Transcript No Slide Title

The nucleus contains the
chromosomes.
Cells have many
parts including the
nucleus
Plants are made of cells
DNA is two strings of ”bases "
Designated “A,” “C,” “G,” or “T”
A gene is part of a
chromosome
and is made of DNA
How much DNA is in cells?
Species
Wheat
Maize
Soybean
Tomato
Rice
Arabidopsis
Mouse
Human
Genome Size (Mbp)
16,000
2,500
1,115
1,000
425
120
3,000
3,000
Genetic Information Processing in a Cell
DNA
Promoter
Exon 1
Intron
Exon 2
Trans.Term.
Transcription
Primary RNA Transcript
Exon 1
Intron
Exon 2
Processing
Functional mRNA
Exon 1
AUG
Exon 2
AAAAAAAAAAAAAAAAA
Stop
Translation
Protein
Folding, Targeting, Decorating
DNA Structure:
Double Stranded, A always bonds to T, G bonds to C.
Chains have polarity, 5’ and 3’ refers to positions on sugar.
DNA sequence is always written in the 5’ to 3’ direction.
5’
P03-0-C
-0-P02-0-C
-0-P02-0-C
0
0
0
0
..
T
...
C
...
T
A
G
C
H0-
0
C-0-P02-0-
G
0
C-0-P02-0-
C-0-P02-0-
C
0
3’
-0H
A
..
0
3’
-0-P02-0-C
= Deoxyribose, in sugar-phosphate backbone
A = Adenine, G = Guanine, T = Thymine, C = Cytosine
0
C-0-P03
5’
Recognition sequences of some restriction enzymes
Enzyme
HindIII
Recognition Site
5’-AA-G-C-T-T- 3’
3’-T-T-C-G-AA- 5’
Type of End
5’ Phosphate
extension
BamHI
5’-GG-A-T-C-C- 3’ 5’ Phosphate
3’-C-C-T-A-GG- 5’
extension
PstI
5’ -C-T-G-C-AG- 3’
3’ -GA-C-G-T-C- 5’
3’ hydroxyl
extension
Sau3aI*
5’-G-A-T-C-- 3’
3’--C-T-A-G- 5’
5’ Phosphate
extension
EcoRV*
5’-G-T-TA-A-C- 3’
3’-C-A-AT-T-G- 5’
Blunt end
* Enzyme won’t cut if methylated.
How often do they cut? In theory,
‘4 cutter’ sites occur on average every 44 =256 nucleotides
‘6 cutter’ sites occur on average every 46 =4096 nucleotides
They cut less frequently if the frequencies of the different bases is not
equal, or if they are sensitive to cytosine or adenosine methylation.
Restriction Fragments can be Separated by Gel Electrophoresis
Discrete bands can be seen after restriction of DNA viruses, phage or plasmids but not
organisms with big genomes.
Lambda (48 Kb)
Maize (2500Mb)
23 kb
Stain
UV light
9.4 kb
6.5 kb
4.0 kb
2.3 kb
2.0 kb
+
0.5 kb
Small fragments run faster
Plasmid DNA
Ligase can join DNA Fragments with compatible ends
---GGTCCTCAATGACCTC
---CCTGGAGTTACTGGAGCTAG
AGCTTCTCGGAATCATGCATGC--AGAGCCTTAGTACGTACG---
BamHI end
HindIII end
No ligation
---GGTCCTCAATGACCTG
---CCTGGAGTTACTGGACCTAG
BamHI end
GATCCCTCGGAATCATGCATGC--GGAGCCTTAGTACGTACG--BamHI end
---GGTCCTCAATGACCTGGATCCCTCGGAATCATGCATGC-----CCTGGAGTTACTGGACCTAGGGAGCCTTAGTACGTACG---
---GGTCCTCAATGACCTG
---CCTGGAGTTACTGGACCTAG
BamHI end
GATCTCTCGGAATCATGCATGC--AGAGCCTTAGTACGTACG--Sau3A end
---GGTCCTCAATGACCTGGATCTCTCGGAATCATGCATGC-----CCTGGAGTTACTGGACCTAGAGAGCCTTAGTACGTACG---
Main features of Plasmid Cloning Vectors e.g. pUC19
Polycloning site:
AAGCTTGCATGCCTGCAGGTCGACTCTAGAGGATCCCCGGGTACCGAGCTCGAATTCA
HindIII
PstI
XbaI
BamHI
KpnI
EcoRI
ß Galactosidase
Intact ß Gal product cleaves
X-gal substrate in media, to
make a blue colony.
palindromes
2.7 Kb
Ori
Amp = ampicillin resistance
Amp
growth on amp media forces the
bacteria to maintain the plasmid.
Ori = origin of DNA
replication which also
controls copy number.
By inserting your sequences in a cloning vector with an Amp resistance gene you
can force E. coli to grow your sequences by growing it on Amp media.
Molecular Cloning:
Cutting of DNA and ligating it into a cloning vector will allow you to
propagate the fragment in some organism and subsequently purify it.
Cut Plasmid Vector
Cut source DNA
Mix and Ligate
Recombinant plasmid
Transform E. coli
How to select recombinant plasmids
Nonrecombinant plasmid
Recombinant plasmid
After transformation of E. coli, only circular plasmids with a single copy of the
vector sequences will replicate (ori needed). Only E. coli with a vector sequences
can grow on the antibiotic. But, both recombinant and nonrecombinant plasmids will
allow E. coli to grow.
How to select recombinant plasmids, cont’
Nonrecombinant plasmids have intact
ß-gal genes and make blue colonies
when grown in media with Xgal.
Blue colony
(non-recombinant)
White colony
(recombinant)
Recombinant plasmids make white colonies
because the ß-gal gene is interrupted
Making Restriction Site Maps: A 3 Kb HindIII fragment was cloned into the the HindIII site of the 2.7 Kb plasmid cloning
vector pUC19 (polycloning site shown below). The fragment sizes following restriction digestion allows you to make a map.
Cloning site
Left
Right
CCAAGCTTGCATGCCTGCAGGTCGACTCTAGAGGATCCCCGGGTACCGAGCTCGAATTCACTG
HindIII
PstI
XbaI
BamHI
KpnI
EcoRI
Fragment size data
Enzyme Digest
HindIII
EcoRI
KpnI
PstI
HindIII + PstI
BamHI
HindIII + BamHI
BamHI + PstI
Fragment sizes (in Kb)
3.0, 2.7
5.7
5.7
4.7, 1.0
2.7, 2.0, 1.0
3.9, 1.8
2.7, 1.8, 1.2
3.9, 1.0, 0.8
H
From pUC19
Restriction Map
B
P
HPBKE
Bacteriophage Lambda can be used as a cloning vector
Phage infects E. coli
Phage Genome replicates
Phage package their DNA, lyse cell and re-infect
Phage makes coat proteins
and packages it genome
Advantages of cloning in Lambda:
•Can clone > 20 Kb in replacement vectors
•Transfection more efficient than transformation
Different types of genomic cloning vectors
Typical insert size
Plasmid
Insert
Vector
<1 to 10 Kb
Lambda phage 10 to 23 Kb
Cosmid
20 to 45 Kb
A plasmid with site that enables it to
be incorporated into phage particles.
BAC
<50 to 150 Kb
Bacterial Artificial Chromosome (a
very big plasmid)
Yeast Artificial Chromosome
100 to 500 Kb
Maintained as a chromosome in yeast
Libraries of Genomic Clones
A library is simply a collection of clones. Genomic clones are made from
chromosomal DNA of some organism. A Genome Equivalent is the
number of clones it would take for the size of the cloned fragments to
equal the size of the genome of the organism. Fox example, consider a
genome equivalent for maize, with a genome size of 2.5 X 109 bases, or 2,
500,000 Kb.
Clone size
Genome equivalent
5 Kb (e.g. plasmid)
20 Kb (e.g. Lambda)
100 Kb (e.g. BAC)
500,000 clones
125,000 clones
25,000 clones
Libraries need to be bigger than a genome equivalent to have a good
chance of containing any given sequence. To have a 95% chance of
containing any given sequence a library needs to be three genome
equivalents. Four genome equivalents should have a 99% chance.
These calculations assume the cuts in the DNA occur randomly which is
never the case when using restriction enzymes.
Libraries of Genomic clones, cont’
Complete genomic libraries, of a specific size fragment, should not be
made from complete enzyme digests. For example, if you want 20 Kb
clones, Sau3a fragments are too small and EcoRI are usually too small
or too big.
20 Kb
S
S SS
E
S
E
S S SSS
E
E
SS
SS S S S SSS S S S S S
E E
E
E
S S S SS SS
S S SSS
S
S
E
Partial Sau3a digest fragments
Digests of maize DNA for 15 min. with different amounts of Sau3a
Increasing Sau3a conc.
20 Kb
2 Kb
Screening libraries by bacterial colony hybridization.
Plate transformed E coli,
bacteria carrying clones
form colonies*
Transfer colonies
to membrane
Expose to X-ray film to
identify colonies that
hybridize to probe
Lyse colonies with
NaOH, neutralize,
bake s.s DNA onto
membrane
Hybridize with labeled
s.s.DNA probe, which
will anneal to homologous
sequences
* Same approach works great with bacteriophage lambda plaques
DNA Polymerase can be used to Radioactively Label DNA
Step 1) Melt DNA into single strands.
5’ GTACGTCTGACTCGTTCATCTGCGCTTAGACTGCATGACGTAGCTGATCGCTGACAATTCGTAAT 3’
3’ CATGCAGACTGAGCAAGTAGACGCGAATCTGACGTACTGCATCGACTAGCGACTGTTAAGCATTA 5’
5’ GTACGTCTGACTCGTTCATCTGCGCTTAGACTGCATGACGTAGCTGATCGCTGACAATTCGTAAT 3’
3’ CATGCAGACTGAGCAAGTAGACGCGAATCTGACGTACTGCATCGACTAGCGACTGTTAAGCATTA 5’
Step 2) Add small random primers.
5’ GTACGTCTGACTCGTTCATCTGCGCTTAGACTGCATGACGTAGCTGATCGCTGACAATTCGTAAT 3’
3’ CTAGCG 5’
3’ CATGCAGACTGAGCAAGTAGACGCGAATCTGACGTACTGCATCGACTAGCGACTGTTAAGCATTA 5’
5’ TAGACT
Labeling DNA with Polymerase, continued...
Step 3) Extend new strand from template by incorporating
A, T, G and labeled C
5’ GTACGTCTGACTCGTTCATCTGCGCTTAGACTGCATGACGTAGCTGATCGCTGACAATTCGTAAT 3’
CATGCAGACTGAGCAAGTAGACGCGAATCTGACGTACTGCATCGACTAGCG 5’
*
*
*
*
* *
*
*
* * *
3’ CATGCAGACTGAGCAAGTAGACGCGAATCTGACGTACTGCATCGACTAGCGACTGTTAAGCATTA 5’
5’ TAGACTGCATGACGTAGCTGATCGCTGACAATTCGTAAT
*
*
*
* *
*
*
Step 4) Re-melt the double-stranded DNA and use to probe single
stranded target DNA.
Libraries can also be screened with immunological assays
Steps: 1) Plate and transfer cells to membrane
2) Lyse cells and bind protein to membrane
3) Hybridize membrane to primary antibody, made from
the protein you are interested in (e.g. in a rabbit)
4) Wash membrane and hybridize to secondary antibody
(e.g. goat anti-rabbit) which is chemically linked to an
enzyme (e.g. a peroxidase or phosphatase).
5) Wash membrane and apply substrate that is modified by
enzyme to produce a colored compound right on the filter.
Construction of cDNA libraries
A cDNA library represents the transcribed sequences in the cells the library
was made from. The content depends on the cell types used and the
treatment of those cells before harvest.
Why make a cDNA library instead of a genomic library?
You may want your gene already processed with no promoter. e.g.
for expression in another organism, or to compare to the genomic clone to
document mRNA processing.
Your gene may be in much higher frequencies in this library than a
genomic library.
You may be looking for genes that are turned on specifically in this
cell type, or trying to demonstrate that your DNA sequence is transcribed.
General process of making a cDNA library
1) purify polyadenylated RNA
mRNA
AAAAAAAAAAAAAAAAA
2) Make first strand DNA with poly-T primer, reverse transcriptase and dNTPs
AAAAAAAAAAAAAAAAA
TTTTTTTTT
3) Degrade RNA (chemically or enzymatically)
TTTTTTTTT
1st strand DNA
4) Make second strand using DNA polymerase & dNTPs. DNA primer used
varies with different protocols
TTTTTTTTT
AAAAAAAAA
5) Clone double stranded DNAs into cloning vector.