FCH 532 Homework 6
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1. Outline the steps for DNA replication. List all enzymes/proteins involved and use
diagrams/pictures to explain your answer.
Helicase (unwinding protein / rep protein) recognizes and binds origin of replication.
Catalyzes separation of the two DNA strands by disrupting H-bonding between base
pairs.
Endothermic reaction is coupled to hydrolysis of ATP.
DNA gyrase (a topoisomerase) assists in unwinding by inducing supercoiling.
Single-stranded DNA binding proteins (SSB) bind exposed strands of DNA.
Protect it from hydrolytic cleavage of phosphodiester bonds.
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3.
Primer synthesis: Short complementary stretch of RNA
synthesized by primase enzyme.
(4-10 bases) is
•Primer with free 3’-OH is required by DNA pol III to start 2nd strand synthesis.
•RNA primer is later degraded by 5’->3’ exonuclease action of DNA pol I, RNaseH
enzymes.
4.
DNA synthesis by DNA pol III begins, extending leading and lagging strands.
•DNA synthesis continues until it meets next fragment.
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5.
RNA primers are removed by 5’3’ exonuclease action of DNA polymerase
I, small gaps are filled in by DNA polymerase I.
6.
Final gap between new strands is closed by DNA ligase enzyme.
•Requires ATP to join 3’ OH on one fragment and 5’ phosphate on second
fragment.
2. Explain how certain mutant varieties of Pol I can be nearly devoid of DNA
polymerase activity but retain almost normal levels of 5’ 3’ exonuclease
activity.
Pol I has 2 structurally independent domains: The larger domain contains its
polymerase and 3’ to 5’ exonuclease functions and the smaller domain contains its 5’
to 3’ exonuclease function. A point mutation or insertion/deletion that alters the
catalytic site on the larger domain need not affect the smaller domain.
3. Why haven’t Pol I mutants been found that completely lack 5’ 3’ exonuclease
activity at all temperatures?
The 5’ 3’ exonuclease activity is essential for the replication of DNA since it
removes the RNA primers and replaces them with DNA. Thus, a mutation that
totally inactivates this function at low temperatures is lethal.
4. The 3’ 5’ exonuclease activity of Pol I excises only unpaired 3’-terminal
nucleotides from DNA, whereas this enzyme’s pyrophophorylysis activity
removes only properly paired 3’-terminal nucleotides. Discuss the mechanistic
significance of this phenomenon in terms of the polymerase reaction.
DNA polymerization results in the formation of base pairs. Since a back reaction
should be the exact reverse of a forward reaction, pyrophosphorolysis, the back
reaction of DNA polymerization, should act only on base paired 3’-terminal
nucleotides. Consequently, there must be two forms of the enzyme-DNA complex.
That which is base paired favors synthesis or pyrophosphorolysis, whereas that
which is unpaired favors hydrolysis. Thus, the enzyme must have 2 at least partially
separate active sites for these activities.
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5. You have isolated E. coli with temperature-sensitive mutations in the following
genes. What are their phenotypes above their restrictive temperatures? Be
specific. (a) dnaB, (b) dnaE, (c) dnaG, (d) lig, (e) polA, (f) rep, (g) ssb, and (h)
recA.
dnaB encodes the replisome’s main helicase. Inactivation of DnaB protein when
the temperature is raised would therefore stop the progression of the replication
fork. This would be lethal
dnaE (aka polC) encodes the subunit of pol III, the DNA replicase.
Inactivation would be lethal (no DNA replication).
dnaG encodes primase. Inactivation of primase stops replication because there is
no synthesis for leading or lagging strand synthesis. Lethal mutation.
lig encodes DNA ligase. Inactivation would result in the synthesis of short
segments of DNA rather than continuous strands. Lethal mutation.
polA encodes PolI. Inactivation prevents the excision of the RNA primers and
excision repair. DNA would be fragmented with RNA leaders on the resulting
Okazaki fragments. This would be lethal.
rep encodes the Rep protein which helps unwind the double helix ahead of the
replication fork. Inactivating Rep slows bud does not prevent DNA replication
(Section 30-2C).
ssb encodes single-strand binding protein (SSB) which prevents separated single
strands ahead of the replication fork from reannealing. Inactivation would be
lethal since it would stop replication.
recA encodes RecA which mediates general recombination and the SOS
response. Inactivation of RecA would render bacteria unable to undergo normal
recombination and they would become hypersensitive to massive DNA damage.
6. Short answers. Define the following terms in two or three sentences. Be specific.
 Leading strand-The continuously synthesized strand during DNA synthesis
from a 3’ to 5’ parental template.
 Lagging strand-Discontinuously synthesiszed strand during DNA synthesis
from the 5’ to 3’ parental template. This template must loop back around to
Pol III and produces Okazaki fragments of DNA which are created by
synthesis from RNA primers. The RNA primers are removed and replaced by
DNA via Pol I. Gaps are sealed by DNA ligase.
 Primase-Enzyme that is responsible for synthesizing the RNA primers.
 Helicase-Enzyme part of replisome that will unwind DNA (DnaB).
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SSB-single stranded binding proteins bind to single stranded DNA and
prevent reannealing during DNA replication.
DnaB- the main helicase.
Rep-Helps to unwind the helix ahead of the replication fork.
PriA-a 3’ to 5’ helicase that is involved in the protein assembly called a
primosome. This helicase is essential for bacteriophage X174 DNA
replication. Binds to a 70-nt sequence containing a 44-nt hairpin called pas
(primosome assembly site).
Ligase-Seals single strand nicks between adjacent Okazaki fragments and also
nicks on circular DNA after leading strand synthesis. Uses either NAD+ to
NMN and AMP or ATP to PPi and AMP.
Okazaki fragment-Fragments of newly synthesized DNA (1000 to 2000 nt)
discontinuously synthesized on the lagging strand during DNA replication.
Theta structure-formed when circular chromosomes are replicating due to the
resemblance of the greek letter theta  and shows that dsDNA replicates by
the progressive separation of its 2 parental strands accompanied by the
synthesis of their complementary strands to yield 2 semiconservatively
replicated duplex daughter strands.
Klenow fragment-The larger fragment of Pol I after cleavage with trypsin that
consists of the C-terminal polypeptide sequence which contains both the
polymerase and 3’ to 5’ exonuclease activities.
Gyrase-Enzyme introduces negative supercoils (Type II topoisomerase) at the
expense of ATP hydrolysis. Essential for replication.
Nucleotide excision repair-repairs relatively bulky DNA lesions. Eliminates
damage to dsDNA by excising an oligonucleotide containing the lesion and
filling in the resulting single-strand gap. ATP dependent process uses UvrA,
UvrB, and UvrC proteins. Cleaves the damaged DNA strand at the seventh
and at 3rd or 4th phosphodiester bonds from the lesion’s 5’ and 3’ sides
respectively. Excised 11 to 12 nt oligo is displaced by the binding of UvrD
(helicase II).
Base excision repair-repairs nonbulky lesions involving a single base.
Usually used to removed deaminated bases via the action of DNA
glycosylases. By making apurinic or apyrimidinic sites, the DNA can be
repaired. The deoxyribose is removed by AP endonuclease and then adjacent
residues are removed by an exonuclease. The gap is filled and sealed by DNA
polymerase and ligase.
Mismatch repair-process by which errors that have eluded the DNA
polymerase editing system may be corrected. In E. coli requires 3 proteins:
MutS binds to the mismatched base pair or unpaired bases as a homodimer.
MutL binds to the MutS-DNA complex as a dimer. MutS-MutL translocates
along DNA in both directions forming a loop in the DNA through ATP
hydrolysis. Once a hemimethylated GATC palindrome is found, the MutSMutL complex recruits MutH (a single strand endonuclease) to make a nick
on the 5’ side of the unmethylated GATC. This GATC is located on either
side of the mismatch up to 1000 bp away. MutS-MutL then recruits UvrD
helicase which separates the strands and with an exonuclease, degrades the
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7.
nicked strand from the nick to beyond the mismatch. If the nick is on the 3’
side use exonuclease I. If on the 5’ side uses RecJ or exonuclease VII. Gap is
filled by Pol III and sealed by DNA ligase.
RecA-a protein that mediates recombination through strand exchange between
single-strand and duplex DNAs. On encountering a dsDNA with a strand that
is complementary to its bound ssDNA, RecA will partially unwind the duplex
and in a reaction driven by RecA catalyzed ATP hydrolysis, exhanges the
ssDNA with the corresponding stand on the duplex.
Holliday junction-a structure that is made when corresponding strands of two
aligned homologous DNA duplexes are nicked, and the nicked strands cross
over to pair with the nearly complementary strands of the homologous duplex
after which the nicks are sealed.
Conjugation-Bacterial transfer of DNA from one cell to another via a
cytoplasmic bridge.
Outline the differences between the 3 main types of DNA polymerases found in
E. coli.
8. Describe an experiment to show bidirectional vs. unidirectional replication of a
circular DNA chromosome.
This can be accomplished by a pulse-chase experiment. Cells are grown for
several generations in a medium that is lightly labeled with [3H] thymidine to that all
of its DNA will be visibile in an autoradiogram. A large amount of [3H] thymidine is
then added to the medium for a few seconds before the DNA is isolated in order to
label only those bases near the replication fork(s). Unidirectional DNA replication
will exhibit only one heavily labeled branch point, whereas bidirectional DNA
replication will exhibit two such branch points.
9. Outline mechanism for the repair of a thymine dimer in E. coli. Be specific.
Thymine dimers are repaired through the action of photolyases that bind to
pyrimidine dimers in DNA (can occur in the dark). A noncovalently bound
chromophore (either N5N10-methenyltetrahydrofolate (MTHF) or 5-deazaflavin)
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aborbs 300-500nm light and transferred the energy to FADH-. This transfers the
energy to the pyrimidine dimer and splits it. The resulting pyrimidine anion
rereduces the FADH and the DNA is released.
10. Draw the mechanism of the ligase enzyme. Show chemical structures.
11. It is generally assumed in general biochemistry that DNA replicates in the 5' 3'
direction. Why do we think this is true? Propose an experiment to prove that 3'
5' replication does not occur.
This can be illustrated by the following figure:
a. The coupling of each nucleoside triphosphate to the growing chain
would be driven by the hydrolysis of the previously appended
nucleoside triphosphate.
b. However, the editorial removal of an incorrect 5’ terminal NTP would
not allow further extension of the DNA.
From this it can be concluded that DNA must be extended in the 5’ to 3’ direction.
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12. Why is uracil so mutagenic in DNA?
Uracil is highly mutagenic because deamination of cytosine results in uracil via
spontaneous hydrolysis. Therefore there would be no indication as to whether U
were a normal DNA base or a deaminated C residue resulting in mismatched GU base pairs that were originally GC or AU.
13. What is the SOS response? What proteins are involved?
This is a response to massive DNA damage in E. coli the SOS response prevents
cell division and increases the capacity to repair DNA. LexA normally represses
the SOS response. RecA plays a central role since it cleaves LexA and allows the
transcription of recA, uvrA, uvrB, and other genes regulated by LexA.
14. Why is the methylation of DNA to form O6-methylguanine mutagenic? The O6methylguanine residue are highly mutagenic since they base pair with
thymine instead of cytosine. This would change a GC to AT.
15. A replication fork encountering a single-strand lesion may either dissociate or
leave a single-strand gap. The latter process is more likely to occur during
lagging strand synthesis than during leading strand synthesis. Explain.
The leading strand is continuously synthesized from origin to terminus, whereas the
lagging strand is discontinuously synthesized as Okazaki fragments. A leading
strand replicase that dissociated from its template on encountering a DNA lesion
would be unlikely to simply resume synthesis immediately past the lesion (leaving a
single-strand gap) because it requires a restart primosome to do so. In contrast, if
the lagging strand replicase dissociated from its template on encountering a lesion, it
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would be likely to start the synthesis of a new Okazaki fragment. This would leave a
single-strand gap between the end of the interrupted Okazaki fragment and the
beginning of the Okazaki fragment that had been previously synthesized.
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