1. SBC 275 Genome Organization I
Dr Marion Burugu
1

2. • Course outline
-The nucleic acids- RNA and DNA
-DNA Structure:
- The Watson-Crick Model
- DNA conformations: B , Z, A and triple
stranded DNA.
- DNA mutations
- DNA Supercoiling : role of enzymes
2

3. - Eukaryotic Nuclei
- Proteins associated with DNA in eukaryotic nucleus:
Histones, Nonhistone Proteins
-Histone Modifications
- Combination of DNA and Histones in Nucleosomes
- Experiments Leading to the Discovery of
Nucleosomes
-DNA endonucleases and nucleosome periodicity
- Nucleosome assembly
- Nucleosome phasing
- Nucleosome alteration during transcription and
replication.
3

4. • Objectives
The learner should be able to
-Define nucleic acids and their requisite building
units
-Describe the DNA structure, the double helix
and the various DNA conformations.
-Explain how DNA can be chemically modified
and reversed and resultant outcomes in cells.
4

5. -Describe the eukaryotic nucleus features
-Describe the Nucleosome structure, its assembly,
how it combines with DNA and histones in the
nucleus, its periodicity and phasing and how it is
altered during DNA replication and transcription.
-Generally describe the DNA replication and
transcription.
-
5

6. References
1. Lewin. Gene Viii. 2004.
2. Lodish. Molecular Biology of the cell. 8th
Edition. 2005.
3. Selected publications.
6

7. Course delivery
-Lectures, group assignments and presentations.
Assessment
- 2 written CAT
-University final examination-70%
7

8. Nucleic acids
-Genome is defined as the complete set of sequences
in the genetic material of an organism that includes the
sequence of each chromosome plus any DNA in
organelles.
-Nucleic acids are molecules that encode genetic
information of an organism. They are made up of a
series of nitrogenous bases joined to ribose molecules
that are linked by phosphodiester bonds.
-There are two types of nucleic acids- Deoxyribonucleic
acid (DNA) and ribonucleic acid (RNA).
-Both are chemically very similar with their primary
structures consisting of linear polymers composed of
nucleotide monomers. 8

9. 9

10. -In cells, RNAs range in length from less than one
hundred to many thousands of nucleotides while DNA
molecules can be as long as several hundred million
nucleotides.
-DNA and RNA each consist of only four different
nucleotides.
-nucleotides consist of an organic base (nitrogenous)
linked to a five-carbon sugar that has a phosphate
group attached to carbon 5.
10

11. - RNA contain the ribose sugar while DNA have deoxyribose
sugar.
-
(a) Adenosine 5'-
monophosphate
(AMP), a
nucleotide present
in RNA.
b) Ribose and
deoxyribose, the
pentoses in RNA
and DNA,
respectively.
11

12. - The are five different bases that are attached
to nucleotides in synthesis of DNA and RNA
i.e The bases adenine (A) and guanine (G) are
purines, which contain a pair of fused rings
and cytosine (C), thymine (T), and uracil (U)
are pyrimidines that contains a single ring.
12

13. 13

14. -Bases A,G and C are contained in both DNA
and RNA but T is found only in DNA, and U
only in RNA.
14

15. -A single nucleic acid strand consist of a backbone composed of repeating pentose-
phosphate units from which the purine and pyrimidine bases extend as side
groups.
-The nucleic acid strand has an end-to-end chemical orientation (directionality): the 5
end has a hydroxyl or phosphate group on the 5 carbon of its terminal sugar while
the 3 end has a hydroxyl group on the 3 carbon of its terminal sugar.
15

16. 16

17. -DNA synthesis proceeds 5' to 3 ' direction.
-Conventionally, polynucleotide sequences are
written and read in the 5 ' 3 ' direction
(from left to right).
17

18. - The chemical linkage between adjacent nucleotides, is
referred to as a phosphodiester bond- it comprises two
phosphoester bonds, one on the 5 ' side of the phosphate
and another on the 3 ' side.
-The resultant linear sequence of nucleotides linked by
phosphodiester bonds constitutes the primary structure of
nucleic acids.
18

19. • Thus, the hereditary nature of every living
organism is defined by its genome, which
comprises a long sequence of nucleic acid that
provides the information needed to construct a
particular organism. The individual subunits
(bases) of the nucleic acid in this case determines
hereditary features.
19

20. -Functionally, the genome is divided into genes.
‘A gene (cistron) is the segment of DNA specifying production
of a polypeptide chain. It includes regions preceding and
following the coding region (leader and trailer) as well as
intervening sequences (introns) between individual coding
segments (exons)’.
-Each gene is a sequence within the nucleic acid that
represents a single protein. As such, an organism genome
may contain a large number of genes e.g <500 genes for a
mycoplasma, a type of bacterium) and as many as >40,000
for Man.
20

21. - Nucleic acids polynucleotides can twist and
fold into three-dimensional conformations
stabilized by noncovalent bonds (function
based folding).
21

22. The nucleic acids macromolecules:
(1) contain the information for determining the amino
acid sequence and hence the structure and function
of all the proteins of a cell
(2) are part of the cellular structures that select and align
amino acids in the correct order as a polypeptide chain
is being synthesized
(3) catalyze a number of fundamental chemical reactions
in cells, including formation of peptide bonds between
amino acids during protein synthesis.
22

23. Link between DNA and RNA
-DNA contains all the information required to
build the cells and tissues of an organism. The
integrity in replication of this information in
any species assures its genetic continuity from
generation to generation and is critical to the
normal development of an individual.
-The information is arranged in hereditary units,
known as genes, that control identifiable traits
of an organism.
23

24. -Transcription is the process through which
the information stored in DNA is copied into RNA.
-RNA has three distinct roles in protein synthesis:
• Messenger RNA (mRNA) carries the instructions
from DNA that specify the correct order of amino
acids during protein synthesis.
24

25. -During translation, the information in mRNA is
interpreted by transfer RNA (tRNA) with the aid
of ribosomal RNA (rRNA), and its associated
proteins.
-Translation is the accurate stepwise assembly
of amino acids into proteins.
-The process of transcription of DNA to RNA and
the subsequent translation to functional proteins
is called the central dogma of life.
25

26. Deoxyribonucleic acid (DNA):
structure and conformations
• It contains all the information required to
build the cells and tissues of an organism.
-The information stored in DNA is arranged in
hereditary units known as genes.
-Genes control the identifiable traits of an
organism.
26

27. Structure of DNA
• The primary structure of DNA is a polymer
composed of monomer called nucleotides.
27

28. • DNA conformations
a. James D. Watson and Francis H. C. Crick proposed
that DNA has a double-helical structure in 1953.
-This proposal was based on analysis of x-ray
diffraction patterns coupled with model building.
- Three notions converged in the construction of the
double helix model for DNA by Watson and Crick :
28

29. • X-ray diffraction data showed that DNA has the form of a
regular helix, making a complete turn every 34 Å (3.4
nm), with a diameter of ~20 Å (2 nm). Since the distance
between adjacent nucleotides is 3.4 Å, there must be 10
nucleotides per turn.
• The density of DNA suggested that the helix must contain
two polynucleotide chains. The constant diameter of the
helix comes about because bases in each chain face inward
and are restricted so that a purine is always opposite a
pyrimidine, avoiding joining of purine-purine (too wide) or
pyrimidine-pyrimidine (too narrow).
29

30. • Irrespective of the absolute amounts of each base,
the proportion of G is always the same as the
proportion of C in DNA, and the proportion of A is
always the same as that of T. Thus the composition
of any DNA can be described by the proportion of its
bases i.e G + C which ranges from 26% to 74% for
different species.
30

31. -DNA consists of two associated polynucleotide strands that
wind together to form a double helix.
-The two sugar phosphate backbones are on the outside of the
double helix while the bases project into the interior.
-The adjoining bases in each strand are stack on top of one
another in parallel planes
-The orientation of the two strands is antiparallel i.e their 5
3 directions are opposite.
‘Antiparallel strands of the double helix are organized in
opposite orientation, so that the 5 ′ end of one strand is
aligned with the 3 ′ end of the other strand.’
31

32. a).Space-filling model of B
DNA. The bases (light
shades) project inward from
the sugar-phosphate
backbones (dark red and
blue) of each strand, but their
edges are accessible through
major and minor grooves.
Arrows indicate the 5’ to 3’
direction of each strand.
Hydrogen bonds between the
bases are in the center of the
structure. The major and
minor grooves are lined by
potential hydrogen bond
donors and acceptors
(highlighted in yellow).
(b) Chemical structure of DNA
double helix showing the two
sugar-phosphate backbones
and hydrogen bonding
between the Watson-Crick
base pairs, AT and GC.
32

33. -The two strands are held precisely by base pairs
between them, where A is paired with T through two
hydrogen bonds and G is paired with C through
three hydrogen bonds (complementary base pairing).
‘Complementary base pairs are defined by the pairing
reactions in double helical nucleic acids (A with T in
DNA or with U in RNA, and C with G)’.
- This base-pair complementarity is a consequence of
the size, shape, and chemical composition of the
bases.
-In addition to hydrogen bonds, further stabilization of
the double-helical structure is by hydrophobic and
van der Waals interactions between the stacked
adjacent base pairs.
33

34. -The diameter of the double helix is 20 Å.It
makes a complete turn every 34 Å, with 10 base
pairs per turn.
-The double helix maintains a constant width
because purines always face pyrimidines in the
complementary A-T and G-C base pairs.
34

35. -The sugar-phosphate backbone is on the
outside and carries negative charges on the
phosphate groups. In vitro when DNA is in
solution, these charges are neutralized by the
binding of metal ions majorly the Na+.
35

36. -Within the cell, positively charged proteins
Neutralize these forces and they play an
important role in determining the organization
of DNA in the cell.
36

37. -The double helix is flexible about its long axis
because it has no hydrogen bonds parallel to
the axis. This allows the DNA to bend when
complexed with a DNA-binding protein. This
bending is vital to the dense packing of DNA
in chromatin.
37

38. - A·T and G·C base pairs associations between a
larger purine and smaller pyrimidine are often
called Watson-Crick base pairs.
-When two polynucleotide strands form such
base pairs , they are said to be complementary.
38

39. B-form DNA
-The most common conformation of DNA in cells
(normal physiological conditions).
-It is a right-handed double helix with 10 base pairs
per complete turn (360°) of the helix.
-The stacked bases are regularly spaced 0.36 nm
apart along the helix axis.
-In this conformation, the helix makes a complete
turn every 3.6 nm; thus there are about 10.5
pairs per turn.
39

40. -On the outside of B-form DNA, the twisting of the two
strands around one another forms a double helix
with a minor groove (~12 Å across) and a major
groove (~22 Å across). (see figures- slide no. 43 and
44 ).
-The atoms on the edges of each base within these
grooves are accessible from outside the helix,
forming two types of binding surfaces.
-DNA binding proteins “read” the sequence of bases in
duplex DNA by contacting atoms in either the major
or the minor grooves.
40

41. Space-filling model of B DNA.
The bases (light shades)
project inward from the
sugar-phosphate backbones
(dark red and blue) of each
strand, but their edges are
accessible through major and
minor grooves. Arrows
indicate the 5’ 3’
direction of each strand.
Hydrogen bonds between the
bases are in the center of the
structure. The major and
minor grooves are lined by
potential hydrogen bond
donors and acceptors
(highlighted in yellow).
41

42. 42

43. • A form DNA
-This is the more compact DNA form that has
11 base pairs per turn and exhibits a large tilt of
the base pairs with respect to the helix axis
-Occurs in very low humidity where the
crystallographic structure of B DNA changes to
the A form.
- RNA –DNA and RNA-RNA helices exist in this
form in cells and in vitro.
43

44. Z form DNA
- This is the left-handed configuration of short
DNA molecules composed of alternating
purine-pyrimidine nucleotides (especially Gs
and Cs).
- In this conformation, the bases seem to zigzag
when viewed from the side.
44

45. Sugar-phosphate backbones of the two strands are on the outside in all structures
(shown in red and blue; the bases (lighter shades) are oriented inward.
45

46. • A triple-stranded DNA structure
-This is formed when synthetic polymers of poly(A)
and polydeoxy(U) are mixed in the test tube.
Also, homopolymeric stretches of DNA composed
of C and T residues in one strand and A and G
residues in the other can form a triple-stranded
structure by binding matching lengths of
synthetic poly(CT).
- Tripple –stranded structures do not occur naturally
in cells but may prove useful as therapeutic
agents.
46

47. Ass 1: Make two teams read and write on DNA
mutations and their outcomes in cells.
47

48. • DNA supercoiling
-Defined as the coiling of a closed duplex DNA in
space so that it crosses over its own axis.
-It occurs only in a closed DNA with no free
ends. A closed DNA can be a circular DNA
molecule or a linear molecule where both
ends are anchored in a protein structure.
48

49. 49

50. - DNA exist in two forms: Linear and circluar.
Linear DNA is extended while a circular DNA
remains extended if it is relaxed
(nonsupercoiled) but a supercoiled DNA has a
twisted and condensed form.
-Supercoling changes the DNA structure by
influencing its conformation in space.
50

51. -The outcome of supercoiling depend on
whether the DNA is twisted around itself in
the same sense as the two strands within the
double helix (clockwise) or in the opposite
sense.
-Twisting in the same sense produces positive
supercoiling. This causes the DNA strands to
wound around one another more tightly
resulting in more base pairs per turn.
51

52. -Twisting in the opposite sense produces
negative supercoiling. This causes the DNA strands
to be twisted around one another
less tightly resulting in fewer base pairs per turn.
-Negative supercoiling can be perceived as creating
tension in the DNA that is relieved by unwinding
the double helix.
-The ultimate effect of negative supercoiling is to
generate a region in which the two strands of
DNA have separated i.e there are zero base pairs
per turn.
52

53. 53
-DNA topology is manipulated in vivo in the following areas-
recombination, replication, and transcription and in the
organization of higher-order structure (the folding of the
DNA thread into a chain of nucleosomes in the eukaryotic
nucleus). These manipulations are central of all its
functional activities.
-These manipulations calls for DNA double strand to be
separated first but the strands do not simply lie side by
side (they are intertwined) and thus their separation
therefore requires the
strands to rotate about each other in space..

54. • DNA strand separation – In vivo
-During replication and transcription (DNA
sythesis) of DNA, the strands of the double
helix separate to allow the internal edges of
the bases to pair with the bases of the
nucleotides to be polymerized into new
polynucleotide chains.
54

55. -During replication the two strands separate
permanently where each reforms a duplex
with the newly-synthesized daughter strand.
- When a circular DNA molecule is replicated,
the circular products may be catenated (one
passed through the other) and they must be
separated in order for the daughter molecules
to segregate to separate daughter cells. i.e.
55

56. 56

57. -In transcription, the movement of RNA
polymerase creates a region of positive
supercoiling in front and a region of negative
supercoiling behind the enzyme and these must
be resolved before the positive supercoils impede
the movement of the enzyme.
-The in vivo manipulations (replication,
transcription and in the folding of the DNA
thread into a chain of nucleosomes) are made
possible by the actions of topoisomerases.
57

58. • DNA Topoisomerases
-These are enzymes relax or introduce supercoils
in DNA. These enzymes catalyze changes in the
topology of DNA by transiently breaking one or
both strands of DNA, passing the unbroken
strand(s) through the gap, and then resealing the
gap.
58

59. -Topoisomerases are divided into two classes,
according to the nature of the mechanisms they
employ:
-Type I topoisomerase changes the topology of DNA
by nicking and resealing one strand of DNA (they
make a transient break in one strand of DNA).
-Type II topoisomerase changes the topology of
DNA by nicking and resealing both strands of DNA
(make a transient a transient double-strand break
).
59

60. -Topoisomerases that introduce negative
supercoils are called gyrases while those that
introduce positive supercoils are called
reverse gyrases.
Topoisomerase enzymes in E. coli
- E. coli has four topoisomerase I, III, IV and
DNA gyrase.
60

61. - DNA topoisomerase I and III are type I enzymes.
- Gyrase and DNA topoisomerase IV are type II
enzymes.
Roles in E.coli
• - The bacterial genome is supercoiled. The overall
level of negative supercoiling in the bacterial
nucleoid is the result of a balance between the
introduction of supercoils by gyrase and their
relaxation by topoisomerases I and IV and these
affects and affects the
initiation of transcription.
61

62. 62
The bacterial
genome
consists of a
large number
of loops of
duplex DNA
(in the form of
a fiber), each
secured at
the base to
form an
independent
structural
domain.

63. -Topoisomerases resolve the problems created
by transcription- gyrase converts the positive
supercoils that are generated ahead of
RNA polymerase into negative supercoils, and
topoisomerases I and IV remove the negative
supercoils that are left behind the enzyme.
These roles are also carried out by the same
enzymes during replication.
63

64. -Removal of precatenanes and decatenations of
any catenated genomes that are left at the
end of replication by topoisomerase IV.
- Precatenation is a stage during DNA
replication where the daughter duplexes can
become twisted around one another.
64

65. • Eukaryote topoisomerases
- Most eukaryotes contain a single topoisomerase I
enzyme.
-Topoisomerase 1 has two roles:
replication fork movement
relaxing supercoils generated by transcription.
-A topoisomerase II enzyme(s) is involved in
unlinking chromosomes following replication.
65

66. • Breaking and resealing strands by
topisomerases
- All topoisomerases commonly function to link
one end of each broken strand to a tyrosine
residue in the enzyme.
- A type I topoisomerase links to the single broken
strand while a type II enzyme links to one end of
each broken strand.
- -Depending on the position of the linkage (5 or 3
prime) phosphodiester tyrosine, topoisomerases
are further divided into types A and B.
66

67. • All the E. coli enzymes are all of type A, using
links to 5 ′ phosphate while all four possible types
of topoisomerase (IA, IB, IIA, IIB) are found in
eukaryotes.
• The linkages involves the use of the transient
phosphodiester-tyrosine bond. This mechanism
where the enzyme transfers a phosphodiester
bond(s) in DNA to the protein, manipulates the
structure of one or both DNA strands, and then
rejoins the bond(s) in the original strand thus
requiring no input of energy.
67

68. - The enzyme binds to a region in which duplex
DNA becomes separated into its single
strands, it breaks one strand, pulls the other
strand through the gap, and finally seals the
gap. Because of the transfer of bonds from
nucleic acid to protein, the enzyme can
function without requiring any input of energy
(energy is conserved through the transfer
reactions).
68

69. 69
Bacterial type I
topoisomerases
recognize
partially unwound
segments of DNA
and pass one
strand
through a break
made in the other.

70. - The type I topoisomerase also can pass one
segment of a single-stranded DNA through
another. This single-strand passage reaction
results in either a knots introduction in DNA
or can catenate two circular molecules .
70

71. -Single-strand passage is a reaction catalyzed by
type I topoisomerase in which one section of
single-stranded DNA is passed through
another strand.
-A knot in the DNA is an entangled region that
cannot be resolved without cutting and
rearranging the DNA.
-To catenate is to link together two circular
molecules as in a chain.
71

72. -Type II topoisomerases relax both negative and
positive supercoils. The reaction requires ATP,
where one ATP is hydrolyzed for each catalytic
event.
-This reaction is mediated by making a double-
stranded break in one DNA duplex. The double-
strand is cleaved with a 4-base stagger
between the ends, and each subunit of the dimeric
enzyme attaches to a protruding
broken end. Then another duplex region is passed
through the break.
72

73. -These steps are followed by religation/release step
where ATP is used when the ends are rejoined and the
DNA duplexes are released. This is why inhibiting the
ATPase activity of the enzyme results in a "cleavable
complex" that contains broken DNA.
-The two strand transfer results in changes of linking
number in multiples of two.
• Linking number is the number of times the two strands
of a closed DNA duplex cross over each other.

74. 74

75. -The topoisomerase II activity also introduces or resolves
catenated duplex circles and knotted molecules. Here ,
the recognition of duplex DNA is nonspecific and the
enzyme binds any two double-stranded segments that
cross each other.
-The hydrolysis of ATP drives the enzyme through
conformational changes that provides the force needed to
push one DNA duplex through the break made in the
other. Due to the topology of supercoiled DNA, the
relationship of the crossing
segments allows supercoils to be removed from either
positively or negatively supercoiled circles.
75

76. In vitro strands seperation
-The invitro unwinding and separation of DNA strands is
referred to as denaturation or “melting,”.
-Experimentally, it is induced by increasing the
temperature of a solution of DNA. Increase in thermal
energy results in increase in molecular motion that
eventually breaks the hydrogen bonds and other forces
that stabilize the double helix; separating the double
strands. This separation is driven apart by the
electrostatic repulsion of the negatively charged
deoxyribose-phosphate backbone of each strand.
76

77. -The melting temperature (Tm) at which DNA
strands will separate depends on several
factors:
a). GC concentration- Molecules that contain
a greater proportion of G·C pairs require higher
temperatures to denature because of the
three hydrogen bonds in G·C pairs that make
these base pairs more stable than A·T pairs,
which have only two hydrogen bonds.
77

78. b). Ion concentration
-the negatively charged phosphate groups in the
two strands are shielded by positively charged
ions. Shielding decreases when the ion
concentration is low, thus increasing the repulsive
forces between the strands and reducing the Tm.
c). Chemical agents- Agents that destabilize
hydrogen bonds e.g. formamide or urea lower
the Tm.
78

79. d). pH- extremes of pH denature DNA at low
temperature. At low (acid) pH, the bases
become protonated and thus positively
charged, repelling each other. At high (alkaline)
pH, the bases lose protons and become
negatively charged, repelling each other.
79

80. -The resultant single-stranded DNA molecules
from denaturation form random coils without
an organized structure.
-For two complementary DNA single strands,
renaturing (reassociation of the single strands
) occurs. This depends on :
a). Temperature- Lowering the temperature
causes the strands to reassociate.
80

81. b). Ion concentration- increasing the ion
concentration favours strands reassociation.
c). pH- neutralizing the pH causes the two
complementary strands to reassociate into a
perfect double helix.
81

82. Eukaryotic nucleus
82
-Eukaryotes have a true nucleus which is
surrounded by double layer membrane.
-The nucleus houses the cell's DNA and directs
the synthesis of proteins and ribosomes, the
cellular organelles responsible for protein
synthesis

83. 83

84. SBC 275 last notes

85. Eukaryotic nucleus
2
-Eukaryotes have a true nucleus which is
surrounded by double layer membrane.
-The nucleus houses the cell's DNA and directs
the synthesis of proteins and ribosomes, the
cellular organelles responsible for protein
synthesis

86. 3

87. • Eukaryotic genomes are composed of a
certain number of pieces of DNA as
chromosomes e.g 46 in humans.
• -Each piece of DNA in eukaryotic genomes is
wrapped around a core histone octamer,
which consists of a pair of the histone
proteins H2A, H2B, H3, and H4, and forms a
nucleosome

88. Nucleosome structure
-At atomic resolution of 1.9 A °, the structure consist of
147 base pairs (bp) of DNA wrapped in 1.7 left-handed
superhelical turns around the histone octamer.
-The core histones have ‘tail’ domains, which are long
disordered structures and play important roles in the
interactions between nucleosomes.
-Each nucleosome particle is connected by linker DNA (
20–80 bp) making up repetitive motifs of 200 bp, which
are de-scribed as ‘beads on a string’ or as a ‘10-nm fiber’

89. -This 10-nm fiber associated with the various
non-histone proteins is called chromatin.
-Chromatin is a negatively charged polymer that
produces electrostatic repulsion between
adjacent regions because only about half of the
DNA negative charges (from the phosphate
backbone) are neutralized by the basic core
histones.

90. • To be folded further, the remaining negative charges are
neutralized by other factors including linker histones,
cations, and other positively charged molecules.
• Thus, the chromatin structure can be dynamically changed
depending on the electrostatic state of its environment.
This change in chromatin structure is critical for gene
expression because it directly governs access to the DNA.
• In eukaryotic nucleus, the chromatin resembles a
disordered assembly of 10-nm fibers.
• Thus, the basic structure of the chromosome is a liquid-like
compact aggregation of 10-nm.

91. • Because chromatin is composed of an irregular
and dynamic 10-nm fiber and does not have a
crystal-like long-range order, chromatin in the cell
is considered ‘liquid-like’ rather than static solid-
like substance. This liquid can take various
structures including extended, folded, interdigi-
tated, bent, looped and columnar structures.
• - The tail domains of histones H3 and H4 play
crucial roles in forming these various structures .
8

92. • In extended fibers, H3 and H4 tails interact
with the DNA in the vicinity of their root
positions while they interact with neighboring
nucleosomes in the folded structure: the H3
tail interacts with DNA, and the H4 tail
interacts with an acidic patch on the H2A/H2B
surface.
• In more condensed, irreg-ular, or
interdigitated structures (e.g., topologically
associating domains, TADs), the tails interact
with distant nucleosomes (long-range
interactions).
9

93. • These schemes of tail-mediated interactions
can change by epigenetic modification of the
tails or other pro-tein factors which overall
contribute to controlling chromatin structure
and its DNA accessibility.
10

94. • Euchromatin and Heterochromatin
-are two distinct types of chromatin that are
revealed through microscopic observations of
interphase nuclei.
- By electron microscopy, heterochromatin is
more electron dense than euchromatin.
- heterochromatin-like staining is often evident
at the nuclear periphery, suggesting that it is
enriched in this part of the nucleus.
11

95. -heterochromatin is highly condensed while
euchromatin have a more open conformation.
-The euchromatin stains less intense than
heterochromatin because heterochromatin has
tighter DNA packaging.
-Euchromatin is lightly packed in the form of DNA,
RNA, and protein, it is definitely rich in gene
concentration and is usually under active
transcription.
- Heterochromatin is only found in eukaryotes but
euchromatin is found in both prokaryotes and
eukaryotes.
-The activities of the euchromatin aid in cell survival.
12

96. -heterochromatin is responsible for gene regulation
and protection of chromosomal integrity. These roles
are made possible because of the dense DNA
packing.
-In mammalian cells, heterochormatin are of two types:
constitutive and falcultative.
13

97. • constitutive heterochromatin is genetically inert
and is composed of tandem repeats (satellites).
- It is defined as the region of the chromosome that
remains condensed throughout the cell cycle.
-Its formation and/or maintenance is influenced by
DNA sequence and DNA methylation.
14

98. • Facultative heterochromatin is euchromatin
that will adopt heterochromatic properties in
a developmentally controlled manner,
suggesting temporal silencing of regions
of the genome.
15

99. • senescence-associated heterochromatic foci
(SAHF)- Recently described.
• It is a distinct heterochromatic structure that
accumulates during cellular senescence.
16

100. Components of the Chromatin Fiber
• A. Histones
1. Core Histones
- The main protein components are the family
of core histones (H2A, H2B, H3, and H4).
the four “core” histones present as an
octameric unit in the nucleosomes
 are positively charged proteins, containing
relatively large amounts of lysine and arginine
17

101. bind directly to the DNA fiber by noncovalent
forces-, the electrostatic interactions between
positively charged residues on the
histonesand DNA phosphates
The C-terminal contains a ‘‘histone-fold’’ motif
that is shared by many proteins, including a
number of transcription factors and is involved
in histone:histone and DNA:histone
interactions.
18

102. The 25–40 amino acid N-terminal domains of
the core histones are highly conserved.
They are mobile and flexible extending out of
the nucleosome core in a manner that frees
them for other interactions.
19

103. 2. Core Histone Variants (CHV)
-Chromatin can be remodeled by replacement of
major histones with specific histone variants
-Mostly occurred by gene duplication
-are often expressed at specific stages during
embryogenesis or during cell differentiation.
E.g. histone H3 variant CENPA-has more than
60% sequence identity to the carboxy-
terminal domain of histone H3. Occurs in
mammalian kinetochores.
20

104. • CENP-A assembles into centromeres in the
absence of DNA replication and can replace H3 in
the nucleosome in vitro.
• H2A.Z histone variant -It is not found in cells of
the inner cell mass but accumulates as the cells
differentiate, being enriched in pericentric
heterochromatin.
• –In euchromatin areas, H3 and H2A are replaced
by H3.3 and H2AZ variants respectively.
• H2AX- represents 2–25% of H2A in a cell and is
phosphorylated on serine-139 at the site of
double-strand DNA breaks. H2AX foci suggest a
mechanism for DNA damage detection and
repair. 21

105. • Linker Histones
-Linker histones (H1, H5) are highly dynamic &
located on the outside of the nucleosome.
-are responsible for the condensation of the
chromatin fiber.
- are constructed from a highly conserved 79-
residue central globular domain with flexible
flanking N- and C-terminal domains.
22

106. -H1 is associated with linker DNA, and provides
partial nuclease protection for 20 bp of linker
DNA.
-It contains a conserved globular region and
extended amino- and carboxy-termini, the
latter being rich in lysines and able to interact
strongly with DNA.
23

107. 24
the core histones and a few of the known variants. Sites of Lys/Arg residue m
ethylation and Ser phosphorylation are indicated. HFD denotes the histone-fold
domain,a structural domain common to all core histones

108. Organization of the histone
octamer
• A histone octamer is the 8 protein complex
found at the centre of a nucleosome core
particle. It consists of two copies of each of
the four core histone proteins.
• Qn: how do histones interact with each other
and with DNA?
-The histone octamer has a kernel of a H32·H42
tetramer associated with two H2A·H2B dimers
i.e.
25

109. 26

110. -In the octamer, each histone is extensively
interdigitated with its partner, H3 with H4, and
H2A with H2B.
-In a symmetrical model for the nucleosome, the
H32- H42 tetramer provides a kernel for the
shape while the H2A-H2B dimers are in the
top and beneath positions i.e.
27

111. 28

112. -The diameter of the octamer is accounted for
by the H3 2 ·H4 2 tetramer.
-Within this structure the protein forms a sort of
spool (a cylindrical str) with a superhelical
structure that forms the path for the binding
site for DNA, which turns twice around the
nucleosome.
-All core histones have the structural motif of
the histone fold (three α-helices are
connected by two loops). 29

113. -These regions of the histones interact to
form crescent-shaped heterodimers; where
each heterodimer binds 2.5 turns of the DNA
double helix (H2A-H2B binds at +3.5 - +6; H3-
H4 binds at +0.5 - +3) (positions are numbered
from 0 to +7) through the phosphodiester
backbones (for easier packaging of DNA
irrespective of sequence).
30

114. -H3-H4 and H2A-H2B pairs in a half nucleosome
i.e.
31

115. 32

116. -Each of the core histones has a globular body
that contributes to the central protein mass of
the nucleosome. Each histone also has a
flexible N-terminal tail, that extend out of the
nucleosome.
-The tail has sites for modification that are
important in chromatin function.
33

117. •Histone modifications
-All the histones are modified by covalently
linking extra moieties to the free groups of
certain amino acids.
-Modification sites are concentrated in the
N-terminal tails but can also occur at the C
terminal tail.
-The modifications have important effects on
the structure of chromatin and in controlling
gene expression.
34

118. -Modifications not only regulate chromatin structure, but
they also recruit remodelling enzymes that utilize the
energy derived from the hydrolysis of ATP to reposition
nucleosomes.
-Histone residues modifications :methylation,
phosphorylation, acetylation, sumolyation,
ubiquitinated and ADP-ribosylation.
35

119. 36
Histone modifications. Each histone protein consists of the
N- and C-terminal tails and a central globular domain (gray
box). The N- and C-terminal tails of core histones can be
chemically modified by methylation (red box), acetylation
(blue box), phosphorylation (green box), or ubiquitination
(Ub) at several residues along the length of the protein.

120. Histone acetylation
- Histone acetylation occurs by the enzymatic addition
of an acetyl group (COCH3) from acetyl coenzyme A.
- The process of histone acetylation is tightly involved
in the regulation of many cellular processes
including chromatin dynamics and transcription,
gene silencing, cell cycle progression, apoptosis,
differentiation, DNA replication, DNA repair, nuclear
import, and neuronal repression.
-Occurs in lysine residues
-highly dynamic and regulated by the opposing action
of two families of enzymes, histone
acetyltransferases (HATs) and histone deacetylases
(HDACs). 37

121. -HATs play a critical role in controlling histone H3 and
H4 acetylation.
-Acetylation neutralize the lysine’s positive charge and
this action has the potential to weaken the interactions
between histones and DNA
- Histone H3 acetylation may be increased by inhibition
of HDACs and decreased by HAT inhibition.
-Histone deacetylaces (HDACs) catalyze the hydrolytic
removal of acetyl groups from histone lysine
residues. An imbalance in the equilibrium of histone
acetylation has been associated with tumorigenesis
and cancer progression.
38

122. 39

123. Histone phosphorylation
-The phosphorylation of histones is highly dynamic.
-It takes place on serines, threonines and tyrosines,
predominantly, but not exclusively, in the N-terminal
histone tails.
The levels of the modification are controlled by kinases
and phosphatases that add and remove the
modification, respectively.
-Histone kinases transfer a phosphate group from ATP to
the hydroxyl group of the target amino-acid side chain.
This modification adds significant negative charge to
the histone that influences the chromatin structure
40

124. Histone methylation
-Histone methylation mainly occurs on the side chains of
lysines and arginines.
-It does not alter the charge of the histone protein.
-lysines may be mono-, di- or tri-methylated, whereas
arginines may be mono-, symmetrically or
asymmetrically di-methylated.
-Lysine is methylated at N terminal tail by histone lysine
methyltransferase ((HKMT) e.g SUV39H1- targets H3K9
41

125. -HKMTs that methylate N-terminal lysines contains a
SET domain (proteins that histone methylation
patterns and preserve them though out cell cyle)
that harbours the enzymatic activity.
-Dot1 enzyme that methylates H3K79 within
the histone globular core and does not contain a SET
domain.
42

126. Arginine methylation
-There are two classes of arginine
methyltransferase, the type-I and type-II
enzymes.
-The type-I enzymes generate Rme1 and
Rme2as, whereas the type-II enzymes
generate Rme1 and Rme2s. Together, the two
types of arginine methyltransferases form a
relatively large protein family (11 members),
the members of which are referred to as
PRMTs (protein/histone arginine
methyltranserases).
43

127. -PRMTs enzymes transfer a methyl group from S
adenosyl methionine (SAM) to the ω-
guanidino group of arginine within a variety of
substrates.
Histone demethylases
- Demethylates both lysine and arginine e.g.
lysine-specific demethylase 1 (LSD1)- the first
lysine demethylase identified in 2004. It
utilizes FAD (flavin adenine dinucleotide) as
co-factor. 44

128. Deimination
- Deimination reaction involves the conversion of an
arginine to a citrulline.
- In mammalian cells, this reaction on histones
is catalysed by the peptidyl deiminase PADI4, which
converts peptidyl arginines to citrulline.
- This reaction effectively neutralizes the positive
charge of the arginine since citrulline is neutral.
-PADI4 also converts mono-methyl arginine to
citrulline, thereby effectively functioning as an
arginine demethylase.
45

129. β-N-acetylglucosamine
- Many non-histone proteins are regulated via
modification of their serine and threonine side chains
with single β-N-acetylglucosamine (O-GlcNAc) sugar
residues.
-Such reactions occurs in histones also. In mammalian cells,
there appears to be only a single enzyme, O-GlcNAc
transferase, which catalyses the transfer of the sugar
from the donor substrate, UDP-GlcNAc, to the target
protein.
- Histones H2A, H2B and H4 have been shown to
be modified by O-GlcNAc
46

130. ADP ribosylation
- Ribosylation is the addition of one or more ADP ribose
moieties to a protein
-Histones are mono- and poly-ADP ribosylated
on glutamate and arginine residues.
- This modification is reversible.
-Poly-ADP ribosylation of histones is performed by the poly-
ADPribose polymerase (PARP) family of enzymes and
reversed by the poly-ADP-ribose-glycohydrolase family of
enzymes.
- These enzymes function together to control the
levels of poly-ADP ribosylated histones that have been
correlated with a relatively relaxed chromatin state
47

131. - Histone mono-ADP-ribosylation is performed by the
mono-ADP-ribosyltransferases and has been detected
on all four core histones as well as on the linker
histone H1.
-These modifications significantly increase upon DNA
damage
48

132. Ubiquitylation
- Ubiquitin itself is a 76-amino acid polypeptide that is
attached to histone lysines via the sequential action of
three enzymes, E1- activating, E2-conjugating and E3-
ligating enzymes.
- The enzyme complexes determine both substrate
specificity (i.e., which lysine is targeted) as well as the degree
of ubiquitylation (i.e., either mono- or polyubiquitylated).
-For histones, mono-ubiquitylation is the most relevant.
• e.g- Ubiquitylation of histones H2A and H2B yields
H2AK119ub1 which is involved in gene silencing and
H2BK123ub1 plays an important role in transcriptional
initiation and elongation respectively.
49

133. - The modification is removed via the action of
isopeptidases called de-ubiquitin enzyme and this
activity is important for both gene activity and silencing.
Sumoylation
- Sumoylation involves the covalent attachment of small
ubiquitin- like modifier molecules to histone lysines via
the action of E1, E2 and E3 enzymes.
-Sumoylation has been detected on all four core histones.
-This modification has mainly been associated with
repressive functions.
50

134. Histone tail clipping
- It is the most radical way to remove histone
modifications by removing the histone N-terminal tail
in which they reside, a process referred to as tail
clipping.
- It exists in yeast and mammals (mouse) where the
first 21 amino acids of H3 are removed.
-The mouse enzyme is Cathepsin L, which cleaves the
N-terminus of H3.
51

135. •Non histone proteins associated with
DNA in eukaryotic nucleus
-In addition to the core histones, the DNA scafford
contain several proteins: large amounts of
histone Hl (located in the interior of the fiber)
and topoisomerase II

136. The presence of topoisomerase II unwinds DNA
during DNA replication. It is crucial for
maintenance of chromatin structure and
inhibitors of this enzyme can kill rapidly dividing
cells.
-Several drugs used in cancer chemotherapy are
topoisomerase II inhibitors that allow the
enzyme to promote strand breakage but not
the resealing of the breaks.

137. -Structural maintenance of chromosome (SMC)
proteins.
-The primary structure of SMC proteins consists of
five distinct domain (see fig below (a)).
-The amino- and carboxyl-terminal globular
domains, N and C, each of which contains part of
an ATP-hydrolytic site are connected by two
regions of a-helical coiled-coil motifs that are
joined by a hinge domain.

138. -The dimeric proteins forming a V-shaped
complex are tied together through the
protein's hinge domains where one N and one
C domain come together to form a complete
ATP-hydrolytic site at each free end of the V
(see fig. below).

140. -Proteins in the SMC family occur in all types of
Organisms from bacteria to humans.
-Eukaryotes have two major types, cohesins and
condensins both of which are bound by
regulatory and accessory proteins

142. -Cohesins play a substantial role in Iinking
together sister chromatids immediately after
replication and keeping them together as the
chromosomes condense to metaphase. This
linkage is essential if chromosomes are to
segregate properly at cell division.
-Together with kleisin, cohesins form a ring ,
(which expands and contracts in response to
ATP hydrolysis) around the replicated
chromosomes that ties them together until
separation is required at cell division
-

144. -condensins are essentiat for the condensation
of chromosomes as cells enter mitosis.
-The cohesins and condensins are essential in
orchestrating many changes in chromosome
structure during the eukaryotic cell cycle

145. B. Chromatin
1. Nucleosomes
- a structural component of the chromatin fiber,
and a modulator of gene expression.
-Positioned across genes in a manner that allows
the key transcription factor binding sites to be
exposed and also for the fiber to be folded in a
stable conformation. The positioning of
nucleosomes can be modulated by minor
alterations in the DNA sequence and by DNA
methylation.
62

146. -The fundamental subunit of chromatin has the
same type of design in all eukaryotes.
-It comprises of nucleosome.
-The nucleosome contains ~200 bp of DNA,
organized by an octamer of small, basic
proteins into a bead-like structure. The protein
components are histones. They form an
interior core; the DNA lies on the surface of
the particle.
63

147. -Nucleosomes are an invariant component of
euchromatin and heterochromatin in the
interphase nucleus, and of mitotic
chromosomes.
64

148. -Both replication and transcription require
unwinding of DNA, and thus must involve an
unfolding of the structure that allows the
relevant enzymes to manipulate the DNA.
-When chromatin is replicated, the nucleosomes
must be reproduced on both daughter duplex
molecules.
65

149. -The mass of chromatin contains up to twice as
much protein as DNA. Approximately half of
the protein mass is the nucleosomes. The
mass of RNA is < 10% of the mass of DNA,
which consists of nascent transcripts still
associated with the template DNA.
66

150. -There is also nonhistones proteins of
chromatin but in lower amounts than the
histones. These functions in control of gene
expression and higher-order structure
67

151. The nucleosome
-This is the subunit of all chromatin
-When interphase nuclei are suspended in a
solution of low ionic strength, they swell
and rupture to release fibers of chromatin.
-Individual nucleosomes can be obtained by
treating chromatin with the endonuclease
micrococcal nuclease. This enzyme cuts the
DNA thread at the junction between
nucleosomes releasing releases groups of
particles first, then single nucleosomes.
68

152. -The nucleosome contains ~200 bp of DNA
associated with a histone octamer that
consists of two copies of the core histones-
H2A, H2B, H3, and H4.
-Core histones are present in equimolar
amounts, with 2 molecules of each per ~200
bp of DNA
69

153. 70

154. -Histones H3 and H4 are conserved proteins
suggesting that their functions are identical in
all eukaryotes.
-The types of H2A and H2B occur in all
eukaryotes, but show species-specific
variation in sequence.
71

155. • -Histone H1 comprises a set of closely related
proteins that show appreciable variation
between tissues and between species ( are
absent from yeast). H1 role is different from
the core histones. It is present in half the
amount of a core histone and located in the
external parts of the nucleosome.
72

156. -The shape of the nucleosome shape is like a
flat disk or cylinder, of 11nm diameter and a
height of 6 nm .
The length of the DNA is roughly twice the ~34
nm circumference of the particle. The DNA
follows a symmetrical path around the
octamer. The DNA makes two turns around the
cylindrical octamer where it "enters" and
"leaves" the nucleosome at points close to one
another, the point where histone H1 is
located .
73

157. 74

158. -one DNA turn around the nucleosome takes
~80 bp of DNA.
75

159. Experiments leading to the
discovery of nucleosomes
-Involves digestion of chromatin with the
enzyme micrococcal nuclease (biased toward
selection of A·T-rich sequences). This cleaves
the DNA into integral multiples of a unit
length.
-The resultant fragments are fractionation by gel
electrophoresis yielding a ~10 steps ‘ladder’
whose increments between successive steps,
is ~200 bp.
76

160. 77

161. - The ladder is generated by groups of
nucleosomes. This shows that in chromatin,
DNA is coiled in arrays of nucleosomes.
- When nucleosomes are fractionated on a
sucrose gradient, they give a series of discrete
peaks that correspond to monomers, dimers,
trimers, tetramers etc.
-
78

162. -When the peak DNA is extracted from the
individual fractions and electrophoresed, each
fraction yields a band of DNA whose size
corresponds with a step on the micrococcal
nuclease ladder.
-The monomeric nucleosome contains DNA
eqivalent of the unit length e.g. the
nucleosome dimer contains DNA of twice the
unit length.
79

163. 80

164. -In the diagram, each step on the ladder
represents the DNA derived from a discrete
number of nucleosomes. In conclusion, the
existence of the 200 bp ladder in any
chromatin indicates that the DNA is organized
into nucleosomes.
-The micrococcal ladder is generated when only
~2% of the DNA in the nucleus is rendered
acid-soluble (degraded to small fragments) by
the enzyme. 81

165. ->95% of the DNA of chromatin can be
recovered in the form of the 200 bp ladder.
Suggesting that almost all DNA is organized in
nucleosomes.
-In their natural state, nucleosomes are closely
packed, with DNA passing directly from one to
the next. Free DNA is probably generated by
the loss of some histone octamers during
isolation.
82

166. -The length of DNA per nucleosome varies for
individual tissues in a range from 154 (in
fungus)-260 bp (in sea urchin sperm).
83

167. Nucleosomes structure
-Nucleosomal DNA is divided into the core DNA
and linker DNA depending on its susceptibility
to micrococcal nuclease.
-The association of DNA with the histone
octamer forms a core particle.
-The core DNA contains 146 bp that produced
by prolonged digestion with micrococcal
nuclease.
84

168. - The core particle DNA is relatively resistant to
digestion by nucleases.
-Linker DNA comprises the rest of the repeating
unit. Its length varies from as little as 8 bp to
as much as 114 bp per nucleosome. Linker
DNA is produced by a reaction in which DNA is
"trimmed" from the ends of the nucleosome.
85

169. -The 8-114bp regions are susceptible to early
cleavage by the enzyme. These changes in the
length of linker DNA account for the variation
in total length of nucleosomal DNA.
-Micrococcal nuclease initially cleaves between
nucleosomes. Mononucleosomes typically have
~200 bp DNA.
-End-trimming reduces the length of DNA first to
~165 bp, and then generates core particles
with 146 bp. 86

170. 87

171. -H1 is associated with linker DNA and may lie at
the point where DNA enters and leaves the
nucleosome. The H1 is lost during the
degradation of nucleosome monomers. It is
retained on monomers that still have 165 bp of
DNA; but is always lost with the final reduction
to the 146 bp core particle. This suggests that
H1 could be located in the region of the linker
DNA immediately adjacent to the core DNA.
88

172. -DNA structure on the surface of the
nucleosome is variable due to to cleavage by
certain nucleases.
-Within the nucleosome, DNAase I and DNAase
II enzymes make single-strand nicks in DNA
where they cleave a bond in one strand, but
the other strand remains intact at this point.
89

173. - These enzymes cleave the DNA only at
regular intervals.
- Using radioactively end-labelling of DNA,
denaturation and electrophoresing the
resultant fragments, a ladder is generated i.e
90

174. 91

175. -Sites for nicking (by either the enzymes ) lie at
regular intervals along core DNA.
-The interval between successive steps on the
ladder is 10-11 bases. The ladder extends for
the full distance of core DNA. The cleavage
sites are numbered as S1 through S13 (where
S1 is ~10 bases from the labeled 5 ′ end, S2 is
~20 bases from it, and so on).
92

176. -The enzymes DNAase I and DNAase II generate
the same ladder, with only some differences in
the intensities of the bands.
-The cutting periodicity (the spacing between
cleavage points) coincides with the structural
periodicity (the number of base pairs per turn
of the double helix) i.e. the distance between
the sites corresponds to the number of base
pairs per turn. E.g. For double-helical B-type
DNA is 10.5 bp/turn.
93

177. -Within the nucleosome, each site has 3-4
positions at which cutting occurs ( the cutting
site is defined ±2 bp). Thus, a cutting site
represents a short stretch of bonds on both
strands, exposed to nuclease action over 3-4
base pairs.
-The variation in cutting periodicity along the
core DNA (10.0 at the ends, 10.7 in the
middle) means that there is variation in the
structural periodicity of core DNA. The DNA
has more bp/turn than its solution value in
the middle, but has fewer bp/turn at the ends.
94

178. -The average periodicity over the nucleosome is
10.17 bp/turn, and for DNA in solution is 10.5
bp/turn.
-The crystal structure of the core particle suggests
that DNA is organized as a flat superhelix, with
1.65 turns wound around the histone octamer.
-Regions of high curvature occur at positions ± 1
and ± 4; corresponding to S6 and S8, and to S3
and S11, which are the sites least sensitive to
DNAase I . 95

179. 96
Two numbering
schemes divide core
particle DNA into 10
bp segments. Sites
may be numbered S1
to S13 from one end;
or taking S7 to identify
coordinate 0 of the
dyad symmetry, they
may be numbered -7
to +7.
Illustration of nucleosomal positions relative to the DNA superhelix

180. - A high resolution structure of the nucleosome
core shows that the structure of DNA is
distorted : with most of the supercoiling
occuring in the central 129 bp, which are
coiled into 1.59 left-handed superhelical turns
with a diameter of 80.
-The terminal sequences on either end makes
only a very small contribution to the overall
curvature.
97

181. - The central 129 bp are in the form of B-DNA,
but has a substantial curvature that is
needed to form the superhelix.
-The major groove is smoothly bent, while the
minor groove has abrupt kinks. These
conformational changes explains why the
central part of nucleosomal DNA is not usually
a target for binding by regulatory proteins (
regulatory proteins typically bind to the
terminal parts of the core DNA or to the linker
sequences).
98

182. Nucleosome assembly
• -During DNA replication, the strands of DNA
are seperated and this disrupt the structure of
the nucleosome, though transiently. The
alteration is confined to the immediate
vicinity of the replication fork, but the
nucleosomes reform more or less immediately
behind the fork as it moves along. Thus, once
DNA has been replicated, nucleosomes are
quickly generated on both the duplicates i.e
the assembly of nucleosomes is directly linked
to the replisome that is replicating DNA.
99

183. • How do histones associate with DNA to
generate nucleosomes?
• Do the histones preform a protein octamer
around which the DNA is subsequently
wrapped?
• Or does the histone octamer assemble on
DNA from free histones?
100

184. -Depending on the conditions , nucleosomes can be
assembled using two invitro pathways:
• a). A preformed octamer binds to DNA i.e the
DNA interacts directly with an intact
(crosslinked) histone octamer. This relates to
what happens in vivo as it reflects the capacity of
chromatin to be remodeled by moving histone
octamers along DNA.
101

185. b). An indirect interaction where a tetramer of
H32 ·H42 binds first, then two H2A·H2B dimers
are added. This is the pathway that is used in
replication. i.e.
102

186. In vitro two paths are predicted
103

187. -- Accessory proteins are involved in assisting
histones to associate with DNA where they act
as "molecular chaperones" that bind to the
histones releasing either individual histones or
complexes ( H3 2·H42 or H2A·H2B) to the DNA in
a controlled manner..
-The assembly reaction occurs preferentially on
replicating DNA. It requires a chromatin
assembly factor (CAF-1) an ancillary factor,
CAF-1, that consists of >5 subunits, with a
total mass of 238 kD.
104

188. -CAF-1 is recruited to the replication fork by
PCNA, the processivity factor for DNA
polymerase. This provides the link between
replication and nucleosome assembly,
ensuring that nucleosomes are assembled as
soon as DNA has been replicated.
-CAF 1 functions by binding to newly synthesized
H3 and H4 (recall new nucleosomes form by
assembling first the H3 2 ·H4 2 tetramer, and
then adding the H2A·H2B dimers) 105

189. - In invitro, nucleosomes are formed in a
repeat length of 200 bp and are devoid of any
H1 histone suggesting that proper spacing can
be accomplished without H1.
-During replication, the histone octamers are
not conserved i.e the old histones are released
and then reassembled with newly synthesized
histones so that when chromatin is
reproduced, a stretch of DNA already
associated with nucleosomes is replicated,
giving rise to two daughter duplexes.
-During replication, H2A·H2B dimers and
H3 ·H4 tetramers are conserved i.e.
106

190. -Replication fork passage displaces histone
octamers from DNA. They disassemble into
H3-H4 tetramers and H2A-H2B dimers. Newly
synthesized histones are assembled into H32-
H42 tetramers and H2A-H2B dimers. The old
and new tetramers and dimers are assembled
with the aid of CAF-1 (an accessory protein) at
random into new nucleosomes immediately
behind the replication fork
107

191. 108

192. 109

193. -It is proposed that nucleosomes are disrupted
and reassembled in a similar way during
transcription.
-In eukaryotic cells, during S phase sufficient
histone proteins to package an entire genome
are synthesised for the duplication of
chromatin. i.e the same quantity of histones
must be synthesized that are already
contained in nucleosomes.
110

194. -The synthesis of histone mRNAs is controlled as
part of the cell cycle and increases enormously
in S phase.
- The pathway for assembling chromatin from
this equal mix of old and new histones during
S phase is called the replication-coupled (RC)
pathway.
111

195. • -When DNA is not being synthesized,
replication-independent (RI) pathway is
utilized for assembling nucleosomes during
other phases of cell cycle. This is necessitated
by damage to DNA or nucleosomes displaced
during transcription.
-RI uses different variants of some of the
histones from those used during replication.
-
112

196. -H3.3 replaces H3 in differentiating cells that do
not have replication cycles. It differs from the
highly conserved H3 histone at 4 amino acid
positions.
-RI uses HIRA accessory protein, not CAF-1.
-Assembly of nucleosomes containing an
alternative to H3 also occurs at centromeres.
Centromeric DNA replicates early during the
replication phase of the cell cycle, in contrast
with the surrounding heterochromatic
sequences that replicate later.
113

197. -Using the nucleosomal digest, each eukaryotic
chromosome contains many replicons. The
incorporation of H3 at the centromeres is
inhibited, and CENP-A protein is incorporated
in higher eukaryotic cells (in Drosophila Cid is
used while in the yeast Cse4p is used). This
occurs by the replication-independent
assembly pathway because the replication-
coupled pathway is inhibited for a brief period
of time while centromeric DNA replicates
114

198. -The replication fork displaces histone octamers,
which then dissociate into H3 2·H42 tetramers
and H2A·H2B dimers. These "old" tetramers
and dimers enter a pool that also includes
"new" tetramers and dimers, assembled from
newly synthesized histones.
-Nucleosomes assemble ~600 bp behind the
replication fork.
115

199. -Assembly is initiated when the tetramers bind
to each of the daughter duplexes assisted by
CAF-1 (an assessory protein). This is followed
by binding of the two dimers (H2A·H2) to each
tetramer to complete the histone octamer.
Thus, the assembly of tetramers and dimers is
random with respect to "old" and "new"
subunits.
116

200. Nucleosome phasing/ positioning
-Nucleosome positioning is defined as the
placement of nucleosomes at defined sequences
of DNA instead of at random locations with
regards to sequence.
-To find out the positioning of nucleosomes in a
particular sequence of DNA, indirect end labeling
technique is employed.
117

201. • Indirect end labeling/ DNA foot printing
-It is a technique for examining the organization
of DNA by making a cut at a specific site and
isolating all fragments containing the
sequence adjacent to one side of the cut. It
reveals the distance from the cut to the next
break(s) in DNA.
118

202. -The DNA is isolated. A section of the DNA is
protected by binding proteins. The
unprotected or susceptible regions are
cleaved by non-specific endonucleases This
results in the removal of the protecting
protein. ( In this case, the restriction nuclease
introduces reference points that are
recognized by complementary polynucleotide
probes for subsequent analysis of the
sequences from the restriction sites towards
the sites at which it was initially cleaved in the
unprotected regions).
119

203. -The fragment is cleaved with a restriction
enzyme that has only one target site in the
fragment thus it cuts at a unique point
producing two fragments, each of unique size.
-The products of the restriction double digest
are separated by gel electrophoresis.
-A probe representing the sequence on one side
of the restriction site is used to identify the
corresponding fragment in the double digest.
120

204. -In the nucleosomal positioning analysis, cleavage
is carried out with micrococcal nuclease which
generates a monomeric fragment that
constitutes a specific sequence.
-The identification of a single sharp band
demonstrates that the position of the
restriction site is uniquely defined with respect
to the end of the nucleosomal DNA
121

205. 122

206. 123
In the absence of
nucleosome
positioning, a
restriction site lies at
all possible locations in
different copies of
the genome.
Fragments of all
possible sizes are
produced when
a restriction enzyme
cuts at a target site
(red) and micrococcal
nuclease cuts at the
junctions between
nucleosomes (green).

207. -the existence of a specific band in the indirect
end-labeling technique represents the distance
from the restriction cut to a preferred
micrococcal nuclease cleavage site.( If there are
preferred sites for micrococcal nuclease in the
particular region, specific bands are produced).
-Nucleosome positioning is both intrinsic and
extrinsic .
124

208. -It is intrinsic because every nucleosome is
deposited specifically at a particular DNA
sequence.
-It is extrinsic because the first nucleosome in a
region is preferentially assembled at a particular
site. This provides a boundary that restricts the
positions available to the adjacent nucleosome,
after which a series of nucleosomes are
assembled sequentially, with a defined repeat
length.
125

209. -Thus, the deposition of histone octamers on
DNA is intrinsic. It is determined by structural
features in DNA extrinsically resulting from the
interactions of other proteins with the DNA
and/or histones.
• Structural features of DNA affect placement of
histone octamers. These include:
a). A·T-rich regions are arranged in such a way
that the minor groove faces in towards the
octamer.
126

210. b). G·C-rich regions are arranged so that the
minor groove points out.
c). Long runs of dA·dT (>8 bp) are not
positioned in the central superhelical turn of
the core.
d). Sequences that cause DNA to take up more
extreme structures may have effects such as
the exclusion of nucleosomes, and thus could
cause boundary effects.
127

211. -Commonly, nucleosomes are positioned near
boundaries especially if the length of the
linker varies by about 10 bp. This results in the
location proceeding away from the first
defined nucleosome at the boundary. In such
cases, the positioning is maintained rigorously
only relatively near the boundary.
128

212. -The location of DNA on nucleosomes can be
described in two ways:
a). translational positioning- describes the
location of a histone octamer at successive
turns of the double helix, which determines
which sequences are located in linker
Regions.
-
129

213. -It describes the linear position of DNA relative
to the histone octamer. Displacement
of the DNA by 10 bp changes the sequences that
are in the more exposed linker regions, but
does not alter the face of DNA that is
protected by the histone surface and which is
exposed to the exterior.
130

214. b) rotational positioning- describes the location
of the histone octamer relative to turns of the
double helix, which determines which face of
DNA is exposed on the nucleosome surface.
-It affects the double helix with regard to the
octamer surface.
-Any movement that differs from the helical
repeat (~10.2 bp/turn) displaces DNA with
reference to the histone surface.
131

215. -Nucleotides on the inside are more protected
against nucleases than nucleotides on the
outside.
132

216. -In particular, it determines which sequences are
found in the linker regions. Shifting the DNA by
10 bp brings the next turn into a linker region.
Thus, translational positioning determines
which regions are more accessible ( e.g to the
micrococcal nuclease).
133

217. -Both translational and rotational positioning can
be important in controlling access to
DNA. E.g. The positioning involving the specific
placement of nucleosomes at promoters.
-Translational positioning and/or the exclusion of
nucleosomes from a particular sequence is
necessary to allow a transcription complex to
form. Some regulatory factors can bind to DNA
only if a nucleosome is excluded to make the
DNA freely accessible, and this creates a
boundary for translational positioning.
134

218. -At times, regulatory factors can bind to DNA on
the surface of the nucleosome, but rotational
positioning ensures that the face of DNA with
the appropriate contact points is exposed.
135