1Which enzyme directly reverses UV-induced pyrimidine dimers using energy from visible light?
Molecular characterization of repair enzymes in direct repair
Easy
A.Endonuclease
B.DNA ligase
C.Photolyase
D.DNA polymerase I
Correct Answer: Photolyase
Explanation:
Photolyase binds to cyclobutane pyrimidine dimers and uses energy from visible/blue light (photoreactivation) to directly reverse the damage without removing any bases.
Molecular characterization of repair enzymes in direct repair
Easy
A.Cutting out the damaged nucleotide
B.Adding a new complementary base
C.Sealing nicks in the backbone
D.Transferring the methyl group to a cysteine residue on itself
Correct Answer: Transferring the methyl group to a cysteine residue on itself
Explanation:
MGMT is a suicide enzyme that removes the methyl group from O-methylguanine and transfers it to one of its own cysteine residues, becoming inactivated in the process.
Incorrect! Try again.
3Direct repair is characterized by:
Molecular characterization of repair enzymes in direct repair
Easy
A.Reversing the damage without excising nucleotides
B.Removing and replacing a stretch of nucleotides
C.Using a sister chromatid as template
D.Cleaving both DNA strands
Correct Answer: Reversing the damage without excising nucleotides
Explanation:
Direct repair mechanisms chemically reverse the lesion in place (e.g., photoreactivation, alkyl transfer) without cutting out or replacing any nucleotides.
Incorrect! Try again.
4Which repair pathway removes small, non-helix-distorting base lesions such as uracil or oxidized bases?
single strand damage repair
Easy
A.Nucleotide excision repair (NER)
B.Mismatch repair (MMR)
C.Non-homologous end joining
D.Base excision repair (BER)
Correct Answer: Base excision repair (BER)
Explanation:
BER corrects small base damage (deamination, oxidation, alkylation) using DNA glycosylases that remove the faulty base to start the repair.
Incorrect! Try again.
5Which enzyme initiates base excision repair by recognizing and removing a damaged base?
single strand damage repair
Easy
A.DNA ligase
B.Helicase
C.Topoisomerase
D.DNA glycosylase
Correct Answer: DNA glycosylase
Explanation:
DNA glycosylases cleave the glycosidic bond between the damaged base and the sugar, creating an apurinic/apyrimidinic (AP) site that is then processed further.
Incorrect! Try again.
6Nucleotide excision repair (NER) is best suited to repair which type of lesion?
single strand damage repair
Easy
A.Small alkyl groups on guanine
B.Double-strand breaks
C.Bulky, helix-distorting lesions like pyrimidine dimers
D.Single mismatched base pairs
Correct Answer: Bulky, helix-distorting lesions like pyrimidine dimers
Explanation:
NER recognizes and removes bulky lesions that distort the DNA helix, such as UV-induced pyrimidine dimers, by excising a short oligonucleotide containing the damage.
Incorrect! Try again.
7The main role of mismatch repair (MMR) is to correct:
single strand damage repair
Easy
A.Deaminated cytosine bases
B.Base-base mismatches missed by proofreading during replication
C.UV-induced dimers
D.Double-strand breaks
Correct Answer: Base-base mismatches missed by proofreading during replication
Explanation:
Mismatch repair corrects replication errors such as mismatched bases and small insertion/deletion loops that escape polymerase proofreading.
Incorrect! Try again.
8In prokaryotic mismatch repair, which protein recognizes the mismatch?
single strand damage repair
Easy
A.RecA
B.DNA gyrase
C.MutS
D.Photolyase
Correct Answer: MutS
Explanation:
In E. coli, MutS binds to the mismatch and, together with MutL and MutH, directs the repair machinery to the newly synthesized strand.
Incorrect! Try again.
9Which two major pathways repair double-strand breaks in DNA?
repair of double strand DNA breaks
Easy
A.Photoreactivation and mismatch repair
B.Base excision repair and direct repair
C.Homologous recombination and non-homologous end joining
D.Nucleotide excision repair and BER
Correct Answer: Homologous recombination and non-homologous end joining
Explanation:
Double-strand breaks are repaired mainly by homologous recombination (HR), which is accurate, and non-homologous end joining (NHEJ), which is faster but error-prone.
Incorrect! Try again.
10Non-homologous end joining (NHEJ) is considered error-prone because it:
repair of double strand DNA breaks
Easy
A.Uses a sister chromatid as template
B.Copies from an intact homolog
C.Directly ligates broken ends without a template
D.Requires extensive strand invasion
Correct Answer: Directly ligates broken ends without a template
Explanation:
NHEJ rejoins broken DNA ends directly without using a homologous template, which can result in small insertions or deletions at the junction.
Incorrect! Try again.
11Homologous recombination repair of double-strand breaks is most accurate because it:
repair of double strand DNA breaks
Easy
A.Reverses the damage chemically
B.Uses an undamaged homologous sequence as a template
C.Removes damaged bases only
D.Simply ligates the broken ends
Correct Answer: Uses an undamaged homologous sequence as a template
Explanation:
HR uses an intact homologous DNA sequence (usually the sister chromatid) as a template, allowing precise, error-free repair of the break.
Incorrect! Try again.
12During which phases of the cell cycle is homologous recombination most active?
repair of double strand DNA breaks
Easy
A.M phase only
B.S and G2 phases
C.G0 phase only
D.G1 phase only
Correct Answer: S and G2 phases
Explanation:
HR is most active in S and G2 phases because a sister chromatid is available to serve as the repair template after DNA replication.
Incorrect! Try again.
13Homologous recombination requires which of the following between the participating DNA molecules?
Homologous and site-specific recombination
Easy
A.Short specific recognition sites only
B.Single-stranded RNA templates
C.Extensive regions of sequence homology
D.No sequence similarity
Correct Answer: Extensive regions of sequence homology
Explanation:
Homologous recombination depends on long stretches of similar or identical sequence to align and exchange strands between the two DNA molecules.
Incorrect! Try again.
14Site-specific recombination differs from homologous recombination because it:
Homologous and site-specific recombination
Easy
A.Occurs only at short specific DNA sequences
B.Requires long homologous regions
C.Needs RecA protein
D.Cannot rearrange DNA
Correct Answer: Occurs only at short specific DNA sequences
Explanation:
Site-specific recombination takes place at defined short recognition sequences recognized by specialized recombinase enzymes, without needing extensive homology.
Incorrect! Try again.
15The cross-shaped intermediate structure formed during homologous recombination is known as the:
Models for homologous recombination - the Holliday Model
Easy
A.Replication fork
B.Primosome
C.Okazaki fragment
D.Holliday junction
Correct Answer: Holliday junction
Explanation:
The Holliday junction is the characteristic four-way, cross-shaped DNA intermediate where two duplexes are joined by crossover strands during recombination.
Incorrect! Try again.
16In the Holliday model, the movement of the crossover point along the DNA is called:
Models for homologous recombination - the Holliday Model
Easy
A.Strand invasion
B.Ligation
C.Resolution
D.Branch migration
Correct Answer: Branch migration
Explanation:
Branch migration is the sliding of the crossover point along the paired DNA molecules, extending the region of heteroduplex DNA.
Incorrect! Try again.
17The final step of separating a Holliday junction into two individual DNA duplexes is called:
Models for homologous recombination - the Holliday Model
Easy
A.Initiation
B.Replication
C.Branch migration
D.Resolution
Correct Answer: Resolution
Explanation:
Resolution is the cleavage of the Holliday junction by resolvases to yield two separate DNA duplexes, producing either crossover or non-crossover products.
Incorrect! Try again.
18The RecBCD enzyme complex of E. coli has which combined activities?
RecBCD pathways
Easy
A.Helicase and nuclease
B.Ligase and polymerase
C.Glycosylase and methyltransferase
D.Topoisomerase and primase
Correct Answer: Helicase and nuclease
Explanation:
RecBCD is a multifunctional enzyme with both helicase (DNA unwinding) and nuclease (DNA cutting) activities that processes double-strand break ends for recombination.
Incorrect! Try again.
19The specific DNA sequence that regulates RecBCD activity and stimulates recombination is called:
RecBCD pathways
Easy
A.TATA box
B.Shine-Dalgarno sequence
C.Chi () site
D.Pribnow box
Correct Answer: Chi () site
Explanation:
When RecBCD encounters a Chi () site (5'-GCTGGTGG-3'), its nuclease activity is altered, promoting the loading of RecA and stimulating recombination.
Incorrect! Try again.
20Which protein, loaded with the help of the RecBCD pathway, promotes strand invasion during recombination?
RecBCD pathways
Easy
A.DNA ligase
B.Photolyase
C.RecA
D.MutH
Correct Answer: RecA
Explanation:
RecA coats single-stranded DNA to form a nucleoprotein filament that searches for homologous sequences and catalyzes strand invasion during homologous recombination.
Incorrect! Try again.
21A bacterial culture is exposed to UV light, forming cyclobutane pyrimidine dimers. When the cells are subsequently exposed to visible light (300–500 nm), the dimers are directly reversed. Which enzyme is responsible for this repair, and what cofactor does it use?
Molecular characterization of repair enzymes in direct repair
Medium
A.DNA polymerase I, using NAD as an electron donor
B.Uracil-DNA glycosylase, using S-adenosylmethionine
C.AlkB dioxygenase, using -ketoglutarate and Fe(II)
D.Photolyase, using a light-absorbing chromophore such as FADH
Correct Answer: Photolyase, using a light-absorbing chromophore such as FADH
Explanation:
Photoreactivation is carried out by photolyase, which absorbs visible/near-UV light through its FADH chromophore (and a light-harvesting cofactor) to directly split pyrimidine dimers, restoring the original bases without excision.
Incorrect! Try again.
22The enzyme -methylguanine-DNA methyltransferase (MGMT) is often called a 'suicide enzyme.' Why is this label appropriate?
Molecular characterization of repair enzymes in direct repair
Medium
A.It cleaves the phosphodiester backbone and then degrades itself
B.It transfers the methyl group to one of its own cysteine residues and is irreversibly inactivated
C.It induces apoptosis in the cell after repairing a single lesion
D.It removes the entire guanine base and cannot be recycled
Correct Answer: It transfers the methyl group to one of its own cysteine residues and is irreversibly inactivated
Explanation:
MGMT directly reverses -methylguanine by transferring the methyl group to a cysteine in its active site. This modification is irreversible, permanently inactivating the enzyme, so each MGMT molecule repairs only one lesion — hence 'suicide' enzyme.
Incorrect! Try again.
23In base excision repair (BER), what is the correct order of enzymatic events after a damaged base is recognized?
single strand damage repair
Medium
A.DNA glycosylase removes base AP endonuclease cleaves backbone polymerase fills gap ligase seals nick
B.Ligase seals nick AP endonuclease cleaves glycosylase removes base polymerase fills
C.Polymerase fills gap glycosylase removes base endonuclease cleaves ligase seals
D.AP endonuclease cleaves glycosylase removes base ligase seals polymerase fills
Correct Answer: DNA glycosylase removes base AP endonuclease cleaves backbone polymerase fills gap ligase seals nick
Explanation:
BER begins with a specific DNA glycosylase excising the damaged base to create an AP site. AP endonuclease then nicks the backbone, DNA polymerase inserts the correct nucleotide, and DNA ligase seals the remaining nick.
Incorrect! Try again.
24A cell has a defect in nucleotide excision repair (NER). Which type of DNA lesion would this cell be LEAST able to remove?
single strand damage repair
Medium
A.A double-strand break caused by ionizing radiation
B.A uracil residue arising from cytosine deamination
C.A single mismatched base pair introduced during replication
D.Bulky, helix-distorting lesions such as pyrimidine dimers and chemical adducts
Correct Answer: Bulky, helix-distorting lesions such as pyrimidine dimers and chemical adducts
Explanation:
NER specializes in recognizing and removing bulky, helix-distorting lesions like UV-induced pyrimidine dimers and large chemical adducts. Mismatches are handled by mismatch repair, uracil by BER, and double-strand breaks by HR or NHEJ.
Incorrect! Try again.
25In E. coli mismatch repair, the MutH endonuclease preferentially nicks the newly synthesized daughter strand. How does the system distinguish the daughter strand from the template strand?
single strand damage repair
Medium
A.The template strand contains uracil, which marks it for protection
B.The daughter strand is methylated at GATC sequences and is therefore cleaved
C.The daughter strand is shorter and recognized by its free 3'-OH end
D.The template strand is methylated at GATC sequences while the transiently unmethylated daughter strand is targeted
Correct Answer: The template strand is methylated at GATC sequences while the transiently unmethylated daughter strand is targeted
Explanation:
In E. coli, Dam methylase methylates adenine in GATC sequences. Immediately after replication the daughter strand is transiently unmethylated, allowing MutH to nick it so the error-containing new strand is corrected, not the template.
Incorrect! Try again.
26Non-homologous end joining (NHEJ) and homologous recombination (HR) both repair double-strand breaks. Which statement best distinguishes them?
repair of double strand DNA breaks
Medium
A.HR operates only in G1 phase, whereas NHEJ operates only in S/G2 phase
C.NHEJ directly ligates broken ends and is error-prone, whereas HR uses a homologous template and is generally error-free
D.NHEJ is error-free because it copies from the homologous chromosome
Correct Answer: NHEJ directly ligates broken ends and is error-prone, whereas HR uses a homologous template and is generally error-free
Explanation:
NHEJ rejoins broken ends directly, often losing or altering nucleotides (error-prone), and works throughout the cell cycle. HR uses an intact homologous sequence (sister chromatid) as a template, making it accurate but restricted mostly to S/G2 phase.
Incorrect! Try again.
27In the initial step of homologous recombination in eukaryotes, DSB ends are processed to generate 3' single-stranded overhangs. What is this process called and why is it essential?
repair of double strand DNA breaks
Medium
A.End capping; it protects ends from further degradation before ligation
B.Phosphorylation; it activates ligase to seal the break immediately
C.End resection; the 3' ssDNA overhang is required for strand invasion into the homologous duplex
D.Blunting; it removes overhangs so ends can be directly joined
Correct Answer: End resection; the 3' ssDNA overhang is required for strand invasion into the homologous duplex
Explanation:
5'3' end resection produces 3' single-stranded tails that are coated by recombinase (Rad51/RecA). These filaments carry out homology search and strand invasion, the defining step that initiates homologous recombination.
Incorrect! Try again.
28Which protein complex in eukaryotes first recognizes a double-strand break and binds the DNA ends during the NHEJ pathway?
repair of double strand DNA breaks
Medium
A.MRN (Mre11–Rad50–Nbs1) followed by Rad51 loading
B.Ku70/Ku80 heterodimer
C.PCNA sliding clamp
D.The Holliday junction resolvase RuvC
Correct Answer: Ku70/Ku80 heterodimer
Explanation:
In NHEJ, the Ku70/Ku80 heterodimer binds the broken DNA ends first, forming a ring around the DNA. It then recruits DNA-PKcs and other factors to process and ligate the ends.
Incorrect! Try again.
29Which statement correctly contrasts homologous recombination with site-specific recombination?
Homologous and site-specific recombination
Medium
A.Both require RecA and both act only at GATC sequences
B.Site-specific recombination requires extensive homology, whereas homologous recombination acts at random sites
C.Homologous recombination requires extensive sequence homology, whereas site-specific recombination acts at short specific target sequences recognized by dedicated recombinases
D.Homologous recombination is always conservative while site-specific recombination always deletes DNA
Correct Answer: Homologous recombination requires extensive sequence homology, whereas site-specific recombination acts at short specific target sequences recognized by dedicated recombinases
Explanation:
Homologous recombination depends on long stretches of shared sequence to align and exchange strands. Site-specific recombination occurs at defined short recognition sequences bound by specialized recombinases (e.g., Cre, integrase), needing little or no general homology.
Incorrect! Try again.
30Bacteriophage integrates into the E. coli chromosome at the attB/attP sites. What type of recombination is this, and which enzyme catalyzes it?
Homologous and site-specific recombination
Medium
A.Direct repair catalyzed by photolyase
B.Site-specific recombination catalyzed by integrase (a tyrosine recombinase)
C.Transposition catalyzed by a transposase
D.Homologous recombination catalyzed by RecA
Correct Answer: Site-specific recombination catalyzed by integrase (a tyrosine recombinase)
Explanation:
integration occurs between the specific attP and attB sites and is mediated by integrase, a tyrosine recombinase, together with host factor IHF. This is a classic example of conservative site-specific recombination.
Incorrect! Try again.
31In the Holliday model, the crossed-strand intermediate can move along the DNA as base pairs are broken on one duplex and reformed on the other. What is this process called?
Models for homologous recombination - the Holliday Model
Medium
A.Branch migration
B.Nick translation
C.Strand invasion
D.End resection
Correct Answer: Branch migration
Explanation:
Branch migration is the movement of the Holliday junction along the DNA, extending the region of heteroduplex DNA by breaking and reforming base pairs. In E. coli it is driven by the RuvAB complex.
Incorrect! Try again.
32A Holliday junction can be resolved in two ways depending on the orientation of the cuts. What determines whether the flanking markers are recombinant (crossover) or non-recombinant (non-crossover)?
Models for homologous recombination - the Holliday Model
Medium
A.Whether the resolvase cuts the crossed (inner) strands or the non-crossed (outer) strands of the junction
B.Whether branch migration occurred in the 5' or 3' direction
C.Whether the junction formed in G1 or S phase
D.Whether the DNA was methylated at the junction point
Correct Answer: Whether the resolvase cuts the crossed (inner) strands or the non-crossed (outer) strands of the junction
Explanation:
Resolution of a Holliday junction depends on which pair of strands the resolvase cleaves. Cutting one pair yields crossover (recombinant flanking markers), while cutting the other pair yields non-crossover products with only patch heteroduplex.
Incorrect! Try again.
33In the classic Holliday model, heteroduplex DNA (a hybrid of strands from two different parental molecules) is formed. Why is heteroduplex DNA biologically significant?
Models for homologous recombination - the Holliday Model
Medium
A.It cannot be replicated and is degraded before cell division
B.It is always cleaved by restriction enzymes to terminate recombination
C.It can contain mismatched base pairs that, if repaired, lead to gene conversion
D.It permanently blocks branch migration from proceeding
Correct Answer: It can contain mismatched base pairs that, if repaired, lead to gene conversion
Explanation:
Heteroduplex DNA joins strands from two parents, so if the parents differ at a site a mismatch results. Mismatch repair of this heteroduplex can convert one allele to the other, producing gene conversion (non-reciprocal allele ratios).
Incorrect! Try again.
34The RecBCD enzyme of E. coli changes its behavior when it encounters a specific sequence called Chi (, 5'-GCTGGTGG-3'). What happens at the Chi site?
RecBCD pathways
Medium
A.RecBCD stops unwinding and dissociates completely from the DNA
B.RecBCD begins degrading both strands with equal vigor
C.RecBCD reverses direction and re-anneals the two strands
D.Nuclease activity is attenuated and RecBCD begins to load RecA onto the 3' single-stranded tail
Correct Answer: Nuclease activity is attenuated and RecBCD begins to load RecA onto the 3' single-stranded tail
Explanation:
Before reaching Chi, RecBCD unwinds and degrades DNA. At the Chi sequence, its strong nuclease activity is down-regulated and it starts loading RecA onto the 3' overhang, generating the recombinogenic filament that promotes strand invasion.
Incorrect! Try again.
35Which set of enzymatic activities does the RecBCD complex possess that makes it suited to initiate recombination at double-strand ends?
RecBCD pathways
Medium
A.DNA polymerase and ligase activities
B.Helicase (DNA unwinding) and nuclease activities
C.Topoisomerase and primase activities
D.Methyltransferase and glycosylase activities
Correct Answer: Helicase (DNA unwinding) and nuclease activities
Explanation:
RecBCD is both a potent helicase, unwinding duplex DNA from a double-strand end, and a nuclease that degrades the strands. These combined activities process the broken end into a recombinogenic 3' overhang for RecA loading.
Incorrect! Try again.
36After RecA forms a nucleoprotein filament on single-stranded DNA generated by the RecBCD pathway, what is RecA's primary role in recombination?
RecBCD pathways
Medium
A.Cleaving the Holliday junction to release products
B.Sealing nicks in the recombinant DNA
C.Catalyzing homology search and strand invasion/exchange with a homologous duplex
D.Methylating the newly formed heteroduplex to protect it
Correct Answer: Catalyzing homology search and strand invasion/exchange with a homologous duplex
Explanation:
RecA polymerizes on ssDNA to form a filament that searches for a homologous duplex, then promotes strand invasion and exchange, creating the joint molecule. Junction resolution and ligation are performed by other enzymes such as RuvABC and ligase.
Incorrect! Try again.
37Site-specific recombinases are grouped into two major families based on the catalytic amino acid that forms a covalent bond with DNA. What are these two families?
Conserved site-specific recombination
Medium
A.Cysteine recombinases and lysine recombinases
B.Glutamate recombinases and arginine recombinases
C.Histidine recombinases and aspartate recombinases
D.Tyrosine recombinases and serine recombinases
Correct Answer: Tyrosine recombinases and serine recombinases
Explanation:
Conservative site-specific recombinases fall into two families: tyrosine recombinases (e.g., Cre, integrase, Flp) and serine recombinases (e.g., resolvases, invertases). Each uses its namesake residue to form a transient covalent phosphotyrosine or phosphoserine link with DNA.
Incorrect! Try again.
38During conservative site-specific recombination, why is no DNA synthesis or high-energy cofactor (like ATP) generally required for strand exchange?
Conserved site-specific recombination
Medium
A.The energy of the cleaved phosphodiester bond is stored in a covalent protein–DNA intermediate and reused for religation
B.The bases are removed and replaced by direct repair enzymes
C.New nucleotides are added by a polymerase to fill any gaps
D.The reaction is driven entirely by ATP hydrolysis at each step
Correct Answer: The energy of the cleaved phosphodiester bond is stored in a covalent protein–DNA intermediate and reused for religation
Explanation:
The recombinase cleaves DNA by forming a covalent phosphotyrosine (or phosphoserine) bond, conserving the bond energy. This stored energy is used to reseal the strands after exchange, so no net loss or gain of nucleotides and no ATP-driven synthesis is needed.
Incorrect! Try again.
39A researcher places two loxP sites in the same (direct/head-to-tail) orientation flanking a target gene. When Cre recombinase is expressed, what is the outcome?
Cre/LoxP recombination
Medium
A.The target gene is amplified into multiple tandem copies
B.The two loxP sites are duplicated with no loss of DNA
C.The intervening DNA is inverted but retained in the chromosome
D.The intervening DNA between the loxP sites is excised as a circle, deleting the target gene
Correct Answer: The intervening DNA between the loxP sites is excised as a circle, deleting the target gene
Explanation:
When two loxP sites are in the same orientation, Cre-mediated recombination excises the intervening sequence as a circular DNA molecule, deleting the flanked gene. If the loxP sites were in opposite orientation, the intervening DNA would instead be inverted.
Incorrect! Try again.
40In a conditional knockout mouse, a gene is 'floxed' and Cre expression is driven by a tissue-specific promoter. What is the main advantage of this Cre/loxP strategy?
Cre/LoxP recombination
Medium
A.The gene is randomly mutated to generate diverse phenotypes
B.The gene is deleted only in the tissue where Cre is expressed, allowing study of gene function in a specific cell type
C.The gene is permanently overexpressed in all tissues
D.The gene is deleted in every cell of the body simultaneously
Correct Answer: The gene is deleted only in the tissue where Cre is expressed, allowing study of gene function in a specific cell type
Explanation:
Because Cre is expressed only under a tissue-specific promoter, recombination between the loxP sites (deletion of the floxed gene) occurs only in that tissue. This enables conditional, spatially restricted knockouts, useful when whole-body deletion would be lethal or uninformative.
Incorrect! Try again.
41The enzyme -methylguanine-DNA methyltransferase (MGMT) removes alkyl groups from guanine. Which feature of its mechanism makes it fundamentally different from a typical catalytic enzyme?
Molecular characterization of repair enzymes in direct repair
Hard
A.It creates a transient double-strand break to flip the base for repair
B.It acts stoichiometrically as a suicide enzyme, becoming irreversibly inactivated after a single transfer
C.It requires a flavin cofactor and visible light to excise the alkyl adduct
D.It regenerates its active cysteine after each reaction cycle using ATP hydrolysis
Correct Answer: It acts stoichiometrically as a suicide enzyme, becoming irreversibly inactivated after a single transfer
Explanation:
MGMT transfers the alkyl group to an active-site cysteine, which permanently inactivates the protein. It is therefore stoichiometric, not catalytic, and each molecule repairs only one lesion.
Incorrect! Try again.
42DNA photolyase repairs cyclobutane pyrimidine dimers using two chromophores. What is the specific role of the MTHF (or 8-HDF) chromophore relative to the FADH cofactor?
Molecular characterization of repair enzymes in direct repair
Hard
A.It binds the dimer and flips it out of the helix into the active site
B.It oxidizes FADH to FAD after the repair reaction is complete
C.It acts as a light-harvesting antenna that transfers excitation energy to FADH
D.It directly injects an electron into the pyrimidine dimer to split it
Correct Answer: It acts as a light-harvesting antenna that transfers excitation energy to FADH
Explanation:
The second chromophore (MTHF or 8-HDF) absorbs blue/near-UV light and funnels the energy to FADH, which then transfers an electron to the dimer to catalyze bond cleavage. FADH is the catalytic cofactor.
Incorrect! Try again.
43In base excision repair (BER), a bifunctional DNA glycosylase differs from a monofunctional one primarily because it:
single strand damage repair
Hard
A.Uses long-patch synthesis exclusively to replace – nucleotides
B.Requires APE1 to incise the phosphodiester backbone before strand displacement
C.Possesses an associated AP lyase activity that nicks the backbone to the AP site
D.Removes the entire nucleotide including the sugar-phosphate in one step
Correct Answer: Possesses an associated AP lyase activity that nicks the backbone to the AP site
Explanation:
Bifunctional glycosylases have an intrinsic AP lyase that cleaves the backbone via - or -elimination immediately after removing the base, whereas monofunctional glycosylases depend on APE1 for incision.
Incorrect! Try again.
44In E. coli mismatch repair, strand discrimination relies on hemimethylation. If a mismatch lies bp from a site and MutH nicks the unmethylated strand, which activity is required to remove the intervening tract back to the mismatch?
single strand damage repair
Hard
A.MutH endonuclease cleaving repeatedly along the entire tract
B.MutS/MutL-activated helicase II (UvrD) and an exonuclease acting from the nick
C.RecA-mediated strand invasion to relocate the mismatch
D.Pol I nick translation displacing the strand toward the methylated
Correct Answer: MutS/MutL-activated helicase II (UvrD) and an exonuclease acting from the nick
Explanation:
After MutH nicks the daughter (unmethylated) strand, UvrD (helicase II) unwinds from the nick and an exonuclease (e.g. ExoI or RecJ) degrades the strand up to and past the mismatch, allowing resynthesis.
Incorrect! Try again.
45Nucleotide excision repair (NER) in E. coli uses UvrABC. What is the precise nature of the dual incision made by UvrC?
single strand damage repair
Hard
A.Makes a single nick to the lesion followed by exonucleolytic degradation
B.Cuts nt and nt of the lesion, releasing a nt oligomer
C.Cuts symmetrically nt on each side, releasing a nt fragment
D.Cuts nt and – nt of the lesion, releasing a – nt oligomer
Correct Answer: Cuts nt and – nt of the lesion, releasing a – nt oligomer
Explanation:
UvrC makes the incision – nucleotides from the lesion first, then the incision nucleotides away, excising a – nt fragment. Human NER excises a larger – nt patch.
Incorrect! Try again.
46Non-homologous end joining (NHEJ) and homologous recombination (HR) compete for double-strand break repair. Which molecular event most strongly commits a break to the HR pathway?
repair of double strand DNA breaks
Hard
A.Binding of the Ku70/Ku80 heterodimer to the blunt ends
B.Ligation by the XRCC4/Ligase IV complex
C. resection of DNA ends generating single-stranded overhangs
D.Recruitment of DNA-PKcs and Artemis to trim the ends
Correct Answer: resection of DNA ends generating single-stranded overhangs
Explanation:
End resection producing ssDNA overhangs commits the break to HR because Ku cannot rebind resected ends, and the ssDNA is required for RPA and then Rad51 filament formation.
Incorrect! Try again.
47During HR, the MRN (Mre11–Rad50–Nbs1) complex initiates resection. Mre11 has a puzzling activity given its role in resection. What is it?
repair of double strand DNA breaks
Hard
A.A DNA-dependent ATPase that only tethers the two ends together
B.A flap endonuclease removing overhangs generated by CtIP
C.A helicase that unwinds the duplex ahead of resection
D.A exonuclease and endonuclease that nicks internally to allow bidirectional processing
Correct Answer: A exonuclease and endonuclease that nicks internally to allow bidirectional processing
Explanation:
Mre11's nuclease activity is , seemingly opposite to the needed resection. Its endonuclease nicks the strand internally, then Mre11 works back toward the break while EXO1/DNA2 extend outward—the bidirectional resection model.
Incorrect! Try again.
48Synthesis-dependent strand annealing (SDSA) is a DSB repair sub-pathway. Why does SDSA produce exclusively non-crossover products?
repair of double strand DNA breaks
Hard
A.The invading strand is displaced after synthesis and anneals to the other end, so no Holliday junction persists
B.Both Holliday junctions are always resolved in the crossover orientation
C.The second end is degraded before capture, preventing junction formation
D.Resolvases cleave the junctions symmetrically to prevent exchange
Correct Answer: The invading strand is displaced after synthesis and anneals to the other end, so no Holliday junction persists
Explanation:
In SDSA the extended invading strand is unwound from the template and reanneals to the resected second end. Because no stable double Holliday junction forms, only non-crossover products arise.
Incorrect! Try again.
49Which statement correctly distinguishes homologous from site-specific recombination at the mechanistic level?
Homologous and site-specific recombination
Hard
A.Homologous recombination requires extensive sequence identity and DNA synthesis, while site-specific recombination uses short defined sequences and no synthesis
B.Homologous recombination is always conservative while site-specific recombination degrades one duplex
C.Site-specific recombination requires a overhang while homologous recombination uses blunt ends
D.Both require RecA-mediated strand invasion but differ in the length of homology
Correct Answer: Homologous recombination requires extensive sequence identity and DNA synthesis, while site-specific recombination uses short defined sequences and no synthesis
Explanation:
HR needs long stretches of homology and involves strand invasion plus DNA synthesis. Site-specific recombination acts at short, specific recognition sites via a recombinase that cuts, exchanges, and religates without new synthesis.
Incorrect! Try again.
50Tyrosine and serine recombinases both catalyze site-specific recombination but form different covalent intermediates. What distinguishes the tyrosine recombinase mechanism?
Homologous and site-specific recombination
Hard
A.A -phosphotyrosine linkage is formed and strands are exchanged one pair at a time via a Holliday intermediate
B.A -phosphoserine linkage is formed and all four strands are cut simultaneously
C.It uses a metal-dependent transesterification without a covalent protein–DNA bond
D.It introduces a double-strand break in both duplexes before strand rotation
Correct Answer: A -phosphotyrosine linkage is formed and strands are exchanged one pair at a time via a Holliday intermediate
Explanation:
Tyrosine recombinases (e.g., Cre, integrase) attack via an active-site tyrosine forming a -phosphotyrosine, exchange one pair of strands to make a Holliday junction, then exchange the second pair. Serine recombinases cut all four strands and rotate.
Incorrect! Try again.
51In the classical Holliday model, branch migration of the crossover point has a key consequence for the DNA involved. What is it?
Models for homologous recombination - the Holliday Model
Hard
A.It generates heteroduplex DNA of increasing length on both participating duplexes
B.It degrades one strand of each duplex to expose complementary sequences
C.It resolves the junction into two crossover products directly
D.It requires DNA synthesis to fill the migrating gap
Correct Answer: It generates heteroduplex DNA of increasing length on both participating duplexes
Explanation:
As the Holliday junction migrates, base pairs are continuously broken and reformed between strands from different duplexes, extending regions of heteroduplex (hybrid) DNA symmetrically on both molecules.
Incorrect! Try again.
52A single Holliday junction can be resolved by cutting either of two strand pairs. How does the choice of cleavage orientation determine the outcome?
Models for homologous recombination - the Holliday Model
Hard
A.Both orientations always yield crossover products regardless of which strands are cut
B.Only cleavage of the outer strands is enzymatically possible, so all products are crossovers
C.Cutting the crossed (inner) strands gives non-crossover flanking markers; cutting the non-crossed (outer) strands gives crossover flanking markers
D.Cutting either pair produces identical patch (non-crossover) products
Correct Answer: Cutting the crossed (inner) strands gives non-crossover flanking markers; cutting the non-crossed (outer) strands gives crossover flanking markers
Explanation:
Resolution in one plane cuts the originally exchanged strands and leaves flanking markers in the parental (non-crossover) configuration; resolution in the perpendicular plane cuts the other strands and produces crossover (recombinant) flanking markers.
Incorrect! Try again.
53The double-strand-break repair (DSBR) model refined the Holliday model. Why can a single double-Holliday-junction intermediate yield either crossover or non-crossover products?
Models for homologous recombination - the Holliday Model
Hard
A.Each of the two junctions can be independently resolved in either orientation, and only specific combinations give crossovers
B.Dissolution by a helicase-topoisomerase always yields crossovers
C.Branch migration alone determines the product without junction cleavage
D.The two junctions always resolve identically, producing only crossovers
Correct Answer: Each of the two junctions can be independently resolved in either orientation, and only specific combinations give crossovers
Explanation:
With two junctions each resolvable in two ways, the combinations of cleavage orientations determine the outcome: certain combinations give crossovers while others give non-crossovers. Dissolution (convergent migration + decatenation) gives only non-crossovers.
Incorrect! Try again.
54The RecBCD enzyme changes its behavior upon encountering a site (-GCTGGTGG-). What is the precise molecular change in its nuclease activity?
RecBCD pathways
Hard
A.It begins degrading the -ended strand more rapidly, exposing a overhang
B.Vigorous degradation is attenuated and a -ended ssDNA overhang is produced for RecA loading
C.It switches from helicase to a pure exonuclease that degrades both strands equally
D.It dissociates completely, halting all further processing of the DNA
Correct Answer: Vigorous degradation is attenuated and a -ended ssDNA overhang is produced for RecA loading
Explanation:
Before , RecBCD preferentially degrades the -terminated strand. Recognition of attenuates this activity (and switches the polarity of degradation), generating a ssDNA tail onto which RecBCD then loads RecA.
Incorrect! Try again.
55The RecB and RecD subunits are both motor proteins that translocate on opposite strands with opposite polarities. Before reaching , which subunit is the faster motor and what is the consequence?
RecBCD pathways
Hard
A.Both move at equal speed, so no ssDNA loop forms before
B.RecD () is faster, so the tail forms a loop ahead of the slower RecB
C.RecB () is faster, so the tail loops out ahead of RecD
D.RecC provides the motor while RecB and RecD are purely structural
Correct Answer: RecD () is faster, so the tail forms a loop ahead of the slower RecB
Explanation:
Before , RecD (translocating ) moves faster than RecB (), causing the -terminated strand to accumulate as a growing loop ('rabbit ear') ahead of RecB. After , RecB becomes the lead motor.
Incorrect! Try again.
56 sites are described as functionally polar hotspots for recombination in E. coli. What does 'polar' mean in this context?
RecBCD pathways
Hard
A. stimulates recombination only when RecBCD approaches it from a specific orientation (entering from the side)
B. only stimulates recombination in the presence of an external electric field
C. functions equally regardless of the direction RecBCD travels
D. activity depends on the local content of surrounding DNA
Correct Answer: stimulates recombination only when RecBCD approaches it from a specific orientation (entering from the side)
Explanation:
A site is recognized and stimulates recombination only when RecBCD enters and translocates toward it from the correct (3') direction. Approaching from the other side has no effect—hence its polarity.
Incorrect! Try again.
57In bacteriophage integration, Int (a tyrosine recombinase) requires the host factor IHF. What is IHF's essential structural role at ?
Conserved site-specific recombination
Hard
A.It supplies energy via ATP hydrolysis to drive strand exchange
B.It methylates to distinguish it from
C.It provides the catalytic tyrosine that attacks the phosphodiester bond
D.It sharply bends the DNA to allow assembly of the higher-order intasome nucleoprotein complex
Correct Answer: It sharply bends the DNA to allow assembly of the higher-order intasome nucleoprotein complex
Explanation:
Integration Host Factor introduces sharp bends (~) in , enabling the assembly of the intasome in which Int bridges the arm and core sites. IHF is architectural and has no catalytic role.
Incorrect! Try again.
58The directionality of recombination (integration vs. excision) is controlled by which factors, and why is excision not simply the reverse of integration?
Conserved site-specific recombination
Hard
A.Only Int is needed for both, and the reaction runs identically in reverse
B.Excision additionally requires Xis; the different attachment sites ( vs ) impose distinct protein requirements
C.Integration needs Xis while excision needs only IHF
D.Excision requires RecA-mediated strand invasion whereas integration does not
Correct Answer: Excision additionally requires Xis; the different attachment sites ( vs ) impose distinct protein requirements
Explanation:
Integration () needs Int + IHF, while excision () also requires the excisionase Xis. Because the substrate sites differ, the reactions are not simple thermodynamic reversals and are regulated separately.
Incorrect! Try again.
59A site is a bp sequence with two bp palindromic arms flanking an bp asymmetric spacer. Two sites in the same (head-to-tail) orientation on a single DNA molecule will be recombined by Cre to give what outcome?
Cre/LoxP recombination
Hard
A.Duplication of the intervening sequence in tandem
B.Inversion of the intervening sequence relative to the flanking DNA
C.Excision of the intervening sequence as a circle, leaving one site behind
D.Translocation of the sequence to a different chromosome
Correct Answer: Excision of the intervening sequence as a circle, leaving one site behind
Explanation:
The asymmetric bp spacer gives directionality. Directly repeated (head-to-tail) sites cause Cre to excise the intervening DNA as a circle, leaving a single . Inverted sites instead cause inversion.
Incorrect! Try again.
60Cre-mediated excision is intramolecular and produces a circular product plus the deletion allele, yet the reaction is reversible in principle. Why does excision dominate over reintegration in practice?
Cre/LoxP recombination
Hard
A.Excision is favored entropically because the released circle diffuses away, making the bimolecular reintegration inefficient
B.The spacer sequence is destroyed during excision, preventing any reverse reaction
C.Reintegration is blocked because the excised circle lacks a functional site
D.Cre only recognizes directly repeated sites during excision but not during integration
Correct Answer: Excision is favored entropically because the released circle diffuses away, making the bimolecular reintegration inefficient
Explanation:
Excision converts one molecule into two (intramolecular, entropically favored), whereas reintegration requires two molecules to collide (bimolecular, concentration-dependent). At low circle concentrations reintegration is inefficient, so excision dominates.
Incorrect! Try again.
Did this save you a night before the exam?
LPU Notes is free, and it stays free. Ads cover part of the server bill.
The rest comes out of a student's own pocket: the domain, the storage,
and keeping the site up through the weeks everyone needs it at once.
The payment button didn't load. An ad blocker or a filtered network is the usual reason.
to try again.
Nothing here is ever locked, and nothing unlocks. Chip in only if it was worth it.
What it pays for →