STANDARD STUDENT PROGRAM 3
STUDENT MANUAL
Molecular Biology of
Nucleic Acids
Preface
This laboratory manual forms an integral part of a unique teaching system. In add
ion to the manual, the complete system consists of electrophoresis equipment and
the chemicals and materials required to perform six key experiments in modern
biology. A description of the electrophoresis apparatus is found in Part Ill of this
manual.
This teaching system was designed principally for students taking a first laboratory
course in modern biology. tt is well suited for secondary level or undergraduate
students enrolled in general biology, genetics, biochemistry or molecular biology.
I have assumed that readers have had an introductory lecture course in the
biological sciences. However, I have attempted to write the manual in such a way
that even a stranger to biology could follow it and perform the exercises by doing
them in sequence.
The laboratory manual is divided into two major sections. The first section provides
basic information on the biology and chemistry of nucleic acids and is also
intended to acquaint students with the principles and techniques of electrophoresis.
Student 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 six exercises in modern biology. The
exercises deal with the structure and function of DNA. Each experiment provides
state-of-the-art information and most can be completed in a 3-hour laboratory
session. Experiments 4 and 6 require three laboratory sessions each to complete.
A background information section is given for each exercise and students should
read this material prior to performing the exercise in the laboratory. I have also
included a number of relevant references from Scientific American other sources
for future study. Each exercise includes study questions, the answers to which
students derive from 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.
MODERNBIO.COM
i
STANDARD STUDENT PROGRAM 3
Table of Contents
Part A. Background Information
I. Nucleic Acid Structure and Function: A Review of the
Basics Nucleotides-Building Blocks of Nucleic Acids.................................. 1
The Polynucleotide Chain............................................................................... 3
DNA Structure............................................................................................... .4
Biological Role of DNA and RNA................................................................. 4
Analyzing Specific Genes and Recombinant DNA Technology.................... 7
II. General Description of Agarose Gel Electrophoresis
III. Practical Aspects of Electrophoresis
Electrophoresis Equipment........................................................................... 12
Electrophoresis Chemicals............................................................................ 14
IV. Electrophoresis Procedures
Pouring the Agarose Gels............................................................................. 15
Sample Application....................................................................................... 16
Electrophoresis............................................................................................. 17
Staining......................................................................................................... 18
Determining the Size of DNA by Electrophoresis........................................ 19
V. Hybridization Analysis
DNA Denaturation and Renaturation............................................................ 21
Hybridization Procedures............................................................................. 24
Application of Southern Blotting in Medicine............................................. 30
VI. Suggested Reading and References for Part A............................................. 33
Part B. Laboratory Exercises
301. Determination of the Length of DNA Molecules....................................... 20
302. Restriction Nuclease Mapping DNA.......................................................... 24
303. Plasmid DNA Structure............................................................................... 30
304. Molecular Cloning
A. Introducing Plasmid DNA into E.coli...................................................... 34
B. Isolating and Analyzing a Plasmid from E.coli........................................ 39
305. Identifying Satellite Sequences .................................................................. 46
306. The Nucleosome Structure of Chromatin
A. General Studies........................................................................................ 50
B. Structure of Plant Chromatin .................................................................. 55
C. Chromatin DNA by Cell Nucleases (Optional)........................................ 57
Suggested Reading and References for Part B.................................................... 60
Appendix - The Metric System........................................................................... 61
MODERNBIO.COM
iii
STANDARD STUDENT PROGRAM 3
Part A. Background Information
I. Nucleic Acids: A Review of the Basics
The concept that chromosomal units known as genes transmit heritable
information from parent to offspring was founded in the late 19th century.
However, a description of genes in terms of their unique structural and functional
properties is relatively new. We now know that genes are composed of a type of
nucleic acid called deoxyribonucleic acid (DNA). The DNA molecule not only
directs its own reproduction but also stores all the information that determines the
types of proteins produced during the lifetime of an organism. In so doing, DNA
orchestrates the complex reactions and structures characteristic of an organism
and its offspring. Ribonucleic acid (RNA), the second major category of nucleic
acids, is involved principally in the transmission of genetic information and in
protein production. The structure and function of DNA and RNA can most easily
be understood by examining the chemical composition of the nucleic acids.
Nucleotides - Building Blocks of Nucleic Acids
Under the proper conditions, nucleic acids can be broken down to low- molecular­
weight products of three types: a pentose (or 5 carbon) sugar; purines and
pyrimidines; and phosphoric acid (Figure 1). The phosphate group is responsible
for the strong negative charge of nucleic acids. The pentose sugar from RNA
is always ribose and that from DNA is 2-deoxyribose. These sugars differ only
by the presence or absence of a hydroxyl group on carbon 2 (so-called 2’). The
numbers assigned to the five carbon atoms are shown in Figure 1. The purines
and pyrimidines are often called nitrogenous bases (or, simply, bases). The major
purine bases in DNA and RNA are adenine (A) and guanine (G), and the major
pyrimidines in DNA are cytosine (C) and thymine (T). RNA contains the base
uracil (U) in place of thymine. The sugars and phosphates are readily soluble
in water. That is, they are hydrophilic. In contrast, the bases are hydrophobic in
that they display limited solubility in water. As will be discussed below, these
differences in water solubility are extremely important for the structure of the
DNA molecule.
A nucleotide consists of a pentose sugar, a nitrogenous base and a phosphate
group structured as shown below. The high-energy storage compound, adenosine
triphosphate (ATP), is a well known nucleotide found in biological systems.
MODERNBIO.COM
1
STANDARD STUDENT PROGRAM 3
Figure 1. The Nucleotide Components
MODERNBIO.COM
2
STANDARD STUDENT PROGRAM 3
The Polynucleotide Chain
Nucleic acids are polynucleotides and have the general structure shown below:
A polynucleotide is composed of repeating nucleotide units linked into chains by
phosphodiester bonds that join the 5’ carbon of one ribose or deoxyribose group
to the 3’ carbon of the next sugar (Figure 2). The sequence or order of nucleotides
in a polynucleotide chain is often abbreviated by a 1-letter code (e.g., G-C-A-T-A)
with the 5’ end of the chain written at the left. A typical RNA molecule is a singlestranded polynucleotide chain. As will be described below, DNA usually contains
two polynucleotide strands coiled around one another to form a double-stranded
helix. The number of nucleotide units in a nucleic acid chain varies tremendously
depending on the nucleic acid type. For example, each chromosome from a
higher organism is thought to contain a single, very long DNA molecule. A DNA
molecule from the largest human chromosome is composed of approximately 5.4
x 108 nucleotides, which corresponds to a molecular weight of the order of 10 11
and a length of about 4cm. On the other hand, transfer RNA molecules generally
contain only 70-80 nucleotides.
Figure 2. Structure of the Polynucleotide Chain
MODERNBIO.COM
3
STANDARD STUDENT PROGRAM 3
DNA Structure
In early physical studies of DNA, a variety of experiments indicated that DNA
molecules occur in long helixes with each helix being formed from two or more
polynucleotide chains bound side by side. Chemical analyses also demonstrated
that the phosphate groups were on the outside of the helix and that the number
of A and T residues in DNA were always equal, as were those of G and C. With
these facts in mind, Watson and Crick in 1953 proposed that the DNA molecule
actually consists of two polynucleotide chains coiled around the same axis to form
a double helix (Figure 3). In this model, the hydrophilic sugar-phosphate groups
follow the outer edges of the molecule where they can interact with water. The
hydrophobic bases face inward toward each other in the molecule’s center and
thus avoid contact with water. The two polynucleotide strands run in opposite
directions (they are anti-­parallel) and are held together primarily by hydrogen and
hydrophobic bonding between the bases, where A is always paired with T, and G
with C. These complementary bases have an affinity for each other such that, when
they are paired, they contribute to the overall stability of the DNA helix. Because
of this complementary base-pairing, the sequence of bases in one polynucleotide
chain determines the sequence in the other. For example, if the bases along
one strand are arranged in the order T-G-C-T-A-G, the opposite bases on the
complementary strand will be A­ C-G-A-T-C. This fact is of extreme biological
significance because it explains how a DNA helix in the chromosome directs the
formation of copies of itself and directs the formation of RNA molecules with its
specific informational content. The B-form DNA shown in Figure 3 is the most
common of the DNA types. It is a right-handed double helix and contains about
10.5 nucleotide pairs per helical turn.
Figure 3. Double-Helical Structure of Common B-DNA
MODERNBIO.COM
4
STANDARD STUDENT PROGRAM 3
Biological Role of DNA and RNA
DNA is an information molecule with two general functions (Figure 4). First,
DNA plays a central role in the propagation of the species and the determination
of the heritable characteristics of the cell and its descendants. Prior to the time of
each cell division, the two strands of the DNA helix separate from one another
and each serves as a pattern or template for the synthesis of a new, complementary
chain. This process of DNA biosynthesis is known as replication. One of the
double helices formed is then transmitted to one daughter cell, and one to the
other. Although the principle underlying DNA replication is straightforward, the
actual mechanism responsible for the replication process in the cell involves an
array of enzymes and regulatory proteins.
The informational content of DNA also determines the types of proteins that are
produced by a cell. In this manner, the DNA molecule functions as a blueprint for
all cellular processes that go on during the lifetime of an organism. In the first step
along the information pathway from DNA to protein, a segment of DNA is copied
into a complementary strand of messenger RNA (mRNA) by a process known
as transcription. Transcription begins when an enzyme called RNA polymerase
binds to a specific sequence on the DNA known as the promoter. At this site, the
enzyme unwinds a small segment of double helix, exposing the bases of the two
single strands of the DNA molecule. One of these strands is then transcribed. As
the polymerase travels along the DNA, ribonucleotides with bases complementary
to the DNA are added to the growing chain. For example, the DNA segment C-GT-A-T­G is transcribed into G-C-A-U-A-C in the mRNA. Each sequence of three
nucleotides in the mRNA is called a “codon,” and codes for one amino acid. Since
most polypeptide chains contain between 100 to 1,000 amino acids, an mRNA
must be at least 300 to 3000 nucleotides long. Therefore, a gene that codes for a
polypeptide chain must contain at least 300 to 3,000 base pairs.
The translation of mRNA into protein is a complex process that occurs on
particles called ribosomes. This process requires ribosomal RNAs (rRNAs) and
transfer RNAs (tRNAs). These RNA species do not specify proteins themselves
but rather take part in decoding the information carried by the mRNAs. At one
end of each tRNA molecule is a nucleotide triplet called the “anticodon,” which
is complementary to an mRNA codon. A specific amino acid is bound to the
opposite end of each tRNA molecule. On the ribosome, tRNAs carrying amino
acids associate with the mRNA by way of complementary base-pairing at the
anticodon-codon sequences. As the ribosome moves along the mRNA, the amino
acids carried by the tRNAs are linked to the growing polypeptide chain. In this
manner, the order of codons along the mRNA directs the amino acid sequence of a
polypeptide chain. The translation process occurs on the surface of the ribosomes.
These particles, which are composed of rRNA and ribosomal proteins, serve
to bring together the mRNAs, the tRNA and other factors that are required for
protein production.
MODERNBIO.COM
5
STANDARD STUDENT PROGRAM 3
Figure 4. Molecular Information Transfer: Complimentary Base Pairing
Replication
The two DNA strands separate and each serves as a template for the synthesis of
a new, complementary polynucleotide.
Transcription
One strand of the DNA serves as a template for the synthesis of complementary
RNA.
Translation
The mRNA associates with ribosomes and its nucleotides are matched three at a
time to a complementary set of three nucleotides in a specific tRNA molecule. The
tRNA molecule carries an amino acid and when it associates with the mRNA, the
amino acid is added to the growing polypeptide chain.
MODERNBIO.COM
6
STANDARD STUDENT PROGRAM 3
Analysis of Specific Genes and Recombinant DNA Technology
A key to one of life’s great mysteries was discovered in 1953 when the double­
helical structure of DNA was perceived by Watson and Crick. Elucidation of the
basic mechanisms of replication, transcription and translation quickly followed, and
by the early 1960’s, the model shown in Figure 4 was generally accepted by most
biologists. However, genes from higher organisms resisted detailed analysis until
the mid
1970’s because of the complexity of the DNA in eukaryotic organisms; a vertebrate
cell contains enough DNA to code for more than 100,000 proteins. In order to
study the structure and function of a single protein coding gene, the gene must be
prepared in a purified form. The isolation of a specific gene from cellular DNA by
conventional biochemical procedures is not practical because of the magnitude of
the purification required (usually 100,000-fold) and because the procedures would
necessitate the use of a large quantity of starting cellular DNA. Herein lies the major
use of recombinant DNA technology, for it permits the amplification and isolation
of specific genes by relatively simple procedures. A basic understanding of these
procedures requires a description of an interesting feature of bacterial physiology.
Plasmids are small circular DNA molecules that exist apart from the chromosomes
in most bacterial species. Under normal circumstances, plasmids are not essential
for survival of the host bacteria. However, many plasmids contain genes that enable
the bacteria to survive and prosper in certain environments. For example, some
plasmids carry one or more genes that confer resistance to antibiotics. A bacterial
cell containing such a plasmid can live and multiply in the presence of the drug.
Indeed, antibiotic-resistant E.coli isolated in many parts of the world contain
plasmids that carry the genetic information for protein products that interfere with
the action of many different antibiotics.
In the laboratory, plasmids can be introduced into living bacterial cells by a process
known as transformation. When bacteria are placed in a solution of calcium chloride,
they acquire the ability to take in plasmid DNA molecules. As illustrated in Figure
5, this procedure provides a means for preparing large amounts of specific plasmid
DNA since one transformed cell gives rise to a clone of cells that contains exact
replicas of the parent plasmid DNA molecule. Following growth of the bacteria in
the presence of the antibiotic, the plasmid DNA can readily be isolated from the
bacterial culture.
Plasmids, as well as certain viruses, are extraordinarily useful tools for the molecular
biologist, because they serve as gene-carrier molecules called cloning vectors.
A basic procedure of recombinant DNA technology consists of joining a gene of
interest to vector DNA to form a hybrid or recombinant molecule that is able to
replicate in bacteria. Thus, cloning vectors contain genes for replication in bacteria.
In addition, vectors generally carry antibiotic-resistance genes so that uninfected
bacteria can be eliminated from the culture. In order to prepare a recombinant DNA
molecule, a procedure is required for cutting cloning vectors and cellular DNA
molecules at precise positions.
MODERNBIO.COM
7
STANDARD STUDENT PROGRAM 3
Figure 5. Introduction of a Plasmid Into Bacteria by Transformation
MODERNBIO.COM
8
STANDARD STUDENT PROGRAM 3
A nuclease is an enzyme that breaks the phosphodiester bonds that connect
the nucleotide units in DNA or RNA. Restriction nucleases are powerful tools
used in recombinant DNA technology because they cut DNA at specific sites.
These enzymes are produced chiefly by bacterial species in which they serve to
degrade invading foreign DNA within the bacterial cell. Most restriction enzymes
recognize a specific sequence of four to six nucleotides in DNA and each will
cut a long DNA double helix into a series of discrete pieces known as restriction
fragments.
Table 1. Properties of Three Restriction Enzymes
MODERNBIO.COM
9
STANDARD STUDENT PROGRAM 3
Typically, the restriction sites for a given enzyme are hundreds to thousands of
base-pairs apart so that the fragments generated are hundreds to thousands of basepairs long. More than 300 different restriction nucleases are now commercially
available. General properties of three of these enzymes are given in Table 1. It
should be noted that some restriction nucleases (e.g., EcoR1 and Bam H1) produce
a staggered cleavage that creates sticky, or cohesive, single-stranded ends on the
cut molecules. These cohesive ends are very important in recombinant DNA
procedures because they enable any two DNA fragments to be linked together by
complementary base pairing at their ends, provided that they were generated with
the same restriction enzyme.
Figure 6 illustrates one basic procedure by which cellular DNA from essentially
any source can be amplified by recombinant DNA techniques. First, a plasmid is
cleaved at a single site by a restriction nuclease, such as EcoR1, that produces
cohesive ends on the plasmid DNA. The cellular DNA to be cloned is cleaved
with the same enzyme, and fragments of the cellular DNA are annealed to the
plasmid DNA by complementary base-pairing at the cohesive ends of the DNA
molecules. The newly formed joints are sealed with an enzyme called DNA
ligase, which forms covalent bonds between the ends of each DNA molecule.
The recombinant DNA molecules are introduced into E.coli by transformation,
and the bacteria are grown in the presence of an antibiotic. The hybrid plasmid
can replicate in the dividing bacterial cells to produce an enormous number of
copies of the original DNA fragment. At the end of the proliferation period, the
hybrid plasmid molecules are purified from the bacteria. Copies of the original
DNA fragments can then be recovered by cleavage of the recombinant plasmid
with EcoR1.
Digestion of DNA from a vertebrate cell with EcoR1 generates about 106 different
DNA fragments. Thus, the DNA cloning procedure described above and outlined
in Figure 6 gives rise to a large number of plasmids, each descended from a single
hybrid DNA molecule. The most difficult step in the procedure is to identify the
hybrid plasmid in this “library” that contains the inserted cellular DNA of interest.
When a particular mRNA can be purified from a tissue, such as the mRNAs for
the polypeptide chains of hemoglobin, the mRNA or a DNA copy of it can be
used to identify its corresponding gene sequence in a recombinant library. In this
procedure, the mRNA, or its DNA copy, is first labeled with a radioactive isotope.
Under the appropriate conditions, the radioactive probe will preferentially stick
or hybridize to the DNA clone of interest because of complementary base-pairing.
By this procedure, the genes for many different proteins in DNA libraries have
been identified.
MODERNBIO.COM
10
STANDARD STUDENT PROGRAM 3
Figure 6. Production of a Recombinant DNA Molecule
MODERNBIO.COM
11
STANDARD STUDENT PROGRAM 3
II. General Description of
Agarose Gel Electrophoresis
Electrophoresis is the movement of charged particles in solution under the influence
of an electric field. In the most common form of electrophoresis, the sample is
applied to a stabilizing medium which serves as a matrix for the buffer in which
the sample molecules travel. The agarose gel is a common type of stabilizing
medium used for the electrophoretic separation of nucleic acids. A diagram of
the essential components of an agarose electrophoretic system is shown in Figure
7. The agarose gel, containing preformed sample wells, is submerged in buffer
within the electro­phoretic gel cell. Samples to be separated are then loaded into
the sample wells. Current from the power supply travels to the negative electrode
(cathode),supplying electrons to the conductive buffer solution, gel and positive
electrode (anode), thus completing the circuit.
At neutral pH, a molecule of DNA or RNA is negatively charged because of the
negative charges on the phosphate backbone. Under these conditions, nucleic
acids applied to sample wells at the negative electrode end of the gel migrate
within pores of the gel matrix towards the positive electrode. The agarose gel
serves as a molecular sieve in that its structure is similar to that of a sponge. The
size of the pores in the gel are generally on the same order as the size of the DNA
molecules that are being separated. As a result, large molecules move more slowly
through the gel than smaller molecules. Thus, the method sorts the molecules
according to size, since it relies on the ability of uniformly charged nucleic acids
to fit through the pores of the agarose gel matrix.
Ill. Practical Aspects of Electrophoresis
Electrophoresis Equipment
The PROCELL Horizontal Electrophoresis unit is composed of an acrylic cell
with central platform, platinum electrodes, four removable gel-casting trays,
four sample well-forming combs and a safety lid with power cords. The four
gels are made in the casting trays and then placed on the central platform of the
electrophoresis cell. Each gel contains 8 separate sample wells. The experiments
described below were designed such that each student uses four sample wells
per experiment. Therefore, the experiments of 8 students can be analyzed in one
electrophoretic run. If the students work in pairs, the system can be used by 16
students.
The Model MB-170 power supply is a general purpose electrophoresis power
source. The unit produces a constant voltage output of 85 or 170 volts. Voltage
selection is controlled by the switch located in the center of the front panel. The
ammeter, also located on the front panel, permits the current to be monitored
during an electrophoretic run. The unit can reach a maximum of 500 mAmp.
MODERNBIO.COM
12
STANDARD STUDENT PROGRAM 3
Figure 7. Components of a Horizontal Electrophoresis System
Electrophoresis Chemicals
The Agarose Gel
Because the agarose gel is an ideal solid support for the separation of nucleic
acids on the basis of size, it is used extensively for this purpose in the molecular
biology laboratory. Agarose is a natural polysaccharide of galactose and 3,6-
MODERNBIO.COM
13
STANDARD STUDENT PROGRAM 3
anhydrogalactose derived from agar, which, in turn, is obtained from certain
marine red algae. Agarose gels are made by dissolving the dry polymer in boiling
buffer, pouring the gels into casting trays and allowing them to set by cooling at
room temperature. The resolving power of an agarose gel depends on the pore
size, which is dictated by the concentration of dissolved agarose. High percentage
agarose gels (e.g., 3%) are used for the separation of small DNA molecules (1021()3 base-pairs in length), while low percentage gels (e.g., 0.6%) are used for large
molecules (104 -105 base-pairs). The 1.2% and 1.5% agarose gels used in the
experiments described in this manual are suitable for separating DNA molecules
that range in length from about 200 to 10,000 base-pairs.
Electrophoresis Buffer
The buffer (0.04M Tris-Acetate-EDTA, pH 8.0) used in the experiments described
below is a common buffer employed in the research laboratory for separating
double­stranded DNA molecules by agarose gel electrophoresis. The buffer is
provided as a 100-fold stock solution and should be diluted with distilled or
deionized water prior to use.
Staining
Nucleic acids are not colored, and therefore it is necessary to make them
visible in some way in order to determine their position in the agarose gel after
electrophoresis. The most common method involves staining the nucleic acids
with ethidium bromide, and then detecting the DNA or RNA-bound dye with
ultraviolet lamps. However, ethidium bromide is a powerful mutagen (and
probably a carcinogen) and UV light can cause serious eye and skin burns.
Therefore, in the studies described below, the agarose gels are stained with the
harmless dye, methylene blue.
The Sample Buffer
The DNA samples in the exercises described below are loaded into the wells of the
agarose gel as 10-20% glycerol solutions. The viscous glycerol ensures that the
samples will layer smoothly at the bottom of the sample wells. The sample buffer
also contains the tracking dye bromophenol blue. As will be described below, this
dye enables the investigator to follow the progress of an electrophoretic run.
Accessories
The accessories listed below are required
in this manual. They are used for sample
analysis of the agarose gels.
*Glass test tubes (25ml)
*Micro tubes (0.5ml)
*Tube holders for the 0.5ml tubes
*Tape
MODERNBIO.COM
14
to perform the experiments described
handling, and for the preparation and
*Gloves
*Gel staining trays with lids
*Macropipets (pipet-syringe)
*Micropipetors and micropipets
STANDARD STUDENT PROGRAM 3
The micropipetors are an important accessory used for electrophoresis procedures.
The micropipetor consists of a stainless steel plunger and 50 micropipets with
calibration lines of 5µ1*, 10µ1, l5µ1, 20µ1, and 25µ1 as shown below.
A Diagram of the Micropipet Apparatus
*One milliliter (ml) = 1,000 microliters (µ1)
To Operate the Micropipetor:
1.
Insert the metal plunger into the end of the glass pipet that is opposite
the calibration lines. The glass pipet can be held between your thumb
and middle finger and the plunger operated with your index finger on the
same hand.
2. Gently push down on the plunger until the plunger handle comes to rest
on the pipet.
3.
Hold the micropipet in a vertical position and place the filling end into
the sample solution.
4.
Draw the sample into the pipet to the appropriate calibration line by
lifting up on the handle of the plunger assembly.
5.
Carefully wipe excess liquid from the outer pipet surface with an
absorbent tissue.
6. Direct the filling end of the micropipet into a tube and slowly eject the
sample.
7.
Rinse the pipet between samples by drawing up and expelling water
three times from the pipetor.
Students should practice using these pipets prior to beginning the experiments.
MODERNBIO.COM
15
STANDARD STUDENT PROGRAM 3
IV. Procedures for the Preparation,
Electrophoresis and Staining of Agarose Gels
The exercises described in this manual were designed such that he samples of two
students are analyzed on one agarose gel. If students work in pairs, four students
will share one gel. Four agarose gels are electrophoresed simultaneously using the
Procell and MB-170 power supply.
Pouring the Agarose Gels
1.
Place the casting tray on a level work surface and place a precleaned
glass slide into the gel support deck.
2. Seal both ends of the gel support deck with tape. The tape must be firmly
pressed against the edges of the deck to ensure a tight seal.
Casting Tray Assembly
3.
* With the macropipetor (pipet-syringe), dispense 15ml of electrophoresis
buffer into a 25ml glass test tube and add 0.18 grams of agarose. The
agarose can be weighed out directly on an appropriate balance. If a
balance is not available, 0.18 grams of agarose can be estimated by
filling a 0.5ml micro tube with agarose until two-thirds full. Gently swirl
the glass tube until the agarose forms a suspension.
4.
Place the test tube into a boiling water bath and allow the agarose
suspension to come to a vigorous boil. After boiling for about two
minutes, remove the test tube from the bath, stir gently with a glass rod,
and cool at room temperature for about 2-3 minutes. At this time, the
agarose solution should be absolutely clear.
*The melted agarose for the four gels (15ml per gel) can also be prepared in one
operation by boiling 0.9g of agarose in 75ml of buffer in a 250-500ml flask over
a Bunsen burner or in a microwave oven. The flask should be rotated periodically
during the heating process in order to prevent damage to the agarose.
MODERNBIO.COM
16
STANDARD STUDENT PROGRAM 3
5.
Pour the melted agarose directly from the test tube onto the casting deck
and return the test tube to the hot (but not boiling) water bath. The small
amount of melted agarose left in the test tube will be used for sample
application (see below). Insert the comb into the casting tray slots and
push down gently on the top of the comb until resistance is encountered.
The teeth of the comb will come to rest in the melted agarose about
0.2mm above the surface of the glass plate.
6. After the gel has cooled for at least 15 minutes, remove the tape strips
and carefully lift the comb straight up and away from the casting tray.
The gel is now ready for sample application. Gels can also be stored
for up to one week before use. For gel storage, the comb is left in place
and the tray containing the gel and comb is wrapped in plastic wrap and
placed in the refrigerator.
Sample Application
Prior to sample application, place the following items on the laboratory bench in
front of you.
*Agarose gel
*Sample for electrophoresis
*Micropipetor and micropipets
*Absorbent tissue (e.g. Kleenex or Kim wipes)
*Small beaker of distilled or deionized water
*Melted agarose - Transfer the melted agarose from the large test tube in the
hot water bath to a small tube. The small tube should then be placed in
a beaker of hot water to ensure that the agarose remains in a liquid state.
1.
Hold the micropipetor in a vertical position and place the filling end of the
micropipet into the sample solution.
2.
Draw the sample into the pipet to the l5µ1 calibration line by lifting up on
the handle of the plunger assembly.
3. Wipe excess liquid from the outer pipet surface with an absorbent tissue.
4. Carefully direct the filling end of the micropipet into the top of the sample
well and slowly eject the 15µ1 of the sample well.
5.
Draw melted agarose into the micropipet to the 20µ1 calibration line, direct
the filling end into the sample well, and slowly eject the agarose onto the
sample until the well is full. Between 10-20µ1 of agarose are required to fill
the well. The agarose will seal the sample in the sample well.
6.
Rinse the pipet by drawing up and expelling water three times from the
pipetor.
7. Wipe excess liquid from the outer pipet surface with an absorbent tissue.
8. Repeat steps 1-7 to load each additional sample.
MODERNBIO.COM
17
STANDARD STUDENT PROGRAM 3
Electrophoresis
1. Transfer the four casting trays with gels to the central platform of the
electrophoresis cell and position them such that the sample wells are
closest to the black (negative) electrode. Upon electrophoresis, DNA will
then migrate from the negative (black) towards the positive (red) electrode.
2. Place the gel tray stabilizing bar parallel to the long axis of the
electrophoresis cell between the gel trays.
3. Slowly fill the electrophoresis chamber with electrophoresis buffer until
the four gels are covered with a 1/4cm layer of buffer. Approximately 2.5
liters of buffer are required.
4. Place the electrophoresis cell lid in position.
5. With the power supply off, connect the cables from the cell to the power
supply, red to red (positive) and black to black (negative).
6. Push the rocker switches on the power supply to “on” and”170V”. The
voltage will now remain constant at 170 volts during the run.
7.
Unless otherwise indicated, electrophorese until the bromophenol blue in
the sample solution has migrated to within 1/4cm of the positive electrode
end of the gel. At 170V, this takes approximately 50 minutes.
8. At the termination of the electrophoretic separation, shut off the power
supply, disconnect the cables and remove the gel casting trays containing
the gels.
9. The buffer should be emptied from the electrophoresis cell and stored
under refrigeration in a separate container until the next electrophoretic run.
The same buffer should be used for at least 3 electrophoretic separations.
However, fresh buffer should be employed for the preparation of all
agarose gels. The electrophoresis cell should be rinsed with deionized or
distilled water and stored in an upright position.
Staining
Normal Procedure (Overnight Staining)
1
Carefully slide the agarose gel out of the casting tray and off of the slide
and place the gel in a staining dish. (Note: The glass slide should not be
placed in the staining dish.)
2. Dilute the gel stain concentrate 1000-fold with distilled water.
MODERNBIO.COM
18
STANDARD STUDENT PROGRAM 3
3. Cover the gel with about 100ml of staining solution, making certain that
the agarose does not stick to the dish and all gel surfaces are exposed to the
stain.
4. Place the staining tray in the refrigerator and allow 3-18 hours for staining.
5. Decant and discard the stain, rinse the gel and dish with distilled water and
add about 100ml of distilled water.
6. Change the water after about 10-20 minutes and hold the staining dish over
a light source such as a desk lamp or light box. Note the position of the
dark blue DNA bands against the light blue gel background. The gel can be
stored in water in the refrigerator for a few days. After about a week or two
in the refrigerator, the dye will diffuse out of the gel and the DNA bands
will no longer be visible.
Rapid Procedure
1. Dilute the gel stain concentrate 1000-fold and stain the gels for 30-60
minutes in 100mls at 37°C.
2.
Decant and discard the stain, rinse the gel and dish with water and add
about 100ml of distilled water.
3. Change the water after about 20 minutes and again after an additional 20
minutes. DNA bands can been seen during these destaining steps.
4. When the background stain has been reduced sufficiently, hold the
staining dish over a light source and carefully observe the stained DNA
bands in the gel.
Gel Storage
1. The gel can be stored in a sealed plastic bag (3x3”) with a few mls of
distilled water for up to one month in the refrigerator.
2. For long term storage, place the gel on a glass slide and smooth with
a gloved index finger to eliminate air bubbles between the gel and the
slide. Allow the gel to dry onto the slide at room temperature for 3-4
days. Cover the dry gel film and glass slide with saran wrap.
V. Suggested Reading and Reference for Part A
Lewin, B. Genes V. Oxford University Press, Oxford NewYork Tokyo, 1994.
Jones, P. Gel Electrophoresis: Nucleic Acids. Chichester, West Sussex, UK, New
York: Wiley 1995.
MODERNBIO.COM
19
STANDARD STUDENT PROGRAM 3
Experiment 302
Determination of the Length of DNA Molecules
Determining the Size of DNA by Electrophoresis
A first step in the analysis of a DNA molecule in the molecular biology laboratory
frequently involves determining its length in nucleotide pairs. Electro­phoresis
in agarose gels has proven to be an extremely useful tool for this purpose, as
well as for separating DNA fragments of different sizes. The smaller a DNA
fragment, the more rapidly it moves during electrophoresis. The length of a given
DNA fragment can be determined by comparing its electrophoretic mobility on
agarose gels with DNA markers of known lengths. As illustrated in Figure 8, DNA
fragments of known lengths (DNA standards or markers) and a DNA fragment
of unknown length, are electrophoresed on adjacent lanes of the same gel. After
electrophoresis, the DNA in the gel is stained and the positions of the standard
and unknown DNA bands are determined. A linear relationship is obtained if the
logarithms of the sizes of the standard DNA fragments are plotted against their
respective electrophoretic mobilities. The length of the unknown DNA fragment
is then estimated from this calibration curve. The length of DNA is frequently
given in base-pairs (bp) for small fragments and kilobase pairs (KB) for large
ones. One kilobase-pair equals 1000 base-pairs.
Figure 8. Determining the Length of a DNA Molecule.
Left Six DNA standards and a DNA fragment of unknown length are electrophoresed on adjacent
lanes of a gel and the distances (in em) that each has migrated during the run are determined.
Right: The distances migrated by the standards are plotted against the logarithms of their
lengths. The length of the unknown is determined by extrapolation from the standard graph. In
the example, the unknown is about 1.5 KB.
MODERNBIO.COM
20
STANDARD STUDENT PROGRAM 3
Table 1-1. Length of Standard DNA Fragments
Fragment NumberFragment Length (KB)
123.13
29.41
36.68
44.36
52.32
62.03
The standard DNAs were obtained by digestion of bacteriophage lambda DNA
with the restriction nuclease Hind III. Fragments smaller than 2.0 kilobase pairs
are not listed in the table above and will not be visible on your gels.
Certain dyes can be used as standards to calibrate an agarose gel. Dyes such as
xylene cyanol, bromophenol blue and orange-G are fairly accurate size markers
for small DNA fragments. Table 1-2 shows the approximate sizes of DNA (in base
pairs) with which these dyes commigrate on a 1.2% gel. Calibration is performed
by including a mixture of these dyes in one of the sample wells of the gel. These
dyes permit the investigator to follow the progress of the separation during an
electrophoretic run.
Table 1-2. Base-Pair Equivalents of Marker Dyes (1.2% Gel)
Dye Color Base-Pair Equivalent
Xylene Cyanol
Blue-Green
Bromophenol Blue
Purple-Blue
Orange-GOrange
2800
250
70
Objective
To determine the length of a DNA fragment.
Materials
A.
The solutions and materials required lor electrophoresis, sample handling
and gel staining (see Appendices 1 and 2).
B.
The three samples provided in the container marked Experiment
301: Dye Mixture: A mixture of xylene cyanol, bromophenol blue and
orange-G. DNA Standards: The sizes of these DNA molecules are given
in Table 1-1, Unknown DNA: DNA fragment of unknown length.
MODERNBIO.COM
21
STANDARD STUDENT PROGRAM 3
Procedure
The procedures for the preparation, electrophoresis and staining of the agarose
gels are described in detail in Section IV of this manual and are briefly outlined
below. This experiments designed so that the samples of two students will be
analyzed on one agarose gel. H the students work in pairs,four students will share
one gel. The samples of each student (or student pair) will be electrophoresed on
four gel lanes.
1.
Prepare the agarose gel as described in Section IV. In this experiment,
1.2% gels will be used.
2.
Load 15μl of each sample into the wells as indicated below.
Sample Well NumberSample
1
Dye Mixture
2
Standard DNAs
3
Unknown DNA
4
Dye Mixture
5
Dye Mixture
6
Standard DNAs
7
Unknown DNA
8
Dye Mixture
3.
Seal the wells with agarose and transfer your gel to the electrophoresis
cell, making sure you note its position.
4.
Electrophorese for 20 minutes and then turn off and disconnect the
power supply.
5.
Remove the lid of the electrophoresis cell and note the relative position
of the three dyes as compared to their point of application at the sample
wells.
6.
Resume electrophoresis until the bromophenol blue (purple dye) in
the dye mixture and DNA samples has migrated to within 1 mm of the
positive electrode end of the gel. Remove the gels from the unit and
measure the distance of the dyes from the sample origin.
7.
Stain and destain the agarose gels as described in Section IV.
8.
Measure the distance of the DNA bands (in em) from the sample origin.
MODERNBIO.COM
22
STANDARD STUDENT PROGRAM 3
Study Questions
1.
On semilog paper,(provided) plot the electrophoretic mobilities of the
standard DNAs as a function of their fragment lengths and determine the
size of the unknown DNA fragment in base-pairs.
2.
The length of DNA helix occupied by one nucleotide pair is 3.4 A
(angstrom). What Is the length of the unknown DNA fragment?
3.
A human has 10” cells and each human cell has about 6.4 x 10° nucleotide
pairs of DNA. What is the length of double helix that could be formed
from this amount of DNA In a human individual?
MODERNBIO.COM
23
STANDARD STUDENT PROGRAM 3
Experiment 302
Restriction Nuclease Mapping of DNA
Viruses are a unique group of parasitic organisms that grow only In the cells of
bacteria, plants and animals. A virus that Infects bacteria is called a bacteriophage,
or simply, a phage. Viruses have proved to be Invaluable tools for the study of
molecular biology because they possess the essential properties of life yet are
simpler than bacteria or eukaryotic cells in their structures and life cycles.
Bacteriophage lambda, which infects E. coN, Is probably the best understood of
the double•stranded DNA phages. The protein component of this phge consists
of a protective coat that forms the tail assembly and the outer shell of the head. A
single molecule of double•stranded DNA Is located In the core of the phage head.
The DNA molecule contains 48,502 base-pairs (molecular weight ~ 3 x 107) that
code for approximately 50 different phage proteins. The sequence of nucleotides
along the entire lambda genome is known and the nucleotide sequences that
comprise the major control regions for transcription and replication have been
identified. Because of the vast amount of Information about the biology of this
phage, lambda has become a common cloning vector In genetic engineering.
Figure 2-1. A Diagram of a Lambda Phage DNA Molecule
The lambda DNA molecule has an unusual structure. Single-stranded segments
of DNA containing 12 nucleotides are found at both ends of the phage DNA
molecule (Figure 2-1). The base sequence of these terminal regions, which are
MODERNBIO.COM
24
STANDARD STUDENT PROGRAM 3
known as cohesive or sticky ends, are complementary to each other. Thus, by
forming base pairs between the cohesive ends, the linear DNA molecule will
circularize,as shown In Figure 2-1. This circularization Is readily demonstrated
in the laboratory. When lambda DNA is incubated at 37°C, the cohesive ends will
anneal (bind) to each other by base-pairing and the circular DNA molecule wil
form. When this circular lambda molecule Is heated to 75°C, the cohesive ends
come apart and the DNA molecule is converted to a linear form. The disruption of
base-pairing by heal treatment is known as denaturation.
The circularization of the lambda DNA molecule also occurs in an infected
bacterial cell. When lambda infects an E.coli host, the phage DNA is injected Into
the bacterium and the cohesive ends anneal or stick together to form a circular
molecule. Transcription and replication of the phage DNA occurs and the newly
formed DNA molecules are packaged into newly synthesized phage particles.
About 100 lambda particles are then released into the medium when the bacterial
cell is broken open (lysed). As a result of this lytic infection, the bacterium dies.
less commonly, the circular DNA phage molecule attaches and Integrates Into the
E.coli DNA. After integration, the bacterial cell behaves normally and the lambda
DNA is replicated as part of the bacterial chromosome. However, when the
bacterium is subjected to an environmental insult, such as exposure to ultraviolet
light, the lambda DNA is excised from the host chromo­some and begins a normal
cycle of viral replication.
Figure 2-2 shows a map of phage lambda DNA. The DNA of phage lambda may be
divided into three regions. The left-hand region includes all the genes (A through
J) whose products are necessary to produce phage head and tail proteins and to
package the DNA into the virus. The central region contains elements involved in
integration of the DNA into the E. coN chromosome. The remaining portion of the
genome Includes the major control region for transcription and replication and the
genes necessary lor cell lysis.
Restriction nucleases are enzymes that cut DNA at specific sites (See Part A-1).
In this exercise, you will study phage lambda DNA using the restriction enzymes
EcoR1 and BamH1. The position of the restriction sites for these enzymes in
lambda DNA Is shown In Figure 2-2. Cleavage of lambda DNA by either EcoR1
or BamH1 produces six DNA fragments, the sizes of which are Indicated in the
figure. The fragments derived from the termini contain the cohesive ends and will
anneal together under the appropriate conditions.
MODERNBIO.COM
25
STANDARD STUDENT PROGRAM 3
Figure 2-2. Genetic and Restriction Map of Phage Lambda DNA
MODERNBIO.COM
26
STANDARD STUDENT PROGRAM 3
Objective
To study the effects of EcoR1 and BamH1 on phage lambda DNA and to Identify
restriction fragments that contain the cohesive ends.
Materials
A. The solutions and materials required lor electrophoresis, sample handling
and gel staining (see Appendices 1 and 2).
B The samples below are provided In the container marked Experiment
302:
1. Phage Lambda DNA.
2. EcoR1: The restriction enzyme to be used In this laboratory should
be made up Immediately before the laboratory session as described In
Appendix 2 of the Instructor Manual and stored in the refrigerator before
use. The solution contains the enzyme suspended in a nuclease digestion
buffer. The solution of enzyme and DNA must not be contaminated, so
use a fresh micropipet whenever you remove the enzyme and DNA from
the stocks.
3. EcoR1 +Bam H1: This mixture contains both restriction enzymes
suspended in a nuclease digestion buffer and should be made up as
described In Appendix 2 of the Instructor Manual.
4. Electrophoresis sample buffer.
Materials Not Provided
1. Water baths lor tube incubation maintained at 37°C and 70”C: A beaker
containing water heated to 70”C with a Bunsen burner can serve as the
latter bath.
2. Ice bath: Ice chips In a beaker can be used.
Procedure
The experiment was designed lor B students working Individually or 16 students
working in teams of two.
A. Preparing the DNA samples
1. Number lour small tubes 1 to 4 with a water-proof marking pen.
2. Place 10μl of distilled water into tube 1, 10μl of the EcoR1-buffer
solution into tube 2 and 10μl of the EcoR1 + Bam H1 mixture Into
tubes 3 and 4.
3. Add 5μl of lambda phage DNA to each tube. Gently tap the tubes
with the tip of your index finger to mix the solutions. Incubate the
tubes tor 50 minutes at 37°C. During this Incubation, the fragments
containing the cohesive ends (the fragments derived from the
termini) should anneal together.
MODERNBIO.COM
27
STANDARD STUDENT PROGRAM 3
4. While the tubes are incubating, prepare 1.2% agarose gels as
described in Section IV.
5. At the end of the 50 minute incubation period, add 5μl of
electrophoresis sample buffer to each of the lour tubes.
6. Transfer tube #4 to the 70°C water bath and, after 5 minutes, place
this tube in an ice bath. This heat treatment will separate the two
fragments containing the cohesive ends. Thus, in tube #4, the
cohesive end fragments should be separated while they should be
together in tubes 1 to 3.
B. Electrophoresis
1. Load 15μl of the following samples from the above section into the
sample wells.
Sample Well Sample
1
Tube 1
2
Tube 2
3
Tube 3
4
Tube 4
5
Tube 1
6
Tube 2
7
Tube 3
8
Tube 4
2.
Seal the wells with agarose and electrophorese until the
bromophenol blue in the samples has migrated to within 2 mm of the positive
electrode end of the gel.
3.
Section IV.
Remove the gels from the unit and stain them as described in
4.
Measure the distance of the DNA bands (in cm) from the sample
wells and draw a picture of the DNA bands in each gel lane. Note the relative
intensity and positions of the DNA bands in lane 3 and lane 4. Note especially the
intensity of the next to the largest DNA band.
Study Questions
1.
The sizes of the DNA bands in lanes 2 and 6 should be 21.2, 7.4, 5.8,
5.6, 4.8 and 3.5 kilobase pairs (see Figure 2-2). From the map of phage
lambda given In Figure 2-2, identify the DNA bands that contain the
genes lor the head proteins and the genes that control lysis of the host
cell.
MODERNBIO.COM
28
STANDARD STUDENT PROGRAM 3
2.
Maps of the restriction sHes lor EcoR1 and BamH1 in lambda DNA are
given in Figure 2-2. With this information:
A. Calculate the length of DNA fragments that should have been
produced when lambda DNA was digested with both EcoR1 and
BamH1. List these values below and indicate the two fragments that
contain the cohesive ends.
DNA Fragment Length
(base-pairs)
1._________________
2._________________
3._________________
4._________________
5._________________
6._________________
7._________________
8._________________
9._________________
10.________________
11.________________
B. Identify the above fragments on lanes 3 and 4 of your gels.
3. A comparison of the result of lanes 3 and 4 on your gels can be used
to identify the DNA terminal fragments of phage lambda without a
knowledge of the map shown in Figure 2-2. Explain.
4.
In DNA of uniform composition, a specific hexanucleotide occurs by
chance once in every 4096 base-pairs. The recognition sequence lor
EcoR1 Is a hexanucleotide (sea Part A-1, Table 1). How many restriction
sites lor EcoR1 would you expect to exist in lambda DNA H the phage
DNA were of uniform composition? There are law restriction sites for a
number of enzymes including EcoR1 and Hpal in the left 20 kilobase of
lambda DNA. Explain. (Hint: The left region of lambda DNA Is rich In
G+C base pairs.)
MODERNBIO.COM
29
STANDARD STUDENT PROGRAM 3
Experiment 303
Plasmid DNA Structure
Plasmids are circular extra chromosomal DNA molecules that replicate in bacteria
(see Section 1-E). A plasmid DNA molecule can exist in three quite different
structural states: linear, relaxed circular and supercoiled circular (see Figure 3-1).
Linear
When a plasmid is treated whit a restriction nuclease that cleaves the DNA in only
one site, the circular forms of the plasmid are converted lo a linear molecule.
Relaxed Circular
The two ends of a linear DNA helix can be brought together and joined lo form
a circle. When this occurs, without rotating or twisting the DNA molecule, the
circle is said to be relaxed because the DNA helix within exists in a nonstrained
state similar to that found in the linear molecule.
Supercoiled Circular
When one end of a double helix, which is itself a twisted structure, is rotated 360
degrees with respect to the other end, and then the two ends are joined, supercoils
are introduced into the resulting circle. A common analogy is to consider the
structures introduced into a rubber band when it is twisted. When a loosely twisted
heavy rubber band (the double helix) is twisted far enough,as when a model
airplane is being prepared for flight, the rubber band crosses over itself in space
and supercoiling takes place. When the DNA is twisted clockwise in the same
direction as the right-handed double helix, positive supercoiling occurs and the
torsional pressure serves lo wind the double helix more lightly. This DNA is said
lo be overwound. In contrast, negative supercoils twist the DNA in the opposite
direction from the turns of the double helix. DNA with negative supercoils is said to
be underwound. The degree of twisting is often referred to as the superhelix density.
A supercoiled DNA molecule Is under strain because it has acquired excess energy
as a result of the supercoiling. In a negatively supercoiled molecule, the strain
can be partially relieved by limited disruption of the DNA double helix to form
single-stranded sites at specific locations along the molecule. Thus,in a supercoiled
plasmid, the fraction of the molecule that Is single-stranded at any given moment is
greater than in a relaxed circular or linear molecule. S1 nuclease is an enzyme that
is isolated from the mold Aspergillus oryzae. This enzyme attacks single-stranded
DNA and RNA but not double-stranded nucleic acid molecules. When a negatively
supercoiled plasmid is treated with S1 nuclease, the enzyme will introduce breaks in
the DNA at regions that have a single-stranded character. As a result of this nicking
action, supercoiling is lost and the plasmid is converted to a relaxed circular form.
MODERNBIO.COM
30
STANDARD STUDENT PROGRAM 3
Figure 3-1. States of a Plasmid DNA Molecule
It is likely that all DNA within a cell is under the influence of negative supercoiling
at some time during the lifetime of the cell. In fact, the natural condition for
plasmid DNA, most viral DNA and the DNA in bacterial and perhaps even
eukaryotic chromosomes,is generally in an underwound state as a result of
negative supercoiling. Although positive supercoiling can be induced in vitro by
treatment with certain chemicals, it probably does not occur as a natural state. A
number of enzymes that remove and introduce supercoils in DNA, called DNA
topoisomerases, have been isolated from different organisms. The degree of
negative supercoiling Is a key factor in determining the higher-order structure of
DNA within a cell. Higher-order DNA structure, In turn, is thought to be important
in the regulation of chromosome structure, gene activity, and DNA replication.
For example, DNA supercoiling facilitates transcription of genes In bacteria and,
perhaps,in eukaryotic organisms as well.
As was noted above, DNA molecules are thought to cycle through various
structural states during the lifetime of many, and perhaps all, organisms. Several
laboratory techniques have been used to distinguish these structural forms.
Electron microscopy is one method that can be used to distinguish between linear,
circular and supercoiled DNA molecules. Another method is gel electrophoresis.
As we have seen in Exercise t, linear double-stranded DNA molecules migrate
through agarose gels at rates that are inversely proportional to the log10 of their
molecular weights. However,supercoiled circular, relaxed circular and linear
DNA fragments containing the same number of base-pairs migrate differently
on agarose gels because of the differences in their shapes (see Figure 3- t ). The
relative mobilities of the three forms are dependent on a number of factors,
including agarose concentration and Ionic strength of the buffer.
MODERNBIO.COM
31
STANDARD STUDENT PROGRAM 3
In today’s experiment, you will compare the electrophoretic migration of
super­coiled circular, relaxed circular and linear forms of the plasmid pUC18.
Plasmid pUC18 is a small E.coli plasmid that contains 2686 base-pairs. This
plasmid contains a single restriction se for EcoR1. Under the conditions that will
be used in this experiment, the supercoiled form of pUC18 will migrate from
slightly faster to slightly slower than the linear form, and it’s migration will vary
somewhat in different preparations of plasmid DNA. The supercoiled and linear
forms of pUC18 will migrate taster than the relaxed circular forms. When pUC18
is isolated from E.coli, most of the plasmid molecules are in the supercoiled and
the relaxed circular forms. Nicking of the supercoiled plasmid with S1-nuclease
will result in the formation of the relaxed circular forms, while treatment of the
plasmid with EcoR1 will convert the supercoiled and relaxed circular forms to the
linear form.
Objective
To compare the electrophoretic mobility of supercoiled circular, relaxed circular
and linear plasmid DNA.
Materials
A. The solutions and materials required for electrophoresis, sample handling
and gel staining (see Appendices 1 and 2 of the Instructor Manual).
B. The samples provided in the container marked Experiment 303:
Plasmid DNA: Plasmid pUC18 is 2686 base-pairs in length and is
cleaved at one site by EcoR1
EcoR1:The enzyme should be made up immediately before the laboratory
session as described in Appendix 2 and stored in the refrigerator before
use
Electrophoresis Sample Buffer,
Dye Mixture: A mixture of xylene cyanol, bromophenol blue and
orange-G.
S1-Nuclease: The enzyme should be made up immediately before
the laboratory session as described in Appendix 2 and stored in the
refrigerator before use.
Materials Not Provided
1. A water bath for tube incubation maintained at 37°C.
Procedure
The experiment was designed tor 8 students working individually or 16 students
working in teams of two.
MODERNBIO.COM
32
STANDARD STUDENT PROGRAM 3
A. Sample Preparation
1.
Number three small tubes 1 to 3 with a waterproof marking pen.
2.
Place 10μl of distilled water into tube 1, 10μl of the EcoR1-buffer
solution into tube 2 and 10μl of S1 nuclease into tube 3.
3.
Add 5μl of plasmid DNA to each tube and gently tap the tubes with
the tip of your index linger to mix the solutions. Incubate the tubes for
50 minutes at 37°C. During this incubation, prepare the agarose gels as
described below.
4.
Add 5μl of electrophoresis sample buffer to each tube and electrophorese
the samples as described below.
B. Electrophoresis
1.
Prepare 1.2% agarose gels and load 15μl of the samples into the sample
wells.
Sample WallSample
1
Tube 1
2
Tube 2
3
Tube 3
4
Dye Mixture
5
Tube 1
6
Tube 2
7
Tube 3
8
Dye Mixture
2.
Seal the wells with agarose and electrophorese until the bromophenol
blue in the samples has migrated to within 1cm of the positive electrode
end of the gel.
3.
Remove the gels from the unit and stain and destain them as described
in Section IV. Measure the distance of the DNA bands from the sample
origin.
Study Questions
1.
Identify the DNA bands that contain linear, relaxed circular and
supercoiled circular forms of pUC18. Explain the rationale for your
identification.
2.
Ethidium bromide is a chemical that will remove negative supercoils and
Introduce positive supercoils in circular DNA molecules. A number of
sequential changes occur in the structure and electrophoretic properties
of supercoiled plasmid DNA when increasing amounts of ethidium
bromide are added. Describe these changes.
MODERNBIO.COM
33
STANDARD STUDENT PROGRAM 3
Experiment 304
Molecular Cloning
A. Introducing Plasmid DNA Into E.coli
Penicillin is one of the most Important anti-Infective agents used In clinical
medicine because n Is Inexpensive, bactericidal and its toxicity lor human cells
Is almost nonexistent. Penicillin Interferes with the synthesis of the bacterial call
wall and will thus cause osmotic lysis of susceptible microbes. Penicillin Is not a
single compound but a group of compounds with related structures and activities.
Many of these compounds are semi-synthetic In that part of each molecule is
made by a mold to which the chemist adds another chemical group. Over 500
semi-synthetic penicillins have been made during the past 30 years. Ampicillin
is abroad-spectrum semi-synthetic penicillin that will kill a number of grampositive and gram-negative bacteria, including Salmonella and Escherichia coli.
Occasionally, E.coli cells are found in nature that are resistant to the toxic effects
of ampicillin. In today’s laboratory, you will create such an ampicillin-resistant
population of E.coli.
Plasmids are small circular DNA molecules that exist apart from the chromosomes
In most bacterial species. Under normal circumstances, plasmids are not essential
for survival of the host bacteria. However, many plasmids contain genes that
enable bacteria to survive and to prosper In certain environments. For example,
soma plasmids carry one or more genes that confer resistance to antibiotics. A
bacterial cell containing such a plasmid can live and multiply In the presence of
the drug. Indeed, antibiotic-resistant E.coli isolated In many parts of the world
contain plasmids that carry the genetic information for protein products that
interfere with the action of many different antibiotics. In this laboratory, you will
introduce a plasmid that contains an ampicillin resistance gene Into E.coli.
Characteristics of pUC18
The plasmid that you will use in today’s laboratory is called pUC18. A map of
plasmid pUC18 is shown in Figure 4-1 and a discussion of some of the features of
this plasmid is given below.
Antibiotic Resistance
Plasmid pUC18 contains an ampicillin-resistance gene that enables E.coli
containing this plasmid to grow In the presence of the antibiotic. Thus,bacteria
lacking this plasmid,or bacteria that lose the plasmid,generally will not grow In
the presence of this antibiotic. The ampicillin-resistance gene of pUC18 codes lor
the enzyme beta-lactamase (penicillinase), which inactivates ampicillin and other
penicillins.
MODERNBIO.COM
34
STANDARD STUDENT PROGRAM 3
Structure
Like all plasmids, pUC18 Is a supercoiled circular DNA molecule (Experiment
303).
Size
A small plasmid, pUC18 contains only 2686 base-pairs (molecular weight = 2 x
106). The small size of this plasmid makes H less susceptible to physical damage
during handling. In addition, smaller plasmids generally replicate more efficiently
in bacteria and produce larger numbers of plasmids per cell.
Restriction Sites
To be useful as a cloning vector, a plasmid should possess a single recognition
site for one or more restriction enzymes into which foreign DNA can be inserted.
One way to Increase the number of such restriction sites Is to add a polylinker, a
segment of synthetic DNA that contains closely spaced restriction sites. Plasmid
pUC18 contains a polylinker and a site for EcoR1 is in this region.
Replication
Replication of plasmid DNA requires a plasmid replication origin and a number
of replication enzymes. The replication enzymes are usually provided by the host
bacterial cell. The replication of some plasmids is coupled to that of the host cell.
As a result of this so-called “stringent control,” only one or at most a few copies
of the plasmid will be present In each bacterial cell. In plasmids under “relaxed
control,” replication of plasmid DNA Is not coupled to replication of the host
chromosome and more than 500 copies of the plasmid may be present In a single
host cell. Plasmid pUC18, as well as other common plasmid-cloning vectors,are
under relaxed control.
Lac Z gene
The lac Z gene, which codes for the enzyme beta-galactosidase, breaks down
lactose and other beta-galactosides Into component sugars. Plasmid pUC18
contains a portion of the lac Z gene from E.coli.
Figure 4-1. Partial Genetic and Restriction Enzyme Map of Plasmid pUC18
MODERNBIO.COM
35
STANDARD STUDENT PROGRAM 3
In this exercise, pUC18 will be Introduced Into E.coli by a process known as
transformation. A common procedure for transformation entails:
1.
The treatment of bacterial cells with calcium chloride in order to enhance
the uptake of plasmid DNA. Such calcium chloride treated cells are said
to be competent.
2.
The incubation of the competent cells with plasmid DNA and the uptake
of the DNA into a small fraction of the cells.
3.
The selection of those cells that have taken up the plasmid DNA by
growth on an antibiotic-containing medium.
Objective
To introduce plasmid pUC18 into E.coli in order to create a population of bacterial
cells resistant to the antibiotic ampicillin.
Materials°
Plasmid pUC18 (180μl) - (300μg/rnl)
*16 Ampicillin-Nutrient Agar plates
Nutrient Broth (10μl)
25 Inoculating loops
19 large sterile Transfer pipets•One pipet should be used for the bacteria, one
for the nutrient broth and 16 for step III.
16 Sterile tubes
CaCl2 solution (5ml) - 100mM CaCl2
E.coli (0.3ml)
* Prepared as described in Appendix 2 of the Instructor Manual - Note:
These plates must be made up at least one day before this lab session.
Materials Not Provided
Water bath maintained at 37°C • If a temperature regulated water bath Is not
available, a beaker containing tap water at 37°C (98°F) will suffice.
Ice bath • Ice chips in a large beaker are suitable.
Air Incubator maintained at 37°C (OPTIONAL - see below)
Procedure
Step I should be performed by the Instructor before or during the laboratory
session. Eight students or eight teams of students should then perform steps II
and III.
MODERNBIO.COM
36
STANDARD STUDENT PROGRAM 3
I. Preparation of Competent Celts
(This step should be performed by the Instructor).
1. Place the vial of CaCl2, and the tube of E.coli in the ice bath.
2. Using a sterile pipet, transfer about 1/2ml of the CaCl2 containing
the bacteria solution to the tube
3. Using the same pipet, transfer the contents of this tube back Into the
larger vial that contains most of the CaCl2, solution.
4. Tap the vial with the tip of your Index finger to mix the solution.
5. Incubate the cells for about 20 minutes on ice. The cells are then
called competent because they can take up DNA from the medium.
If desired, the cells can be stored in the CaCl2 solution for up to 12
hours on ice before use.
II. Uptake of DNA by competent cells
1. One member from each of the eight groups should label one small
tube “plus DNA” and one tube “minus DNA”.
2. Place the two tubes in an ice bath.
3. Use an inoculating loop to add the plasmid DNA to the tube labeled
“plus” DNA. Immerse the loop into the plasmid DNA solution and
transfer one loop full (about 10μl) of DNA to the “plus” DNA tube.
4. Gently tap the vial of competent cells with the tip of your Index
finger to ensure that the cells are In suspension. Then, using a sterile
transfer pipet, add 5 drops (5 drops ~ 100μl) of the competent cells
to each of the two tubes. Tap each of these tubes with the tip of your
Index finger to mix these solutions and store both tubes on ice lor 20
minutes. The competent cells, which are suspended In CaCl2, will
now begin to take up the plasmid DNA.
5. Transfer the tubes to a water bath, preheated to 37°C, lor 5 minutes.
This heat shock facilitates the uptake of plasmid DNA.
6. Add about 0.7ml of nutrient broth to each tube and incubate at 37°C
lor 30 minutes. The nutrient broth should be dispensed with the
sterile pipet provided with the Experiment package. This incubation
period allows the bacteria time to recover from the CaCl2 treatment
and to begin to express the ampicillin-resistance gene on the plasmid.
MODERNBIO.COM
37
STANDARD STUDENT PROGRAM 3
III. Selection of cells that have taken up the plasmid by growth on an
ampicillin-containing medium.
1. Obtain two ampicillin-nutrient agar plates from your Instructor and
label one plate “plus DNA” and the other plate “minus DNA”.
2. Using a sterile pipet remove 0.25ml of the mixed bacterial suspension
from the “minus DNA” tube. remove the lid from the “minus DNA”
plate and dispense the bacteria onto the agar. Use an Inoculating
loop to spread the bacteria evenly onto the agar surface.
3. Transfer 0.25ml of the bacterial suspension from the “plus DNA”
tube to the “plus DNA” plate and spread these cells onto the agar
surface as described above.
4. Replace the lids on the plates and leave the platesat room temperature
until the liquid has been absorbed (about 10-15 minutes).
5. Invert the plates and incubate at 37°C or room temperature.
6. Colonies should appear in about 15 hours at 37°C or 48 hours at
room temperature. Count the number of visible colonies on each
plate and record these values below.
Number of colonies on “minus DNA” plate ___________
Number of colonies on “plus DNA” plate
___________
Study Questions
1.
Given that each colony is the progeny of a single bacterial cell, determine
the number of cells in the transformation mixture that express ampicillin
resistance in the original “plus DNA” tube.
2.
Why are these cells resistant to ampicillin?
MODERNBIO.COM
38
STANDARD STUDENT PROGRAM 3
Experiment 304
Molecular Cloning
B. Isolating and Analyses of a Plasmid
A strain of E.coli containing the plasmid pUC18 is provided for this exercise
where you will grow the bacterial cells, isolate tho plasmid from the bacterial
culture, digest the plasmid with EcoR1 and analyze h by electrophoresis.
Bacterial Growth
Growth of E.coli in a liquid medium can be divided into four stages (Figure
4-2). When nongrowing cells are introduced into a liquid growth medium, they
require time to adapt before cell division occurs. This period,called the lag phase,
usually lasts 10-90 minutes at 37°C. The cells in the culture will then enter tho
exponential phase in which the cell concentration Increases as an exponential or
logarithmic function of time. The length of time (generation time) necessary for
the culture to double remains constant during this growth phase. In rich medium
at 37°C, approximately 20 minutes is required for most E.coli strains to double
in cell number. Exponential growth continues until a lack of oxygen becomes a
limiting factor or until a change in the composition of the medium occurs. At this
time,growth ceases and the culture enters the stationary phase. The growth rate of
the cells and tho density of the population in the stationary phase are dependent
on a number of factors, including the medium, temperature, degree of aeration
and E.coli strain.
The three major techniques used to quantity the growth of a bacterial population
in a liquid culture are total cell count, viable count and cell mass. A total cell count
is determined from microscopic enumeration of bacteria in a set volume of the
culture. This method does not distinguish between live (viable) and dead cells.
The viable count is based on the premise that each bacterial colony that results
when a given volume of the liquid culture Is placed on a solid culture media,is
the progeny of a single bacterial cell. The cell mass of the culture can easily be
calculated by measuring the optical density (OD) of tho culture at a wavelength of
600nm. This measurement provides an index of the turbidity (cloudiness) of the
culture solution which is related to cell mass and, more indirectly,to cell number.
As a general guide;
10D600 _ 8 x 108 cells/ml
Plasmid Isolation
Bacterial cells that have been transformed by plasmid pUC18 can grow on an
ampicillin-containing medium (Experiment 304 A). This observation provides
evidence, at the phenotypic level, that tho cells contain plasmid pUC18 and express
the ampicillin-resistance gene. In this exercise, you will verify plasmid uptake at the
molecular level by isolating the plasmid from transformed bacterial cells.
MODERNBIO.COM
39
STANDARD STUDENT PROGRAM 3
Figure 4-2. Growth of E.coli in a Liquid Culture
Although there are many different methods available for plasmid isolation, most
have the following steps In common:
1.
Growth and Harvesting of Plasmid-Containing Bacterial Cells - The
cells from an antibiotic-resistant colony are usually grown overnight in
a liquid culture to stationary phase and then harvested as a pellet by
centrifugation.
2.
Lysis of the Bacteria-Lysis (breaking open) of the bacteria is frequently
brought about by agents that destroy the bacterial cell wall and plasma
membrane.
3.
Purification of the Plasmid DNA•Isolating the plasmid DNA from the
lysate often involves differential precipitation, where tho bacterial
components ( DNA, proteins, and membrane fragments) are removed by
centrifugation,while the smaller plasmid DNA remains in solution.
4.
Concentrating Plasmid DNA•Nucleic acids including plasmid DNA
can be precipitated by alcohol (e.g. ethanol, isopropanol), and alcohol
precipitation is fre­quently used to concentrate plasmid DNA following
purification.
MODERNBIO.COM
40
STANDARD STUDENT PROGRAM 3
Materials Provided
Electrophoresis Sample Buffer
Dye Mixture: A mix of xylene cyanol, bromophenol blue and orange-G,
*EcoR1: This restriction enzyme should be made up as described in the
Instructor Guide and stored in the refrigerator before use.
Plasmid DNA: Plasmid pUC18 (from Experiment 304 A)
Nutrient Broth Plus Ampicillin (100ml),
Solution I - Tris Buffer and Ribonuclease A
Solution II - Sodium Dodecyl Sulphate and Sodium Hydroxide
Solution III - Guanidinium Chloride and Acetic Acid
Equilibration Buffer - Sodium Hydroxide
Buffer HB - Guanidinium Chloride and Isopropanol
*Wash Buffer - Tris Buffer, Sodium Chloride and Ethanol Note: The ethanol
must be added to this buffer prior to use.
Elution Buffer - Tris Buffer
HiBind DNA Columns
Collection Tubes
**E.coli -pUC18 (1ml)
Calibrated 1ml transfer pipets - The pipets used in Experiment 304 A should
be washed with soapy water and then with distilled or deionized water.
Alternatively, 5ml glass pipets can be used.
* Made up as described in the Instructor Guide
** This culture must be started about one day before the laboratory.
Materials Not Provided
1. An incubator maintained at 37°C for bacterial growth. It is preferable
that the incubator be of the shaking type.
2. A microcentrifuge and 1.5 ml centrifuge tubes.
3. Spectrophotometer and accessories (optional).
4. A 37°C water incubator
Procedure (Laboratory Session 1)
A. Growing E.coli Containing the Plasmid pUC18
One bacterial culture will be used by the entire class and the instructor should start
the culture one day prior to the class meeting.
1. Carefully pour the entire (1ml) E.coli -pUC18 starter culture into
the flask containing the I 00 Nutrient Broth and replace the cap.
Alternatively, E.coli containing pUC18 can be obtained from a plate
containing transformed colonies (Exercise 304 A). If this option is
chosen, touch an inoculating loop to a single, well-isolated bacterial
colony and transfer the colony to the I 00 ml of liquid medium.
Ampicillin has been added to the broth to ensure that the bacteria
will not lose the plasmid during the growth phase.
MODERNBIO.COM
41
STANDARD STUDENT PROGRAM 3
2. Pour a few mls of the inoculated culture into a test tube and measure
the OD at 600nm. Omit this step if a spectrophotometer is not
available.
3. Incubate the flask at 37°C for 14-24 hours. Samples may be removed
at various times after inoculation to monitor cell growth. The flask
should be shaken during the incubation, if possible, since aeration
will increase the final cell density.
4. At the end of the incubation period, remove a few mls of the culture
and read the OD at 600nm. It will probably be necessary to dilute this
sample by about I 0 fold in order to obtain an accurate measurement.
5. Divide the culture into eight portions, each containing about 6-10
mls, and place each portion into tubes. The eight tubes will be
processed by eight groups of students.
B.
Harvesting the Bacteria
1. Transfer about 1.5 ml of the culture to a labeled 1.5 ml microcentrifuge
tubes and centrifuge at maximal speed for 1-2 minutes to pellet the
bacteria.
2. Discard the supernatant, fill the tube again with the culture and
centrifuge for 1-2 minutes. Repeat this step one time.
3. Carefully remove and discard the supernatant fluid. Invert the tube
and allow the cell pellet to dry for a minute.
C.
Preparing the Plasmid DNA
1. Add 250μl of Solution I to the tube and suspend the pellet in the
solution by vigorous shaking or by using a vortex mixer. This
solution contains RNAse, which destroys RNA. Complete resuspension (no visible clumps) is important in order to have high
plasmid yield.
2. Incubate the tube at room temperature for 2 minutes.
3. Add 250μl of Solution II, cap the tube and mix by inversion several
times until the contents are mixed thoroughly. The solution should
become clear and viscous within 1-2 minutes. Solution II contains
the detergent sodium dodecyl sulfate (SDS) and sodium hydroxide,
which dissolve the bacterial membranes and causes cell lysis. The
sodium hydroxide also denatures the DNA. Upon neutralization in
the step below, the plasmid DNA strands renature because of their
proximity while the bacterial DNA does not.
MODERNBIO.COM
42
STANDARD STUDENT PROGRAM 3
4. Add 350μl of Solution III to the tube, cap the tube and shake until
the contents are mixed thoroughly. A white flocculent precipitate
should form. The acetic acid in Solution III will neutralize the
solution. The Guanidinium Chloride in Solution III and the SDS
from the previous step produce a tangled network of long strands of
bacterial chromosomal DNA and cell debris, and this matrix can be
separated from the smaller plasmid DNA by centrifugation.
5. Centrifuge the tube for 10 minutes at maximal speed. After
centrifugation, plasmid DNA and small amounts of cellular proteins
are in the supernatant, and bacterial chromosomal DNA and most
of the proteins are in the pellet. This supernatant is known as a
“cleared lysate”. A common procedure that is used to remove the
proteins, SDS and salts that are present in the cleared lysate involves
the use silica, which binds to the DNA but not to the contaminants.
The silica is packed into the bottom of a column that you will use,
and the lysate is added to the column over the silica. The column is
then washed by centrifugation to remove the contaminants and the
plasmid DNA is eluted by using a low salt buffer.
6. Obtain a blue column, and a collection tube, label each, and insert
the column into the collection tube. Add 100μl of Equilibration
Buffer to the column and centrifuge for 1 minute at maximal speed.
7. Carefully pour the cleared lysate from step 5 into the column
making sure that the pellet is not dislodged. Centrifuge the column
with collecting tube for 1 minute at maximal speed.
8. Discard the liquid in the collection tube, place the column back
into the collection tube and add 500μl of HB Buffer to the columns.
Centrifuge the column/collecting tube for I minute.
9. Discard the liquid in the collection tube, place the column back into
the collection tube and add 700 ul of Wash Buffer to the column.
Centrifuge the column/collecting tube for 1 minute.
10. Discard the liquid in the collection tube, place the column back into
the collection tube and centrifuge for 2 minutes to dry the column.
11. Discard the collection tube and insert your column into a clean­
labeled 1.5-2 ml centrifuge tube. Place 25μl of Elution Buffer onto
the center of the silica in the column and incubate for 2 minutes.
Centrifuge for 2 minutes, discard the column and cap the tube that
contains the eluted purified plasmid DNA. The plasmid DNA can
be stored in the freezer until the next laboratory or processed as
described below.
MODERNBIO.COM
43
STANDARD STUDENT PROGRAM 3
D. Preparing DNA Samples (Laboratory Session 2)
The plasmid DNA that you have isolated should consist of circular molecules.
In order to characterize the DNA by electrophoresis, the plasmid should be in a
linear form. Therefore, you will digest some of the isolated pUC18 with EcoRI
which will generate a linear DNA molecule containing 2686 base pairs.
1. Number two small tubes 1 and 2 with a waterproof marking pen.
2. Place 20μl of distilled water into tube #I and 20μl of the EcoR1
buffer solution into tube #2.
3. Add 10μl of your plasmid to each tube and tap the tubes with the tip
of your index finger to mix the solutions. *
4. Incubate the tubes for 60 minutes at 37°C. During this incubation,
prepare 1.2% agarose gels.
E. Electrophoresis
1. Add 70μl of sample buffer to the pUC18 DNA tube supplied with
Experiment 304 A.
2. Load 15μl of the following samples into the sample wells of the
agarose gels.
Sample WellSample
1.
Dye Mixture
2.
DNA Sample from tube #1
3.
DNA Sample from tube #2
4.
pUC18 DNA
5
Dye Mixture
6.
DNA Sample from tube #1
7.
DNA Sample from tube #2
8.
pUC18 DNA
3. Electrophoreses until the bromophenol blue in the samples has
migrated to within 2 mm of the positive electrode end of the gel.
4. Remove the gels from the unit and measure the distance of the
xylene cyanol (blue-green) dye from the sample origin. This marker
dye should comigrate with DNA fragments that are about 2800 basepairs in length which is the approximate size of the linear pUC18
molecule. Stain and destain the gels and measure the distance of the
DNA bands from the sample origin.
MODERNBIO.COM
44
STANDARD STUDENT PROGRAM 3
Study Questions
1.
Describe the stained nucleic acid pattern you observe in lanes 2 and 3
and 6 and 7. What do these patterns represent?
2.
The SDS-Ammonium Acetate step and the RNAse in the sample buffer
serve to remove bacterial nucleic acids. Why do you think these steps
were necessary? Discuss your answer in terms of the electrophoretic
patterns you might have observed had these steps been omitted.
3.
Construct a plot of the log of the OD at 600nm vs. time on semi-log paper
and compute the generation time. Calculate the number of cells in your
culture at each time point. Assuming that each cell contains 100 pUC18
molecules, determine the number of plasmid DNA molecules present in
the culture at the beginning and at the end of the culture period.
MODERNBIO.COM
45
STANDARD STUDENT PROGRAM 3
Experiment 305
Identifying Satellite Sequences
Genetic and biochemical studies suggest that there are only about 25,000-50,000
essential genes in the eukaryotic genome. However, there is enough DNA In the
vertebrate genome to code for over one million different proteins. There are three
major reasons for this apparent discrepancy. First, most eukaryotic genes contain
introns which are Insertions of DNA that do not code for mRNA or protein. During
transcription, these genes give rise to RNA species that are precursors to mRNAs.
The intron sequences are then removed from the precursor and destroyed and the
remaining segments of the precursor are spliced together to form mRNA which
can then be translated Into protein. Second, many protein-coding genes are found
in multiple copies within the genome. For example, there are approximately 10-20
genes that code for the protein,growth hormone, in humans. Third, a large fraction
of the DNA in eukaryoties does not contain protein-coding information and these
DNA stretches are often repeated thousands or more times in the genome. Satellite
DNA is one class of these repeated DNA sequences. The two methods that are
commonly used to detect satellite DNA are described below.
Equilibrium Ultracentrifugation
A common method used in the molecular biology laboratory is equilibrium
centrifugation in a density gradient. In this procedure, macromolecules having
different densities can be separated. The procedure Is the method of choice for
separating DNA fragments that have different base compositions because such
fragments also differ in their densities.
If E.coli DNA is fragmented into 500-1000 pieces and then centrifuged in a
density gradient, the DNA forms a single, narrow zone (Figure 5-1). The reason
is that the buoyant density of DNA is proportional to the (G+C) content, and the
(G+C) content of most of the E.coli DNA fragments is similar. On the other hand,
if DNA from the crab Cancer borealis is subjected to the procedure,two bands
are observed (Figure 5-1). The major band contains the bull of the crab DNA
and has a base composition of about 58% (A+T). The minor band, which is 97%
(A+T), is the (A+T)-rich satellite DNA of the crab. This DNA is called satellite
DNA because it can be separated from the bulk of the cellular DNA as a minor
component (or “satellite”).
MODERNBIO.COM
46
STANDARD STUDENT PROGRAM 3
Figure 5-1. Equilibrium Density Gradients of E.coli and Crab DNA
Separation of Restriction Fragments by Electrophoresis
There are approximately 106 restriction sites lor the enzyme EcoR1, in the genome
of a typical vertebrate. Therefore, about 106 DNA fragments,which usually
differ In length, are generated following digestion of vertebrate DNA with this
enzyme. When this DNA is separated by electrophoresis according to size, the
large number of differently-sized fragments will appear as a background smear
on the gel lane. However,fragments derived from highly repeated sequences such
as satellite DNA will often form discrete bands. For example, the (G+C)-rich
satellite DNA from cow, which comprises about 20% of the total cow genome,
can be detected as discrete bands following the electrophoresis of EcoR1-digested
cow DNA. This experiment will be performed in today’s laboratory.
Satellite DNA is not found in bacteria but is found In the genomes of most
eukaryotic organisms. In soma vertebrates, (e.g., chickens and humans), satellite
sequences occur in very low concentrations and are difficult to detect by the
standard methods described above. In other species (e.g.,kangaroo rats), these
sequences can make up as much as 50% of the genome.
The function of satellite DNA sequences is not known. However, n is known that
most satellites,including those from cow, are not transcribed into RNA. In addition,
most satellites are localized in discrete regions along the chromosomes. For example,
many of them are found In the centromeres of each chromosome. Centromeres are
the sites of attachment of the chromosomes to the mitotic spindle. However, not all
satellite DNA sequences are found within the centromere regions and there is no
universally accepted hypothesis available lor the function of these sequences.
Objective
To identify the major satellite sequence in cow DNA and to characterize the
effects of EcoR1 on DNA from cow and chicken.
MODERNBIO.COM
47
STANDARD STUDENT PROGRAM 3
Materials
A. The solutions and materials required lor electrophoresis, sample handling
and gel staining (see Appendices 1 and 2).
B. The samples below are provided in the container marked Experiment 305:
Calf DNA: from calf thymus,
Chicken DNA: from chicken erythrocytes,
EcoR1: this restriction enzyme should be made up as described in
Appendix 2 and stored in the refrigerator before use, Electrophoresis
sample buffer,
Dye mixture: A mixture of xylene cyanol, bromophenol blue, and
orange-G.
Materials Not Provided
Water bath for tube incubation maintained at 37°C.
Procedure
The experiment was designed for 8 students working individually or 16 students
working in teams of two.
A. Preparation of DNA samples
1. Number three small tubes 1-3 with a waterproof marking pen.
2. Place 10μl of distilled water into tube 1 and 10μl of the EcoR1buffer solution into tubes 2 and 3.
3. Add 5μl of calf DNA to tubes 1 and 2 and 5μl of chicken DNA
to tube 3. Tap the tubes with the tip of your index finger to mix
the solution. Incubate the tubes for 50 minutes at 37°C. During this
incubation, prepare the gels as described below.
4. Add 5μl of electrophoresis sample buffer to each tube and
electrophorese the samples as described below.
B. Electrophoresis
1. Prepare 1.2% agarose gels as described in Section IV and load 15μl
of the following samples from the above section into the sample
wells.
Sample Well Number Sample
1
DNA Sample from Tube 1
2
DNA Sample from Tube 2
3
DNA Sample from Tube 3
4
Dye Mixture
5
DNA Sample from Tube 1
6
DNA Sample from Tube 2
7
DNA Sample from Tube 3
8
Dye Mixture
MODERNBIO.COM
48
STANDARD STUDENT PROGRAM 3
2. Seal the wells with agarose and electrophorese until the bromophenol
blue in the sample has migrated to within 1 em of the positive
electrode end of the gel.
3. Remove the gels from the unit and mark the position of the xylene
cyanol(blue­green dye) and bromophenol blue (purple-blue dye) in
the dye mixture with a razor blade or similar sharp object. Stain and
destain the gels as described in Section IV and measure the distance
of the DNA bands (in em) from the sample origin.
Study Questions
1.
The major satellite band from cow is 1400 base-pairs In length. Identify
this band on your gel. Note that the band migrates between the xylene
cyanol and bromophenol blue. Can you identify other discrete bands In
cow or chicken DNA?
2.
The 1400 base-pair satellite band can be recovered from the agarose gel
after electrophoresis. Outline the steps that you would follow to amplify
this satellite DNA in E.coli
MODERNBIO.COM
49
STANDARD STUDENT PROGRAM 3
Experiment 306
The Nucleosome Structure of Chromatin
A. General Studies
In the nucleus of a typical eukaryotic cell, Individual chromosomes can be
Identified during cell division. Each chromosome probably contains a single, very
long molecule of DNA folded In a highly ordered fashion. The chromosome also
contains chromosomal proteins that are Involved In this folding process.
A eukaryotic cell spends most of the life cycle in Interphase, which Is the stage
between cell divisions. In Interphase, the chromosomes are highly diffuse and
the chromosomal material Is called chromatin. Chromatin contains the general
compo­nents listed in Table 6-1.
Table 6-1. Chromatin Components
ComponentsRelative Amount by Weight
DNA 1
RNA0.5-3
Histone: Basic Proteins
1
Nonhistone Proteins
0.5-2
In addition to DNA, the chromatin contains RNA as well as chromosomal
proteins in two types. The histone chromosomal proteins are basic in nature and
are associated with the acidic DNA backbone in the chromatin. There are 5 major
histones in eukaryotes (histones H1, H2A, H2B, H3 and H4), four of which are the
most highly conserved proteins known. With few exceptions, the same 5 histone
proteins are found In all cell types in both plants and animals. The nonhistone
chromosomal protein group contains at least 500 different proteins and is thought
to play a variety of roles including the regulation of gene activity.
If the DNA double helix from a single chromosome were to be stretched out, l
would be about 4cm long. The histone proteins are the major elements responsible
for packing this long DNA molecule Into chromosomes and chromatin. There are
several levels of DNA packaging, each of which serves to condense the DNA In
an ordered manner. The primary, or lowest, level of compaction occurs when the
DNA double helix is wrapped around histones into particles called nucleosomes
(Figure 6-1). In the nucleosome, 145 base pairs of DNA are folded about 1.7 times
around eight histone molecules (2 each of histone H2A, H2B, H3 and H4). The
DNA between adjacent nucleosomes, called the nuclaosomal linker, is thought to
be associated wh a single molecule of histone H1. Each linker segment usually
contains between 25-60 base pairs of DNA. More than 95% of the DNA in all
eukaryotic cells is packaged in this manner, and a single human cell contains
MODERNBIO.COM
50
STANDARD STUDENT PROGRAM 3
about 3x107 nucleo­somes. An average gene, which contains about 10,000 base
pairs of DNA, is associated with about 50 nucleosomes, since there is about 1
nucleosome for every 200 base pairs of DNA in chromatin.
Figure 6-1. The Nucleosome Structure of Chromatin
DNA is periodically folded around cores of histone particles to form nucleosomes
There are two major lines of evidence for the nucleosome model of chromatin.
The first comes from studies of the unfolded chromatin fiber observed with the
electron microscope (Figure 6-2). When the nucleus of essentially any eukaryotic
cell is broken open and its chromatin content examined, electron microscopy
reveals a “beads-on-a-string” structure. Each nucleosome “bead” Is a disc-shaped
particle with a diameter of about 10 nanometers.
Figure 6-2. Electron Micrograph of Chromatin Strands
MODERNBIO.COM
51
STANDARD STUDENT PROGRAM 3
Additional evidence for the nucleosome model of chromatin is derived from
nuclease digestion studies. A nuclease is an enzyme that breaks down DNA or
RNA. Micrococcal nuclease is a bacterial enzyme that is often used for chromatin
studies. This enzyme, unlike restriction nucleases, does not recognize a specific
DNA sequence, but cleaves DNA that is not protected from digestion by proteins
such as nucleosomal core histones. When nuclei are Isolated and Incubated
with this nuclease, the enzyme crosses the nuclear membrane and cleaves the
nucleosomal linker DNA but not the nucleosomal core DNA. If large amounts
of this enzyme are added, essentially all of the linker regions are broken, but the
nucleosomal cores remain Intact. When the DNA isolated from these nuclei Is
electrophoresed on agarose gels, a single 145-base-pair band derived from these
nucleosomal core particles is observed. If, in contrast, the nuclei are incubated
with low amounts of micrococcal nuclease,only some of the linkers are cleaved.
As a result, the DNA will exhibit ladder pattern on the gel, with the length of each
band representing a multiple of the nucleosomal unit of about 200 base-pairs.
The sequential breakdown of chromatin by micrococcal nuclease is illustrated in
Figure 6-3. The analysis of DNA from micrococcal-nuclease-digested nuclei Is
diagrammed In Figure 6-4.
Figure 6-3. Sequential Breakdown of Chromatin by Micrococcal Nuclease
MODERNBIO.COM
52
STANDARD STUDENT PROGRAM 3
Figure 6-4. Analysts of DNA from Micrococcal-Nuclease-Digested Nuclei
Objective
To examine DNA obtained from micrococcal-nuclease-digested nuclei. The
digested DNA Is provided in the Experiment Package. The nuclei were obtained
from calf thymus. In Parts B and C (below) you will analyze nucleosomal DNA
in nuclei that you prepare.
Materials
A. The solutions and materials required for electrophoresis, sample handling
and gel staining (see Appendices 1 and 2).
B. The samples below are provided in the container marked Experiment
306:
Dye mixture: A mixture of xylene cyanol, bromophenol blue, and
orange-G.
Nucleosome 1,
Nucleosome 2,
Nucleosome 3.
These three DNA samples ware prepared from calf thymus nuclei,as
outlined in Figure 6-4 using low (nudeosome 1), medium (nucleosome
2) and high (nucleo­some 3) concentrations of micrococcal nuclease.
MODERNBIO.COM
53
STANDARD STUDENT PROGRAM 3
Procedure
1.
Prepare agarose gels as described In Section IV and Experiment 301 and
load 15μl of the following samples Into the sample wells.
Sample Well Number Sample
1
DNA Sample from Nucleosome Tube 1
2
DNA Sample from Nucleosome Tube 2
3
DNA Sample from Nucleosome Tube 3
4
Dye Mixture
5
DNA Sample from Nucleosome Tube 1
6
DNA Sample from Nucleosome Tube 2
7
DNA Sample from Nucleosome Tube 3
8
Dye Mixture
2.
Seal the wells with agarose and electrophorese until the orange-G has
migrated to within 2cm of the positive electrode end of the gel.
3.
Remove the gels from the unit, mark the positions of the marker dyes
with a razor blade and slain and destain the gels as described In Section
IV.
4.
Measure the distance of the DNA bands (in em) from the sample origin
and note the positions of the DNA bands.
Study Questions
1.
Identify the DNA bands derived from one nucleosome
(mononucleosomes), two nucleosomes (dinucleosomes) and three
nucleosomes (trinucleosomes).
2.
Note that the mononucleosomes bands become slightly smaller with
Increasing micrococcal nuclease digestion. Why? (Hint: see Figure 6-3).
B. Structure or Plant Chromatin
The wheat kernel consists of three parts: (1) The embryo or germ that produces
the new plant; (2) the starchy endosperm, which serves as a food source for the
embryo; and (3) covering layers which protect the grain. The average wheat
grain is composed of about 85% endosperm, 13% covering layers and 2% germ,
by weight. The endosperm Is the raw material for flour production,and the
wheat germ Is used in food products as a source of vitamins. Wheat germ Is
also frequently used In the molecular biology laboratory as a source of protein
synthesizing extracts as well as plant cell proteins and nucleic acids. In today’s
laboratory, you will analyze the structure of wheat germ chromatin.
MODERNBIO.COM
54
STANDARD STUDENT PROGRAM 3
Objective
To Isolate nuclei from wheat germ,to digest the nuclear chromatin with micrococ­
cal nuclease and to analyze the chromatin DNA by electrophoresis. This exercise
requires two sessions of two hours or one session lasting three to four hours.
Materials
A. The solutions and materials required for electrophoresis,sample handling
and gel staining (see Appendices 1 and 2).
B. The samples below are provided In the container marked Experiment
306: Nuclear Incubation buffer: This buffer should be made up before
the laboratory session as described In Appendix 2 and stored in the
refrigerator,
Sample buffer,
Dye mixture: A mixture of xylene cyanol, bromophenol blue, and orange-G,
Wheat germ,
Micrococcal nuclease: The enzyme should be made up before the laboratory
session as described In Appendix 2 and stored In the refrigerator,
Cheese cloth,
Graduated pipets.
Materials Not Provided
A water bath for tube Incubation maintained at 37°C.
Ice bath.
Refrigerated centrifuge and small (10ml capacity) centrifuge tubes (A small
clinical centrifuge placed in the refrigerator is suitable for this exercise).
Four small beakers.
Procedure - Laboratory Session 1
Preparing the DNA Samples
This procedure is designed such that the preparation of the samples will be
performed by four groups of students. Unless otherwise indicated, all steps should
be performed at 4’C.
1. Place 3g of wheat germ into a beaker, add 10 ml of pre-cooled nuclear
incubation buffer and stir for a few minutes The buffer contains the
detergent NP-40, which will disrupt the plasma membrane and cause
cell lysis.
2.
Filter the material through three layers of cheese cloth into a centrifuge
tube and squeeze the cheese cloth to ensure that all of the liquid is
collected in the tube. Centrifuge the tube at 4’C for 5 minutes at 5001000 x G to pellet the nuclei.
3. Carefully pour off and discard the supernatant. Add 5ml of nuclear
incubation buffer to the nuclear pellet and mix well.
MODERNBIO.COM
55
STANDARD STUDENT PROGRAM 3
4.
Centrifuge at 4°C for 5 minutes at 500-1000 x G.
5. Carefully pour off and discard the supernatant. Add 1ml of nuclear
incubation buffer to the nuclear pellet using a graduated pipet provided
with the Chemical Package. Draw the nuclei into the pipet several times
until a homogeneous suspension is formed.
6.
Divide the nuclear suspension into 3 approximately equal portions
containing about 0.3 ml each and place each portion into a small
centrifuge tube located in an ice bath.
7. Label the tubes 1-3 with a waterproof marking pen.
8.
Place 5μl of micrococcal nuclease into tube 2 and 30μl of the enzyme
into tube 3.
9.
Transfer the tubes to a 37°C incubator.
10.
After 10 minutes, remove the tubes from the incubator and centrifuge for
5 minutes at 500-1000 x G. Small chromatin fragments produced by the
action of micrococcal nuclease, should be liberated into the supernatant
fractions.
11.
Carefully pour the supernatant from each tube into clean tubes labeled
1-3. These supernatants should contain soluble chromatin particles
(mono-, di-, and tri­nucleosomes).
12.
Each student or student pair should transfer 40μl of these solutions into
3 small tubes provided with the Chemical Package.
13. Add 20μl of sample buffer to each tube and incubate the tubes for 15
minutes at 37°C. This sample buffer contains RNase which serves to
destroy RNA in the samples. The sample buffer also contains the
detergent, Sarkosyl, which will disrupt the histone-DNA interactions,
liberating the DNA in a form that can be analyzed by electrophoresis.
14. The samples can then be stored at -20’C until the next laboratory session
or analyzed by electrophoresis as described below.
MODERNBIO.COM
56
STANDARD STUDENT PROGRAM 3
Laboratory Session 2
Electrophoretic Analysis of DNA
1.
Prepare 1.5% agarose gels as described in Section IV.
2.
Load 15μl of each sample into the sample wells as indicated below.
Sample Well Number Sample
1
Tube 1
2
Tube 2
3
Tube 3
4
Dye Mixture
5
Tube 1
6
Tube 2
7
Tube 3
8
Dye Mixture
3.
Electrophorese until the orange-G has migrated to within 2 em of the
positive electrode end of the gel.
4. Remove the gels from the unit mark the positions of the marker dyes, and
stain and destain the gels as described in Section IV.
C. Chromatin DNA Breakdown by Cell Nucleases (Optional)
The nuclei of most, if not all, eukaryotic cells contain a number of different
nucleases. The physiological role of most of these enzymes is pearly understood.
One such enzyme, or group of enzymes, is similar to micrococcal nuclease in
that it attacks linker DNA between nucleosomes. The enzyme in rat liver nuclei
has been characterized extensively. When liver nuclei are incubated at 37°C for
various times up to 30 minutes, the progressive breakdown of chromatin occurs.
DNA from this chromatin will exhibit the typical nucleosome “ladder” appearance
upon electro­phoresis. Although nuclei isolated from vertebrate liver contain large
amounts of this enzyme, the enzyme activity is also found in nuclei from other
animal and plant tissues.
Objective
To study the sequential breakdown of DNA In chromatin during Incubation of
Isolated nuclei. This Is an optional exercise that requires two laboratory sessions.
In the first session, you will, prepare nuclease-digested chromatin. In the second
session, you will use electrophoresis to analyze the DNA from this chromatin.
This experiment Is optional because a tissue must be provided for the preparation
of nuclei.
MODERNBIO.COM
57
STANDARD STUDENT PROGRAM 3
Materials
A. The solutions and materials required for electrophoresis,sample handling
and gel staining (see Appendices 1 and 2).
B. The samples below are provided in the container marked Experiment
306:
Nuclear Incubation Buffer: This buffer should be made up before
the laboratory session as described In Appendix 2,and stored in the
refrigerator,
Sample Buffer,
Dye Mixture: A mixture of xylene cyanol, bromophenol blue, and
orange-G.
Materials Not Provided
Water Bath for Tube Incubation Maintained at 37 •c
Ice Bath
Refrigerated centrifuge and small (3-10ml capacity) centrifuge tubes. A small
clinical centrifuge placed in the refrigerator is suitable for this exercise.
A device for tissue disruption. A commercial glass homogenizer is preferred.
However, a pre-cooled mortar with pestle is a suitable alternative.
Four grams of tissue. Although nuclei from vertebrate liver contain large
quantities of the desired enzyme activity, nuclei from other tissue sources
can also be used. Possible tissues include:
a. Calf or chicken liver obtained from animals recently butchered at you
local slaughter house.
b. Rodent liver, frog liver.
There are sufficient materials provided with the Chemical Package for
you to perform this exercise with more than one tissue type.
Procedure - Laboratory Session 1
Preparing the DNA Samples
The procedure Is designed so that the preparation of the samples with be performed
by the entire class as a group. However, there is sufficient buffer provided lor
4 groups of students to conduct the experiment. Unless otherwise indicated, an
steps should be performed at 4°C.
1.
Place 4g of tissue into a pre-cooled mortar and add t 0 ml of pre-cooled
nuclear incubation buffer.
2.
Cut the tissue into small (about 1mm2) sections with scissors and grind
the sections with the pestle until a homogeneous suspension is formed.
The mechani­cal action of the pestle as well as the chemical action of the
detergent, NP-40, which is present In the buffer, will serve to disrupt the
plasma membranes and to cause cell lysis.
MODERNBIO.COM
58
STANDARD STUDENT PROGRAM 3
3.
Transfer this cell homogenate to a centrifuge tube and centrifuge at 4°C
for 5 minutes at 500-1000 x G to pellet the nuclei.
4.
Carefully pour off and discard the supernatant. Add 3ml of nuclear
digestion buffer to the nuclear pellet and mix well.
5.
Divide the nuclear suspension into 3 approximately equal portions and
place each portion into a small centrifuge tube located in an ice bath.
6.
Label the tubes 1-3 with a waterproof marking pen.
7.
Transfer tubes 2 and 3 from the ice bath to a 37°C incubator.
8. After 5 minutes, remove tube 2 from the incubator and return it to the ice
bath.
9. After 30 minutes,remove tube 3 from the incubator and return it to the
ice bath.
10. Centrifuge tubes 1-3 for 5 minutes at 500-1000 x G. Small chromatin
fragments produced by the action of the nuclear enzymes, should be
liberated Into the supernatant fractions.
11.
Carefully pour the supernatant from each tube into clean tubes labeled
1-3. These supernatants should contain soluble chromatin particles
(mono-,di-, and tri- nucleosomes).
12.
Each student or student pair should transfer 40μl of these solutions into
3 small tubes provided with the Chemical Package.
13. Add 20μl of sample buffer to each tube and Incubate the tubes for 15
minutes at 37°C. This sample buffer contains RNase which serves to
destroy RNA in the samples. The sample buffer also contains the
detergent, Sarkosyl,which with disrupt the histone-DNA Interactions,
liberating the DNA In a form that can be analyzed by electrophoresis.
14.
The samples can then be stored at -20°C until the next laboratory session.
MODERNBIO.COM
59
STANDARD STUDENT PROGRAM 3
Laboratory Session 2
Electrophoretic Analysis of DNA
1.
Prepare 1.5% agarose gels as described in Section IV.
2.
Load 15μl of each sample into the sample wens as indicated below.
Sample Well Number Sample
1
Tube l
2
Tube 2
3
Tube 3
4
Dye Mixture
5
Tube 1
6
Tube 2
7
Tube 3
8
Dye Mixture
3.
Electrophorese until the orange-G has migrated to within 2 em of the
positive electrode end of the gel.
4.
Remove the gels from the unit, mark the positions of the marker dyes and
stain and destain the gels as described in Section IV.
Suggested Reading and References for Part B
Laboratory Methods:
Hames, B.D. and Richwood, D. Gel Electrophoresis of Nucleic Acids. Oxford:
IAL Press, 1982.
Hackel, P.B., Fuchs, J.A. and Messing, J.W. An Introduction to Recombinant DNA
Techniques. Menlo Park: The Benjamin/Cummings Publishing Co., 1984.
Manialis, T., Fritsch, E.F. and Sambrook, J. Molecular Cloning, A Laboratory
Manual. Cold Spring Harbor Laboratory Publications, 1982.
General References:
Alberts, B., Bray,L., lewis,J., Ra1f, M., Roberts,A. and Watson. Molecular
Biology of the Cell. New York and London: Garland, 1983.
Freifelder, D. Molecular Biology. Portola: Science Books Int., 1983.
Metzler, D.E. Biochemistry. New York: Academic Press, 1977.
Lewin, B. Genes. John Wiley and Sons, Inc.: New York, 1983.
MODERNBIO.COM
60
STANDARD STUDENT PROGRAM 3
Appendix
The Metric System
LengthEnglish Equivalent
1 meter (m)
= 100 centimeters
1 centimeter (em)
= 10 millimeters (mm)
2.5 centimeter
= 0.3937 inches
= 1.0 inch
Volume
1 liter (1)
1 milliliter (ml)
1 microliter (μl)
= 1000 milliliter
= 1000 microliters
= 10-3ml
= 1.06 quarts
= 1000 grams
= 10-3kg
= 10-3g
= 10-6g
= 10-9g
= 2.2046 pounds
Mass
1 kilogram (kg)
1 gram (g)
1 milligram (mg)
1 microgram (μg)
1 nanogram (ng)
MODERNBIO.COM
61
STANDARD STUDENT PROGRAM 3
MODERN BIOLOGY
3700 East 700 South
Lafayette, IN 47909
Call Toll Free: 1-800-733-6544
FAX: 1-765-523-3397