Molecular Biology Working with DNA

Topics Genomic vs. Vector DNA Purifying plasmid DNA Restriction enzymes Basics of restriction mapping

DNA Genomic Extra-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

Plasmid DNA isolation by alkaline lysis (E.coli )

Solutions Used Sol. I – Resuspension buffer Sol. II – Lysis solution 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 palindromes

5’-G G A T C C-3’ 3’-C C T A G G-5’ Palindromes The same sequence is read in the 5’ » 3’ direction on both strands 5’-G G A T C C-3’ 3’-C C T A G G-5’

5’-G 3’-C C T A G G A T C C-3’ G-5’ The same phosphodiester linkages are cleaved on both strands! 5’-G 3’-C C T A G G A T C C-3’ G-5’

Different ends are generated 3’-C C T G A A G T C C-3’ G-5’ Blunt ends

Different ends are generated 3’-C C T A G G A T C C-3’ G-5’ 5’ overhangs

Different ends are generated 3’-C 5’-G G A T C C-3’ C T A G G-5’ 3’ overhangs

Compatibility of ends Blunt ends HO P OH O P Compatible

Compatibility of ends Overhangs HO P OH HO P O Incompatible

Compatibility of ends Overhangs HO P OH HO P O Incompatible

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

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

Mapping Restriction Sites

Is the DNA digested? ND D1 D2 Must compare to an undigested control

Is the digestion complete Must compare to an undigested control ND D1 D2 Partial Complete

Partial Digestion All the target molecules are not cut at all the possible sites! Generates different intermediate products

Partial Digestion 1 2 3 1 Product of a complete digestion 1 + 2 Product of a partial digestion (intermediate product) 2 + 3 Product of a partial digestion (intermediate product)

Complete Vs Partial A complete digestion results in a stoechiometry of 1:1:1… The number of molecules of the different fragments is the same! The stoechiometry of a partial digest is variable: Ex. 1:3:2…

Ex. 1 2 3 How many molecules of each of the fragments would be generated following a complete digest? 3; therefore a stoechiometry of 3:3:3 =1:1:1 How many molecules of each of the fragments would be generated following a partial digest?

Assessing the stoechiometry on an agarose gel Basis The amount of ethidium bromide that binds to DNA is proportional to the size of the DNA The amount of ethidium bromide that binds to DNA is proportional to the amount of DNA With a complete digestion, the amount of ethidium bromide that binds DNA is only a function of the size of the DNA Why?

Ex. D1 D2 Stoechiometry not respected Stoechiometry not respected Stoechiometry respected Stoechiometry respected

Examples

Examples

1st step: Determine the number of cuts Was the DNA digested? No- No mapping required Yes Is the digestion complete? Is the digestion partial? Which products are intermediates How many times was the DNA cut The cut sites are in the vector No mapping required The cut sites are in the insert Mapping required

How many restriction sites are there? Linear DNA (ex. Human Chromosome) The number of cuts is equal to the number of fragments minus one. Circular DNA (ex. Plasmid) The number of cuts is equal to the number of fragments.

The cut sites are in the vector or the insert? M ND V B P E insert B E ? S Size of vector 2.5kpb P B cuts 2 times in the vector and 0 times in the insert E cuts 1 times in the vector and 1 times in the insert P cuts 1 times in the vector and 2 times in the insert

2nd step: What are the positions of the restriction sites? Relative mapping is done Restriction sites are mapped as a function of a reference point The position of the reference point must be known

Must determine size of the insert ND V B P E insert B E ? S Size of vector 2.5kpb P Determine the size of the fragments resulting from a complete digestion Determine the total size of the plasmid (Sum of sizes) Determine the size of the insert (Size of plasmid - Size of vector)

Determining Sizes P 0.6 0.9 2.1 3000 1100 900 E 0.3 1.9 4500 1000 L P Enz Distance (cm) Size bp P 0.6 0.9 2.1 3000 1100 900 E 0.3 1.9 4500 1000 Thus: Size of plasmid 5Kbp Size of insert 5000-2500=2500

Mapping Site P 1.1kbp ou P 1.1kbp P Impossible since the second site must be in the insert! P Insert (2.5kbp) 0.9kbp ou P 0.9kbp P Impossible since the second site must be in the insert! P Insert (2.5kbp)

Mapping Site P 0.9kbp 1.1kbp P P P Insert (2.5kbp) Or Insert (2.5kbp)

Determining Orientation 1.0kbp E E P Insert (2.5kbp)

Determining Orientation 0.9kbp 1.1kbp P P P Insert (2.5kbp) 1.0kbp E E Or Insert (2.5kbp) P 0.9kbp 1.1kbp 1.0kbp E E

Determining Orientation (Cont’d) L UD P E P+E P+E