w4 slides fc Flashcards
initiator proteins
recognizes origin of replication and binds there
sliding clamp function
hold polymerase onto dna and synthesizes dna
dna ligase
nick sealing and stitches tg okazaki fragments
initiator proteins for rep in e.coli
- binds to origin
- helps helicase bind
- requires atp
–>for regulation of dna rep
–>until they get atp, CAN’T start new cycle
what happens after atp is hydrolyzed
no other initiator protein can bind
What does the AT-rich sequence attract
initiator proteins
initiator proteins
require atp
-don’t hydrolyze atp until dna synthesis begins
-after hydrolysis, can’t start new
unwinding dna
- 2 types of helicases exist
-the predominant one moves 5’ - 3’ along the LAGGING strand template - helicase requires ATP (has 6 subunits, quaternary structure)
single strand binding proteins
-following the action of helicase, ssBPs keep DNA strands separated (coat LAGGING strand)
- separates strands by: binding ssDNA
- prevents strands from: H-bonding
-contains single stranded region of DNA template with short regions of base-paired “hairpins”
rna primers
-made by primase
1) in order to begin, DNA polymerase requires: bound primer
2) what is the purpose of primase in replication
–>synthesize an rna primer
3) primase proceeds in which direction?
–>3’ to 5’ along template strand
primase able to synthesize primer with JUST template and building block ribonucleotides (*unlike dna polymerase)
process of primase making rna primers
primase (type of rna polymerase) joins tg 2 ribonucleotides
primase synthesizes in 5’ to 3’ direction
–synthesizes primerrrrrrr
rna primer=temp. placeholder in dna rep.
–needs to be compl. and antiparallel to template
primase proceeds
steps in bacterial dna replication
- origin of replication
- binding of initiator proteins
–to rep. origin and destabilization of AT-rich sequence - unwinding by helicase
- binding of SSBP
- rna primers made by primase
dna polymerase
-dna made in complementary and anti-parallel way using base pairing
-nucleotides are always added onto the 3’ OH side
–grows DNA 5’ to 3’
-dna polymerase takes 2 phosphate from the nucleoside triphosphate to make pyrophosphate
what role does dna polymerase play in growing the new strand
incoming nucleoside triphosphate pairs with a base in the TEMPLATE strand
dna polymerase catalyzes covalent linkage of nucleoside triphosphate (takes the 2 P from template and uses it in new strand)
sliding clamps in the role of new strand
holds polymerase onto dna
-doesn’t impede progree
ssBP
keeps everything separated
-need them on LAGGING strand
how are okazaki fragments on the lagging strand linked together?
a special dna repair system is responsible for removal of the rna primer and replacing it with a correctly matched dna sequence
-goal is to remove rna primer so okazaki fragments can link tg (bc rn its in the way of them linking)
process of getting okazaki fragments to link together
new rna primer synthesized by primase
dna polymerase adds nucleotides to 3’ end of new rna primer to synthesize okazaki fragment (until it reaches rna primer thats there)
dna polymerse finishes okazaki fragment
-phosphodiester bonds broken and
previous rn primer removed by NUCLEASES and replaced with dna by repair polymerase
-(nuclease is an enzyme that breaks phosphodiester bonds in nucleic acids)
rna primer removed = “nick” (gap)
-nick sealed with dna ligase
leading strand is synthesized _____ from _____ primer (s)
continuously; single rna
lagging strand is synthesized ________ from _____ primer(s)
discont; multiple
okazaki fragments are made up of
rna primer and dna
primosome made up of
helicase and primase
which strand is the predominant helicase on
lagging
when labelling 5’ and 3’ ends with one side of the helix open and one side twisted where do u start from
opening side
issues in dna rep and dna repair mechanisms
- dna topisomerases
- telomerase and telomeres
- dna proofreading and repair of repl.
errors (3’-5’ exonuclease activity of dna
polymerase mismatch repair) - DNA damage
- other DNA repair mechanisms
what happens when dna is unwound?
as helicase unwinds dna, supercoiling and torsional strain increases
problem in circular chromosomes and large linear eukaryotic chromosomes
solved by dna topoisomerase
unwinding problem
dna helicase uses energy to force helix open
–in the absence of topoisomerase, dna can’t rapidly rotate, and torsional STRESS builds up –> opens up helix (causes SUPERCOIL)
some torsional stress relieved by dna supercoiling
torsional stress ahead of the helicase relieved by free rotation of dna around the phosphodiester bond opposite the single strand break; the same DNA topoisomerase that produced the break reseals it
–dna topoisomerase creates transient single-strand break (cut allows it to freely rotate + seal nick –> works as ligase)
what happens at the ends of eukaryotic linear chromosomes during replication?
replication fork reaches end of chromosome
rna primers replaced by dna; gaps sealed by ligase = missing sequence
lagging strand incompletely replicated
=major problem for lagging strand
—shortening of the 5’ end of daughter dna is a problem due to loss of sequence info
telomerase function
-prevents sequence from being lost
-the repetitive sequence that’s added to the 3’ end of the PARENTAL strand (lagging) is determined by the rna template in telomerase
-able to add dna nucleotides to 3’ end of parental dna in the absence of a primer or dna template
telomere replication
- rna template –>telomere binds to template
-has telomere repeat sequences + incomplete, newly synthesized LAGGING strand - resembles: reverse transcriptase
-telomerase adds additional telomere repeats to template strand - generates: G-rich ends
- adds nucleotides to: 3’ ends of parental strand template
-completion of lagging strand by dna polymerase
extended template strand
telomeres and cancer
-telomeres abundant in stem and germ-line cells, not somatic
-loss of telomeres, which occurs normally during dna rep. limits number of rounds of cell division
-most cancer cells produce high levels of telomerse
issues in dna rep
in semiconservative rep
-mistake can occur during replication of bottom strand
=new strand with error along with og parent strand
-rep w/o repair
=strand with error + new strand (dna w/ permanent mutation)
+
=og parent strand +new strand (dna with og sequence)
dna polymerase has proofreading exonuclease activity
the 3’-5’ exonuclease (exo= OUT, digests nucleotiddes from edges):
-removes misincorporated nucleotide
-polymerase adds an incorrect nucleotide
-mispaired nucleotide removed by proofreading (cut off by exonuclease)
-correctly paired 3’ end allows addition of next nucleotide (base pairing fixed)
-synthesis continues in 5’ to 3’ direction
strand-directed mismatch repair
this is a dna rep error REPAIR process (if proofreading fails)
initiated by detection of distortion in the geometry of the double helix generated by mismatched base pairs
-error in newly made dna, sliding clamp attaches and single strand break
-MutS protein recognizes and locks onto dna mismatch (doesn’t know which strand)
-MutL knows which strand and gets recruited to scan dna
-sliding clamp encountered, MutL nuclease is activated and cuts strand
-repair dna synthesis by repair polymerase and ligase seals it
dna damage
even after synthesis dna can get damaged and need repair
dna can be damaged by:
oxidation, radiation, heat and chemicals
thymine dimer is the result of uv radiation and sticks tg so polymerase can’t get through
-within one base, shouldn’t have dimer
-2 adjacent pyrimidine bases
base excision repair
Uracil shouldn’t be where it is
-gets cut off = dna helix w/ missing base
AP endonuclease and phosphodiesterase remove sugar phosphate =dna helix with single-nucleotide gap
dna polymerase adds new nucleotide using bottom strand as template (adds in C to pair w/ G thts there) dna ligase seals back
nucleotide excision repair
for longer repairs
pyrimidine dimer
-excision nuclease, removes it
-dna helicase opens things up= dna helix with 12 nucleotide gap(=bacteria, euk. would have 30)
-dna polymerase adds new nucleotides using bottom strand as template; dna ligase seals break
nonhomologous end joining
accidental double-strand break
-processing of dna end by nuclease, cells digests ends
-end joining by dna ligase
-deletion of dna sequence
-break repaired with some loss of nucleotides at repair site
homologous recombination
homologous dna molecules with one damaged and one undamaged dna molecule (2 separate strands)
processing of broken ends by recombination-specific nuclease
double strand break accurately repaired using undamaged dna as template
break repaired with NO loss of nucleotides at repair site
pros and cons of dna repair of double stranded breaks
non-homol.
-faster, some loss of nucleotides
homol.
-slower and NO loss of nucleotides
you have identified a patient with cancer, performed a biopsy of the cancerous tumour and have sequenced the genome in some of the cells from the biopsy. When you analyze the genome sequences from the cancer cells, you may find many more A/T bases where you have expected G/C bases (compared with healthy cells). What’s the most likely issue?
a) problem with nucleotide excision repair
b) problem with strand directed mismatch repair
c) problem with base excision repair
d) problem with 3’ to 5’ exonuclease repair activity
c) problem with base excision repair
deamination C–>U, adds base pairs
What is the major issue caused by helicase activity during DNA replication?
(A) Helicase introduces positive supercoiling ahead of the replication fork, leading to torsional strain.
(B) Helicase slows down replication to prevent errors.
(C) Helicase only unwinds DNA in prokaryotes.
(D) Helicase breaks hydrogen bonds but does not affect DNA topology.
Answer: (A) (Unwinding DNA creates supercoils, which must be resolved by topoisomerases.)
How does the cell prevent separated DNA strands from reannealing after helicase action?
(A) RNA primers permanently separate the strands.
(B) DNA polymerase keeps the strands apart.
(C) Ligase prevents strand separation.
(D) Single-strand binding proteins (SSBs) stabilize the unwound DNA and prevent reannealing.
d
Why do bacterial cells typically have a single origin of replication?
(A) Their genomes are small and circular, allowing replication to complete efficiently from one origin.
(B) Prokaryotic DNA is too unstable for multiple origins.
(C) They lack DNA helicase.
(D) Their DNA polymerase can start replication without an origin.
Answer: (A) (Bacterial genomes are compact, making a single origin sufficient.)
What is the function of initiator proteins at the bacterial origin of replication?
(A) They synthesize primers.
(B) They unwind DNA.
(C) They recognize and bind to the origin, recruiting helicase and other replication proteins.
(D) They terminate replication.
c - Initiator proteins trigger the replication process by recruiting helicase and polymerase.)
What fundamental limitation of DNA polymerase necessitates the use of primase?
(A) DNA polymerase requires a pre-existing 3’-OH group to initiate synthesis.
(B) DNA polymerase can only replicate circular DNA.
(C) DNA polymerase synthesizes DNA in the 3’ to 5’ direction.
(D) DNA polymerase is unstable without helicase.
a
What role does the sliding clamp play in DNA replication?
(A) It unwinds DNA.
(B) It holds DNA polymerase in place on the template strand, increasing processivity.
(C) It removes RNA primers.
(D) It links Okazaki fragments together.
b
Why does the lagging strand require multiple RNA primers?
(A) DNA polymerase needs help proofreading the lagging strand.
(B) The lagging strand lacks a template.
(C) DNA polymerase synthesizes DNA discontinuously in short Okazaki fragments.
(D) The lagging strand is synthesized before the leading strand.
c
Why is replication of the ends of linear chromosomes problematic?
(A) DNA polymerase cannot fully replicate the 5’ end of the lagging strand, leading to sequence loss.
(B) The leading strand has no issue, but the lagging strand cannot initiate synthesis.
(C) The 3’ end of the template cannot be replicated.
(D) Telomeres shorten due to primase activity.
a
How does telomerase solve the end-replication problem?
(A) It functions only in prokaryotes.
(B) It removes excess telomeric DNA.
(C) It synthesizes Okazaki fragments.
(D) It extends the 3’ overhang of the parental DNA using an internal RNA template.
d
Which repair pathway corrects mismatches that escape DNA polymerase proofreading?
(A) Strand-directed mismatch repair
(B) Base excision repair
(C) Homologous recombination
(D) Nucleotide excision repair
Answer: (A) (Mismatch repair detects distortions in DNA geometry and corrects replication errors.)
What is the main function of base excision repair?
(A) To repair double-strand breaks.
(B) To remove chemically altered single bases, such as deaminated cytosine.
(C) To correct large DNA lesions.
(D) To remove pyrimidine dimers caused by UV light.
Answer: (B) (Base excision repair removes small, non-helical-distorting lesions.)
How does loss of telomere function contribute to cancer?
(A) Telomere shortening triggers chromosomal instability and increased mutation rates.
(B) Short telomeres prevent DNA replication.
(C) Telomerase inactivation allows indefinite division.
(D) All cancer cells lack telomerase activity.
Answer: (A) (Critically short telomeres lead to genomic instability, a hallmark of cancer.)
What kind of DNA damage does UV light primarily cause?
(A) Double-strand breaks
(B) Formation of pyrimidine dimers, leading to DNA distortion.
(C) Mismatched base pairs
(D) Oxidative damage
b
Why is DNA polymerase highly accurate in DNA replication?
(A) It has 3’ to 5’ exonuclease activity that removes incorrectly paired nucleotides.
(B) It can synthesize DNA in both directions.
(C) It replaces damaged bases during replication.
(D) It prevents mismatched bases from being incorporated.
Answer: (A) (DNA polymerase proofreading corrects errors by excising mismatched bases before continuing synthesis.)
How does strand-directed mismatch repair recognize which strand to correct?
(A) The original strand is single-stranded.
(B) The damaged strand is more tightly wound.
(C) The newly synthesized strand is not yet fully methylated in prokaryotes or has nicks in eukaryotes.
(D) The correct strand always contains more adenine bases.
Answer: (C) (Mismatch repair relies on differences in methylation or strand nicks to distinguish the new strand.)
What is the role of topoisomerase in DNA replication?
(A) It relieves supercoiling tension ahead of the replication fork.
(B) It unwinds DNA at origins of replication.
(C) It synthesizes Okazaki fragments.
(D) It repairs mismatched bases.
a
What is the key difference between topoisomerase I and topoisomerase II?
(A) Only topoisomerase II functions in eukaryotes.
(B) Topoisomerase I cuts a single strand of DNA, while topoisomerase II cuts both strands.
(C) Topoisomerase I requires ATP.
(D) Topoisomerase II removes RNA primers.
b
What is the main difference between homologous recombination (HR) and non-homologous end joining (NHEJ)?
(A) HR uses a homologous template for accurate repair, while NHEJ directly ligates broken DNA ends.
(B) HR occurs only in bacteria.
(C) NHEJ is always error-free.
(D) HR removes damaged bases before repair.
Answer: (A) (HR ensures high fidelity by copying an intact sequence, while NHEJ can introduce mutations.)
Why is homologous recombination preferred over non-homologous end joining in dividing cells?
(A) It is faster.
(B) It requires less energy.
(C) It is more accurate because it uses a sister chromatid as a repair template.
(D) It does not require enzymes.
c
How does nucleotide excision repair (NER) correct UV-induced DNA damage?
(A) It removes a segment of damaged DNA and synthesizes a new complementary strand.
(B) It directly reverses thymine dimers.
(C) It replaces mismatched bases.
(D) It fuses damaged DNA strands.
a
How might CRISPR-Cas9 be used to repair genetic mutations?
(A) By introducing targeted DNA breaks and correcting sequences via homologous recombination.
(B) By blocking DNA replication.
(C) By degrading damaged DNA entirely.
(D) By enhancing telomerase activity.
a
What is a potential therapeutic strategy for targeting cancer cells with defective DNA repair pathways?
(A) Inducing more mutations
(B) Increasing telomerase expression
(C) Blocking all DNA polymerases
(D) PARP inhibitors, which selectively kill cells lacking homologous recombination repair
d
