Recombinant%20DNA%20cloning%20technology

1
Recombinant DNA cloning technology

• Chapter 7

2
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.

3
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

4
Restriction Enzymes

• Are found naturally in bacteria as a defense
  against vital DNA.
• Restriction sites are methylated in bacteria, and
  thus protected.

5
Restriction Enzymes

• Are denoted by three letter names derived from
  the bacterial strain they originate from.

6
Restriction Enzymes

• They usually are palindromes of 4-, 6- or 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.

7
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.

8
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.

9
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.

10
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.

11
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

12
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
  40GC.

13
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) 0.2x0.2x0.3x0.3x0.2x0.20.000144
• Number of sites in the genome 0.000144 x (3 x
  109) 432,000.

14
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
15
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.

16
Cloning Vectors

• Plasmid
• Phage
• Cosmid
• Shuttle
• Yeast Artifical Chromosomes (YACs)
• Bacterial Artificial Chromosomes (BACs)

17
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.

18
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
  (B-galactosidase). 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.

19
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.

20
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.

21
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)

22
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.

23
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.

24
Other plasmid vectors

• They contain
• 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.

25
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.

26
Phage l (lambda) as a Cloning Vector

• Plasmids are limited in the size of DNA that can
  be easily introduced into bacteria, about 5-10 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.

27
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).

28
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.

29
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.

30
Yeast Artificial Chromosomes (YACs)

• Function as artificial chromosomes in yeast.
• 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.

31
YACs

• Several hundred kb of insert DNA can be cloned in
  a YAC. YAC clones are made by
• 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).

32
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.

33
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.

34
Recombinant DNA libraries

• Genomic library
• Chromosome library
• cDNA library

35
Genomic libraries

• Are constructed by digesting genomic DNA
• Complete digestion
• Mechanical shearing
• Partial digestion

36
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.

37
Partial Digest for Producing Clonable Fragments
Enzymes with compatible Sticky ends are used.
38
How many clones are needed to contain all
sequences in genome?

• Depends on
• Size of the genome being cloned
• The average size of the DNA fragments inserted
  into the vector.

39
How many clones are needed to contain all
sequences in genome?

• The probability of having at least one copy of
  any DNA sequence in the genomic library is
• Nln(l-P)/ln(1-f),
• 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

40
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 f40,000/(3x109)
• Nln(l-P)/ln(1-f) 172,693 fragments

41
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

42
cDNA synthesis

• A short oligo(dT) primer is used. It anneals to
  the mRNAs poly(A) tail, allows reverse
  transcriptase to synthesize the cDNA (DNA-mRNA
  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.

43
cDNA from a Polyadenylated mRNA
Annealing Reverse transcription RnaseH
degradation DNA polymerase I DNA ligase
44
cDNA cloning

• 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

45
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.

46
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).

47
Klenow fragment PolI

• The large or Klenow fragment of DNA polymerase I
  has DNA polymerase and 3' -gt 5' exonuclease
  activities, and is widely used in molecular
  biology.

48
cDNA synthesis making probe

• DNA polymerases require a primer
  (oligonucleotides, 6-20 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.

49
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.

50
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.

51
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.

52
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?

53
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

54
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.

55
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?

56
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.

57
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

58
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.

59
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.

60
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.

61
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.

62
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
E
H
B
B
1.3
0.6
1.1
0.9
0.5
63
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
B
B
1.1
1.2
1.3
64
Question 7.16 southern genomic
E
H
B
Size Kb
7.8
7.4
6.1
3.6
2.0
1.4
1.3
65
Question 7.16 northern cDNA
Liver
Brain
Size Kb
4.4
3.6
66
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.

67
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.

68
Answer 7.16

• Why do the cDNAs have different restriction maps?

E
H
E
H
B
B
1.3
0.6
1.1
0.9
0.5
E
H
B
B
1.1
1.2
1.3
69
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
E
H
B
B
1.3
0.6
1.1
0.9
0.5
E
H
B
B
1.1
1.2
1.3
70
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

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Techniques and how they work

• http//lifesciences.asu.edu/resources/mamajis/inde
  x.html