1
Chapter 24
Takusagawa’s Note©
Chapter 24: Replication (DNA Biosynthesis)
-
If you forget the nucleic acid structures, you should review Chapter 23.
DNA is replicated in a semiconservative fashion
- DNA can be replicated in three possible ways.
1. Conservative fashion --- Parental strands stay together and newly synthesized strands stay
together. H+H and L+L (H = heavy DNA; L = light DNA)
2. Semi-conservative fashion --- New DNA has one parental strand, i.e., parental strands are
separated. H+L and H+L
3. Dispersive fashion --- New DNA has parental and new-DNA in both strands, i.e., each strand
contain parental and new-DNA. H+L and H+L
= Parental strand containing 15N bases (Heavy DNA)
= New strand containing 14N bases (Light DNA)
Semiconservative
Conservative
Dispersive
2(H+L)
(H+H) &(L+L)
2(H+L)
Initial double
helical DNA
After one
replication
After two
replication
(H+H) & 3(L+L)
2(H+L) & 2(L+L)
1
4(H+L)
Chapter 24
-
2
Takusagawa’s Note©
Conservative and dispersive syntheses (1 and 3) are eliminated by the following
experimental.
- E. coli are grown in 15N medium. Therefore, nitrogen atoms in bases of the E. coli DNA
are all 15N. (These DNAs are heavier than DNAs synthesized in normal N medium).
- The first and second generations of the E. coli were grown in the normal medium. Their
DNAs were collected and were analyzed by a density centrifugation.
- DNA after one generation show one band (see (d) below figure) that is corresponding
to the hybrid DNA (H + L) or dispersive mixture. This result eliminates the
conservative
fashion synthesis.
- DNA after two generation show two bands (see (e) below figure) that are corresponding
to the hybrid DNA (H + L) and the light DNA (L+L). This result eliminates the
dispersive fashion synthesis.
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Takusagawa’s Note©
3
Chapter 24
DNA polymerase
- New strand is synthesized from 5’→ 3’ direction.
- Thus, the old strand (template) is read from 3’ → 5’.
5’
3’
Leading strand
Lagging strand
3’
-
5’
In the elongation process, a tri-deoxynucleotide (such as dATP, dCTP, etc.) is attached.
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Chapter 24
4
Takusagawa’s Note©
DNA replication can be three different ways:
- Continuous synthesis on both strands.
- Continuous synthesis of one strand and discontinuous synthesis of the other strand.
- Discontinuous synthesis on both strands.
-
As described above, DNA is composed of two strands. One strand is called “leading-strand”
(5’ → 3’) and the other is “lagging-strand” (3’ → 5’).
The leading-strand is continuously synthesized whereas the lagging-strand is discontinuously
synthesized since the direction of DNA synthesis is 5’ → 3’. See (b) below figure.
Discontinuously synthesized DNA fragments are called “Okazaki fragments”.
The following experiment was carried out to prove the above conclusion:
1. DNA synthesis in labeled dNTP’s for very short time (2 - 10 seconds) --- ~50% DNA are
short fragments (Okazaki fragments).
2. DNA synthesis in labeled dNTP’s for longer time (1 - 2 minutes) --- much larger DNA
fragments are found, indicating that the short fragments are ligated.
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5
Takusagawa’s Note©
- Each Okazaki fragment contains a small amount of RNA --- indicates that RNA is a
primer in replication.
-
RNA primers are removed, the gaps are filled with DNA, and the fragments are ligated.
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Chapter 24
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Takusagawa’s Note©
Replication of E. coli
- Replication of the E. coli chromosome initially leads to a replication “eye” or “bubble”.
- The replication eye becomes larger; at this stage, the replication chromosome is referred to as
a theta (θ) structure.
θ structure
-
Growth fork --- growth point of DNA replication.
There are two different modes of DNA synthesis at the growth fork(s).
1. Unidirectional replication --- minor replication mode
2. Bi-directional replication --- major replication mode.
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Takusagawa’s Note©
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Chapter 24
DNA polymerase I (E. coli)
DNA polymerase I (Pol I) is a repair enzyme, and has
- 5’ → 3’ polymerizing activity.
- 3’ → 5’ exonuclease activity.
- 5’ → 3’ exonuclease activity.
Note: 3’→5’ exonuclease cleaves the 3’-end residue of DNA.
3'
OH
P
P
P
P
5' O
Base
O
O
-
O
Base
P O
O
O
O
-
O
Base
P O
O
O
O
3' end cleavage
-
O
P O
Base
O
O
3' O
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Takusagawa’s Note©
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Chapter 24
-
Pol I has an editing function --- If Pol I erroneously incorporates a wrong (unpaired)
nucleotide at the end of a growing DNA chain, the polymerase activity is inhibited and the
3’→5’ exonuclease function is activated by a unpaired 3’-terminal nucleotide with a free OH
group.
-
Pol I 5’→3’ exonuclease binds to duplex DNA at single-strand (5’-end of Okazaki
fragment), and cleaves either mononucleotides or oligonucleotides (up to 10 residues).
Pol’s polymerase and two exonuclease functions each occupy separate active sites
- Proteases such as subtilisin or tripsin cleave Pol I into two fragments.
- Small fragment (residues 1 - 323) has 5’→3’ exonuclease activity.
- Large fragment (residues 324 - 928) has polymerase and 3’→5’ exonuclease activities.
This fragment is called “Klenow fragment”.
1
323
928
N-ter
C-ter
324
Klenow fragment
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Chapter 24
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-
9
Takusagawa’s Note©
Crystal structure of Klenow fragment shows that the fragment looks like a right hand shape,
and DNA synthesis (polymerase activity) occurs between thumb and index finger whereas
the editing (3’→5’ exonuclease) is taken place on palm.
Unpaired nucleotide (wrong nucleotide) goes to the palm area (E below figure) and is
removed by 3’→5’ exonuclease.
Pol I catalyzes nick translation
- As shown in Fig. 31-13, the nicked 5’-end section is removed and the radioactive strand is
added when the radioactive NTP’s are used.
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Takusagawa’s Note©
10
Chapter 24
Major roles of Pol I
- Although Pol I has a polymerase activity, its activity is very weak. Major roles of Pol I are:
1. Remove the damaged DNA section from the 3’-end (3’→5’ exonuclease) and make the new
DNA (polymerase) on the section.
2. Remove the primer RNA (5’→3’ exonuclease) and make the new DNA (polymerase) on the
section.
DNA polymerase III (Pol III)
DNA polymerase III (Pol III) is a replicase, and has
- 5’ → 3’ polymerizing activity.
- 3’ → 5’ exonuclease activity (Editing activity).
Structure of Pol III
- The holoenzyme consists of at least 10 different subunits.
- Its core is consisted of 3 subunits (αεθ) which has a polymerase activity.
- Pol III β subunit is a dimer and forms ring-like structure.
- Double stranded DNA is passed through the center of dimeric β-subunits.
β-subunits
DNA
-
Pol III core has a processivity of 10-15 residues, but the holoenzyme (or core + β subunits)
has unlimited processivity (>5000 residues).
core
core
10-15
unlimited
αεθ
subunits
10
β subunits
Chapter 24
11
The other important enzymes for E. coli
replication are:
- Ligase --- joins Okazaki fragments
- This enzyme uses NAD+ as energy
source.
- The reaction mechanism is shown in right.
- NAD+ is required as the substrate.
DNA1 + DNA2 + NAD+ →
DNA12 + NMN+ + AMP
-
Topoisomerase --- relax positively
supercoiled DNA and/or generate negative
supercoil.
- Topo II (DNA gyrase) of E. coli
prevents positive supercoil, i.e., induces
negative supercoil.
-
Primase --- make primer RNA
-
SSB --- single strand binding protein.
- keeps strands apart in order to prevent
re-annealing.
-
DnaA --- recognizes the oriC site and bind
there to form a complex negatively
supercoiled oriC DNA.
-
DnaB (helicase)--- unwinding DNA helix.
-
DnaC binds on DnaB.
11
Takusagawa’s Note©
Chapter 24
12
Initiation of replication
Bacterial chromosome is a circular DNA, which
has a unique 245-bp segment called oriC
(Origin of chromosome).
1.
DnaA proteins activated by ATP bind to the 4
segments of 9-bp, and forms a negative
supercoil. Thus, a complex of DnaA proteins
is wrapped by oriC DNA. The histone-like
HU proteins facilitate this wrapping process.
2. Three segments of 13-bp (composed of ATrich sequence) located near oriC melt to two
strands (separate to two strands). These
segment sites are P1 endonuclease sensitive if
DNA is single-stranded. Thus, called P1
insensitive (double strands) ↔P1 sensitive
(single strand).
(P1 is a endonuclease that acts on a single
strand DNA).
3. DnaB binds on a single strand DNA with aids
of DnaC. This process requires ATP as
energy source. DnaB is a helicase that
unwinds double helical DNA to two single
strand DNAs.
4. Primase and RNA polymerase bind to the
single strand DNA, and synthesize a primer
RNA on both leading and lagging strands.
For the lagging strand, primers are attached on
each Okazaki fragment. Primase and RNA
polymerase complex is called primosome.
5. Two holo Pol III bind on each strand. Two
holo Pol III are linked together and oriented to
the same direction. Thus, both leading and
lagging strands are synthesized to the same
direction.
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Takusagawa’s Note©
Chapter 24
13
Takusagawa’s Note©
6. The templates of leading and lagging strands are antiparallel to each other at the replication
point. Therefore, the template of the lagging strand is bent to form a loop so that a section
of the loop is parallel to the template of the leading strand.
-
These protein complex is called “replisome”.
13
Chapter 24
14
Overall replication process of E. coli is shown below.
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Takusagawa’s Note©
Takusagawa’s Note©
15
Chapter 24
Control of initiation
1. Level of DnaA controls the initial complex formation.
2. Methylation of adenine (makes m6A = N6-methyl adenine)
- OriC contains sequence GATC 14 times. The sequence GATC is recognized by a
methylating enzyme, and the adenine is methylated (m6A).
- Methylated oriC is efficiently initiated, i.e., methylation required for good initiation.
Termination
- ~350-bp region flanked by six nearly identical 23-bp terminator sites:
- TerE, TerD and TerA on one side and TerF, TerB and TerC on the other side.
- A replication fork traveling counterclockwise passes through TerF, TerB and TerC, but stops
encountering TerA (TerD and TerE are backup terminator sites).
- Similarly, a replication fork traveling clockwise passes through TerE, TerD and TerA, but
stops at TerC (TerB and TerF are backup terminator sites).
- This arrangement guarantees that the two replication forks will meet in the replication
terminus.
- Tus protein from tus gene (terminator utilization substance gene) binds to a Ter site and
prevents the helicase action of DnaB.
OriC
TerE
TerD
TerA
~350-bp
termination
TerC
TerB
site
The final step of DNA replication is unlinking of the catenated parental DNA strands by
topoisomerases.
TerF
-
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Takusagawa’s Note©
16
Chapter 24
Fidelity (error rate) of replication
- One mispairing (error) occurs per 108 to 1010 base pair replications.
- Such high replication accuracy arises from five sources:
1. Cells maintain balanced levels of dNTP.
2. Pol III has very high fidelity by use of base-pairing (open conformation) and shape
recognition of the base (closed conformation).
3. 3’→5’ exonuclease removes mispairing and re-synthesizes the correct pair.
4. Cells contain repair mechanisms for newly synthesized DNA as well as damaged DNA
(i.e., Pol I activity).
5. Use of primer RNA. Most errors occur at the initiation stage. The RNA primers are
removed and paired DNAs are newly synthesized later. Therefore, the errors at the
initiation stage are completely eliminated.
Why both DNA strands are synthesized to 5’→3’ direction ?
- Why not 5’→3’ on one strand and 3’→5’ on other strand ?
Ans. if the DNA was synthesized to the 3’→5’ direction, the chain elongation reaction is
stopped after editing mispaired 5’-terminal nucleotide.
Normal case: 5’→3’ elongation and 3’→5’ exo-hydrolysis by removing error residue.
- After hydrolysis, a correct dNTP can be added to the 3’-OH by usual 5’→3’ elongation
procedure. Note that elongation reaction is carried out by a hydrolysis of phosphoanhydride
bond between α and β-phosphate of dNTP. Therefore, without PPP, the polymerization
reaction does not proceed.
O
P
O
5'
B
B
O
3'
PPi
O
O
O
O
OH
O
O
P
O
P
O PPi
O
C
O
O
O
O
B
B
O
H 2O
OH
C
C
O
B
B
O
P
O
O
C
O
B
O
P
O
O
C
O
O
O
P
O
O
C
O
B
O
P
O
O
C
O
O
P
O
O
C
O
O
O
O
C
O
P
O
B
OH
OH
OH
Wrong base
16
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Takusagawa’s Note©
17
Chapter 24
Abnormal case: 3’→5’ elongation and 5’→3’ exo-hydrolysis by removing error residue
- After hydrolysis, a correct dNTP cannot be added to the 5’-OPO3 by usual 3’→5’ elongation
procedure since there is no PPP on the 5’-end, whose hydrolysis energy is required to
proceed the polymerization reaction.
Wrong base
O
O
PPi
O
O
O P
O
C
B
O
PPi
O
O P
O
C
O
B
5'
OH
O
O
P
O
O
C
B
O
PPi
O
O
O
H2O
O
O
3'
O
O
O
O
O P
O
C
O P O
O
O
O
O
O P
O
C
B
B
B
O
O P
O
C
OH
O
O P
O
C
B
O
O P
O
C
PPi
B
O
PPi
O
C O
B
OH
-
Thus, if the DNA was synthesized to the 3’→5’ direction, the chain elongation reaction is
stopped after editing mispaired 5’-terminal nucleotide.
-
Therefore, both leading and lagging strand DNAs must be synthesized to the 5’→3’ direction
for editing processes.
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Chapter 24
18
Replication in eukaryotes
- Processes are quite similar to those in E.
coli.
- There are several differences because:
- Chromosomes are much larger and
linear except those in mitochondria.
- Multiple replication sites because the
DNA is very long. As shown in right,
when the radioactive dNTPs are used,
several regions along the template DNA
are labeled. This experiment indicates
that the eukaryotic DNA is replicated at
multiple sites.
-
Five different type polymerases (α, β, γ,
δ, ε) are involved:
α and δ are in nucleus and replicate
chromosomal DNA.
α is lagging stand polymerase, and δ is
leading stand polymerase.
γ is in mitochondrial DNA polymerase.
β is a repair enzyme, and ε is probably
repair enzyme.
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Takusagawa’s Note©
Chapter 24
19
Replication of γ mitochondrial polymerase
is quite unique.
- Mitochondrial chromosomes are circular
DNAs.
- The origins of leading strand and
lagging strand are located at different
places.
1. Replication is started at the leading
strand origin.
2. The template of the lagging strand is
displaced to form a displacement or Dloop.
3. When replication of the leading strand
reaches to the replication origin of
lagging stand, the replication of lagging
stand is started.
4. Consequently, replication of leading
stand is firstly completed, and
replication of lagging stand is completed
lately.
Note: Okazaki fragments are produced
in the lagging replication.
Takusagawa’s Note©
D-loop
This figure is wrong. Replication
of the lagging strand must move
the same direction with the leading
strand.
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Chapter 24
20
Takusagawa’s Note©
Reverse transcriptase
- In retroviruses, DNA is synthesized using RNA as a template.
- This reaction (RNA-directed DNA synthesis) is catalyzed by enzyme reverse transcriptase
(RT).
1. Use a host tRNA as a primer (e.g. HIV reverse transcriptase uses lysine-tRNA as the primer).
2. RNase H degrades the RNA on the newly synthesized RNA:DNA hybrid, and produces a
single strand DNA.
3. The second strand DNA is synthesized on the first DNA strand (DNA-directed DNA
synthesis) to yield double strand DNA.
-
RT has no 3’→5’ exonuclease activity (editing function), indicating high rate of error.
Thus, retroviruses have high mutation rate.
-
RT is very useful tool in biological studies. For example, RT is used to make cDNA (Note:
c stands complementary).
RT
⎯→ cDNA
mRNA ⎯ ⎯
- cDNA can be used to probes in Southern transfer analysis, and sequencing of mRNA by
cDNA.
20
21
Chapter 24
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Takusagawa’s Note©
Some nucleoside analogues inhibit the RT activities. These nucleoside analogues do not
have 3’-OH. Thus, the chain elongation is stooped when the analogue is incorporated in the
DNA chain.
Telomeres and telomerase
- A lagging strand is initially synthesized for many short Okazaki fragments on the template
DNA. In this process, each Okazaki fragment requires an RNA primer.
-
At the 3’-end of the template, an RNA primer cannot be produced because no template DNA.
5'
3'
Okazaki fragments
-
Primer
Cannot make prime
Therefore, 3’ end of chromosomes have additional DNA called telomeres or telomeric
DNA.
3'
5'
telomere
Okazaki fragments
Primer
21
Can make primer
Chapter 24
22
Takusagawa’s Note©
-
Telomeric DNA is synthesized by telomerase that contains a short RNA template as shown
below.
-
The telomeric DNA has an unusual sequence, i.e., it consists of up to 1000 or more tandem
repeats of a simple G-rich sequence (such as TTGGGG and TTAGGG).
-
Without telomerase action, a chromosome would be shortened at both ends with every cycle
of DNA replication and cell division. Otherwise immolate cells are die off if telomerase is
eliminated by mutagenesis.
Somatic cells have no telomerase activity. Thus, chromosomes of somatic cells are
gradually shortened upon aging, and finally the cells die.
However, ~80% of somatic cancer cells have telomerase activity. Thus, these cancer cells
are immortal and therefore cause serious problem. Thus, it is important to find out potential
inhibitors for telomerase.
-
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Takusagawa’s Note©
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Chapter 24
Repair of DNA
DNA is not inert substance.
- is modified by UV and ionization radiation (such as X-ray, γ-ray).
- UV (200 ~ 300 nm) promotes the dimerization of thymine rings that are adjacent to each
other.
-
Exposure of DNA to alkylating agents (such as MNNG) yields O6-alkylguanine as shown
below.
H3C
NH
H
N C N
O
H
H2N
N
N
N
NO2
O N
N-methyl-N'-nitro-N-nitrosoguanidine
(MNNG)
N
O
Guanine residue
N
N
H2N
CH3
N
N
O6-Methylguanine residue
23
-
Deamination of adenine and cytosine
NH2
N
deaminase HN
N
N
NH2
O
N
N
N
Adenine (A)
-
Takusagawa’s Note©
24
Chapter 24
N
N
deanimase
N
O
Hypoxanthine (Hy)
O
H
N
O
Cytosine (C)
N
Uracil (U)
The glycosidic bonds of DNA are broken quite frequently.
O
O
O P O3'
O P O3'
O5'
O5'
O
Base
OH
O
O
O
+
Base
O
P O3'
O
O5'
P O3'
O5'
Thus, DNA repair systems are constantly necessary.
There are several repair mechanisms.
1. Direct reversal of damage
- Pyrimidine dimers produced by UV irradiation may be restored to their monomeric forms by
photoreactivating enzymes or DNA photolyases which require cofactor a flavin or 5N, 10Nmethenyl-THF under 300 ~ 500 nm light.
Thymine dimer
TT
Photoreactivating
enzyme
+ visible light
AA
-
TT
AA
O6-alkylguanine frequently form base-pair with T instead of C. Thus, mutation rate is
increased.
The methyl group of O6-alkylguanine is removed by a DNA methyltransferase.
24
Chapter 24
Takusagawa’s Note©
25
2. Excision repair
a) Remove of a damaged section of DNA
requires nuclease activity.
- Seven upstream and four down stream
sites from the unpaired base(s) are
usually cleaved by UvrABC
endonucleases, and removed. After
removal of the damaged DNA, new
DNA is synthesized on the template.
- This is one of reason why DNA is
composed of double strands.
b) Glycosylases remove altered bases
- Methylated bases and deaminated
bases whose rings are cleaved by
glycosylases, and leaves
deoxyribose residue.
Fig. 31-39
c) AP nuclease removes segment. (AP = apurinic & apyrimidic residue)
25
Takusagawa’s Note©
26
Chapter 24
- AP sites yielded by glycosylase activities are recognized by AP nuclease, and AP nuclease
removes a segment containing the apurine or apyridine.
Uracil N-glycosylase
- Cytosine is deaminated under physiological condition.
deamination
C ⎯⎯⎯⎯⎯⎯→ U (occurs naturally)
NH2
OH
N
O
H
N
N
C (keto form)
O
O
N
U (enol form)
N
O
N
U (keto form)
- Also, dTTP and dUTP both are used by Pol III.
- Therefore, U can be found in DNA with relatively high rate.
- Since U in DNA is highly mutagenic (because U can make base-pair either G or A), presence
of U in DNA must be reduced. There are two mechanisms:
1. keep [dUTP] low by converting dUTP to dUMP.
dUTP diphosphohydrolase
dUTP ⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ dUMP + PPi
2. Uracil N-glycosylase removes U ring if U is entered or present in the DNA.
Xeroderma pigmentosum
- is a rare inherited disease in humans.
- inability of skin cells to repair UV-induced DNA lesions.
- Thus, these individuals are extremely sensitive to sunlight and suffer skin cancers.
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Chapter 24
27
Takusagawa’s Note©
3. Recombination repair
- When damaged DNA underwent replication before the lesion was eliminated by either a
direct reversal or nucleotide excision mechanism, the recombination repair mechanism is
applied.
1. For example, DNA containing a pyrimidine dimer in a strand is used as a parental DNA.
2. Replication produces one normal duplex DNA and a damaged duplex DNA whose newly
synthesized strand (daughter strand) has gap since the parental strand has a pyrimidine dimer.
3. The gapped site is repaired by cutting out the same section of the parental strand of the
normal duplex DNA and pasting it on the gapped site of the damaged duplex DNA.
4. The gapped parental strand of the normal duplex DNA is filled and ligated. The section of
pyrimidine dimer is then repaired by either a direct reversal or nucleotide excision
mechanism.
27
Chapter 24
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-
28
Takusagawa’s Note©
4. SOS response (in E. coli)
E. coli chromosomes have lexA gene that produces LexA protein.
LexA is a repressor of SOS genes (including lexA, recA, uvrA, uvrB). LexA binds on
operators of SOS genes. [Note: names with italic letter indicate the DNA gene whereas
names with non-italic letters represent proteins. LexA = protein, lexA = DNA gene.]
The SOS proteins produced from the SOS genes involve in the repair of the DNA damage.
During normal growth, LexA largely represses SOS gene expression. When DNA damage
has been sufficient to produce post-replication gaps, this single-stranded DNA binds to RecA
so as to stimulate LexA cleavage. The cleaved LexA is inactive. Consequently, the LexA
repression is released, and the SOS proteins are synthesized in order to repair the damaged
DNA.
28
Takusagawa’s Note©
29
Chapter 24
DNA methylation
- m6A, m5C and m4C are the only types of modifications to which DNA is subjected in cellular
organisms.
H
N
CH3
N
N
N
-
CH3
N
N
5-Methylcytosine (m5C)
N
CH3
N
O
N
O
N6-Methyladenine (m6A)
-
H
NH2
N
N4-Methylcytosine (m4C)
These methyl groups are on the surface of the major groove of DNA and, thus can interact
with the DNA binding proteins.
In some bacteria, methylation protects against restriction enzymes by preventing the binding
of restriction enzymes.
In eukaryotes, only m5C is present, and methylation appears to control transcription.
Examples
1. Globin gene in erythroid cell are less methylated than in other cells, suggesting that
methylation may turn off expression.
- The specific methylation of the control region in a recombinant globin gene inhibits its
transcription in transfected cells.
2. 5-Azacytosine inhibits methyltransferases, and also stimulates some protein syntheses in
cells.
NH2
N
O
Cannot methylate on N
N
N
OH
O
OH
OH
5-Azacytosine (5-azaC)
- because, 5-azaC binds at the active site of a methyltransferase and inhibits the enzyme’s
methylation activity since the enzyme cannot methylate the N5 position of 5-azaC. Thus,
overall activity of methyltransferase is reduced, and consequently DNA are less methylated.
Therefore, genes of some proteins are more expressed, and the protein syntheses are
stimulated.
-
Certain genes are differently expressed by degree of methylation.
29
Chapter 24
30
Takusagawa’s Note©
Where is methylated?
- Methylation occurs in CG in certain palindromic sequences.
e.g.,
-AGCT-TCGA-, where methylation could occur at the C.
Fragile X syndrome
- In affected individuals, the tip of the X chromosome’s long arm is connected to the rest of
the chromosome by a slender thread that is easily broken.
- The area of the tip of the X chromosome’s long arm has long (CCG)n (n < 1000) sequence.
- In fragile X cases, (CCG)n are totally methylated.
- Fragile X syndrome is a syndrome of X-linked mental retardation.
30