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Published byAdrian Marsh Modified over 9 years ago
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Molecular Biology Working with DNA
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Topics Genomic vs. Vector DNA Purifying plasmid DNA Restriction enzymes Restriction maps
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DNA Genomic Prokaryote vs. eukaryote Circular or linear One or more chromosomes Extra-genomic Vectors Plasmids
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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
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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
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Characteristics of plasmid vectors Restriction sites for cloning Origin of replication (Ori) Selection marker Genes conferring resistance to antibiotics
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DNA Isolation Goals Isolation of DNA of interest Chromosomal or plasmid? Eliminate other components Chromosomal or plasmid DNA? Proteins RNA Chemicals Salts, detergents, etc.
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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
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Plasmid DNA isolation by alkaline lysis (E.coli )
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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
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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??
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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
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Restriction enzymes Endonuclease Cleaves internal phosphodiester linkages. Recognize specific double stranded DNA sequences Different endonucleases recognize different sequences Recognize palindrome sequences
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Palindromes The same sequence is read in the 5’ » 3’ direction on both strands 5’-GGATCC-3’ 3’-CCTAGG-5’
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The same phosphodiester linkages are cleaved on both strands! 5’-G 3’-CCTAG GATCC-3’ G-5’
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Different ends are generated 5’-G 3’-CCT GA AG TCC-3’ G-5’ Blunt ends
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Different ends are generated 5’ overhangs 5’-G 3’-CCTAG GATCC-3’ G-5’
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Different ends are generated 3’ overhangs 3’-C 5’-GGATCC-3’ CTAGG-5’
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Compatibility of ends O P O P Blunt ends HO P OH P Compatible
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Compatibility of ends Overhangs HO P OH P HO PO P Incompatible
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Compatibility of ends Overhangs P-CTAG HO GATC-P OH Compatible P-CTAG OGATC-P O Annealing
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Compatibility of ends Overhangs P-TCCA HO GATC-P OH Incompatible P-TCCA HO GATC-P OH Annealing
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Restriction Maps
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Restriction maps Determining the positions of restriction enzyme sites Linear DNA maps Circular DNA maps (plasmids) Maps of inserts within vectors
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Approach 1. Determine whether the DNA has digested 2. Is the digestion complete or partial? 3. How many cuts? 4. Determine the relative positions
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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
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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
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Complete digestion Digestion
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Partial digestion: Partial undigested Digestion Non digested
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Partial digestion Digestion partial
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Is the digestion complete or partial? Compare to control Verify the intensity of the bands Verify the sizes Ladder Control 1234
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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
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Linear DNA maps EnzymeFragments (Kb) HindIII3 and 4 SalI2 and 5 HindIII + SalI2 and 3 3.0 4.0 HindIII 7.0 HindIII + SalI 2.0 3.0
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Circular DNA maps (plasmids) EnzymeFragments (Kb) BamHI2, 3 and 5 HindIII1 and 9 BamHI + HindIII1, 1.5, 2, 2.5 and 3 10.0 7.0 10.0 1.0 9.0 3.0 2.0 1.0 1.5 2.5
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Insertion maps Recombinant plasmid Insertion site Vector MCS
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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
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Insertion maps EnzymeFragments BamHI7.7Kb EcoRI1.0, 3.0, 3.7Kb PstI2.0 and 5.7 XbaI2.7 and 5.0 1. 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
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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 5.7 2 11 XbaI2.7 and 5.0 2 Insertion site 0 Sites to map
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Map of PstI : 2 and 5.7Kb 5.0 5.7 Kb 2.0 Kb
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Map of EcoRI: 1, 3 and 3.7Kb 1.0 3.0 3.71.0 3.01.0 P
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