1. Molecular Biology
Restriction enzymes

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

3. 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’

4. 5’-GGATCC-3’ 3’-CCTAGG-5’ Palindromes
The base at one position determines the corresponding base at the same position on the other strand
5’-GGATCC-3’
3’-CCTAGG-5’

5. Palindromes CAC_ _ _ GT _ TA _ AT _ _ AT _ _ CACGTC GTATAC ATATATAT
Complete the following palindromes
CAC_ _ _
GT _ TA _
AT _ _ AT _ _
CACGTC
GTATAC
ATATATAT

6. Palindromes Problem How many palindromes of 4 bases are possible?
N1N2N3N4
Position 1: 4 choices (A, G, C, or T)
Position 2: 4 choices (A, G, C, or T)
Position 3: 1 choice (determined by position 2)
Position 4: 1 choice (determined by position 1)
Therefore: 4 X 4 X 1 X 1 =16

7. How many palindromes does the enzyme AccI recognize?
Problem
How many palindromes does the enzyme AccI recognize?
GTMKAC (M= A or C; K = G or T)
1 X 1 X 2 X 1 X 1 = 2

8. Frequency of Occurrence
Probability of finding a given palindrome
Depends on the base distribution
Ex. 50% G or C or 1/4 for each base
What is the probability of occurrence of GGATCC?
1/46 ou 1 divided by (0.25)-6 = 1/4096
How many times would this enzyme be expected to cut in a 10 kb genome?
1/4096 X = 2.44
Round off to nearest whole number. Therefore 2X

9. Frequency of Occurrence
Problem
What is the probability of finding any 4 base palindrome in a genome which is 50% G or C
Number of different sequences of 4 bases
44 = 256
Number of different palindromes of 4 bases
4 X 4 X 1 X 1 = 16
Probability of finding a 4 base palindrome
16/256 = 1/16

10. Frequency of Occurrence
Unequal distribution of bases
Determine the sequence of each possible palindrome
Independently determine probability of each palindrome according to base distribution
Determine the sum of probabilities
Ex. AccI cuts GTMKAC (M= A or C; K = G or T)
Base distribution: 70% G/C, 30% A/T
2 palindromes possible
GTATAC: (0.35)(0.15)(0.15)(0.15)(0.15)(0.35) = 1/16125
GTCGAC: (0.35)(0.15)(0.35)(0.35)(0.15)(0.35) = 1/2962
Probability: 1/2502

11. 5’-G 3’-C C T A G G A T C C-3’ G-5’ Cleavage
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’

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

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

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

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

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

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

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

19. Restriction Maps

20. Compaible ends
Which restriction sites in List 1 would generate compatible ends with those generated from List 2?
List 1
GA▼TATC
AAAT ▼ TT
CGC ▼ CGC
A ▼ CGCGT
List 2
CCAT ▼ GGC
CAC ▼ GTG
CCGC ▼ GG
CCATAT ▼ ATGG
2+A & 2+D

21. Restriction maps Determining the positions of restriction sites
Linear DNA maps
Circular DNA maps (plasmids)
Maps of inserts within vectors
Mapping genomes
Southern analysis

22. Approach Determine whether the DNA has digested
Is the digestion complete or partial?
How many cuts?
Determine the relative positions

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

24. 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

25. Complete digestion
Digestion

26. Partial digestion: Partial undigested
Non digested
Digestion

27. Partial digestion
partial
Digestion

28. Is the digestion complete or partial?
Ladder
Control 1 2 3 4
Compare to control
Verify the intensity of the bands
Verify the sizes

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

30. Linear DNA maps HindIII HindIII + SalI Enzyme Fragments (Kb) HindIII
3 and 4
SalI
2 and 5
HindIII + SalI
2 and 3
7.0
3.0
4.0
HindIII
HindIII + SalI
2.0
3.0

31. 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
3.0
2.0
1.0
1.5
2.5
7.0
10.0
1.0
9.0
10.0

32. Recombinant plasmids
MCS X
Digested with X
Vector

33. Recombinant plasmid +
Insert X
Recombinant
Vector + insert

34. Determining restriction map of a DNA insert
Step 1: Determining Insertion Site +
Insert X
Cut with X

35. Insertion maps Total size Insert size Insertion site 7.7Kb
Generates 2 fragments one of which is the size of the vector
XbaI
Enzyme
Fragments
BamHI
7.7Kb
EcoRI
1.0, 3.0, 3.7Kb
PstI
2.0 and 5.7
XbaI
2.7 and 5.0

36. Determining Relative Orientation
Insertion site delimits Right & Left
Ex. If Pst is the insertion site: Bam is to the left.
If Xba is the insertion site, then Bam is to the right

37. Determining Relative OrientationX A X
Orientation 2 A X A X

38. Relative mapping Sites to map Enzyme Fragments Total cuts
Sites in vector
Sites in insert
BamHI
7.7Kb 1
EcoRI
1.0, 3.0, 3.7Kb 3 2
PstI
2.0 and 5.7
XbaI
2.7 and 5.0
Insertion site

39. Map of PstI : 2 and 5.7Kb
5.7 Kb
2.0 Kb
5.0

40. Map of EcoRI: 1, 3 and 3.7Kb P
3.7
1.0
3.0
1.0
1.0
3.0

41. Restriction analysis of large DNA
Purpose
Find the position of a sequence in a large DNA fragment
Find similar sequences in other organisms
Analyze non-sequenced DNA

42. Digestion of Large DNA
Number of fragments makes analysis impossible

43. Southern Analysis Separate DNA fragments according to size Denature
Hybridize with single stranded probe representing region of interest
Probe allows visualization of only the region of interest

44. Southern
Probe
Agarose gel
Hybridize to probe

45. Problem
What fragment sizes would be observed with probes 1 and 2 following sigle digests with each enzyme?

46. Solution Enzyme Probe 1 Probe 2 C 2.5 & 11kb 11kb K >11kbX
>9.8 & 10.2kb
>9.8kb