1. Molecular Biology
Working with DNA

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

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

4. 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

5. 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

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

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

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

9. Plasmid DNA isolation by alkaline lysis (E.coli )

10. 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

11. 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??

12. 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

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

14. 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’

15. 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’

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

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

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

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

20. Compatibility of ends
Overhangs HO P OH HO P O
Incompatible

21. Compatibility of ends
Overhangs HO P OH HO P O
Incompatible

22. Compatibility of ends Annealing Compatible Overhangs P-CTAG HO GATC-POH
Annealing
GATC-P O
P-CTAG O
Compatible

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

24. Mapping Restriction Sites

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

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

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

28. 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)

29. 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…

30. 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?

31. 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?

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

33. Examples

34. Examples

35. 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

36. 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.

37. 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

38. 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

39. Must determine size of the insertND 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)

40. 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 =2500

41. 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)

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

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

44. 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

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