Download Recombinant DNA cloning technology

Recombinant DNA cloning
technology
Chapter 7
DNA cloning

To obtain large amounts of pure DNA

Procedure




Isolate DNA
Use restriction enzymes to cut DNA
Ligate fragments into a cloning vector
Transform recombinant DNA into a host to
replicate the DNA and pass copies into progeny.
Restriction Enzymes



Recognize a specific DNA sequence
(restriction site)
Break a phosphodiester linkage
between a 3’ carbon and phosphate
Used for


Create DNA fragments for cloning
Analyze positions of restriction sites in
cloned or genomic DNA
Restriction Enzymes


Are found naturally in bacteria as a
defense against vital DNA.
Restriction sites are methylated in
bacteria, and thus protected.
Restriction Enzymes

Are denoted by three letter names
derived from the bacterial strain they
originate from.
Restriction Enzymes


They usually are palindromes of 4-, 6or 8-base pairs.
Based on the probability, a specific
short DNA sequence occur more
frequently than a long one.


In 50% GC content, each base has a ¼
chance of occurring at a position.
The frequency of a particular restriction
site is (1/4)n.
Restriction Enzymes: EcoRI
• EcoRI (“Echo R one”) is a
commonly used enzyme. It
was the first (one)
restriction enzyme isolated
from the “R” strain of E.
coli. It demonstrates the
usual type recognition site, a
palindrome (the same on
both strands, reading in
opposite directions) EcoRI
leaves a four base, 5’
overhang, sticky end.
Restriction Enzyme Sites
• Sma I (from Serratia
marcescens) cuts a
palindrome to give blunt
ends.
• BamHI (from Bacillus
amyloliquefaciens H)
cuts to give a 5’
overhang.
• PstI (from Providencia
stuartii) cuts to give a 3’
overhang.
Restriction Enzymes


Blunt ends: both strands are cut at the
same position.
Sticky ends: overhanging regions (3’ or
5’) are useful in cloning. They are
complementary, thus anneal, DNA
ligase can covalently link them.
Question 7.2

A new RE from a bacterium cuts DNA into
fragments that average 4096 bp long. This RE
has a two-fold rotational symmetry. How
many based pairs of DNA constitute the
recognition site for the new enzyme?

Two-fold rotational symmetry: axis of symmetry
through the midpoint. The base sequence from 5’
to 3’ on one DNA strand is the same as the base
sequence from 5’ to 3’ on the complementary
strand.
Answer 7.2

The average length of fragments produced
indicates how often the RE site appears, in
this case every 4096 bp, there is an RE in the
genome.

If DNA is composed of equal amounts of A, T, C,
and G, then the chance of finding one specific
basepair (AT, TA, GC, or CG) at a particular site is
¼.


Finding n number of specific basepairs is (1/4)n
1/4096=(1/4)6
Question 7.3
An endonuclease called AvrII cuts DNA
whenever it finds the sequence
5’-CCTAGG-3’
3’-GGATCC-5’

About how many cuts would AvrII make
in human genome, which is about 3x109
bp long and about 40%GC.
Answer 7.3
5’-CCTAGG-3’
3’-GGATCC-5’
The enzyme recognizes a sequence that has
two GC bp, two CG bp, one AT bp, and one
TA bp in a particular order. Since genome
has 40% GC, the chance of finding a GC or a
CG pair is 0.2 while the chance of finding an
AT or a TA pair is 0.3. The chance of finding
the 6 base pairs in this sequence is:
P(AvrII site) = 0.2x0.2x0.3x0.3x0.2x0.2=0.000144
Number of sites in the genome = 0.000144 x (3 x 109) = 432,000.
How far apart they are within
the genome?
Enzyme
Recognition
Seq.
Probability
Distance
between
Sites
BamHI
5’-GGATCC-3’
3’-CCTAGG-5’
(0.2)4(0.3)2 =
0.000144
1/0.000144 =
6,944 bp
EcoRI
5’-GAATTC-3’
3’-CTTAAG-5’
(0.2)2(0.3)4 =
0.000324
1/0.000324 =
3,086 bp
NotI
5’-GCGGCCGC-3’
3’-CGCCGGCG-5’
(0.2)8 =
0.000000256
1/0.00000256
= 390,625 bp
HaeIII
5’-GGCC-3’
3’-CCGG-5’
(0.2)4 = 0.0016 1/0.0016 = 625
bp
Cloning DNA
with the
Restriction
Enzyme EcoRI
• A typical DNA cloning experiment requires that the DNA to be
cloned (often called the “insert DNA”) and the vector (often a
plasmid) both be cut with the same enzyme (or with two
enzymes which produce compatible ends). The insert DNA and
the vector are then mixed, and DNA ligase is used to join the
molecules.
Cloning Vectors




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Plasmid;
Phage;
Cosmid;
Shuttle;
Yeast Artifical Chromosomes (YACs)
Bacterial Artificial Chromosomes (BACs)
Plasmid Cloning Vectors


Derived from natural plasmids.
Plasmids are circles of dsDNA that include
origin sequences (ori) needed for replication
in bacterial cells.

E.coli plasmid vectors contains



An ori sequence required for replication in E.coli
A selectable marker, e.g., antibiotic resistance, to allow
selection of cells harboring the plasmid.
At least one unique restriction enzyme cleavage site, so
that DNA sequences cut with the RE can be spliced into
the plasmid.
pUC19 (2,686-bp)





High copy number in E.coli, ~100 copies per
cell (makes it easy to purify the plasmid);
Has a selectable marker ampR (indicates the
success of transformation).
A cluster of unique restriction sites, called
polylinker (multiple cloning site makes it easy
to use different RE sites for cloning).
Polylinker is part of lacZ gene (Bgalactosidase). If pUC19 plasmid is put in a
lacZ- E. coli, cell will become lacZ+.
If DNA is cloned into the polylinker, lacZ is
disrupted, no complementation occur.
Plasmid pUC 19
•
The commonly used plasmid pUC19
(“puck 19”) is a small plasmid with the
essential elements for a vector: An origin
of DNA replication A dominant selectable
marked (resistance to an antibiotic,
ampicillin) And a cloning site, usually a
polylinker with recognition sites for
numerous restriction enzymes.
How to detect if DNA is cloned
into the polylinker?

X-gal, a chromogenic analog of lactose,
truns blue when B-galactosidase is
present, and remains white in its
absence, so blue-white screening can
indicate which colonies contain
recombinant plasmids.
Blue white screen
•
•
•
A portion of the lactose utilizing gene (b-galactosidase) is interrupted by the polylinker cloning site.
Insertion of a DNA fragment prevents expression of the gene.
Growing the E. coli containing the plasmid on petri plates containing a substrate for the enzyme allows
you to tell those which express the lac-z gene (no insert, blue color) form those colonies that do not
(contain an insert, white color)
How to insert the DNA?

First digest then ligate:




Cut pUC19 with an RE that has a unique site in
the polylinker.
Cut the DNA to be cloned (insert DNA) with the
same enzyme.
Mix insert DNA and pUC19 DNA
Transform plasmids into E. coli, either through
chemical treatment of cells or electroporation.


Incubate the recombinant DNA plasmids with E.coli cells
treated chemically to take up DNA
Electroporation requires that an electric shock is
delivered to the cells causing temporary disruptions of
the cell membrane and letting the DNA enter.
How to insert the DNA?


Grow cells on media containing ampicillin
and X-gal.
Ampicillin-resistant colonies will grow.


Blue colonies contain only the vector
White colonies contain vector with insert.
Other plasmid vectors

They contain



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different unique restriction enzyme sites
Phage promoters (e.g., T7, T3, SP6) for
transcription of the cloned DNA.
Available for prokaryotic and eukaryotic
organisms
Size of the insert is limited; plasmids
carrying more than 5-10 kb are not stable.
Phage Lambda Vectors


Versions of bacteriophage lambda with
sequences for lysogeny removed, so that only
lytic infection is possible.
Lambda replacement vector:

Has a chromosome with a ‘left’ arm and a ‘right’
arm, that contain all the genes needed for lysis.
Between two arms, there is a disposable segment
since it does not contain any lytic cycle genes.
These two regions, the arms and the disposable
region is separated by EcoRI sites. The lambda
chromosome central region is replaced with the
insert DNA (~15kb), using RE digestion and
ligation.
Phage l (lambda) as a Cloning Vector
•
•
Plasmids are limited in the size of DNA that
can be easily introduced into bacteria, about 510 Kb cloned DNA (transformation). By
cloning into a phage, the viral entry system can
be exploited to introduce the DNA into
bacteria. Phage l allows insertion of 15-30 Kb
DNA, with efficient introduction into E. coli.
Subcloning: transfer of a DNA insert from the
phage clone into a plasmid by having a special
bacterial strain do the work.
Only phages with DNA insert
between the two arms can
replicate; Why?


The phage needs both arms to be
together for reproduction and lysis.
Each DNA fragment is cloned by
repeated rounds of infection and lysis.
Eventually culture becomes transparent
as all the bacteria has been lysed, and a
population of progeny lambda phages is
produced (1010 to 1011 phages/mL).
Phage vectors


When ligated DNA is mixed with phage
lambda proteins, phage heads assemble, and
DNA is packaged, forming virus particles.
Only phages with both arms of the phage
lambda chromosome and a properly sized
(37-52 kb) central insert sequence are able to
replicate by infecting E. coli.
Shuttle Vectors

A cloning vector capable of replicating
in two or more types of organism (e.g.,
E.coli and yeast) is called a shuttle
vector. They replicate autonomously in
both hosts or integrate into them.
Yeast Artificial Chromosomes
(YACs)

Function as artificial chromosomes in yeast.


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Linear structure with a yeast telomere (TEL) at each end.
A yeast centromere sequence (CEN)
A marker gene on each arm that is selectable in yeast
(e.g., TRP1, URA3).



Tryptophan and uracil independence in trp1 and ura3 mutant
strains, respectively.
Unique restriction sites for inserting foreign DNA that can
be up to 500kb long. This size of inserts are important
in generating physical maps of genomes.
An origin of replication sequence-ARS (autonomously
replicating sequence)-that allows the vector to replicate
in yeast.
YACs

Several hundred kb of insert DNA can
be cloned in a YAC. YAC clones are
made by:

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Generating YAC arms by restriction digest
Ligating with insert fragments up to 500 kb
in length
Transforming into yeast
Selecting for markers (e.g., TRP1 and
URA3).
Question 7.8

Why one might want to clone DNA in an
organism other than E.coli.

One can transform yeast as well as plant
and animal cells. This can be useful in
studying cloned eukaryotic genes in a
eukaryotic environment, commercial
production of gene products (e.g., drugs,
antibodies), developing gene therapy,
genetic engineering.
Question 7.8

Why use shuttle vectors?

They can replicate in two or more host
systems, e.g., replicate in E.coli (it is easy
to do the initial cloning), then be
transferred to yeast. A yeast shuttle vector
contains selectable markers for both
systems (ura3 for yeast; ampR for E. coli);
autonomous replication sites to replicate as
a plasmid.
Recombinant DNA libraries



Genomic library
Chromosome library
cDNA library
Genomic libraries

Are constructed by digesting genomic
DNA


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Complete digestion
Mechanical shearing
Partial digestion
Partial digestion



They are selected in a certain size range by
density gradient centrifugation or agarose gel
electrophoresis
DNA fragments from sticky ends can be
cloned directly.
Genomic sequences are not equally
represented in the library


Regions of DNA with relevant restriction sites very
close together or far apart are removed at the
selection stage
Some regions prevent vector replication so
eliminated.
Partial Digest for Producing Clonable Fragments
Enzymes with compatible
Sticky ends are used.
How many clones are needed
to contain all sequences in
genome?

Depends on:

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Size of the genome being cloned
The average size of the DNA fragments
inserted into the vector.
How many clones are needed
to contain all sequences in
genome?
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The probability of having at least one copy of
any DNA sequence in the genomic library is:
N=ln(l-P)/ln(1-f),



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Where N = the necessary number of recombinant
DNA molecules;
P = the probability desired;
F = average size of the fragments divided by the
genome size
Ln = natural algorithm
Question 7.10



Within the human genome (3 x 109
bp), how many 40-kb pieces would you
have to clone into a library if you
wanted to be 90% certain of including a
particular sequence?
P = 0.90; f=40,000/(3x109)
N=ln(l-P)/ln(1-f) = 172,693 fragments
cDNA libraries


cDNA drives from mature mRNA, no
introns.
polyA tail at the 3’ is useful for:


Isolating mRNA from cell lysates
Priming the synthesis of cDNA providing a
known 5’ sequence
cDNA synthesis

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A short oligo(dT) primer is used. It anneals to
the mRNA’s poly(A) tail, allows reverse
transcriptase to synthesize the cDNA (DNAmRNA hybrid).
Rnase H degrades the mRNA strand, creating
small fragments that serve as primers
DNA polymerase I makes new DNA
fragments, DNA ligase connects them to
make a complete chain.
cDNA from a Polyadenylated mRNA
Annealing
Reverse transcription
RnaseH degradation
DNA polymerase I
DNA ligase
cDNA cloning



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Introduction of restriction site linkers to the
ends of the cDNA by blunt end ligation
Digestion with cognate restriction enzyme to
create sticky ends
Mixing cDNA with vector DNA cut with the
same restriction enzyme in the presence of
DNA ligase
Transforming into an E. coli host for cloning

Use polylinkers engineered with appropriate
ssDNA overhangs so do not digest
Linkers for Cloning DNA
• Any DNA fragment can have a
specific restriction site added
to the ends by ligating on a
“linker”.
• Linkers are small, synthetic
(made in the lab, or ordered
from a company) DNA
fragments which contain the
recognition sequence for one
or more restriction enzymes.
• After ligating on linkers, the
DNA is cut with the
appropriate restriction enzyme
to produce ends for cloning.
Random Primed DNA Synthesis for Making a
“Probe”
• One common technique for
making a probe to detect a
specific DNA sequence is
called “random primed DNA
synthesis”.
• Short (8 bases) primers of
random sequence are annealed
to the heat denatured DNA (of
the sequence for the probe),
and DNA polymerase is used
to synthesize copies of the
DNA with one of the
nucleotides incorporated
carrying a detectable label
(such as radioactive
phosphate).
Klenow fragment PolI

The large or Klenow
fragment of DNA
polymerase I has
DNA polymerase
and 3' -> 5'
exonuclease
activities, and is
widely used in
molecular biology.
cDNA synthesis & making
probe

DNA polymerases require a primer (oligonucleotides, 620 bases) to provide a free 3' hydroxyl group for initiation
of synthesis. Mixed single-stranded template (usually
denatured double-standed DNA), primers and the
enzyme in the presence of an appropriate buffer. As
Klenow proceeds, it can displace primers downstream
and continue synthesizing new DNA.
Colony Lift Hybridization to Find a Cloned Sequence
in a Plasmid (or Cosmid) Library
• The presence of a clone
containing a specific
sequence can be determined
by making a “lift” of the
colonies, lysing the cells on
the surface of a “membrane”,
and hybridizing a labeled
(radioactive) “probe” of the
sequence being searched for.
Finding a Cloned Sequence: Plaque Lift
Hybridization in a Lambda Library
• If a library is made in
phage l, the desired
sequence can be found
by hybridizing a probe
to a “lift” of the plaques.
This is detected with a
probe, as in the colony
lift.
Antibodies
• Antibodies can be (in theory, at least) be produced
which react with any molecule.
• If a protein is injected into a rabbit (or goat, or sheep,
etc.) the blood isolated from the injected animal will
have antibodies against the injected protein.
• Mono-clonal antibodies are produced from cells
grown in tissue culture, and can be “made” to have
antibodies to any protein sequence.
• Sometimes the goal of cloning is to express protein
for the production of antibodies.
Question 7.12

A researcher wants to clone the genomic
sequences that include a human gene for
which a cDNA has already been obtained.
The researcher has a cDNA probe and a
variety of genomic libraries.

How many clones should the researcher screen for
being 95% sure at least one clone is hybridized by
the probe?



If screening a plasmid library with inserts on average 7
kb
If screening a lambda library with inserts on average 15
kb
If screening a YAC library with inserts on average 350
kb?
Answer 7.12

N = ln(1-p)/ln(1-f)



Ln(1-0.95)/ln(1-7000/3x109) = 1.3 x 106
plasmids
Ln(1-0.95)/ln(1-15000/3x109) = 6 x 105
phages
Ln(1-0.95)/ln(1-350000/3x109) = 2.6 x 104
YACs
Question 7.12

What advantages/disadvantages are
there to screening different libraries?


Larger average inserts, less number of
clones must be screened.
It is difficult to later analyze large inserts,
e.g., by using restriction enzyme mapping.
Question 7.15

You are given a genomic library of yeast
prepared in a bacterial plasmid vector.
You are also given a cloned cDNA for
human actin, a conserved protein. How
would you identify the yeast actin
gene?
Answer 7.15





Label the human actin cDNA to use it as a
heterologous probe.
Plate the genomic library on bacterial media
Overlay colonies with a positively charged
membrane, and lift it off
Lyse bacterial cells using alkaline soln. So that
single stranded plasmid DNA binds; hybridize
with the probe and detect
Identify the clone on the membrane with that
on the plate for further analysis.
Restriction Enzyme Analysis of
Cloned DNA sequences


Cloned DNA can be cut with restriction
enzymes and electrophoresed on agarose
gels and visualized with ethidium bromide, in
order to map its restriction sites
DNA cut with several enzymes, each loaded
in a lane of an agarose gel.

Electrical currents drives the negatively charged
DNA fragments through the gel. Small molecules
move much faster, so the fragments are separated
Restriction enzyme analysis of
cloned DNA sequences


DNA is stained with ethidium bromide, which
fluoresces under UV when complexed with
DNA. The gel is photographed, and the
distance migrated by each band of identical
DNA molecules is measured and compared
with a calibration curve.
Restriction mapping may be done with a
circular plasmid, a cloned sequence, or a
fragment of plasmid prepared by gel cutting.
Gel Electrophoresis
• A common technique in a molecular genetics lab is gel
electrophoresis. Several types of gel can be used
(agarose and acrylamide are the most common). All
work similarly: a gel matrix is formed, the DNA is
loaded into a “well” or slot in the gel. The gel is put
between the electrodes of a power supply, the DNA
moves through the gel toward the positive electrode
(since the phosphates are negatively charged). Small
fragments of DNA move faster (and farther) than large
fragments.
Restriction Mapping
• Once a fragment of DNA is cloned,
a restriction map is often made to
characterize the clone. A restriction
map is a diagram showing where
various restriction enzymes cut the
DNA. If the sequence of the cloned
fragment is known, the map can
verify that the right fragment of
DNA is cloned. If the sequence is
not known, the map provides a way
of identifying the fragment, and
information about possible
additional cloning steps.
Southern Blot
• A powerful technique for identifying a
specific sequence of DNA on a gel is the
“Southern” blot (named for Dr. Southern).
• DNA from an organism is cut with a
restriction enzyme, then separated by
electrophoresis.
• The DNA is then transferred to a
“membrane” and hybridized to a labeled
probe containing the sequence of interest.
• The position of the DNA complementary
(having the same sequence) to the probe is
detected by seeing the “label” in the probe.
Question 7.16

A cDNA library is made with mRNA
isolated from liver tissue, and digested
with the enzyme EcoRI (E), HindIII (H),
and BamHI (B).
E
H
1.1
E H
0.9
0.5 0.6
B
B
1.3
Question 7.16

When the cDNA from liver is used to screen a
cDNA library made with mRNA from brain,
three identical cDNA with the following
restriction map were produced.
E
H
1.1
B
1.2
B
1.3
Question 7.16: southern
genomic
E
H
B
Size
Kb
7.8
7.4
6.1
3.6
2.0
1.4
1.3
Question 7.16: northern cDNA
Liver Brain
Size
Kb
4.4
3.6
Answer 7.16

Do these cDNAs derive from the same
gene?

Since both cDNAs hybridize to the same
bands on a genomic Southern blot, they
are copies of mRNAs transcribed from the
same sequence.
Answer 7.16

Why are different sized bands seen on
the northern blot?

The primary mRNA may be alternatively
spliced in brain and liver tissue. It is
possible that the 0.8-kb difference between
two bands corresponds a 0.8-kb intron that
is spliced out in brain tissue that is not
spliced out in liver tissue.
Answer 7.16

Why do the cDNAs have different
restriction maps?
E
H
1.1
E
E H
0.9
0.5 0.6
H
1.1
B
B
1.3
B
1.2
B
1.3
Answer 7.16

Why do the cDNAs have different restriction
maps?

The two cDNAs are copies of mRNAs from two
different tissues. The northern indicates that
there are some differences between the mRNA in
their size. So it is not surprising that the RE maps
are also different.
E
H
1.1
E
E H
0.9
0.5 0.6
H
1.1
B
B
1.3
B
1.2
B
1.3
Answer 7.16

Why are some of the bands seen on the
whole-genome Southern blot different sized
than the RE map?

Genomic Southern blot gives an indication of the
gene organization at the DNA level, while cDNA
maps give an indication of the mRNA structure.
When cDNA used to probe genomic DNA
sequences, it will hybridize to transcribed
sequences that are connected to nontranscribed
sequences
Techniques and how they work

http://lifesciences.asu.edu/resources/m
amajis/index.html