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Learning Objectives : •^ Understand the basic differences between genomic and cDNAlibraries •^ Understand how genomic libraries are constructed •^ Understand the purpose for having overlapping DNA fragments ingenomic^ libraries and how they are generated •^ Understand how cDNA libraries are constructed and the use ofreverse^ transcriptase for their construction •^ Understand the rationale for library screening •^ Understand the method of plaque hybridization •^ Understand the four methods for library screening and when theyare put into^ use
Molecular cloning inbacterial cells….This strategy can beapplied to genomicDNA as well as cDNA
NNG^ GATCCNNNNCCTAG^ GNN^ cut with Bam HI(6-base cutter) internal fragmentremove internalfragment“left arm” “right arm”
cut with Sau 3A (4-base cutter)which has ends compatiblewith Bam HI:NNN^ GATC
NNNNNNCTAG NNN isolate ~20 kbfragments human genomic DNA^ (isolated from
many cells) Bam HI sites:
“left arm”^ “right arm” combine and treatwith DNA ligase^ “left arm”
“right arm”^ package into bacteriophageand infect E. coli 2 3 14 5 6 7 •^ genomic library of human DNA fragmentsin which each phage contains a differenthuman DNA sequence
All possible sites: Results of a partial digestion:
= uncut^ = cut
Genomic Library making… The partial digest is one of themost important steps.
Why??? Due to the production of overlapping DNA fragments
Construction of a cDNA library^ •^ reverse transcriptase makes a DNA copy of an RNAThe life cycle of a retrovirus depends on reverse transcriptase^ retrovirus^ 1. virus enters celland looses envelope
- the capsid is uncoated, releasing genomicRNA and reverse transcriptase
- reverse transcriptasemakes a DNA copy 4. then copies the DNA strand tomake it double-stranded DNA,removing the RNA with RNase H 5. the DNA is then integratedinto the host cell genomewhere it is transcribed byhost RNA polymerase II
- it is translated into viral proteins,and assembled into newvirus particlesnew viruses
-^ cDNA library construction
AAAAA 5’^
3’ mRNA(all mRNAs in cell) anneal oligo(dT) primers of 12-18 bases in length
AAAAATTTTT 5’^
3’ 5’ add reverse transcriptase
and dNTPs^ AAAAATTTTT 5’ 3’
3’ 5’ cDNA add RNaseH^ (specific for the RNA strand of an RNA-DN
A
hybrid) and carry out a partial digestion
AATTTTT 5’^ short RNA fragments serve as primers for
3’ 3’ second strand synthesis using DNA polymerase I
AAAAATTTTT
5’ 3’^
NNNNNNNNG EcoRI linkers are ligated to both ends NNNNNNNNCTTAA
using DNA ligase AAAAANNNNNNNNGTTTTTNNNNNNNNCTTAA AATTCNNNNNNNN 5’ GNNNNNNNN 3’ • double-stranded cDNA copies of mRNA with EcoRI cohesive ends arenow ready to ligate into a bacteriophage lambda vector cut with EcoRI
EcoRI sites “left arm” “right arm” combine cDNAs withlambda arms and treatwith DNA ligase^ “left arm”^
“right arm” package into bacteriophageand infect E. coli 2 3 14 5 6 7 •^ cDNA library in which each phage containsa different human cDNA
cDNAs
Genomic DNA complexity • To screen for a clone in a library usually want a99% probability that your clone is found there. • Frequency is the size of the DNA fragment in thelibrary/the size of the haploid genome. For alambda library 17 kb (1.7 x 10
4 ) is the average size
of library. The size of the genome is^9 3 x 10bp^4 F = 1.7 x 10/ 3 x 10
9 bp N = ln (1-.99) / ln (1- [1.7 x 10
4 9 / 3 x 10]) N = ln .01 / ln (1 - 0.56 x 10
-5 ) N = -4.6061702 / -0.0000056 N = 822,351 clones
Genome equivalents • How many genome equivalents are therein this library? How do you calculate this?^4 822,351 x 1.7 x 10bps = 1.40 x 10
10 bps
Divide by the genome size 3.0 x 10
9 bps
= 4.67 times the genome equivalent How many positives will you get if youscreen for a single copy gene?
Insertional mutagenesis II • The use of the beta-galactosidase gene for aninsertional mutagenesis target allowed thescreening of all clones for those that containedinserts by a simple blue white color assay.
This
gene cleaves X gal (chromagen) to give rise to ablue dye that colors the bacteria or phageplaque. This allows the screening thoseplasmids or phage particles that contain DNAdisrupting the target gene.
Insertional mutagenesis III • In addition suppressor tRNA genes can beused to identify YAC that contain an insert.The suppressor tRNA can suppress theeffects of a^ Ade
ochre mutation. This
gives a white yeast colony. When thetRNA gene is disrupted the colonies arepink due to the accumulation of aprecursor of Adenine. Pink colonies arewhat is desired. See Figure 4.
