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BASIC LABORATORY COURSE 3
Chromatographic
Separation of
Proteins
Preface
This laboratory manual forms an integral part of a complete teaching
system. In addition to the manual, the system consists of the chemicals and
materials required to perform a series of experiments in modem biology.
This teaching system was designed principally for students taking a
laboratory course in general biology, physiology, or basic biochemistry.
I have assumed that most readers have had an introductory lecture course
in the biological sciences. However, I have attempted to write the manual
such that even a stranger to biology could follow it and perform the
exercises.
The laboratory manual is divided into two major sections. The first section
provides basic information on the biology and chemistry of proteins and
is also intended to acquaint the student with the principles and techniques
of chromatography­. Students should be familiar with this material before
advancing to the second section of the manual. The second part of the
manual shows students how to apply what they have learned to perform
five exercises in modem biology. The exercises deal with the structure,
function, and isolation of proteins, and stress the relevance of molecular
biology to animal physiology. Each experiment provides state-of-theart information and each experiment can be completed within a 2 hour
laboratory session. A background information section is given for each
exercise and students should be required to read this material prior to
performing the exercise in the laboratory. Study questions are found at
the end of each exercise. Students derive answers to these questions by
integrating the background material found in the text with the results of
the experiments.
John N. Anderson, Ph.D. Professor of Biology
Department of Biological Sciences
Purdue University
West Lafayette, Indiana 47907
© 2013 by John N. Anderson, All rights reserved.
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Table Of Contents
Part A. Background Information
I. Protein Composition and Structure: A Review of the Basics
Amino Acids-Building Blocks of Proteins..................................................... 1
The Peptide Bond........................................................................................... 2
Hierarchies of Protein Structure..................................................................... 3
II.Chromatography
Chromatographic Separation of Small Biomolecules..................................... 6
Chromatographic Separation of Proteins........................................................ 7
Part B. Laboratory Exercises
Introductory Remarks.......................................................................................... 10
Experiment 1.
Separation of Molecules by Gel
Filtration Chromatography...................................................... 12
Experiment 2.
Determination of the Molecular Weight
of Human Hemoglobin............................................................ 19
Experiment 3.
Dye Binding of Serum Albumin............................................. 24
Experiment 4.
A Comparison of the Properties of a-Amylase from
Human Saliva and Pancreas.................................................... 27
Experiment 5.
Isolation and Immunological Analysis of a Protein
from Egg White....................................................................... 37
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BACKGROUND INFORMATION
I. Protein Composition and Structure:
A Review of Tile Basics
Proteins occupy a central position in the structure and function of all living
organisms. Some proteins serve as structural components while others function in
communication, defense,and cell regulation. The enzyme proteins act as biological
catalysts which control the pace and nature of essentially all biochemical events.
Indeed, although DNA serves as the genetic blueprint of a cell, none of the life
processes would be possible without the proteins.
Amino Acids - Building Blocks of Proteins
The fundamental unit of proteins is the amino acid. The common amino acids
have the general structure shown in Figure I. Each amino acid has an amino group
(NH2 and a carboxylic acid group (COOH) attached to a central carbon atom
called the alpha carbon. Also attached to the alpha carbon are a hydrogen atom
and an R-­group or side chain.
Figure 1. General Structure of Alpha-Amino Acids.
The C stands for a carbon atom; C* is the alpha carbon; His hydrogen;
N is nitrogen, O is oxygen, -NH2 an amino group and -COOH is the
carboxylic acid group. R is a general term for any one of several
different side chains that determine the nature of different amino acids.
There are 20 amino acids commonly found in proteins and these differ from each
other in the nature of the R-groups attached to the alpha carbon. A convenient
classification of amino acids depends on the number of acidic and basic groups
that are present. Thus, the neutral amino acids contain one amino and one
carboxyl group. The acidic amino acids have an excess of acidic carboxyl over
amino groups. The basic amino acids possess an excess of basic amino groups.
Table I lists the major amino acids found in proteins.
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Table 1. Amino Acids Found in Proteins.
Neutral Amino Acids
GlycineAlanine
ValineLeucine
IsoleucineSerine
ThreonineCysteine
MethioninePhenylalanine
TyrosineTryptophan
Proline
Acidic Amino Acids
Aspartic Acid
Glutamic Acid
Basic Amino Acids
Arginine
Lysine
Histidine
The Peptide Bond and the Primary Structure of Proteins
Proteins are composed of amino acids linked into chains by peptide
bonds as shown in Figure 2. Two amino acids joined by a single peptide
bond form a di­-peptide; three amino acids form a tri-peptide, and a large
number of amino acids joined together constitute a polypeptide. A protein
is a polypeptide chain that contains more than 50-100 amino acids. The
monomer units in the chain are known as amino acid residues. The average
protein contains about 350 amino acid residues although proteins with as
many as 1000 residues and those with as few as 100 are not uncommon.
Figure 2. Formation of a Peptide Bond.
Peptide bonds are enclosed in the dotted boxes. The dotted circle shows
how a peptide bond is formed with the production of H2O.
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The sequence or order of amino acids along a polypeptide chain is referred to as the
primary structure of the protein. The primary structure of the protein myoglobin
is given in Figure 3. This protein serves to bind and store oxygen in muscle. The
primary structure of over 500 different proteins is now known.
Figure 3. The Primary Structure of Whale Myoglobin.
[Amino or N-terminus]
Val-Leu-Ser-Giu-gly-Giu-Trp-Gin-Leu-Val-Leu-His-Val-Tyr-Aia-Lys-Val-­
Giu-Aia-Asp-Val-Ala-Giy-His-Gly-Gin-Asp-lle-Leu-lle-Arg-Leu-Phe-Lys­Ser-His-Pro-Glu-Thr-Leu-Glu-Lys-Phe-Asp-Arg-Phe-Lys-His-Leu-Lys-Thr-­
Giu-Ala-Glu-Met-Lys-Aia-Ser-Giu-Asp-Leu-Lys-Giy-His-His-Giu-Ala-Giu­
Leu-Thr-Ala-Leu-Giy-Aia-lle-Leu-Lys-Lys-Lys-Giy-His-Giu-Ala-Giu-Lys­
Leu-Lys-Pro-Leu-Ala-Gln-Ser-His-Ala-Thr-Lys-His-Lys-lle-Pro-lle-Lys­-TyrLeu-Giu-Phe-lle-Ser-Giu-Aia-lle-lle-His-Val-Leu-His-Ser-Arg-His­-Pro-GiyAsn-Phe-Giy-Aia-Asp-Aia-Gin-Giy-Aia-Met-Asn-Lys-Aia-Leu-Giu­-Leu-PheArg-Lys-Asp-lie-Ala-Ala-Lys-Tyr-Lys-G lu-Leu-G ly-Tyr-G ln-Gly
This protein consists of 153 amino acid residues. Each residue in the protein
is given a 3-letter abbreviation (thus: lysine = Lys); valine = Val). Proteins are
always written with the free amino or N-terminus toward the left.
Three-Dimensional Protein Structure
In the cell,the polypeptide chain is folded into a highly ordered shape or con­
formation. Most proteins are globular in shape and these proteins are usually
soluble in water or in aqueous media containing salts. This group includes the
enzymes, antibodies, and a variety of other proteins. Less frequently, proteins are
long and fibrous and most of these elongated molecules are insoluble in water and
serve a role in the maintenance of cell structure.
The three-dimensional structure of a protein is due to the type and sequence of
its constituent amino acids. Since the amino acid sequence of each protein is
unique, it follows that different proteins assume different shapes. Thus, there is a
remarkable diversity of three-dimensional protein forms. The conformation of a
protein is usually of critical importance in the protein’s function. For example, a
protein can be unfolded into a polypeptide chain that has lost its original shape.
In general, proteins such as enzymes are rendered nonfunctional upon unfolding
because functional activity is dependent on the proteins native shape. This process
is called denaturation. Most proteins can be denatured by heating, by certain
detergents, and by extremes of pH. The ionic detergent, sodium dodecyl sulfate
(SDS), is often used to denature proteins. The denaturing treatment can frequently
be reversed, for example by removing the detergent or by neutralizing the pH.
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During this renaturing process, the polypeptide chain spontaneously refolds
into its original conformation and the protein regains its biological activity. A
similar folding process occurs in the cell for when a polypeptide is constructed
on the ribosomes, it folds into a biologically active conformation. Thus, the threedimen­sional folding of a protein and its biological properties are directed by the
sequence of amino acid residues along the polypeptide chain.
Biochemists have identified three structural levels that define the three-di­mensional
shape of a protein. These levels of organization are secondary structure, tertiary
structure,and quaternary structure. Figure 4 shows examples of these levels of
organization. The major force involved in the formation and maintenance of these
structures are various types of weak, noncovalent bonds that are formed between
the amino acid residues and between the amino acid residues and water. Although
a noncovalent bond typically has less than 1/20 the strength of a covalent bond,
a large number of noncovalent bonds participate in the folding of a single protein
into its native conformation.
Figure 4. Levels of Structural Organization.
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Secondary Structure
The spatial arrangement of the protein backbone that is generated from the
folding of the polypeptide chain is called the secondary structure of the
protein. The secondary structures of proteins are stabilized by hydrogen bonds
in which a hydrogen serves as a bridge between oxygen and nitrogen atoms
(-C=O••••••HN-). A common secondary structure is the a-helix which consists
of a single polypeptide chain coiled into a rigid cylinder. In the a-helix, each
peptide bond along the polypeptide is itself hydrogen bonded to other peptide
bonds. Many enzymes contain small regions of a-helices, while long sections of
the a-helix are often found in proteins involved in cell structure. Another type
of secondary structure of proteins is the b-sheet, which is a central organizing
feature of enzymes, antibodies, and most other proteins that perform nonstructural
functions. Here, a single polypeptide chain folds back and forth upon itself to
produce a rather rigid sheet. Hydrogen bonds between neighboring polypeptide
chains are a major stabilizing force for the b-sheet conformation.
Tertiary Structure
The tertiary structure of a protein describes the detailed features of the three­
dimensional conformation of the polypeptide chain. It is brought about by the
interactions between the amino acid side chains which cause the folding and bending
of a-helix and b-sheet segments of the protein. One very important interaction at
this level of organization involves the hydrophobic and hydrophilic side chains
of the amino acid residues. Hydrophobic amino acids, such as phenylalanine and
leucine, show limited solubility in water. Thus, these hydropho­bic residues in a
protein tend to cluster on the inside of the protein in order to avoid contact with
the aqueous environment. Hydrophilic amino acids such as glutamic acid and
lysine are readily soluble in water, and thus these amino acids arrange themselves
on the surface of the protein molecule, where they can interact with water and
with other hydrophilic side chains. The consequence of these interactions is that
a polypeptide chain typically folds spontaneously into a stable, usually globular
structure, with the hydrophobic side chains packed into the central core of the
protein and the hydrophilic side chains forming the irregular, external surface.
Quaternary Structure
Some proteins contain more than one polypeptide chain. For example, each
molecule of human hemoglobin consists of four polypeptide chains which are held
together by a variety of noncovalent bonds. The arrangement of the polypeptides
in such proteins is called the quaternary structure.
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II. Chromatography
A. Chromatographic Separation of Small Biomolecules
In 1906, the Russian botanist Tswell introduced the term chromatography (“to
write with color”) as a result of his studies on the separation of plant pigments
using a column of calcium carbonate. Chromatography, like many other important
discoveries, remained largely unnoticed for almost one-half century but in the
past 30 years has become one of the most important tools for the rapid separation
of biological molecules.
Paper chromatography is used to separate small molecules such as amino acids,
sugars, and dyes. In a common form of this technique, the sample is applied as
a spot to a sheet of adsorbent paper and a solvent is allowed to flow through the
sheet from one edge. As the solvent travels across the paper by capillary action,
it picks up those sample molecules that are soluble in it and not bound to the
paper. Consequently, sample molecules that move the fastest are those that are
most soluble in the solvent and have the lowest affinity for the adsorbent paper. A
simple optional experiment, described in Table 2, illustrates the separation of the
compo­nents of ink by paper chromatography.
Table 2. Separation of the Components of
Writing Ink by Paper Chromatography
1.
(An Optional Experiment)
Place samples of different inks as small spots along the shorter edge of a
piece of Whatman No. I filter paper (20x30cm), 2cm from the edge.
2. After the samples have dried, staple the two longer edges together to
form a cylinder.
3. Place the end of the cylinder containing the samples in a jar containing I
em of water. As the water rises, the different colored components of the
inks will migrate at different speeds, which are related to their solubility
in water and adsorption to the paper.
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B. Chromatographic Separation of Proteins
A major goal in the modern biology laboratory is the isolation of specific
proteins from complex mixtures. Such isolation procedures frequently involve
chromatography. Proteins are most often separated by column chromatography,
in which a mixture of different proteins is passed through a column or tube that
contains a matrix material. The separation occurs because proteins interact with
various matrices in different ways. In ion-exchange chromatography, small beads
that carry either a positive or negative charge form the column matrix,and proteins
are separated according to their charge. An acidic protein that carries a negative
charge will bind to a positively charged matrix. Thus, such a matrix can be used
to separate negatively charged acidic proteins (which will interact with the matrix
and be retained) from positively charged proteins (which will flow through the
matrix). The acidic proteins can then be removed or eluted from the column by
disrupting their interactions with the matrix. In hydrophobic chromatography,the
columns are packed with beads that contain hydrophobic side chains. Proteins
with exposed hydrophobic regions bind to the matrix, but proteins without such
regions do not.
Gel filtration, another type of chromatographic method, will be used in the
laboratory exercises described in this manual. Gel filtration is the technique of
separating molecules of different size by passing them through a gel column.
In the modem biology laboratory, this method is most commonly used for the
separation of proteins, although the technique can be used for the separation of
other types of biomolecules as well.
The steps in the separation of two proteins of different size by gel filtration are
shown in Figure 5 and described below.
1. Preparation of the Gel Column
The two most common polymers used to form the gel are polyacrylamide and
dextran. The polymers are cross-linked to themselves to form small beads which
swell upon addition of water to form structures that resemble microscopic porous
sponges. The hydrated beads,which are typically about 0.1mm in diameter,are
then packed into a column and washed with buffer.
2. Sample Application
A sample containing a mixture of proteins of different sizes is carefully introduced
to the top of the gel bed. Buffer is added to the top of the column and the proteins
enter the gel.
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3. Sample Separation
The porous (sponge-like) nature of the hydrated gel beads forms the basis of this
separation method. Molecules larger than the largest pores of the gel cannot enter
the pores (they are excluded) and therefore pass rapidly down the column between
the beads. The path of smaller molecules through the gel bed is much longer
than that of the large molecules because they move in and out of the pores of
the gel (they are included). Thus, the ability of the small molecules to penetrate
the pores in the gel results in the retardation in the rate of their migration down
the column. The porosity of the beads is an important feature in gel filtration,
since it determines the size range of proteins that can be separated by the method.
Different types of beads are commercially available which differ in the extent of
cross-linking and thus pore size. Beads with large pores are used for separating
large protein molecules while small proteins and peptides are separated on gels
with small pores. Sephadex G-100 will be used in the experiments described
in this manual and this material permits separation of proteins in the molecular
weight range of 5,000-100,000.
4. Samnle Collection
The column eluate is allowed to run dropwise into a series of test tubes and the
amounts and types of the proteins in each tube (fraction) are determined.
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Figure 5. Stages of a Separation of Two Proteins
of Different Size on a Gel Filtration Column
1. PREPARATION OF TilE GEL COLUMN- Hydrated beads (large open
circles) are packed into a cylindrical tube or column to form the gel bed.
2. SAMPLE APPLICATION - A sample containing a mixture of a large
protein (large black dots) and a small protein (small black dots) is added to
the top of the gel bed.
3. SAMPLE SEPARATION - Buffer is applied to the top of the column and
the proteins enter the gel bed. The smaller proteins penetrate the gel beads
and their movement down the column is retarded. The larger proteins are
excluded from the gel beads and thus move rapidly down the column.
4. SAMPLE COLLECTION - The two proteins are now separated and can be
collected in the column eluate.
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PARTB LABORATORY EXERCISES
Introductory Remarks
A. FEATURES OF THIS LABORATORY
1. This laboratory course was designed for 8 groups of students working
in teams of 14. Each group will be responsible for the preparation and
maintenance of one gel filtration column. We encourage full cooperation
between members of each group and an equitable distribution of work
and responsibility.
2. Each exercise consists of a background information section, an
experimental procedure, and study questions. Before each laboratory,
the student should read the background information section preceding
the description of the exercise and then study the directions for doing the
experiment. During this study, the student should understand the reason
for each step in the procedure.
3. After completing the experiment, the student should clean the materials
he/she has used in the laboratory.
B. QUANTITIES AND THEIR MEASUREMENT
To successfully perform the experiments in this manual, the student must be
familiar with the following metric units of measurements.
Length
1 meter(m)
= 100 centimeters
1 centimeter (em) = 10 millimeters (mm)
Volume
1 liter (l)
= 1.06 quarts
milliliter (ml)
= 1000 microliters
1 microliter (µ1)
= 10-3ml
Mass
1 Kilogram (Kg) 1 gram (g)
1 milligram (mg) 1 microgram (mg) = 1000 grams
= 10-3Kg
= 10-3g
= 10-6g
English Equivalent
= 0.3937 inches
= 1000 milliliters
= 2.2046 pounds
A solution is a homogenous mixture whose composition may be varied. The
solute is the component present in lesser quantity and the solvent is the constituent
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present in greater quantity. Because the composition of a solution may vary, it is
necessary to specify the concentration of the solutes to describe the solution. The
concentration of a solute is often expressed as the amount of solute per amount
of solution. In this manual,concentration is frequently expressed as the mass of
solute per unit volume of solution. For example, 10µg protein/ml means that 10µg
of protein is dissolved in I ml of solution.
The basic tools you will have available for measuring liquids are the macropipetors
and two types of transfer pipets. The macropipetors are calibrated in mls and should
be used for measuring volumes greater than 2 mls. A diagram of the transfer pipets
is given in Figure 6 showing the approximate volumes of liquid that are dispensed
if the pipets are filled to the indicated positions. The large transfer pipets should
be used for measuring volumes between 0.5-1.ml and the small transfer pipets
for volumes less than 0.5 mi. Thus, to transfer 0.1ml (100µl) from tube A to tube
B, place the tip of a small transfer pipet in tube A, depress the bulb slightly with
your thumb and index finger, slowly relax your index finger until the solution
has filled the pipet to the 100µl level,transfer the pipet to tube Band depress the
bulb. A separate small pipet should be used by the entire class for dispensing
each solution provided with this Experiment Package. The small transfer pipets
will also be used for the application of samples to the chromatographic columns.
Students should practice using these devices prior to beginning the experimental
exercises.
Figure 6. Transfer Pipets.
Large Type
Small Type
Approximate
Volume
1.0ml (1,000µl)
Approximate
Volume
0.4ml (400µl)
0.75ml (750µl)
0.1ml (100µl)
0.5ml (500µl)
0.05ml (50µl)
0.25ml (250µl)
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0.01ml (10µl)
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Experiment 1. Separation of Of Molecules
by Gel Filtration Chromatography
Background Information
In order to characterize the behavior of a molecule in gel filtration, its. elution
volume (Ve) is measured. The elution volume of a molecule is defined as the
volume of buffer required to elute the molecule from the column. The elution
volume of a molecule is,of course,dependent on its size. As we shall see in
Exercise 2, this value can be used to estimate the molecular weight of an unknown
protein. In order to standardize the elution volume, the void volume (Vo) of the
column is determined. The void volume is the elution volume for a substance
that is completely excluded from the gel. A common material for determining the
void volume is blue dextran,a very large colored polysaccharide with a molecular
weight of about 2,000,000. This substance does not enter the pores in the gel
beads because of its large size and therefore flows only between the beads during
its journey down the column. Thus, Vo is equal to the volume of liquid between
the beads in the column bed. The ratio Ve/Vo is a common expression used in
gel filtration chromatography and, in general, this value is dependent solely on
the size of the molecules under analysis. Figure 1-1 shows the measurements
of the elution volumes of three molecules: blue dextran; the protein myoglobin;
and the small molecular weight dye, phenol red. The molecular weights of these
molecules are 2,000,000,17,000, and 354, respectively. In this example, the
Ve values for these molecules are 5ml, 15ml and 20ml respectively; VeNo for
myoglobin is 3 and for phenol red is 4.
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Figure 1-1. Elution Volumes of Blue Dextran,
Myoglobin and Phenol Red
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Objective To separate a mixture of blue dextran, myoglobin and phenol red by
gel filtration chromatography and to determine their elution volumes. Note that
this exercise requires about two hours to complete. About one hour is required to
pour the column and one hour to perform the separation. If necessary, these two
sections can be performed during different lab periods.
Materials Provided
7 grams of Sephadex G-100 (fine grade). This material must be allowed to
swell in excess column buffer before the columns are poured. For this
purpose, add about 260ml of column buffer to the 7 grams of dry beads,
stir briefly and let stand at room temperature for at least three hours.
Seven grams of the dry gel will form about 90ml of hydrated column
bed volume. The 90ml of hydrated beads is more than enough to prepare
8 columns.
Column Buffer - The buffer is provided as a 50X concentrate and must be
diluted before use as described in the instructor manual. The diluted
buffer contains
0.lM NaCL,10mm Tris-HCl, pH 7.4. The buffer should be made up at leastl
day
before the laboratory and stored at room temperature.
8 columns. Each column (8mm by 200mm) consists of a 20ml reservoir, a
9ml column tube, a lower fitting with cap and a bed support, which is a
porous polyethylene disc.
8- 24 well fraction collection plates.
8 small transfer pipets
8 large transfer pipets
2 macropipets (12ml).
Blue Dextran - Myoglobin mixture - Let this tube warm to room temperature
and then shake the tube to ensure that the material is in solution.
Phenol red - The phenol red is in a 20ml bottle and will be used in all
subsequent exercises.
8 small tubes.
8 tube racks.
Materials Not Provided
8 ring stands and clamps such as burette clamps to hold the columns.<optionalsee below).
Metric rulers
8 small (50ml) beakers or other similar containers.
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Procedure
1. Assembling the chromatographic
columns
Figure 1-2.
CHROMATOGRAPHIC COLUMN
A. As shown in Figure 1-2,
the column consists of a 20 ml
reservoir, a 9 ml column, a lower
column fitting with blue cap and
a porous disc that is used as a gel
bed support. Assemble the column
as indicated in the Figure.
B. Secure the column in a vertical
position with blue cap end down,
by clamping the column to a ring
stand or by taping the column
2. Packing the Column Bed
Proper packing of the column bed is very important for good performance.
In addition, the gel bed that is prepared today will be used for all subsequent
experiments in this manual so it is imperative that it be prepared correctly.
A. Place a beaker or the collection plate under the column to catch the
buffer that will flow through, remove the column reservoir and place
the blue bottom cap on the column.
B. Place 25ml of the mixed gel slurry into a small beaker.
C. Fill the column to about 20% of its length with column buffer.
Thoroughly mix the gel slurry to make sure that it is uniformly
suspended,and add part of the slurry to fill the column.
D. Allow a portion of the gel to settle.
E. As thegel is settling you will observe two layers forming: a bottom
opaque layer of gel and a top layer of buffer. A 2-5 em gel bed will
form in 5-10 minutes. At this time, place the reservoir on the column,
fill it with the mixed gel slurry and then remove the blue cap. The
buffer will now flow out of the column and the gel will slowly settle.
F. Add more gel slurry to the reservoir in small portions until the gel
bed is 3.5 em from the top of the column tube.
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G. Place the blue cap on the column outlet to halt the flow of buffer and
fill the column reservoir with buffer. When not in use, the column
should be stored in this fashion to prevent the gel from drying. If
the gel bed does dry out, it should be removed from the column,
rehydrated and repacked into the column as described above. The
same procedure should be performed if the column flow rate slows
substantially during the course of the experiments described in this
manual.
3. Determining the Column Flow Rate
The rate at which buffer will flow through your column should be roughly
8-l0ml/hour. However, this value is subject to column variation and must be
determined experimentally for your column.
A. Fill the column reservoir with column buffer and place the dry
collection plate under the column to catch the buffer that will flow
through.
B. Remove the blue cap from the column and allow the buffer to collect
in one well of the plate for exactly 10 minutes.
C. Measure the volume of the buffer in the well by drawing the solution
into a small transfer pipet to the 0.4ml level (see Figure 6).
D. In the space provided below, record the flow rate of your column.
FLOW RATE
______________________ml/10 minutes
______________________ml/hour
*_____________________ml/3 minutes
*This value will be used as the volume of each column fraction.
4.
Separating Phenol Red, Myoglobin and Blue Dextran.
A. Using a small transfer pipet (see Figure 6), one student from each
of the 8 groups should place 0.lml of the Myoglobin-Blue dextran
mixture and 0.05ml (about 4 drops) of phenol red into a small tube
and place the tube in a tube rack.
B. Remove the column reservoir and bottom blue cap and let the buffer
drain down to the level of the gel bed.
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C. Transfer the 0.15ml sample to the column. In this manipulation,
place the tip of the pipet about l-2mm above the gel bed and carefully
layer the sample onto the upper bed surface.
D. Let the sample drain into the bed and then add about 0.2ml of
column buffer to wash the sample into the bed.
E. Wash out the pipet with buffer, add buffer to the column until it is
filled, and then attach and carefully fill the reservoir with buffer.
F. Place the collection plate under the column and begin to collect the
first fraction.
G. After 3 minutes, move the collection plate so that the second well is
under the column and collect the second fraction for 3 minutes.
H. Repeat the above process until the 3 colored substrates have been
eluted from the column. Approximately 15-20 fractions should
be collected. During collection, watch the separation of the three
substances as they progress down the column. White paper held
behind the column will facilitate visualization of the colored bands.
I. Examine the column fractions in the collection plate and identify
those fractions that contain the highest concentrations of blue
dextran, myog­lobin and phenol red. These can be seen most clearly
when the collection plate is placed on a sheet of white paper. In the
space provided below, record this information.
FRACTION NUMBER
Blue Dextran
__________________
Myoglobin
__________________
Phenol Red
__________________
Since you have already determined the volume of the column fractions (see
section 3D. on column flow rates) you can now calculate the elution volumes of
the 3 molecules.
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Study Questions and Analysis
1.
Record, in the spaces provided below, the elution volumes (Ve) for blue
dextran, myoglobin and phenol red and the void volume of the column.
Also, calculate the Ve/Vo values for myoglobin and phenol red. These
values will be used in the next exercise.
Blue Dextran Ve
=_______________
Myoglobin Ve
=_______________
Phenol red Ve
=_______________
Void volume Vo
=_______________
Myoglobin Ve/Vo =_______________
Phenol red Ve/Vo =_______________
2.
From the results of your experiment, what can you conclude about the
size of myoglobin in relation to the sizes of phenol red and blue dextran?
3.
Gel filtration chromatography is frequently used to remove salts such as
NaCl from a protein preparation. Why?
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BASIC LAB 3
Experiment 2. Determination of the Molecular
Weight of Human Hemoglobin
Background Information
Table 2-1 lists the molecular weights of the colored proteins that you will study in
this exercise and the molecular weights of phenol red and blue dextran.
Table 2. Molecular Weights of Molecule Used in This Exercise
MOLECULE
CHEMICAL NATURE
COLOR
MOLECULAR* WEIGHT
HORSE MYOGLOBIN
PROTEIN
BROWN
17,000
COW (BOVINE) HEMOGLOBIN
PROTEIN
BROWN
64,000
RABBIT HEMOGLOBIN
PROTEIN
RED
TO BE DETERMINED IN
THIS EXERCISE
Phenol RED
DYE
RED
354
BLUE DEXTRAN
POLYSACCHARIDE
BLUE
2,000,000** (100,000)**
*The terms “molecular weight” and “dalton” are used interchangeably in
this manual. For example, a 20,000-dalton protein has a molecular weight
of 20,000. A dalton is a unit equal to 1.0000 on the atomic mass scale; this
unit is very nearly equal to that of a hydrogen atom. The average amino
acid residue in a protein is 120 daltons. Thus, a protein with a molecular
weight of 20,000 contains 167 amino acid residues (20,000/120=167).
**The column packing material used in this experiment is Gl00 which
means that all macromolecules that have molecular weights that are
greater than 100,000 are excluded from the beads and have the same
elution volumes (The void volume). Consequently, the molecular weight
used for blue dextran should be 100,000.
A brief description of the functions and properties of the proteins listed in the
table is given below.
Myoglobin - Myoglobin and hemoglobin also have an iron containing
heme group and the iron is involved in oxygen binding. Myoglobin binds
and stores oxygen in muscle and hemoglobin is involved in the transport
of oxygen in blood.
Hemoglobin - Red blood cells, or erythrocytes, carry the protein
hemoglobin in the circulation. This protein serves to transport oxygen
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BASIC LAB 3
from the lungs to the tissue. Hemoglobin is a globular protein made up
of 4 subunits, each of which contains a polypeptide chain attached to
heme. Each heme group contains one iron atom, which can bind one
02 molecule. The polypeptide chains of hemoglobin are referred to as
the globin portion of the molecule. Normal adult hemoglobin has •fow
sub-units made up of two alpha (a) and two beta (b) polypeptide chains.
When oxygen is bound to hemoglobin, the protein is bright red in color.
However, when oxygen is lost from the protein, or when the protein is
exposed to certain chemical agents, it becomes a dark blue or brown. The
cow hemoglobin that will be analyzed in today’s experiment is brown
while the hemoglobin from the blood of the rabbit should be oxygenated
and thus red.
The determination of the molecular weight of a protein is often of critical
importance for with this information, one can compare proteins and calculate the
number of their amino acids residues. Gel filtration chromatography has proven
to be a useful tool for protein molecular weight determinations as well as for the
separation of proteins of different sizes. In this exercise, we will determine the
molecular weight of human or rabbit hemoglobin.
The molecular weight of a protein such as hemoglobin can be estimated by
comparing its elution profile with the elution patterns of standard proteins of
known molecular weights. As shown in Figure 3, a linear relationship is obtained
if the logarithms of the molecular weights of proteins are plotted against their
respective elution volumes. To determine the molecular weight of an unknown
protein, its elution volume is determined and compared to the elution volumes
of the standard proteins as shown in Figure 3. In practice, the void volume of the
column is also measured and the ratios of Ve/Vo are plotted against log molecular
weights for the series of reference proteins.
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BASIC LAB 3
Figure 2-1. Determination of the Molecular
Weight of an Unknown Protein
To determine the molecular weight of an unknown, protein standards of
known molecular weight (A= 80,000; B = 55,000; C = 14,000) are separated
on a column and their elution volumes (Ve) are plotted against the
logarithms of their molecular weight The elution volume of the unknown
protein is used to determine its molecular weight by extrapolation from
the standard graph. In !he example shown above, the unknown has a
molecular weight of about 28,000.
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BASIC LAB 3
Objective To determine the molecular weight of hemoglobin from your blood or
from the blood of the rabbit.
Materials Provided
8 gel filtration columns packed with gel, collection plates, column buffer,
small transfer pipets and test tubes
Erythrocyte lysis buffer
Cow hemoglobin Rabbit blood Phenol red
Materials Not Provided (Optional)
Sterile finger lancets
70% alcohol
Sterile cotton
Procedure
I. Separation of cow hemoglobin and phenol red.
A. Using a small transfer pipet, one student from each of eight groups
should place 0.1ml of cow hemoglobin and 0.05 ml (about 4 drops)
of phenol red into a small tube.
B. Remove the column reservoir and bottom blue cap and let the buffer
drain down to the level of the gel bed.
C. Layer the sample onto the column gel bed as described in Exercise
I and collect fractions every 3 minutes until the three colored
substances have been eluted from the column.
D. In the space provided below, record the fractions that contain the
highest concentration of cow hemoglobin, cytochrome C and phenol
red, and calculate their elution volumes.
FRACTIONELUTION
NUMBER VOLUME
Bovine Hemoglobin
___________ __________
Phenol Red
___________ __________
II. Human Hemoglobin -(optional)- If this section is not performed proceed
to Section III (below).
A. Place about 601-11 (5-6 drops) of erythrocyte lysis buffer into a
small test tube.
B. Disinfect your finger with 70% alcohol, allow it to dry and puncture
it with a sterile finger lancet. CAUTION: Do not exchange lancets
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BASIC LAB 3
with other students. Allow 1-3 drops of blood to fall into the test
tube and mix the contents of the tube thoroughly. The erythrocyte
lysis buffer contains the detergent Nonidet P-40 which serves
to lyse (break open) the eryth­rocytes which in turn liberate their
hemoglobin.
III.
C. Using a small transfer pipet, add 0.05ml (about 4 drops) of phenol
red and layer the entire sample onto your chromatographic column.
Collect fractions every 3 minutes and identify the fractions that
contain the hemoglobin and phenol red.
FRACTIONELUTION
NUMBER VOLUME
Human Hemoglobin
___________ __________
Phenol Red
___________ __________
Rabbit Hemoglobin
If human hemoglobin was not analyzed in the preceding section, perform the
experiment described above using the rabbit blood that is provided with the
Chemical Package. Erythrocyte lysis buffer has been added to the rabbit blood
so you should add phenol red and then subject the mixture to chromatography as
described above.
Study Questions aud Analysis
1.
On semilog paper, plot the elution volumes of cytochrome C, myoglobin,
and bovine hemoglobin as a function of their molecular weights (see
Table 2-1) and determine the size of human or rabbit hemoglobin. The
elution volume of myoglobin should be found on page 17.
2.
Interferon is a protein with a molecular weight of 25,000. What elution
volume would it have if it had been included in this experiment?
3.
Give the approximate number of amino acids found in human (or rabbit)
hemoglobin (see footnote -Table 2-1).
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BASIC LAB 3
Experiment 3. Dye Binding of Serum Albumin
Background Information
A. Interaction Between Molecules in Biological Systems
One of the most important principles of modem biology is that two molecules
with complementary surfaces tend to bind or stick together, whereas molecules
without such surfaces do not. This principle can be illustrated by considering
the specificity of enzyme action. Enzymes accelerate the velocity of virtually all
reactions that occur in biological systems, including those involved in breakdown,
synthesis and chemical transfers. In so doing, they are responsible for performing
essentially all the changes associated with life processes.
The general expression frequently used to describe an enzyme reaction is:
Certain important features of the nature of enzyme reactions are evident from this
diagram:
l.
The term substrate refers to the compound that is acted upon by the
enzyme. In general, enzymes exhibit a high degree of substrate specificity
in that they usually catalyze only a single chemical reaction.
2.
The enzyme binds to the substrate to form an enzyme-substrate complex.
This interaction is responsible for the specificity of enzyme action, since
only those compounds that “fit” into the substrate binding site can be
acted upon by the enzyme.
3.
The enzyme is not destroyed during the reaction but rather is set free
after the formation of the end product. Thus, the liberated enzyme is
available to combine with more substrate to produce more product.
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BASIC LAB 3
The binding of an enzyme to its substrate is only one example of the many
specific molecular interactions that occur in biological systems. An analogous
binding process occurs when an antibody binds antigen and this interaction will
be described in more detail in Exercise 5. In fact, the selective interaction of
biomolecules with each other often permits the self-assembly of these molecules
into geometrically regular aggregates such as structural fibers, tubes, spheres
and even functional cellular organelles. In today’s laboratory, you will study the
specificity of the binding of a protein from blood for small dye molecules and
determine whether the native structure of the protein is required for its binding
activity.
B. Serum Albumin
Blood is a remarkable tissue containing cellular elements (erythrocytes, leu­
kocytes and platelets) suspended in a liquid medium called plasma. Whole blood,
or plasma, clots upon standing and if the clot is removed, the remaining straw­
colored fluid is called serum. Serum has basically the same composition as plasma
except that it lacks certain proteins that are involved in the clotting process.
Serum contains a variety of small molecular weight components as well as
hundreds of different serum proteins. The major protein in serum is albumin,
which functions as a carrier molecule for the transport of certain small molecular
weight compounds in blood. Molecules that bind to serum albumin include
bilirubin, fatly acids, hormones, and some synthetic dyes.
Objectives To determine the molecular weight of serum albumin and to study the
binding specificity of this protein for small synthetic dye molecules using a gel
filtration assay. In addition, you will examine the effects of the detergent sodium
dodecyl sulfate (SDS) on the binding activity of serum albumin. This detergent is
a strong protein denaturant (see page 3).
Materials Provided
Bovine (cow) serum albumin (5%) Phenol red
Bromophenol blue
24 small test tubes and 8 tube racks
8 gel filtration columns, collection plates, column buffer and small transfer
pipets.
SDS (Sodium Dodecyl Sulfate)
Procedure
The small transfer pipets should be used in this exercise.
1.
One member from each group should place 0.1 ml of bovine serum
albumin into two tubes and place the tubes in a rack.
2.
Add 3 drops of phenol red and 2 drops of bromophenol blue to each tube.
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BASIC LAB 3
3.
Add I drop of SDS to one tube and mark the tube +SDS.
4.
Apply the mixture from the tube lacking SDS to your gel filtration
column and collect fractions every 3 minutes as described in Exercise I.
Record the position of the two dyes in the elution profile.
5.
Apply the protein-dye mixture from the tube containing SDS to the
column and record the position of the two dyes in the elution profile as
described above.
Study Questions and Analysis
1.
Which of the dye molecules binds to serum albumin? Which does not?
2.
Estimate the molecular weight of bovine serum albumin by comparing
its elution volume to the elution volumes of hemoglobin, myoglobin and
cyto­chrome (Exercises I and 2).
3.
How would you determine the molecular weight of serum albumin in
your blood?
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BASIC LAB 3
Experiment 4. A Comparison of the Properties of α-Amylase
from Human Saliva and Pancreas
Background Information
Gel filtration chromatography has the advantage of being able to yield the
molecular weight of a specific protein in a complex mixture, provided the
protein bas a biological activity that can be measured. For example, a crude cell
exttact contains hundreds of different enzymes yet it is possible to determine the
molecular weight of a single enzyme protein in this extract by passing the extract
through a column and determining the elution volume of the enzyme activity in
the column eluates. In today’s laboratory, this procedure will be used to estimate
the molecular weight of an enzyme that is involved in the breakdown of complex
carbohydrates.
A. Basic Carbohydrate Chemistry
Carbohydrates, together with lipids and proteins, play a fundamental role in the
life of aU organisms. The carbohydrates include such substances as sugars, starch,
and cellulose. These compounds are composed of carbon, hydrogen, and oxygen.
The simplest type of carbohydrates are the monosaccharides, simple sugars that
contain 3-7 carbon atoms. There are about 20 different monosaccharides that are
found in living organisms. One of the most common monosaccharides is glucose,
whose structure is shown in Figure 4-1.
Figure 4-1. The Structure of Glucose.
In nature, glucose occurs in a ring form like that shown in the figure.
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BASIC LAB 3
Polysaccharides are macromolecules (giant molecules) that contain many
monosaccharide units linked together by glucosidic bonds. The major
polysaccharides digested in the human gastrointestinal tract are glucose polymers
that comprise the plant starches and animal glycogens. The starches consist of two
components called amylose and amylopectin (Figure 4-2). Amylose, which turns
blue when stained with iodine, is made up of glucose molecules arranged in long
straight chains. Amylopectin, which turns purple to red when stained with iodine,
is a branched polymer of glucose. Animal glyocogen, which is a storage form of
glucose in liver and muscle, is similar in structure to amylopectin in that it is a
complex branched carbohydrate.
B. Amylase
Digestion of carbohydrate is initiated in the mouth by the action of a hydrolytic
enzyme in saliva called a-amylase. This enyzme is a major protein found in saliva.
The optimal pH for this enzyme is about 7, so its action is inhibited when food enters
the stomach by the acidic stomach juice. In the small intestine, polysaccharide
digestion is reinitiated by pancreatic a-amylase, an enzyme that is similar but not
identical to salivary a-amylase. Both enzymes cleave the glucosidic bonds that
link glucose molecules into starches and glycogen (Figure 4-2). The a­amylase first
hydrolyzes the polysaccharides to products called dextrins, which contain several
to several hundred simple sugar molecules. The dextrins, in turn, are broken down
under the influence of α-amylase to di, tri- and oligosaccharides. These sugars are
subsequently broken down to glucose by intestinal oligosaccharidases and the
glucose molecules cross the intestinal wall and enter the blood. Figure 4-3 shows
the breakdown of starch under the influence of a-amylase.
Figure 4-2. Structure of the Components in Starch Amylose.
Amylose and amylopectin are glucose polymers. The portion of amylose
shown above contains 5 glucose residues.
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BASIC LAB 3
Figure 4-3. Hydrolysis of Starch by α-Amylase.
C. Detection and Measurement of Enzyme Activity
In this exercise, you will study the hydrolysis of starch by the enzyme, α­-amylase.
The enzyme reaction can be represented in a simplified form as follows:
α-AMYLASE
H 2O
STARCHOLIGOSACCHARIDES
(SUBSTRATE)
The reaction can be detected either by following the decrease in the amount of
starch present in an enzyme reaction mixture or by measuring the appearance of
oligosaccharides. Iodine staining is frequently used to follow the course of starch
disappearance because starch turns blue-black when stained with iodine, while
lhe oligosaccharides do not
Two parameters that influence the amount of starch breakdown are time and
amylase concentration. As shown in Figure 4-4, if starch is in excess, !hen there is
an increase in the amount of starch breakdown with increasing reaction time and
with increasing amylase concentration. If time is held constant, then the amount of
starch breakdown is directly related to the amylase concentration. This feature of
the reaction enables one to determine the amount of amylase activity in a sample.
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BASIC LAB 3
Figure 4-4. Effects of Time and Amylase Concentration on Starch
Breakdown
In this experiment α-amylase activity will be quantified by the method outlined
below:
1.
An agar gel containing starCh is cast in a petri dish.
2.
Amylase or cellular extracts containing amylase are placed in holes
made in the gel.
3.
The gel is incubated at room temperature for one day. Amylase diffusion
into the gel is accompanied by digestion of the starch.
4.
The starCh in the gel is detected by iodine staining, and enzyme activity
is noted by transparent rings around the sample wells in a blue-black gel.
Figure 4-5 illustrates the relationship between ring diameter and a-amylase
concentration applied to the gel. To determine the amount of amylase in an
unknown sample, the sample is applied to the gel and the diameter of the resulting
ring around the sample well is compared to the diameters of the rings around
the wells containing known concentrations of α-amylase. For example, a ring
diameter of 13mm means that the sample contains 10μg of amylase activity (see
dotted lines on the graph in Figure 4-5).
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BASIC LAB 3
Figure 4-5. Determination of Amylase Activity.
Amylase solutions (50μl) containing (A) 0, (B) 0.8, (C) 4, (D) 20 (E) 100 and (F)
(500)μg amylase per ml were incubated for 24 hours in an agar gel which
was then stained with iodine solution.
The diameters of the rings from the above experiment were measured
and plotted as a function of amylase concentration. Note that the data are
presented as a semi­logarithmic plot
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BASIC LAB 3
Objective To determine the amounts and molecular weights of amylase in saliva
and in pancreatic extracts.
Materials Provided
8 columns containing Sephadex G-100
8 collection plates
16 small test tubes and 16 large test tubes
8 tube racks
*Column buffer
Macropipet
Agar
Starch solution
16 Petri dishes
8 Glass pasteur pipetes
14 small transfer pipets: one pipet per student group, one for the starch
solution
and one for each of the 5 amylase solutions
•Amylase (500μg/ml)
•Amylase (100μg/ml)
•Amylase (20μg/ml)
•Amylase (4μg/ml)
•Amylase (0.8μg/ml)
Pancreatic extract
*I2Kl solution
Phenol red
*Prepared as described in Appendix 2 of the instructor manual.
Materials Not Provided
Balance- Desirable but not absolutely necessary (see below)
Rulers, marking pens
A container such as a 600ml beaker to boil water
Bunsen burner with wire gauze on a tripod
Procedure:
I. Preparation of the Sample from Saliva
Add one drop of your saliva to 0.4ml of column buffer in a small tube and mix
thoroughly.
II. Chromatography
In this exercise, 4 groups of students will study a-amylase in saliva and the other
4 groups will study a-amylase in the pancreatic extract provided in the Chemical
Package. At the end of the experiment, students should compare their results.
MODERNBIO.COM
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BASIC LAB 3
I.
One member from each group should place 0.1ml of the saliva sample
(4 groups) or 0.1 ml of the pancreatic extract (4 groups) into a small test
tube.
2.
Add 50μl (about 4 drops) of phenol red to the tube and apply the sample
to the column.
3.
Collect fractions every 3 minutes until the phenol red has been eluted
from the column and record the fraction number that contains the phenol
red.
III. Determination of Amylase Activity in Column Fractions
A. Preparation of the Agar Gels
1.
Each group of students should obtain two large glass test tubes. To each
tube, add 20ml of column buffer and 0.32 grams of agar. The agar can
be weighed out directly on an appropriate balance. If a balance is not
available, 0.32 grams of agar can be estimated by filling one of the small
0.5ml tubes provided with the package with agar until full.
2.
Add 0.4ml of the mixed starch solution to each tube and swirl the tubes
until the agar forms a suspension.
3.
Place the test tubes into a boiling water bath and allow the agar suspension
to come to a vigorous boil. The level of the water in the bath should be
about equal to the level of the solution in the tubes.
4.
Pour the contents of each tube into the bottom section of a petri dish.
5.
Let lhe agar cool for at least 15 minutes. The agar gels can be used
immediately or they can be stored covered in the refrigerator in a sealed
plastic bag for up to one week.
6.
Place lhe pattern shown in Figure 4-6 beneath the dishes and use the
Pasteur pipets provided in the Chemical Package to cut the wells in the
agar according to the pattern. Push the wide end of the Pasteur pipet
into the agar to cut the wells. Withdraw the pipet and remove the agar
segments from the wells by piercing them with the narrow end of the
pipet Repeat the process using the pattern until all sample wells on both
plates are formed.
7.
With a marking pen, place the letter A on one plate and the letter B on the
other. Turn the plates over and number the wells #1-#9 with a marking
pen as shown in Figure 4-6.
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BASIC LAB 3
Figure 4-6. Pattern for Arrangement of Wells in the Petri Dish.
B. Sample Application
Using the small transfer pipets, carefully fill the wells as indicated below. Draw
a small amount of sample (about 01ml = 100μl) into the pipet and slowly eject
it into the well until the well is almost full. At this time,the well should contain
about 0.05ml (50μl). Five pipets should be used by the entire class to dispense the
amylase solutions. Each group should then use a single pipet to add the saliva,
pancreatic extracts and portions of the column fractions to the sample wells. Rinse
this pipet between samples by drawing up and expelling water three times.
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BASIC LAB 3
PLATE A
WELL NUMBER
PLATE B
SAMPLE
WELL NUMBER
1
2
3
4
5
6
7
8
9
SAMPLE
(Column Fraction Number)
Column buffer
1
2
Amylase0.8µg/ml2
4
Amylase4µcg/ml3
6
Amylase201µg/ml4
8
AmylaseIOOµg/ml5
10
Amylase500µg/ml6
12
Saliva- sample
7
14
Saliva- sample
8
16
Pancreatic Extract
918
After all samples have been loaded into the wells, place the lid on the dish. The dish
should not be moved at this time. The plates should remain at room temperature
for 12-24 hours before iodine staining. If longer times are desired, the plate should
be placed in the freezer after one day. C. Detection of Amylase Activity
1.
Place about 5ml of the diluted iodine solution onto the agar in the dishes.
2.
After 10-20 minutes, discard the iodine and fill the petri dishes with
water.
3.
After 10-20 minutes, discard the water, measure the diameter (in mm) of
the clear rings around each well and record your results on the following
page. The rings can be seen most clearly by holding the dish over a light
source such as a desk lamp.
4.
The plastic plates and pipets should be washed with a mild detergent and
rinse with water.
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BASIC LAB 3
PLATE A
PLATE B
WELL NUMBER RING DIAMETER (mm) WELL NUMBER RING DIAMETER (mm)
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8 8
9 9
Study Questions and Analysis
1.
On semilog paper (provided below) plot the diameters of the rings from
wells #2-#6 (PLATE A) as a function of the concentration of a-amylase.
2. Determine the amount of amylase activity in the column fractions and
estimate the molecular weight of amylase by comparing its elution
volume with the elution volumes of serum albumin, hemoglobin,
myoglobin and cytochrome C (Exercises 1-3).
3. Explain the basis for the formation of the rings in the agar-starch gels.
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BASIC LAB 3
Experiment 5. Isolation and Immunological
Analysis of a Protein from Egg White
Background Information
A. Isolation of the Protein Ovalbumin from Egg White
A knowledge of the solubility of proteins is frequently useful in the isolation of
these macromolecules. For example, the major protein found in chicken egg white
is ovalbumin (also called egg albumin) and this protein is soluble in distilled
water. Egg globulins, in contrast, are proteins that are not soluble in water but are
soluble in dilute salt solutions. Thus, if egg white is added to water, the globulins
precipitate and can readily be removed from the soluble albumin preparation
by centrifugation or filtration. This differential solubility of ovalbumin forms
the basis of the first step in the isolation of ovalbumin that will be performed in
today’s laboratory.
In most cases, the purification of a single protein involves a series of steps with
each step exploiting differences in protein solubility, size or charge. Gel filtration
chromatography is frequently the method of choice for the size fractiona­tion step
in a protein purification scheme and this procedure will be used as the second step
in today’s ovalbumin purification procedure.
In order to determine the elution volume of ovalbumin, and thus its molecular
weight, a method must be used to detect this specific protein in fractions from
the gel filtration column. However, ovalbumin is a colorless protein that does
not possess enzymatic activity. For these reasons,an immunological assay will be
used to identify the position of ovalbumin in the elution profile of the gel filtration
column in today’s experiment.
B. Antibodies as Tools in Modern Biology
The immune system consists of a variety of cells, tissues and organs and about
1012 antibody molecules. The major function of the immune system is to protect
the organisms from viruses,bacteria, protozoans and larger parasites. When
these organisms enter the body, macromolecules on their surfaces induce the
production of specific antibodies that appear in the serum of the infected animal.
The antibodies, in turn, combine with these foreign macromolecules, thereby
rendering the invading organisms inactive and noninfective.
The macromolecules that elicit antibody production are called antigens and are
most often proteinaceous in nature. Although antigens are frequently compo­nents
of foreign organisms, purified foreign proteins will serve as antigens in that they
will stimulate the formation of antibodies when injected into a suitable test animal
such as the rabbit. The antibodies are recovered in the serum of the immunized
animal and this serum is referred to as an antiserum.
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BASIC LAB 3
Antibodies against specific proteins are extremely important tools used in the
modem biology laboratory for they permit one to identify and quantify specific
protein antigens. In the experiment described in this section, you will perform
the double-diffusion technique to identify the egg white protein ovalbumin. The
technique is based on the observation that specific antibody-antigen complexes
undergo precipitation reactions that are readily visualized in an agar gel. The
method was developed by Organ Ouchterlony more than 40 years ago and is still
in wide use today. In this method, an agar gel is prepared in a petri dish and
a small amount of antigen and antibody solution are placed separately in small
wells cut into the agar. The antigen and antibody diffuse outward toward each
other at rates that are related to their concentration, size and shape. A milky line
of precipitate forms where one antigen encounters its antibody. This technique is
illustrated in Figure 4-1.
Figure 4-1. Double-Diffusion Technique for
Analysis of Antigen-Antibody Complexes.
1.
An agar gel is cast in a dish and six small wells are produced by
cutting circular holes in the agar.
2.
Antiserum prepared against the egg white protein ovalbumin is
placed in the center well and ovalbumin (A), serum albumin (B) a-amylase
(C), total egg white (D) and serum (E) are placed in wells on the periphery
of the plate. Following time to permit the diffusion of antibodies and
antigens, the plate is viewed for the presence of milky white precipitation
lines. Note that a single line is formed when purified ovalbumin or total
egg white is placed in the wells.
Objective: To isolate ovalbumin from the chicken eggs.
Materials Provided
8 columns containing BIO-GEL P100
8 collection plates
Small test tubes, tube racks, and large test tubes
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BASIC LAB 3
Column buffer
Agar
16 Petri dishes
8 glass Pasteur pipets
Macropipets
Cheese cloth
14 small transfer pipets
x Ovalbumin 20 mg/ml
x Ovalbumin 4 mg/ml
Ovalbumin 0.8 mg/ml
x Ovalbumin 0.16 mg/ml
xx Anti-ovalbumin antibodies
Phenol red
x Prepared as described in Appendix 2 of the instructor manual.
xx This antibody preparation was produced by rabbits that were injected
with purified ovalbumin.
Materials Not Provided
One fresh egg
One 500ml beaker
Procedure
I. Extraction of Ovalbumin from the Chicken Egg
This section should be performed by the instructor or by the entire class as a
group.
1.
Carefully break the egg and separate the yolk from the white by passing
the yolk between the two egg shell halves while allowing the white to
flow into a 500ml beaker or another suitable container.
2.
Slowly add l00ml of distilled or deionized water to the egg white and stir.
A white precipitate composed of the egg globulins will form.
3.
To remove the precipitate either:
A. Centrifuge a portion (about 20%) of the extract and pour off and save
the supernatant ovalbumin fraction. Label this fraction “ovalbumin
extract”.
B. Filter a portion (about 20%) of the extract through 8 layers of
cheesecloth provided with the Chemical Package and label the
filtrate “ovalbumin ex­tract”.
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BASIC LAB 3
II. Chromatography
1.
One member from each group should place 0.15ml of the ovalbumin
solution into a small test tube containing 50μ1 (about 4 drops) of phenol
red.
2.
Apply the sample to the column and collect fractions every 3 minutes
until the phenol red has been eluted. Record the fraction number that
contains the highest concentration of phenol red.
Ill. Identification or Ovalbumin in tbe Column Fractions
A.
Preparation of the Agar Gels
1. Prepare agar gels in two petri dishes as described in Experiment 4
except omit the starch from the gel.
2. Label one plate A and one plate B.
3. Place the pattern shown in Figure 5-2 beneath the dishes and use the
Pasteur pipets to cut the wells in the agar according to the pattern.
4. Tum the plates over and number the wells #l-#9 with a marking pen
as shown in Figure 5-2.
Figure 5-2. Pattern for the Arrangement of Wells in the Petri Dish.
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BASIC LAB 3
B. Sample Application and Ovalbumin Detection
1.
Using the small transfer pipets, carefully fill the wells as indicated below
and as detailed in Experiment 4.
PLATE A
WELL NUMBER
1
2
3
4
5
6
7
8
9
SAMPLE
Anti-ovalbumin antibodies
Column buffer
Ovalbumin 20 mg/ml
Ovalbumin 4 mg/ml
Ovalumin 0.8 mg/ml
Ovalbumin 0.16 mg/ml
“Ovalbumin Extract”
“Ovalbumin Extract”
“Ovalbumin Extract”
PLATE B
WELL NUMBER
1
2
3
4
5
6
7
8
9
SAMPLE
Anti-ovalbumin Antibodies
Column Fraction 4
Column Fraction 6
Column Fraction 8
Column Fraction 10
Column Fraction 12
Column Fraction 14
Column Fraction 16
Column Fraction 18
2.
Place the lids on the dishes.
3.
Observe the dishes daily for about 3 days for the development of
precipitation lines.
4.
After 3-6 days, examine the gel carefully for precipitation lines
and describe your results. The intensity of the precipitation lines is
proportional to the concentration of ovalbumin in the sample.
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BASIC LAB 3
Study Questions and Analysis
1.
Estimate the elution volume, the molecular weight, and the number of
amino acid residues for ovalbumin.
2.
Estimate the amount of ovalbumin present in one chicken egg.
3.
Lysozyme is an egg white protein. This protein is also an enzyme that
hydrolyzes the carbohydrates in the capsules of bacteria, which results
in the lysis (disintegration) of the bacteria. Describe two methods or
procedures that you would use to determine the amount of lysozyme that
is present in one egg.
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BASIC LAB 3
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