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 FACTORDo 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’-C C T A G

G A T C C-3’

G-5’

Different ends are generated

5’-G

3’-C C T

G A

A G

T C C-3’

G-5’Blunt ends

Different ends are generated

5’ overhangs5’-G

3’-C C T A G

G A T C C-3’

G-5’

Different ends are generated

3’ overhangs3’-C

5’-G G A T C C-3’

C T A G G-5’

Compatibility of ends

OPO

P

Blunt ends

HOPOH

P

Compatible

Compatibility of ends

Overhangs

HOPOH

P

HOPO

P

Incompatible

Compatibility of ends

Overhangs

P-CTAGHOGATC-P

OH

Compatible

P-CTAGOGATC-P

O

Annealing

Compatibility of ends

Overhangs

P-TCCAHOGATC-P

OH

Incompatible

P-TCCAHO

GATC-POH

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

1.1. Is the DNA digested?Is the DNA digested?

Compare to the Compare to the undigested controlundigested control Which samples were Which samples were

not digested?not digested? 1 and 41 and 4

Which samples were Which samples were digested?digested?

2 and 32 and 3

Ladder

Control

1 2 3 4

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

DigestionNon digested

Partial digestion

Digestionpartial

partial

Is the digestion complete or partial?

Compare to control

Verify the intensity of the bands

Verify the sizes

Ladder

Control

1 2 3 4

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 The fragment sizes represent the distances between the sitesbetween the sites

Linear DNA mapsLinear DNA maps

Enzyme Fragments (Kb)

HindIII 3 and 4

SalI 2 and 5

HindIII + SalI 2 and 3

3.0 4.0 HindIII

7.0

HindIII + SalI2.0 2.03.0

Circular DNA maps (plasmids)Circular DNA maps (plasmids)

Enzyme Fragments (Kb)

BamHI 2, 3 and 5

HindIII 1 and 9

BamHI + HindIII 1, 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

Insertion mapsInsertion maps

Recombinant plasmid

Insertion site

Vector

MCS

MCS

Approach

1. Determine the total size2. 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

Enzyme Fragments

BamHI 7.7Kb

EcoRI 1.0, 3.0, 3.7Kb

PstI 2.0 and 5.7

XbaI 2.7 and 5.0

1.1. Total sizeTotal size• 7.7Kb7.7Kb

2.2. Insert sizeInsert size• 7.7 – 2.7 = 5.0Kb7.7 – 2.7 = 5.0Kb

3.3. Insertion siteInsertion site• Generates 2 Generates 2

fragments of which fragments of which one is the size of the one is the size of the vectorvector

• XbaIXbaI

Insertion maps

Enzyme

FragmentsTotal cuts

Sites in vector

Sites in insert

BamHI 7.7Kb 1 1 0

EcoRI 1.0, 3.0, 3.7Kb 3 1 2

PstI 2.0 and 5.7 2 1 1

XbaI 2.7 and 5.0 2 Insertion site

0

Sites to map

Map of PstI : 2 and 5.7Kb

5.0

5.7 Kb2.0 Kb

Map of EcoRI: 1, 3 and 3.7Kb

1.0 3.0

3.7 1.03.0 1.0

P