive form
MAP kinases were first discovered in somatic cells (any cell that is not a germ cell),
where they have very different functions than in egg cells. They received their name
because they were activated by a variety of stimuli (“mitogens”) that stimulate cells
to enter mitosis and begin dividing—the opposite of the role they play in germ cells.
In somatic cells, MAP kinase is inactivated by a protein phosphatase that removes
the phosphate groups on Thr 202 and Tyr 204.
After fertilization, the sea urchin egg (now a zygote) enters the cell cycle and begins
to divide. In order for this to happen, MAP kinase must become inactivated. Unfertilized sea urchin eggs can be “tricked” into entering the cell cycle by treating them
with agents that inhibit MEK activity and prevent phosphorylation of MAP kinase
on Thr 202 and Tyr 204. Is the mechanism of MAP kinase inactivation at fertilization
the same as in somatic cells (i.e., dephosphorylation via a phosphatase)? The different role of MAP kinase in oocytes vs. somatic cells suggests this mechanism could
potentially be very different. At fertilization, MAP kinase could be inactivated by
removing all the MAP kinase protein (rapid degradation of MAPK), or by dephosphorylating it through the activity of a phosphatase. It is also of interest to know
what the signal is that triggers the inactivation of MAP kinase. The calcium influx
is one of the central signaling events to occur at fertilization; however, other signal
transduction pathways also become activated (see Lab 13). The other two questions
you will answer in this project are:
2. Is a cytoplasmic calcium influx sufficient to inactivate MAP kinase in sea urchin
eggs?
3. At fertilization, is MAP kinase inactivated by dephosphorylation or by
degradation?
C. Experimental Design
Notice that we use the term “sufficient” in questions 1 and 2 to ask about effects that
are produced by the calcium influx. When designing experiments, it is often helpful
to consider whether the experiment will show if a particular stimulus is sufficient
to produce the effect in question, and/or if it will show the stimulus is necessary to
produce the effect. An experiment that shows a stimulus is sufficient answers if it can
cause the effect or not, but does not answer if the stimulus is an absolute requirement
for the effect or if different stimuli can also cause the effect. An experiment that
shows a stimulus is necessary answers if it is an absolute requirement for the effect,
but does not answer if other stimuli are also needed to cause the effect (a stimulus
98
LAB 9A • Sea Urchin Fertilization: Preparation of Samples and Examination of Egg Activation and Cell Division
can be necessary but not sufficient). You will be creating an artificial calcium influx
in unfertilized sea urchin eggs to answer questions 1 and 2. Since the calcium influx
will occur in the absence of sperm binding, these experiments will tell you if calcium
alone is sufficient to cause MAP kinase inactivation and cell division. Keep in mind
that many things occur when the sperm binds to the egg, and your experiments will
not tell you about the involvement of other signal transduction pathways or cellular
events in producing these effects.
You will use the Western blot that you will perform in the next two labs to answer
questions 2 and 3. It is easier to answer question 3 first, and then determine the
role of calcium once you know how MAP kinase gets inactivated. In performing the
Western blot, your lab group and your partner group will use two different antibodies
that bind to MAP kinase. One antibody will specifically detect the phosphorylated
form of MAPK, and this will tell you whether active MAP kinase is present. The
other antibody will bind to both phosphorylated and dephosphorylated forms of the
protein, and will be used to detect the total amount of MAPK protein present. To
help us answer question 2, we will use the calcium ionophore mentioned in section
A, a compound called A23187. An ionophore is a compound that facilitates diffusion
of an ion across cell membranes. A23187 is a lipid soluble molecule that enters the
membrane and allows calcium to move across the membrane down its concentration gradient. By adding A23187 we can artificially stimulate the release of Ca2+ from
the endoplasmic reticulum in the absence of sperm. The A23187 is dissolved in the
organic solvent, DMSO (dimethylsulfoxide). DMSO is miscible in both aqueous and
nonpolar solutions, and is commonly used for adding hydrophobic compounds to
cells. The A23187 is prepared as a 1 mM stock in DMSO, and 5 μl of this stock solution is added to 1 ml of eggs for a working concentration of 5 μM.
DMSO is not a compound that sea urchin eggs normally encounter, and is toxic at
high concentrations, so how can you be sure that any effect produced with the calcium
ionophore is not really due to the DMSO? This dilemma highlights the importance
of having a correctly controlled experiment. The key to having a correctly controlled
experiment is: For any experimental condition (e.g., eggs in seawater, eggs in seawater
plus A23187), you should have another experimental condition where not more than
one variable changes between the two samples. For example, in going from an egg
sample with seawater alone to a sample with A23187, you are really changing two
variables, the A23187 and the 5 μl of DMSO it is dissolved in. Therefore we need to
have a control condition, referred to as a solvent control, where we add only the 5 μl
of DMSO to the eggs. Other important considerations when designing an experiment
are keeping all of the samples uniform (same size, developmental stage, handling,
etc.) and having enough replicates of each sample to ensure that any effect observed
is not due to random variation in the data.
The table below lists the experimental conditions that will be performed and
99
LAB 9A • Sea Urchin Fertilization: Preparation of Samples and Examination of Egg Activation and Cell Division
designates the groups that will perform them. Each of these treatments is performed
on a 1 ml suspension of eggs in seawater (seawater alone means that nothing is added
to the sample). Each group will perform their designated treatments for the induction
of cell division (Exp. 1) and preparation of the Western blot samples (Exp. 2), and
also preparation of the ELISA samples (Exp. 3) if your class is doing that experiment.
Group
1
2
3
4
5
6
7
8
9
10
11
12
100
Sample
Experimental Condition
Treatment
(per 1 ml of eggs)
1
Seawater alone
None
2
Fertilized
Add 1 μl sperm
3
5 μM A23187
Add 5 μl of A23187
stock in DMSO
4
DMSO
Add 5 μl DMSO
5
Seawater alone
None
6
Fertilized
Add 1 μl sperm
7
5 μM A23187
Add 5 μl of A23187
stock in DMSO
8
DMSO
Add 5 μl DMSO
9
Seawater alone
None
10
Fertilized
Add 1 μl sperm
11
5 μM A23187
Add 5 μl of A23187
stock in DMSO
12
DMSO
Add 5 μl DMSO
13
Seawater alone
None
14
Fertilized
Add 1 μl sperm
15
5 μM A23187
Add 5 μl of A23187
stock in DMSO
16
DMSO
Add 5 μl DMSO
17
Seawater alone
None
18
Fertilized
Add 1 μl sperm
19
5 μM A23187
Add 5 μl of A23187
stock in DMSO
20
DMSO
Add 5 μl DMSO
21
Seawater alone
None
22
Fertilized
Add 1 μl sperm
23
5 μM A23187
Add 5 μl of A23187
stock in DMSO
24
DMSO
Add 5 μl DMSO
LAB 9A • Sea Urchin Fertilization: Preparation of Samples and Examination of Egg Activation and Cell Division
D. Procedure
Safety Note: DMSO and compounds dissolved in it (A23187) are readily
absorbed through the skin. You should take extra care when working with
these compounds. Wear gloves, and remove the glove immediately if it comes
in contact with these reagents.
Observation of Sea Urchin Eggs and Egg Activation
1. Take a clean microscope slide up to the front bench where the gametes are stored
on ice. Gently swirl the tube of eggs to be sure they are evenly suspended in the
seawater. With plastic transfer pipette, place one large drop of the egg suspension onto the slide, and take this to the microscope at your workstation (note
that no coverslips are used).
2. Look at the eggs under low magnification on the microscope. They are spherical,
opaque cells that should be easy to see. Ask your TA for help if you have trouble
finding the correct focal plane.
3. After each group member has seen the eggs, have your TA bring the tube of
sperm to your workstation. Take a 200 uL pipette tip in your fingers and dip it
into the tube of sperm. With the slide still on the microscope stage, gently stir
the sperm into the drop of eggs without dispersing the drop.
4. Immediately look through the microscope to observe the eggs becoming activated. Take turns with your lab partners so everyone can see the egg activation.
The sperm are much smaller than the eggs but are still visible under low magnification. The sperm show up particularly well using phase contrast, if your
microscope has this.
5. Rinse the microscope slide with distilled water when you are finished, and save
the slide for scoring egg activation in Experiment 2.
Experiment 1: Induction of Cell Division
1. Get two 2 ml microfuge tubes and label them, one for each experimental condition you will be doing (see table). You will need 1 ml of unfertilized eggs for each
condition. Use a P-1000 micropipette to transfer the eggs, but cut ~5 mm off
the 1000 uL tip before pipetting; the wider diameter will prevent the eggs from
being damaged as they enter and leave the tip. The eggs are very large cells and
will settle in the bottom of the tube quickly. Be sure that the egg suspension
is well resuspended before pipetting so that you get the right amount of eggs.
Gently swirl the tube of eggs until the suspension looks homogeneous, then
pipette 1 ml into each of your labeled test tubes.
2. Incubate the tube in the 16°C water bath for 10 minutes to equilibrate.
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LAB 9A • Sea Urchin Fertilization: Preparation of Samples and Examination of Egg Activation and Cell Division
3. When your eggs are done equilibrating, you can begin your experimental treatments. Refer to the table in section C for the treatments that your group will
perform. For fertilized samples only: Pipette 1 μl of sperm into the appropriate
tube of eggs and mix well by gently stirring with the pipette tip. Remove the
tip from the micropipette so that you do not accidentally introduce sperm into
unfertilized conditions. Add 5 μl of A23187 or DMSO to the appropriate tubes
if your conditions include them. Add these reagents slowly with gentle stirring
so the eggs will not be exposed to concentrated DMSO.
4. Place your tubes in the 16°C water bath and incubate them for 2½ hours. Gently
invert the tubes every 20 minutes or so to keep the eggs suspended. This will
allow maximum contact with the sperm and other reagents. You will need to
perform Experiments 2 and 3 during this long incubation.
5. Retrieve your tubes after 2½ hours, resuspend the eggs, and place one large drop
on your microscope slide. Observe the eggs/zygotes under low magnification.
Have they undergone cell division? Cells that have undergone first cleavage will
have a clear cleavage furrow delineating the two cells that now occupy the sphere
of the fertilization envelope. Has anything reached the 4-cell stage? Allow your
neighboring lab group to examine your samples while you look at theirs. This
way everyone gets to see samples from each of the four experimental conditions.
6. Score both of your samples for the percentage of eggs/zygotes that have reached
first cleavage (see Experiment 2, step 6). Try to count at least 30 to 40 eggs/zygotes.
Write this data on the board and record the data from the other lab groups.
Experiment 2: Preparation of Western Blot Samples
1. For this experiment, you will need 2 ml of egg suspension for each sample. Take
two more 2 ml microfuge tubes and label them with your experimental conditions. Place 2 ml of egg suspension in each tube, taking the same precautions
that you did for Experiment 1, step 1.
2. Incubate the tubes in the 16°C water bath for 10 minutes to equilibrate.
3. Add the egg treatments, this time adding twice the volumes indicated in the
table on page 100 (i.e., 2 μl sperm, 10 μl A23187, 10 μl DMSO).
4. Incubate at 16°C for 35 minutes. Mix the tubes every five minutes by gently
inverting them.
5. During the incubation, prepare the MAPK lysis buffer. Take 90 μl of MAPK lysis
buffer and place it in a microfuge tube. Add 10 μl of the protease/phosphatase
inhibitor cocktail and vortex to mix (look at the list of inhibitor compounds on
page 105—why are these necessary?).
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LAB 9A • Sea Urchin Fertilization: Preparation of Samples and Examination of Egg Activation and Cell Division
6. At the end of the incubation period, one of your group members will need to
score your egg samples for activation while the other group members prepare
the egg lysate. Invert your egg tubes to resuspend and be sure there are no eggs
clumped at the bottom. Using a plastic transfer pipette (this has a wide mouth
and won’t shear the eggs), place one drop of egg suspension onto a glass slide
for both samples. Proceed to step 7 with the remaining egg samples. To score
the eggs for activation, use the microscope to examine 100 eggs (or as many as
you have in your drop) for elevation of the fertilization envelope. Calculate the
percent activation by dividing the number of activated eggs by the total eggs
counted. When counting, move the microscope stage in an “S” pattern to avoid
counting the same field twice. Do not count eggs that look damaged, or anything that looks unusual (sometimes the eggs will come out of their fertilization
envelopes). In which samples do you expect to see egg activation? Consult your
TA if these samples have less than 85% of the eggs activated; if this is the case,
you may need to repeat the sample beginning with step 1.
7.
You will now pellet the eggs in your benchtop microcentrifuge, but first you
need to transfer the suspension to a 1.5 ml microfuge tube so that you will get
a more compact pellet. Use a plastic transfer pipette to place approximately
1 ml of egg suspension in a clean 1.5 ml tube. Centrifuge for 30 seconds at full
speed. Remove most of the seawater, leaving a little behind to cover the pellet,
and then add the remaining egg suspension from the 2 ml tube. Centrifuge for
30 seconds at full speed.
8. Carefully remove all the seawater from the pelletted eggs; use a micropipette with
a small tip to remove the last of it from around the pellet. Keep your samples
on ice from this point on.
9.
Add 50 μl of MAPK lysis buffer with inhibitors to the egg pellet. Using a 200 μl
pipette tip, stir and pipette up and down to homogenize the eggs. Remember
that the fertilization envelope makes a tough shield around the eggs in activated
samples, so you will need to work at mashing the eggs with the pipette tip. Just
be careful not to cause foaming in the sample. While you are doing this, be sure
to keep the sample cold (on ice) to avoid degradation of MAPK by proteases and
phosphatases.
10. Centrifuge your samples in the cold-room microfuge, full speed for 10 min. This
will pellet the nuclei and insoluble cell debris, while the cytosol (where MAPK is)
will be in the supernatant. The supernatant should have a light yellowish-brown
tint to it, if the eggs were lysed well. Consult your TA if you do not see this; you
may need to mash the pellet more and spin again. Remove the supernatant to a
clean, well-labeled microfuge tube. It is a good idea to pre-chill this tube on ice
before adding your supernatant. Keep the egg lysate cold at all times.
11. Store your samples in the freezer in the box marked “Western blot samples.”
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LAB 9A • Sea Urchin Fertilization: Preparation of Samples and Examination of Egg Activation and Cell Division
Experiment 3: Preparation of ELISA Samples
Complete the preparation of your Western blot samples before beginning this
experiment.
1. Take two 1 ml egg samples in seawater, as you did for Experiment 1. Label them
carefully; give the experimental conditions and clearly indicate that they are
ELISA samples.
2. Add 100 μl of 0.5 M LiCl to your ELISA egg samples. This will give a diluted
LiCl concentration of 45 mM. Gently swirl to mix.
3. Incubate the tubes in the 16°C water bath for 10 minutes.
4. Add the egg treatments as you did in step 3 of Experiment 1.
5. Incubate at 16°C for 35 minutes. Mix the tubes every five minutes by gently
inverting them.
6. During the incubation take 110 μl of the ELISA lysis buffer. Do not add the
inhibitor cocktail.
7.
At the end of the incubation period, transfer all of your two ELISA egg samples
to separate, labeled 1.5 ml microfuge tubes. Centrifuge in your benchtop microcentrifuge for 30 seconds. Carefully remove all of the seawater from the pelleted
eggs. Keep your samples on ice from this point on.
8. Add 50 μl of ELISA lysis buffer to each egg pellet. Homogenize by pipetting
up and down, and mashing with the pipette tip. (Note: You will not centrifuge
these samples.)
9.
104
Keep your ELISA lysates on ice and place them in the freezer box marked “ELISA
samples” as soon as possible. Be sure all of your tubes for the day are accurately
labeled with the sample number or treatment and whether they will be used for
the Western blot or the ELISA. Note that there are separate freezer boxes for
the ELISA and Western blot samples.
LAB 9A • Sea Urchin Fertilization: Preparation of Samples and Examination of Egg Activation and Cell Division
Recipes
MAPK Lysis Buffer
• 1% NP-40 (a detergent)
• 20 mM HEPES, pH 7
• 15 mM EGTA
• 150 mM NaCl
Inhibitor Cocktail (10×)
• 2 mM Pefabloc
• 10 μg/ml pepstatin
• 100 mM β-glycerophosphate
• 4 mM NaF
• 2 mM Na3VO4
End-of-Lab Checklist:
❏ Two Western blot samples in freezer
❏ Two ELISA samples in freezer
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106
LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
9B
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LABORATORY
FGF Signaling: Experimental Design
and Preparation of Cell Lysates
Examining Signal Transduction from the Fibroblast Growth Factor Receptor in
Cultures of NIH 3T3 Fibroblasts
In this project, you will examine the intracellular signals that are produced by fibroblast growth factors and will tackle some real-life research questions concerning this
family of growth factors. You will work with NIH 3T3 cells and add the growth factor
to the culture medium in which these cells are grown. This type of cell culture is
different from bacterial cultures or cultures of other single-celled microorganisms
you may have worked with. In the early 20th century, it was found that cells could
be isolated from the tissues of mammals and other animals, and under the right
conditions of pH, salt concentration, and nutrient availability, could be grown in
vitro (under glass, in a Petri dish). It was found that some cell types grew very well
in vitro while others would not grow and that infecting cell cultures with viruses
that cause tumors in animals could cause the cells to grow continually in culture.
The ability to grow animal cells in culture has been and continues to be critical to
research because it allows experiments to be performed on a single cell type outside
of the complex environment of the whole animal. While you will not actually get
to perform the sub-culturing, you will be given dishes of NIH 3T3 cells to perform
your experiments.
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
The particular growth factor that you will be adding to your NIH 3T3 cultures is
fibroblast growth factor-2, or FGF-2. The types of biochemical signals that can be
produced when FGF-2 binds to its cell-surface receptor have been well defined.
However, the mechanism by which these signals are coordinated to produce the
different cellular effects of FGF-2 is not well understood, particularly in the context
of a whole organism (in vivo). When FGF-2 is added to cultures of NIH 3T3 cells, it
is known to stimulate specific cellular effects, which will be described below. You
will investigate which biochemical signals are associated with these specific effects.
To do this, you will perform two types of procedures: a Western blot (described in
Labs 10 and 11) and an ELISA (described in Lab 12), and you will learn about these
techniques in the process of carrying them out. The exciting part of this project,
however, is that you will get to design the Western blot and ELISA experiments.
This will require some careful thinking, planning, and working together with the
other students in your lab section. To begin, read the following sections carefully to
understand the system you will be working with.
A. Introduction to Fibroblast Growth Factors and Their Signal Transduction
Growth factors and cytokines are soluble proteins that are secreted from cells and
then act on those same cells (autocrine signaling) or on neighboring cells in the same
tissue (paracrine signaling) to produce specific biological effects. Growth factors
are distinct from cytokines in that one of their biological effects is to stimulate cell
division in target cells, although the terms growth factor and cytokine are sometimes
used interchangeably. The fibroblast growth factors (FGFs) are a family of 22 structurally related proteins encoded on different genes. They were originally discovered
in the mid-1970s as the source of the mitogenic (growth stimulatory) activity in
bovine pituitary gland extracts for stimulating cell division of murine fibroblasts.
Two mitogenic proteins were purified from these pituitary extracts—using techniques
similar to those in your LDH purification but following mitogenic activity rather
than catalytic activity—and were originally named acid FGF and basic FGF, due to
the different numbers of charged amino acids they contained. With the discovery
of many more FGFs, a numbering system was implemented and they were renamed
FGF-1 and FGF-2, respectively. The different FGFs have an extraordinarily diverse
range of biological activities (not all of them are mitogenic for fibroblasts). They play
a critical role in embryonic development by regulating the differentiation of different
cell types such as neural, bone, and muscle cells, and also by establishing pattern
formation by controlling cell growth vs. targeted cell death (apoptosis). For instance,
FGF-2 will induce the differentiation of neuronal cells in culture but will inhibit the
differentiation of cultured osteoblast (bone) and myoblast (muscle) precursors. In
adult animals, FGFs are required for tissue homeostasis through processes like wound
healing, where they help to regulate the migration and proliferation of cells. FGF-2
released during an injury will stimulate the proliferation of fibroblasts in adjacent
connective tissue so that new extracellular matrix can be laid down at the injury
site. It will also cause endothelial cells to migrate to the injury site and proliferate
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
to form new blood vessels, a process called angiogenesis. FGF-2 cannot, however,
stimulate the growth of epithelial cells, such as keratinocytes (skin cells), that need
to grow in and cover the damaged tissue. FGF-7, also known as keratinocyte growth
factor, is required for this. Hence, the FGFs appear to be necessary for many aspects
of intercellular coordination. They have been most widely studied in mammals, but
are found in all animals, with homologs in C. elegans and Drosophila.
FIGURE 9B.1 Activation of FGF Receptor Complexes. FGFs (shown in green) can bind
independently to FGFR and heparan sulfate. No signal transduction occurs, however, until an
extracellular FGF concentration is achieved that allows two FGF-FGFR complexes to be brought
together through binding to a single heparan sulfate chain.
A
FGFR
FGF
Heparan sulfate chain
Extracellular
ligand binding
domains
Proteoglycan
Plasma membrane
Cytoplasmic
domain
B
Cross-phosphorylation of
cytoplasmic domains leading
to signal transduction
©Hayden-McNeil, LLC
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
This array of biological processes is mediated by only four cell-surface receptors that
bind and elicit the effects of the entire FGF family. Fibroblast growth factor receptors 1–4 (FGFR-1–4) are type 1 membrane proteins, meaning that a single length
of the polypeptide chain spans the plasma membrane with the amino-terminus
extending outside the cell and the carboxy-terminus extending into the cytoplasm.
Differential mRNA splicing can produce two isoforms of each of the FGFRs, and
these receptor isoforms have different FGF ligand specificity. For instance, FGF-2
binds to FGFR IIIc isoforms but not to FGFR IIIb isoforms (IIIb vs. IIIc designating the presence of alternative exons b or c in the third “III” immunoglobulin-like
domain), whereas FGF-1 binds equally well to both isoforms. This differential ligand
specificity is, however, not enough to account for the variation in biological effects
by different FGF ligands.
FGFRs are typical growth factor receptors in that ligand (FGF) binding to the extracellular part of the receptor causes two receptor proteins to dimerize (form a complex
of two), which produces activation of an intrinsic tyrosine kinase catalytic activity
in the cytoplasmic domain (see Figure 9B.1). The mechanism by which dimerization is produced, however, is unique for FGFRs. An FGF ligand binds to a single
FGFR and also binds to a heparan sulfate polysaccharide chain. A single heparan
sulfate chain forms a bridge that brings two FGF/FGFR complexes together. Heparan sulfate is a glycosaminoglycan, a family of polysaccharides formed of repeating
amino-disaccharide units that can contain sulfate groups. They are present in high
abundance in the extracellular matrix (ECM) around cells and on the surface of cells
as polysaccharide chains extending from membrane proteins called proteoglycans.
All FGFs will bind to heparan sulfate and heparin (shorter fragments of heparan
sulfate) with relatively high affinity in the absence of FGFRs. FGFs, particularly
FGF-2, bound to heparan sulfate in the ECM may form an important reservoir of
FGFs that can be released when a tissue becomes damaged. Heparin is not abundant
in the ECM but can be released in large quantities from mast cells at sites of tissue
damage. Heparin-affinity chromatography is a classic method for purifying FGFs.
The biochemical signals that can be relayed from the dimerized receptor complexes
are thought to be the same for each of the four FGFRs. Figure 9B.2 shows how these
signals are produced, using FGFR-1 as an example. When the cytoplasmic tails of
two FGFRs are brought into close juxtaposition by ligand binding, a weak tyrosine
kinase activity in each receptor is able to phosphorylate key tyrosine amino acids
in the other receptor. This phosphorylation event fully activates the tyrosine kinase
activity of the receptor, allowing it to phosphorylate other tyrosines on the opposing receptor along with many other cytoplasmic proteins. This begins the signal
transduction cascade that propagates the signal from ligand binding throughout the
cell. FGFRs primarily activate two well-known signal transduction pathways, the
Ras-MAP kinase pathway and the phospholipase C (PLC) pathway.
Activation of the Ras-MAP kinase pathway by FGFRs is mediated through an adaptor
protein called FRS2. Adaptor proteins do not have any catalytic activity but serve
to bind other signal transduction proteins to recruit them to an active signaling
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
complex. FRS2 binds constitutively (always bound) to the cytoplasmic domain of
FGFRs and becomes phosphorylated by the FGFR tyrosine kinase when the receptor is activated by ligand binding. Tyrosine phosphorylation of FRS2 promotes
the binding of another adaptor protein called Grb2, which in turn recruits the G
protein Ras and its transactivator SOS to the signaling complex. Ras initiates the
sequential activation of the serine-threonine protein kinases Raf, MEK, and MAP
kinase, each phosphorylating and activating the next. At the level of MAP kinase,
the signal branches to a plethora of proteins that produce the biological effects
of FGFs. MAP kinase phosphorylates a variety of different transcription factors,
altering their activity and, therefore, the gene expression program in the cell. The
mitogen-activated protein (MAP) kinases are actually a family of serine/threonine
kinases that include p42, p44 Erk (extracellular-signal regulated kinase), and p38
kinase, among others—note that in this nomenclature “p” signifies a protein and
the number following designates its molecular weight, so Erk has 42 kDa and 44
kDa isoforms. Erk and p38 kinase appear to be the only downstream/effector MAP
kinases activated by FGFs. Erk is also simply referred to as “MAP kinase,” which can
be a source of confusion; read more about Erk/MAP kinase in section B of Lab 9A.
FIGURE 9B.2 Signal Transduction from FGFR-1. The cytoplasmic domain of FGFR-1 is shown
with the numbering indicating the amino acid position. Tyrosine (Y) 653 and 654 are in the
activation loop of the kinase domain and their phosphorylation is required for kinase activity.
For the Ras/MAP kinase pathway, adaptor proteins are shown in gray, G-proteins and guanine
nucleotide exchange factors (GEFs) are shown in green, and serine/threonine kinases are shown
in red. Phospholipase C (PLC) binds to phosphorylated Y766 and cleaves the phospholipid PIP2 to
form IP3 and DAG.
Transmembrane
Juxtamembrane
376
476
397
V427
T428 FRS2 Grb
2
SOS
Ras
PKC
Kinase domain
Raf
Ca2+ influx
Phosphorylation
DAG
Y653
Y654
IP3
PLC 766Y
MEK
Phosphorylation
757
822
C-terminus
Erk
©Hayden-McNeil, LLC
Phosphorylation of tyrosine 766 by the receptor tyrosine kinase facilities the binding and activation of PLC. The lipase activity of PLC cleaves the head group from
the phospholipid PIP2, forming IP3 and DAG. Read section A of Lab 12 for more
information on PLC. The net effect of PLC activation in FGF signaling is to activate
protein kinase C (PKC), which is a serine/threonine kinase that regulates many
downstream proteins. FGF stimulation of NIH 3T3 cells activates only the Ras-MAP
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
kinase pathway and the PLC pathway. In other cell types, however, the interaction
of FRS2 and Grb2 with other signaling proteins can activate a third major signaling
branch, the PI3-kinase pathway.
The experiments you will perform in this project will allow you to measure signaling down the Ras-Erk pathway and the PLC pathway.
Select Bibliography
•
Dailey, L., Ambrosetti, D., Mansukhani, A., Basilico, C. (2005) Mechanisms
underlying differential responses to FGF signaling. Cytokine and Growth Factor
Reviews 16:233–247.
•
Harmer, N.J. (2006) Insights into the role of heparan sulphate in fibroblast growth
factor signaling. Biochemical Society Transactions 34:442–445.
B. The Effects of FGF-2 on NIH 3T3 Cells
The particular FGF you will be working with in this project is FGF-2, and you will
be looking at the signal transduction that it produces in NIH 3T3 fibroblasts. NIH
3T3 cells are a cell line that was derived from taking normal mouse embryo fibroblasts, growing them through many population doublings in culture, and allowing
a spontaneously arising “immortal” subpopulation to grow out. Normal, or primary,
cells taken directly from the tissue of an animal will grow for only limited number
of population doublings in culture before they stop dividing and become senescent.
A cell “line,” by definition, has acquired mutations that have abrogated the checks
to continued proliferation in culture and, thus, has an immortal phenotype.
NIH 3T3 cells express high levels of FGFR-1 and much lower levels of FGFR-2 on their
cell surface (Li et al., 1994), and activation of either receptor is thought to activate
both the Ras-MAP kinase and PLC pathways. The addition of FGF-2 to NIH 3T3
cultures produces multiple, distinct effects in these cells, including the stimulation
of cell division and changes to the cellular morphology (shape). The experiments
that measured theses effects are described on the following pages. Read through
these experiments carefully, paying particular attention to the experimental details.
Keep what you learned about FGF signaling in mind as you do this. Once you have
interpreted the data from these experiments, you will need to form a hypothesis
to explain how these effects might be signaled from the FGF receptor. Then you
will design your own experiment to test your hypothesis.
When reading through this section, be sure you understand how each technique
works, and what specific information each experiment provides. The boxed Experimental Technique sections will explain how the experiments work. It is important
to note that these are not the techniques you will perform for your experiment. You
need to know how they work, however, to understand the data.
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
One last detail regarding the NIH 3T3 cells: It is important to note that these cells
express a significant amount of endogenous* FGF-2. Interestingly, FGF-2 lacks a
signal sequence to direct it through the endoplasmic reticulum/Golgi apparatus for
secretion. It is not clear how much of the endogenous FGF-2 makes it outside of the
cell where it is available to bind receptors.
Measurement of Growth Stimulation by 3H-Thymidine Uptake
When FGF-2 is added to the culture medium of NIH 3T3 cells, it stimulates them
to undergo cell division and proliferate. As is the case with almost all animal cell
cultures, these cells require stimulation by mitogenic factors such as growth factors and hormones in order to grow—merely providing the essential nutrients is
not enough. For routine culturing of NIH 3T3 cells (and many other cell lines), this
mitogenic stimulation is provided by the addition of bovine serum to the culture
medium. Serum is the liquid fraction of blood with the cells and clotting factors
removed, and it is rich in mitogens. In order to test if FGF-2 will stimulate NIH 3T3
cells to divide, it is necessary to remove the serum from the medium.
FIGURE 9B.3 Growth Simulation of NIH 3T3 Cells by FGF-2. Cell cultures were serum starved
for 24 hours (0.1% calf serum; cells are routinely grown in 10% calf serum), followed by the
addition of increasing concentrations of FGF-2 (0.1–10 ng/ml) to the culture medium for 24
hours. 3H-thymidine was added during the last four hours. Bars represent the average amount of
radioactivity retained on filters from triplicate culture wells and error bars indicate standard error
of the mean.
100,000
50,000
©Hayden-McNeil, LLC
3H-Thymidine
incorporation (cpm)
150,000
0
0
0.1
0.5
1
FGF-2 (ng/ml)
5
10
Critical Examination of Data: In addition to the no FGF control, what other
control would be useful in interpreting the data presented in Figure 9B.3?
* Endogenous means “from within.” In your experiments you will be adding exogenous (“from the outside”)
FGF-2.
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
Experimental Technique—Measurement of Cell Growth Stimulation by
Incorporation of Tritiated Thymidine into Newly Synthesized DNA
Cell cultures are rendered quiescent (nondividing) by withdrawing serum from the culture medium.
Cell
DNA
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Thymidine is the deoxy-nucleoside of thymine and deoxyribose. Tritiated or 3H-thymidine has one hydrogen
atom replaced with the radioactive hydrogen isotope, tritium (3H), and can easily be measured in the laboratory.
3
H-thymidine is added to the culture medium of the cells to be studied.
3H-thymidine
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Cells that are being stimulated to divide are actively synthesizing deoxynucleotide triphosphates for
DNA synthesis. 3H-thymidine is taken up by these cells, phosphorylated to TTP, and then incorporated
into the DNA as the cell replicates its genome. Cells that are not stimulated to divide will not incorporate
3
H-thymidine. After incubating the cells with 3H-thymidine, they are lysed, giving a homogenate that contains
soluble cellular debris, unincorporated 3H-thymidine, and newly synthesized DNA that has been labeled with
3
H-thymidine.
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Each cell lysate is suctioned through a glass fiber filter on a vacuum manifold. Small molecules such as the
unincorporated 3H-thymidine and most of the cell debris and proteins will pass through the filter. Chromosomal
DNA is large enough, however, that it is retained on the filter. The filters are then removed from the vacuum
manifold, and the amount of radioactivity retained on each filter—the 3H-thymidine incorporated into newly
synthesized chromosomal DNA—is measured in an instrument called a scintillation counter. Each nuclear
disintegration of tritium releases one beta particle, which is detected by the scintillation counter giving one
“count.” The counts per minute (CPM) indicate the amount of radioactivity on each filter.
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
Experimental Technique—Measurement of Cell Growth Stimulation by
Incorporation of Tritiated Thymidine into Newly Synthesized DNA, continued
Low CPM
Vacuum
©Hayden-McNeil, LLC
Filter disk
High CPM
3
H-thymidine incorporation is a very sensitive method for directly measuring DNA synthesis. Cell proliferation
can also be measured with formazan dye-based assays, or directly counting the number of cells in culture
dishes over time (a growth curve). In the dye-based assays, a tetrazolium salt or related compound is reduced
to a colored formazan by enzymes and NADH/NADPH present in the cells being tested. These are the MTT, XTT,
and MTS assays (the spot test you used to detect LDH works in a similar way). These assays are fast and more
convenient, but the amount of color produced is based on mitochondrial activity only (from dividing and nondividing cells) and are therefore not as sensitive. DNA synthesis can also be measured through the incorporation of bromodeoxyuridine (BrdU), a thymidine analog. An anti-BrdU antibody with a fluorescent tag is used to
detect BrdU incorporated into DNA, and the fluorescence is measured using microscopy or cell-flow cytometry.
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
Changes to Cellular Morphology
FIGURE 9B.4 FGF-2 Alters the Cellular Morphology of NIH 3T3 Cells. Cells (3 × 105) were
seeded in 10 cm culture plates and allowed to attach overnight, and then were left in standard
growth medium alone (A) or were treated with 10 ng/ml FGF-2 (B) for 24 hours. Photographs
were taken using phase contrast microscopy at an original magnification of 100×. The red arrow
indicates a cell that is considered to be refractile.
A. Growth medium alone
B. Plus FGF-2
As shown in Figure 9B.4, FGF-2 induces a distinct change in the shape, or morphology, of NIH 3T3 cells, causing them to become more rounded. These cells are not
detaching from the culture dish and are not dying; they continue to proliferate in
this condition. The optics of the microscopy used to observe the cells gives the most
rounded cells a halo of reflected light. These cells are termed “refractile,” and their
appearance was used to quantify the effect of FGF-2 on cell shape. The results from
an experiment where the morphology shift was quantified over different FGF-2
concentrations are shown in Table 9B.1.
TABLE 9B.1 The Effect of FGF-2 on NIH 3T3 Cell Morphology. The morphology shift at different
concentrations of FGF-2 was quantified by determining the percentage of refractile cells in
micrographs (100×) of cells treated for 24 hours with or without FGF-2. Refractile cells are defined
as those with a significant reflective halo (see Figure 9B.4). The results shown are from a single
experiment. Repeat experiments for some FGF-2 concentrations are shown in Table 9B.2.
FGF-2 Concentration (ng/ml)
Cells per Frame
Percent Refractile Cells
0
281
5.7
1
313
15.3
2
245
20.4
5
318
24.8
10
337
50.4
20
318
42
Experiment 1
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
Critical Examination of Data: Any figure or table that presents quantitative
data should give some indication of how repeatable the result is. This should
include stating the number of measurements included in the average and
the range of error or variation in the data. In some types of experiments, the
baseline for an effect may fluctuate between experiments, making it impossible to determine a meaningful average value for a repeated measurement
(as in the morphology shift induction shown in Table 9B.1); however, the trend
or fold-induction of the effect can still be consistent between experiments. In
this case, it should be clear how many times the experiment was repeated.
Induction of Urokinase Plasminogen Activator Gene Expression
FGF-2 also induces gene expression of a proteolytic enzyme called urokinase-type
plasminogen activator, or uPA. uPA has only a single substrate protein that it cleaves;
it converts the inactive pro-enzyme, plasminogen, to the active serine protease,
plasmin. Plasminogen is present in the serum of all mammals, and upon cleavage
to plasmin by uPA, plasmin can hydrolyze fibrin, the major component of blood
clots, along with a variety of other serum and extracellular matrix (ECM) proteins.
In endothelial cells, induction of uPA expression is part of the angiogenic response
that is stimulated by FGF-2. This allows endothelial cells to migrate through fibrin
clots and form new blood vessels in damaged tissue. The function of FGF-2-induced
uPA expression in fibroblasts is not as well understood, but could potentially play
a role in increased fibroblast motility in response to FGF-2—the morphology shift
depicted in Figure 9B.4 may also reflect this. Figure 9B.5 shows a Northern blot that
illustrates the induction of uPA by FGF-2 in NIH 3T3 cells.
FIGURE 9B.5 Induction of uPA Expression by FGF-2 in NIH 3T3 Cells.
Increasing concentrations of FGF-2 were added to the culture medium of NIH 3T3 cells for
24 hours. RNA was extracted and analyzed by Northern blot analysis using a probe for uPA
(top panel). The top band is the uPA transcript, and the lower band is nonspecific hybridization to
the 28S ribosomal RNA. The blot was stripped and then rehybridized with a probe to GAPDH as a
loading control (bottom panel).
Critical Examination of Data: You should be critical when examining the
bands in figures of Northern blots, Western blots, etc. If many background
bands are present, does the author confirm the band of interest is migrating
to the correct size or molecular weight? Are nonspecific or background bands
sufficiently explained? Is an appropriate loading control shown?
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
Experimental Technique—Measurement of RNA Transcript Levels
by Northern Blot
It is often easier to measure the level of expression for a particular gene by looking at the abundance of its
mRNA transcripts rather than protein levels, which requires an antibody that can specifically detect the protein
gene product. Northern blotting allows the determination of how many mRNA transcripts for a specific gene
are present in the cell.
Induced cell
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Un-induced cell
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Various mRNA
transcripts
Transcript for
gene of interest
RNA is extracted from cells treated under the experimental conditions and then separated by agarose gel electrophoresis. This is done under denaturing conditions to remove any secondary structure from the RNAs and
ensure that they migrate strictly based on their length. The 18S and 28S ribosomal RNAs are highly abundant
in all cells, and they produce distinct bands that can be seen if the gel is stained with ethidium bromide. Following the electrophoresis, the RNA bands are transferred to a nitrocellulose or nylon filter by electroblotting
or capillary blotting. The migration pattern of the RNA bands is retained on the filter, or “blot.”
The transcripts for the gene of interest are detected with cDNA probe for that gene. The probe is a cloned DNA
fragment of the gene that contains all or part of the mRNA sequence. The cDNA is double stranded, so one of
the strands will be complementary to the mRNA sequence and can anneal to it, forming a RNA–DNA hybrid.
Gel
Filter (blot)
Filter (blot)
©Hayden-McNeil, LLC
118
Film/imaging instrument
©Hayden-McNeil, LLC
LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
Experimental Technique—Measurement of RNA Transcript Levels
by Northern Blot, continued
32
P-labeled cDNA probe fragments
©Hayden-McNeil, LLC
Single-strand fragments of the cDNA probe are prepared so that they are labeled with phosphorous-32 (32P).
The probe fragments are synthesized from the cDNA using deoxynucleotide triphosphates that have 32P for the
alpha phosphate, so that the radioactive label is incorporated into the newly formed probe fragments. The filter
is then incubated in a solution of the labeled probe fragments under the right conditions of salt concentration
and temperature to favor hybridization of the probe fragments to the complementary RNA band on the filter. The
filter is washed to remove unbound probe. To visualize the probe-hybridized RNA bands, the filter is exposed
to a piece of film or an imaging instrument that can detect the radioactive emission of the 32P. Both of these
detection methods will produce a hard-copy image that shows the migration position (length in nucleotides)
and intensity (number of copies of the mRNA) of the mRNA band on the filter.
RNA degrades very easily. To ensure that each lane of the Northern blot was loaded with an equivalent amount
of total cellular RNA, it is necessary to include a loading control. After the filter is probed and visualized for the
mRNA of interest, it is stripped of that probe fragment and then hybridized with a control probe for an RNA
that is not expected to change in abundance with the different experimental conditions. These are typically
mRNAs for “housekeeping genes” whose gene products carry out basal physiologic functions in the cell, such
as cytoskeletal proteins or glycolytic enzymes like glyceraldehyde 3-phosphate dehydrogenase (GAPDH).
While Northern blotting is both sensitive and quantitative, it is being replaced more and more by quantitative
(Q) PCR. Once a good Q-PCR method is established for a particular mRNA, it is much simpler and faster than
Northern blotting.
Modulation of FGF-2’s Effects on NIH 3T3 Cells by Suramin and Heparin
Heparin is free, soluble fragments of heparan sulfate that are 15 to 90 repeating
disaccharide units in length (see Figure 9B.6). The disaccharide units carry many
sulfate groups, although not every unit is sulfated and the overall pattern of sulfation varies between heparin molecules. All of the sulfate and carboxylate groups on
heparin give it an extremely high density of negative charge, and charge repulsion
causes heparin chains to assume an extended, rod-like conformation. The presence
of heparin will enhance FGF-2 receptor binding and activation. Soluble heparin will
increase the affinity of FGF-2 for FGFR by stabilizing its binding, and will facilitate
FGFR activation by providing the polysaccharide bridge to bring together the receptor dimer (see Figure 9B.1).
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
FIGURE 9B.6 The Chemical Structures of Heparin and Suramin. The chemical structure and
space-filling conformational models are shown for heparin (A and B) and for suramin
(C and D). (A) shows a single disaccharide unit for heparin, while the space-filling model (B)
shows six repeating disaccharide units. The colors for the atoms in the space-filling models are:
carbon, gray; oxygen, red; nitrogen, blue; sulfur, yellow; hydrogen, white.
A. Heparin (chemical structure)
O−
O
S
O
B. Heparin (space-filling conformational model)
OH
O
O
O
O
O
HN
OH
O
©Hayden-McNeil, LLC
O
S
O−
O−
O
O
O
O
S
O
O−
C. Suramin (chemical structure)
D. Suramin (space-filling conformational model)
O
O
O
N
N
H
N
H
H
H
O−
H
NO
O
S
O−
O
O
O−
S
O
O−
O
S
©Hayden-McNeil, LLC
H
O N
O
O−
S
O
N
S
O
O
O
O−
S
O
O
Suramin is a pharmacologic compound (a drug) that has been used to study FGF
signaling. It inhibits FGF-FGFR interactions, and it is known to block some of the
cellular effects stimulated by FGF-2. The chemical structure of suramin is a chain
of eight benzene rings that carries six sulfate groups (see Figure 9B.6). The structure
of suramin is similar to that of heparin in that it is an extended chain of six-carbon
units with a high density of negative charge. Indeed, suramin binds to FGFs on the
amino acids that form the heparan sulfate/heparin binding site. The bound suramin molecule extends into the FGFR-binding site, thus preventing the FGF from
binding its receptor. Curiously, it has been reported that suramin will block some
FGF-2-mediated effects but not others (Rusnati et al., 1996). As suramin appears to
prevent the FGF/heparan sulfate/FGFR complex from forming, it is not understood
why the inhibitor does not block all the effects of FGF-2. Another important note
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
about suramin is that it is very nonspecific. It is known to block the actions of other
cell-surface receptors besides FGFRs, as well as affecting some intracellular biochemistry. The experiments described below examine how the growth stimulation,
morphology shift, and uPA induction effects of FGF-2 are affected when heparin
and/or suramin are added to culture medium of NIH 3T3 cells along with FGF-2.
The assays used to determine these effects are the same as those described earlier
(i.e., 3H-thymidine uptake and Northern blotting). The Northern blot experiment
includes another fibroblast growth factor, FGF-1. However, you will not use this FGF
in the experiments you will design.
FIGURE 9B.7 Modulation of FGF-2-Induced Effects by Heparin and Suramin.
Growth stimulation and induction of uPA by FGF-2 in NIH 3T3 cells were examined in the
presence of heparin and/or suramin. A) Growth stimulation was measured by 3H-thymidine
incorporation as described in Figure 9B.3. Cells were incubated in medium alone (control), or in
0.5 ng/ml FGF-2 in the presence or absence of 100 µM suramin. Bars show the average CPM from
triplicate wells ± standard error. B) uPA induction was determined by Northern blot analysis, as
in Figure 9B.5. Cells were treated for 24 hours in standard growth medium alone (control), or in
medium supplemented with the indicted concentrations of FGF-2 or FGF-1.
Some cultures were additionally treated with 1 µg/ml heparin and/or 100 µM suramin. Northern
blot filters were probed for uPA and then stripped and probed for GAPDH.
A.
B.
40,000
30,000
20,000
©Hayden-McNeil, LLC
3H-thymidine
incorporation (cpm)
50,000
10,000
0
Control
FGF-2
FGF-2
+
suramin
Critical Examination of Data: For complex figures like Figure 9B.7B, they
will be easier to interpret if you examine them one condition at a time. If any
of the conditions are repeated from an earlier figure, look at these first to
establish the baseline of what you already know. Find the control and then
follow how it changes through each condition, one at a time.
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
TABLE 9B.2 Modulation of the FGF-2-Induced Morphology Shift by Heparin and Suramin. The
extent of the FGF-2-induced morphology shift in NIH 3T3 cells was determined by counting the
percent refractile cells, as in Table 9B.1. Suramin was used in Experiment 2 at a concentration of
100 µM, and heparin was used in Experiment 3 at a concentration of 1 µg/ml.
Treatment
Cells per Frame
Percent Refractile Cells
No treatment
264
5.7
10 ng/ml FGF-2
209
23.9
Suramin
252
8.7
10 ng/ml FGF-2 plus suramin
247
22.7
No treatment
128
7.0
0.5 ng/ml FGF-2
136
5.9
Heparin
79
3.8
0.5 ng/ml FGF-2 plus heparin
121
16.5
Experiment 2
Experiment 3
Select Bibliography
•
Kathir, K.M., Kumar, T.K.S., Yu, C. (2006) Understanding the mechanism of the
antimitogenic activity of suramin. Biochemistry 45:899–906.
•
Li, Y., Basilico, C., Mansukhani, A. (1994) Cell transformation by fibroblast
growth factors can by suppressed by truncated fibroblast growth factor receptors. Molecular and Cellular Biology 14:7660–7669.
•
Rusnati, M., Dell’Era, P., Urbinati, C., Tanghetti, E., Massardi, M.L., Nagamine,
Y., Monti, E., Presta, M. (1996) A distinct basic fibroblast growth factor (FGF-2)/
FGF receptor interaction distinguishes urokinase-type plasminogen activator
induction from mitogenicity in endothelial cells. Molecular Biology of the Cell
7:369–381.
C. Form a Hypothesis to Explain How the Effects of FGF-2 Are Signaled from the
Receptor
The scientific method is at the heart of how research is able to help us figure out
how biological processes work. It consists of making observations about a system,
theorizing about how the system works, and then designing and performing experiments to test specific questions. This process is cyclical in nature; as information
is gained from experimentation it modifies the theory about how the system works
and allows new questions to be formed and tested.
122
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
im
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Pr gn
( De
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©Hayden-McNeil, LLC
How the three effects of FGF-2 in NIH 3T3 cells are signaled from the FGF receptor is a real, unanswered research question, and you will carry out each of the four
steps of the scientific method in addressing this question. The observations you
are beginning with are the experiments described in section B. What questions
are brought to your mind by these data? Based on what you know about the signal
transduction pathways that can be activated by the FGF receptor in NIH 3T3 cells,
develop a hypothesis to explain how the three effects of FGF-2 (growth stimulation,
morphology shift, and uPA induction) are signaled. Think of your hypothesis as a
model for how this signaling works in fibroblasts under normal circumstances in
the body. You may wish to diagram your hypothesis/model.
In the next section you will design experiments to test your hypothesis. Regardless
of whether your results support your hypothesis or contradict it, you will have provided new insights into FGF signaling in NIH 3T3 cells that will allow you to form
a more accurate, revised form of your initial hypothesis. This is how science works.
Coming up with a good hypothesis is the most important part of this process, so
put some careful thought into it. Your hypotheses should work to explain as many
as possible of the “facts at hand.” What are the relevant facts that you know about
this system? The Ras-Erk and PLC signal transduction pathways are the only ones
documented to be stimulated by FGF-2 in NIH 3T3 cells. This does not mean,
however, that there can’t be other, unknown signaling factors involved. The other
facts at hand are the data presented in section B. Your hypotheses should attempt
to explain the signal transduction in a way that accounts for the differences in how
FGF-2 produces these effects. Keep in mind, however, that heparin and suramin are
pharmacologic agents used to alter normal FGF signaling to conduct experiments.
Since your hypothesis is a model for how this signaling works under normal circumstances in cells, it should not include heparin or suramin.
Your TA will arrange for your lab group to work with a small number of other groups
on this project. This will be your research team. You will work together to come up
with your hypothesis, and design and carry out the experiments to test it.
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
D. Design Your Western Blot and ELISA Experiments
Once you know what you want to test, you can begin designing your experiments. You
will perform a Western blot that will measure activation of the Ras-Erk pathway,
and you will perform an ELISA that will measure activation of the PLC pathway.
Start by making a prediction about how you think one or more of the reagents listed
below will affect signaling down the Ras-Erk and PLC pathways. Then you can think
of specific experimental conditions that will test these predictions. Be sure you have
sufficient control conditions to let you know that the system is working correctly.
Control conditions will reproduce what you already know to be true. For instance,
based on what you know, you should expect very little or no signaling down either
of the signal transduction pathways if no FGF-2 is added to the cells.
In Lab 9B, you will be provided with dishes of NIH 3T3 cells that you will treat with
different experimental conditions, and then prepare as cell lysates for the Western
blot and ELISA. In Labs 10 and 11, you will perform the Western blot and in Lab
12, you will perform the ELISA. The Western blot and ELISA procedures are fairly
standard, so all of your experimental design will be carried out in Lab 9B. You will
be provided with the following reagents to carry out your experiments:
•
•
•
•
•
Recombinant human FGF-2; stock solution of 5 μg/ml in PBS* with 0.5% BSA
Calf serum; 100%
PBS with 0.5% BSA
Suramin; stock solution of 25 mM in dH2O
Heparin; stock solution of 0.5 mg/ml in dH2O
You will work in a research team of four groups for this project (groups 1–4, 5–8, etc.).
Each team will receive 8 plates of NIH 3T3 cells from which 8 Western blot samples
can be prepared. You will also get 8 wells of NIH 3T3 cells in a 24-well plate to
prepare four ELISA samples in duplicate. Design your Western blot experiment first,
and then design the ELISA experiment. For the Western blot, you should consider
how many replicates to include for each experimental condition—the experiment
may not work perfectly and some samples might not give an interpretable result.
While you don’t necessarily need to replicate every condition, you need to determine
which conditions are critical to answering the question asked in your experiment.
These conditions should be run with at least duplicate samples, if not more. Be sure
you include all the controls that will be necessary to accurately interpret your experiment. A well-designed experiment does not assume an effect will automatically
occur just because it has been reported previously. Fill in the table below for your
research team’s Western blot experiment.
124
* PBS is phosphate buffered saline, and BSA is added as a “carrier protein” to prevent loss of FGF-2 from
sticking to the walls of the tube.
LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
Western Blot Experiment
Group
Sample #
Experimental Condition
1
2
3
4
5
6
7
8
It is necessary to have more replicate samples of the different experimental conditions in the ELISA due to the considerable well to well variation that can occur
during the experiment. Each group will prepare two replicate samples of a single
experimental condition for the ELISA. Fill in the table below for your research team’s
ELISA experiment.
ELISA Experiment
Group
Sample #
Experimental Condition (each group prepares two
samples that are replicates of the same condition)
1
2
3
4
5
6
7
8
It is important to consider what the results of your experiments will tell you and
exactly what you can interpret from them. Will your data provide evidence as to
whether MAP kinase and/or PLC signaling is causing the effects produced by FGF-2?
Or will your data provide evidence that would only establish a correlation between
activation of a pathway and the FGF-2 effects? Correlative data (data that shows
similar patterns in the occurrence of things) is also useful, and it is often necessary
to gather this kind of data first before more focused experiments can be designed.
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
E. Procedures
Determine how you will prepare your dilutions of FGF-2, heparin, and suramin for
the Western blot and ELISA samples (To be done prior to Lab 9B): A step-by-step
procedure for exactly how you will prepare your experimental conditions should
be written in your lab notebook and checked by your TA on a lab day prior to the
experiment day. You will be graded on the accuracy of these calculations in a lab
notebook check. Exactly how you will prepare your samples depends on the experimental conditions your group will perform. Use the guidelines that follow.
For the Western Blot Samples
• You will receive two plates of NIH 3T3 cells (one for each experimental condition), and you will add 10 ml of cell stimulation medium (CSM) containing
your experimental reagents at the right concentrations to each plate. Therefore, your final volume for your dilution calculations is 10 ml.
• Do not pipette anything in a volume that is less than 2 μl—it is too difficult to
do this accurately. If necessary, make a dilution of the reagent in CSM so that
you can pipette a larger volume.
• Keep the amount of reagents you use to a minimum—they are very expensive!
For the ELISA Samples
• You will prepare two ELISA samples by adding 0.5 ml of ELISA stimulation
buffer with your experimental reagents to each of two wells of NIH 3T3 cells.
The two wells will be the same (duplicate samples) experimental condition;
therefore, you need to prepare 1 ml of stimulation buffer with the appropriate
concentrations of experimental reagents for your condition. Your final volume
for your dilution calculations is 1 ml.
• Except for the different final volume for your dilution calculations, all of the
same guidelines for preparing the Western blot samples also apply to your
ELISA samples. You will use ELISA stimulation buffer to carry out any serial
dilutions.
Prepare Culture Media for Experimental Conditions
1. Prepare the experimental medium for your ELISA samples. Label a single
microfuge tube for your experimental condition and add 1 ml of ELISA stimulation buffer. Following the plan that you have previously written in your lab
notebook for your experimental condition, add the required reagents. Vortex
briefly to mix. Place the tube in the 37°C water bath until you are ready to use
it. You will add half (0.5 ml) of this to each of two wells of cells.
2. Prepare the experimental media for your Western blot samples. Take two 15 ml
Falcon tubes and label them with your sample numbers and experimental conditions. Following the plan written in your lab notebook, prepare exactly 10 ml
of CSM (cell stimulation medium) with your experimental conditions for each
sample. Add the amount of CSM you will need directly to each Falcon tube,
and then add your experimental reagents. Secure the caps and invert the tubes
several times to mix, and then place them in the 37°C water bath until you are
ready to use them.
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
Treatment of ELISA Samples and Preparation of Cell Lysates
1. Retrieve your experimental medium from the 37°C water bath. The NIH 3T3
cells are grown in 24-well plates for this assay, and it is not possible to separate
the wells. Your research team will get one plate with 8 wells of cells. Obtain the
24-well plate of cells from your TA and label the wells with your sample numbers.
2. You will remove the medium from the wells of the 24-well plate using the vacuum
aspirator on your bench. Attach a Pasteur pipette to the end of the vacuum line
(inserting it no more than 1 cm or it will get stuck) and turn on the vacuum.
Place a small pipette tip on the end of the Pasteur pipette. This will produce a
gentler vacuum and give you more control. Do not begin aspirating the medium
until all four groups are ready to begin.
3. In order, each of the four groups should aspirate the buffer from their two wells
and quickly add 0.5 ml of their stimulation buffer with the reagents for their
experimental condition to each well. Do not let the cells dry out! Place the plate
in the 37°C incubator for one hour.
4. During the incubation, take 450 μl of the ELISA lysis buffer from the front table.
Do not add the protease/phosphatase inhibitor cocktail to the ELISA lysis buffer.
5. After the one-hour incubation, aspirate the stimulation buffer and add 200 μl of
ELISA lysis buffer to each well. Rock the plate to spread the lysis buffer on the
bottom of the well. Incubate at 37°C for 30 minutes.
6. Prepare two clean microfuge tubes and label them with your sample numbers and
experimental conditions—you will have two replicate samples, one for each well.
7.
Pipette the lysis buffer out of the wells and place it in the labeled microfuge tubes.
As you do this, scrape the pipette tip rapidly across the bottom of the well to be
sure all of the cells are lysed and off of the plastic. Place on ice and immediately
give to your TA to place in the freezer.
Treatment of Western Blot Samples and Preparation of Cell Lysates
1.
Take 100 μl of MAPK lysis buffer and add 11 μl of the protease/phosphatase
inhibitor cocktail. Vortex to mix. Place on ice. You will not use this until step 13.
2. Obtain your two cell culture plates of NIH 3T3 cells from your TA. Move them
carefully to avoid spilling any of the growth medium. Observe your cells under
the inverted microscope to be sure your cultures are healthy.
3. Label the top of each plate with the sample number and experimental condition
for that plate.
4. Retrieve your experimental media from the 37°C water bath.
127
LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
5. When everything is prepared, aspirate the medium from your plates and add
your experimental media by gently pouring it from the Falcon tube. Do this one
plate at a time—do not let the cells dry out. Place the plates in the 37°C incubator for 10 minutes.
6. During the 10-minute incubation, label two microfuge tubes with your sample
numbers.
7.
Aspirate the medium from the plate and then quickly add 1 ml of PBS. Rock the
plate to spread the PBS so that the cells will not dry out.
8. Use a cell scraper to scrape the cells from the surface of the plates. You will see
a layer of whitish film coming off of the plate—this is what you want to collect.
Push the cells down into the corner of the plate while holding the plate at an
angle to allow the cells to accumulate there. Be sure to scrape all the cells from
the plate.
9.
Using a narrow stem plastic transfer pipette, transfer the scraped cells into the
correct microfuge tube that you labeled in step 6. Rinse the remaining cells in
the dish down to the corner with an additional 0.4 ml of PBS and transfer this
to the microfuge tube as well.
10. Centrifuge at 5000 RPM in your bench-top microfuge for three minutes to pellet the cells.
11. Remove the PBS supernatant by pipetting, first with your P1000 and then with
a smaller pipette to remove the residual PBS around the cell pellet. Be sure to
remove all of the PBS.
12. Immediately place the tube on ice. Keep everything cold on ice from this point
on. Do not let the cells or cell lysate reach room temperature.
13. Add 50 μl of MAPK lysis buffer with the inhibitor cocktail to each cell pellet and
gently pipette up and down and stir with the pipette tip to dissolve the cells. Be
sure to keep the tube cold while you are doing this to avoid degradation of MAP
kinase by proteases and phosphatases.
14. Centrifuge the tubes at full speed for 10 minutes in the cold-room microfuge.
This will pellet the nuclei and insoluble cell debris while the cytosol (where MAP
kinase is) will be in the supernatant.
15. Remove the supernatant to a clean microfuge tube and label it with the sample
number and experimental condition. This is your cell lysate for your two experimental conditions. Keep on ice.
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LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
16. You now need to perform a Bradford assay to determine the protein concentration
of your cell lysates (review section F in Lab 6 for how to perform the Bradford
assay). You need to prepare three samples, a blank and one sample for each of
your two cell lysates. Remember: do not use your plasticware for Bradford—it
will stain. Use the 5 mL Bradford pipette to put your 3 mL aliquot into a dedicated 50 mL glass beaker.
a.
Prepare your two cell lysate samples by adding 5 μl of lysate to 95 μl of dH2O
in a clean, labeled microfuge tube. Then add 900 μl Bradford reagent to each
tube, and vortex.
b. The blank is 900 μl Bradford reagent plus 100 μl dH2O; vortex.
You do not need to prepare standards this time; the standard curve has been
done for you. Once you have measured the absorbance for your samples, use the
standard curve posted above each spectrophotometer to determine the protein
concentration of your cell lysates in μg/μl (μg protein per tube from the standard curve divided by 5 μl—there is no dilution factor here). If your absorbance
numbers do not fall within the range of absorbance values on the standard curve
you will need to repeat the assay, changing the amount of cell lysate that you
add, and keeping the total volume with dH2O at 100 μl. Calculate the volume of
your cell lysates that contains 5 μg of protein. This is the volume that you will
run in the SDS-PAGE gels in Lab 10.
17. Aspirate all tubes and cuvettes containing Bradford reagent into the Bradford
waste flask on the middle bench. Place empty cuvettes in the lab trash bucket,
then rinse the 50 mL beaker and return it to the bin box.
18. As soon as you have a good Bradford assay, give your cell lysates to your TA to
store in the freezer.
End-of-Lab Checklist:
❏ Two Western blot samples in freezer
❏ Two ELISA samples in freezer
129
LAB 9B • FGF Signaling: Experimental Design and Preparation of Cell Lysates
Recipes
NIH 3T3 Standard Growth Medium (SGM)
Dulbecco’s Modification of Eagles Medium (DMEM) supplemented with:
• 10% calf serum
• 100 I.U./ml penicillin
• 100 μg/m streptomycin
Western Blot Cell Stimulation Medium (CSM)
DMEM supplemented with:
• 0.1% calf serum
• 25 mM HEPES (pH 7.3)
MAPK Lysis Buffer
• 1% NP-40 (a detergent)
• 20 mM HEPES (pH 7.0)
• 15 mM EGTA
• 150 mM NaCl
Inhibitor Cocktail (10×)
• 2 mM Pefabloc
• 10 μg/ml pepstatin
• 100 mM β-glycerophosphate
• 4 mM NaF
• 2 mM Na3VO4
ELISA Stimulation Buffer
• 10 mM HEPES (pH 7.4)
• 1 mM CaCl2
• 0.5 mM MgCl2
• 4.2 mM KCl
• 146 mM NaCl
• 5.5 mM glucose
• 50 mM LiCl
ELISA Lysis Buffer
• 2.5% Triton-X100
• 50 mM HEPES (pH 6.8)
• 0.1% BSA
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LAB 10 • SDS-PAGE and Western Blotting
LABORATORY
10
SDS-PAGE and Western Blotting
In this lab, you will take the cell lysates you made in Lab 9A or 9B, and separate the
proteins by size using an SDS-PAGE. You will then transfer the proteins out of the
gel and onto a nitrocellulose filter. Together, these procedures are referred to as
Western blotting. In the following lab, you will use antibodies to specifically detect
phosphorylated and total MAP kinase on the nitrocellulose filter.
A. (For Sea Urchin Fertilization Project) Separation of Sea Urchin Egg Lysates on
SDS-PAGE Gels
Here you will run the egg lysates you prepared in the last lab on an electrophoresis
gel. All of the proteins in the egg lysates will separate by size, and each protein will
migrate to a specific position in the gel based on its molecular weight. To save time,
we will use premade 10% polyacrylamide gels. TAs will demonstrate how to prepare
these. Two groups will share one gel. This gel will have samples from four different
groups and will be probed using the antibody listed in table 10.1. Your two samples
will be run on two different gels—yours and one other. This is so that each sample
will be probed with each of the two antibodies. Each gel will have eight experimental
samples. One well will be used for the prestained standards. There are 10 wells total
in the gel so there is one empty lane.
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LAB 10 • SDS-PAGE and Western Blotting
TABLE 10.1 Gel Loading Guideline.
Lab Groups
Gel
Samples
Antibody Used
1 and 2 (or 5 and 6)
(or 9 and 10)
P-MAPK
1–8 (or 9–16) (or
17–24)
Detects (binds to) MAKP
phosphorylated on
Thr202/Tyr204
T-MAPK
1–8 (or 9–16) (or
17–24)
Detects (binds to)
both phosphorylated
and unphosphorylated
MAPK
3 and 4 (or 7 and 8)
(or 11 and 12)
1. Set up the gel in the electrophoresis chamber and add the buffer to both the upper
and lower chambers. The buffer in the upper chamber should come almost to
the top of the upper plate to ensure that the wells do not run dry. You will still
need to keep checking the buffer levels as the gel runs to make sure that there
is enough buffer. (Buffer stock is 10×, make 600 ml 1×.)
2. Move your electrophoresis tank to a location near a power supply. Once your
gel is loaded, you will not be able to move it.
3. Clean out the wells by squirting buffer into each well with a Pasteur pipette.
4. Thaw your samples and gently mix them. Be sure to keep them on ice.
5. Before preparing your samples to run on the gel, you need to do a Bradford
assay to determine the protein concentration of your egg lysates. Review part
F in Lab 6 for how to perform the Bradford assay. You do not need to prepare
a standard curve again, use the standard curve you created for Lab 6. Just be
sure to use the same spectrophotometer for your egg lysate samples. Do not use
your plasticware for Bradford—it will stain. Use the 5 mL Bradford pipette to
put your 3 mL aliquot into a dedicated 50 mL glass beaker.
a.
For each egg lysate, add 5 μl lysate to 95 μl dH2O in a clean microfuge tube.
Label the tubes, then add 900 μl Bradford reagent to each tube. Vortex to
mix.
b. Prepare a blank that is 900 μl Bradford reagent and 100 μl dH2O; vortex.
6. Measure the absorbance of your samples at 595 nm, and use the standard curve
posted above the spectrophotometer to calculate the protein concentration of
your egg lysates in μg/μl (μg protein in the tube divided by 5 μl—there is no
dilution factor here). If your absorbance numbers do not fall within the range
of the standard curve, repeat the assay and change the amount of egg lysate that
you add, keeping the total volume with dH2O at 100 μl.
7.
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Aspirate all tubes and cuvettes containing Bradford reagent into the Bradford
waste flask on the middle bench. Place empty cuvettes in the lab trash bucket,
then rinse the 50 mL beaker and return it to the bin box.
LAB 10 • SDS-PAGE and Western Blotting
8. Calculate the volume of your egg lysates that contains 10 μg total protein.
9.
Prepare two SDS-PAGE samples for each of your two egg lysates, one for you
to run and a second that you will give to the group running that sample on the
duplicate SDS-PAGE gel. To prepare your SDS-PAGE samples, add the volume
of egg lysate that contains 10 μg protein to a new tube and then add water (if
necessary) to achieve a volume of 16 μl. Add 4 μl of 5× SDS-PAGE sample buffer; this gives you a loading volume of 20 μl. Stir with the pipette tip to mix—do
not foam the SDS.
Safety Note: The 5× sample buffer contains 25% 2-mercaptoethanol. This
compound is toxic and is hazardous by inhalation and skin contact. Wear
gloves and remove the glove immediately if you come in contact with the 5×
sample buffer. Add the 5× sample buffer to your samples in the fume hood,
being sure to use a new tip for each sample. It is okay to take your tubes out
of the hood once the 5× sample buffer has been diluted into your samples.
10. Swap samples with the other lab groups in your experiment group so that you
will have all 8 samples to run on your gel (see Table 10.1). IMPORTANT: Decide
on the order that you will load the samples on the gel (look at the wells on the
gel). In addition to your samples, designate one lane for the molecular weight
markers. Each of you should record this in your lab notebook.
11. Heat the samples for 3 minutes at 95°C, let them cool to room temperature, and
spin them briefly in your benchtop microfuge to bring down any condensation.
Load 20 μl of each sample into the gel. Be very careful to avoid spillover from
one lane to the next. Put 5 μl of the pre-stained standards into the appropriate
well (see Table 10.1—you must load in that order).
12. Place the lid on the chamber (red to red and black to black) and run at 120 V.
13. Run the gel until the tracking dye is within 1 cm of the bottom of the gel. Check
periodically to be sure the buffer level in the upper chamber hasn’t fallen below
the upper surface of the gel. While the gel is running, collect the materials you
will need for Western blotting.
A. (For FGF Signaling Project) Separation of NIH 3T3 Lysates on SDS-PAGE Gels
In this lab, you will run the cell lysates you prepared in Lab 9B on an SDS-PAGE
gel. You will pair up with the group next to you to run the gel. You will run all 8
samples for your research team on the gel. See the following table. The other groups
in your research team will run the identical samples on their gel so that duplicate
Western blots are done for all the samples. This allows one blot to be probed with
the phospho-Erk antibody and the other to be probed with the total-Erk antibody,
giving you a picture of both the total amount of Erk-MAP kinase protein present
and how much of it is active.
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LAB 10 • SDS-PAGE and Western Blotting
Lab Groups
Gel
Samples
Antibody Used
Groups that
prepared samples
1–4
P-MAPK
1–8
Detects (binds to) MAKP
phosphorylated on
Thr202/Tyr204
1–8
Detects (binds to)
both phosphorylated
and unphosphorylated
MAPK
Groups that
prepared samples
5–8
T-MAPK
1. Make 600 ml of 1× tris-glycine running buffer from the 10× stock. Set up the gel
in the electrophoresis unit and add buffer to both the upper and lower chambers.
The buffer in the upper chamber should come almost to the top of the upper
glass plate to ensure the top of the gel will be in contact with the buffer. Slow
leakage from the upper chamber to the lower chamber will sometimes cause the
buffer level to drop, so you will need to check this periodically during the run.
2. Move your electrophoresis unit near the power supply you will use. You must
do this before you load the gel.
3. Clean out the wells by squirting buffer into each well with a Pasteur pipette.
4. Thaw your samples, briefly vortex to mix, and spin briefly in your microfuge to
move all the sample to the bottom of the tube. It is critical that you continue
to keep the samples cold on ice.
5. Prepare two SDS-PAGE samples for each of your two cell lysates, one for you
to run and a second that you will give to the group running that sample on the
duplicate SDS-PAGE gel. Calculate the volume of your cell lysates that contains
5 μg total protein. To prepare your SDS-PAGE samples, add the lysate volume
that contains 5 μg protein to a new tube and then add water (if necessary) to
achieve a volume of 16 μl. Add 4 μl of 5× SDS-PAGE sample buffer; this gives you
a loading volume of 20 μl. Stir with the pipette tip to mix—do not foam the SDS.
Safety Note: The 5× sample buffer contains 25% 2-mercaptoethanol. This
compound is toxic and is hazardous by inhalation and skin contact. Wear
gloves and remove the glove immediately if you come in contact with the 5×
sample buffer. Add the 5× sample buffer to your samples in the fume hood,
being sure to use a new tip for each sample. It is okay to take your tubes out
of the hood once the 5× sample buffer has been diluted into your samples.
6. Swap samples with the other lab groups in your research team so that you will
have all 8 samples to run on your gel (see table). IMPORTANT: Decide on the
order that you will load the samples on the gel (look at the wells on the gel). In
addition to your samples, designate one lane for the molecular weight markers.
Each of you should record this in your lab notebook.
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LAB 10 • SDS-PAGE and Western Blotting
7.
Heat the samples for 3 minutes at 95°C, let them cool to room temperature (do not
place them on ice), and spin briefly in microfuge to bring down any condensation.
8. Load all 20 μl of each sample into the appropriate well, as designated in your
lab notebooks. Load 5 μl of pre-stained molecular weight markers into the
appropriate well.
9.
Place the lid on the electrophoresis unit and run at 120 V. Run until the tracking
dye is within 1 cm of the bottom of the gel. Begin setting up for the electroblotting procedure while the gel is running.
B. Introduction to Western Blotting
The technique of Western blotting is used when you need to specifically detect one
protein in a mixture of proteins. For Western blotting, the proteins are first separated according to MW on a polyacrylamide gel using SDS-PAGE. The separated
proteins are then transferred electrophoretically from the polyacrylamide gel onto a
hydrophobic membrane such as nitrocellulose (NC) or PVDF (polyvinyl-difluoride).
How well the proteins move out of the gel and how well they bind to the membrane
are affected by various factors as summarized below. (E.M. Southern first developed
a method for transferring DNA to filters following electrophoresis, which is called
Southern blotting. When a method for transferring RNA to filters was developed to
detect complementary DNA sequences, it was called Northern blotting. To continue
the tradition, the method for transferring proteins to filters following electrophoresis
was called Western blotting.)
Factors Influencing Transfer
The buffer. A buffer that is commonly used for transfer, and the one we will use, is
Tris-Glycine, pH 8.3 with a low concentration of SDS (0.0013 M; 0.0375%) and 20%
methanol.
1. Glycine is used instead of Cl because it has a lower charge density—less current
means less heat production (H = I2R; I is current, R is resistance).
2. SDS favors elution from the gel, but if the concentration of SDS is too high the
proteins will not bind to the membrane. (Why?)
3. Methanol:
a.
helps remove SDS from the proteins, which uncovers hydrophobic sites on
the proteins that can then bind to the NC or PVDF membrane. If there is
no methanol, elution from the gel is favored, but binding to the membrane
is poor. If there is too much methanol, the SDS could come off the proteins
before they leave the gel, and the exposed hydrophobic regions could aggregate forming large protein aggregates that would be trapped in the gel.
b. helps prevent swelling of the gel during electrotransfer—this would result
in distortion of the bands.
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LAB 10 • SDS-PAGE and Western Blotting
The membrane. Membranes are available with different pore sizes. 0.45 μm pores
are commonly used, but smaller pore sizes (0.22 or even 0.1 μm) are necessary to
efficiently trap proteins of <20 kDa.
Voltage during transfer. It is the current that is more critical than the voltage. Heating
is one factor (item 1 on the previous page). Another consideration is that if the current is too high, low molecular weight proteins can move through the membrane so
fast that they do not have a chance to bind to the membrane.
Time of transfer. Larger proteins transfer more slowly than smaller proteins. Some-
times two layers of membrane are placed next to the gel to trap smaller proteins that
might pass through the first membrane.
Protein charge. Since removing SDS from the proteins decreases their negative charge,
it can be difficult to transfer basic (positively charged) proteins since their isoelectric points (pH at which the protein has no net charge) can be similar to the pH of
the buffer system. This problem can be solved by increasing the pH of the buffer
system to 9.5 or 10 (e.g., use 25 mM CAPSO buffer instead of Tris-Glycine. CAPSO
is 3-N-cyclohexylamino-2-hydroxypropane sulfonic acid.).
C. Western Blotting with the Semidry Blotting Apparatus
1. During the electrophoresis run, obtain a sheet of nitrocellulose and two transfer
stacks. The individual transfer stacks are separated by blue sheets of paper.
2. Label the nitrocellulose with the gel designation (A–D) in the lower right-hand
corner of the non-protein side (you will transfer the proteins to the other side
of the filter). Use a VWR marker to do this (located on front bench—please
return after use!).
3. When the electrophoresis is done, turn off the power supply, unplug the leads,
and dispose of the electrophoresis buffer in the sink (this is not hazardous waste).
Disassemble the electrophoresis unit.
4. Separate the gel plates. With the gel sticking to one plate, use a razor blade to
remove the stacking gel by crimping the gel at the top of the running gel (use
one of the blue sheets separating the transfer stacks as a template to find the
boundary between the running and stacking gels). Do not drag the razor blade
across the gel as this will tend to tear the polyacrylamide. Scrape the stacking
gel off the plate with the razor blade.
5. Remove the gel from the plate and place it in one of the trays on the front table
that contains electroblotting buffer.
6. Use the other tray to thoroughly wet the nitrocellulose and the two transfer
stacks with electroblotting buffer. Do not let the transfer stacks separate into
individual sheets.
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LAB 10 • SDS-PAGE and Western Blotting
7.
Open the cassette of the semi-dry electroblotting unit (Trans-Blot Turbo Transfer System from Bio-Rad). Place one of the wet transfer stacks onto the bottom
electrode plate. This is the positive electrode (see diagram).
8. Place the wet nitrocellulose onto the first transfer stack so that it aligns precisely.
Place the gel on top of the nitrocellulose, followed by the second transfer stack.
All the edges should neatly align.
9.
Place the lid on the cassette (this contains the negative electrode), lock it into
place, and place the cassette into the electroblotting unit. Run the unit using one
of the preprogrammed protocols. The transfer will use 1.3 amps constant current per gel stack in the cassette; the transfer will be complete in 3–10 minutes.
10. After the run is complete, remove and open the cassette. Peel off the blotting
stack and gel and remove the nitrocellulose. Place gel in gel waste bucket on
middle bench. Filter paper should go in the lab trash bucket.
11. Place the nitrocellulose into the pre-labeled ("T" or "P") container designated
for that blot. Be sure that the protein side of the blot is facing up. Add 20 ml of
the blocking buffer. The blocking buffer is TBST buffer with 5% bovine serum
albumin added as the blocking agent. This will help to cover any surface of the
nitrocellulose not already bound by protein and, thus, prevent nonspecific sticking of the antibodies to exposed nitrocellulose. Use tape to label the box with
your section and group numbers.
TBST (Tris-Buffered Saline with Tween-20): 10 mM Tris, pH 7.4, 150 mM
NaCl, 0.1% Tween-20 (this is a mild, non-ionic detergent). This buffer maintains
physiologic pH and salt content to keep the antibody proteins safe during the
immunodetection steps. The low concentration of Tween-20 detergent reduces
nonspecific binding of antibodies to the nitrocellulose blot.
12. Incubate overnight at 4°C with gentle agitation.
13. This step will be done for you. Discard blocking buffer and add 15 ml of TBST/
BSA + rabbit anti-MAPK (T or P) antibody diluted 1:1,000. Incubate at 4°C with
gentle agitation (rotator in cold room) until next lab. If you are doing the FGF
signaling project, an anti-GAPDH loading control antibody (1:50,000 dilution)
will be added with the anti-MAPK antibodies.
Side view
−
Transfer stack
blotting sheets
Gel
Nitrocellulose
©Hayden-McNeil, LLC
+
Direction of
transfer
Transfer stack
blotting sheets
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LAB 10 • SDS-PAGE and Western Blotting
138
LAB 11 • Immunodetection
11
Floris Slooff/Shutterstock.com
LABORATORY
Immunodetection
Now that all of the proteins from your SDS-PAGE gel are transferred to the nitrocellulose membrane, you can specifically detect MAP kinase using antibodies. The
nitrocellulose membranes will be probed with one of two different antibodies against
MAP kinase. One antibody recognizes only MAP kinase that is phosphorylated on
Thr 202 and Tyr 204 (phospho-MAPK). The other antibody recognizes both phosphorylated and unphosphorylated MAP kinase (total-MAPK).
A. Background on Antibodies
A major strategy used by vertebrates to protect themselves from foreign entities is
the production of antibodies. The substances that the antibodies bind to are called
antigens. The antibodies used in biochemistry are proteins made by the immune
systems of mammals or chickens. Antibodies are a type of protein called globulins,
which are produced by cells of the immune system; hence they are referred to as
immunoglobulins or Ig for short. There are different classes of immunoglobulins
that function in defense against pathogens. The G-class immunoglobulins, or IgG,
are the most prevalent Ig in mammalian serum, and it is IgG class antibodies that
are used as tools in biology. IgG class antibodies have a common structure. They
are tetramers composed of two heavy chains and two light chains that are held
together by disulfide bridges. The amino acid sequence in the base of the “Y” does
not vary for a given species, and this is termed the constant region. The amino acid
sequence at the ends of the Y does vary, and this variable region is the part of the
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LAB 11 • Immunodetection
IgG that binds to the antigen. The different amino acid sequence of the variable
regions in different IgG molecules creates specific binding sites that will bind to different antigens. Since each IgG molecule has two such binding sites, in either prong
of the Y, it can potentially bind two antigen molecules, assuming it is not sterically
prevented from doing so.
FIGURE 11.1 An IgG class antibody.
Antigen binding site
N
N
Disulfide linkages
Heavy chains
C
Light chains
C
C
C
N
N
Constant region
Variable
region
Antigen binding site
©Hayden-McNeil, LLC
Antibodies are produced when an animal’s immune system reacts to an antigen,
which is a compound foreign to the animal making the antibody. Hence, a rabbit will
not make antibodies to rabbit serum albumin. However, if it is repeatedly injected
with human serum albumin, which has a slightly different amino acid sequence
than rabbit albumin, the rabbit will produce antibodies to human serum albumin.
Because of this, we often talk in terms of “raising” antibodies to the antigen of interest. It is possible to raise antibodies to a tremendous variety of compounds, although
some compounds may require special preparation. Antibodies can also be raised to
antibodies; for example, it is possible to use a goat to raise antibodies against rabbit immunoglobulins. This is done frequently to make what are called “secondary
antibodies.” By definition, a secondary antibody binds to the constant region of a
“primary antibody.” The primary antibody is what binds directly to the protein of
interest (what you want to detect), and the secondary antibody allows you to visualize
the bound primary antibody by producing a signal. The secondary antibody usually
has a fluorescent tag or peroxidase enzyme covalently linked to its constant region to
produce this signal. To create the secondary antibody, an animal is injected with the
constant region portion of IgG from the animal in which the primary antibody was
raised. Obviously, the animal producing the secondary antibody must be different
from the animal that produced the primary antibody, so that it will recognize the
IgG as foreign. Therefore, if the primary antibody is a rabbit IgG, then anti-rabbit
IgG raised in a goat is typically used as the secondary antibody.
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LAB 11 • Immunodetection
FIGURE 11.2
Gel
Nitrocellulose
1° antibody
MAPK
Peroxidase
conjugate
Other proteins
2° antibody
Antibodies made in response to a given antigen will be highly specific for that antigen.
They will bind to, or “recognize,” a specific region of that antigen molecule called the
epitope. It is this property that makes antibodies so useful in biochemistry. They can
be used to help determine where in a cell an antigen may be concentrated (immunocytochemistry), whether a cell expresses a certain molecule on its surface, or whether
a molecule is present in a sample. Antibodies can also be used for quantifying the
amount of antigen in a sample and for purifying an antigen from a complex mixture.
Because some molecules, particularly proteins, may have similar structures, it is
possible for an antibody to recognize and bind an antigen other than the one for
which it was originally created. An antibody that does this is said to “cross-react”
with another compound. For example, consider the rabbit producing antibodies to
human serum albumin. It is possible that the antibodies raised against human serum
albumin might also bind chimpanzee serum albumin. Sometimes this cross-reaction
is desirable. An antibody raised against human albumin could be used in a Western
blot to detect albumin from a variety of species, including humans, primates, rats,
and mice. When antibody cross-reaction causes binding to different proteins in
the same species, this is typically not desirable. For instance, the antibody against
human albumin might also bind to human hemoglobin if some region on each of
these proteins shared a similar amino acid sequence. This type of cross-reaction can
cause nonspecific, background bands in a Western blot, which can make the experiment difficult to interpret. The possibility of cross-reactions should be considered
in any use of antibodies.
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LAB 11 • Immunodetection
Monoclonal vs. Polyclonal Antibodies
Although these terms do imply structural and functional differences in the IgG
molecules, it is more accurate to say that they define how that batch of antibodies
was produced. Polyclonal means that the antibody is a mix of different IgG molecules
that will bind to different epitopes on the antigen. The term “polyclonal” refers to
the fact that multiple, different “clones” of immune cells called B-lymphocytes produced the IgG, each clone producing one specific IgG molecule that binds a specific
epitope. Different clones of B-lymphocytes have different genetic rearrangements
of the heavy and light chain genes that encode the subunits of the IgG molecule;
these different genetic rearrangements are what produce the varying amino acid
sequence in the variable region of the IgG. Polyclonal antibodies are produced by
injecting the antigen (the protein of interest or a fragment of that protein) into an
animal to produce an immune response. In the immune response, multiple clones
of B-lymphocytes become activated and begin secreting IgG that will recognize the
antigen. The IgG molecules are then purified from the blood of the animal. What
type of affinity chromatography might you do to efficiently purify IgG proteins?
(Take a look back at Table 4.1 on page 51.)
Monoclonal means that the batch of antibody consists of a single IgG molecule
(with a single variable region amino acid sequence) that will bind only a single epitope of the antigen. To produce a monoclonal antibody, the antigen is injected into
an animal to produce an immune response just as with production of polyclonal
antibody. However, in this case, the B-lymphocytes themselves, and not the IgG,
are taken from the animal. This is done by removing the spleen, which is filled with
B-lymphocyte clones that are proliferating in response to the antigen. This mix of
different clones is then placed into culture. These cells are genetically manipulated so
that they will grow continuously in culture, and then single clones (hence monoclonal)
are isolated out of this mix. This is the production of “hybridoma clones” and this
will not be explained here—see an immunology textbook or Google “production of
monoclonal antibodies” for more information on this. In the end, what is obtained
are cells derived from a single clone of B-lymphocytes. These cells secrete the IgG
molecules into the culture medium in which they are grown, and the IgG is easily
purified from the culture medium.
Clearly, it is much more time consuming and expensive to produce a monoclonal
antibody (most researchers purchase antibodies from companies that specialize in
making them). Polyclonal antibodies, however, by virtue of the fact that they consist of many different IgG molecules that bind to different epitopes, are much more
likely to have undesirable cross reactions with proteins other than the antigen. The
more clones, the greater the chance one will recognize an epitope that is structurally
similar between different proteins. Thus, monoclonal antibodies are generally more
specific for the protein of interest and will produce fewer background bands from
non-specific binding in a Western blot. Polyclonal antibodies with a high degree of
specificity are also used for Western blots, however.
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LAB 11 • Immunodetection
Primary Antibody against MAP Kinase/Erk
The antibodies used to detect MAPK/Erk in your Western blot experiment are polyclonal antibodies that were raised in a rabbit against a fragment of human MAPK/
Erk. Remember that you are using two different primary antibodies, one that detects
only the phosphorylated (active) form of MAPK/Erk and one that detects both the
phosphorylated and dephosphorylated forms of MAPK/Erk (review section B on page
97 for how the activity of MAPK/Erk is regulated by phosphorylation). The antigen
used to produce the antibody against phospho-MAPK/Erk was a short (approximately
10-amino acids) fragment of the MAPK/Erk polypeptide that contained Thr 202, Glu
203, Tyr 204, with Thr 202 and Tyr 204 in their phosphorylated form. Because this
antigen is much smaller than the entire MAPK/Erk protein, far fewer B-lymphocyte
clones react against it and fewer IgG molecules are produced. These IgG molecules
will only bind to this short region of the MAPK/Erk polypeptide, and they will only
bind when MAPK/Erk is phosphorylated. Therefore, the anti-phospho-MAPK/Erk
antibody (“anti” meaning against, or what it binds to) binds only to active MAPK/
Erk and not to inactive MAPK/Erk.
The anti-total MAPK/Erk antibody you are using binds to both the phosphorylated/
active and dephosphorylated/inactive forms. The antigen used to produce this antibody is another short fragment of the MAPK/Erk polypeptide, but this fragment is
from a region away from Thr 202, Tyr 204. Therefore, the anti-total MAPK/Erk IgG
binds to a region away from that which is phosphorylated, and phosphorylation does
not affect its binding. The use of small peptide fragments of the greater MAPK/Erk
protein, for both the total and phospho-specific antibodies, allows the generation
of polyclonal antibodies that are relatively specific for only MAPK/Erk, and which
will not produce as many non-specific background bands on the Western blot. Both
of the peptide fragments used as antigens are amino acid sequences that are unique
to MAPK/Erk. The amino acid sequence in other regions of MAPK/Erk is not as
unique. All protein kinases share a conserved “kinase domain,” which is the amino
acid sequence that forms the ATP binding site in the active site of these enzymes.
What type of cross reactivity might you expect from a polyclonal antibody produced
by using the entire MAPK/Erk protein as the antigen?
Secondary Antibody against the Anti-MAPK/Erk IgG
The secondary antibody you will use is anti-rabbit IgG raised in a goat. This antibody
will bind to the constant region of the anti-MAPK/Erk primary antibody, and will produce a signal that will allow you to visualize your MAPK/Erk bands on the Western
blot. The constant region of the goat secondary antibody has a fluorescent molecule
attached to it. When this fluorescent tag is excited by UV light it will emit visible
light that will allow you to capture an image of the MAPK/Erk bands on your blot.
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LAB 11 • Immunodetection
B. Immunodetection of MAP Kinase
1. Discard the primary antibody solution and add 20 ml phosphate-buffered saline
(PBS). Keep the filter protein side up. Agitate gently, and discard the PBS into
a beaker (later this can be discarded into the sink). Tilt the tray slightly so any
liquid on the side walls will also be rinsed off.
2. Follow this quick rinse with three additional washes with PBS. For each wash, add
20 ml PBS to the tray. Gently agitate for five minutes at RT to rinse off unbound
antibody. Don’t let the filter dry in air between washes.
3. Before you pour off the third rinse, get 10 ml of the goat anti-rabbit IgG fluorescent-tagged antibody (2° antibody) solution from the TA. (This antibody is
diluted 1:500 in PBS with nonfat dried milk proteins as a blocking agent.)
4. Pour off the third rinse and immediately pour on the 2° antibody solution.
5. Incubate for one hour at RT on the rotator at slow speed.
6. Wash four times, 20 ml PBS each, as in the steps above, 10 minutes each wash
(longer washes, to remove all excess antibody).
7.
After the last wash, pour off the PBS and add deionized water to rinse the filter.
Leave the filter in the deionized water as you take it to the imager (E-gel Imager
from Invitrogen-Life Technologies).
8. Place the filter protein side down onto the UV transilluminator of the imager.
The “top-edge” of the filter, which corresponds to the top of the gel, should be
pointing away from you so that the image you collect will match the orientation
of your SDS-PAGE gel.
9.
Your instructor or TA will collect a digital image of the fluorescent bands on
this filter, and this will be made available to you as a .jpeg file.
10. After you have captured the images you need, ask your TA whether you should
keep the nitrocellulose filter or dispose of it in the lab trash container. Rinse the
incubation tray, remove any labeling tape, and return it to the bin box.
C. (For Sea Urchin Fertilization Project) Analyzing the Western Blot Images
Find your MAP kinase bands on your images, and confirm their position corresponds
to the correct molecular weight based on the positions of the MW standards. Sea
urchin MAP kinase will be a single band at 43 kDa. Once you are confident you are
looking at the correct bands, carefully examine the band pattern on your total-MAPK
blots first, looking for differences in the intensity of the bands between lanes. This
band represents the abundance of MAP kinase protein (both active and inactive)
present in the cells you were working with in Lab 9A—the darker the band, the more
MAP kinase that was present. Differences in band intensities between lanes could
mean that there was a difference in the amount of MAP kinase protein present in
the eggs under different treatments or that different amounts of total protein were
loaded onto the SDS-PAGE gel for those samples. Problems with unequal loading of
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LAB 11 • Immunodetection
egg lysate protein between lanes could occur from unequal numbers of eggs going
into those samples. In this case, the phospho-MAPK blot will be affected the same
way. Or the problem could have occurred during the process of actually loading the
sample onto the gel (spillage), in which case the phospho-MAPK blot would not be
affected. During sea urchin fertilization, it is still an open question as to whether
MAPK kinase protein becomes degraded. Examine the films to answer this question,
but keep the points about potential unequal loading in mind.
Next, examine the films from your phospho-MAPK blots. This band represents
the amount of Thr202/Tyr204-phosphorylated, and therefore active, MAP kinase
present in the cells. Look for a band in the different lanes to determine under which
conditions MAP kinase was active. The absence of a band indicates that no active
MAP kinase was present in those cells—it does not tell you, however, if inactive, dephosphorylated MAP kinase was present. The band may not be completely absent for
samples in which MAP kinase is inactive; a significant difference in band intensity
can also indicate a real effect. Keep in mind, though, that differences in amount of
total protein loaded on the gels will also affect band intensity. Your Western blot
results may not be perfect—this is a technically challenging experiment!—but it is
usually possible to answer your questions with what you have. Look at all of the lanes
on the films from each blot, and try to determine the overall pattern of changes.
What Is a Loading Control?
Due to the difficulty in finding antibodies that recognize sea urchin proteins, we do
not run a loading control with this experiment. However, Western blots typically
include a loading control that indicates the amount of total protein loaded into each
lane and serves to identify lanes where too much or too little sample was loaded. A
loading control is done by including a second primary antibody that recognizes a
protein whose abundance is known not to change with the experimental conditions.
Such proteins are sometimes referred to as “housekeeping proteins” because they
carry out basal, or routine, functions in the cell. Cytoskeletal proteins such as actin
and tubulin are often used as loading controls. If all the lanes of the SDS-PAGE
gel were equally loaded, the loading control antibody should yield bands of equal
intensity for every lane.
C. (For FGF Signaling Project) Analyzing the Western Blot Images
Find the .jpeg files on Canvas for the two Western blot images from your research
team. Open the .jpeg files. First, identify which end of the images are the top of the
blot; look at the MW standards to help you do this. Examine the blot probed with the
total-Erk antibodies first. They will have bands for Erk at 42 and 44 kDa, representing
the p42 and p44 Erk isoforms, respectively. You will also see a single band at 37 kDa
for the GAPDH loading control. It is possible that some background bands that result
from non-specific binding (or cross reaction) of the anti-Erk antibodies to other cellular proteins will occur. These background bands are typically more prominent in
the higher molecular weight-region of the blot. It is important to carefully establish
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LAB 11 • Immunodetection
which bands are p42 and p44 Erk and 37 kDa GAPDH by comparison to the MW
standards. The p42 and p44 bands indicate the total amount of Erk protein, both
active and inactive, present in the NIH 3T3 cells—the darker the band, the more
protein that is present. Are the bands equivalent in intensity across each lane of the
blot? The treatments you performed on the NIH 3T3 cells are not likely to alter Erk
protein levels in the short, 10-minute incubation time. However, this serves as an
important control. If the band intensity does not vary across the lanes of the blot,
you can be confident that any changes in band intensity you see on the phosphospecific Erk blots are due to differences in the amount of Erk activation, and not due
to differences in the amount of Erk protein present in the cells.
Now examine the blot probed with the anti-phospho Erk antibodies. On these
images, you should see some lanes with bands for p42 and p44 Erk, and all of the
lanes should have a band at 37 kDa, for GAPDH. Begin by identifying each band
at the correct MW and discriminating any non-specific, background bands. The
p42 and p44 bands show the amount of phosphorylated and, therefore, active Erk
present in the cell—the darker the band, the more Erk activation that occurred. The
GAPDH band should be the same intensity for every lane on the blot, if each lane
was accurately loaded with 5 μg of total protein.
What Is a Loading Control?
Western blots typically include a loading control that indicates the amount of total
protein loaded into each lane, and which serves to identify lanes where too much or
too little sample was loaded. A loading control is done by including a second primary
antibody that recognizes a protein whose abundance is known not to change with the
experimental conditions. Such proteins are sometimes referred to as “housekeeping
proteins” because they carry out routine metabolic or structural functions. GAPDH,
which is the glycolytic enzyme glyceraldehyde 3-phosphate dehydrogenase, is found
in all cells and is frequently used as a loading control. Cytoskeletal proteins such as
actin and tubulin are also frequently used. If all the lanes of the SDS-PAGE gel were
equally loaded, the loading control antibody should yield bands of equal intensity
for every lane.
Now that you understand what the various bands on the blot images represent, follow the steps below to label your blots and interpret the results of your experiment.
1. Import your .jpeg images into MS PowerPoint so that you can work with them.
Adjust the image size if necessary, and then label each lane of the blot. To make
it easier to analyze the data, label the lanes with actual experimental condition
(e.g., 5 μg/ml FGF-2 + suramin) rather than the sample number. Write the label
in a text box and then rotate the text box so that all of the lane labels will fit
above the gel. You will also want to show the positions of at least two or three
of your MW standards. Place a line at the position that the standard migrated
to and then add a text box to give the MW next to the line. Ask your instructor
or TA for help if you are not familiar with MS PowerPoint.
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LAB 11 • Immunodetection
2. Now examine your phospho-specific Erk blot to determine which experimental
conditions produced activation of the Ras-Erk pathway and which did not. You
should have at least one lane where the NIH 3T3 cells are “unstimulated,” meaning that neither FGF-2 nor calf serum were added. Find these lanes and look at
the band intensity to see what “no FGF signaling” down the Ras-Erk pathway
looks like. There may be faint bands in these lanes representing a basal level of
phosphorylated Erk. Activation of Ras-Erk pathway is indicated by an increase
in the band intensity from this basal level.
3. Next, examine the other lanes to see which, if any, conditions activated the
Ras-Erk pathway. For each lane, look at the GAPDH loading control. Was the
total protein loaded into that lane equivalent to the other lanes on the blot?
How much weight you put on a particular lane for answering your experimental
question depends on how confident you are that the lane was accurately loaded.
Disregard lanes that appear to be significantly under- or overloaded. Finally,
for each lane/condition, look at the lane on the corresponding total-Erk blot in
which the same sample was loaded. Is the amount of Erk protein equivalent to
the other experimental conditions?
D. Different Signal-Producing Conjugates for Secondary Antibodies
Peroxidase Conjugates
P7 peroxidases are enzymes that catalyze the oxidation of a variety of small molecule
substrates. Luminol is one compound that will serve as a substrate for peroxidase.
The oxidation of luminol is a chemi-luminescent reaction, meaning that light is
emitted. Peroxidase-conjugated (meaning that the peroxidase enzyme is attached
to the constant region) secondary antibodies are commonly used for Western blots.
The detection procedure for visualizing the protein bands using these secondary
antibodies is called ECL, for enhanced chemi-luminescence. After incubation with
the antibodies and washing, a substrate solution containing luminol is added to the
blot. The band where the antibody complexes are bound to the protein of interest
then emits light, and an image is captured using film or an imaging system.
Fluorescent Conjugates
The other method for visualizing the antibody complexes bound to the protein of
interest in Western blotting is fluorescence. Small molecule fluorescent dyes, such as
Fluorescein isothiocyanate (FITC), are used to tag secondary antibodies for a variety
of techniques such as immunofluorescence staining of tissue sections or cells, and flow
cytometry. However, these small molecule dyes are not suitable antibody conjugates
for Western blotting because they require excitation by wavelengths of light in the
visible range that would interfere with collecting an image of the protein bands on the
blot (see section E of Lab 16 for more information on absorption and fluorescence).
The fluorescent conjugate on the secondary antibody used in your Western blot is
not a small molecule dye, but is something called a quantum dot. Quantum dots are
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LAB 11 • Immunodetection
inorganic nanocrystals that are fluorescent. The fluorescent properties of quantum
dots are size dependent, meaning that the excitation and emission spectra will shift
as the diameter of the nanocrystal changes due to what is called the “quantum size
effect.” The quantum dots that are conjugated to our secondary antibodies have a
broad excitation band (band refers to a particular range of wavelengths) in the UV
part of the spectrum and a narrow emission band in the visible light range. This
allows the detection to be performed with a standard laboratory UV transilluminator and a digital camera. Quantum dots range from about 5–25 nm in diameter.
This is quite large in comparison to the size of the antibody it is attached to. Several
fluorescent dye molecules, such as FITC, are typically conjugated to the constant
region of an antibody. With quantum dots, however, several antibody molecules are
typically linked to single nanocrystal.
FIGURE 11.3
A. Quantum dot—secondary
antibody conjugate.
B. Small molecule fluorescent
dye—secondary antibody conjugate.
IgG
IgG
Fluorescent dye
Quantum dot
©Hayden-McNeil, LLC
Reference for Quantum Dots: Resch-Genger, U., Grabolle, M., Cavaliere-Jaricot, S.,
Nitschke, R., and Nann, T. 2008. Quantum dots versus organic dyes as fluorescent
labels. Nature Methods 5:763–775.
E. Quantifying and Normalizing Western Blot Band Intensities Using ImageJ
Western blot results are indeed quantitative; the darkness or intensity of the band
is directly proportional to the amount of antigen present in the sample. However, it
is difficult to put a number to differences in band intensities between lanes by eye
(for example, is one band 5 times darker than another band or 10 times darker?).
Additionally, if you run a loading control you would like to be able to use it to adjust
how you are quantifying the bands for your protein of interest based on differences in
the loading between lanes. If you can generate numbers that quantify the intensities
for your protein of interest bands and your loading control bands, you can normalize
the data to remove any differences from uneven loading between lanes. To do this
you simply divide the value for your protein of interest band in each lane by the value
for that loading control band. This sort of analysis is referred to as densitometry,
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LAB 11 • Immunodetection
and there are a variety of programs that allow one to quantify band intensities from
digital Western blot images. Digital imaging systems (such as the Life Technologies
E-gel system that we use) typically have built-in software for doing this.
ImageJ is an open source, image analysis software package that runs on Java. It is
a versatile platform for quantifying and manipulating digital images from a wide
variety of applications in biology. For Western blot analysis, it provides a simple and
easily accessible way to quantify band intensities. ImageJ is available on the computers in the Biochemical Techniques lab rooms; however, you may wish to install the
program on your own computer. It is free and easy to install.
Download ImageJ
1. Go to https://imagej.nih.gov/ij/index.html.
2. Click Download and select the appropriate version for your computer. You can
select from options that will install both Java and ImageJ. If you already have a
recent version of Java, then select Platform Independent, or ImageJ under Mac
OS X.
To check if you already have Java and which version, find Java in your programs
list and open About Java. For Windows, if you are installing Java you will need
to check if you should install 64-bit or 32-bit Java. Go to Settings or Control
Panel, select System, and click About. Look under System type to match the Java
version to your system.
Quantifying band intensities
1. You should have the image file for your Western blot saved on your computer
before beginning. ImageJ can analyze a wide variety of digital image file types.
2. Open the ImageJ application file (the one with the microscope icon). You will
see only one small menu bar appear.
3. Before you open the image file, define the type of measurement you wish to
make. Go to Analyze and select Set Measurements. The default setting with
which ImageJ opens will have the “Area,” “Min and max gray value,” and “Mean
gray value” boxes checked. Uncheck the Area and Min and max boxes and click
okay.
4. Open your Western blot image file (File, Open, and browse to select your image).
Examine the blot image. You can zoom in and zoom out in the image using the
up and down arrow keys.
5. Move the cursor across the image and examine the values displayed on the menu
bar. The x and y coordinates give the position of the cursor on the image. Value
gives the “darkness,” or number of contrasting pixels at those coordinates.
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LAB 11 • Immunodetection
Note that these instructions are written assuming you collected an “inverted”
image (dark bands on a white background) from the Life Technologies E-gel
imager. In this case the value will vary from 255–0 as the darkness increases.
If you did not invert the image from the E-gel imager (white bands on a black
background) the value will vary from 0–255. ImageJ measures light on dark so
the analysis is actually more straightforward if your image is not inverted.
Move the cursor across the bands and look at how the value changes. If most of
the band has a value of zero that means that band is “burned in” and the image
is too overexposed to accurately quantify that band.
6. Define an analysis area that you will place around each band. First, zoom in on
your bands so that the row of bands you wish to analyze fills a large portion of
your screen. Find the largest band in the row and draw a box around it (select
the rectangle from the menu bar and click and drag to draw). The largest band
should just fit inside the box. You can move the box by clicking inside it (the
cursor is an arrow) and dragging, or by using the arrow keys.
If you click off of the box it will disappear and you will need to redraw it. Save
the box; File, Save As, Selection, and name the file with protein band you are
quantifying. Once it is saved, you can open the box (find the file and open using
ImageJ) at any time and it will reappear on the image.
Note that if you are measuring p42 and p44 Erk you should measure the two
isoforms separately. Draw one box for p42 and another for p44, and treat them
as separate proteins of interest. A separate data column should be generated for
each different MW position on the blot.
7.
Measure the intensity of each band. Move the box over the protein band in lane
1 and carefully center it over the band. Go to Analyze in the menu bar and click
Measure. A results window will appear with the lane number and the mean
darkness of the band. You can also use ctrl + m to make a measurement.
8. Center the box over the protein band in lane two and take another measurement.
Proceed to the next lane until the bands in each lane have been measured. If
no band is apparent for the protein, center the box over the position where the
band would be and take a measurement. Be careful not to accidentally change
the size of the box between lanes. If this happens, delete the box (click anywhere
away from the box) and open it again from the saved selection file.
9.
150
Once you have measured each band, copy and paste the data from the Results
window into a Microsoft Excel file. Close the ImageJ Results window. You do
not need to save the measurements in ImageJ.
LAB 11 • Immunodetection
10. Take a background measurement for each band. To adjust for any background
signal on the blot near your bands you must make a background measurement
for each band to subtract from the band darkness value. Use the same box that
you used to measure the protein bands. Going from lane to lane in the same
order that you measured your protein bands, place the box just above or below
the protein band so that it is not touching any bands and make a measurement.
Copy and paste these into your Excel spreadsheet. See the example below for
how to set up your spreadsheet.
If your blot is very clean and has no significant background, the value above and
below each band may be 255. If this is the case you can omit this step.
11. Now you will need to “invert” the data in your Excel spreadsheet. This will give
you an increasing positive value as the darkness of each band increases (skip this
step if your E-gel Imager blot image was light bands on a dark background).
Create another column and subtract the value for each band from 255; click
on the cell where you want perform the calculation and type the following not
including the quotation marks: “=255-C5”. Instead of C5 type the cell number
for your un-inverted band value, or simply click on that cell. Then hit enter. You
can copy and paste the formula from this cell for all your other inverted value
cells. Do this for all your band of interest and background values.
12. Now subtract the inverted backgrounds from the inverted band values for each
lane. Select a cell in a new column and type the following not including the quotation marks: “=D5-H5” where D5 and H5 are your inverted band and inverted
background cells, respectively. Now you have final values for each band that you
can plot with a bar graph. Plot one bar for each lane and do not average lanes
with replicates of the same experimental condition.
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LAB 11 • Immunodetection
Normalizing band intensities to a loading control
1. Measure the band intensities for your protein of interest, and calculate the
inverted, background-subtracted values for each lane as described in the previous section.
2. Open the blot image that contains your loading control bands in ImageJ as
described above. This does not have to be the same image file that you used
to quantify your protein of interest. It is important to use the image from an
exposure where the bands are not burned in, so you may want to use different
exposures for your protein of interest and loading control.
3. Define the analysis area for your loading control bands by drawing a new box
around the largest band. Save this box as a selection and name the file with your
loading control.
4. Measure the value of the loading control band for each lane, in the same order
you used for your protein of interest. Paste the values into your spreadsheet and
calculate the inverted, background-subtracted values for each lane.
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LAB 11 • Immunodetection
5. Divide the actual (inverted, background-subtracted) values for the protein of
interest band in each lane by the actual loading control values. This gives you
the normalized values for your protein of interest bands, that are adjusted for
differences in loading between lanes.
You should present your normalized band intensities as a bar graph with one
bar for each lane. Do not take the average of lanes that are replicates of the same
experimental condition. Normalizing to the loading control controls only for
error that arises from differences in the amount of total protein loaded into each
lane. It does not control for error that may have occurred in the preparation of
the sample. Therefore, it is important to see the normalized values for each lane
of the blot to see how these other sources of error may be affecting your results.
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LAB 11 • Immunodetection
154
LAB 12 • ELISA Determination of IP1 Levels
LABORATORY
12
ELISA Determination of
IP1 Levels
Since antibody–antigen binding is so specific and sensitive, it forms the basis for a
wide variety of tests in both the clinic and the laboratory. The ELISA (Enzyme Linked
Immunosorbent Assay) is a widely used test in which the antigen–antibody complex
is detected with the help of an enzyme that converts a substrate to a colored product.
In this lab, you will use an ELISA to further investigate the initial signal transduction
that occurs during sea urchin fertilization, or in activation of the FGF receptor. The
antigen detected by the ELISA is a small molecule called inositol monophosphate
(IP1), which is the breakdown product of a critical second messenger called inositol
triphosphate (IP3). IP3 is produced by the action of a signal transduction enzyme
called phospholipase C, which is turned on immediately after sperm binding in sea
urchin fertilization or by phosphorylation of the FGF receptor. The three common
types of ELISA tests are the antibody capture assay, the competition assay, and the
two site capture assay. We will use the competition assay to detect IP1.
A. Introduction to the Phospholipase C Signal Transduction Pathway
Several signal transduction pathways become activated when the sperm binds and
fuses with the egg. The initial events that activate these downstream pathways are
the least understood part of the overall signaling cascade. It is clear that material
diffuses from the sperm acrosome, a large vesicle in the head of the sperm, into the
egg cytoplasm. However, the precise sequence of events, the ligands and receptors
involved, and their location in the egg remain the subject of much investigation and
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LAB 12 • ELISA Determination of IP1 Levels
debate. During sea urchin fertilization (there may be differences between species)
it is relatively well accepted that activation of the enzyme phospholipase C (PLC)
and subsequent production of inositol triphosphate (IP3) is the trigger for the cytoplasmic Ca2+ influx.
The PLC signal transduction pathway is a well-understood mechanism for relaying
signals from the cytoplasmic face of the plasma membrane into the interior of the
cell. It is known to function in a wide variety of organisms, from sea urchins to
humans. PLC belongs to a family of enzymes called lipases, which carry out hydrolysis reactions on lipids to cut them into smaller pieces. PLC catalyzes the cleavage
of the phosphorylated “head group” from a phospholipid called phosphatidyl-inositol
4,5-bisphosphate (PIP2). PIP2 is an abundant phospholipid in the inner leaflet of the
plasma membrane.
FIGURE 12.1 Phospholipase C signal transduction.
IP3
H
O
OH
H
OH
HO
P
H
P
PIP2
H
O
O
OH
H
OH
HO
O
H
P
P
−
O
olysis
Hydr
P
Phospholipase C
O
H2C
O
H
H
O
CH
O
O
C
C
CH2
OH
O
H
P
H 2C
H
H
H
O
O
CH
O
O
C
C
DAG
CH2
O
Membrane inner leaflet
When a biological signal is relayed across the plasma membrane, the proteins involved
interact with PLC and turn on its catalytic activity. PLC then rapidly cleaves the IP3
head group on adjacent PIP2 from the membrane-embedded lipid component. This
can produce large quantities of IP3 and diacylglycerol (DAG)—the now decapitated
lipid—in a short period of time. IP3, now freed from the plasma membrane, diffuses
throughout the cytoplasm where it interacts with specific receptors. Some of these
receptors are found on the surface of the endoplasmic reticulum, and binding of IP3
triggers the opening of Ca2+ channels and, thus, produces the cytoplasmic calcium
influx. The DAG lipid remains in the hydrophobic environment of the inner plasmamembrane leaflet, but it rapidly diffuses away from the site of cleavage and acts to
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LAB 12 • ELISA Determination of IP1 Levels
initiate a second arm of the PCL signaling pathway. Protein kinase C (PKC) is a serine/
threonine kinase that is associated with the inside of the plasma membrane, and
DAG-binding stimulates the activity of this enzyme (note that most PKC isoforms
also require Ca2+ ions for activity). The ELISA assay we will be using determines
the amount of IP3 that has been produced by measuring the concentration of its
breakdown product IP1, and thus indicates how much PLC activation has occurred
in the fertilized eggs.
Fertilization
Activation of
phospholipase C
Hydrolysis of
PIP2 to DAG and IP3
DAG
PKC
Phosphorylation of
protein targets
IP3
Ca2+ release from ER
Once it is cleaved from PIP2, IP3 does not survive long in the cytoplasm. It has a
half-life of less than 30 seconds and is rapidly dephosphorylated by phosphatases
to free inositol. As with any signaling second-messenger, it needs to be short-lived
so that the “on” and “off” states of the biological signal that it relays can be precisely
controlled. To produce a prolonged signal, PLC must continue to be stimulated by
whatever is upstream of it so that it will produce more IP3. Once PLC activity stops,
the IP3 is rapidly broken down and the signal is turned off. The ELISA we are going
to use detects IP1 (resulting from the cleavage of two of the three phosphate groups
from inositol). In order to prevent IP1 from being broken down to free inositol, you
added lithium chloride to your egg samples.
B. Background on Enzyme-Linked Immunosorbent Assays
In the antibody capture assay, the antigen is coated onto the surface of the wells. The
wells are made from a special plastic (polystyrene) that proteins stick to. Following
incubation with the antigen, the excess is washed off and an antibody specific for
that antigen is added (primary antibody). The antibody has to be specific because
many substances are usually present in the test sample and will bind to the plastic.
Excess antibody is washed off, and then a secondary antibody, which will bind to
the immunoglobulin type of the first antibody, is added. The secondary antibody
has an enzyme covalently linked to it. Commonly used enzymes are peroxidase and
alkaline phosphatase. Following washing to remove the excess secondary antibodyenzyme complex, the enzyme substrate is added and a colored product is produced.
The intensity of the color is a measure of how much antigen was present in the well.
If no antigen was present, no antibodies bind and no color will develop.
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LAB 12 • ELISA Determination of IP1 Levels
FIGURE 12.2 Antibody capture assay.
Product
©Hayden-McNeil, LLC
Substrate
Antigen
1° antibody
2° antibodyperoxidase
In the two-site capture assay, two different antibodies that both bind to the antigen
are used. The pregnancy tests that are available at pharmacies are derived from this
type of assay. The wells are coated with an antibody specific for one region of the
antigen, then the test solution containing antigen is added. Following washing, the
second (not a “secondary”) antibody, which recognizes a different epitope of the antigen, is added. This antibody has the enzyme coupled to it. In some pregnancy tests,
the first antibody recognizes a portion of the beta chain of the pregnancy hormone,
HCG (human chorionic gonadotropin). The beta chain is unique to that hormone.
The second enzyme-conjugated antibody recognizes the alpha chain of HCG, which is
also found in LH (luteinizing hormone), present in nonpregnant women and men (in
women LH causes ovulation and in men LH stimulates cells in the testes to produce
testosterone). The two-site capture assay cannot be used to detect small antigens like
IP1 or steroid hormones that are too small to have more than one epitope.
FIGURE 12.3 Two-site capture assay.
Substrate
Product
©Hayden-McNeil, LLC
Antigen
Antibody #1
Antibody #2
plus peroxidase
We will use the competition assay to detect IP1 in today’s experiment. The competition assay is the only one of the three types of ELISA that can accurately quantify
the amount of antigen present. In the antibody capture method, the amount of
antigen that will bind to the wells can vary considerably depending on what else
is in the sample, so this is not a good method for quantifying the antigen. In the
competition assay we will use, the wells are coated with antiserum against mouse
immunoglobulins (this is the secondary antibody). Then a defined amount of pure
IP1 coupled to peroxidase is added to each well, followed by a monoclonal antibody
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LAB 12 • ELISA Determination of IP1 Levels
against IP1 (this is the primary antibody). The IP1-peroxidase conjugate competes with
the IP1 present in egg lysate samples to bind to a fixed number of primary antibodies
present in the well. Complexes formed with the primary antibody then bind to the
secondary antibody coating the wells. After incubation, all the unbound molecules
are washed out and substrate for the peroxidase is added. The peroxidase will convert the substrate to a colored product. Therefore a high concentration of IP1 in the
egg lysate will result in less color development, since the lysate IP1 outcompetes the
IP1-peroxidase conjugate for the fixed amount of primary antibody. A low concentration of IP1 in the lysate will result in more color development, because more of the
IP1-peroxidase conjugate is binding to the primary antibody.
FIGURE 12.4 Competition assay.
Lysate IP1
Substrate
Product
©Hayden-McNeil, LLC
IP1
peroxidase
2° antibody
1° antibody
This is the second time you have encountered peroxidase in the techniques you are
performing for this project. Peroxidases are a large, diverse family of enzymes that
all catalyze reactions where hydrogen peroxide is a substrate. Recall that you added
H2O2 to your substrate solution for the ECL detection of MAP kinase in Lab 11.
The P7 subfamily of peroxidases carries out a wide variety of functions in plants. A
general equation for the reactions they catalyzed can be written as:
H2O2 + AH2
$
2H2O + A
where A is any of a wide variety of small molecules that become oxidized in the
reaction. Peroxidase from horseradish (abbr. HRP) is a 44 kDa enzyme that is widely
used as a tool in the molecular biology laboratory as a way of producing a visible
signal to detect antibody binding. Because it is an enzyme, continuous conversion
of the substrate to its oxidized form serves to amplify the signal from the binding of
a small number of antibody molecules. HRP is convenient because it has relatively
loose substrate specificity and can catalyze the oxidation of many aromatic compounds. Oxidation of different substrates can, in turn, produce different types of
signals, from luminescent to colorimetric (color producing). The substrate you will
use with your competition ELISA is TMB (3,3’,5,5’-Tetramethylbenzidine). Oxidation of TMB shifts its absorption max to give the product a blue color. The reaction
is stopped by the addition of sulfuric acid, which converts the product to a yellow
color that can be measured at 450 nm.
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LAB 12 • ELISA Determination of IP1 Levels
C. ELISA Procedure
Each group will use four wells. You will either run both of your samples in duplicate,
or you will run one sample in duplicate, one well of the other sample, and one of
the control wells.
TABLE 12.1 (If you are doing the FGF signaling project, your TA will provide this table.)
Group
Sample #
Control
1
1 (×2)
2
Blank
2
3 (×2)
4
TA
3
5 (×2)
6
NSB
4
7 (×2)
8 (×2)
5
9 (×2)
10 (×2)
6
11 (×2)
12 (×2)
7
13 (×2)
14 (×2)
8
15 (×2)
16 (×2)
9
17 (×2)
18 (×2)
10
19 (×2)
20 (×2)
11
21 (×2)
22 (×2)
12
23 (×2)
24 (×2)
1. Collect the following materials:
a.
ELISA wells; your group and a neighboring group will share one set
b. IP1-peroxidase conjugate; 100 μl*
c.
Anti-IP1 monoclonal antibody; 100 μl*
d. Your two egg lysate samples
e.
Plastic Petri dish and clay balls
f.
Dilution buffer; 300 μl
*Note: Your TA will provide you with aliquots of these.
Two neighboring groups will each use 4 wells of an 8-well strip (they cannot be
broken apart without shattering). The ELISA wells are pre-coated with a secondary antibody against the constant region of the anti-IP1 monoclonal antibody.
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LAB 12 • ELISA Determination of IP1 Levels
2. Thaw your lysates by warming them with your fingers and mix by gently flicking
the tube or stirring with a pipette tip.
•
•
If you are doing the sea urchin fertilization project, add 60 μl of dilution
buffer to each of your two egg lysates, and mix.
If you are doing the FGF signaling project, do not dilute the lysates. You
will use them as is.
Keep the diluted lysates on ice until you are ready for them.
3. Examine Table 12.2 to understand what you will be adding to each ELISA well,
and then label the wells that you will use. Pipette 50 μl of the lysate (diluted
or undiluted, depending on your project) into the appropriate wells. Group 3
should add 75 μl of dilution buffer and 25 μl of only the IP1-peroxidase conjugate
to the NSB well.
4. Add 25 μl of IP1-peroxidase conjugate to each well.
5. Add 25 μl of anti-IP1 monoclonal antibody to each well. Gently pipette up and
down two or three times to mix, being sure to expel all the liquid from the
pipette tip when you are done.
6. Place your wells into the plastic Petri dish and secure them with the clay balls.
Add strips of moistened paper towels beside the wells (your TA will demonstrate).
Cover with the lid and incubate the wells on the orbital shaker for two hours at
room temperature.
In addition to a blank, the two controls that are being run are the “nonspecific binding” (NSB) control and the “total activity” (TA) control. The table below summarizes
the reagents that are added to the wells for the lysate samples and the controls. What
information will the two controls provide?
TABLE 12.2
IP1 and
antibody
binding
Samples
NSB
Blank
TA
Standard solutions
---
---
---
---
Cell lysates
50 μl
---
---
---
IP1-peroxidase conjugate
25 μl
25 μl
---
---
Anti-IP1 Mab
25 μl
---
---
---
Buffer
---
75 μl
---
---
2-hour incubation on orbital shaker—Wash wells
Peroxidase
reaction
TMB substrate
100 μl
100 μl
100 μl
100 μl
IP1-peroxidase conjugate
---
---
---
5 μl
100 μl
100 μl
100 μl
20-minute incubation
Signal
production
Stop solution
100 μl
161
LAB 12 • ELISA Determination of IP1 Levels
7.
During the incubation collect 6 ml of wash buffer, 410 μl of TMB substrate solution, and 410 μl of stop solution.
8. Once the incubation is complete, remove the wells from the Petri dish. You will
need to wash the wells before adding the substrate solution. You will use the
vacuum aspirator on your bench to remove the liquid from the wells. Insert a
glass Pasteur pipette into the vacuum hose (no more than 1 cm) and then turn
on the vacuum. Add a 200 μL micropipette tip to the end of the Pasteur pipette;
it will be held in place by the vacuum. When aspirating the wells you should
gently touch the tip into the bottom corner of the well without scraping down
the side of the well. Be sure to remove all of the liquid from the well. Touch the
tip to the bottom on the same side of the well each time you aspirate so that you
will minimize disruption or removal of the antibody complexes attached to the
surface of the plastic.
9.
Remove the lysate/antibody solution from the wells and add 250 μl of wash buffer
per well. Gently tap the wells and then remove the wash buffer. Repeat this five
more times. Do not allow the wells to dry out between washes.
10. Remove the last wash and add 100 μl of the TMB substrate solution to each well.
Incubate for exactly 20 minutes. The color is produced by the enzymatic oxidation of the TMB substrate and, therefore, the amount of color produced is time
dependent. Place the wells back in the Petri dish and cover. Cover the Petri dish
with a paper towel during the incubation to protect it from light.
Safety Note: The stop solution contains 0.2 M sulfuric acid. Wear your PPE
and gloves while pipetting it.
11. Do not aspirate the TMB. Add 100 μl of stop solution to each well. Bring your
wells to the front and your TA will supervise you transferring the contents to a
96-well plate. If you are doing the sea urchin fertilization project, use Table 12.3
as your guide for transferring samples. If you are doing the FGF signaling project,
your TA will provide the table for transferring the samples. Begin the samples
in the upper-left well of the plate. Transfer only 175 μl from each well and leave
the rest. This will ensure that each well of the 96-well plate has precisely the
same volume (why is this important?). You will now measure the absorbance of
each sample at 450 nm with the plate reader.
12. Aspirate the remaining 25 μL from each well into the TMB/Stop Solution waste
flask on the middle bench. Place the empty 8-well strip in the lab trash bucket.
Return the incubation dish and clay to the middle bench.
162
LAB 12 • ELISA Determination of IP1 Levels
TABLE 12.3 96-Well Plate Loading.
Seawater-Alone Samples
All replicates
Fertilized Samples
All replicates
A23187 Samples
All replicates
DMSO Samples
All replicates
Controls
NSB
TA
NSB—nonspecific binding
Blank
TA—total activity
D. Analysis of ELISA Data
In order to generate IP1 concentrations from the absorbance values for your sample
wells, you will need to use a standard curve, where the absorbance has been measured for samples with known IP1 concentrations. The standard samples will be run
for you so that you will have more wells available to run replicates of your unknown
samples. The standard data will be provided on Canvas, and you will use it to generate the standard curve. The data will include the absorbance values for the following
IP1 concentrations, plus a blank.
Standard
Stn. 0
Stn. 1
Stn. 2
Stn. 3
Stn. 4
Stn. 5
IP1 Concentration
in Well
0 nM
17 nM
52 nM
208 nM
833 nM
5000 nM
1. Obtain the standard data from Canvas. Subtract the blank from each absorbance
value and use the adjusted values to plot your standard curve. Plot the log10 of the
nanomolar IP1 concentration on the x-axis and the corresponding absorbance
on the y-axis. Do not plot the 0 nM IP1 standard. This standard is included only
to provide a reference absorbance value for samples that contain no IP1.
2. The data will follow a sigmoid or S-shaped trend. Add a best-fit curve using the
Trendline function in Excel (Chart Tools; Layout; Trendline; More Trendline
Options). Select a trend line equation that is polynomial to the third order, and
check the boxes to display the equation and the R-squared value. The R-squared
value indicates how well the curve generated by an equation fits a particular set of
data points; the closer the R-squared value is to 1, the better the fit. Be sure that
the trend line fits the standard data points with an R-squared of 0.97 or better.
Note: MS Excel has limited graphing and curve-fit capability. While the polynomial third-order equation will accurately give log IP1 concentrations as a
function of absorbance inside the range of our IP1 standard concentrations,
the equation does not describe the relationship outside of this range—you
will see when you plug the polynomial equation into a graphing calculator.
Graphing and analysis software such as Prism GraphPad or Matlab is better
suited for this type of analysis. Feel free to use an alternate software package
if that is an option for you.
163
LAB 12 • ELISA Determination of IP1 Levels
3. Subtract the blank from each of your experimental sample absorbance values.
Use the blank from your experimental ELISA plate, not the blank from the
standards.
4. You can now use the standard curve to determine the log nM IP1 for your
unknown samples. Use your graphing calculator or a free, online graphingcalculator package to do this. An easy-to-use graphing calculator can be found
at https://www.desmos.com/calculator. Enter the equation for your standard
curve into the calculator. Determine the log IP1 from the absorbance values for
the four ELISA wells done by your group (for the Desmos calculator, click and
drag along the curve to see the x, y-coordinates). Only use the x values that fall
within the range of the log10 IP1 concentrations for your standards—this is the
part of the curve that matches your MS Excel standard curve. Take the anti-log
to get the nanomolar IP1 concentration. You will share the IP1 concentrations
for your wells with the rest of your class.
5. You may not be able to determine the log IP1 for absorbance values greater than
that of the lowest concentration IP1 standard (17 nM). Use the 0 nM IP1 standard
to verify the absorbance value that corresponds to no IP1 present in the well.
Absorbance values that fall between those of the 17 nM and 0 nM standards can
simply be taken as zero IP1. Don’t be alarmed if many of your wells give zero IP1
concentration. Some of your experimental conditions should hopefully not be
stimulating phospholipase C (PLC)!
6. Be sure that the IP1 concentration for each well is determined independently (i.e.,
don’t average the absorbance values). Once you have all the IP1 concentrations,
calculate the average and standard deviation for all of the replicates of each
experimental condition. The higher the average IP1 concentration, the greater
the PLC activation that occurred under that condition. In which conditions
did you predict you would see activation of PLC? Do your results match your
prediction?
164
LAB 13 • Isolation of Fluorescent Protein Plasmid DNA
LABORATORY
13
Isolation of Fluorescent Protein
Plasmid DNA
In this project, you will express the genes for fluorescent proteins in bacterial cells,
and then purify the proteins from the cells. By doing this, you will learn how foreign
proteins can be expressed in a host organism (the E. coli bacterial cells) in order to
more easily obtain large quantities of the protein. Proteins produced in this manner
are referred to as recombinant proteins because the gene for the protein has been
inserted into a plasmid vector that is taken up and carried by the host organism.
Thus, the word recombinant refers to recombination of the gene encoding the protein, not the protein itself. It is typically easier to purify the proteins from bacterial
cells than it is from the organism and tissue in which it is naturally produced. Large
amounts of recombinant proteins can be expressed and purified from bacterial
cells for research, medical, and commercial use. Insulin, used to treat patients with
diabetes, is produced as a recombinant protein.
You will be assigned two unknown fluorescent proteins that you will have to identify.
In Lab 13, each group will be given two different cultures of the bacterium E. coli
strain DH5α that contain a plasmid with the gene for a fluorescent protein. Strain
DH5α is used to produce large amounts of the plasmid but is unable to express the
fluorescent protein because it lacks the molecular machinery (T7 RNA polymerase)
required to drive transcription. In Lab 13, you will lyse the DH5α bacteria and purify
the plasmid DNA. In Lab 14, you will learn how to digest a plasmid with restriction
enzymes and analyze the DNA fragments by agarose gel electrophoresis. In Lab 15,
you will transfer the plasmid DNA into a different E. coli strain (JM109-DE3) that is
able to express the fluorescent proteins (this strain produces T7 RNA polymerase).
The bacteria with the plasmid will be plated out to grow and synthesize the proteins. In Lab 16, the proteins will be purified, and you will begin to analyze them by
165
LAB 13 • Isolation of Fluorescent Protein Plasmid DNA
measuring the excitation (absorption) and emission spectra of their fluorescence.
You will then perform SDS-PAGE to determine the subunit composition of your
unknown fluorescent proteins in Lab 17. In Lab 17, you will also analyze the results
of the Coomassie-stained gel.
A. Background on Fluorescent Proteins
Fluorescent proteins (FPs) are found in light- and color-producing cells of many
coelenterates (e.g., jellyfish and corals). Fluorescence is the process by which light
is absorbed by a molecule and then reemitted at a longer wavelength, producing a
particular color. The function of FPs in these marine organisms remains an intriguing mystery. When excited by light, the proteins fluoresce without requiring ATP or
any other cofactor. A fluorophore is defined as the chemical structure in a fluorescent molecule that absorbs and emits the photons of light (the term chromophore
is sometimes also used, although this more precisely describes the active chemical
group in compounds that only absorbs light and does not fluoresce). In FPs, the
fluorophore is typically a double ring structure formed by three amino acids. It is
enclosed in a barrel formed by 11 strands of beta-pleated sheet, with additional amino
acid sequence closing the top and bottom. For this reason the structure is described
as a beta barrel (see the figures at the end of this section).
FPs are of great interest because they can be used as “molecular tools” to carry
out experiments. They can often be attached to other proteins without interfering
with their functions. This makes it possible to label and detect molecules, cells, and
organisms while they go about their normal activities. Currently there is an explosion of knowledge in areas such as cell and developmental biology, neurosciences,
and ecology where FPs are making important contributions. FPs are also being used
for screening drugs, evaluating viral vectors for human gene therapy, biological pest
control, and monitoring genetically altered microbes in the environment, among
other applied efforts.
You will receive two unknowns from a panel of 12 different fluorescent proteins. All
of these were derived from one of two naturally occurring “parent” proteins, either
green fluorescent protein or DsRed. The derivative FPs were created by introducing
point mutations into the genes on which they are encoded. These point mutations
produce specific changes in the amino acid sequence, referred to as “amino acid
substitutions.” By changing the amino acid sequence of a protein with amino acid
substitutions, you can change its biochemical properties and alter its function. Green
Fluorescent Protein (GFP) is produced by the jellyfish, Aequorea victoria. The GFP
protein is composed of a single beta-barrel polypeptide. EGFP, EYFP, and ECFP
were derived from the wild-type GFP (see Table 13.1). The amino acid substitutions
in EGFP increased (“E” for enhanced) the fluorescent light output. The color of the
emitted light was changed in ECFP (cyan) and EYFP (yellow).
We also have nine different red fluorescent proteins. The wild type red fluorescent
protein, DsRed, is from the coral, Discosoma. DsRed is a bulky tetramer formed from
four identical beta-barrel subunits. dTomato is a dimer (two subunits), and mRFP1
is a monomeric (single subunit) form with mutations that eliminate aggregation.
166
LAB 13 • Isolation of Fluorescent Protein Plasmid DNA
There were 33 amino acid substitutions that were needed to derive mRFP1, the
first functional monomer red protein, from DsRed. mRFP1 was further modified to
produce mOrange, mStrawberry, mGrape1, and the other FPs. The “td” prefix indicates those proteins are tandem dimers. These are not true dimers, but are created
by joining the ends of two single beta-barrel polypeptides by splicing together the
gene-coding sequences. The resulting polypeptide is twice as long, with either end
folding into a separate beta barrel. tdTomato is a single polypeptide of two identical
red-protein beta barrels. tdGO was created by fusing the polypeptides for one green
protein and one red protein.
TABLE 13.1 Fluorescent Proteins.
Parent
Protein
GFP
DsRed
Green–Red
Hybrid
Fluorescent
Protein
Subunit
Composition
Excitation Max
Emission Max
EGFP
monomer
488 nm
507 nm
ECFP
monomer
434 nm
(452 nm)
476 nm
(505 nm)
EYFP
monomer
514 nm
527 nm
DsRed
tetramer
558 nm
584 nm
dTomato
dimer
554 nm
581 nm
tdTomato
tandem dimer
554 nm
581 nm
mRFP1
monomer
584 nm
607 nm
mOrange
monomer
548 nm
562 nm
mGrape
monomer
595 nm
620 nm
mStrawberry
monomer
574 nm
596 nm
mCherry
monomer
587 nm
610 nm
tdGO
tandem dimer
?
?
Select Bibliography
Baird, G.S., Zacharias, D.A., Tsien, R.Y. (2000) Biochemistry, mutagenesis, and
oligomerization of DsRed, a red fluorescent protein from coral. Proc Natl Acad
Sci USA 97(22): 11984–11989.
Campbell, R.E., Tour, O., Palmer, A.E., Steinbach, P.A., Baird, G.S., Zacharias, D.A.,
Tsien, R.Y. (2002) A monomeric red fluorescent protein. Proc Natl Acad Sci USA
99(12): 7877–7882.
Lippincott-Schwartz, J., Patterson, G.H. (2003) Development and use of fluorescent
protein markers in living cells. Science 300: 87–91.
Shaner, N.C., Campbell, R.E., Steinbach, P.A., Giepmans, B.N., Palmer, A.E., Tsien,
R.Y. (2004) Improved monomeric red, orange, and yellow fluorescent proteins
derived from Discosoma sp. red fluorescent protein. Nat. Biotechnol 22(12):
1567–1572.
167
LAB 13 • Isolation of Fluorescent Protein Plasmid DNA
FIGURE 13.1 Protein folding and beta-barrel conformation of GFP.
23
25
48
57
217
208
147
171
176
100
105
128
Amino acid number
Beta strand
1
2
11
3
36
*
41
11
71
10
227
N term
7
199
8
153
9
161
4
187
5
92
6
115
118
C term
©Hayden-McNeil, LLC
*Fluorophore S65 Y66 G67
Alpha helix
secondary structure
©Hayden-McNeil, LLC
Beta-sheet
secondary structure
N
C
©Hayden-McNeil, LLC
168
LAB 13 • Isolation of Fluorescent Protein Plasmid DNA
FIGURE 13.2 Structure of the GFP fluorophore.
HO
tyr
CH
O
O
O
gly
N
O
N
CH2
val
C
NH
CH
C
CH
N terminus
NH
CH
C
NH
CH
H3C
ser
phe
CH2
CH2OH
CH3
NH
CH
CO
CH2
gln
C terminus
CH2
C
O
F–S–Y–G–V–Q
NH2
B. Background on the Plasmid Vector
The fluorescent protein genes are carried in the plasmid vector pRSET-B, in the
host bacterium E. coli strain DH5α. Plasmids are circular, double-stranded DNA
molecules capable of autonomous replication within a host bacterium. They are not
part of the bacterial chromosome. Although they can replicate independently of
the host chromosome, plasmids require host proteins to do so. Plasmids are found
in bacteria in nature, where they can carry genes that confer antibiotic resistance
or the ability to metabolize a particular nutrient. They have been exploited by
molecular biologists for use as vectors. A vector is a vehicle for maintaining and
propagating cloned genes in a host organism, such as E. coli. Foreign genes can be
inserted into a plasmid vector and are then replicated along with the plasmid DNA
inside the bacterial cell. Cloning is the production of multiple identical copies of
an organism, cell, or gene (clon is Greek for twig—some plants can be cloned just
by growing cuttings). If a gene is isolated from the genetic material of an organism
and is inserted into a vector where identical copies can be replicated, it is referred to
as having been cloned. Thus, the fluorescent protein genes were previously “cloned”
into the pRSET plasmid.
The plasmid vector pRSET-B has been specifically designed for the expression and
purification of proteins from E. coli. The plasmid was purchased from Invitrogen, a
biotech company that specializes in tools for molecular biology. Many specialized
genetic elements have been engineered into pRSET-B to make it easy to express and
purify foreign proteins.
All plasmid vectors have some common elements that are necessary for cloning
genes into the vector, and for the plasmid’s stable replication in E. coli cells. These
include the following elements on pRSET-B.
169
LAB 13 • Isolation of Fluorescent Protein Plasmid DNA
•
Origin of Replication (pUC): Plasmids remain autonomous of the bacterial chro-
mosome, and require an origin of replication for the E. coli cells to replicate
the plasmid DNA. The pUC sequence is about 700 bp long, and it is a common
replication origin in E. coli plasmids. The E. coli DNA polymerase binds to this
sequence to replicate the plasmid DNA.
•
Selectable Marker (AmpR): This allows E. coli cells that carry the plasmid to survive
in the presence of the antibiotic ampicillin, while cells that have not taken up
the plasmid are killed off. The AmpR gene (referred to as BLA in E. coli genetics)
encodes an enzyme called β-lactamase, which breaks down ampicillin.
•
Multi-Cloning Site (MCS): This is a series of different DNA sequences called restric-
tion sites. These sites can be recognized and cut by different endonuclease
enzymes (or restriction enzymes). The restriction enzymes that can cut pRSET-B
at the MCS (BamHI–HindIII) are given in Figure 13.3. Each of these enzymes
cuts a different restriction site in the MCS. This tandem series of restriction sites
allows the gene encoding the protein of interest to be inserted into the vector.
The DNA fragment containing the gene of interest and the plasmid vector are
both cut with the same restriction enzymes and the two are then joined at the
restriction sites.
The genes for the FPs we will be working with have been cloned into the MCS of
pRSET-B. The 5’ end of the FP gene was cut with the restriction enzyme BamHI and
then inserted into the BamHI restriction site on pRSET-B. The 5’ end of the gene
encodes the amino-terminal end of the protein. The 3’ end of the FP gene (encoding
the carboxyl-terminal end of the FP) was inserted into the EcoRI restriction site. The
intervening sequence between the BamHI and EcoRI sites on pRSET-B was removed
during the process of adding the FP gene.
There are several genetic elements in the promoter region of pRSET-B that facilitate
high expression levels of the protein of interest in E. coli cells, and that also provide
an efficient way to purify the protein. Note that the 5’ end of the FP gene was inserted
at the BamHI site in such a way that the coding sequence of the gene is in the same
reading frame with some additional coding sequence added by pRSET-B (see Figure
13.4). This allows some amino acids encoded by pRSET-B to be added on to the aminoterminal end of the FP. This additional amino acid sequence contains two epitope
tags that can be used to rapidly purify the FP by affinity chromatography. The term
epitope tag comes from the fact that the first affinity resins used for this purpose
had covalently bound antibodies that would recognize a specific protein epitope
(see Lab 11 for the definition of an epitope). These antibody-containing resins can
purify any protein to which that epitope has been added. These days, an epitope tag
can be any amino acid sequence added to a protein that allows purification with an
affinity resin. You will use the His-tag to purify your FPs, and this does not utilize
an antibody–epitope interaction.
170
LAB 13 • Isolation of Fluorescent Protein Plasmid DNA
FIGURE 13.3 Plasmid map and cloning of FP gene into pRSET-B.
EcoR I
BamH I
mOrange gene
723 bp
BamHI
XhoI
BglII
PstI
NcoI
EcoRI
HindIII
Promotor region
T7 Promoter
pRSET-B
2897 bp
MCS
AmpR
pUC replication
origin
T7 Promoter
BamHI
Insert gene
pRSET-B with
mOrange gene insert
3580 bp
EcoRI
HindIII
AmpR
pUC
C
LL
il,
cNe
n-M
e
d
y
a
©H
171
LAB 13 • Isolation of Fluorescent Protein Plasmid DNA
Because additional amino acid sequence is added by pRSET-B, this means that the
start codon (ATG) to initiate translation is in pRSET-B vector sequence and not
the FP gene-insert sequence. You should understand where transcription starts
and stops, and where translation starts and stops when you examine Figure 13.4.
You should also understand the role of each genetic element in the promoter region
in either driving high levels of protein expression, or for manipulating the protein
once it has been produced. Consider the level of gene-to-protein information flow
at which each element functions to accomplish its role.
•
T7 Promoter: This DNA sequence is the binding site for T7 RNA polymerase, which
allows it to initiate transcription. Wild-type E. coli does not produce this RNA
polymerase. Only E. coli strains to which the gene for T7 RNA polymerase has
been added (such as JM109-DE3) will produce the polymerase and allow those
cells to transcribe the gene inserted into the MCS. The T7 promoter sequence
was “borrowed” from some of the genes of T7 bacteriophage. You will note that
a few genetic elements in pRSET-B have been taken from this bacteriophage.
•
T7 Terminator: This sequences forces the T7 RNA polymerase to disengage from
pRSET-B, terminating transcription.
•
Ribosome Binding Site (RBS): This sequence facilitates efficient binding of the
bacterial ribosome to the mRNA transcript of the gene. Once the ribosome is
attached to the mRNA, it starts scanning for a start codon.
•
T7 Gene 10 Leader: The position of this sequence just after (or “downstream” from)
the vector ATG increases the probability that the ribosome will begin translating
from that start codon on the mRNA (AUG). If the ribosome missed the vector
start codon, are there other start codons at which translation could begin? Are
they all in the desired reading frame?
•
Stop Codons: The FP gene insert has a stop codon at the end of its coding sequence,
and this is where translation of the FPs is terminated. Note that pRSET-B provides a stop codon just past the end of the MCS in case the inserted gene does
not contain a stop codon.
•
Polyhistidine Tag (His-Tag): This run of six histidine amino acids is an epitope tag
that can be bound by nickel-NTA affinity resin (see Lab 16).
•
XpressTM Epitope Tag: This is a traditional epitope tag that can be bound by an
affinity resin containing a covalently attached antibody. This sequence of amino
acids (see Figure 13.4) forms the epitope that is bound by the antibody.
•
Enteropeptidase (EP) Recognition and Cleavage Site: This amino acid sequence is
recognized and cut by the EP protease. This provides a way to remove the additional amino acid sequence encoded by pRSET, after the protein of interest has
been purified using one of the two epitope tags. An interesting side note, the
small intestines of all mammals produce EP. This is the protease that cleaves and
converts inactive trypsinogen to the active digestive protease, trypsin.
172
LAB 13 • Isolation of Fluorescent Protein Plasmid DNA
FIGURE 13.4 DNA sequence-level map of the pRSET promoter region (blue) with the mOrange
gene insert (orange) at the BamHI and EcoRI restriction sites. The amino acids encoded by the
vector and FP gene insert are shown below the DNA sequence.
T7 promoter
AATACGACTC
RBS
ACTATAGGGA
GACCACAACG
GTTTCCCTCT
AGAAATAATT
TTGTTTAACT
Polyhistidine tag
TTAAGAACGGA
T7 gene 10 leader
GATATACAT ATG CGG GGT TCT CAT CAT CAT CAT CAT CAT GGT ATC GCT AGC ATG ACT GGT
Met
Arg
Gly
Ser
His
His
His
His
His
His
Gly
Met
Ala
Ser
Met
Thr
Gly
BamHI
Xpress™ Epitope tag
GGA CAG CAA ATG GGT CGG GAT CTG TAC GAC GAT GAC GAT AAG GAT CCG ATG GTG AGC
Gly
Gln
Gln
Met
Gly
Arg
Asp
Leu
Tyr
Asp Asp Asp
Asp
Lys
Asp
EP recognition site
Pro
Lys
Gly
Glu
678 bp
Val
Ser
EP cut site
EcoRI
AAG GGC GAG GAG
Met
HindIII
GAG CTA TAC AAG TAA GAA TTC GAA GCT TGA TCCGG
Glu
Glu
Leu
Tyr
Lys stop
stop
T7 terminator
CTGCT
AACAAAGCCC
GAAAGGAAGC
TGAGTTGGCT
GCTGCCACCG
CTGAGCAATA
ACTAGCATAA
CCC
©Hayden-McNeil, LLC
C. Alkaline Lysis Mini Plasmid Prep
Overnight cultures of E. coli DH5α containing the pRSET-B plasmid with one of
the twelve different fluorescent protein genes will be prepared ahead of time. Each
group will receive two tubes with 1.5 ml bacterial culture for two different stocks
(two tubes for each culture, so four tubes total). The tubes will be labeled with code
numbers, FPx and FPy. Record these code numbers and use these numbers while
labeling tubes/recording information in your lab manual.
You will now purify the plasmid DNA from the bacterial cells using an alkaline
lysis plasmid prep. “Alkaline lysis” refers to the NaOH that is used to break open the
cells and precipitate the bacterial proteins. This is an efficient and very widely used
method for extracting plasmid DNA from bacteria (even commercially available
plasmid extraction kits begin with an alkaline lysis step). The alkaline lysis involves
the successive addition of three solutions:
Solution I: 50 mM glucose, 25 mM Tris-HCl pH 8, 10 mM EDTA, 50 μg/ml RNase A.
The glucose increases the solute concentration outside the cell, so that once the cells
are broken open with solution 2 the osmotic pressure will draw water out of the cell
and hopefully carry the plasmid along with it. EDTA is a chelator of divalent cations,
meaning it binds up Mg2+, Mn2+, and Ca2+ and removes them from solution. Its main
173
LAB 13 • Isolation of Fluorescent Protein Plasmid DNA
function in solution 1 is to remove Mg2+ and inactivate the DNA nucleases for which
this ion is a cofactor. RNase A is a ribonuclease that will degrade contaminating
bacterial RNA.
Solution II: 0.2 M NaOH, 1% SDS. SDS lyses cells by dissolving the cell membrane.
The high pH created by the NaOH will denature the bacterial proteins, causing them
to precipitate from solution. The extreme pH will also denature DNA (separating
the two strands of the double helix). Both the bacterial chromosome and plasmid
DNA will denature into single strands, but the small interlocked plasmid strands
will stay together.
Solution III: 3 M Kacetate, pH 5.5 (29.4 g KOAc + 11.5 ml glacial acetic acid/100 ml).
The acetate buffer brings the solution to a more neutral pH. The plasmid DNA strands
reanneal. The chromosomal DNA is too big and complex to reanneal under these
conditions. The K+ ions complex with the dodecyl-sulfate ions from the SDS. KDS
is insoluble and precipitates from solution.
Protocol
The following steps will be carried out on each of the four tubes (for example, two
tubes “FPx” and two tubes “FPy”). Note: your samples are biohazardous until the
cells are lysed. Be sure to use the proper waste streams. You will use sterile tips for
this experiment. These will be available at the front of the lab.
1. Centrifuge one minute at 13,000 rpm; discard supernatant into your 10% bleach
beaker. Invert tube to drain residual LB. Collect any residual liquid on a kimwipe
and place in the solid biohazard waste can.
2. Resuspend pellet completely in 100 μl solution I (RT); use vortex if necessary.
3. Add 200 μl solution II. Mix by gently inverting the tube five or six times. Do
not vortex. Vortexing may shear (break into small pieces) the bacterial chromosome, causing fragments of bacterial chromosomal DNA to purify with your
plasmid. The solution should turn clear during the mixing as the bacterial cells
lyse. Incubate RT for five minutes.
4. Add 150 μl cold solution III. Mix by gently inverting. A large amount of white
precipitate will appear; this is the KDS coming out of solution. Incubate on ice
for five minutes.
5. Centrifuge at 13k rpm for 10 minutes. The denatured bacterial proteins, KDS,
and chromosomal DNA will all come down in the pellet. The chromosomal
DNA is still in solution; however, the strands are so large that it gets caught up
in the precipitate and is pulled down into the pellet. Only the plasmid DNA
and residual bacterial proteins will remain in the supernatant. How might you
remove the residual proteins if you wanted to make your prep cleaner?
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LAB 13 • Isolation of Fluorescent Protein Plasmid DNA
6. Remove supernatants to four clean tubes. Avoid contamination with the pellet.
As you remove them, determine the volume of the supernatant transferred to
each of the four tubes.
7.
Add an equal volume (1×) of isopropanol to each tube. Vortex. Incubate 15
minutes in the freezer. (K+ from Solution III neutralizes charge on DNA and
isopropanol precipitates the DNA.)
8. Centrifuge 13k rpm for 10 minutes. Place the microfuge tube in the centrifuge
with the hinge facing out, so you will know where your pellet should be in case
it is very small. If you do not see the pellet, consult your TA. They may have you
add more isopropanol and repeat the precipitation.
9.
Remove supernatants completely with pipette tip.
10. Wash pellets with 1 ml 70% ethanol: add the ethanol and invert the tube a couple
of times to rinse the pellet and the inside of the tube. Do not try to resuspend
the pellet. The ethanol wash removes remaining salts and detergent.
11. Centrifuge 13k rpm 10 minutes. Remove supernatants completely with fine tips.
12. Let pellets air dry. If the pellet isn’t dry enough, it won’t dissolve (too much
ethanol).
13. Use sterile pipette tips to handle your plasmid from this point on. Dissolve each
pair of pellets in 20 μl nuclease free H2O. Combine both “FPx” pellets into one
tube and both “FPy” pellets into another. Be careful. Do not combine “FPx”
and “FPy” supernatants. Dissolve the first pellet, and then transfer that 20 μl
to the second tube of the same FP and dissolve that pellet. Mix by pipetting up
and down until the pellets are completely dissolved.
D. Quantify the Plasmid DNA and Test for Purity
1. Take three UV cuvettes. Put 1 ml dH2O into each one.
2. Add 1 μl of the “FP1” DNA solution to one cuvette, cover it with a small square
of Parafilm and mix by inverting several times. Add 1 μl of the “FP2” to the
second cuvette and mix. Use the third cuvette to blank the spectrophotometer.
3. Read A260 for the two samples. If for either sample the A 260 reading is too close
to 0 to be meaningful (below about 0.015), add another μl from your plasmid
prep, mix, and try again.
4. Read A280 for the two samples.
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LAB 13 • Isolation of Fluorescent Protein Plasmid DNA
5. Using your absorbance readings, calculate the concentration of your plasmid
DNA samples, and the total yield from the plasmid prep.
Concentration: A260 = 1.0 for 50 μg/ml double-stranded DNA,
so
(A value for sample) 50 μg/ml = V μg/ml
For example, suppose your A260 was 0.057
0.057 (50 μg/ml) = 2.85 μg/ml
But you diluted your sample 1:1000 before reading the absorbance, so you must
multiply the concentration of the diluted sample by 1000 to find the concentration
of your plasmid. So in the example above, you would multiply (2.85 μg/ml) (1000)
= 2850 μg/ml or 2.85 μg/μl.
Yield: To find your yield, take the volume and multiply by the concentration. If you
have 20 μl and the concentration is 0.4 μg/μl, then your yield is 8 μg.
6. Estimate the purity of your plasmid sample:
Purity: Calculate the ratio A260/A280. If the number is ≥1.6, the purity is considered
to be adequate. ≥1.8 is better. Protein contamination reduces the ratio.
FIGURE 13.5 Nucleotides absorb UV light with a maximum at 260 (254) nm. The amino acids
tyrosine and tryptophan absorb with a maximum at 280 nm, and phenylalanine at 260 nm. Tyrosine
absorbs more strongly than tryptophan and phenylalanine. The peptide bond absorbs at shorter
wavelengths and is usually measured at 220 nm.
1.0
Relative absorptivity
BSA
©Hayden-McNeil, LLC
DNA
220
240
260
280
Wavelength (nm)
176
300
320
LAB 13 • Isolation of Fluorescent Protein Plasmid DNA
E. Prepare Diluted Stock Solutions of Your Plasmids
For both plasmids, prepare 50 μl of a 1 μg/μl stock. If your resuspended pellet is
already at or less than 1 μg/μl concentration you should not dilute it further. Consult
your TA if you are uncertain how to set up the dilutions.
End-of-Lab Checklist:
❏ Two µg/µl plasmid stocks in freezer
❏ Two concentrated plasmid stocks in freezer
F. Supplementary Material—Precipitation of Nucleic Acids with Alcohols
Alcohol precipitation is used to 1) concentrate DNA or RNA and/or 2) remove
detergents, salts, and other low molecular weight contaminants. It can also be used
if one wants to change the buffer composition. The nucleic acids will not precipitate
unless there is a monovalent cation to neutralize the negative charges and permit
the DNA to aggregate. Na acetate (or Cl), is commonly used to precipitate DNA. The
advantage of acetate is that it can also serve as a buffer (~pH 5.2; pKa of acetic acid =
4.76). Acidic pH favors precipitation of DNA (why?). NH4 acetate is preferred when
one wants to remove free deoxynucleotides, as they co-precipitate less with NH4 than
with Na. LiCl is commonly used to precipitate RNA, since Li is more soluble in ethanol
than Na, and efficient precipitation of RNA requires larger amounts of ethanol than
precipitation of DNA. When precipitating nucleic acids in the presence of SDS, K+
is used when one wants the SDS to precipitate also. (SDS or sodium dodecyl-sulfate
is soluble, and KDS or potassium dodecyl-sulfate is insoluble.) The salts are added
from concentrated stock solutions to give a final concentration of ≥0.2 M.
After the salts are added, two volumes of ethanol or 0.7–0.8 volumes of isopropanol
are added to precipitate the DNA. (Slightly larger volumes are added to precipitate
RNA, e.g., 2.5 volumes of ethanol.) One advantage of using isopropanol is that the
precipitation reaction can be carried out in smaller tubes. A disadvantage is that
protein contamination tends to be greater than with ethanol precipitation.
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LAB 13 • Isolation of Fluorescent Protein Plasmid DNA
178
LAB 14 • Restriction Digest and Agarose Gel Electrophoresis
LABORATORY
14
Restriction Digest and
Agarose Gel Electrophoresis
You will now do a restriction enzyme digest of your plasmid DNA, and run the resulting DNA fragments on an agarose gel. By examining the pattern of bands obtained
on your gel, you can confirm that you have isolated the correct plasmid (and not
bacterial genomic DNA) from the DH5α cells. You will also be able to estimate the
length in base pairs of the FP gene insert being carried in pRSET-B. The fluorescent
protein genes were inserted into the pRSET-B vector at the BamHI and EcoRI sites
of the MCS (multi-cloning site). Therefore, if your plasmid contains the inserted
gene, digestion with these two enzymes (we are using HindIII instead of EcoRI)
should yield a piece of DNA the size of the insert and another piece the size of the
vector. What size DNA fragments do you expect to obtain by cutting the plasmid
with BamHI and HindIII? Refer back to the plasmid map in Figure 13.3.
A. Background on Restriction Enzymes
Restriction enzymes (endonucleases) are produced by bacteria to protect them from
viruses that infect bacterial cells, called bacteriophage. Restriction enzymes recognize
a specific sequence of nucleotides in the bacteriophage chromosome and produce a
double strand cut in the DNA that prevents the phage from replicating. These specific
nucleotide sequences are usually in inverted repeats, called palindromes because they
read the same backward and forward. The bacterium protects its own DNA from
being cut at these sequences by methylating adenosine or cytidine nucleotides in the
179
LAB 14 • Restriction Digest and Agarose Gel Electrophoresis
sequence, which prevents the endonuclease from binding. Restriction enzymes are
named with the first letter of the genus, followed by the first two letters of the species, followed by the strain or serotype if any, and a Roman numeral if the bacterium
contains more than one type of restriction enzyme.
BamHI (Bacillus amiloliquefaciens H)
HindIII (Haemophilus influenzae)
B. Background on Agarose Gel Electrophoresis
Agarose is a linear polymer of alternating D-galactose and 3,6-anhydro-L-galactose
isolated from red algae. It is dissolved in a buffer by heating, and then forms a gel
when it cools. The gel can be thought of as a “maze” of different sized holes. The
higher the concentration of agarose, the smaller the range of hole sizes. The following
table shows the DNA size ranges that are best separated with different concentrations of agarose.
Gel Percentage
DNA Size Range (bp)
0.5
1,000 – 30,000
0.7
800 – 12,000
1.0
500 – 10,000
1.2
400 – 7,000
1.5
200 – 3,000
2.0
50 – 2,000
The migration of a molecule in an electric field depends on its size, shape, and charge
density. DNA is uniformly negatively charged due to the acidic phosphodiester groups.
Linear DNA molecules have approximately the same shape and therefore a similar
“drag.” Thus the separation will be primarily based on size. However, supercoiled
circular DNA is more compact and will migrate faster than a linear DNA of the
same size. If only one strand is cut (“nicked circle”), the DNA remains circular but
uncoils and, therefore, migrates more slowly (usually more slowly than linear DNA
of the same size).
The two most common buffer systems are Tris-acetate-EDTA (TAE) and Tris-borateEDTA (TBE). TBE has a better buffering capacity than TAE, but TAE gives better
resolution and is preferred when the DNA is to be purified from the gel after electrophoresis. Both buffers contain EDTA which chelates Mg2+, the cofactor of DNase.
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LAB 14 • Restriction Digest and Agarose Gel Electrophoresis
The 10× loading solution contains 50% glycerol, 50% dH2O, and 0.2% bromophenol
blue. With glycerol, the DNA solution is denser than water, so it will fall to the bottom of the well. Bromophenol blue is included as a tracking dye. In a 1% agarose gel
it runs at about the same position as 300 bp DNA.
The gel is loaded with a DNA ladder which contains linear fragments of known sizes.
We are using the 1 kb ladder from New England Biolabs (N3232).
FIGURE 14.1 The photo shows 0.5 μg of 1 kb ladder loaded on a 0.8% TAE agarose gel.
DNA mass (ng)
Kilobases
42
10.0
42
8.0
50
6.0
42
5.0
33
4.0
125
3.0
48
2.0
36
1.5
42
1.0
42
0.5
©Hayden-McNeil, LLC
We can compare the intensity of the bands in the ladder with the intensity of our
sample bands to approximate how much DNA we have in each lane. How close was
your calculation of DNA concentration based on the A260 reading to what you see
on your gel?
To visualize the DNA, the gel contains the fluorescent dye, SYBR Safe. In many
research labs, ethidium bromide is used instead. Ethidium bromide is a powerful mutagen and should be handled with care. SYBR Safe is safer and is almost as
sensitive as ethidium bromide. Both dyes slip in between the stacked base pairs of
double-stranded DNA. In this environment, ethidium bromide fluoresces intensely
(560 nm) when excited by UV light (260–360 nm), permitting the detection of as
little as 20 ng of DNA or less. SYBR Safe fluoresces at 500 nm and will detect down
to 50 ng of DNA. Binding to single stranded DNA and RNA also increases the fluorescence of the dye, but the intensity is less than with dsDNA.
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LAB 14 • Restriction Digest and Agarose Gel Electrophoresis
FIGURE 14.2 Structure of ethidium bromide.
NH2
Br−
N+
H2N
CH2CH3
Ethidium bromide is a powerful mutagen, so it must be handled with gloves, and all
waste disposed of properly.
EtBr is soluble in ethanol, which is used to clean up spills.
C. Restriction Digest of pRSET
1. You will digest both DNA samples with the restriction enzymes HindIII and
BamHI. For each DNA sample, you will run the following three reactions: cut
with HindIII, cut with HindIII and BamHI, and uncut (no restriction enzyme).
For the DNA digest, you will make a master mix. A master mix contains all the
common reagents needed for the experiment. Making a master mix saves time
and more importantly reduces variation due to pipetting errors. You will have
two master mixes, one for each DNA sample. You will make enough of each
master mix for four reactions even though you will only do three, in order to
allow for possible pipetting errors. Note: you should use sterile tips and tubes
for the restriction digest. These will be available at the front of the lab.
TABLE 14.1 Master Mix Preparation.
Reagent
For 1 Reaction
For 4 Reactions
Nuclease-free H2O
12 μl
48 μl
1 μg/μl stock DNA
4 μl
16 μl
10× restriction enzyme buffer
2 μl
8 μl
2. Mix master mix well and centrifuge briefly (1–2 seconds) if necessary to bring
down liquid on the side of the tube.
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LAB 14 • Restriction Digest and Agarose Gel Electrophoresis
3. For both DNA samples, label three new tubes FP#-HB (for HindIII + BamHI),
FP#-H (for HindIII), and FP#-U (for uncut). Your TA will make enzyme aliquots,
giving you twice the amount you should need. Then add:
Tube
Nuclease-free
H2 O
Master Mix
Restriction
Enzyme(s)*
Final Volume
Should Be
HB
none
18 μl
1 μl BamHI +
1 μl HindIII
20 μl
H
1 μl
18 μl
1 μl HindIII
20 μl
U
2 μl
18 μl
none
20 μl
*It is extremely important that you do not submerge the pipette tip in the restriction enzyme solution. This
solution contains glycerol and will adhere to the outside of the tip causing you to remove a large excess of
enzyme. Barely touch the tip to the meniscus to withdraw the enzyme. Note also that when you add the
enzyme to the digest tube, it will fall to the bottom because glycerol is denser than water; therefore, it is also
essential that you mix well after adding the enzymes.
Each of the three tubes you make will have 18 μl of the master mix (containing 4 μg
of plasmid DNA and 2 μl of 10× buffer) and 2 added μl of restriction enzymes or
water, for a final volume of 20 μl.
4. Mix by gently stirring with your pipette tip. Centrifuge if necessary to bring
down drops on the sides. Look at the tube to be sure there is not a drop of glycerol at the bottom.
5. Cap well and incubate one hour at 37°C.
Note: One unit of a restriction enzyme is defined as the amount required to
digest 1 μg of phage λ DNA in one hour in a volume of 50 μl. Usually 1–5
units are used in a 20 μl reaction like ours.
D. Preparing and Running the Agarose Gel
1. Prepare a 1% agarose gel in 1× TAE buffer. (Add 0.5 g agarose to 50 ml 1× buffer in a 250 ml flask.)
Make 300 ml 1× TAE by diluting the 50× stock solution. (Have your TA check
the calculations! Note that this is a 50× stock, not just 10×.) The 50× stock
contains 242 g Trizma base, 57.1 ml concentrated acetic acid, and 100 ml 0.5 M
EDTA, pH 8, per liter. You will use the remaining 250 ml to run the gel.
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LAB 14 • Restriction Digest and Agarose Gel Electrophoresis
2. Melt the agarose in the microwave oven at high for about one minute. Leather
gloves are available by the microwave to remove the hot flask.
Don’t overdo it or the solution will be all over the oven.
3. Let it cool to around 60°C (hot to touch) then the TA will add 5 μl of SYBR
Safe dye. Swirl to mix. (If you do not add SYBR Safe, you will not be able to see
anything on your gels!)
4. Prepare the gel chamber:
Slide the baffles into the grooves next to the gel bed, handles to the outside. The
baffles should rest on the small rectangular baffle blocks, and remain extended
above the edges of the gel bed. Use a Pasteur pipette to lay an agarose bead (a
thin line) between the baffles and the edges of the gel bed to form a leakproof
seal. Note: do NOT use a micropipette for this!
Before pouring the gel, place the comb/bridge assembly over the gel bed. Loosen
the white screws holding the comb to the bridge, and adjust the comb depth.
Gently retighten the adjusting screws. The teeth of the comb should be just above
the bottom of the gel casting plate and should not touch the plate—otherwise
the samples will leak out when they are loaded in the wells. Also, the comb needs
to be at the end of the gel that is next to the negative electrode—decide which
end this is before you pour the gel (see step 9).
Pour all of the agarose solution onto the gel bed. Quickly position the gel comb
at the desired location. Immediately after pouring, rinse the flask with water
to remove any residual agarose before it hardens onto the flask. Let the agarose
solidify for 25 minutes. It will turn slightly opaque when it is solid.
5. Remove the baffles from their slots so they will not interfere with gel/buffer
contact. Fill the gel chamber with 1× TAE to a level that covers the gel by a few
millimeters (~250 ml). If there is too much buffer, less current will run through
the gel and more time will be required to separate the DNA.
6. Carefully remove the comb so the walls of the wells don’t tear.
7.
Add 2 μl 10× loading dye to each of the six digest tubes and mix.
10× loading dye: 50% glycerol, 50% dH2O, 0.2% bromophenol blue.
8. Load the samples (15–20 μl each) into the agarose gels. Load 6 μl (0.5 μg) 1 kb
DNA ladder into one well of the gel (get aliquot from your TA). Be sure to record
which sample is in each well. To load the samples, just touch the pipette tip to
the top of the well. Do not punch it through the bottom of the well.
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LAB 14 • Restriction Digest and Agarose Gel Electrophoresis
9.
Put the lid on the chamber. By convention, black is negative, so the black leads
from the lid will plug into the black jacks on the power supply. The wells containing your samples need to be on the negative electrode/black side. Plug the
leads into the power supply, and then turn it on. Set the power supply to run at
a constant voltage of 120 volts, and start the current.
Safety Note: Always make sure the current is not running when you plug in
the leads or unplug the leads from the power supply. You can start and stop
the current as many times as you need to during the run without harming the
experiment. Never run an electrophoresis unit without the lid.
10. Run until the bromophenol blue is at least two-thirds down the length of the gel.
While the gel is running, do the exercise on analyzing DNA digests (section E).
After the gel is done, turn off the power supply. Remove the gel to a piece of plastic
wrap. Leave the top of the gel uncovered for the photograph.
Photograph your gel and analyze the results. The TA will take a picture of your gel
using the UV transilluminator and camera on the front table. Calculate the sizes
of the fragments in each sample and construct plasmid map of pRSET with your
FP gene insert (see page 181 for sizes of DNA fragments in the standard ladder).
End-of-Lab Checklist:
❏ Return plasmid stock tubes to freezer
❏ Dispose of gel in the gel waste bucket (with plastic wrap)
185
LAB 14 • Restriction Digest and Agarose Gel Electrophoresis
E. How to Analyze a Plasma Restriction Digest
Here is a photo of a restriction digest experiment. All lanes (except for the DNA
ladder) contain the same plasmid DNA that has undergone different treatments.
Lane D is undigested DNA, while lanes B, C, E, and F have been digested with one
or more enzymes. Lane A has a different DNA ladder than the one shown previously.
The sizes of the bands in this DNA ladder are in kb (from the top): 10, 8, 6, 5, 4, 3,
2.5, 2, 1.5, 1 (brightest band), 0.750, 0.500.
FIGURE 14.3 Restriction digest experiment.
A
B
C
D
E
F
©Hayden-McNeil, LLC
Answer the questions using the information below.
1. What is in the upper band in lane D?
2. What is in the lower band in lane D?
3. What is in lanes B, C, and F? Can you estimate how many bp are in these bands
from the DNA ladder?
4. What is in lane E? How many bp are in each of these two bands? Which would
be the “vector” band and which would be the “insert” band?
5. (Graduate-level question) If the vector and insert bands contain the same number
of DNA fragments (and they must), why is the insert band fainter?
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LAB 14 • Restriction Digest and Agarose Gel Electrophoresis
Definitions
Super-coiled: Circular, super-coiled DNA migrates more rapidly than linear DNA of
the same size. Plasmid DNA is normally super-coiled.
Nicked, circular: One of the two strands of the double-stranded DNA is broken. DNA
uncoils, but remains circular (open circular) and usually migrates more slowly than
linear DNA of the same size.
Digest with one enzyme: One piece of linear DNA.
Digest with two enzymes: Two pieces of linear DNA; one is the size of the gene inserted
between the two enzyme sites in the multi-cloning site (insert band), and the other
is the size of the rest of the plasmid (vector band).
(Note: The standards in the DNA ladder are linear fragments.)
Is there any RNA in your gel? You can usually tell by the appearance of very bright,
wide bands in every lane at the bottom of the gel—there is none in Figure 14.3. If so,
how might this affect your estimation of the plasmid yield and purity?
187
LAB 14 • Restriction Digest and Agarose Gel Electrophoresis
188
LAB 15 • Preparation of Competent Cells and Transformation
LABORATORY
15
Preparation of Competent Cells
and Transformation
A. Background to Transformation of Bacterial Cells
Transformed bacteria are those that have taken up and expressed foreign DNA.
Cells that have been treated so they are temporarily able to take up DNA are called
competent cells. The method used in this lab to make competent cells involves cold
CaCl2 treatment of bacteria that are dividing rapidly, followed by a brief heat shock.
How this treatment allows DNA molecules to cross the plasma membrane isn’t
known, but it is proposed that they cross through adhesion zones where there are
pores formed by the fusion of the inner and outer membrane bilayers. The observation that adhesion zones only form in growing cells is consistent with the fact that
transformation is most successful with cells in early-mid log phase. Ca2+ would
shield the negative charges on the DNA and plasma membrane, permitting them to
come into close contact. This association may be favored by the low temperature.
The heat shock could briefly destabilize the membrane so that somehow the DNA
is allowed to get across.
A culture is started with diluted bacteria and their growth is followed by light scattering (see The Spectrophotometer, page 7) at 590 nm. (The more bacteria are
present, the more light is scattered and fails to reach the detector in the spectrophotometer. The detector reports this as increased absorbance.) E. coli in early-mid
log phase will give an absorbance reading of about 0.4. Overgrowth (>0.4) reduces
transformation efficiency. However, even with optimal conditions, the expected yield
of transformants is only ~3–10%.
189
LAB 15 • Preparation of Competent Cells and Transformation
JM109 (DE3) bacteria will be grown to early-mid log phase before lab. These are a
strain of E. coli carrying the gene for T7 RNA polymerase. This polymerase binds
to the T7 promoter in the pRSET-B plasmid and synthesizes mRNA coding for the
regions marked in the pRSET-B plasmid map. JM109 (DE3) cells are preferred to
the more commonly used BL21 (DE3) strain because JM109 (DE3) cells grow more
slowly. This gives the fluorescent proteins time to mature to their full fluorescence
(e.g., see Baird et al., 2000).
The typical procedure for growing the cells is to pick a single colony from a culture
on agar plates and culture it in 50 ml LB medium (without antibiotic) overnight at
37°C. One ml of this culture is then added to 100 ml LB (Luria Broth) in a 500 ml
flask (so there is a good air supply) and shaken at 37°C until A590 ≃0.4 (usually about
three hours). This will be done before lab.
B. Procedure
Be sure to follow all times as written for maximum transformation efficiency. You
will use sterile tips and tubes for this entire lab. They are available at the front of the
lab. Please return the boxes at the end of your lab. Also, it is important to use the
biohazard waste streams for contaminated liquids, solids, and sharps. These will be
explained at the start of the lab.
Make competent cells. All steps must be performed carefully to preserve sterility.
1. Transfer 1.2 ml of the culture to a sterile Eppendorf tube and centrifuge for one
minute at high speed.
2. Discard supernatant and resuspend pellet in 0.6 ml ice cold 0.1 M CaCl2. Do not
vortex. Resuspend the pellet of bacterial cells by flicking or gently stirring with
pipette tip. Incubate 30 minutes on ice.
3. Centrifuge 45 seconds at high speed (minimize exposure to room temperature),
discard supernatant, and gently resuspend in 120 μl ice cold 0.1 M CaCl2. Keep
on ice until use.
Transformation
1. For each of your two plasmids, transfer 25 μl competent cells to a fresh sterile
thin-walled 0.5 ml tube and add 5 μl of 1 μg/μl plasmid DNA. If your plasmid
DNA stock is more dilute than this, add 10 μl. Mix by gently stirring with the
pipette tip.
In each lab, one control sample of competent cells without plasmid will be prepared.
TA will assign volunteer group. This control will be placed on an agar plate without
ampicillin in section C. Why is this control useful? One other group will do an
additional control of plating cells without plasmid on LB + ampicillin plates. What
does this control demonstrate?
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LAB 15 • Preparation of Competent Cells and Transformation
2. Incubate on ice 30 minutes.
3. Thermal shock: Place tubes in 42°C water bath for one minute. Rapidly transfer
to ice. Incubate 1–2 minutes.
4. Remove from ice and rapidly add 150 μl LB medium without ampicillin (RT).
(Ampicillin is omitted here to give the transformed bacteria time to synthesize
β-lactamase.)
5. Incubate 30 minutes at 37°C.
Why did we use thin-walled tubes?
C. Culture
Spread all of each sample onto an agar plate with LB + ampicillin (50 μg/ml). Spread
the transformed cells evenly over the entire surface of the plate. Let the sample soak
into the plate for a few minutes. Invert the plates and put them into the incubator. (Do
not seal them with Parafilm.) Incubate at 30°C for two days. Why 30° and not 37°?
BE CAREFUL! Make sure you don’t combine the suspensions from the two
different plasmids. Always flame your spreader before and after each use!
191
LAB 15 • Preparation of Competent Cells and Transformation
192
LAB 16 • Purification and Analysis of Fluorescent Proteins
LABORATORY
16
Purification and Analysis of
Fluorescent Proteins
Now that the JM109-DE3 cells have expressed the FP genes carried in the pRSET-B
vector, you will be able to see the proteins from the colors they produce in the bacterial cells. A visual assessment of the color of the bacterial colonies is the first piece of
data you have to determine the identity of your unknowns. The absorption spectrum
of the purified FP will be your second piece of data. In the next lab, you will run your
purified FPs on an SDS-PAGE gel to observe their MW under heated and unheated
conditions, giving you your last piece of evidence. From these you should be able to
determine the identity of your two unknown FPs.
A. Visualization of Fluorescent Proteins
Take a moment to look at all the plates in your section to see the range of colors that
are produced. You should be able to distinguish the GFP derivatives from the DsRed
derivatives by eye. To see the true colors of the GFP proteins, you will need to shine
a handheld UV lamp on the colonies. Place all of the GFP protein plates in the light
box, remove the lids, and shine the UV lamp directly onto the colonies. Can you tell
which is EGFP, EYFP, and ECFP? In your notebook, record the color of the colonies
on your two plates and give a rough estimate of the number of colonies.
B. Purification of 6-histidine-containing Proteins with Ni-NTA Resin
The His-tag that has been added on to the N-terminal end of the fluorescent proteins
(see map of pRSET-B vector) doesn’t interfere with the function of the protein and
provides a method for purifying it on a nickel affinity column. The nickel is chelated
193
LAB 16 • Purification and Analysis of Fluorescent Proteins
to a nitrilotriacetic acid (NTA) spacer molecule attached to agarose beads. The nickel
atom has six spatially defined coordination positions that can participate in charge
interactions (similar to hydrogen bonding).
R
O
C
C
N
HC
C
CH
N
−O
C
NH
−O
CH2
O
OH
H
N
C
H
C
C
H
H
CH2
Ni2+
C
HC
C
H
NH
H
H
C
CH2
NH
O
H
−O
O
H
H
CH2
C
H
H
H
H
O
©Hayden-McNeil, LLC
HN
Nitrilotriacetic acid
C
O
Agarose
bead
N
HN
Histidines
R
H3+N
CH
COO−
CH2
N
N
NH
NH
Imidazole
Histidine
Binding to the NTA fills four of the coordination positions. The side chains of two
adjacent histidines (more precisely the imidazole rings of the side chains) have just
the right spatial orientation needed to fill the remaining two coordination positions
of the nickel. A single histidine in a protein could fill one of the coordination positions on its own, but this binding would be very weak. A single pair of histidines
forms a more stable interaction, but the binding affinity (see Lab 4) would still be
relatively low. However, three tandem pairs of histidines (a His-tag) can form bonds
with three nickel ions that are next to each other on the same bead, and the binding
affinity of this interaction is very high. Six histidine amino acids in a row are never
found on naturally occurring proteins.
The lysis and column wash buffers contain a low concentration of free imidazole
(10 mM), which competes out low affinity binding to the resin of a single or two
adjacent histidines in a protein. In the elution buffer, the imidazole concentration is
raised to 250 mM, high enough to compete successfully for the nickel and displace
the His-tagged protein.
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LAB 16 • Purification and Analysis of Fluorescent Proteins
As with any purification procedure, you should check both supernatants and pellets
to see where the desired fraction (fluorescence in this case) is going. If it does not go
to the right fraction, you may have to modify the procedure.
To purify the protein from our bacterial cells, we need to remove the cells from the
plate by making a slurry using LB then transferring the slurry to a microfuge tube.
To make sure that you do not mix the cells from the two different cultures, remove
the cells from one plate first, then do the second plate.
C. Purification Procedure
1. For each plate, pipette 700 μl LB medium onto the surface and gently scrape off
bacteria with a spreader. Push the suspension together on one side of the plate and
transfer it to a microfuge tube using a disposable, wide-mouth transfer pipette.
Repeat with a second 700 μl aliquot of LB and add this suspension to the first
tube (~1.5 ml cell suspension). Avoid digging up and transferring pieces of agar.
BE CAREFUL! Make sure you don’t combine the suspensions from the two
different plasmids. Always flame your spreader before and after each use!
2. Centrifuge one minute at high speed (13,000–14,000 rpm) to pellet the bacteria.
3. Discard the supernatant into a beaker with some bleach. Estimate the size of the
pellet* and resuspend it in five volumes of lysis buffer. Use the vortex to resuspend
the pellet. Make sure the pellet is completely resuspended before proceeding.
Lysis Buffer
• 50 mM NaH2PO4
• 300 mM NaCl
• 10 mM imidazole
• pH adjusted to 8.0 with NaOH
4. Add 10 μl of 10% SDS per 500 μl of lysis buffer. Do not exceed this amount of
SDS—if there is too much SDS, the bacteria will release their DNA and a gel will
form, making further separations impossible. Gently stir in the SDS with your
pipette tip. Avoid any vigorous mixing, as this will cause the SDS (a detergent)
to foam, which will make your sample difficult to work with.
5. Incubate the tubes at room temperature on the platform rocker for ≥30 min. Lay
them on their side on a Kimwipe, so that the suspension will be better agitated.
* Fill an empty microfuge tube with water to about the same volume as your pellet and measure the volume
of the water.
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LAB 16 • Purification and Analysis of Fluorescent Proteins
6. Centrifuge the lysate at high speed for one minute to pellet the debris. Transfer
supernatant to fresh tube. Save the pellet. Is the supernatant fluorescent* (colored)? If not, add 50% more SDS and gently stir with the pipette tip to mix and
resuspend the pellet, then repeat from step 5.
7.
Save 10 μl of each of the lysate supernatant to run on the gel in Lab 17. Label it
“crude lysate.” Put each aliquot in a well-labeled microfuge tube and store it in
your section’s box in the freezer.
8. Cut off a pipette tip so it is wide enough to take up the beads of the Ni-NTA
resin, mix the resin slurry well,** and add 50 μl to each tube of supernatant
(change pipette tip).
9.
Incubate on the rocker table ≥30 min. at room temp.
10. During this incubation, begin re-extracting your bacterial cell pellets; add
100 μl lysis buffer and 2 μl of SDS, and stir to mix and resuspend the pellet.
Place this on the rocker. This will give you some additional extracted FPs to
help in obtaining a strong absorption spectrum if you do not get enough from
the nickel-resin purification.
11. After the 30-minute incubation of the lysate supernatant with the resin, centrifuge at 5,000 rpm for 20 seconds to pellet the resin. Remove the supernatant
and save 20 μl of it in a new tube labeled “unbound lysate.” You will run this on
the SDS-PAGE gel in Lab 17. Discard the remaining supernatant.
12. Wash the pellet (resin + bound protein) by adding 100 μl of lysis buffer (without
SDS) then inverting the tube several times to mix well. Centrifuge to pellet the
resin. Discard the supernatant.
13. Elute the protein from the resin by adding 30 μl elution buffer (lysis buffer
plus 250 mM imidazole). Stir with the pipette tip and mix on the rocker for
~5 minutes. Centrifuge to pellet the resin. Save the supernatant solution, label
it as elution 1, and use it for section D.
14. Elute again with another 30 μl of elution buffer, mix on rocker for ~5 minutes,
centrifuge to pellet the resin, and remove the supernatant to a new tube (label
as elution 2). It will be used in the SDS-PAGE. Place these tubes in the room
temperature box on the TA desk.
End-of-Purification Checklist:
❏ Two crude lysate samples in freezer
❏ Two unbound lysate samples in freezer
❏ Both elution 2 samples to TA
* Sometimes the green protein fluorescence is weak, so you can get a stronger signal by looking at the sample
in the UV light box set up in the lab. (Wear UV safe goggles!)
** The beads will fall rapidly to the bottom of the tube, so mixing is important to be sure you get the right
amount of resin.
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LAB 16 • Purification and Analysis of Fluorescent Proteins
D. Collect Excitation and Emission Spectra
Fluorescent proteins absorb and emit light at characteristic wavelengths, based on
the chemical structure of their fluorophore. You will now determine the excitation
(absorption) and emission spectra for your unknown proteins. Compare the λmax
values for the two spectra to the known values for each fluorescent protein given in
Table 13.1. This is your best evidence yet as to the identity of your unknown proteins!
1. For each of your unknown proteins, place 1 ml of dH2O in a cuvette and add
all of your elution 1 sample. Cover with Parafilm and invert to mix. Prepare a
third cuvette that is a blank by adding 1 ml of water plus 30 μl of elution buffer.
Important:
Do not throw away the sample in your cuvette after taking the
excitation spectrum. You will also use this for your emission spectrum.
2. You will measure the absorbance of your fluorescent proteins from 400 to 700
nm. The instrument will print out the spectrum and indicate the peak λ max
values. If you are required to plot your excitation spectra with MS Excel for
your lab write-up, then you should also print the tabulated absorbance values
for each wavelength.
3. Your instructor will collect the emission spectra for your FPs using the platereader fluorometer. Your TA will supervise the loading of your samples into a
96-well plate, with separate rows for the green vs. red FPs. You must first determine if your FP concentration is too high for the fluorometer, and dilute it if it
is. Look at your excitation spectrum for that FP, and determine the absorbance
at the top of the peak (lambda max). If the absorbance value is greater than 0.03,
dilute the FP solution in your cuvette to achieve an absorbance of 0.03 at lambda
max (for example, if the absorbance was 0.09 you would perform a 1:3 dilution).
4. Load 150 μl from the cuvette with your diluted elution 1 sample into an appropriate well. Be sure to record the well number; this is how you will access your data
from Canvas. Depending on time constraints you may only be able to collect the
emission spectrum for one of your unknown FPs. If this is the case, you should
run the unknown for which you are less sure of the identity.
5. The amount of fluorescence is quantified in Relative Fluorescence Units (RFU).
The RFU for each wavelength will be uploaded to Canvas as an MS Excel file;
use these to plot the emission spectrum for your unknown.
Prepare your excitation and emission spectra, and then examine your data to determine the identity of your unknowns. The λmax values for some of the fluorescent
proteins are very close together and it may be difficult to discriminate between them
if you don’t have strong peaks in your spectra. The SDS-PAGE gel you will run in the
next lab will help you to determine some of these. For mRFP1 and mCherry, both
the excitation and emission λmax are only 3 nm apart. An overlay of the excitation
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LAB 16 • Purification and Analysis of Fluorescent Proteins
spectra for both proteins is shown in Shaner et al., 2004 (see reference on page 167).
Examining this overlay of the excitation spectra will provide a useful clue as to which
of the two proteins you may have.
E. Supplemental Information: Light Absorption and Fluorescence
Ultraviolet (UV) and visible light are electromagnetic waves that oscillate at frequencies
unique to those regions of the electromagnetic spectrum (see page 6). Light waves
visible to us oscillate with a frequency between 4 and 7.5 × 1014 per second. The speed
of light (3 × 1010 cm/sec.) divided by the oscillation frequency gives the wavelength:
λ =
3 × 1017 nm/sec
4 × 1014/sec
= 750 nm red light
Light also has the properties of a particle, and can be considered to occur in discrete
packets of energy called photons. Electromagnetic radiation carries different amounts
of energy based on the frequency of oscillation. More energy is contained in higher
frequency, and therefore shorter wavelength radiation, and less energy is contained
in longer wavelength radiation. Consider the difference in energy between X-rays
(0.1–10 nm) and visible light (400–700 nm)! This means that individual photons will
contain an amount of energy that is characteristic for light of that wavelength. The
amount of energy in a photon (E) is equal to the product of the Planck Constant (h)
times the speed of light (c), divided by the wavelength (λ):
E=
hc
λ
Light can interact with matter in different ways. It can be reflected back, diffracted
(bent), or scattered in such a way that the photons striking the matter are identical in wavelength to the photons leaving the matter. This is referred to as Rayleigh
scattering, or simply light scattering. No energy is transferred from the light to the
matter. Light can also interact with matter in a way where some or all of its energy is
transferred to the molecules composing the matter, in a process known as absorption.
In absorption, the energy present in a photon of light is absorbed by an electron in an
atom or chemical group causing it to become excited, and to temporarily occupy a
higher energy orbital. When this happens the electron is said to be in excited state.
The electron does not remain in excited state very long (only a few billionths of a
second), and quickly drops back down to its original orbital, referred to as ground
state. The energy that boosted the electron to excited state is dissipated in some
fashion when the electron returns to ground state, most commonly in the form of
kinetic energy (heat) and/or photons of light that are emitted.
The movement of an electron between ground state and excited state when light is
absorbed is called an electronic transition. Electronic transitions in organic and
biomolecules typically occur with the absorption of light between 200 and 700 nm,
which is the UV and visible light part of the electromagnetic spectrum. Absorption
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LAB 16 • Purification and Analysis of Fluorescent Proteins
at these wavelengths occurs primarily by excitation of an electron from a π molecular
orbital (π = ground state) to a higher energy molecular orbital (π* = excited state—
this is an antibonding orbital). Electronic transitions of non-bonding electrons to
π* also contribute to absorption at these wavelengths, although to a lesser extent.
FIGURE 16.1
Energy
π*
Quantum energy
difference
e−
π
©Hayden-McNeil, LLC
You can also think of electronic transitions in terms of a dipole that can be made
to resonate, or to increase the amplitude of an oscillation. A nucleus (pos.) and an
electron (neg.) pair in a molecule can form a dipole, and this dipole oscillates at a
certain frequency. A molecule will absorb light only when it contains a dipole that
oscillates at the same frequency as the incident light. When the dipole and the light
are oscillating at the same frequency they will resonate and increase the amplitude
of the dipole. This means that more energy is carried by the electron of the dipole
and it is in excited state.
The energy that was transferred to the molecule, assuming no fluorescence occurs,
is dissipated through molecular vibrations and collisions as kinetic energy, or heat
(see Figure 16.3). As this occurs the excited electron returns to ground state. Think
of what happens when you park your car in the sun on a hot day. All of the energy
absorbed from the sunlight can make the outside too hot to touch, let alone the
inside. Why do cars with dark paint and interiors get hotter than cars with lighter
colored paint and interiors? The energy absorbed from photons can also break bonds
and cause chemical reactions to occur.
What determines which wavelengths of light will be absorbed by a molecule? The
energy difference between ground state and excited state is distinct in different
molecules and is determined by the particular structure and arrangement of bonds
in a molecule. An electron in a π bond can only occupy the π orbital (ground state)
or the π* orbital (excited state). It cannot exist in any state of energy that is intermediate to the ground state and excited state orbitals, and cannot exist in an energy
state that is higher than π* (there are actually multiple energy levels of excited state,
but the point remains—the electron cannot exist at an energy state between that
of these defined orbitals). The energy difference (ΔE) between ground and excited
states is a quantum energy difference. This means that the excitation of an electron
to excited state is an “all-or-none” event; it can only become excited if it acquires
the precise amount of energy that is ΔE. Therefore, only a photon that contains an
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LAB 16 • Purification and Analysis of Fluorescent Proteins
amount of energy that is equal to ΔE can be absorbed. Photons of other wavelengths
are reflected by light scattering.
hc
E(π*) – E(π) = ΔE =
λ
The idea of ΔE as a single energy difference corresponding to only a single electronic
transition of unchanging energy is somewhat misleading, however. This is because all
molecules at a temperature above absolute zero are vibrating. The kinetic energy of a
vibrating molecule bends and stretches the bonds between atoms in that molecule,
and this affects the energy level of π and π* orbitals.
Energy in molecule
FIGURE 16.2
Vibrational energy
levels of π orbital
Distance between atoms in a
molecule as vibrations bend and
stretch molecular bonds
©Hayden-McNeil, LLC
As a molecule vibrates you can think of the energy level of ground state and excited
state as oscillating through a range of energy that makes up each state. This range
of energy is not a continuum, but rather a series of discrete sub-energy levels that
are referred to as vibrational energy levels. Therefore, depending on the vibrational
state of the molecule, an electronic transition can occur from any of the vibrational
energy levels of ground state to any of the vibrational energy levels of excited state.
This means that photons of slightly different wavelengths can be absorbed. When
looking at an absorption spectrum, λmax is the wavelength that contains the energy
equal to the most common energy difference between ground and excited states,
and the shoulders of the absorption peak represent wavelengths that match less
common energy differences.
Energy
FIGURE 16.3 Absorption.
A
B
C
Excited state (π*)
A. Higher-energy electronic transition
from shorter-wavelength photon
B. Lower-energy electronic transition
from longer-wavelength photon
C. Energy lost as heat when electron
returns to ground state
©Hayden-McNeil, LLC
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Ground state (π)
LAB 16 • Purification and Analysis of Fluorescent Proteins
The production of heat is by far the most frequent mode of energy dissipation from
excited state (even in “fluorescent” molecules, there is a greater probability to lose
energy by heat rather than fluorescence). However, in some molecules the energy
taken up by absorbing a photon can also be lost by the emission of a photon. This is
called fluorescence. The emitted or fluorescent photon is always longer wavelength
than the photon that excited the chemical group (the fluorophore) to begin with.
This is because some of the energy in the excited electron is first dissipated as heat
by an electronic transition from a higher energy level to a lower energy level within
excited state.
Aside from the vibrational energy levels, there are multiple levels of excited state in
which the electron can exist. Most excitatory electronic transitions are to the S1 and
S2 (S for singlet) states; ground state is also referred to as S0. Both S1 and S2 contain
multiple vibrational energy levels (see Figure 16.4). When the fluorophore absorbs
a photon of light (the λmax of its absorption spectrum is referred to as its excitation
max), an electronic transition occurs that boosts the electron from ground state
to one of the possible energy levels of excited state. Before a photon is emitted, the
electron drops to the lowest possible energy level of excited state (lowest vibrational
energy level of S1). This drop can be between electronic energy levels (S2 to S1) and/
or vibrational energy levels. The energy dissipated by the electronic transition takes
the form of increased vibrations in the molecule and collisions with solvent molecules, which translates to kinetic energy or heat. The electron can then drop back
down to ground state, and this electronic transition releases energy in the form of
an emitted photon. Note that this electronic transition is from the lowest vibrational
energy level of excited state to any of the possible vibrational energy levels in ground
state. Therefore, photons of slightly different wavelength can be emitted, with the
electronic transition of most frequent energy corresponding to the emission max
of the fluorophore.
FIGURE 16.4 Fluorescence.
Energy lost as heat
S2
Excited state
S1
Absorbed photon
Emitted photon
or
S0
©Hayden-McNeil, LLC
Ground state
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LAB 16 • Purification and Analysis of Fluorescent Proteins
Now that you know how absorption and fluorescence work, you can make predictions about whether a molecule will absorb and/or fluoresce UV/visible light based
on its structure. Look at the structures of Coomassie Blue, the fluorophore of GFP,
and ethidium bromide on pages 77, 169, and 182, respectively. How are they
similar? They all contain aromatic rings with lots of π molecular orbitals. Since
it is the excitation of electrons in π orbitals that causes UV and visible light to be
absorbed, the more double bonds a compound contains, the greater the chance that
it will absorb light, and possibly also fluoresce.
F. Supplemental Information: FRET
FRET is a technique that is widely used to study the interactions of proteins with
other proteins. FRET is an acronym for Förster (or fluorescence) resonance energy
transfer. FRET has been used for many years to study the interactions of proteins
that are labeled with small-molecule fluorescent tags. More recently the availability
of different fluorescent proteins with a variety of emission wavelengths has vastly
expanded the utility of FRET to study a number of dynamic cellular processes in
living cells. The ability to express proteins of interest linked to one or more fluorescent protein tags in eukaryotic cell cultures—this is done by combining the coding
sequences of the genes in a plasmid vector—allows researchers to synthesize the
tagged proteins and observe their behavior in situ (“in place” in living cells), something that was much more difficult with the small molecule fluorescent tags.
FRET occurs when one fluorophore (the FRET donor) is excited by light at its excitation max, and rather than a photon being emitted the energy is transferred to a
second fluorophore (the FRET acceptor), causing it to become excited (see Figure
16.5). Emission of a fluorescent photon then occurs from the second fluorophore.
Energy is transferred directly between the two fluorophores; each contains a dipole
that oscillates at the same frequency and energy is transferred from the excited
fluorophore to the ground state fluorophore when the two dipoles resonate. The two
fluorophores must be positioned very close to each other (typically less than 1 nm)
in order for this transfer of energy to occur.
FRET Molecular Rulers
The distance between the FRET donor and acceptor determines how much energy
from the excited donor can be transferred by FRET. As the distance (d) increases,
FRET decreases rapidly by a proportion of 1/d6. This means that the two fluorophores
(donor and acceptor) can be used as “molecular rulers” to measure the distance
between two tagged molecules. A distance in angstroms can be generated using
the equation
E=
1
1 + ^ d/R 0 h6
where E is the amount of FRET transfer (or quantum yield) that occurs, d is the
distance between the fluorophores, and R0 is the “Förster distance,” a constant
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LAB 16 • Purification and Analysis of Fluorescent Proteins
determined by the properties of that FRET donor/acceptor pair. This has been an
important tool for structural biochemists to measure distances between proteins
in a complex.
It is important to note that an electron is not transferred between the fluorophores;
this is not a redox reaction. The electronic transition in the FRET donor where the
electron returns to ground state must be the same magnitude as the excitation
electronic transition in the FRET acceptor. This means that, in terms of the two
fluorophores apart from each other when FRET is not occurring, the emission max
of fluor-1 has to be near to the same wavelength as the excitation max of fluor-2.
FIGURE 16.5 FRET.
FRET Donor
Absorbed
photon
FRET Acceptor
FRET
Emitted
photon
©Hayden-McNeil, LLC
One exciting use of FRET has been the creation of “biosensors” that can measure
changes in the concentrations of important molecules like glutamate, Ca 2+, and
cyclic-AMP (cAMP). We know that Ca2+ is an important signal transduction second
messenger (see Lab 9A), and the ability to measure real-time fluxes of Ca2+ from the
ER in living cells would be very useful. Figure 16.6 describes a Ca2+ biosensor that
allows researchers to do exactly this. The biosensor is a fusion protein, created by
linking together the amino acid sequences from different proteins in a single polypeptide. Calmodulin is a signal transduction protein that relays a calcium signal
to other proteins in the cell. Ca2+ binds to calmodulin, forcing it to shift its conformation such that it can bind to and activate other proteins. The Ca2+ biosensor
links two fluorescent proteins, ECFP and EGFP, to the ends of calmodulin. There is
also a fourth amino acid sequence (an M13 domain) in the fusion protein between
calmodulin and EGFP (not shown in Figure 16.6). This fusion protein is expressed
in cell cultures and its fluorescence is observed with fluorescence microscopy using
the excitation wavelength for ECFP (440 nm). In the absence of Ca2+, the calmodulin
component is in a relaxed conformation, the fluorophores of ECFP and EGFP are
some distance apart, and FRET cannot occur. In this condition the cells fluoresce
blue light from ECFP (480 nm). When a stimulus is provided and Ca2+ is released into
the cytoplasm, the calmodulin component assumes a contracted conformation (by
wrapping around the M13 domain) that brings the fluorophores of ECFP and EGFP
much closer together. Now FRET can occur and the cells fluoresce green light (510
nm). The strength and duration of the Ca2+ influx can be monitored by measuring
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LAB 16 • Purification and Analysis of Fluorescent Proteins
the ratio of blue to green light emitted. For more information see: Miyawaki et al.,
1997, Nature 388:882–887 (calcium biosensor); Zaccolo and Pozzan, 2002, Science
295:1711–1715 (cAMP biosensor).
FIGURE 16.6 Calcium biosensor.
480 nm
Ca2+
440 nm
440 nm
510 nm
Calmodulin
P
P
ECF
EG
F
EGFP
FP
EC
©Hayden-McNeil, LLC
FRET is actually a broader physical phenomenon that occurs widely in nature. It is
often more simply referred to as resonance energy transfer (RET) or exciton transfer.
It turns out that many molecules that can become “excited” (such as fluorophores,
chromaphores, and sometimes even enzyme active sites), under the right closelyspaced structural arrangement, can directly transfer energy to another “excitable”
molecule. In photosynthesis, photons of sunlight are absorbed by “antenna” chlorophyll molecules and the energy of those photons is then moved to the reaction
center of the photosystem by RET. In another example, RET may be responsible for
the bioluminescent glow of GFP in the Aequorea victoria jellyfish. There is no light
to excite the fluorophore of GFP in the dark ocean, and excitation of the fluorophore
is thought to occur through a luminescent reaction catalyzed by an enzyme called
aequorin. Purified aequorin can catalyze luminescence that is blue light, but the jellyfish only emits green light. While it is still being worked out how aequorin forms
a complex with GFP, aequorin is thought to excite the fluorophore of GFP through
RET, allowing the jellyfish to glow green in the absence of light to excite GFP. (You
can read more about this in: Gorokhovatsky et al., 2004, Biophysical and Biochemical Research Communications 320:703–711.)
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LAB 17 • SDS-PAGE of Fluorescent Proteins
17
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LABORATORY
SDS-PAGE of Fluorescent Proteins
In today’s lab, you will run your elution 2 samples from Lab 16 on an SDS-PAGE gel
to separate all the proteins present in the sample by size. In this experiment, you
will deviate from the standard SDS-PAGE procedure in that you will heat half of
your elution 2 sample and leave the remaining half unheated. Remember that heating the sample is necessary to denature the proteins before running them on the
gel, and this breaks apart any protein complexes allowing everything to run as a
single polypeptide. The heating will separate potential dTomato dimers and DsRed
tetramers into their constituent monomer subunits. In the unheated samples, the
subunits of these proteins will remain together, and these proteins will run on the
SDS-PAGE at the higher molecular weight of the dimer and tetramer.
A. Running the Gel
We will be using pre-cast 12% polyacrylamide gels for this electrophoresis. Each
group will run their own SDS-PAGE gel. For each fluorescent protein, you will run
the following samples on the gel:
a.
b.
c.
d.
FP-heated;
FP-not heated;
the crude lysate from the purification in Lab 16; and
the unbound lysate from the purification in Lab 16.
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LAB 17 • SDS-PAGE of Fluorescent Proteins
Therefore, including the size standards you will run nine lanes on the gel.
1. Prepare 600 ml of 1× electrophoresis buffer.
2. Secure the gel in the electrophoresis tank and add the buffer. Fill the inner
chamber almost to the top.
3. Remove the comb and use a Pasteur pipette to rinse out the wells.
4. Obtain your elution 2 samples from Lab 16. Measure the volume of each sample
(you should have about 30 μl) and divide it evenly in two, placing each half in a
clean tube. Prepare each tube as a separate sample by adding 4 μl of 5× sample
buffer. Add water, if necessary, to a final loading volume of 20 μl.
Safety Note: The 5× sample buffer contains 25% 2-mercaptoethanol. This
compound is toxic and is hazardous by inhalation and skin contact. Wear
gloves and remove the glove immediately if you come in contact with the 5×
sample buffer. Add the 5× sample buffer to your samples in the fume hood,
being sure to use a new tip for each sample. It is okay to take your tubes out
of the hood once the 5× sample buffer has been diluted into your samples.
5. Take one tube for each FP, and heat at 95°C for 10 minutes. Keep the other tube
at room temperature; it will not be heated.
6. For both the crude lysate and unbound lysate, prepare 16 μl of sample plus water
in separate tubes with 4 μl of 5× sample buffer. Mix gently. Heat these tubes for
10 minutes at 95°C.
7.
After heating, cool the tubes briefly on ice (not too long or the SDS will precipitate), and then spin the tubes for 30 seconds at full speed in a microfuge to
bring down any condensation.
8. Load 5 μl of the molecular weight standards into one well. Don’t forget to load
your gel near a power supply so that you will not have to move it once it is loaded.
9.
Load 20 μl (or as much as will fit without spilling over the well) of each of the
other samples.
10. Put the cover on the tank, connect the leads to the power supply and run the gel
at 120 V for about an hour, or as much time as is necessary for the blue dye front
to move all the way to the bottom of the gel. Always check the current after you
start the power supply. It should read at least 20 mA at the beginning of the run.
If not, something is wrong, most likely the level of buffer in the upper reservoir
has fallen below the top of the short inner plate. If this happens, disconnect your
gel from the power supply, add more buffer and start again.
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LAB 17 • SDS-PAGE of Fluorescent Proteins
B. Staining the Gel
1. When the run is finished, turn off the power supply, pour off the buffer, and
remove the gel cassette.
2. Carefully pry off one of the plates (TA will demonstrate).
3. Look for fluorescence in your gel (it is easier to see with one of the plates removed).
If you see anything, make a note or a sketch in your lab notebook indicating
where you see it.
4. Add the gel to a staining box with 15 ml of the rapid stain Coomassie blue. Use
tape to label the box with your group number.
5. Incubate the gel in the staining solution for 30 minutes with gentle agitation on
the orbital shaker.
6. After 30 minutes, check to be sure you can see your bands. Leaving the gel in
the staining container, aspirate the staining solution into the waste flask on the
middle bench. Add distilled water from the sink to gel to rinse it, and leave the
gel in the water.
7.
Examine your gel and take a picture.
8. Rinse the staining box and return it to the bin box. Place the gel in the gel waste
bucket on the middle bench (NOT the lab trash bucket).
C. Analysis of the SDS-PAGE Gels
Examine your bands to see how your FPs migrated in the gel. Did heating the sample
cause the FP to migrate to a lower MW position? If so, you may have an FP that is
a complex of more than one subunit. Looking at only the heated samples, was the
MW of the FPs in the gel the same as the “true” MW of the proteins? If not, what
is a possible explanation?
One consideration for interpreting the results is how denatured the proteins were.
The beta-barrel structure of the FPs is quite stable, so it is possible that it wasn’t
completely unraveled in a sample. This could affect both the shape of the protein and
the amount of SDS bound, factors that influence the relative mobility. For this reason,
you may have a range of band migrations distances that all represent a monomer,
just in various states of denaturation. The large shift in MW between the heated and
unheated samples for the FPs that form a complex should be much more distinct.
Examine your crude lysate and unbound lysate lanes. The samples are controls to
monitor the effectiveness of the His-tag/nickel-resin purification. Refer back to Lab
16 to see where you took these samples. What specifically do they tell you?
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LAB 17 • SDS-PAGE of Fluorescent Proteins
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LAB 18 • Protein Crystallization
18
ynse/Shutterstock.com
LABORATORY
Protein Crystallization
In this lab you will make crystals of the protein lysozyme. Like other organic molecules, protein molecules can crystallize under the right conditions. This is particularly important for proteins because it allows researchers to do X-ray diffraction
to determine their structure. You will not do X-ray diffraction on your lysozyme
crystals; this requires a powerful, focused X-ray beam that can only be produced
at specialized installations like a synchrotron (which is the size of a football field!).
You will, however, explore the properties that allow protein crystals to form. You
will perform two rounds of crystallization, using the information you gain from the
first round to improve your crystals in the second round.
A. X-Ray Crystallography and Determining Protein Structures
Do you ever wonder how all the 3-D structures of proteins that you see in your textbooks have been figured out? Most of these structures have been determined using
X-ray crystallography. The protein of interest must first be crystallized. Crystals form
when molecules leave solution and solidify as an ordered lattice where each molecule
is in the same orientation. In protein crystals, this means that each chemical group
on every amino acid in the polypeptide is aligned the same way in three dimensions.
To do X-ray crystallography, a very pure sample of the protein that is completely
free of contaminants is placed in conditions that favor crystallization (versus general
precipitation where protein molecules form disordered aggregates). These conditions must produce crystals that are large enough for X-ray crystallography. A single
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LAB 18 • Protein Crystallization
crystal is then placed in a focused beam of X-rays. When the X-rays interact with
the protein molecules in the crystal, they are diffracted or bent. Because each atom
on every protein molecule in the crystal is in the same position relative to the other
protein molecules, the X-rays are diffracted in a consistent set of angles based on the
structure of that protein. The diffracted X-rays are picked up by a detector, which
records them as a pattern of spots. This pattern of spots is called the diffraction
pattern. Each spot in the diffraction pattern will be at a particular angle relative to
the direction of the X-ray beam. Imagine a focused beam of light being reflected off
a mirror. From the position of the reflected light spot relative to the mirror, you can
see how it is possible to calculate the angle of the beam of light. The same is true with
X-rays being refracted (reflected) by the atoms in a crystallized protein. Once the
angle for each spot has been determined, it is then possible to calculate the relative
position of the atom in the protein molecule that reflected the X-ray to produce the
spot. This is essentially a map of the electron density of the protein molecule. For
a large and complex molecule like a protein, it requires a complex diffraction pattern from which the positions of many of the atoms can be determined. This being
said, it is not necessary that the diffraction pattern have a spot for every atom in
the protein. Based on the positions of the known atoms, the remaining atoms and
overall conformation of the polypeptide are modeled into the structure. Back in the
1950s when John Kendrew elucidated the structure of myoglobin, the first protein
structure determined by X-ray crystallography, the modeling calculations were done
by hand and were enormously time consuming. Modern computing has made this
much easier and allowed structures to be determined for much larger proteins. In
the end, what is produced is a set of three-dimensional coordinates that give the
position of each atom in the protein. The coordinate files that contain this information are used by various types of viewing programs to produce three-dimensional
images of a protein.
X-ray crystallography has also been used to figure out the structures of other biological molecules. The X-ray diffraction pattern of crystallized DNA provided the
missing information that allowed Watson and Crick to determine the structure of
the DNA double helix. Watson and Crick were not crystallographers, and the X-ray
diffraction information that was crucial to elucidating the structure of DNA was
actually provided by Rosalind Franklin, an unsung hero of this story.
While the large majority of protein structures have been determined by X-ray crystallography, other techniques are making increasingly important contributions to the
field. Nuclear magnetic resonance (NMR) spectroscopy has contributed to solving
the structures of many proteins. NMR determines the relative alignments of nuclei in
the atoms of the protein molecule, and has the advantage that it is done on proteins
in solution. NMR can capture “snapshots” of the protein as it moves through multiple
low-energy conformations and thus can provide a more realistic picture of the protein in solution. NMR can accurately determine the configuration of atoms in small
to mid-range molecules, protein fragments, and even some low MW proteins. For
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LAB 18 • Protein Crystallization
large molecules with MWs above 25 kDa, however, its utility is limited because these
molecules contain thousands of atoms and the NMR spectra become too complex
and noisy to interpret. Therefore, NMR and X-ray crystallography are considered
complementary techniques, each having some advantages that can be combined to
produce a more complete and accurate picture of a protein’s conformation.
Electron microscopy has also been used to obtain low-resolution images of proteins
for some time. It takes a resolution of about 3.5 Å (symbol for angstrom; 1 Å = 0.1
nm) to determine the positions of individual atoms in a protein structure. Standard electron microscopy techniques can only produce images with a resolution of
20–40 Å, only good enough to see large lobes of a protein as fuzzy blobs. More
recently, though, cryo-electron microscopy techniques have produced structural
information with a resolution down to 4 Å. In these techniques, the sample is flash
frozen in liquid ethane. The low temperature reduces the damage to the protein molecule caused by the electron beam of the microscope, which has been the limiting
factor to the resolution that can be achieved. While still not as good as the resolution
that is achieved with X-ray crystallography (down to 1.5 Å), cryo-electron microscopy
has the advantage of being able to image proteins in environments that can’t be done
with crystallography. Integral membrane proteins such as receptors and channels
have essentially been off limits to X-ray crystallography because it is not possible to
crystallize these proteins. Cryo-electron microscopy techniques, however, can image
the structures of proteins while they are embedded in a cell membrane.
B. Crystallizing Proteins
Before X-ray diffraction can be done, it is necessary to obtain high-quality crystals of
the protein. Rarely is this an easy task! Proteins are large, complex macromolecules
that don’t easily aggregate into the organized lattice required to form a crystal. Proteins are flexible and often have more than one stable conformation. Having more
flexible segments of the polypeptide that move around in solution makes it more
difficult to produce high-quality crystals.
FIGURE 18.1 Hanging drop method.
Protein solution
Seal
Coverslip
Reservoir
©Hayden-McNeil, LLC
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LAB 18 • Protein Crystallization
The most common method used to produce protein crystals is the hanging drop
method. In this method, a drop of protein solution is placed in the center of a coverslip, which then inverted over a sealed chamber that contains a buffer of higher ionic
strength. As water evaporates, it tends to collect in the buffer below the drop, so that
the protein solution in the hanging drop gradually becomes more concentrated. If
the buffer pH, composition, and concentration are right, the protein molecules will
aggregate in an organized lattice and form a crystal. If not, an unorganized precipitate
will form, or there may be no precipitate. As you can imagine, crystal formation is
an art and there are few rules—it is mostly achieved by trial and error.
A salt is the principle agent in the crystallization buffer that promotes aggregation of the protein molecules. The principle is the same as with ammonium sulfate
precipitation of proteins like you did in Lab 3; a high concentration of salt ions
preferentially hydrogen-bond with solvent water molecules, reducing solvation of
the protein molecules and thus promoting interactions between protein molecules.
However, a milder precipitating effect is needed to promote crystal formation as
opposed to precipitation in the form of disorganized protein aggregates done for
protein purification purposes. For crystallization, lower concentrations are used, and
sodium chloride is typically favored over ammonium sulfate. The concentration of
the protein in solution is also an important factor. The interplay of the precipitant
salt concentration and the protein concentration in initiating the formation of a
crystal, termed nucleation, can be conceptualized with a phase diagram (Figure
18.2). The phase diagram represents the solubility state of the protein as a function
of the protein concentration (y-axis) and precipitant or salt concentration (x-axis).
The goal is to achieve a concentration of both the protein and the precipitant that
moves the protein molecules into the nucleation zone where they will begin to form
crystals. Too high a concentration of protein or precipitant will result in precipitation of the protein as a disordered aggregate that is not a crystal (precipitation zone).
Too low a concentration of either will result in the protein molecules remaining in
solution (undersaturation zone). The metastable zone occurs when concentrations
are too low for crystal nucleation but high enough for a pre-existing crystal nucleus
to grow (add more protein molecules to an already existing lattice).
FIGURE 18.2 Phase Diagram
Supersaturation
Protein
Precipitation
zone
Nucleation
zone
Undersaturation
Precipitant
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©Hayden-McNeil, LLC
Metatable
zone
LAB 18 • Protein Crystallization
For the formation of large crystals that are suitable for X-ray crystallography, only a
limited amount of nucleation is desired. Too much nucleation can result in a “shower”
of microcrystals that are very small. Therefore, the optimal protein and precipitant
concentrations will create an environment that is just inside the nucleation zone
from the metastable zone. Under these conditions, limited nucleation will occur
followed by the growth of the crystals in three dimensions.
The pH of the crystallization buffer can also affect the ability of protein molecules to
nucleate and form crystals. By affecting the charge on individual R-groups, the pH
affects how protein molecules may interact. Non-ionic polymers such as polyethylene
glycol (PEG) may also be added to the crystallization buffer. Similar to salt precipitants, PEG serves to further reduce the concentration of available water. However,
because PEG has no charge, it reduces the dielectric constant of the crystallization
buffer, meaning the solution is less able to insulate charges on different molecules
from one another. This increases the probability that opposite charges on protein
molecules will find each other, thus increasing protein-protein interactions and
potentially aiding in crystal nucleation. For some proteins, however, increasing the
potential for charge interactions between molecules may promote the formation of
disordered aggregates over crystal nucleation.
You will attempt to crystallize the protein lysozyme, an anti-bacterial protein produced by mammals. Lysozyme will form crystals with NaCl as a precipitant at a
concentration of 0.25–0.5 M. Your job is to optimize the crystallization conditions
to produce the largest crystals possible. You can try different concentrations of NaCl
inside the range given above, and you may also use different lysozyme concentrations and/or add PEG. You will perform two rounds of the crystallization, taking
what you learn from the first round to improve your crystals in the second round.
C. Procedure
You have the following solutions available. All solutions have 50 mM acetate pH 4.8.
Lysozyme Solutions
• Lysozyme solution 1: 50 mg/ml lysozyme
•
Lysozyme solution 2: 100 mg/ml lysozyme
Reagents to Prepare Crystallization Buffer
• NaCl stock for crystallization buffer: 3 M NaCl
• PEG stock for crystallization buffer: 30% PEG
• Diluent for crystallization buffer: This contains only the acetate buffer solution. You will dilute the NaCl and/or PEG stocks with this to get the desired
concentration.
You will prepare 3 hanging drops with different crystallization buffer conditions and/
or different lysozyme concentrations. Before you do this, you need to decide on your
conditions. You will make up the crystallization buffer with your desired NaCl and
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LAB 18 • Protein Crystallization
PEG concentrations. Each hanging drop will have 3–5 μl of both the crystallization
buffer (CB) and the lysozyme solution (LS) so that to total volume of the drop is 8 μl.
4 μl LS + 4 μl CB or
5 μl LS + 3 μl CB or
3 μl LS + 5 μl CB
1. Decide on the crystallization buffer conditions for your three hanging drops.
Each crystallization buffer should have a NaCl concentration of 0.5–1 M. You
may also add PEG to get a concentration of 10–20%. The NaCl and PEG stocks
will be diluted with the acetate buffer diluent to achieve the desired concentration. You don’t necessarily need to have different crystallization buffers for
your three drops; it is okay to vary the conditions simply by adding different
volumes to the drop. For your first round of crystallization, it is all somewhat
arbitrary, so don’t sweat what might or might not work at this point. Just try a
range of conditions so that you will have enough information to modify your
conditions in the second round. You should include at least one condition that
does not contain PEG.
2. Prepare your crystallization buffer(s). You will have one to three of these. You
will need 1 ml of crystallization buffer for each hanging drop condition (if two
drops have the same crystallization buffer, prepare 2 ml, etc.). Use the table
below to help you prepare your crystallization buffers, and reproduce this table
in your lab notebook. Add the NaCl stock, PEG stock (if any), and diluent to a
microfuge tube in the amounts necessary to achieve your desired concentrations. If you need to prepare 2 or 3 ml of crystallization buffer, prepare it in a
5 ml Falcon tube.
Crystallization Buffer (CB)
Final concentration and volume of stock to get that concentration in 1 ml
NaCl
concentration
Drop 1
Drop 2
Drop 3
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volume
NaCl
stock
PEG
concentration
(if any)
volume
PEG stock
volume
diluent
LAB 18 • Protein Crystallization
3. Decide on the lysozyme stock concentrations you will use (50 mg/ml or 100 mg/
ml) and record these in the table below. Take 8 μl of the lysozyme solution per
drop. For example, if you have two drops that use lysozyme solution 1 (50 mg/
ml) and one drop that uses lysozyme solution 2 (100 mg/ml), take 16 μl of solution 1 and 8 μl of solution 2. Place your lysozyme solutions in microfuge tubes.
Reproduce the table below in your lab notebook to document the composition
of your hanging drops.
Drop Composition
μl CB in drop
LS conc.
μl LS in drop
Drop 1
Drop 2
Drop 3
4. Collect 3 siliconized cover slips from your IA. You will prepare your drops on
these coverslips. Handle them with gloves to keep them clean.
5. Get a 24-well plate. Be sure the plate is clean and free of grease before you start.
Clean it with 70% ethanol, and remove any old markings or tape.
6. Select three wells in which to prepare your hanging drops. You will fill these
wells with crystallization buffer and then place the cover slips with the drop
hanging on the underside over the well. You don’t want to bump the cover slips
after they have been placed on the wells, so choose wells that are away from the
edge of the plate, and not directly adjacent to each other. Label the lid of the
plate to indicate which wells have which drops/conditions.
7.
Use a syringe with vacuum grease to place a small bead of grease around the
rim of the three wells you will use. This will create an air-tight seal to prevent
evaporation from the well.
8. Spin the tubes with your crystallization buffers in your benchtop microcentrifuge
for 5 minutes at full speed. This will pellet any dust particles or microscopic
fibers that could inhibit crystal nucleation.
9.
Place 975 μl of each crystallization buffer into the appropriate wells. Retain 25
μl for your hanging drops.
10. It is now time to prepare your hanging drops. You may handle the cover slips
with your fingers (with gloves!) or with the forceps, whichever is easier. Use the
squirt bottle “air guns” to blow any dust off the surface of the cover on which
you will place your drop.
11. Take one of the clean cover slips and pipet the lysozyme solution (3–5 μl) directly
in the center. You want a single, round drop; don’t spread the solution across the
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LAB 18 • Protein Crystallization
surface of the coverslip. Now add the crystallization buffer (3–5 μl). Change the
tip on the micropipette and bring up the crystallization buffer into the tip. Gently
touch the tip to the drop of lysozyme solution on the cover slip and gently press
the plunger. You want to inject the buffer into the preexisting drop.
12. Turn the cover slip over without letting the drop slide off or to one side. You
want a single well-rounded drop hanging from the underside of the cover slip.
Place the cover slip over the well on the 24-well plate. Press gently to be sure you
get an airtight seal. Prepare your other two hanging drops in the same manner.
13. Once all three hanging drops have been prepared, place a small clay ball in each
corner of the plate. This raises the lid enough that it will not touch the cover
slips. Place the lid on the plate, resting on the clay balls, to protect the cover slips.
14. Gently move the plate to the location in which you will keep it while your crystals
form (your IA will designate a quiet spot in the lab where your section’s plates
can rest in peace). It is critical that you don’t bump the plate in a way that will
disrupt your hanging drops.
D. Analysis of Crystals
To look at your crystals, remove the lid of the 24-well plate, and place the plate
onto the stage of the dissecting microscope. Leave the cover slips on the wells. It
can sometimes be difficult to find the right focal plane. If you don’t see anything
that looks like a crystal, ask your IA for help. Record the following data in your lab
notebook for each drop.
1. Examine your hanging drops to see if any crystals have formed. Which of the
following did you obtain?
•
•
•
No crystals present
Shower of microcrystals that are very small, close together, and too many
to count
Individual crystals where you can see each side of the crystal. Approximately how may crystals are in the drop? What shape are they?
Take a picture of each drop to document your results. If you have individual
crystals, proceed to the next two steps.
2. Get a rough estimate of the size of individual crystals. To do this, use the
microgrid slide that is provided. Place the microgrid slide on top of the 24-well
plate so it will be in a focal plane close your hanging drops. Focus on the grid
pattern. Each small square is 1 mm on a side. Look at the microgrid squares
and then look at your crystals. Measure the crystal size for one crystal in each
drop; select the largest crystal in the drop and measure the length of its longest
dimension. An estimate is okay. Is your crystal about 1 mm, 0.5 mm, or 0.25
mm in diameter?
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LAB 18 • Protein Crystallization
3. Test to be certain your crystal is a protein crystal and not a salt crystal. To do
this you will add a small of amount of the dye methylene blue to the hanging
drop, and see if the dye diffuses into the crystals. The molecular lattice of a
protein crystal has pores that are large enough for the dye molecules to enter
the crystal, whereas the atoms in a salt crystal are too close together. Protein
crystals will turn blue and salt crystals will remain clear. To add the dye, gently
remove the coverslip and turn it over, keeping the drop in the center. You will
use a small micropipette tip as a capillary tube to add the dye to the drop; do not
use the micropipette. Touch the tip to the dye solution and hold it until some
dye is drawn up into the tip by capillary action. Now touch the tip to the surface
of the drop. Hold it there until the dye diffuses out of the tip and into the drop.
Invert the coverslip over the well again and gently press to reseal. Let the drop sit
for 30 minutes to give the dye time to diffuse into the crystal. After 30 minutes,
look at the drop to see if the crystals have turned blue; leave the coverslips over
the wells. Take a picture to document your results.
4. Once you have pictures of your crystals, place the cover slips in the green sharps
bucket. Return the clay balls to the clay box. Wipe the grease from the well rims,
empty the wells in the sink, and rinse the 24-well plate/lid well with DI water.
Leave the plate on the middle bench to dry.
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LAB 18 • Protein Crystallization
218
LAB 19 • Computational Tools for Biochemistry
LABORATORY
19
Computational Tools for
Biochemistry:
Sequence Analysis, Bioinformatics,
and Examination of Protein Structures
A. Identification and Characterization of an Unknown Protein Involved in
Melanoma
Some of the online tools you will use in this tutorial do not function well with
the Mac OS Safari browser. Please do not use this browser.
Melanocytes are pigment producing cells that reside in the basal (bottom) layer of
the epidermis of the skin. These cells produce the pigment melanin, which provides
protection from UV radiation and is what produces the particular skin coloration
for an individual. Melanoma is a type of skin cancer that occurs when normal
melanocytes acquire genetic mutations that cause them to lose their normal growth
regulation. The UV radiation in sunlight can damage the DNA of the cells in the
skin and can produce mutations. Melanoma is more frequently associated with
sun-exposed areas of the skin, and frequent sun burns early in life are considered to
be a risk factor. When caught early enough, a melanoma can be removed from the
skin and will pose no further health risk. However, melanoma is the deadliest of the
three types of skin cancer (the other two, basal-cell and squamous-cell carcinoma,
arise from keratinocytes) because it tends to rapidly metastasize to distant parts
of body. Metastatic melanoma has a relatively low survival rate compared to other
types of cancer.
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LAB 19 • Computational Tools for Biochemistry
Much effort has been made by researchers to identify the genes that become mutated
in the development of melanoma, and to understand how the functions of the proteins they encode become altered. One such gene has just been discovered! A gene on
chromosome 7 has been identified as a mutational hotspot. In a screen for melanoma
driver mutations (the mutations responsible for causing the cancer), a thymidine at
a specific base pair position of this gene was found to be changed to an adenosine
in 20 out of 34 melanoma samples. It is now your job to characterize the gene that
contains the T to A mutation. As you go through the module, answer the questions
on the perforated answer pages at the end of this section. You will turn in the answers
for credit at the end of the lab, but be sure to make a copy of your answers that you
can refer to later in the module.
Go to Canvas and open up the text file that contains the DNA sequence for this
unknown gene. This is a cDNA sequence, meaning that this gene was cloned by
making a DNA copy from an mRNA. Therefore, this sequence does not contain the
introns present in the genomic DNA sequence of the gene on chromosome 7.
Do you understand the difference between the cDNA sequence and the
genomic DNA sequence for a gene?
The T to A mutation that was identified occurs at nucleotide number 1860 in this
sequence; this is the sequence from normal human tissue, so it is not mutated.
Highlight the thymidine at position 1860.
Does this gene actually encode a protein? The first step in characterizing the gene
is to identify any open reading frames (ORF) present in the sequence that could be
translated to a polypeptide. An ORF is simply all of the codons that make up the
coding sequence of a gene. Most cDNA clones of a gene will contain some untranslated region (UTR) at the 5' and 3' ends of the ORF. You will use a program at NCBI
that will determine all of the possible ORFs by identifying the sequences flanked
by a start (ATG) codon and a stop (TAA, TAG, or TGA) codon that are in the same
reading frame. The program will identify ORFs in all three reading frames on both
the “sense” DNA strand (this is the sequence that is shown) and the “antisense”
strand (not shown), for six reading frames total.
Go to the NCBI homepage and click Resource List (A–Z). Under O, click ORF Finder.
Copy and paste your DNA sequence into the box. It’s okay to leave the numbers; the
sequence is in FASTA format and will be read by the program as is. The Minimum
ORF Length can be used to set the minimum length in nucleotides (nt) for identified ORFs. An ORF of less than 75 nt would encode less than 24 amino acids—too
small to be a protein! Look at the Genetic Codes dropdown menu list. Mitochondrial
genes and the genomes of some prokaryotic organisms use a slightly altered version
of the genetic code. Keep the 1–Standard genetic code and click the Submit box at
the bottom.
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LAB 19 • Computational Tools for Biochemistry
The graphic at the top of the results page shows all of the ORFs found. The rows in
the table below display information about each ORF. The Strand and Frame columns
indicate the reading frame for each ORF. The +1 frame is set relative to the first
nucleotide of the sense input sequence, where that nucleotide would be the first of
the three base pair codon, and the –1 frame is set by first nucleotide of the antisense
sequence. +2 begins with the second nucleotide of the sense sequence, etc. Click on
different ORF rows to see the translated amino acids sequence in the box to the right.
Examine the possible ORFs and determine which is most likely to be the actual ORF
for an encoded protein. Consider the T to A mutation at position 1860.
1.
List the start and stop positions and the length of the ORF. What is its reading
frame?
2. How many amino acids in length is the protein encoded by this ORF?
Note the slide bar at the top of the graphic. Sliding it to the right lets you zoom in
on the sequence. Click the ATG button next to it to zoom in all the way and see the
DNA sequence (you can slide the bar to the left to zoom back out). You will see the
sequence of both the sense (on top) and antisense strands.
Now click the Tools button and select Go To from the dropdown menu. Enter 1860
and click okay. This will take you to the position of the melanoma mutation.
-UUUU Phe
U-- UUC
UUA
Leu
UUG
CUU
C-- CUC Leu
CUA
First
CUG
base
(5' end)
AUU
A-- AUC Ile
AUA
or
AUG Met
start
GUU
G-- GUC
GUA
GUG
Val
Second base
-C-AUCU
UAU Tyr
UCC Ser UAC
UCA
UAA Stop
UCG
UAG Stop
-GUGU Cys
UGC
UGA Stop
UGG Trp
CCU
CCC
CCA
CCG
CGU
CGC
CGA
CGG
ACU
ACC
ACA
ACG
GCU
GCC
GCA
GCG
Pro
Thr
Ala
CAU
CAC
CAA
CAG
AAU
AAC
AAA
AAG
GAU
GAC
GAA
GAG
His
Gln
Asn
Lys
Asp
Glu
AGU
AGC
AGA
AGG
GGU
GGC
GGA
GGG
Arg
Ser
Arg
Gly
--U
--C
--A
--G
--U
--C
--A
Third
--G base
--U (3' end)
--C
--A
--G
--U
--C
--A
--G
©Hayden-McNeil, LLC
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LAB 19 • Computational Tools for Biochemistry
3. What amino acid is encoded by the codon in which nucleotide 1860 resides?
Write out the full name. Use the codon translation table in your lab manual.
4. At what position in the protein is this amino acid (from N-terminus)? (Do the
math here; the number of nucleotides in the cDNA up to and including this
codon, minus the number of nucleotides before your ORF start, divided by?)
5. Would this amino acid be changed by the T to A mutation at this position?
6. If the amino acid is changed, write out the amino acid substitution using the
correct nomenclature (wild-type amino acid: position: mutant amino acid).
Examine the translated amino acid sequence for this ORF in the box to the left. The
sequence is given with the single letter amino acid codes. Refer to your lab manual
for the key for the amino acid each letter stands for. Copy the amino acid sequence
for the protein into a text file so that you will be able to access it later. When you
do this, copy the header (the top line beginning with >) along with the amino acid
sequence. The header will be required for some of the tools you will use later. You
will need to access the amino acid sequence on your second bioinformatics lab day,
so save the text file to a USB drive or email it to yourself.
You have identified a protein whose function may become altered in the development
of melanoma, but you don’t yet know anything about what this function might be. It
is possible to get an idea of what a protein might do by looking for function-specific
domains. A domain is a part of a protein that folds with a particular conformation
that has a specific function associated with it. Both the amino acid sequence of the
domain and the folding pattern it produces will be conserved in different proteins
with similar functions, and across species. Therefore, if a protein has the amino acid
sequence for a certain type of domain, it is likely to have that function. Go to the
NCBI home page and click Domains and Structures from the left-hand list. Read the
text under Conserved Domain Database and then click the link. Click CD-Search
under CDD tools.
Copy and paste the amino acid sequence for your unknown protein into the entry
box (include the header). In the Options box to the right, find the “Search against
database” drop-down menu and select the Pfam database. Click submit. It may take
a few moments as the program compares the input sequence against the consensus
domain sequences in its database. The results page will contain a graphical summary
of where any matching domain resides in your protein, and then below that a list
of each conserved domain in the database that matched your protein. Examine the
list of domain hits, then click zoom to residue level on the graphical summary. Use
the slide bar to scroll along the length of the protein.
7.
222
Look at the table of domain hits below the graphical summary. What is the
description of the first domain hit? What is the position of this domain in the
protein (give the amino acid position at which it starts and stops)? You can
find this by looking at the graphical summary or under Interval in domain
hit table.
LAB 19 • Computational Tools for Biochemistry
8. Based on the second domain hit, what other protein might this protein bind
to?
9.
Thinking about what might happen when this protein becomes mutated in
melanoma and what you have learned about its functional domains, what
sort of protein is this and what role might it play in normal cells? There is
no right or wrong answer here, but you must provide a one- or two-sentence
description.
You now know quite a bit about the function of your unknown protein and may have
some idea of its identity. Is there a way to more precisely identify this protein? The
amino acid sequences of all known proteins (from humans and many other animals)
are retained in the protein database at NCBI. You will now use a program called
BLAST (for Basic Local Alignment Search Tool) to match the amino acid sequence
of your unknown protein against those of all the proteins in the NCBI database.
Go to the NCBI homepage and select BLAST from the Popular Resources list to
the right, then click Protein Blast. Copy your unknown amino acid sequence into
the query field. Since the amino acid sequence was translated from a human gene,
you are looking for a human protein, so enter Homo sapiens in the Organism field
(select from the drop-down menu as it auto-fills). This will cut down on the number
of extraneous matches with the NCBI protein database. Click the BLAST button at
the bottom; it may take a couple minutes for your search to be processed.
The top of the results page gives a summary of your search details, and a box to the
right where you could enter information if you wanted to refine the search. Below
this is a table that lists each hit from the protein database, along with the values that
describe the degree of matching and the statistical validity of the match.
•
•
•
•
•
Max Score and Total Score—these are the raw numbers generated by the
BLAST algorithm for the highest matching region and matching from all
regions combined, respectively. The higher the score the better the match.
Query Coverage—the percentage of the unknown (query) sequence that can
be aligned to the protein from the database hit.
E Value—A statistical measure of the validity of the match that is similar to a
p-value. It represents the number of hits you would expect to get by random
chance from the database with the degree of matching for that hit.
Identity—For the aligned portions of the sequence designated by Query
Coverage, the percentage of amino acids in the query sequence that match the
protein from the database.
Accession Number—The database identification number for that matching
protein.
You will notice that there are hits with many redundant entries in the protein database
that are all the same protein. These are entries from different sources that may be
sequence variants or partial sequences. In addition to this there are hits with entries
for related but distinct proteins.
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10. What is the identity of the unknown protein?
11. What is the catalytic activity of this enzyme? Be as specific as possible.
12. Write down the accession number for the best matching protein from the
database.
Now that you have identified the unknown protein, let’s take a closer look at one
of its functional domains. Click the Graphic Summary tab at the top of the results
table. You will see an image titled “Putative conserved domains that have been
detected.” Click on this image. This will open a new page where you can investigate
different elements inside each domain (do not close the BLAST results page; you will
come back to it in a few minutes). The bar on top represents the linear amino acid
sequence. The domains are shown below the bar, with the different layers indicating multiple hits in the domain database for the same domain, but with decreasing
specificity from top to bottom. Specific hits, non-specific hits, and super-families
are entries in the domain database of increasing sequence divergence. Specific hits
are a match to a domain unique to a small number of proteins. Non-specific hits are
matches to different proteins with similar, but not identical, biochemical functions.
Super-families span broader groups of protein families with more widely ranging
function. You can look at the list of domain hits below for more information on the
different database entries.
13. With which domain is the amino acid substitution from the melanoma
mutation associated? Use the more general name for this domain that you
identified earlier. This domain is the active site of the enzyme.
The small triangles below the bar represent specific amino acids that make up the
different subcomponents of a domain. Click zoom to residue level. Scroll to the
right and examine the different components that make up the active site. Touch the
cursor to a triangle to view a pop-up box with information about that element of
the domain. The triangles are aligned under the residue they indicate in the amino
acid sequence above.
14. What are the first two amino acids associated with the polypeptide substrate
binding site? Give the full name and position for both amino acids.
15. How many amino acid residues total make up the consensus sequence for the
polypeptide substrate binding site? (Touch the cursor to the triangle or count
the triangles in that row.)
16. With which particular feature of this domain is the melanoma mutation
associated? Look at the bottom row of triangles at the position for this amino
acid (you may need to scroll more to the right). What range of amino acids
make up this feature of the domain? Make note of this; it will be important
later on!
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Before you return to the BLAST results page, examine the very first entry in the list
of domain hits. Click the [+] for this entry and read the paragraph of information
provided.
17. There are three members of this protein family to which your newly identified
protein belongs (referred to as isoforms). Write the names of all three isoforms.
Return to the BLAST results page and go back to the descriptions tab. Scroll through
the list of hits to examine some of the matches in the NCBI protein database that
have less sequence similarity to your newly identified protein. The other two members of this three-member protein family are also represented in the list of hits. Can
you identify them? In this list of hits one of the family members is named after an
oncogenic (cancer causing) version of the homologous mouse protein. This oncogenic
homolog is found in a retrovirus that causes tumors in mice. The normal, non-cancer
causing version of the protein in mice, and its homolog in humans, is referred to as
the “proto-oncogene.”
18. Identify the other two members of this protein family using the accession
numbers given in the table on the perforated answer pages, and then complete
the table by filling in the required information.
The three members of this protein family have very similar catalytic activities and
have similar, overlapping functions in the cell. From the information you entered
in the answer table you can see that the overall sequence is fairly similar between
these proteins. For the differences in the amino acid sequence, do these differences
occur more in one domain of the proteins? Or are they evenly spread throughout
the proteins?
To answer this, you will now perform an alignment of the amino acid sequences for
all three protein family members. There are various amino acid sequence alignment
tools available for free on the internet, including many here at NCBI. We, however,
will use a popular sequence alignment program at the European Bioinformatics
Institute (EBI) called Clustal-Omega. Follow this link to the Clustal-Omega site:
http://www.ebi.ac.uk/Tools/msa/clustalo/. Keep the browser page open for your
protein blast results.
You will enter all three protein sequences into the sequence input field at the top
of the page. Before you do this you need to access all three sequence and convert
them to the correct format. First, paste in the sequence from your newly identified
protein from the text file, including the header (beginning with >). You want to enter
the next sequence on the line immediately beneath the sequence you just pasted. To
access the other two proteins (NP 001645.1 and NP 001341618.1) go back to the blast
results page and click on those accession numbers. This will take you to the protein
database entry for those proteins. Click on the blue “FASTA” tab near the top left of
the page to get the FASTA formatted sequence. Copy the sequence including all of
the header into the sequence input field starting on the next line down from your
first sequence. You want to copy and paste everything from the > at the beginning
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of the header through the last amino acid. Do this with the last sequence, entering
it and all of the header on the next line down from the second sequence. Be sure
there are no extra, blank lines between the sequences; this will cause the alignment
program to give you an error message.
In the dropdown menu for Output Format, select “ClustalW with character counts”
so that you can identify specific amino acids in the alignment. Click submit.
Examine the results page. The sequence for your newly identified protein will be on
top (labeled ORF1), with the sequences of the other two family members aligned
beneath it. The ORF1 protein is considerably longer (greater MW) than the other
two members of the family.
•
•
•
•
•
A dash (-) below the ORF1 sequence indicates that amino acid or region is
missing from the aligned related proteins.
An asterisk (*) indicates the same amino acid is present in all three of the
aligned proteins.
A colon (:) indicates that a different amino is present at that position in one or
both of the aligned proteins, but that the amino acids still have very similar
chemistry in their R-groups (a conserved substitution).
A single dot (.) indicates the chemistry of the substituted amino acid has weak
similarity (semi-conserved substitution).
No symbol means the R-group of the different amino acid has completely different chemistry.
Click the Show Colors button at the top of the results page. This colors the amino
acids according to their class of R-group. Carefully examine the differences between
the aligned proteins.
19. What class of R-group are the amino acids colored in blue? (See the Amino
Acids and R-Group pKa values section in the beginning of the lab manual.)
20. Which halves of the two aligned proteins are most different from the ORF1
protein sequence, the amino-terminal half or the carboxy-terminal half?
21. Relative to other regions of the protein, is there more divergence or less divergence in the kinase domain? Refer back to your answer to question 7 to find
the kinase domain.
22. Is the amino acid at the position that is potentially affected by the melanoma
mutation (refer to the answer to questions 4–6) the same or different in the
three aligned proteins?
23. You will see that the three amino acids at positions 594–596 of the ORF1
protein sequence are the same in all three proteins. Write out the names of
these amino acids. Make note of this. This motif plays an important role in
the catalytic mechanism of this enzyme.
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Answer Page for Section A
NAME: ______________________________________________________________________________
SECTION:____________________________________________________________________________
1. List the start and stop positions and the length of the ORF. What is its reading
frame?
2. How many amino acids in length is the protein encoded by this ORF?
3. What amino acid is encoded by the codon in which nucleotide 1860 resides?
Write out the full name.
4. At what position in the protein is this amino acid (from N-terminus)? (Do the
math here; the number of nucleotides in the cDNA up to and including this
codon, minus the number of nucleotides before your ORF start, divided by?)
5. Would this amino acid be changed by the T to A mutation at this position?
6. If the amino acid is changed, write out the amino acid substitution using the
correct nomenclature (wild-type amino acid: position: mutant amino acid).
7.
Look at the table of domain hits below the graphical summary. What is the
description of the first domain hit? What is the position of this domain in the
protein (give the amino acid position at which it starts and stops)? You can find
this by looking at the graphical summary or under Interval in the domain hit
table.
8. Based on the second domain hit, what other protein might this protein bind to?
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9.
Thinking about what might happen when this protein becomes mutated in
melanoma and what you have learned about its functional domains, what sort
of protein is this and what role might it play in normal cells? There is no right or
wrong answer here, but you must provide a one- or two-sentence description.
10. What is the identity of the unknown protein?
11. What is the catalytic activity of this enzyme? Be as specific as possible.
12. Write down the accession number for the best matching protein from the
database.
13. With which domain is the amino acid substitution from the melanoma mutation associated? Use the more general name for this domain that you identified
earlier. This domain is the active site of the enzyme.
14. What are the first two amino acids associated with the polypeptide substrate
binding site? Give the full name and position for both amino acids.
15. How many amino acid residues total make up the consensus sequence for the
polypeptide substrate binding site? (Touch the cursor to the triangle or count
the triangles in that row.)
16. With which particular feature of this domain is the melanoma mutation associated? Look at the bottom row of triangles at the position for this amino acid
(you may need to scroll more to the right). What range of amino acids makes
up this feature of the domain? Make note of this; it will be important later on!
17. There are three members of this protein family to which your newly identified
protein belongs (referred to as isoforms). Write the names of all three isoforms.
18. Identify the other two members of this protein family using the accession numbers given in the table, and then complete the table by filling in the required
information.
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Accession Number
Title of Hit
Percent Query
Coverage
Percent
Identity
NP 001645.1
NP 001341619.1
19. What class of R-group are the amino acids colored in blue? (See the Amino Acids
and R-Group pKa values section in the beginning of the lab manual.)
20. Which halves of the two aligned proteins are most different from the ORF1
protein sequence, the amino-terminal half or the carboxy-terminal half?
21. Relative to other regions of the protein, is there more divergence or less divergence in the kinase domain? Refer back to your answer to question 7 to find the
kinase domain.
22. Is the amino acid at the position that is potentially affected by the melanoma
mutation (refer to the answer to questions 4–6) the same or different in the three
aligned proteins?
23. You will see that the three amino acids at positions 594–596 of the ORF1 protein
sequence are the same in all three proteins. Write out the names of these amino
acids. Make note of this. This motif plays an important role in the catalytic
mechanism of this enzyme.
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B. Structure and Function of B-Raf Protein Kinase
The protein you discovered in the previous section of the module is B-Raf, a serine/
threonine kinase that functions in Ras-MAPK signaling from many tyrosine kinase
cell-surface receptors (see lab manual Figure 9B.2 in Lab 9B). Mutations that produce
the Val600Glu substitution are indeed found in about 60% human melanomas. This
mutation constitutively activates the kinase activity of B-Raf (meaning it can no longer
be regulated and is always on) and leads to continuous cell proliferation signaling.
Targeted drug design has led to the development of two B-Raf-specific inhibitors that
compete for ATP binding at the active site. Vemurafenib was approved for clinical
use by the Food and Drug Administration in 2011, followed by dabrafenib in 2013.
These drugs have remarkable response rates in patients whose tumors harbor the
V600E mutation. Unfortunately, the B-Raf inhibitors are typically only effective
in melanoma patients for 5–7 months. Drug-resistant progression of the disease
frequently occurs where subsequent mutations in the cancer cells lead to alternate
modes of activating growth stimulatory Ras-MAP kinase signaling.
The first Raf serine/threonine kinase to be identified was the product of the v-raf
gene from murine sarcoma virus, a retrovirus that causes tumors in mice. The gene
was named for its tumor forming activity, rapidly accelerating fibrosarcoma, or
raf. The normal, proto-oncogene version of the gene was cloned from mice a short
time later, and the role of the Raf kinases in Ras-MAPK signaling and regulation of
growth control was eventually elucidated. There are three Raf kinases in humans,
Raf-1 (the homolog of the viral oncoprotein, also called C-Raf), A-Raf, and B-Raf.
Each can bind to Ras when it becomes activated by a receptor tyrosine kinase. The
bound, active Raf phosphorylates MEK, which then goes on to phosphorylate and
activate the downstream MAP kinases, p42 and p44 Erk, and p38.
Raf is composed of two different functional regions connected by a flexible segment
of the polypeptide that acts as a hinge (see Figure 19.1). The C-terminal half of Raf is
the catalytic region. This includes the kinase domain, with the active site that carries
out the enzymatic transfer of a phosphate group from ATP to the substrate MEK
protein. The N-terminal half of Raf is the regulatory region. When Raf is inactive,
it is free in the cytosol, detached from any protein complexes at the cell membrane.
In this state, the regulatory region is folded across the kinase domain and blocks
access to the ATP and MEK binding pockets in the active site. In part A of this tutorial, you used a Domain Database search to identify two conserved domains in the
regulatory region, a Ras-binding domain and a C1 (diacyl glycerol-binding) domain.
When Ras becomes activated by cell-surface receptor signaling, it recruits Raf to
the receptor signaling complex at the membrane by binding to the regulatory region
(Ras-binding domain). When the regulatory region is bound by Ras, it is pulled away
from the kinase domain, allowing access to the active site. To stabilize Raf in its
active conformation in the receptor signaling complex, the C1 domain binds to the
inside of the cell membrane. This segment of Raf is also called the diacylglycerolbinding domain. Diacylglycerol is the glycerol backbone and two fatty acid tails of
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phospholipid, minus the head group (see Figure 12.1 in Lab 12). Proteins that can
bind diacylglycerol can anchor themselves in the cell membrane this way.
FIGURE 19.1 The regions and functional domains of human B-Raf. For Phe468 and Val600 the
asterisked numbers are the corresponding amino acid positions in the Protein Data Bank 1UWH
structure that you will examine in the next section of the module.
Regulatory region
Kinase domain
Ras binding
domain
Activation
loop
P loop
N
C
156 227
458
621
GSGSFGTV
464
D F G L AT V K
471 594
Phe468
(*467)
716
601
Val600
(*599)
A more detailed look at kinase domains
As you have learned, a kinase domain is the conserved region found in any protein
kinase that has the catalytic function of the enzyme. There are multiple elements to
a kinase domain that are reflected in stretches of similar amino acid sequence and
similar folds of the polypeptide. These elements come together to create the chemical
structure in the active site that can catalyze the transfer of phosphate from ATP to
the substrate protein. One of these elements is called the P-loop. The P stands for
“phosphate-binding,” and it has amino acids that interact with the ATP phosphates
to help position the molecule. The P-loop has the consensus sequence GxGxxG,
where G is glycine and x is another amino acid (a consensus sequence is always the
same between different proteins with that domain). Look at Figure 19.1 to see if you
can find this consensus sequence.
Another conserved element in a kinase domain is the activation loop. Movement
of the activation loop turns the catalytic activity of the kinase domain on and off
by affecting access of ATP to its binding pocket and altering the positions of amino
acids that participate in the catalysis. A hydrophobic interaction between the activation loop and the P-loop typically stabilizes the structure in an inactive conformation. The V600E mutation that you investigated in part A of the tutorial occurs in
the activation loop (see Figure 19.1). How might valine 600 affect the position of
the activation loop? When a protein kinase becomes activated, the activation loop
becomes phosphorylated and this disrupts the hydrophobic interaction with the
P-loop, allowing the activation loop to shift to an active conformation.
At one end of the activation loop is a conserved sequence of three amino acids called
the DFG motif* (Asp, Phe, Gly). The negatively-charged aspartate of the DFG motif
chelates (binds) a Mg2+ ion. Charge interactions between this Mg2+ and the phosphate
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groups on the ATP substrate are critical for positioning the gamma phosphate (the
one on the end) so that it can be transferred to the substrate protein. If you would
like to learn more about how the different parts of the kinase domain work together
in catalyzing phosphate transfer, see the interactive site Proteopedia: Eukaryotic
Protein Kinase Catalytic Domain (Google the site name to find it).
*In general a motif is a folding pattern that spans two or more sections of
ordered secondary structure. In this context it refers to a very short sequence
of amino acids of functional significance that are found in a loop of loose coil
between segments of α-helix or β-sheet.
References
H. Davies et al. (2002) Mutations of the BRAF gene in human cancer. Nature
417:949–954.
J. A. Lo and D. E. Fisher (2014) The melanoma revolution: From UV carcinogenesis
to a new era in therapeutics. Science 346:945–949.
C. M. Udell et al. (2011) Mechanistic principles of RAF kinase signaling. Cell. Mol.
Life Sci. 68:553–565.
P. T. C. Wan et al. (2004) Mechanism of activation of the RAF-ERK signaling pathway
by oncogenic mutations of B-RAF. Cell 116:855–867.
C. Examining the B-Raf Protein Structure
As you read in Lab 18, Protein Crystallization, protein structures can be determined
by X-ray crystallography. The X-ray diffraction pattern that is achieved is used to
develop a set of coordinates for the position of each atom in the folded protein. The
coordinate files that are produced can be used to view a three-dimensional model
of the protein structure by using a viewing program that translates the coordinates
into a visual representation. The Protein Data Bank (PDB) is a repository for protein
coordinate files, and has over 100,000 protein structures available for viewing. In
addition, it contains various sequence analysis and other tools for analyzing proteins.
The PDB and all of its resources are available for free on the internet. This means
that any time you like you can look up and view the structure of just about every
known protein! You will now use the resources at the PDB to further analyze the
amino acid sequence and then view the structure of B-Raf.
Go to the PDB homepage (http://www.rcsb.org/pdb/home/home.do). Try to find
human B-Raf in the PDB database by typing some key words into the search field at
the top of the page. You will see a list of hits where every entry starts with a 4-digit
code of letters and numbers. This is the accession number for that PDB entry. Find
5FD2 and click on it to look at that entry (if you don’t quickly find this entry simply
type the accession number into the search field). In addition to the coordinate files
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each entry contains information on the protein sequence and some of the details
for how the coordinates were determined.
1.
Look under Literature. What is the first author’s last name and the year of
publication for the journal article that describes this structure?
2. Look under the entry information at the top. What is the Method that was
used to determine this structure?
3. What is the Resolution of this structure in angstroms (Å)? One Å is 1/10th
of a nanometer.
Note the title for this entry, just under the accession number at the top. B-Raf was
crystalized with an inhibitor bound to the active site for this structure. It is often
easier to get enzymes to crystalize when binding an inhibitor. This is because the
unbound active site of an enzyme is flexible and can assume different conformations, making the protein less ordered and less likely to crystalize well. When an
inhibitor binds it will hold the active site in its induced-fit conformation, making
the protein more ordered and more likely to produce crystals from which a good
X-ray diffraction pattern can be obtained. Why can’t this be achieved by simply adding the substrate for the enzyme? Scroll down to Small Molecules and click on the
structure of the inhibitor. Which part of the inhibitor structure looks most similar
to the structure of ATP?
Scroll back to the top and click on the Sequence tab. The graphic on this page shows
different kinds of information about the B-Raf amino acid sequence and the quality
of the structural model for this coordinate file along different segments of the polypeptide. The row labels on the left indicate the different features examined and the
symbols in that row indicate the position of those features along the length of the
polypeptide (the scale at the top is the polypeptide length in number of amino acids).
Place the cursor over the graphic and use the scroll wheel to zoom in and zoom out.
You can click and drag to move along the length of the polypeptide. You can also use
the dropdown items from the function icon at the top right corner (circle with three
bars) to perform these functions. Touch the cursor to the blue bar on the UNIPROT
P15056 row and examine the popup window. The crystals used to determine this
structure included only the isolated kinase domain of B-Raf without the regulatory
region. There are not actually any structures of the entire B-Raf protein because the
flexible connection between the kinase domain and regulatory region does not allow
for the formation of ordered crystals.
4. What is the range of amino acids from the B-Raf kinase domain that were
crystalized to determine this structure? Give the amino acid sequence numbers. Open the B-Raf amino acid sequence you saved from Lab 19 part A. The
numbering to the right in the popup window is the same as for this sequence,
and matches the numbering in Figure 19.1 in part C.
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Remember that only regions of the crystalized polypeptide that are ordered will
produce a consistent X-ray diffraction pattern. Therefore, only the atoms of ordered
amino acids will be represented in the coordinates and show in the protein structure.
Segments of the polypeptide that are very flexible can be in different conformations in
different crystalized protein molecules. These are disordered, meaning they will not
give a consistent X-ray diffraction pattern and are not represented in the structure.
The amino- and carboxy-terminal ends of a crystalized polypeptide are typically
disordered and are not represented in the protein structure—they are flexible because
they stick out from the folded protein. Examine the Unmodeled row in the graphic.
This tells you which parts of the B-Raf kinase domain are ordered and appear in the
structure and which do not.
5. What is the range of B-Raf amino acids that are ordered and appear in the
protein structure? Note that there is a gap of disordered amino acids in the
middle of the sequence. Give the ranges of all the ordered amino acids.
6. Some additional amino acid sequence was added to the N-terminus of the
kinase domain to make it easier to purify as a recombinant protein from E.
coli. Zoom in so that you can see the amino acid sequence. What are the six
histidine residues at the N-terminus? You may need to ask your TA for help
on this if you have not yet done the fluorescent protein project.
7.
Look at the Secondary Structure row. How many separate segments of the
polypeptide have beta-sheet secondary structure?
8. Find the DFG motif in the graphic (zoom into sequence level and touch the
cursor to individual amino acids in the UNIPROT row to see their position).
Refer to your answers from part A and the diagram in part C. The DFG motif
begins as loose coil (no secondary structure), and goes into what?
9.
The unordered sequence from 605–613 falls into what specific component of
the kinase domain? Does it make sense that this sequence is unordered?
10. Find valine 600. To what sort of secondary structure is it assigned? Does
this make sense considering the fact that this valine is in the kinase domain
activation loop? (Hint: Activation loops need to be flexible.)
Click on the 3D-View tab at the top to view protein structure. Find the “Select a different viewer” dropdown menu at the lower right to see the different viewing programs
available. Select the JSmol program to view the structure. The default mode to view
the protein is in “cartoon” style where alpha-helix is displayed in pink, 3/10-helix in
purple, beta-sheet in yellow, and loose polypeptide as the white line. The inhibitor
bound at the active site is shown as a ball and stick model.
Click and drag (slowly) to rotate the structure through three dimensions; use the scroll
wheel to zoom in and out. Note that this structure is a pair of two kinase domains
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bound together. It is common for the coordinates from an X-ray diffraction pattern
to include two protein molecules. This does not necessarily imply a biochemically
important interaction between them.
With one kinase domain above and one below, twirl the structure through 360
degrees and examine the bottom domain as you turn it. You will see two lobes to
the kinase domain, one with mostly yellow beta-sheet on the bottom and one with
mostly pink alpha-helix on the top. The active site is the cleft between these two lobes;
the inhibitor is bound more or less to the ATP binding site. You can see the amino
acid position in the structure by touching the cursor to the backbone at that point.
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Answer Page for Section C
NAME: ______________________________________________________________________________
SECTION:____________________________________________________________________________
1. Look under Literature. What is the first author’s last name and the year of publication for the journal article that describes this structure?
2. Look under the entry information at the top. What is the Method that was used
to determine this structure?
3. What is the Resolution of this structure in angstroms (Å)? One Å is 1/10th of
a nanometer.
4. What is the range of amino acids from the B-Raf kinase domain that were crystalized to determine this structure? Give the amino acid sequence numbers. Open
the B-Raf amino acid sequence you saved from Lab 19 part A. The numbering
to the right in the popup window is the same as for this sequence, and matches
the numbering in Figure 19.1 in part C.
5. What is the range of B-Raf amino acids that are ordered and appear in the protein
structure? Note that there is a gap of disordered amino acids in the middle of
the sequence. Give the ranges of all the ordered amino acids.
6. Some additional amino acid sequence was added to the N-terminus of the kinase
domain to make it easier to purify as a recombinant protein from E. coli. Zoom in
so that you can see the amino acid sequence. What are the six histidine residues
at the N-terminus terminus? You may need to ask your TA for help on this if
you have not yet done the fluorescent protein project.
7.
Look at the Secondary Structure row. How many separate segments of the
polypeptide have beta-sheet secondary structure?
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8. Find the DFG motif in the graphic (zoom into sequence level and touch the cursor to individual amino acids in the UNIPROT row to see their position). Refer
to your answers from part A and Figure 19.1. The DFG motif begins as loose
coil (no secondary structure), and goes into what?
9.
The unordered sequence from 605–613 falls into what specific component of the
kinase domain? Does it make sense that this sequence is unordered?
10. Find valine 600. To what sort of secondary structure is it assigned? Does this
make sense considering the fact that this valine is in the kinase domain activation loop? (Hint: Activation loops need to be flexible.)
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D. Examining the B-Raf Structure with PyMOL
While the protein structure viewing programs that run through the browser are nice
for looking at the overall secondary and tertiary folding of a protein, you probably
noticed that it is difficult to identify specific amino acids or to look at fine details
involving R-groups. PyMOL is a much more powerful viewing program that makes it
easy to examine these details. It is the primary modeling program used by biochemists
studying protein structure. It has a wide range of functions that allow interactions
between different amino acids in the protein to be identified and detailed images
to be produced. The program is designed to be efficient and functional, and it is not
particularly user friendly. It is most easily used with command line script, but it
does have some point and click functions. This tutorial will walk you through a few
of the features to further examine the V600E mutation in B-Raf.
Find the PyMOL shortcut on your desktop and open the program. The top half of
the PyMOL screen has a dropdown menu bar and displays the command line script.
Command line prompts are entered in this window in the field next to PyMOL>.
The bottom half of the screen has the viewing window where the structure will be
displayed and a field to the right that lists each object displayed there, with a series
buttons for manipulating them.
1. Open a B-Raf coordinate file in PyMOL. In the command line field, enter the
prompt:
fetch 1uwh
“Fetch” is the prompt to load a coordinate file from the Protein Data Bank
(PDB) and open it, and 1UWH is the PDB accession number. Alternatively, your
instructor may have the coordinates file available on Canvas. In this case, find
the file on Canvas, click it, and select “open” from the download options.
2. You will see the B-Raf structure in the viewer window, displayed in ribbon
format. Note that this structure is also a pair of two kinase domains. You can
use the mouse to manipulate the structure; left-click and drag to rotate. On a
PC you can right-click and drag away from the center of the viewing field to
zoom in; right-click and drag toward the center to zoom out. On a Mac, go to
Display in the menu bar, click Zoom, and select one of the options to zoom (also
in PC). The options give you the number of Angstroms across the viewer; 20 or
12 Angstroms will provide a good amount of detail without zooming in too far.
Select All under Zoom to return to the starting view.
3. Look at the field to the right of the viewing window. The structure (1UWH) is
now listed as an object. The five buttons to the right are:
•
•
•
•
•
A (action): Manipulate or reset the structure in different ways.
S (show): Turn on different ways of displaying the structure.
H (hide): Turn off different ways of displaying the structure.
L (label): Label different elements of the structure.
C (color): Add different colors to different elements of the structure.
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For now, follow the instructions below carefully so that you can identify different amino acids in the activation loop and the P-loop. You will need to save two
images for credit. Once you have accomplished this you can explore the various
feature of this program on your own.
4. On the 1UWH object line, click S (show), lines. You will see the bond lines for
all of the amino acids and their R-groups superimposed over the ribbon diagram
of the polypeptide backbone. Turn off the bond lines by clicking H (hide), lines.
5. The ribbon diagram shows only polypeptide backbone and how it is folded, it
does not show the R-groups that actually make up the bulk of the protein. The
Van der Waals radius of atoms on the outside of the protein form its surface.
An image of this surface more accurately depicts what that protein would “look
like” with all its atoms, and how it might interact with other molecules. On the
1UWH line, click S (show), surface. As you can see, this makes it difficult to
view details of the molecular structure. Turn off the surface view by clicking H
(hide), surface.
6. This coordinate file has two kinases domains that were adjacent in the protein
crystal. Rotate the structure so that you can see the two kinase domains and
where they meet. Add different colors to the two kinase domains; C (color), by
chain, and select the third “by chain” option from the top.
7.
Click anywhere on the structure without dragging. By doing this you have
selected an amino acid in the structure and it will show as “(sele)” in the object
list. You don’t yet know which amino acid this is, so deselect it by going to that
row and clicking A (action), delete selection.
8. You selected an amino acid residue in the previous step, but you can control what
is selected when you click on the structure, from individual atoms to the entire
polypeptide chain. Examine the box in the lower right of the viewing window
and find “Selecting Residues.” Residue here means you are selecting individual
amino acids. Click once on the word Residue, and now you have set it to select
polypeptide chains. Now click on the word Chains to change the selection option
again. Click through each of the selection options to see what is available.
Let’s remove one of the duplicate Raf kinase domains to make the structure
easier to look at. Set the selection option to Chains. Click on the green chain to
select it. In the (sele) row, click H (hide), everything. The purple dots that remain
are the positions of the individual atoms for that polypeptide. Click anywhere
off the structure to make them disappear. These atoms are still retained in the
structure, and you can view the green polypeptide again if desired.
What is the green molecule that was left behind when you hid the green kinase
domain? This B-Raf coordinate file was also produced from B-Raf that was
crystalized with an inhibitor bound to it. The green molecule is the inhibitor
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bound to the green kinase domain. Remove it by setting the selection option to
Molecules, selecting the green inhibitor, and then (sele), H (hide), everything.
9.
Rotate the blue kinase domain to look at it from different angles. If you need to
center it, select it and use A (action), center. You will notice several small crosses
around the polypeptide. These are water molecules. To turn them off, go to the
1uwh row and click A (action), remove waters.
10. Find the inhibitor molecule bound to the blue B-Raf kinase domain. Give it
a different color so it stands out from the blue polypeptide backbone. Set the
selection option to Molecules, select the inhibitor, and then color it red (C, reds,
red). Spend some time rotating the structure and zooming in and out to examine where the inhibitor is bound between the two lobes of kinase domain. The
inhibitor is bound to the ATP-binding site of the kinase domain. Remember that
ATP is one of the substrates for B-Raf, and that the catalytic activity of B-Raf is
to transfer a phosphate group from ATP on the substrate/target protein, MEK.
11. Now let’s look at where some of the important amino acids you identified in
Part A of the tutorial lie in the structure. First turn on the amino acid sequence
numbering. Go to the top menu bar above the command line script, and click
Display, and then Sequence. The amino acid sequence will be displayed at the
top of the viewing window. Amino acids 447–722 of the green kinase domain
(now hidden) are listed first; scroll to the right to see the amino acid sequence
of the blue kinase domain. The amino acid sequence numbering is shifted by
one residue for this structure, so D447 is D448 in the sequence you determined
in part A, V599 is V600, etc.
12. Set the selection option to Residues. Find V599 in the blue sequence and click on
it. You will see purple dots appear at the positions of the atoms for V599. Rotate
the structure so that you can see them clearly. Find (sele) in the object list and
click S, sticks. Then click C, by element, and select one of the options that will
make carbon a different color than the blue chain. Click the black background
and the purple dots will disappear. Examine valine 599 and look at its hydrophobic R-group (−CH(CH3)2). Hydrogen atoms are not shown.
You will see that V599 is at edge of a gap in the chain, at the end of a section
of loose coil. Remember that V599 is in the activation loop of the B-Raf kinase
domain (see Figure 19.1). Most of the activation loop is disordered in this structure, and therefore the positions of these residues are not indicated in the coordinates. Activation loops tend to be flexible by nature, so this is not uncommon.
This means that in the structure image most of the activation loop is not shown.
The other end of the gap is adjacent to V599; Figure 19.2 has the polypeptide
chain drawn in with a dashed line to help you see where the activation loop
should be.
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13. Look at figure 19.1 on page 231 and find the P-loop. Select phenylalanine (F)
467 on the kinase domain structure, and fill in the colored stick representation
of F467 as you did for V599. Use the same color pattern by element that you
used for V599.
Look at the positions of the R-groups for V599 and F467. How close are they to
each other? To see the electron density of the R-groups and whether they might
be interacting, select them (you can simply click on them in the structure) and
S (show) spheres. Do this for both V599 and F467. What kind of interaction are
they having? How might this interaction affect the position of the activation
loop?
When the B-Raf kinase is turned off, a hydrophobic interaction between V599
and F467 helps to stabilize the activation loop in an inactive conformation where
it is blocking access of ATP to its binding cleft. Save your current image before
continuing. Go to File in the top menu bar, Export Image As, and save it as a
PNG file to the desktop. After you have saved the image, hide the spheres on
V599 and F467 so that you can more easily see the polypeptide backbone again.
Select them (you can select both at the same time) and on the (sele) row click
H, spheres.
14. Find the DFG motif. The DFG motif is just a few residues toward the N-terminus
from V599 (Figure 19.1). Select the aspartate (D) of the DFG motif, and fill in
a colored stick representation of this residue. Examine the position of this
aspartate relative to the red inhibitor. The negatively charged carboxylate of the
aspartate R-group binds a Mg2+ ion that positions the phosphates in the ATP
substrate. Identify the carboxylate in the aspartate R-group (refer to the amino
acid structures on page 13 if needed). Is the carboxylate pointed toward the
ATP-binding cleft (more or less where the red inhibitor is bound) or away from
the cleft? What would you predict should the conformation be in a catalytically
active kinase domain when B-Raf is turn on?
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When V599 is interacting with F467 and the activation loop is in its inactive
conformation, D593 of the DFG motif is not correctly aligned for catalysis (Wan
et al., 2004). When B-Raf is turned on, the activation loop becomes phosphorylated. This disrupts the hydrophobic interaction between the activation loop and
P-loop, which changes the position of the activation loop. This allows D593 to
become correctly aligned (pointing toward the ATP) so that its chelated Mg2+
can coordinate the substrate ATP phosphates. This movement of the activation
loop also opens access for ATP to its binding cleft.
Consider the valine to glutamate substitution that occurs with B-Raf mutation in
melanoma. What would happen if V599 became mutated to a bulky, negatively
charged amino acid that could mimic the structure of a phosphate group?
15. PyMOL has the ability to model amino acid substitutions to see how they might
affect the protein conformation. You will now model the valine to glutamate
amino substitution caused by the melanoma mutation. First, zoom in on V599
so you can see small details. Go to Wizard in the top menu bar, click Mutagenesis, and select Protein. A Mutagenesis box will appear in the field to the right
of the viewer. Click the box that says No Mutation and select glutamate (GLU,
not GLUH). Then click on V599 in the blue amino acid sequence above. You will
see the R-group of glutamate superimposed on top of the valine.
You are substituting the valine with an amino acid whose atoms are not specified
in the coordinate file, so the glutamate can be in several different conformations
in the structure. These different conformations are called rotamers. Rotate the
glutamate through its different rotamers using the buttons at the bottom of the
right-hand field.
Move through each of the different rotamers by rotating the R-group back and
forth with the arrows. You will see red disks appear for conformations where
there would be steric interference from nearby atoms. The bigger the red disks
the greater the interference. Find a rotamer with minimal steric interference and
then click Apply in the Mutagenesis box. Click Done to remove the Mutagenesis
box. V will have been replaced with E at position 599 of the blue amino acid
sequence.
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Select F467 and E599 and show spheres to see the electron density for the atoms.
Look at the atoms for both R-groups. Would these R-groups still have a hydrophobic interaction? Keep in mind that the activation loop is flexible, and the
negatively charged carboxylate on the glutamate would want to form H-bonds
with water molecules. How might this amino acid substitution affect the position of the activation loop? Save another png image of the B-Raf kinase domain
with the amino acid substitution showing spheres.
The V599E substitution causes B-Raf to be constitutively active (always on) in
melanoma tumor cells; the activation loop cannot go into the inactive conformation. This causes continual signaling down the Ras-MAP kinase signal
transduction pathway that produces continual growth stimulation in these cells.
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E. Supplementary Information
Some PyMOL shortcuts:
• On PCs, you can center-click and drag to move the structure to a different
center point.
• Click S in the bottom row of buttons (to the right of the rotamer arrows) to
toggle on and off the amino acid sequence.
• When you first open a structure, you can quickly turn on preset viewing options without having to go through all the individual steps. Click A
(action), preset, and then try some of the different options. The Publication
preset option shows the ribbon diagram polypeptide and ligands, with the
N-terminal and C-terminal halves in different, transitioning colors.
How to Make a Protein Image with a Transparent Surface that Shows the
Polypeptide Backbone Inside
1. Load a coordinate file into PyMOL using the fetch prompt. We will use the
accession number 2FGF in this example.
command line:
fetch 2FGF
2. Select the polypeptide you want to work with. Change the Selection function to
select “chains.” On the structure, click the polypeptide with the cursor to select it.
3. Show surface for the selected polypeptide. Under (sele), click S, and select surface.
4. Change the surface to a neutral color: (sele), click C, and select a single gray
color from the bottom of the menu. You can use whichever color you wish but
the image will look better if the entire surface is a single color.
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5. Use the command line prompt to set the surface transparency. Use the prompt
set transparency followed by a value of 0.0–1.0 to set the degree of transparency. 1.0 will be completely transparent and 0.0 will be completely opaque. For
example, to set the surface to 50% transparency:
command line: set transparency, 0.5
You can change the level of transparency using the same command.
6. You can set whichever representation you wish for the polypeptide backbone
and/or R-groups. In this example, the ‘cartoon’ ribbon diagram is the default.
To remove it, select it by clicking it inside the transparency and under (sele),
click H and click cartoon.
7.
To show the polypeptide backbone and R-groups as sticks, click the backbone
to select it. Under (sele), click S and “sticks.”
8. If you wish to highlight particular residues, under Selection, go back to “residues.”
9.
Display the amino acid sequence at the top by clicking S in the lower-right control
panel. You can toggle it on and off with this function.
10. Select a residue by clicking it in the amino acid sequence or on the structure.
To color it, (sele), C, by element, and choose an option that makes carbon a different color than the rest of the structure. Or choose whichever color you want
to highlight that residue.
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