The genetic material of any organisms is the substance that stores information about structure, function and Development of various characteristics of a living organisms.

1. 1

2. 2

3. Dr. Yanal Ahmad Al-kuddsi
Biotechnology Researcher
Department of Biotechnology
General Commission for
Scientific Agricultural Research
Damascus – Syria
alkuddsi.strikingly.com-yanal-http://dr

4. -Introduction.
-DNA Structure.
-DNA Repair Mechanism.
4
DNA Replication

5. 5

6. All living organisms contains DNA as genetic material
except virus where RNA is the genetic material.
There are two types of nucleic acids are found in the
cells of all living organisms. These are:
1- Deoxyribonucleic acid -DNA.
2- Ribonucleic acid - RNA.
1-Introduction.
6

7. The number of DNA molecules = The number of
chromosome per cell.
In bacteriophages and viruses there is a single molecule
of DNA, which remains coiled and is enclosed in a
protein coat.
The main role of DNA in the cell is the long- term
storage of information.
The DNA segments that carry this genetic information
are called genes.
7

8. DNA is found in combination with proteins
forming nucleoproteins or the chromatin
material.
Evidences That DNA Is
The Genetic Material
. The genetic material of any organisms is the substance
that stores information about structure, function and
Development of various characteristics of a living
organisms.
8

9. Experimental Evidence:
.DNA was not considered to be the possible genetic
material .
Bacterial Recombination Bacterial Transformation
Bacteriophage infectionTransduction
9

10. 1928Frederick Griffith,
studied Streptococcus pneumoniae ,
apathogenic bacterium causing pneumonia
there are 2 strains of Streptococcus:
- S strain is virulent.
- R strain is nonvirulent.
Griffith infected mice with these strains
hoping to understand the difference between
the strains.
10

11. Transformation Of Bacteria
Two Strains Of Streptococcus
Capsules
Smooth Strain
(Virulent)
Rough Strain
(Harmless)
Timothy G ,Standish,200011

12. Griffith’s results:
- live S strain cells killed the mice.
-live R strain cells did not kill the
mice.
-heat-killed S strain cells did not kill
the mice.
-heat-killed S strain + live R strain
cells killed the mice.
12

13. Griffith's experiment to demonstrate genetic transformation in
Diplococcus bacteria
13

14. Griffith’s conclusion:
- information specifying virulence
passed from the dead S strain cells
into the live R strain cells.
- Griffith called the transfer of this
information transformation.
14

15. DNA = DeoxyribosePentose (5-carbon) sugar
Nitrogenous base
RNA = Ribose
(Compare 2' carbons)
PURINES
-Adenine A
Guanine G-
PYRIMIDINES
-Cytosine C
-Thymine(DNA) T
-Uracil(RNA) U
Rastogi,V.
OH
O
OH
OHOH
OH
O
OH
OH
Ribose:Deoxyribose:
2- DNA STRUCTURE
15

16. nucleotide structure
The nucleotide structure consists of:
-the nitrogenous base attached to the 1’ carbon of
deoxyribose.
-the phosphate group attached to the 5’ carbon of
deoxyribose.
-a free hydroxyl group (-OH) at the 3’ carbon of
deoxyribose.
Phosphate group attached to 5’ carbon
16

17. OH
O
CH2
Sugar
H
H
H
A Nucleotide
Adenosine Mono Phosphate (AMP)
OH
NH2
N
N N
N
BaseP
O
OH
HO O
Phosphate
2’3’
4’
5’
1’
Nucleotide
Nucleoside
H+
-
2000Timothy G. Standish,17

18. The Role of Nucleotides
Energy
Carrying Genetic Information
Essential Chemical Links
Components of Enzyme
18

19. Histones
The family of five basic proteins thatHistones :
associate tightly with DNA in the chromosomes of all
eukaryotic cells.
Properties of Histones
- There are five classes of them: H1, H2A, H2B, H3,H4
- Generally small polypeptides 11 - 21 kd.
- Positively charged
􀂾 high content of positively charged amino acids
􀂾 About one in four residues is either lysine or
arginine.19

20. - Basic pI values
􀂾 pI values of Giardia lambia histones:
H2a = 10.48, H2b = 9.38, H3 = 10.58, H4 = 10.79
􀂾 Facilitates interactions with negatively charged
phosphate backbone of DNA.
The core histones into the nucleosome .20

21. Nucleosome:
Nucleosome:
- Proposed by Roger Kornberg in 1974
- Structural unit for packaging DNA
- Consists of histone core (octamer) plus ~ 200 bp of DNA
-DNA content of nucleosome is different in different
organisms, ranging from 160-240 .
Nucleosome core particle:
- histone core + 146 bp of DNA associated with the
octamer
- Nucleosome core is nearly the same in all organisms .
Chromatosome:
-Nucleosome + H121

22. 22

23. Pyrimidines
- One ring only.
- The ring is hexagonal and heterocyclic.
- It is formed of four carbon and two nitrogen atoms.
- The atoms in the ring are numbered clockwise.
- Nitrogen atoms are present at the first and third position and
rest of the positions are occupied by carbon atoms .
- Cytosine and Thymine found in DNA. But in RNA they are Cytosine
and Uracil.
23

24. HIStructural formulae of the pyrimidines in DNA and RNA
24

25. PURINES
- Purines are two –ringed nitrogen compounds.
-- Each purine molecule is dicyclic, formed of nine atoms.
-- There is one pyrimidine ring of six atoms and an
imidazole ring of five atoms.
- The atoms are numbered anticlockwise in pyrimidine
ring but clockwise in imidazole ring.
- The two rings share carbon atoms present on 4th and
5th positions. The nitrogen atoms in purine ring are four in
number present at first ,third ,seventh and ninth positions.
25

26. Structural formulae of the purines
26

27. -+
+
+
-
-
Base Pairing
Guanine And Cytosine
27

28. +
- Thymine
-
+
Adenine
Base Pairing
Adenine And Thymine
28

29. Base Pairing
Adenine And Cytosine
-
+
-
29

30. Base Pairing
Guanine And Thymine
+
+
-
30

31. Question:
-If there is 30% Adenine,
how much Thymine and
Cytosine is present?
Answer: Chargaff's Equimolar Base Ratio))
31

32. 1- The purine and pyrimidine components occur in
equal amounts in a DNA molecule.
2- The amount of adenine (A) is equivalent to the
amount of thymine (T) and of cytosine (C) is equivalent
to that of guanine (G).
3- The base ratio between A = T pair and G = C pair
may vary in the DNA of different groups of animals but
is constant for a particular species.
32

33. - In a body or somatic cell:
A = 30.3%
T = 30.3%
G = 19.5%
C = 19.9%
-There would be 20%
Cytosine
Adenine (30%) = Thymine
(30%)
Guanine (20%) = Cytosine
(20%)
Therefore, 60% A-T and
40% C-G.
33

34. Nucleotides are Covalently Linked to form
Polynucleotide Chain
- DNA is a macromolecule formed by the linking of
several thousand of nucleotides. These are called
monomerse or building blocks of DNA.
- The phosphoric acid molecule is attached to fifth
carbon atom (C-5') of the deoxyribose molecule.
-The phosphoric molecule of a nucleotide is joined
to the third carbon atom of the deoxyribose of the
next nucleotide by covalent bond. .Such 5'-3'
linkages between adjacent nucleotides are called
phosphodiester bonds.
34

35. -The adjacent nucleotides are connected together
forming the sugar-phosphate-sugar-phosphate chain
in which sugar and phosphate molecules are
arranged in alternate fashion.
-The nitrogenous bases are attached to the first
carbon atom of the deoxyribose and are directed at
right angle to the long axis of the polynucleotide
chain. They are stacked one above the other like the
steps in a ladder.
Each polynucleotide chain has marked ends:-
Its top end has a sugar residue with free 5' carbon
atom which is not linked to another nucleotide. The
triphosphate group is still attached to it. This end is
called the 5' end or 5'-P terminus.
35

36. The other end of the chain ends in a sugar residue with C-
3' carbon atom not linked. It bears 3' –OH group. This
end of polynucleotide chain is called 3' end or 3' –OH
terminus.
36

37. A polynucleotide chain37

38. 38

39. - James Watson and Francis Crick, 1953
deduced the structure of DNA using
evidence from Chargaff, Franklin, and
others proposed a double helix structure.
-The double helix consists of:
-2 sugar-phosphate backbones
-nitrogenous bases toward the interior of
the molecule.
-bases form hydrogen bonds with
complementary bases on the opposite
sugar-phosphate backbone.
39

40. -The two strands of nucleotides are
antiparallel to each other one is oriented 5’
to 3’, the other 3’ to 5’.
-The two strands wrap around each other to
create the helical shape of the molecule.
Nitrogenous
Base (A,T,G or C)
“Rungs of ladder”
“Legs of ladder”
Phosphate &
Sugar Backbone40

41.  helical structure of DNA
 major & minor groves
 10Å radius & 20Å diameter
 3.4Å between nucleotide
base pairs
 34Å / 360° turn
 10 nucleotide base pairs /
360° turn
Minor
groove
Major
groove
41

42. Biological significance of Watson and Crick’s
Double Helical Model OF DNA
Self Replication
Information Transfer
Information Storage
Variation
DNA Repair
Transcription
Translation
42

43. Rosalind Franklin and Maurice Wilkins
Rosalind Franklin and Maurice
Wilkins
-Franklin performed X-ray
diffraction studies to identify
the 3-D structure
-discovered that DNA is helical
-discovered that the molecule
has a diameter of 2nm and
makes a complete turn of the
helix every 3.4 nm.43

44. Forms of the Double Helix
1 nm
Major
groove
Minor
groove
A T
T A
G C
C G
C G
G C
T A
A T
G C
T A
A T
C G
0.34 nm
3.9 nm
B DNA
0.26 nm
2.8 nmMinor
groove
Major
groove
1.2 nm
A DNA
0.57 nm
6.8 nm
0.9 nm
Z DNA
2000Timothy G. Standish,44

45. DNA-ZDNA-BDNA-AForm
Left-handedRight-handedRight-handedHelical Sense
Å18~Å20~~26ÅDiameter
dimmers)6)121011.6Base pairs per helical
turn
9° for pyrimidine-
purine steps; 51° for
purine – pyrimidine
steps
º36°31
Helical twist per base
pair
44ÅÅ34Å34Helical pith(rise per
turn)
7.4Å per dimer3.4 Å2.9 ÅHelix rise per base
7°6°20°Base tilt normal to the
helix axis
FlatWide and deepNarrow and deepMajor groove
Narrow and deepNarrow and deepWide and ShallowMinor groove
C2' –endo for
pyrimidines;C3 –
endo for pyrines.
C2' –endoC3' –endoSugar pucker
Anti for pyrimidines
;syn for purines.
AntiAntiGlycosidic bond
45

46. Denaturation and Renaturation
- Heating double stranded DNA can overcome the
hydrogen bonds holding it together and cause the
strands to separate resulting in denaturation of the
DNA
- When cooled relatively weak hydrogen bonds between
bases can reform and the DNA renatures
TACTCGACATGCTAGCAC
ATGAGCTGTACGATCGTG
Double stranded DNA
TACTCGACATGCTAGCAC
ATGAGCTGTACGATCGTG
Double stranded DNA
TACTCGACATGCTAGCAC
ATGAGCTGTACGATCGTG
Denatured DNA
46

47. - DNA with a high guanine and cytosine content has
relatively more hydrogen bonds between strands
- This is because for every GC base pair 3 hydrogen bonds
are made while for AT base pairs only 2 bonds are made
- Thus higher GC content is reflected in higher melting or
denaturation temperature
Intermediate melting temperature
Low melting temperatureHigh melting temperature
67 % GC content -
TGCTCGACGTGCTCG
ACGAGCTGCACGAGC
33 % GC content -
TACTAGACATTCTAG
ATGATCTGTAAGATC
TACTCGACAGGCTAG
ATGAGCTGTCCGATC
50 % GC content -
47

48. Determination of GC Content
-Comparison of melting temperatures can be used to determine the
GC content of an organisms genome.
-To do this it is necessary to be able to detect whether DNA is melted
or not.
-Absorbance at 260 nm of DNA in solution provides a means of
determining how much is single stranded.
-Single stranded DNA absorbs 260 nm ultraviolet light more strongly
than double stranded DNA does although both absorb at this
wavelength.
-Thus, increasing absorbance at 260 nm during heating indicates
increasing concentration of single stranded DNA.
48

49. 49

50. DNA Replication
Common features of DNA polymerases
1. Polymerization occurs only 5' to
3‘
2. Requires a template to copy
3. Requires 4 dNTPs (dATP, dGTP,
dCTP and dTTP)
4. Requires a pre-existing primer
50

51. DNA replication is
semi -conservative
Parent
molecule
Replication
Daughter
molecules
51

52. Common features of DNA replication
1) Semi conservative
2) bidirectional
3) synthesis in a 5’ to 3’ direction
4) Semi discontinuous:
leading strand, continuous; lagging strand,
discontinuous (in Okazaki fragments).
52

53. 1. Initiation
2. Elongation
3. Termination
DNA replication in E. coli
53

54. 1. Initiation
- Replication origin oriC
- DnaA
- SSB
- DnaB helicase
- DnaC primase
Factors involved:
54

55. Replication initiation at the E. coli oriC
55

56. 2. Elongation
Factors involvedd:
1. DNA pol III holoenzym
2. DNA pol I
3. DNA gyrase
4. DnaB helicase
5. DnaG primase
6. SSB
7. DNA ligase
56

57. subunits  and 
not shown
DNA pol III holoenzyme
57

58. Elongation
process
58

59. 59

60. 3. Termination
60

61. 3-DNA Repair Mechanism.
Error
Mutant
DNA
Normal DNA
Damage
Repair
61

62. Agents that Damage DNA
1. Spontaneous (self-generated)
2. Environmental
62

63. 1. Spontaneous (self-generated) DNA
damaging sources
 Hydrolytic attack.
 Nonenzymatic methylation.
 Oxidative attack.
 Replication error.
63

64. 64

65. 2. Environmental DNA damaging sources
 Chemical exposure (e.g. tobacco smoking,
food chemicals, industrial pollutants,
chemotherapeutics)
 Ionizing radiation (e.g. X-ray, -ray)
 UV (sunlight) exposure
65

66. DNA is damaged by Alkylation, Oxidation, and
Radiation
Often mispqir with thymine
G:C –A:T
Reactive oxygen species
O2-, H2O2, OH•
G modification ( alkylation & oxidation)
66

67. Types of DNA Damage
can be(A, T, C, G)All four of the bases in DNA-1
covalently modified at various positions.
One of the most frequent is the loss of an amino group
)“ deamination ") resulting, for example, in a C being
converted to a U.
of the normal bases because of a failureMismatches-2
of proofreading during DNA replication.
Common example: incorporation of the pyrimidine U
instead of T.
67

68. .Breaks in the backbone-3
Can be limited to one of the two strands (a single-
stranded break, SSB) or
on both strands (a double-stranded break (DSB).
Ionizing radiation is a frequent cause, but some
chemicals produce breaks as well.
4- Crosslinks Covalent linkages can be formed between
bases
on the same DNA strand ("intrastrand") or
on the opposite strand ("interstrand").
68

69. Biochemical Mechanism of
DNA Repair
1- Repair of damage caused by
Incorrect base Insertion (Mismatch
Repair).
- During DNA replication ,both the enzymes polymerase I
and polymerase III occasionally incorporate wrong or
incorrect base in the daughter strand under synthesis.
- A wrong base fails to form a hydrogen bond with the
template base in the parental or template strand .
69

70. - MMR system is an excision/ resynthesis system that can be
divided into 4 phases:
(i) recognition of a mismatch by MutS proteins
(ii) recruitment of repair enzymes
(iii) excision of the incorrect sequence
(iv) resynthesis by DNA polymerase using the parental strand as
a template.
- Such errors are usually corrected by the editing function
of these enzymes
-The protein MutS recognises and binds to the
mismatched base.
-MutH binds to DNA at the hemimethylated GATC sites
and cleaves the unmethylated strand.
-MutL binds to MutH and MutS and helps in final repair steps.
70

71. Mismatch repair71

72. Repair of Thymidine Dimers-2
a - Formation of Pyrimidine Dimers
• UV radiations produce intrastrand photodimers
between two adjacent pyrimidine bases of the same
strand.
•The dimers result from the formation of Cyclobutyl ring
by the joining of 5th and 6th carbon atoms of two
pyrimidine rings.
72

73. 73

74. b- Effect of Dimers
-The presence of thymine dimer causes distortion of
DNA helix because the two thymines are pulled closer .
-This produces a 'kink' in the polynucleotide strand in
which dimer is formed.
- The kink causes inhibition of the advancement of
replication fork and thus interferes with DNA replication.
74

75. GA
T
T
A
C
G
A
T
A
T
T
A
A
T
G
C
C
G
C
G
G
C
A
A
T
A TC
A A
C
C TA G
A T T G
UV
Thymine dimer
Distortion of DNA helix and formation of 'Kink' caused by
thymine dimer
75

76. C- Mechanism of Dimer Repair
Photo – reactivation or light induced repair.
-The energy to break the cyclobutane ring is derived from
visible light .
-The repair of pyrimidine dimers by photoreactivation by
the enzyme photolyase ,is common to many
prokaryotes and eukaryotes.
-The enzymes absorb radiant energy of wavelengths
300 to 600 nm and split the cyclobutyl ring between
two pyrimidine molecules of the dimer.
76

77. -The cyclobutyl thymidine dimers are converted to
pyrimidine monomers.
77

78. Base Excision Repair
Base excision repair pathway (BER).
(a) A DNA glycosylase recognizes a
damaged base and cleaves between the
base and deoxyribose in the backbone.
(b) An AP endonuclease cleaves the
phosphodiester backbone near the AP
site.
(c) DNA polymerase I initiates repair
synthesis from the free 3’ OH at the
nick, removing a portion of the
damaged strand
(with its 5’→3’ exonuclease
activity) and replacing it with
undamaged DNA.
(d) The nick remaining after DNA
polymerase I has dissociated is sealed
by DNA ligase.
.
AP= a purinic or a pyrimidinic
(a=without)
78

79. UvrA recognizes
bulky lesions
Structural distortion = signalUvrB and
UvrC make cuts
(a) Two excinucleases (excision endonucleases) bind DNA at the site of bulky lesion.
(b) One cleaves the 5’ side and the other cleaves the 3’ side of the lesion, and the
DNA segment is removed by a helicase. (c) DNA polymerase fills in the gap and (d)
DNA ligase seals the nick
Nucleotide excision repair
79

80. Recombinational Repair
5’
5’
3’
3’
3’
5’
Thymine
dimer
5’
5’
3’
3’
3’
5’
Thymine
dimer
Undamaged parental strand
‘recombines’ into the gap
opposite the dimer, leaving
a gap on the other parental
strand.
Recombination is dependent
RecA protein
80

81. The gap in the undamaged
parental strand is filled by DNA
pol I and ligase. The thymine
dimer can now be repaired by
excision repair.
5’
5’
3’
3’
3’
5’
Thymine
dimer
5’
5’
3’
3’
3’
5’
Thymine
dimer
81

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