- An E. coli strain possesses a temperature-sensitive mutation in the lexA gene, rendering the LexA repressor non-functional at restrictive temperatures. Which of the following phenotypes is most likely to be observed when this strain is cultured at the restrictive temperature, even in the complete absence of external DNA-damaging agents?
Answer: A significant increase in the spontaneous mutation rate.
The LexA protein is a repressor of the SOS response genes, which include error-prone DNA polymerases like Pol V (encoded by umuDC). If LexA is non-functional, the SOS response becomes constitutively active. This leads to the persistent expression of Pol V, which has low fidelity and can introduce errors even during normal DNA replication, thereby significantly increasing the spontaneous mutation rate. Extreme sensitivity to UV (B) is characteristic of strains lacking repair capacity, whereas this strain has a constitutively active (albeit error-prone) repair system. NER (C) is regulated by LexA but not the sole pathway, and its components would be expressed, not inhibited. Cell cycle arrest (D) via SulA is part of the SOS response, but the most direct and prominent phenotype related to genetic stability is the elevated mutation rate.
- Three isogenic E. coli mutant strains, with knockouts in uvrA, mutS, and recA respectively, are grown under optimal conditions without the presence of any external mutagens. Which strain is expected to display the highest spontaneous mutation rate?
Answer: The *mutS*⁻ strain.
When you encounter questions about DNA repair systems and mutation rates, focus on understanding what happens when each repair mechanism is disabled and which types of errors they normally prevent.
The mutS gene encodes a key component of the mismatch repair (MMR) system, which corrects replication errors like base mismatches and small insertional/deletional loops that escape DNA polymerase proofreading. Without functional MutS protein, these spontaneous replication errors accumulate at dramatically high rates during normal cell division. Since DNA replication occurs continuously during bacterial growth, the *mutS*⁻ strain experiences the highest spontaneous mutation rate among these three knockouts.
Let's examine why the other options are incorrect: Option A is wrong because these three repair systems have vastly different impacts on spontaneous mutation rates - they're not equivalent. Option B (*uvrA*⁻) is incorrect because UvrA is part of nucleotide excision repair, which primarily fixes bulky DNA lesions caused by UV radiation and chemical mutagens. Since no external mutagens are present, this deficiency has minimal impact on spontaneous mutation rates. Option C (*recA*⁻) is also wrong because RecA primarily facilitates homologous recombination and SOS response activation. While RecA deficiency affects DNA repair capacity, it doesn't directly cause the high spontaneous mutation rates seen with mismatch repair defects.
Remember this hierarchy: mismatch repair defects cause the most dramatic increases in spontaneous mutations because they affect the fundamental accuracy of DNA replication, while other repair system defects primarily impact responses to external DNA damage.
- A microbiologist creates a double mutant strain of E. coli that is deficient in both photoreactivation (*phr*⁻) and nucleotide excision repair (*uvrA*⁻). How will the phenotype of this double mutant most likely compare to the respective single mutants?
Answer: It will exhibit extreme sensitivity to UV radiation under both light and dark conditions.
When you encounter questions about DNA repair mechanisms, focus on understanding which pathways target specific types of damage and whether they work independently or overlap.
Both photoreactivation and nucleotide excision repair (NER) primarily fix UV-induced DNA lesions like pyrimidine dimers, but they operate under different conditions. Photoreactivation requires visible light and directly reverses UV damage using photolyase enzyme. NER works in both light and dark conditions by cutting out damaged DNA segments and resynthesizing the correct sequence.
The correct answer is B because these two repair systems represent the main pathways E. coli uses to handle UV damage. A double mutant lacking both mechanisms would be extremely vulnerable to UV radiation regardless of lighting conditions - it can't use photoreactivation (no light-dependent repair) nor NER (no excision repair). This creates severe UV sensitivity under all conditions.
Answer A incorrectly suggests photoreactivation dominates over NER. While photoreactivation is highly efficient when light is available, NER serves as a crucial backup, especially in dark conditions. Answer C misunderstands the pathways' function - these systems specifically target UV-induced lesions, not spontaneous mutations occurring without UV exposure. Answer D overstates the defect; these mutants lose UV damage repair capability but retain other repair systems like mismatch repair (MutS/MutL) and double-strand break repair (RecA pathway).
Remember: when analyzing repair mutants, consider both the specific damage types each pathway addresses and the environmental conditions where they function. Multiple repair systems often provide overlapping protection against the same DNA lesions.
- In an actively transcribed gene within an E. coli cell, RNA polymerase encounters a cyclobutane pyrimidine dimer (CPD) on the template DNA strand. Which of the following represents the most likely initial event leading to the repair of this lesion?
Answer: The stalled RNA polymerase is recognized by the Mfd protein, which then recruits the UvrA protein.
When you encounter DNA repair questions involving transcription-blocking lesions like cyclobutane pyrimidine dimers (CPDs), think about transcription-coupled nucleotide excision repair (TC-NER). This specialized repair pathway specifically handles bulky DNA lesions that block RNA polymerase during active transcription.
The correct answer is D because transcription-coupled repair begins when RNA polymerase stalls at a DNA lesion. The Mfd protein (mutation frequency decline) acts as a transcription-repair coupling factor that recognizes the stalled polymerase, removes it from the DNA, and recruits the UvrABC nucleotide excision repair machinery. This coupling ensures that actively transcribed genes—which are essential for cell survival—receive priority repair.
Choice A is incorrect because MutS is part of the mismatch repair system that fixes base-pairing errors, not bulky lesions like CPDs. Choice B represents a common misconception: while RecA does respond to DNA damage, the SOS response is typically triggered by extensive single-stranded DNA from replication problems, not the limited unwinding at a stalled transcription complex. Choice C is wrong because DNA glycosylases initiate base excision repair for small base modifications, but CPDs are bulky lesions requiring nucleotide excision repair—glycosylases cannot cleave the covalent bonds between dimerized pyrimidines.
Remember this key distinction: transcription-coupled repair (Mfd-mediated) handles transcription-blocking lesions, while global nucleotide excision repair handles similar lesions throughout the genome. The coupling to transcription machinery makes TC-NER both faster and more specific for essential genes.
- The central regulatory event in the E. coli SOS response is the autocatalytic cleavage of the LexA repressor, which is stimulated by an activated form of the RecA protein. The direct molecular signal that triggers the conformational change in RecA to its activated, co-protease form is:
Answer: the binding of RecA to long stretches of single-stranded DNA.
When you encounter questions about the SOS response, focus on the molecular cascade: DNA damage → RecA activation → LexA cleavage → gene expression. The key insight is understanding what directly activates RecA protein.
The SOS response begins when DNA damage creates single-stranded DNA (ssDNA) regions through replication fork stalling or repair processes. RecA protein has a specific binding affinity for ssDNA and forms nucleoprotein filaments when it encounters these stretches. This binding induces a crucial conformational change that converts RecA from its inactive form to an activated co-protease. The activated RecA then stimulates LexA's autocatalytic cleavage, releasing repression of SOS genes. Answer B correctly identifies this direct molecular trigger.
Answer A is incorrect because RecA doesn't directly recognize double-strand breaks. While such breaks can lead to ssDNA formation during processing, it's the resulting ssDNA, not the breaks themselves, that RecA binds. Answer C misrepresents the signal - pyrimidine dimers are a type of DNA damage, but RecA responds to the ssDNA consequence of this damage, not the dimers' concentration directly. Answer D describes a hypothetical mechanism that doesn't occur in the SOS system. RecA activation is purely through ssDNA binding, not post-translational modification by kinases.
Remember this sequence for microbiology exams: DNA damage creates ssDNA → RecA binds ssDNA → conformational change activates RecA → LexA cleavage → SOS response. The direct trigger is always ssDNA binding, regardless of what initially caused the DNA damage.
- A bacterium with a functional ada gene is exposed to an alkylating agent, leading to the formation of O⁶-methylguanine. Which statement best describes the primary repair mechanism for this specific DNA lesion?
Answer: A methyltransferase protein transfers the methyl group to one of its own cysteine residues, becoming inactivated.
When you encounter questions about DNA alkylation damage, focus on the specific type of lesion and which repair pathway evolution has optimized for it. The ada gene codes for a highly specialized repair protein that handles O⁶-methylguanine through direct reversal rather than excision.
The correct mechanism involves a methyltransferase enzyme that directly transfers the problematic methyl group from the O⁶ position of guanine to one of its own cysteine residues. This elegant repair strategy completely restores the original guanine base, but comes at a cost—the methyltransferase becomes permanently inactivated after the transfer, making it a "suicide enzyme." This direct reversal is incredibly efficient because it requires no DNA cutting, gap filling, or ligation.
Option A describes what happens when O⁶-methylguanine goes unrepaired—it pairs with thymine during replication, causing mutations. This isn't a repair mechanism. Option B incorrectly suggests base excision repair (BER), but DNA glycosylases don't recognize O⁶-methylguanine as their substrate. Option C describes nucleotide excision repair (NER) via the UvrABC system, which handles bulky lesions like UV-induced pyrimidine dimers, not simple alkylation products like O⁶-methylguanine.
Remember that bacteria have evolved specific, targeted repair mechanisms for common DNA damage types. O⁶-methylguanine is highly mutagenic because it pairs with thymine, so the direct reversal mechanism prevents mutations more effectively than excision-based repairs. When you see alkylation damage questions, consider whether the cell has a direct reversal option before thinking about more complex excision pathways.
- Which of the following microbial processes is most accurately classified as a DNA damage tolerance mechanism as opposed to a DNA damage repair mechanism?
Answer: Bypass of a thymine dimer by DNA Polymerase V during replication.
The key distinction is that repair mechanisms physically remove or reverse the DNA lesion, restoring the original DNA sequence and structure. Tolerance mechanisms allow the cell to proceed with processes like replication despite the presence of the lesion, which may be repaired later or diluted out. Translesion synthesis (TLS) by Pol V (A) is a classic tolerance mechanism; it fills in bases opposite the dimer, allowing replication to continue, but the dimer itself remains in the template strand. Options B, C, and D all describe processes where the lesion is physically removed or reversed, classifying them as true repair mechanisms (NER, photoreactivation, and BER, respectively).
- A population of wild-type E. coli is briefly exposed to a low dose of UV light, inducing the SOS response. The culture is then transferred to a fresh, damage-free medium and allowed to grow for many generations. What is the expected status of the SOS system in the descendants?
Answer: The SOS genes will be fully repressed due to the accumulation of intact LexA protein.
The SOS response is a bacterial stress response system that activates when DNA damage is detected. Understanding how this system turns on and off is crucial for predicting long-term cellular behavior after damage exposure.
When E. coli experiences DNA damage from UV light, single-stranded DNA activates RecA protein, which then stimulates the LexA repressor to cleave itself. This self-cleavage removes LexA from SOS gene promoters, allowing expression of DNA repair enzymes and error-prone polymerases. However, this is a temporary emergency response, not a permanent state change.
Once the bacteria are moved to damage-free medium and DNA repair is completed, the triggering signal (single-stranded DNA) disappears. RecA returns to its inactive state, stopping LexA self-cleavage. New LexA protein accumulates and rebinds to SOS gene promoters, fully repressing the system. This explains why answer C is correct.
Answer A is wrong because maintaining high levels of error-prone polymerases would be detrimental - these enzymes introduce mutations and are only beneficial during active damage repair. Answer B incorrectly suggests bacteria have epigenetic memory systems like those in eukaryotes; the SOS response is purely regulated by current cellular conditions, not past experiences. Answer D is incorrect because once error-prone polymerases are repressed and DNA repair is complete, mutation rates return to normal baseline levels.
Remember: Bacterial stress responses are typically temporary and self-limiting. They activate in response to current conditions and shut down when those conditions resolve, unlike permanent genetic changes.
- The 5'→3' exonuclease activity of DNA Polymerase I is a unique feature crucial for its role in several DNA repair pathways. This activity is essential for the process of:
Answer: nick translation, where a nick is moved along the DNA during repair synthesis.
DNA Polymerase I is unique among DNA polymerases because it possesses both 3'→5' exonuclease activity (like other polymerases) and 5'→3' exonuclease activity. This 5'→3' exonuclease activity allows it to remove nucleotides ahead of its synthesis direction, making it essential for specific repair processes.
The correct answer is B because nick translation is the signature process of DNA Pol I's 5'→3' exonuclease activity. During DNA repair, when there's a nick (break in one strand), DNA Pol I binds and simultaneously removes nucleotides ahead of it (5'→3' exonuclease) while adding new nucleotides behind it (polymerase activity). This effectively "translates" or moves the nick along the DNA while replacing damaged sections with new DNA.
Option A describes 3'→5' exonuclease activity (proofreading), which DNA Pol I shares with other polymerases—this isn't its unique 5'→3' function. Option C involves proteins like RecA and strand exchange mechanisms, not DNA Pol I's exonuclease activity. Option D describes direct repair by enzymes like O⁶-methylguanine-DNA methyltransferase, which directly reverse damage without requiring DNA synthesis or exonuclease activity.
When studying DNA repair mechanisms, focus on which specific enzymatic activities each polymerase contributes. DNA Pol I's dual exonuclease activities make it the "repair specialist"—remember that its 5'→3' exonuclease literally clears the path ahead while its polymerase activity fills in behind, making nick translation possible.
- A bacterial population is exposed to a mutagen that creates lesions which block replicative DNA polymerases. The resulting strong induction of the SOS response allows many cells to survive. What is the fundamental trade-off of activating the UmuD'₂C (DNA Polymerase V) complex in this survival strategy?
Answer: Survival is enabled by completing replication, but at the cost of introducing a high frequency of mutations.
When you encounter questions about DNA damage responses, focus on understanding the biological trade-offs between survival and genetic fidelity. The SOS response is bacteria's emergency system for surviving severe DNA damage that would otherwise be lethal.
The UmuD'₂C complex (DNA Polymerase V) is specifically designed as an error-prone "rescue" polymerase. Unlike normal replicative polymerases that stall at DNA lesions, Pol V can bypass these blocking lesions through translesion synthesis. This allows replication to continue and cells to survive, but the polymerase lacks proofreading ability and frequently incorporates incorrect bases. The fundamental trade-off is survival versus mutation rate—cells live through otherwise lethal damage but accumulate many mutations in the process.
Looking at the wrong answers: Choice A incorrectly suggests lesion reversal, but translesion synthesis bypasses lesions without removing them. Choice B has the mechanism backwards—Pol V decreases fidelity rather than increasing it, and the purpose isn't accurate repair but damage tolerance. Choice C mischaracterizes the process as creating metabolic dormancy, when actually the SOS response involves active mutagenic replication.
Choice D correctly captures this essential trade-off: survival through continued replication comes at the cost of dramatically increased mutation frequency.
Remember that error-prone DNA polymerases represent evolution's "better to be alive and mutated than dead and perfect" solution. When you see SOS response questions, think about the balance between immediate survival and long-term genetic consequences.
- A researcher discovers a novel bacterium that thrives in an environment with high levels of oxidative stress. Genomic analysis reveals a gene encoding a protein with high sequence similarity to a known apurinic/apyrimidinic (AP) endonuclease. The presence of this enzyme is strong evidence for an active pathway involved in:
Answer: base excision repair.
AP endonuclease is a key enzyme in the base excision repair (BER) pathway. BER is initiated by a DNA glycosylase that recognizes and removes a specific damaged base (e.g., an oxidized base common in oxidative stress), creating an AP site. The AP endonuclease then cleaves the phosphodiester backbone at this AP site, allowing for subsequent removal of the sugar-phosphate and synthesis of new DNA. The other pathways listed use different key enzymes for incision or processing.
- Which statement most accurately describes the action and specificity of the family of DNA glycosylases that initiate base excision repair in a bacterium?
Answer: A diverse set of glycosylases exists, with each enzyme specialized to recognize and remove a specific type of damaged base.
Base excision repair (BER) is a critical DNA repair mechanism that fixes damaged or modified bases while leaving the DNA backbone initially intact. Understanding how this process begins will help you tackle similar questions about DNA repair specificity.
The correct answer is C because DNA glycosylases operate on a "one enzyme, one substrate" principle. Each glycosylase has evolved to recognize specific types of base damage through precise molecular interactions. For example, uracil-DNA glycosylase specifically removes uracil bases that shouldn't be in DNA, while 8-oxoguanine glycosylase targets oxidatively damaged guanine. This specificity allows cells to efficiently address the diverse array of DNA lesions they encounter daily.
Let's examine why the other options are incorrect. Option A describes the mechanism of nucleotide excision repair (NER), not base excision repair – NER enzymes do recognize general helix distortions first. Option B is wrong on two counts: no single glycosylase handles all base damage types, and glycosylases specifically do NOT cleave the phosphodiester backbone – that's the job of AP endonucleases that act later in BER. Option D confuses BER with mismatch repair (MMR) – MutS proteins are involved in MMR, not BER, and they recognize mismatched base pairs rather than damaged bases.
When studying DNA repair mechanisms, focus on the specificity principle: each repair pathway has evolved distinct recognition strategies. BER uses highly specific enzymes for specific lesions, while other pathways like NER cast a broader net for general distortions.
- A dam methylase deficient (*dam*⁻) mutant of E. coli is exposed to a chemical mutagen known to cause base pair substitutions. How will the cell's ability to correct the resulting mismatches be affected compared to a wild-type strain?
Answer: The mismatch repair system will be unable to distinguish the template strand from the newly synthesized strand.
The E. coli mismatch repair (MMR) system (MutS-L-H) identifies the newly synthesized strand by its transient lack of adenine methylation at GATC sequences. Dam methylase is responsible for this methylation. In a *dam*⁻ mutant, neither strand is methylated. Consequently, the MMR system cannot distinguish the parental template strand from the new daughter strand, leading to random correction and potentially fixing the mutation into the template strand. MutS binding (B) is not dependent on methylation. NER (C) repairs bulky lesions, not mismatches. While proofreading (D) is a primary defense, MMR is a critical secondary system, and its failure results in a hypermutator phenotype, not just reliance on proofreading.
- Spontaneous deamination of 5-methylcytosine at Dcm methylation sites in E. coli DNA results in a T:G mismatch. This specific lesion is preferentially repaired by a system distinct from the canonical mismatch repair pathway. This system is known as:
Answer: Very Short Patch (VSP) repair.
When you encounter questions about DNA repair in bacteria, focus on matching the specific type of damage to its corresponding repair mechanism. Different lesions require specialized repair pathways that have evolved to handle particular problems.
The key here is understanding that 5-methylcytosine spontaneously deaminates to thymine, creating a T:G mismatch specifically at Dcm methylation sites. This creates a unique problem: canonical mismatch repair systems can't determine which strand contains the error, since both thymine and guanine are normal DNA bases. E. coli has evolved Very Short Patch (VSP) repair specifically for this situation. VSP repair recognizes the sequence context around Dcm sites and preferentially removes the thymine, restoring the original cytosine. This makes option B correct.
Option A is incorrect because SOS-mediated translesion synthesis deals with severe DNA damage that blocks replication, not simple base mismatches. Option C is wrong because transcription-coupled nucleotide excision repair handles bulky lesions like UV-induced pyrimidine dimers that distort the DNA helix, not point mismatches. Option D fails because homologous recombination repair addresses double-strand breaks and stalled replication forks, not single-base mismatches.
For microbiology exams, remember that bacterial DNA repair systems are highly specialized. When you see methylation sites and deamination mentioned together, think about the unique challenges this creates for normal repair mechanisms and look for specialized pathways like VSP repair that have evolved to handle these specific problems.
- A bacterial strain with a null mutation in the recA gene is exposed to a high dose of ionizing radiation. Which cellular process is most critically impaired, leading to the observed high rate of lethality in this mutant?
Answer: Homologous recombination-mediated repair of double-strand breaks.
Ionizing radiation primarily causes double-strand breaks (DSBs) and oxidative damage. The RecA protein is the central enzyme in homologous recombination (HR), which is the main high-fidelity pathway for repairing DSBs in bacteria. A *recA*⁻ mutant cannot perform HR and is also unable to induce the SOS response, making it extremely sensitive to DSB-inducing agents like ionizing radiation. NER (B) primarily repairs bulky lesions like pyrimidine dimers, BER (C) handles smaller base modifications, and photoreactivation (D) is specific to UV-induced dimers.
- In the nucleotide excision repair (NER) pathway of E. coli, after the UvrBC complex has made incisions flanking a DNA lesion, what is the specific function of the UvrD protein?
Answer: It acts as a helicase to displace the single-stranded fragment containing the lesion.
When you encounter questions about DNA repair mechanisms, focus on the sequential steps and the specific role each protein plays in the process. The nucleotide excision repair (NER) pathway in E. coli follows a precise order: recognition, incision, removal, filling, and ligation.
The UvrD protein functions as a helicase that unwinds and displaces the damaged single-stranded DNA fragment after UvrBC has made the flanking cuts. This 3' to 5' helicase activity is essential because simply cutting the DNA isn't enough—the damaged segment must be physically removed from the double helix to allow repair synthesis. UvrD accomplishes this by breaking the hydrogen bonds between the damaged strand and its complement, effectively "peeling away" the excised fragment.
Looking at the incorrect options: (A) describes the role of UvrA, which recognizes DNA distortions and helps assemble the UvrABC complex at damage sites. (B) incorrectly assigns ligase function to UvrD—DNA ligase (not UvrD) seals the final nick after repair is complete. (D) describes the function of DNA polymerase I, which fills the single-stranded gap using 5' to 3' polymerase activity and removes RNA primers with its 5' to 3' exonuclease activity.
For microbiology exams, remember that DNA repair pathways involve multiple specialized enzymes working in sequence. Focus on learning what each protein does and when it acts in the pathway—UvrD always comes after the cutting step to remove the damaged fragment.
- A bacterial culture containing a mutant strain that lacks a functional uracil-DNA glycosylase (*ung*⁻) is exposed to an agent that causes the deamination of cytosine. What is the most likely molecular outcome for a C:G base pair that undergoes this damage in the *ung*⁻ mutant after two complete rounds of DNA replication?
Answer: One of the four resulting DNA duplexes will contain a T:A base pair at the original site.
Deamination of cytosine creates uracil (U), forming a U:G mispair. Uracil-DNA glycosylase is the enzyme in base excision repair (BER) that removes U from DNA. In an *ung*⁻ mutant, the U persists. In the first round of replication, the strand with G templates a new C strand (C:G duplex), while the strand with U templates a new A strand (U:A duplex). In the second round, the C:G duplex replicates to two C:G duplexes. The U:A duplex replicates to one U:A duplex and one T:A duplex. The question asks about the final state of the original C:G pair, but the logic is about progeny. A better phrasing is the consequence after replication. After one round, one daughter cell gets the C:G, the other gets U:A. After the second round, the C:G cell makes two C:G daughters. The U:A cell makes one U:A daughter and one T:A daughter. Therefore, one of the four granddaughter cells has a fixed T:A mutation. NER (C) does not recognize U:G mispairs. The mispair does not typically stall replication (D).
- An experiment compares wild-type E. coli and a *phr*⁻ mutant (lacking photolyase) for survival after UV irradiation. After irradiation, one set of plates for each strain is immediately placed in the dark, and another set is exposed to a strong source of visible light before dark incubation. The greatest difference in survival between the two strains is expected under which condition?
Answer: The wild-type will show significantly higher survival than the *phr*⁻ mutant only on plates exposed to visible light.
The phr gene encodes photolyase, an enzyme that uses energy from visible light to directly reverse UV-induced pyrimidine dimers. This process is called photoreactivation. The wild-type strain can use this highly efficient pathway when exposed to light, greatly increasing its survival. The *phr*⁻ mutant cannot. In the dark, both strains must rely on other, light-independent pathways like nucleotide excision repair (NER). Therefore, the functional advantage of photolyase in the wild-type strain, and thus the greatest difference in survival compared to the mutant, will only be observed under the visible light condition.
- During replication, an E. coli replisome stalls at a pyrimidine dimer on the template strand. Which of the following describes a plausible damage tolerance mechanism that allows replication to resume without immediately repairing the dimer?
Answer: The replication fork reverses, allowing the nascent lagging strand to be used as a template for the nascent leading strand.
When DNA replication encounters damage like pyrimidine dimers, the replisome can't simply read through the lesion. This creates a critical problem: how does the cell continue replication without losing essential genetic information? The key is distinguishing between damage repair mechanisms (which fix the lesion) and damage tolerance mechanisms (which bypass the problem temporarily).
Replication fork reversal is a sophisticated tolerance mechanism where the stalled fork backs up and the DNA strands rearrange. The nascent lagging strand, which was synthesized from the undamaged template, can then serve as a template for the stalled leading strand. This allows replication to continue past the dimer without actually fixing it—the damage is dealt with later by repair systems.
Option A describes the mismatch repair system (MutS/L/H), which handles replication errors, not UV damage like pyrimidine dimers. Option B is incorrect because DNA Pol III's 3'→5' exonuclease activity is a proofreading function that removes misincorporated nucleotides from the growing strand—it cannot excise lesions from the template strand. Option D confuses pyrimidine dimers with alkylation damage; Ada protein removes methyl groups, not the covalent bonds linking pyrimidines in a dimer.
Remember that damage tolerance mechanisms are about bypassing problems temporarily, while repair mechanisms fix the actual lesion. Fork reversal, translesion synthesis, and template switching are your key tolerance pathways—they keep replication moving when the machinery hits roadblocks.
