Molecular Biology Working with DNA

Topics  Genomic vs. Vector DNA  Purifying plasmid DNA  Restriction enzymes  Restriction maps

DNA  Genomic  Prokaryote vs. eukaryote  Circular or linear  One or more chromosomes  Extra-genomic  Vectors  Plasmids

Vectors Vs Plasmids  Vector:  DNA vehicle that allows the cloning, maintenance and amplification of a DNA sequence  Plasmids  Virus  Chromosomes  All plasmids are vectors  Not all vectors are plasmids

Plasmids  Small circular DNA molecules maintained and amplified in eukaryotic or prokaryotic cells  Amplification in bacteria  Used as vector for cloning or expression of DNA of interest

Characteristics of plasmid vectors  Restriction sites for cloning  Origin of replication (Ori)  Selection marker  Genes conferring resistance to antibiotics

DNA Isolation  Goals  Isolation of DNA of interest  Chromosomal or plasmid?  Eliminate other components  Chromosomal or plasmid DNA?  Proteins  RNA  Chemicals  Salts, detergents, etc.

DNA isolation (cont’d)  Cell lysis  Cell wall and membrane  Enzymatic  Chemical  Mechanical  Isolation of DNA of interest  Differential sedimentation  Chromatography  Removing other components  Enzymatic  Differential sedimentation  Chromatography

Plasmid DNA isolation by alkaline lysis (E.coli )

Solutions Used  Sol. I – Resuspension buffer  Tris HCl – Buffer that protects nucleic acids  EDTA - Chelates Mg++, prevents nucleases from working  Sol. II – Lysis solution  NaOH - ^pH lyses cells, denatures DNA  SDS – Dissolves membranes, denatures and binds proteins

Solutions Used (Cont’d)  Sol. III- Potassium acetate  Renaturation of DNA  Precipitates SDS  Precipitates genomic DNA and proteins  Isopropanol / Ethanol  Precipitates nucleic acids (plasmid and ?)  Salts remain soluble  TE-RNase - Tris & EDTA again; RNase??

Quantification of DNA  Determining Conc. of DNA  A260 of 1.0 = 50µg/mL or 50ng/µL  Determining Amount of DNA  1mL of a solution with an A260 of 1.0 contains 50µg DNA  1µL of a solution with an A260 of 1.0 contains 50ng DNA  Do not forget to account for the DILUTION FACTOR

Restriction enzymes  Endonuclease  Cleaves internal phosphodiester linkages.  Recognize specific double stranded DNA sequences  Different endonucleases recognize different sequences  Recognize palindrome sequences

Palindromes  The same sequence is read in the 5’ » 3’ direction on both strands 5’-GGATCC-3’ 3’-CCTAGG-5’

 The same phosphodiester linkages are cleaved on both strands! 5’-G 3’-CCTAG GATCC-3’ G-5’

Different ends are generated 5’-G 3’-CCT GA AG TCC-3’ G-5’ Blunt ends

Different ends are generated 5’ overhangs 5’-G 3’-CCTAG GATCC-3’ G-5’

Different ends are generated 3’ overhangs 3’-C 5’-GGATCC-3’ CTAGG-5’

Compatibility of ends O P O P Blunt ends HO P OH P Compatible

Compatibility of ends Overhangs HO P OH P HO PO P Incompatible

Compatibility of ends Overhangs P-CTAG HO GATC-P OH Compatible P-CTAG OGATC-P O Annealing

Compatibility of ends Overhangs P-TCCA HO GATC-P OH Incompatible P-TCCA HO GATC-P OH Annealing

Restriction Maps

Restriction maps  Determining the positions of restriction enzyme sites  Linear DNA maps  Circular DNA maps (plasmids)  Maps of inserts within vectors

Approach 1. Determine whether the DNA has digested 2. Is the digestion complete or partial? 3. How many cuts? 4. Determine the relative positions

1.Is the DNA digested? Compare to the undigested control Which samples were not digested? 1 and 4 Which samples were digested? 2 and 3 Ladder Control 1234

2.Is the digestion complete?  Complete digestion  All the DNA molecules are cleaved at all the possible sites  Partial digestion  A fraction of the molecules are not digested  Partial undigested  A fraction of the molecules were digested, but not at all the possible sites  Partial digestion

Complete digestion Digestion

Partial digestion: Partial undigested Digestion Non digested

Partial digestion Digestion partial

Is the digestion complete or partial? Compare to control Verify the intensity of the bands Verify the sizes Ladder Control 1234

3.How many cuts?  Number of sites  Circular DNA = number of bands  Linear DNA = Number of bands – 1 4. Determine the relative positions  The fragment sizes represent the distances between the sites

Linear DNA maps EnzymeFragments (Kb) HindIII3 and 4 SalI2 and 5 HindIII + SalI2 and HindIII 7.0 HindIII + SalI

Circular DNA maps (plasmids) EnzymeFragments (Kb) BamHI2, 3 and 5 HindIII1 and 9 BamHI + HindIII1, 1.5, 2, 2.5 and

Insertion maps Recombinant plasmid Insertion site Vector MCS

Approach 1. Determine the total size 2. Determine size of the insert  Total size – size of vector 3. Determine the insertion site within the MCS 4. Determine which enzymes cut wihin the insert 5. Relative mapping in relation to the sites at known positions

Insertion maps EnzymeFragments BamHI7.7Kb EcoRI1.0, 3.0, 3.7Kb PstI2.0 and 5.7 XbaI2.7 and Total size 7.7Kb 7.7Kb 2. Insert size 7.7 – 2.7 = 5.0Kb 7.7 – 2.7 = 5.0Kb 3. Insertion site Generates 2 fragments of which one is the size of the vector Generates 2 fragments of which one is the size of the vector XbaI XbaI

Insertion maps EnzymeFragmentsTotal cuts Sites in vector Sites in insert BamHI7.7Kb 1 10 EcoRI1.0, 3.0, 3.7Kb 3 12 PstI2.0 and XbaI2.7 and Insertion site 0 Sites to map

Map of PstI : 2 and 5.7Kb Kb 2.0 Kb

Map of EcoRI: 1, 3 and 3.7Kb P