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BIOL 202
Monday, February 8th, 2010
Prof. Chevrette
Lecture #16
LECTURE 16 - Gene Cloning: Using the Tools
Required readings:
9th ed. Ch. 20; pp 715-727;
8th ed. Ch. 11 pp 341-354
Announcements: The rooms for the midterm is posted on WebCT and there will be
about two questions per lecture on the midterm.
Cloning
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
Figure 2
Last week, we looked at the tools we can use to
analyze DNA clearly.
o Usually this is done by cutting the vector and
the DNA with a restriction enzyme and reanneal the vector and the DNA. This gives us a
recombinant (hybrid) plasmid. (Figure 1)
 The insertion of the gene of interest (donor
DNA) in vector requires complementary
sticky ends/overhang from cutting with
restriction enzymes.
 Vector and donor DNA can be cleaved
with different restriction enzymes as long
Figure 1
as complementary overhangs are generated, but usually the same
enzyme is used as that guarantees complementarity.
o This method will work if you have a single fragment that you want to clone in
this vector.
o It will also work if you have many different fragments that your want to clone.
This requires a different type of vector. Note that the restriction enzyme
fragments do not have to be of the same size, however there is a limit in the
size of the vector itself.
The goal, is not to isolate the DNA fragment, but to get huge quantities of the
fragment in the plasmid.
o In Figure 2, notice that the vectors are
the same but there are different
fragments inserted, thus generating
different recombinant vectors.
 When the plasmid enters the bacteria, it is
then able to bring in the DNA fragment
(as it is part of the plasmid itself now).
 After transformation, the plasmid is found
inside the bacteria.
o Transformation is the bacterial uptake
of DNA from environment.
o The bacteria will now replicate the plasmid as the vector contains the origin
of replication (as seen in the previous lecture). Thus replication occurs,
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independent of the genomic DNA. This results in many copies of the plasmid
within a single bacterium.
o When the bacterium divides, it will also divide with the plasmid. By doing so,
you’ll get millions of bacterial cells
Selecting Plasmid
 How do you determine which plasmid has the inserts that you desire? Note that
transformation is not 100% effective, therefore you must select for the successful
transformations.
o As seen in Figure 3, pUC18 vector has lacZ gene containing polylinker.
o Functional lacZ gene creates β-galactosidase, which forms blue dye in
presence of X-Gal substrate.
o Therefore, successful insertion of donor DNA in polylinker region disrupts
lacZ gene so β-galactosidase is
not formed, and colonies are
white in the presence of X-Gal
 After transformation with pUC18
vector (contains genes for ampicillin
resistance and lacZ), bacteria are
incubated and plated on agar
containing ampicillin and X-Gal.
o All colonies that actually grow
(on the agar with ampicillin)
contain pUC18 vector because
they must be ampicillin-resistant
Figure 3
o Blue colonies do NOT contain
recombinant plasmids, because the lacZ region is functional, meaning there is
a donor DNA has NOT been inserted to disrupt the lacZ gene.
o White colonies DO contain recombinant plasmids -> However, you still do
not know if the donor DNA in the recombinant plasmid is actually the gene of
interest (or YMWG= Your Most Wanted Gene).
o Also note that such plasmid can only incorporate DNA inserts of 10-15kB.
This is a relatively small DNA insert compared to the size of the human
genome.
Clicker Question: If you has used the pBR322 plasmid (figure4)
instead of the pUC18 in the previous experiment:
1. All colonies will be blue
2. All colonies will be white
3. We will obtain the same results
Figure 4
Answer: The question is not clear since it did not state where the DNA
fragment was inserted into the plasmid. So let’s assume that it was
inserted in the tetracycline resistant gene. Also notice that from the
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Lecture #16
figure 4, that pBR322 does not have the lacZ gene. So the answer is 2All colonies will be white.
Clicker Question: If you clone in BamH1 site of pBR322 and plate the bacterial
suspension on agar containing ampicillin and tetracycline:
1.
2.
3.
4.
Colonies containing an insert will be white
Colonies containing an insert will be blue
Only colonies with an insert will survive
Only colonies without an insert will survive
Answer: 4 – Only colonies without an insert will survive as the
colonies are not tetracycline resistant.
Phage λ



Bacteriophage is a virus capable of infecting bacteria. For example, a
bacteriophage (48, 502 bp) can infect E.Coli.
o Bacteriophages, like plasmid, can be used as vectors and are capable of
prolific replication within a cell. One third of its genome is not required for
lytic growth, and can be replaced by exog enous DNA.
o DNA of varying amounts (depending on the bacteriophage of choice) is
packaged into the capsid (head) of the virus and injected into bacteria, which
replicate the DNA
Bacteriophage Lambda heads are capable of packaging DNA up to 50 kB long;
however, this is NOT to say that an insert is necessarily 50 kB.
o Parts of the phage DNA that contains genes which encode functional
information (like replication machinery, packaging information, and structural
proteins) must be conserved, therefore only a limited amount of DNA can be
removed in order to create a vector.
o Inserts (exogenous DNA) can be no longer than 20kB and are inserted into
the middle of the phage DNA, which does not encode any of the necessary
genes described above.
Process of Insertion (figure 5):
1. The total genomic DNA is digested
partially (i.e. not cut at every
restriction site so that you do not
end up with tiny fragments) with a
restriction enzyme, and segments
that are about 15kB are isolated.
Fragments that are too small can
not be inserted into the virus.
2. A compatible restriction enzyme
is then used to remove the central
portion of the phage DNA, and this
portion is discarded resulting in two
‘arms’ of DNA.
Figure 5
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3. The genomic DNA is then ligated at random with the arms from the phage
DNA generating a large strand of recombinant DNA.
4. The virus machinery then cuts the large piece of DNA (at the end of the left
and right arm) and this DNA is then ‘stuffed’ into phages (viral capsids) in
vitro and allowed to infect E.Coli. Note that the original size of the phage
lambda DNA is restored.
5. When plated on agar, the clear areas represent E.Coli cells that have lysed
releasing viral entities. These areas are called Phage Plaques.
o Plaques signal that an effective recombinant virus has been formed.
o Note that there is no ampicillin on this agar plate, or else all the bacteria
would be killed.
6. The bacterial lawn can then be screened using a nucleic acid probe (ssDNA)
and autoradiography to expose the plaques that contain the gene of interest, as
opposed to other similar-sized DNA that may have been cut and inserted
along with the DNA of interest.
Cosmids


Figure 6
In order to replicate larger segments of DNA, special vectors have been
engineered which far exceed both the basic plasmid and the phage λ.
Cosmids are plasmids containing cos sites. (Cos sites determine what fits in the
head of the virus.) Cosmids are engineered hybrids of phage λ
DNA and bacterial plasmid DNA that are capable of carrying 3545 kB inserts.
 The plasmid component provides:
o Origin of Replication: allows the host bacteria to recognize
where it will start transcribing the
DNA
o ampR gene: confers resistance to ampicillin so that cosmid
carrying bacteria can be selected for.
o Polycloning site (aka polylinker): an area concentrated with
restriction sites.
 The phage λ provides:
o Cos sites: signals to the virus where it should cut the DNA in
order to inject it into the host bacteria
 The process (figure 6):
1. A cos site is inserted into a plasmid containing an ampR and an
ori.
2. The plasmid is linearized via cutting with restriction enzymes at a site directly
next to the introduced cos site.
3. The linearized plasmid is ligated to genomic DNA fragments forming a long
string of DNA.
4. The recombinant DNA is then inserrted into a phage head in vitro, note thos
phage does not replicate itself (the only viral DNA is the cos site).
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Lecture #16
5. The phage is allowed to infect bacterial cells. The phage will cut the inserted
DNA at the cos sites which are complementary thus once injected, it will form
a circular extra-nuclear, self-replicating plasmid within the cell.
6. Infected cells are then separated by growing on an ampicillin medium. Cells
that grow have been infected with the exogenous DNA. This is different from
the plaques seen in phage λ because here we are
looking for cells which have incorporated the
DNA as opposed to phages that have incorporated
the DNA).
BAC
 Bacterial Artificial Chromosome (figure 7)are
vectors derived from the F plasmid, which can carry
inserts ranging from 150-300 kB.
 Figure 8 is a summary of studied vectors. Note that
the size of the target DNA is the crucial determinant of
which vector will be used.
Figure 7
Figure 8
Construction of Genomic Libraries
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
A genomic library is a collection of either phages or bacterial cells that include
plasmids repressing the entire genome of a given organism. These are created by
using vectors.
Rather than using restriction enzymes to create fragments of DNA, a process
known as DNA shearing is much more efficient. (figure 9)
o Shearing is a mechanical means whereby the DNA is subjected to harsh
measures which break it into longer fragments. The DNA is pushed through a
small needle.
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Lecture #16
 Following shearing, the DNA is subjected to a lambda exonuclease (meanwhile a
cosmid/other vector has been treated with an endonuclease to remove a segment
which will later be replaced
with the sheared fragments).
 Complementarity between the
cosmid and the foreign sheared
DNA is generated in the:
o cosmid: by mixing it with a
terminal transferase and
ATP (adenine
triphosphate) to generate a
poly A sequence.
o DNA: by mixing it with a
terminal transferase and
TTP (thymine
triphosphate) generating a
Figure 9
poly T sequence.
 These complementary segments anneal and any overhanging poly A or poly T
ends are removed with exonuclease III
 The gaps are filled in with a DNA polymerase and a DNA ligase generating a
complete plasmid, without the use of the restriction enzymes.
 How many clones are needed for a representative
genomic library?
o It depends of: 1) the size of the genome and 2) the
average size of the DNA inserts.
o N = Required number of recombinant
clones
P = Probability for any clone to be
present in the library
f = Fraction of the genome present in
an average size clone
Retrieving DNA of Interest
 Once a genomic library has been generated, in this case within bacteriophage
lambda, one might want to isolate the phage with DNA encoding the gene of
interest so that it could be replicating to produce a large quantity of the gene to do
this:
1. The phages are allowed to infect a lawn bacteria
2. A nitrocellulose filter is used to ‘soak up’ some phages from each plaque.
3. The nitrocellulose filter then incubated with a radioactive probe. At this
point (during incubation), the DNA becomes denatured (i.e. single stranded)
which allows the probe (which is also single stranded) to hybridize with it.
4. Autoradiography is then used, wherein a film is developed by the
radioactivity of the bound probe, producing a map that indicates where the
phage containing the gene of interest is located on the lawn of bacteria.
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5. This phage is removed from the lawn and inoculated into a pure sample of
bacteria allowing mass production of this single gene.
 However, there is a problem->genomic libraries do not only contain the genes!
o If we want only genes (genomic libraries contain DNA coding for many other
things as well as genes), then we need
cDNA, which is DNA complementary to a
portion of mRNA isolated from a cell.
o mRNA contains no introns and is the
finalized version of the DNA from the
nucleus that will be transcribed and
eventually produce a protein,
o Since no vector can incorporate a piece of
RNA, reverse transcriptase is used to
convert the RNA strand of interest into a
piece of cDNA.
1. An OligoT primer is used to create a
starting point for the reverse
transcriptase on the single-stranded
mRNA template
2. Once at the end of the template, the
reverse transcriptase generates a
hairpin loop, creating a second
primer, that can be used by DNA
polymerase
3. DNA polymerase uses this loop as a
starting point to create a second strand
of DNA that is complementary to the
initial piece of RNA that we started
Figure 10
with
4. S1 nuclease is used to cut off the loop this generates a small piece of
DNA that is a concentrated version of the large nucleic DNA (this
version is free of introns and is capable of directly encoding a protein).
Isolating Genes of Interest from a cDNA Library
 In order to isolate the RNA of the gene of interest, a
process similar to the one above is performed (figure
10). This involves a complementary strand of DNA
labeled with a radioactive probe – note that it is singlestranded.
Screening for “Your Most Wanted PROTEIN”
 If a complementary probe cannot be generated, an
antibody can be used to screen for a protein. (figure 11)
o A phage called λgt11 is used and a cDNA gene is
incorporated into it, disrupting the reading frame of
the gene of the phage.
Figure 11
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o This disruption allows the creation of a fusion protein (a piece of protein that
is attached to the protein of interest, because they were transcribed as one, that
can be targeted by an antibody).
 An antibody is a protein that will bind to a specific protein, thus an
antibody for the protein of interest is introduced.
o Again, a similar process to the seen previously is used, but in this case the
nitrocellular filter is incubated with the antibody for the fusion of protein.
o This antibody is then targeted by a second radiolabeled antibody, and the
resulting film is autoradiographed generating a map.
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