DNA Biosynthesis and DNA Damage Repair

1
Chapter 15 DNA Biosynthesis and DNA Damage
Repair
2

• Section One
• The General Features of
• Replication of Chromosomal DNA

3

• DNA replicates semiconservatively.
• ?. Watson and Crick predicted that DNA
• might semiconservatively replicate.

4

• the hypothesis
• ___________
  p
• ___________
  d
• ___________ p
• ___________ p
• ____________
  d
• ____________
  p

5

• ?. In 1958 Meselson and Stahl demons-
• trated the semiconservative nature of DNA
• replication in E coli.

6

• The experiment
• CsCl equilibrium gradient density ultracentri-
• fugation of 15N labeled E coli DNA.

7

• The density of DNA was increased by labeling it
  with 15N, a heavy isotope of
• nitrogen.
• This was done by growing E coli 15 ge-
• nerations in a medium that contained 15NH4Cl as
  its only nitrogen source.

8

• 15N DNA was extracted and subjected to
• CsCl equilibrium gradient density ultracen-
• trifugation.
• The DNA band position was recorded.
• There was one band of 15N DNA.

9

• The bacteria were transferred to an 14NH4Cl
• medium and grown for one generation.
• The DNA density was determined again.
• The position of the DNA band was compared
• with that of the 15N DNA .

10

• What was the result ?

11

• There was one DNA band. It had a lower
• density than 15N DNA because its position
• was above on that of 15N DNA.
• It was 15N/14N hybrid DNA.

12

• After another generation growing in the
• 14NH4Cl medium the bacterial DNA density
• was determined.
• There were two DNA bands.

13

• One half of the DNA was 14N DNA, and
• another half was hybrid DNA.
• In succeeding generations the ratio of 14N
• DNA to hybrid DNA increased gradually.
• The hybrid DNA became less and less..

14

• summary
• DNA replicates in a semiconservative man-
• ner. When the two parental strands sepa-
• rate, each serves as the template for making
• a new , complementary strand.

15

• 2. The point at which separation of the
• strands and synthesis of new DNA takes
• place is known as the replication fork.
• The replication fork is Y-shaped. Two
• arms (V) are separated strands which
• act as the template and DNA synthesis
• is actively taking place. The body (I) is
• the parental DNA.

16

• 3.DNA replication is usually bidirectional.
• ?.Replicon Any piece which replicates as
• a single unit is called a replicon.
• All bacterial chromosomes and many
• phage and virus DNA molecules are
• circular and comprise single replicons.

17

• In contrast eukaryotic chromosomes consist
• of multiple replicons.
• ?. Origin The initiation within a replicon
• always occurs at a fixed point known
• as the origin.

18

• ?.Terminus In a circular replicon there is
• a single termination site roughly 180
• opposite the unique origin.

19

• Summary
• In a circular replicon replication begins
• from the fixed origin and forms two
  repli-cation forks.
• The two replication forks proceed bidirec-
• tionally away from the origin and the strands
  are copied as they separate until the terminus is
  reached.

20

• 4. DNA replication is semidiscontinuous.
• ?.The mechanism of DNA replication allows only
  for synthesis in a 5?3
• direction.
• ?. The two strands of DNA are antiparallel.

21

• Question
• How is the parental strand that runs 5?3
• past the replication fork copied ?
• The answer is semidiscontinuous replica-
• tion.

22

• At each replication fork one strand ( the lead-
• ing strand ), whose template runs 3?5 past
• the replication fork, is synthesized as one con-
• tinuous piece, while the other strand ( the lag-
• ging strand ), whose template runs 5?3 past
• the replication fork is made discontinuously as
• short fragments in the reverse direction.

23

• These short fragments are called Okazaki
• fragments.
• They are joined by DNA ligase and form the
• lagging strand.

24

• 5. Origins contain short AT-rich repeat se-
• quences.
• ?. Prokaryotic and eukaryotic origins have
  common features
• a. They consist of multiple unique short
  repeat sequences.
• b. These sequences are recognition and
  binding sites of multi-subunit initiation
• factors.

25

• c. These sequences are usually AT- rich.
• ?. E coli s origin is called oriC.
• It is 254bp long and contains three 13-bp
• direct repeats and four 9-bp inverted re-
• peats.

26

• 6. DNA replication needs priming.
• ?. DNA polymerases cannot initiate DNA
• replication by starting a new DNA chain.
• They can only add nucleotides to the 3 end of
  an existent piece of RNA or DNA
• under the direction of the template.
• The existent piece of RNA or DNA are called
  primer.

27

• ?. The leading strand and all Okazaki frag-
• ments are primed by synthesis of a short
• piece of RNA ( an RNA primer ), which is
• then elongated with DNA by DNA poly-
• merase.
• ?. There are also DNA priming or nucleotide
• priming.

28

• 7.Multi-enzymes and proteins participate in
• DNA replication.
• ?. Topoisomerases regulate the type and
• level of supercoiling of dsDNA.
• ?. Helicases unwind the dsDNA.
• ?. SSBs bind and stabilize the single DNA
• strand.
• ?. Primase synthesizes the RNA primer.

29

• ?. DNA polymerases elongate DNA chains.
• ?. DNA ligase joins Okazaki fragments.

30

• 8. DNA replication is of high fidelity.
• ?. There are two types of replication errors.
• a. base ( nucleotide ) substitution.
• b. nucleotide insertion or deletion.
• ?. There are two types of error controls
• a. presynthetic error control.
• b. proofreading control
• ?. mismatch repair.

31

• Section Two
• Features of DNA Polymerases

32

1. The substrates of DNA polymerases are

33

• 2. The active center of DNA pols catalyzes
• DNA synthesis.
• ?. The active center can differentiate dNTP
• from NTP.
• ?. DNA pols can choose the right nucleotide
• for base-pairing with the template nucleo-
• tide.

34

• 3. The semi-closed right-handed structure
• of the DNA pol. is composed of three
• domains.
• ?. thumb domain, fingers domain and palm
• domain
• ?. two active centers polymerase active
• center and 3-5exonuclease active center.
• They are located in the palm domain.

35

• ?.The palm domain has three functions
• a. polymerase activity
• b. to check the newly formed base pair
• c. proofreading control to remove the
• mis-paired nucleotide.
• ?. The fingers domain binds the template
• strand and interacts with the nucleotide
• that enters the polymerase active center.

36

• ?.The thumb domain keeps the primer-template
• junction in position in the active center
  and
• makes the polymerase bind the substrates
• tightly.

37

• 4. The protein of sliding clamp, that
  encircles the DNA and interacts with the DNA pol,
  is responsible for the processivity of the DNA
• polymerase.
• Processivity of the DNA pol means DNA pol
  going on synthesis of DNA rapidly without
  stopping.

38

• Section Three
• DNA Replication in E coli

39

• 1.E coli DNA replication initiates at oriC in a
  process mediated by a multi-protein com-
• plex.
• ?. Protein factors participate in initiation at
• oriC include DnaA, DnaB, DnaC, HU, to-
• poisomerase II ( gyrase ) and SSB.

40

• ?. DnaA protein forms a complex of 20-40
• molecules, each bound to an ATP mole-
• cule, around which the oriC DNA with
• four 9-bp repeats becomes wrapped.
• ?. This facilitates melting of three 13-bp
• repeats which open to allow binding of
• DnaB protein.

41

• ?. With the help of DnaC, DnaB binds the
• opened DNA. DnaB is a helicase and
• can unwind dsDNA by using the energy
• of ATP hydrolysis.
• ?. SSB binds the single DNA strand.
• ?. Gyrase introduces negative supercoils
• into the dsDNA ahead of the replication
• fork.
• ?. The prepriming complex is formed.

42

• 2. Primase synthesizes RNA primer.
• ?. DnaG is a primase. It binds the
  template and is activated by DnaB.
• ?. The activated primase synthesizes
• RNA primer.

43

• 3. DNA pol III elongates DNA and DNA pol
• I removes the primer.
• ?. DNA pol III is the principal enzyme in
• elongation of DNA.
• ?. The structure of the holoenzyme is
  composed of 10 different subunits in total
• number of 16.
• (ae?)2?2?2dd??ß2

44

• Two core enzymes (ae?)2 are held together by a
  ?complex.
• asubunit DNA synthesis
• esubunit proofreading
• ßsubunit the sliding clamp
• A single holoenzyme is responsible for
• the synthesis of both leading strand and
• Okazaki fragments of lagging strand.

45

• The holoenzyme of DNA pol III has two
• core enzymes. One is responsible for
• the synthesis of leading strand. The other
• for the synthesis of Okazaki fragments.
• Because the template of the lagging strand
• is looped out both leading and lagging strand
• synthesis move in the same direction.

46

• When the lagging strand core enzyme
• completes an Okazaki fragment, it re-
• leases the strand.
• Then the primosome ( DnaB-DnaG com- plex )
  synthesizes another primer and
• the core enzyme elongates and completes
• another Okazaki fragment.

47

• ?. DNA pol I has only one polypeptide.
• It has three enzyme activities
• a. the polymerase activity
• b. the 3? 5 exonuclease activity
• c. the 5? 3 exonuclease activity.
• Subtilicin can cut it into two fragments.

48

• The large fragment is called klenow fragment.
• It has the polymerase activity and the 3?5
  exonuclease activity.
• The 5?3 exonuclease removes the primer.

49

• The polymerase function simultaneously
• fills the gap with DNA by elongating the
• 3-end of the adjacent Okazaki fragment.
• The final phosphodiester bond between the
  fragments is made by DNA ligase.

50

• 4.Tus protein recognizes and binds to the
• TER site.
• That prevents the replication fork advancing.
• DNA replication terminates.

51
5. Only at the full-methylated oriC can
initiate replication. There are 11 copies of
sequence GATC in oriC. The dam methylase that
recognizes the sequence GATC and places a
methyl group on the A.
52
GATC is a palindrome. The opposite strand also
reads GATC in the 5?3 direction.
5GATC3 3CTAG5 Only at the
full-methylated oriC can initiate
replication.
53
During DNA replication oriC is also re-
plicated. The parental strand is methylated, but
the newly synthesized daughter strand
isnt. Although GATC in the daughter strand is
also destined to become methylated, about 10
minutes elapse before this can happen.
54
6. Two types of topoisomerases are required in
DNA replication. DNA unwinding at the
replication fork can generate positive
supercoiling of the dsDNA ahead of the
replication fork. DNA gyrase uses the energy of
ATP hydrolysis to introduce negative su-
percoiling into DNA hence removing
supercoiling..
55
E coli chromosome is circular. When DNA
replication is completed, there are two daughter
circular DNA linked together ( catenane ). E
coli topoisomerase IV can unlink the catenane.
56
Section Four DNA
Replication in Eukaryotes
57

• There are five types of common
• eukaryotic DNA polymerases a,ß,?,d,e.
• 2. Eukaryotic and prokaryotic enzymes
• and protein factors that participate in
• DNA replication at the replication fork
• are comparable.

58
3. After the polymerasea/ primase complex
initiates replication polymerase d starts
elongation. 4 .There are two mechanisms of
removing primers. a. RNase HI and
FEN1-dependent mechanism. b. helicase Dna2 and
EFN1-dependent mechanism.
59
5. The eukaryotic chromosome repli- cates
only once in a cell cycle. a. pre-RC foprms in
the G1-phase and is activated in the
S-phase. b. Cdk controls the formation and
activation of pre-RC.
60
6. Telomerase participates in the replica-
tion of telomere DNA. a. the problem of
replicating the ends of linear chromosomes
The ends of linear chromosomes cant be fully
replicated by semidiscontinuous replication as
there is no DNA to elon- gate to replace the RNA
removed from the 5-end of the lagging strand.
61
Thus genetic information could be lost from the
DNA. 2. Telomere and telomerase solve the
problem. To overcome this, the end of eukaryotic
chromosomes ( telomeres ) consists of hundreds
of copies of a simple non- informational repeat
sequence ( TTA GGG ) with the 3-end overhanging
the 5-end.
62
The enzyme telomerase contains a short RNA
molecule, part of whose sequence is
complimentary to this repeat. This RNA acts as a
template for the ad- dition of these repeats to
the 3-over- hang by repeated cycles of
elongation. The complimentary strand is then
synthe- sized by normal lagging strand
synthesis leaving a 3-overhang.
63

• Section Five
• DNA Replication in Mitochondria and Phages
• mtDNA replicates in D-loops.
• 2.Phages circular DNA replicates in rolling
  circle.

64
Section Six The Repair
of DNA Damage
65
1. Physical or chemical agents may cause DNA
damage. There are replication errors. There
are several DNA repair systems in both
prokaryotic and eukaryotic cells.
66
2. Mismatch repair system repairs the
replication errors. Replication errors that
escape proof- reading have a mismatch in the
daug- hter strand. Hemi-methylation of the DNA
after repli- cation allows the daughter strand
to be distinguished from the parental strand.
67
The mismatched base is recognized and bound by
MutS. MutS and DNA complex recruits MutL. MutH,
an endonuclease , joins them and makes a nick in
the newly synthesized DNA strand.
68
Helicase UvrD and an exonuclease remove a piece
of ssDNA that contains the error. DNA
polymerase III fills the gap and DNA ligase
seals the nick.
69
3. DNA Damage Repair Systems a . Direct
Repair System The most common DNA damage
is formation of thymine dimer by UV
radiation. In the TT dimer there is a
cyclobutane ring between the two neighbouring T
residues in the same DNA strand.
70
The photoreactivation repair system is a direct
repair system. Direct repair can repair DNA
damage without removing a base or
nucleotide. In the photoreactivation repair the
photolyase is activated by visible light and
makes use of the light energy to break the
cyclobutane ring. The TT dimer are restored to
the original structure.
71
b. Base Excision Repair System Single base
damages such as deamination of C, depurination,
and depyrimidination are also very
common. Glycosidases can recognize the damaged
base and remove it. That results in an AP site (
apurinic or apyrimidinic site ).
72
AP endonuclease hydrolyzes the phos- phordiester
bond at the 5-end of the AP site. AP
exonuclease cleaves the phosphor- diester bond
at the 3-end of the AP site. and the
deoxyribose-phosphate is re- moved.
73
The gap left is filled with a nucleotide
complementary to the template. DNA ligase makes
the final phosphor- diester bond.
74
c. Nucleotide Excision Repair System Nucleotide
excision repair system recog- nizes the
distortion of the DNA double helix. The
distortion may be caused by TT,CT or CC dimer.
75
In E coli the NER components consist of UvrA,
UvrB, UvrC, and UvrD proteins. UvrA recognizes
the distortion of DNA and combines with UvrB and
ATP to separate the dsDNA. UvrB recruits UvrC ,
an endonuclease.
76

• UvrC makes nicks at both sides of the damage
  on the DNA strand.
• UvrD, a helicase , removes the damage
• contained fragment.
• DNA polymerase I fills the gap.
• DNA ligase links the 3-OH and 5-P by
• phosphordiester bond.

77
Xeroderma Pigmentosum ( XP ) Human NER
system consists of a number of XP proteins
including XPA, XPB, XPC, XPD, XPF, XPG and
ERCC1. The complex of XP proteins scans the DNA,
recognizes the lesion and removes a piece of
damaged DNA about 25 nuc- leotides in length.
78
The gap is repaired by DNA polymerase and DNA
ligase. Patients having recessive mutations in
XP protein genes cannot repair UV- induced DNA
damages in the skin be- bause of lacking of XP
protein activities. That causes cancer and is
known as xeroderma pigmentosum.
79
d. Recombinational repair ( RR ) system is
responsible for repairing DNA double- strand
breaks. The exchange of homologous regions
between two DNA molecules is called homologous
recombination ( HR ).
80
Homologous recombination ( HR ) plays an
important part in organisms. One of the HR
functions is the repair of DNA double-strand
breaks. The hypothetical mechanism of RR is as
follows
81
broken DNA 5-----------------------
------------------------3 3-------------------
---- ------------------------5, 5--------------
------------------------------------3 3---------
-----------------------------------------5
intact homologous DNA The broken region
undergoes 5-exonuclease digestion.
82
5-------------------3 5-----------------3
3------------5 3------------------------
-5 5-------------------------------------------
----3 3-----------------------------------------
------5 unwinding and strand
invasion
83
5----------------3 5--------------------3
3------------5 -----
-------------------5
5-------------- 3 -------------------
--3 3-------------------------------------------
----5 DNA synthesis
84
5------------------?------------------------3 3
-------------5 ---------- ---------------5 3
------------- ?-------
-------------5 5--------------------------------
--------------3 resolution of Holiday
junctions
85
5-------------------?----------------------3 3
--------------------------------------------5 3
--------------?---------------------------5 5---
-----------------------------------------3
86
In E coli RecBCD protein complex has both
nuclease and helicase activities. RecA protein
participates in the formation of the Holliday
junction. DNA polymerase synthesizes DNA. RuvA,
RuvB, and RuvC proteins resolute the Holliday
junction.
87
e. Translesional DNA polymerases can synthesize
DNA translesionally. Translesional DNA pols
require template for DNA synthesis, but they do
not strictly obey the base-pairing rules in
incorporation of nucleotides. Hence translesion
DNA synthesis ( TLS ) is error-prone. A high
mutation rate will occur.
88
In E coli TLS is a part of a cellular stress
response to extensive DNA damage known as SOS
response. The SOS system involves a lot of
genes which are related to DNA replication and
repair. SOS system is regulated by LexA and RecA
proteins.
89
LexA binds to the control region of SOS genes
encoded for repair enzymes and shuts off the
genes. Severe DNA damages activate the RecA
mediated cleavage and destruction of LexA. That
results in efficient expression of SOS response
genes.
90
DNA repair occurs with error-prone properties.
Section Seven
Reverse Transcription
91

• The discovery of reverse transcriptase
• develops the central dogma put forth
• by Crick.
• the central dogma
• replication replication
• DNA-----?RNA---------?protein
• transcription translation

92
In 1964 Temin put forth the hypothesis virus
RNA--?provirus DNA --?virus RNA In 1970 Temin
and Baltimore found the enzyme responsible for
synthesis of DNA directed by RNA template in RNA
virus independently. It is the reverse
transcriptase.
93
2. Reverse transcriptase makes use of RNA as
template, tRNA as primer and synthesizes
double-stranded cDNA. Reverse transcriptase has
three enzy- matic activiyies a. RDDP
activity It transcribes the RNA template onto a
complimentary DNA strand to form an RNA-DNA
hybrid.
94
RNase H activity The enzyme then degrades the
RNA template. DDDP activity It replicates the
resulting single-stranded DNA to form the
duplex DNA.
95
3. The life cycle of retroviruses Retroviruses
have a diploid, positive sense RNA ( mRNA )
genome. The genome replicates via a dsDNA in-
termediate, the provirus. The provirus is
inserted into the host cells genome.
96
The provirus genome is expressible. The gene
products are mRNA, the retrovirus genome, and
various pro- teins, including reverse
transcriptase. The retrovirus particle is
packaged in the host cell and the mature virion
buds from the host cell surface.
97

• Homework
• Explain the following terms
• a. semiconservative replication
• b. replication fork
• c. leading strand
• d. lagging strand
• e. Okazaki fragment
• f. replicon
• g. primase
• h. telomere
• i. telomerase

98

• k. mismatch repair
• l. photoreactivation
• m. replication origin
• answer the following questions
• What are the substrates of the DNA polymerase
  during DNA synthesis?
• b.In what form are the deoxynucleotide
• units incorporated in the newly synthesized
  DNA chain?

99
3. What is the direction of DNA chain growth
during DNA synthesis? 4. What are the features
of the DNA polymerase? 5. Describe the features
of oriC. 6. State the process of DNA
synthesis at the replication fork in E coli.
100
7. Give examples of agents that can damage
DNA. 8. Give examples of mutation. 9. Tell the
story of NER. 10. Describe the process of
RR. 11. Describe the features of reverse
transcriptase.