AP Biology Lab 6: Molecular Biology Lab
Objectives: At the end of this lab students should:
 Understand the principles and practice of agarose gel electrophoresis
 Demonstrate the separation of molecules based on charge
 Examine the rate of separation of molecules based on size
 Determine which component is missing in a mixture of several dyes
 Investigate transformation as a mechanism of genetic exchange
 Create competent cells by chemically and thermally treating E. coli cells
 Insert a plasmid containing antibiotic-resistant genes into competent E. coli cells
 Screen the transformed cells to determine which have been genetically altered
 Calculate the efficiency of the transformation reaction
 Using models, be able to design a recombinant DNA molecule
 Become familiar with common restriction enzymes used to build recombinant DNA molecules
Background
Concepts in molecular biology are often difficult to grasp. When trying to visualize the invisible
world of the molecule, students are too often confronted by abstract theory.
But like everything else in the field of biotechnology, things are changing fast. In 1950, a scientist
named Oliver Smithies invented gel electrophoresis. Gel electrophoresis is an advancement in
biotechnology that actually allows students to separate and visualize DNA, RNA, proteins, and other
polypeptides and nucleotide sequences. Genes control all life processes. Therefore, the separation of DNA
and gene products provides the potential to examine all of life’s processes.
What exactly is gel electrophoresis? It is a separation technology that is the sum of its parts: gel, a
substrate (think of gelatin); electro, referring to electricity; and phoresis, from the Greek verb phoros,
meaning “to carry across.”
Gel electrophoresis, then, refers to the technique in which molecules are
forced across a gel by an electrical current; activated electrodes at either end of the gel provide the driving
force. Arne Tiselius, a Swedish biochemist, won the Nobel Prize for chemistry in 1940 for his work with
electrophoresis.
Exploring the “Bar Codes of Life”
By using gel electrophoresis, students can gain a unique perspective on DNA structure and
function. Nucleic acids, both DNA and RNA, can be separated on the basis of size and charge by means of
electrophoresis to identify structural forms of plasmid DNA, as well as determine the size of DNA
fragments of genes. Furthermore, the size of unknown DNA fragments can be determined be constructing a
standard curve using the migration distances and sizes of a known DNA marker.
Band patterns of separated DNA on a gel resemble a “bar code,” that familiar pattern used to
identify consumer products. Each band is a unique “signature” revealing a recorded and identifiable DNA
fragment. (See below)
Scientists and clinicians regularly use scanning instruments to glean vital information from
samples separated by electrophoresis. This information is critical in a range of applications: pinpointing
cancer types, identifying diseased tissue, characterizing genetic dysfunctions, assessing coronary risk, and
even reading and compiling nucleotide sequences to map the genomes of many life-forms, including our
own.
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How does the technique work?
Separation of large (macro) molecules depends on two forces: charge and mass. When a
biological sample is mixed in a buffer solution and applied to a gel, these two forces act in concert. The
electrical current from one electrode repels the molecules, while the other electrode attracts them, and the
frictional force of the gel material acts as a “molecular sieve,” separating the molecules by size and charge.
Negatively charged molecules will migrate toward the positive pole, while positively charged molecules
will migrate toward the negative pole. The net negative charge of the phosphate backbone results in the
DNA fragments having a slightly negative charge and thus will always migrate toward the positive pole.
The material is roughly analogous to that of a thoroughly wetted sponge, except that in this case, the
“pores” are submicroscopic. During electrophoresis, macromolecules are forced to migrate through the
“pores,” away from the closest electrode and toward the farther electrode when electrical current is applied.
After staining, the separated macromolecules in each lane can be seen; they appear as a series of bands
spread from one end of the gel to the other.
What’s in a Gel
There are two basic types of materials used to make gels: agarose and polyacrylamide.
Polyacrylamide is a material similar to that found in soft contact lenses and is primarily used to separate
proteins. Agarose is a natural colloid extracted from seaweed. In the plant, it helps to resist dessication
when exposed to air for extended periods of time Agarose gels have a very large “pore” size and are used
primarily to separate very large molecules such as DNA with a molecular mass greater than 2,000 kdal.
Kdal is the abbreviation for kilodalton or 1,000 daltons. A Dalton is a unit of molecular weight very nearly
equivalent to the mass of a hydrogen atom, or 1.000 on the atomic mass scale. When agarose is heated to
about 90°C it melts, but solidifies again when cooled below 45°C. During the solidification process,
agarose forms a matrix of microscopic pores. The size of these pores depends on the concentration of
agarose used. Typically the concentration varies from 0.5% to 2.0%. The lower the concentration, the
larger the pore size of the gel and the larger the nucleic acid fragments that can be separated. During
electrophoresis, DNA molecules wind through the pores in the gel, similar to trying to pass a thread
through a stack of gauze pads. As the pore size decreases (by increasing agarose concentration), it is harder
for longer DNA fragments to orient properly. Smaller DNA fragments can thread their way through the
pores, thus migrating faster.
Separation Techniques
While there are other separation techniques, the two most prominent in biotechnology are
chromatography and gel electrophoresis. Both processes can simultaneously separate many common
molecular entities from molecular mixes. Chromatography works best for small molecules (leaf pigments,
for example), or for large batch-processing requirements. Electrophoresis is primarily used as an analytical
tool to separate large macromolecules ranging in size from 20 to 2,000 kdal. This technique can separate
hundreds, even thousands of macromolecules from one another while using less than a thousandth of a
gram of sample material. Chromatography typically uses larger samples and can, at best, separate about
twenty common molecular groups at one time.
Lab Activity Prep: Casting the Agarose Gel
1.
2.
3.
4.
5.
Place the gel casting tray on a flat surface and tape the ends of the trap with masking tape, which
will seal the ends of the tray.
Insert the gel comb, using the 8-well side, into the center slots of the tray.
Pour approximately 25 mL of the melted agarose into the tray, until it reaches a depth of about 2
mm
Allow the gel to solidify for approximately 20-30 minutes. Do not disturb the gel tray or comb.
When the agarose has solidified, it will turn opaque.
After the gel has solidified, carefully remove the comb from the gel. Remove the tape from the
ends of the dams. Carefully place your gel in a zip loc baggie labeled with your group’s id and
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period, along with a small amount (1-2 mL) of TBE running buffer. Await further instructions for
tomorrow’s procedure.
Important Safety Notes:
6. The power supply running the electrophoresis unites produces a high enough voltage to cause
severe electrical shock if handled improperly. For safe operation, follow all directions and
precautions.
7. Do not come in personal contact with or allow metal or any conductive material to come in contact
with the reservoir buffer or the electrophoretic cell while the power supply is on.
8. You should wear protective gear, such as safety goggles and aprons, while loading and running
gels.
9. ALWAYS call your teacher over to examine the apparatus before you turn it on, and apply power
to the gel.
Part A: A Process to Dye for: Gel Electrophoresis
Materials Needed Per Group
Dyes to be tested:
Malachite Green
Orange G
Safranin O
Alizarin Red S
m-Cresol Purple
Agarose gel (2.0%) on gel tray
TBE running buffer 1X, 350 mL
Micropipets, 10µL
Metric ruler
Calculator
Electrophoresis chamber
Power Supply
Procedure
Scenario
You are a lab technician mixing up a batch of stain (a mixture of five dyes) for one of the researchers in the
lab. You had started making a large batch when you were interrupted during the procedure. When you
returned to the stain, you could not remember if you had added all of the components. Rather than waste
the whole batch of stain and start over, you will employ agarose electrophoresis to determine if all of the
dyes are present in the mixture.
Loading and Running a Gel
(You may choose to use the dry, or submarine methods to load your dyes. The following (steps 1-5) describes the dry method. The
submarine method follows that.)
The Dry Method
Place the tray with the gel on the lab bench. Here is the order you will load the gel:
a. Lane #1: Malachite Green
b. Lane #2: Orange G
c. Lane #3: Safranin O
d. Lane #4: Alizarin Red S
e. Lane #5: m-Cresol Purple
f. Lane #6: Unknown (mixture)
2. Use a micropipette to load 10µL of each dye sample into the corresponding lane. Do not pierce the
bottom of the wells with the micropipette tip. Do not overload wells.
3. Place the tray with the loaded gel in the center of the electrophoresis chamber.
1.
When filling the electrophoresis chamber with buffer, avoid pouring the solution directly
onto the gel and be sure to pour it VERY SLOWLY. If the buffer is poured too quickly, it
may wash away the dye samples.
3
4.
5.
Add approximately 350 mL of 1X TBE running buffer to the chamber: SLOWLY pour buffer
from a beaker into one side of the chamber until the buffer is level with the top of the gel. Add
buffer to the other side of the chamber until the buffer is level with the top of the gel. Continue to
SLOWLY add buffer until the level is approximately 2-3 mm above the top of the gel.
Continue with number 4 below (submarine method)
The Submarine Method (use the same dye order and arrangement)
1. Place the gel on the gel tray, in the center of the electrophoresis chamber.
2. Add 1X TBE running buffer to the chamber: SLOWLY pour buffer from a beaker into one side of
the chamber until the buffer is level with the top of the gel. Add buffer to the other side of the
chamber until the buffer is level with the top of the gel. Continue to SLOWLY add buffer until
the level is approximately 2-3 mm above the top of the gel.
Loading the gel may be easier to accomplish over a dark background, such as the lab table
provides. Placing a piece of black paper under the electrophoresis chamber will increase
the contrast and make the wells easier to see and fill, if no dark lab table is available.
3.
4.
5.
Use a micropipette to load 10µL of each dye sample into the corresponding lane. Do not pierce
the bottom of the wells with the micropipette tip. Do not overload wells.
Making sure that the cover, as well as the female jacks and the plugs, are completely dry, slide the
cover onto the electrophoresis chamber. Wipe off any spills on or around the apparatus before
proceeding to the next step.
Making sure that the patch cords attached to the cover are completely dry, connect the red patch
cord to the red electrode terminal on the power supply. Connect the black patch cord to the black
electrode on the power supply.
***Have your teacher check the connections before proceeding to the next step.***
Running the Gel:
1.
Plug in the power supply and set it to the desired voltage.
It is recommended that you set the power supply between 75-150V. The system will run at a
lower voltage but will increase the running time of the gels. (The higher the voltage, the
faster the running time)
2.
3.
4.
5.
6.
7.
Turn on the power supply. The red power light will illuminate, and bubbles will
form along the platinum electrodes in the chamber.
Observe the migration of the samples across the gel toward the electrodes. Turn
off the power when the fastest-moving sample has neared the end of the gel.
Unplug the power supply.
Wait approximately 10 seconds, and then disconnect the patch cords from the
power supply. Remove the cover from the electrophoresis chamber.
Carefully remove the gel, on the casting tray, from the electrophoresis chamber
The dye samples are not fixed on the gel and will continue to diffuse over several
hours, making the bands less distinct. You must take measurements on the same
day the dye was cast. If you have the means, take a photograph to place in your
quadrille of your gel.
Sketch your gels, and take measurements on the dye samples. Record results in a
data table similar to the one in the analysis section.
The dry method has the advantage of ease of loading the gel. It is easy to see the wells when they are not covered
with buffer solution. The disadvantage is that one must exercise extreme caution when adding the running buffer
to the chamber so as not to wash the samples from the wells.
The submarine method has the advantage in that the buffer is already added, and you don’t have to worry about
the dyes being washed out of the wells. The disadvantage is that it harder to see the wells underneath the buffer.
Did you know?
There are several factors that can affect the speed of migration of a molecule through an agarose gel. The most
important are the size of the molecule (how physically large it is) and its isoelectric point (the pH at which the
molecule has no net charge). The concentration of the agarose is also a key factor.
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Molecular Genetics Lab 6: Part B: Pop Bead Recombinant Plasmid Model
Background Information
The activity described in this material was designed to help you understand more about making
recombinant DNA molecules. You will be using models of DNA molecules made from plastic pop beads.
As you build your DNA models, there are a couple of features of DNA that you need to have in mind.
1.
DNA is a double-stranded molecule. Each strand is made up of a sugar-phosphate backbone with
nitrogenous bases attached to the sugars. The two strands associate because the bases form
specific pairs: adenine with thymine and cytosine with guanine.
2. The two sugar-phosphate backbones of DNA have opposite orientations. This is possible because
the individual sugar-phosphate backbones have unequal ends. One end is called the 5’
(pronounced 5 prime) end, while the other is the 3’ end. These ends are chemically different. In a
double-stranded molecule, one backbone is arranged 5’ to 3’ from left to right, and the other is
arranged 3’ to 5’ from left to right.
You will use the pop beads to build double-stranded DNA molecules. The protruding connector on the
beads will represent the 5’ end of the backbone, and the hole will represent the 3’ end. You will see how 5’
and 3’ ends are important in building recombinant DNA as you work with the models.
Restriction Enzymes
Restriction enzymes are a group of enzymes that each recognize a specific sequence of bases along a DNA
molecule. When an individual restriction enzyme finds its recognition sequence, it binds to the DNA
molecule and cuts it in a characteristic manner. For example, the enzyme HindIII (pronounced hin dee 3)
recognizes the double-stranded DNA sequence shown below and cuts each of the DNA backbones between
the two adenines:
5’-AAGCTT-3’
3’-TTCGAA-5’
5’-A
3’-TTCGA
+
AGCTT-3’
A-5’
Look carefully at the HindIII recognition site (also called a restriction site). If you read the base sequence
from 5’ to 3’ on each strand (the bottom strand will be read from right to left), you will see that the
sequences are identical. The fancy word for this kind of base sequence is palindrome. The recognition
sites of nearly all restriction enzymes used in gene cloning are palindromes. Here are the recognition
sequences of two more restriction enzymes. Both sequences are palindromes; check this for yourself.
Enzyme BamHI
5’-GGATCC-3’
3’-CCTAGG-5’
Enzyme EcoRI
5’-GAATTC-3’
3’-CTTAAG-5’
When HindIII cuts its recognition site, it makes a break in a double-stranded molecule as shown below (in
this example, the X’s stand for any DNA base):
5’-XXXXXAAGCTTXXX-3’
3’-XXXXXTTCGAAXXX-5’
5’-XXXXXA-3’
3’-XXXXXTTCGA-5’
+
5’-AGCTTXXX-3’
3’AXXX-5’
Here is a representation of how the enzyme BamHI cuts at its recognition sequence:
5’-XXXXXGGATCCXXX-3’
3’-XXXXXCCTAGGXXX-5’
5’-XXXXXG-3’
3’-XXXXXCCTAG-5’
+
5’-GATCCXXX-3’
3’-GXXX-5’
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DNA Ligase
Look at the products from the HindIII digestion above. If you imagine putting the two pieces back
together, you can see that the two single-stranded regions of the ends would form correct DNA base pairs
with each other (adenine with thymine; cytosine with guanine). All that would be required to connect the
two products back into one DNA molecule would be a mechanism to re-form the chemical bonds that the
restriction enzyme broke in the backbones.
Cells contain an enzyme that has just this ‘re-gluing” ability. The enzyme is called DNA ligase. If two
pieces of DNA like the fragments shown above can line up with correctly paired bases, DNA ligase can
form bonds between sugars and phosphates to seal up the chemical backbone of the molecule. Would the
two products of BamHI digestion also line up correctly for DNA ligase to glue the pieces together?
When ligase joins two DNA fragments into one, it must always join a 5’ end to a 3’ end. Likewise, when
you join the pop beads together, you must always join a protruding connector (5’ end) to a hole (3’ end).
Restriction enzymes, DNA ligase, and recombinant DNA
Restriction enzymes and DNA ligase are important tools in genetic engineering because they allow
scientists to cut and paste pieces of DNA. Together, they can be used to fuse pieces of DNA that
previously belonged to two different molecules into one new molecule. Here’s an example to show how.
Suppose you had DNA molecule X and DNA molecule Y. Both contain a HindIII site somewhere in the
middle. Assume you use the enzyme HindIII to cut both molecules, as shown below (the X’s and Y’s stand
for any DNA bases again):
XXXXXAAGCTTXXX
XXXXXTTCGAAXXX
XXXXXA
XXXXXTTCGA
AGCTTXXX
AXXX
YYYYYAAGCTTYYY
YYYYYTTCGAAYYY
YYYYYA
YYYYYTTCGA
AGCTTYYY
AYYY
Now look carefully. If you line up the right-hand “X” fragment with the left-hand “Y” fragment, the two
single-stranded regions will make perfectly good base pairs. Since the two single-stranded regions form
complementary base pairs, they are often called complementary ends. Add DNA ligase, and you will get a
molecule in which half comes from molecule X, and the other half from molecule Y:
YYYYYA
+
YYYYYTTCGA
AGCTTXXX
AXXX
YYYYYAAGCTTXXX
YYYYYTTCGAAXXX
Anything made up of pieces that originally came from different sources can be called a recombinant
“thing”. Therefore, the molecule you just made is a recombinant DNA molecule.
Here’s a very important point for understanding how to make recombinant DNA molecules. Fragments of
and DNA molecule cut with HindIII could be ligated to any other DNA molecule that was cut with HindIII,
because their single-stranded ends are complementary. Likewise, and DNA fragments generated by
BamHI digestion could be joined by DNA ligase, because their single-stranded ends would also be
complementary. However, if two DNA molecules had ends generated by BamHI and HindIII, respectively,
the BamHI ends would not be complementary to the HindIII ends, and so DNA ligase would not join them.
When scientists make recombinant DNA molecules, they usually purify the starting molecules and work
with them in test tubes. They digest the starting molecules with restriction enzymes, mix the resulting
pieces together, and add DNA ligase. The ligase forms bonds between fragments with complementary
ends. How does the DNA ligase know which fragments the scientist wants to paste together? The answer
to that question is, it doesn’t. Ligase is just an enzyme. All it does is join DNA backbones when two
molecules line up with correct base pairs between them. If this is the case, how does a scientist find the
molecule he or she wants out of all the possible molecules DNA ligase might form? In many cases, the
scientist uses a process called transformation to help.
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Transformation
Transformation is a method of introducing new DNA into cells by causing the calls to absorb DNA from
their environment. Some types of cells require more persuading to do this than others. The point of
transforming cells with a particular DNA molecule is to have that DNA become part of the cell’s genome.
When this happens, the cell and all its descendants will copy the molecule before every cell division, giving
the scientist a continuing source of the DNA (by recovering it from the transformed cells).
Sounds neat, doesn’t it? Make a recombinant DNA molecule, put it into a cell, and that cell makes many
copies of it for you as it divides. Of course, there are a couple of catches. Many organisms, E. coli among
them, will not maintain and copy just any old DNA molecule. A newly introduced DNA molecule must
meet some conditions to become part of E. coli’s genetic makeup. Those conditions are:
1. The DNA usually must be a circular molecule (plasmid). E. coli, like many other cells, contains
enzymes that break down linear pieces of DNA.
2. The DNA must have an origin of replication. A DNA molecule has to be copied if it is to be
passed to daughter cells. In order to be copied, a DNA molecule must have an origin of
replication. An origin of replication is a special sequence of bases where the copying (replication)
begins.
If a DNA molecule is circular and has a replication origin, it will be stable inside an E. coli cell and will be
passed down to the descendants of that cell. All of these cells will express any genes on the molecule. So,
if an E. coli cell is transformed with a circular DNA molecule containing an origin of replication and a gene
for, say, resistance to the antibiotic ampicillin drug. Of course, there is a special name for circular DNA
molecules with origins of replication. They are called plasmids. Many plasmids exist in nature; others
have been created through recombinant DNA technology, mostly by recombining pieces of natural
plasmids.
There is one other key fact about transformation you need to know to understand how it helps sort out the
products of a ligation reaction: transformation is a rare event. If you attempt to transform a large number
of E. coli cells at the same time, only a small fraction of them will take up new DNA. Each of these
transformations will usually have taken up only one DNA molecule. You can purify copies of that DNA
back out of these descendants of a single E. coli cell and examine the structure of the molecule to find out if
it is the recombinant product you wanted in the first place.
This Activity
This activity was designed to give you an idea of the number of different recombinant DNA molecules that
can be made in a single ligation reaction, and to let you think about what you might expect if you
transformed E. coli with such a mixture of molecules.
You will use plastic beads to make models of two different (but real) plasmids: pAMP and pKAN. (The
small “p” in their names stands for “plasmid”; most plasmid names start with it.) Plasmid pAMP contains a
gene for resistance to the antibiotic ampicillin. Plasmid pKAN contains a gene for resistance to the
antibiotic kanamycin.
You are going to simulate the activities of a scientist who wishes to recombine parts of these plasmids into
one new plasmid that will contain genes for resistance to both antibiotics. The tools you have to work with
are the two starting plasmids, the restriction enzymes HindIII and BamHI, DNA ligase, and a plasmid-free
E. coli strain that is sensitive to both antibiotics.
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Instructions
Use the following color code for beads, unless instructed otherwise:
Green: gene for resistance to the antibiotic kanamycin
Purple: gene for resistance to the antibiotic ampicillin
Orange: origin of replication
Yellow: HindIII restriction site
Red: BamHI restriction site
Blue: other plasmid DNA
Building the DNA models
Unless instructed otherwise, you will be working in pairs. One member of each pair will build the plasmid pAMP. The other member
of the pair will build pKAN. Make two identically colored DNA strands for each plasmid, but make sure one strand is put together 5’
to 3’ (protrusion to hole), while the other is 3’ to 5’ (hole to protrusion) as you look from left to right. Before you form circles from
the strands, line the two strands of beads up perfectly and tie the two strands together by wrapping twist ties around them every two or
three beads. Then connect the circles. An example is shown below. IN real life, DNA strands are held together by interactions
between the base pairs. They are bonded weakly with hydrogen bonds.
Bead model of a
segment of DNA
Calculating the size of the plasmid in DNA base pairs
Each bead represents approximately 200 bases of DNA, except for the restriction sites (the red and yellow beads). If you want to
calculate the approximate size of a plasmid in base pairs, count all the bead pairs except the red and yellow ones, and multiply by 200.
Restriction digestion and ligation
Each restriction site will be represented by two adjacent beads of the same color (for example, two yellow beads for a HindIII site).
When you simulate the action of the enzyme, you will separate the yellow beads on both chains by popping them apart. The yellow
beads now at the ends of your DNA molecule represents the HindIII ends. The protruding connector is the 5’ overhanging single
strand, and the hole is the 3’ recessed strand (look at the figure above. Remember that these ends are complementary to other HindIII
ends, but not BamHI ends. You will simulate the action of BamHI by separating the red beads, as above.
When you simulate the activity of DNA ligase, you may connect yellow beads (HindIII ends) to any other yellow beads, but not to any
other color (because HindIII ends are complementary to each other, as shown above). Likewise, you may connect red beads (BamHI
ends) to any other red beads, but not to any other color.
Procedure
1.
2.
3.
4.
Following the general instructions for building DNA models above, construct your parent plasmid.
Look at the diagrams to build a model of the parents pAMP or pKAN. Your partner will build the
other model. Each bead represents 200 bases of DNA, except for the restriction sites. Each
restriction site is represented by a pair of uniquely colored beads - yellow for HindIII and red for
BamHI. Each plasmid has one recognition site for HindIII and one for BamHI, a replication
origin, and the designated antibiotic resistance gene. Don’t put twist-ties between the two red or
the two yellow beads, since you’ll be separating them in the next steps.
Digest each plasmid with HindIII. Symbolize this by popping apart the two yellow bead pairs
representing the HindIII recognition site. Each plasmid should give a single linear chain of beads
with yellow at either end, representing the HindIII ends.
Now digest each plasmid with BamHI by separating the two red beads. Together, you and your
partner should now have four DNA fragments: two from each molecule. Each fragment should
have a pair of yellow beads at one end and a pair of red beads at the other. There should be a 5’
(protrusion) and a 3’ (hole) end at each end of your fragments.
Put your fragments on a desk or tabletop together with fragments made by your partner. The set
of fragments represents what you would have in a test tube if you mixed the digested plasmids
together.
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5.
Now simulate the activity of DNA ligase. DNA ligase will only re-form HindIII sites (two yellow
beads) and BamHI sites (two red beads), because only they will form complementary base pairs.
Also, bear in mind that DNA ligase will join whichever complementary ends line up correctly
first. If you re-join two HindIII ends, you should have a molecule with two free BamHI ends. It is
more likely that these two complementary ends will happen to find each other and be joined first
than it is likely that new complementary pieces will find their correct places. In fact, it is unlikely
that any product DNA will be formed from more than 4 starting pieces.
Parent Plasmid Diagrams
B=Blue; P=Purple; Y=Yellow; R=Red; G=Green; O=Orange. Remember, you will make two bead strands
for each plasmid, one 5’ to 3’, and the other 3; to 5; to model the double-stranded structure of DNA.
Questions
Fill in the answers to questions 1-4 on a chart you recreate in your quadrille that looks like the one provided
on the next page. Make sure you leave yourself enough room to fully answer each question in that chart.
You may want to leave it open ended at first, so it can be added to. An example is given on the chart.
Answer question 5 separately.
1.
2.
3.
4.
5.
Focus on all the different circular products you could make from joining two of your pooled
fragments. There are ten different products. Working as a team, draw them, being sure to show
any replication origins, antibiotic resistance genes, and restriction sites.
What is the approximate size of each of the 10 products, in base pairs?
Assume you transform the mixture of 10 products into plasmid-free E. coli cells that are normally
killed by both kanamycin and ampicillin. Which of the products you predicted in the answer to
Question 1 would be replicated by the host cell and passed on to its descendants?
If each of the E. coli transformants from Question 3 received only one recombinant plasmid,
which of the ten plasmids would render the transformed E. coli and its descendants able to grow in
the presence of ampicillin? Of kanamycin? Of both?
After E. coli cells have undergone a transformation procedure, they are usually spread on agar
medium so that each individual E. coli cell and its descendants can grow into an isolated colony
of cells. Can you propose a method in which only those E. coli cells that received the amp/kan
double-resistance plasmid could grow?
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10
AP Biology Lab 6 Part C: Molecular Genetics
Transformation of E. coli with pUC8
Background
The ability to exchange genes within a population is a nearly universal attribute of all living
things. Among prokaryotes, there is no known case where genetic exchange is an obligatory step (as it
often is in eukaryotes) in the completion of an organism’s life cycle. Rather, genetic exchange in
prokaryotes seems to be an occasional process that occurs through three different mechanisms in various
prokaryotes. These three mechanisms are transformation, transduction, and conjugation.
In transformation, DNA is released from cells into the surrounding medium, and recipient cells are
then able to incorporate it into themselves from the medium. Transduction occurs when a phage viron
attaches to a bacterial cell and transfers some or all of its DNA into the bacteria. Conjugation is controlled
by plasmid-borne genes and occurs between cells that are in direct contact with one another. Usually only
the plasmid itself is transferred, although sometimes chromosomal genes can be transferred as well.
Although the existence of plasmids was inferred from genetic studies in the 1950s, it has only
been during recent decades, with the advent of more accurate means of detection, that the impact of
plasmids on the biology and ecology of prokaryotes has been fully appreciated. Plasmids are circular
molecules of double-stranded DNA and generally function as small chromosomes. They are selfreplicating and can encode for a variety of cellular functions. However, unlike chromosomes, they are
dispensable. The types of functions they encode only benefit the cell in a limited set of environments, and
none are known to encode for essential cellular functions. Many, but not all, bacteria may contain
anywhere from one to several dozen plasmids and
although they are capable of autonomous replication,
the number of plasmids remains fairly constant from
one generation to the next. These bits of DNA vary in
size from a few to several hundred Kb (kilobase) pairs
in length.
Many plasmids, called R factors, carry genes
that confer resistance to antibiotics on the host cell.
First discovered in 1955, R factors have spread rapidly
among pathogenic bacteria in recent years, profoundly
affecting medical science by causing many strains of
pathogenic bacteria to be highly resistant to antibiotics.
The number of inhibitory substances for R-mediated
resistance has grown to almost all antibiotics, many
other chemotherapeutic agents, and a variety of heavy
metals. The mechanisms of resistance conferred by
these genes tend to be different from those that are
chromosomally determined.
Plasmid genes often
encode enzymes that chemically inactivate the drug or
eliminate it from the cell by active transport. In
contrast, chromosomal mutations usually modify the
cellular target of the drug, rendering the cell resistant to
the drug’s action.
Transformation (Uptake of DNA from the
environment)
The first mechanism of bacterial genetic
exchange to be discovered was transformation. In
1928, a now-famous experiment demonstrated that
injecting mice with a non-pathogenic (not capable of
causing disease) strain of Streptococcus pneumoniae,
together with heat-killed cells of a pathogenic (diseasecausing) strain killed mice, while injecting these strains
separately did not. This and subsequent experiments
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established that the surviving cells were recombinant, meaning that they exhibited certain properties
(including the ability to cause disease) that were typical of the killed cells and others that were typical of
the non-pathogenic culture. A genetic exchange of the DNA dissolved in the external medium has occurred
between the dead cells and the live ones. At the time, it was thought that a particular substance, a
“transforming principle”, caused the exchange to take place. Thus the word “transformation” came to be
used to describe genetic exchange among prokaryotes.
When cells are able to be transformed by DNA in their environments they are called “competent”.
In a significant number of bacteria, entry into a competent state is encoded by chromosomal genes and
signaled by certain environmental conditions. Such bacteria are said to be capable of undergoing natural
transformation. Many other bacteria do not become competent under ordinary conditions but can be made
competent by exposing them to a variety of artificial treatments, such as exposure to high concentrations of
divalent cations. A divalent cation is missing two electrons as compared with the neutral atom. Example:
iron(II) or Fe2+ is the divalent cationic form of iron.
E. coli cells, which do not possess a natural system for transformation, are capable of being
artificially transformed. They become competent only after the cultured cells are exposed to calcium
chloride solution. These newly-competent cells are now receptive to an insertion of foreign DNA
contained in a plasmid.
Objectives
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Investigate transformation as a mechanism of genetic exchange
Create competent cells by chemically and thermally treating E. coli cells
Insert a plasmid containing antibiotic-resistance genes into competent E. coli cells
Screen the transformed cells to determine which have been genetically altered
Calculate the efficiency of the transformation reaction
Materials
2 Luria agar plates (petri dishes)
2 Luria agar plated with ampicillin
2 Microcentrifuge tubes
1 inoculating loop
2 Bacti-spreaders
4 Sterile graduated pipettes
Micropipette w/tips
Sterile disposable graduated syringe
Calcium Chloride
Plasmid pUC8
Water bath
Starter plate of E. coli
Procedure
Before the experiment, be certain a 42° C water bath is available and that the calcium chloride is in an ice
bath and kept cold throughout the experiment.
1.
2.
3.
4.
5.
Obtain two microcentrifuge tubes and mark one tube “+”, the other “-“. The “+” tube will have
the plasmid added to it.
Using a sterile pipette, add 0.25 ml (250 µl) ice-cold calcium chloride to each tube.
Obtain a starter plate. Use a sterile inoculating loop to transfer a large colony of bacteria from the
starter plate to each tube of cold calcium chloride. Be sure not to transfer any agar to the tube.
To remove the bacteria from the transfer loop, place the loop into the calcium chloride and twirl
rapidly. Gently tapping the loop against the side of the tube may help dislodge the bacteria.
Dispose of the loop according to your instructor.
Using the micropipette and appropriate tip, add 10 µl of the plasmid pUC8 solution, which carries
the antibiotic resistance gene, to the “+” tube. Discard tip in container of ethanol after use.
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6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
Gently tap the tube with your finger to mix the plasmid into the solution.
Incubate both tubes on ice for 15 minutes.
While the tubes are incubating, obtain two plain Luria agar plates and two Luria agar plates with
ampicillin. Label one Luria agar plate “+”, the other “-“. Do the same for the Luria agar plates
with ampicillin. Be sure to label all four plates with your group letter, and period number. Both
time and temperature are critical in the following heat-shock protocol. Be sure your water bath is
at 42º C, and do not exceed 90 seconds in the water bath.
The bacterial cells must be heat shocked to allow the plasmid to enter the cell. Remove the tubes
from ice and immediately place in a 42º C hot water bath for 60 to 90 seconds.
Remove the tubes from the 42º C water bath and immediately place on ice for two minutes.
Remove the tubes from the ice bath and add 0.25 ml (250 µl) of room temperature Luria broth to
each tube with a sterile disposable pipette. Gently tap the tube with your finger (creating a finger
vortex) to mix the solution. The tubes may now be kept at room temperature.
Add 0.1 ml (100 µl) of the “+” solution to the two “+” plates with a sterile disposable graduated
syringe. Add 0.1 ml (100 µl) of the “-“ solution to the two “-“ plates with a different sterile
disposable graduated syringe.
Using a sterile Bacti-spreader, spread the cells over the entire surface of the Luria agar “-“ plate.
Then, using the same Bacti-spreader, spread the liquid on the Luria agar with the ampicillin “-“
plate.
Using a new Bacti-spreader, repeat the procedure for both of the “+” plates. Spread the liquid on
the Luria agar “+” plate first followed by the Luria agar with ampicillin “+” plate. Dispose of the
Bacti-spreaders according to your instructor.
Let the plates sit for five minutes to absorb the liquid. Place the plates in a 37º C incubator,
inverted, overnight.
The next day, remove the plates from the incubator. Count and record the number of colonies on
each plate. If the bacteria have grown over the entire surface so that individual colonies cannot be
distinguished, write “lawn”. Record your results in a data table of your own design.
Address the analysis questions for this part of the lab in your quadrille.
Analysis Questions
Part A:
1. Make an accurate and precise sketch of your gel results in your quadrille. (Use
appropriate colored pencils, and a metric rulers to do this)
2. For lanes 1-5, measure the distance each dye has traveled. Measure the distance
from the center of each well to the center of each dye sample. Record your results
in a table such as the one below. Also note which electrode each dye migrated
toward and whether the dye is positively or negatively charged.
Lane
Migration Distance (mm)
Migration Direction
(+ or -)
Molecule Charge
(+ or -)
1
2
3
4
5
Migration distance will vary based on gel size and the amount of time the gel is allowed to
run.
3. Why did certain dyes migrate toward the positive electrode and others toward the
negative electrode?
4. Examine the dye mixture you placed in lane 6. Based on your results, can you
determine whether or not any of the dyes were absent from the mixture?
5. How does varying the concentration of agarose used in a gel affect the ability of
the gel to separate molecules?
6. The gel used in this activity was prepared with 2.0% agarose. Many DNA
separations are performed on agarose gels varying from 0.7% to 1.2% agarose.
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7.
What does this tell you about the size of these DNA molecules in comparison to
the dyes you used in this lab?
If you were called away during the electrophoresis procedure and were not able to
monitor your electrophoresis run, what do you think would happen if the
electricity were to remain running in your absence?
Part C:
8. A cell must be competent for transformation to occur. Not all cells in the solution become
competent and therefore never receive the gene for antibiotic resistance. Transformation
efficiency is the number of resistant colonies per microgram of plasmid. Using the directions
below, calculate transformation efficiency.
a. Total mass of plasmid used:
(Total mass = volume x concentration)
b. Total volume of suspension:
c.
Fraction of cell suspension put on plate:
(µL on plate/total volume)
d. Total mass of plasmid in fraction:
(Mass of plasmid x fraction on plate)
e. Number of colonies per µg of plasmid:
(# of colonies counted/mass of plasmid put on plate)
9. Based on your experimental results, did transformation occur? Why or why not?
10. Which of the following statements is/are not true about bacterial transformation?
a. Cells can only be transformed when they are in a competent state
b. Transformation may only be performed using plasmid DNA containing antibioticresistance genes (R factors)
c. Transformed cells are capable of passing their newly acquired traits onto succeeding
generations
d. Transformation was first discovered in the bacterium Streptococcus pneumoniae.
e. Cells must first be treated artificially in the laboratory before they are capable of
undergoing transformation.
11. Your expected transformation results for each of the four plates are listed below. Briefly explain
why the listed results occurred.
Luria “-“: Bacterial Lawn
Luria “+”: Bacterial Lawn
Luria w/amp “-“: No Growth
Luria w/amp “+”: Several colonies
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