MCB 720: Molecular Biology - PowerPoint PPT Presentation

1 / 18
About This Presentation
Title:

MCB 720: Molecular Biology

Description:

Recombinant DNA cloning procedure. Recombinant DNA cloning ... that confers resistance to ampicillin (source plasmid pBR322); (3) region ... Ampicillin (Amp) ... – PowerPoint PPT presentation

Number of Views:64
Avg rating:3.0/5.0
Slides: 19
Provided by: show63
Category:

less

Transcript and Presenter's Notes

Title: MCB 720: Molecular Biology


1
MCB 720 Molecular Biology
  • DNA cloning and restriction enzymes
  • Cloning vectors
  • Gene (genomic) libraries
  • Genomic library screening

2
Recombinant DNA cloning procedure
3
Recombinant DNA cloning procedure Animation
http//bcs.whfreeman.com/lodish5e/pages/bcs-main.a
sp?vcategorys00010n09000i09010.05o00510
006100052000530005400056000570005900060000
700007100001000020000300004000050010000200
00300004000050000600007000080000900010000
110001200013000140001500016000170001800019
0002000021000220002300099000ns456
4
Restriction enzymes DNA methylation
5
Recognition sequences of some REs
6
Mapping of restriction enzyme sites
7
Cloning vectors and their insert capacities
8
Plasmid cloning vectors
  • Three important features
  • Cloning site
  • Ori-an origin of replication
  • A selectable marker (ampr)

9
pBR322
ori
The plasmid pBR322 is one of the most commonly
used E.coli cloning vectors. pBR322 is 4361 bp in
length and contains (1) the replicon rep
responsible for the replication of plasmid
(source plasmid pMB1) (2) rop gene coding for
the Rop protein, which promotes conversion of the
unstable RNA I RNA II complex to a stable
complex and serves to decrease copy number
(source plasmid pMB1) (3) bla gene, coding for
beta-lactamase that confers resistance to
ampicillin (source transposon Tn3) (4) tet
gene, encoding tetracycline resistance protein
(source plasmid pSC101).
10
pUC18/19
pUC18 and pUC19 vectors are small, high copy
number, E.coli plasmids, 2686 bp in length. They
are identical except that they contain multiple
cloning sites (MCS) arranged in opposite
orientations. pUC18/19 plasmids contain (1) the
pMB1 replicon rep responsible for the replication
of plasmid (source plasmid pBR322). The high
copy number of pUC plasmids is a result of the
lack of the rop gene and a single point mutation
in rep of pMB1 (2) bla gene, coding for
beta-lactamase that confers resistance to
ampicillin (source plasmid pBR322) (3) region
of E.coli operon lac containing CAP protein
binding site, promoter Plac, lac repressor
binding site and 5-terminal part of the lacZ
gene encoding the N-terminal fragment of
beta-galactosidase (source M13mp18/19). This
fragment, whose synthesis can be induced by IPTG,
is capable of intra-allelic (alfa)
complementation with a defective form of
beta-galactosidase encoded by host (mutation
lacZDM15). In the presence of IPTG, bacteria
synthesize both fragments of the enzyme and form
blue colonies on media with X-Gal. Insertion of
DNA into the MCS located within the lacZ gene
(codons 6-7 of lacZ are replaced by MCS)
inactivates the N-terminal fragment of
beta-galactosidase and abolishes
alfa-complementation. Bacteria carrying
recombinant plasmids therefore give rise to white
colonies.
11
pGEM-3Z
12
Cloning foreign DNA into a plasmid vector
Alkaline phosphatase-removes 5 phosphate (P)
groups of DNA molecules BAP is more stable but
less active than CIP
T4 DNA ligase joins 5 phosphate (P) groups of
DNA molecules to 3 hydroxyl (OH) groups of DNA
13
Some antibiotics commonly used as selective agents
14
Eukaryotic gene organization
enhancers silencers
15
Genomic library construction
16
Screening a genomic library using DNA
hybridization to a (radio-)labeled DNA probe
17
Production of a (radio-)labeled DNA probe by the
random primer method uses the Klenow fragment of
DNA polymerase
5
3
5
3
5
3
18
How many genomic clones must be screened to find
your gene?
  • Theoretically, you will need to screen N clones
    where Nln(1-P)/ln(1-f) where Pthe probability
    of finding your gene and fthe average size of
    the cloned genomic sequence in your vector
    divided by the total genome size.
  • How many clones must you screen to find your gene
    in a human gene library packaged in EMBL 3 with
    99 certainty?
  • Nln(1-0.99)/ln(1-20kb/2.8 x 106kb) 6.4 x 105
    clones
Write a Comment
User Comments (0)
About PowerShow.com