Genetics Quiz: Dna Repair Pathways

What this quiz covers

This quiz focuses on Dna Repair Pathways, giving you a quick way to practice the rules, question types, and explanations that matter most for Genetics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The MMR system would be unable to distinguish the template from the new strand, leading to random repair and an increased mutation rate. (correct answer)
2. The overall mutation rate would decrease because rapid methylation signals more efficient repair of mismatches.
3. The MMR system would become hyperactive, repairing even correct base pairs and thereby introducing mutations.
4. Nucleotide excision repair would increase its activity to compensate for the resulting functional loss of MMR.

Explanation: When you encounter questions about DNA mismatch repair (MMR), focus on the critical timing of methylation that allows the system to identify which strand contains the error. In E. coli, the MMR system depends on GATC sequences being hemimethylated (only the template strand is methylated) immediately after replication, creating a temporary window where the system can distinguish old DNA from newly synthesized DNA. The correct answer is A because hyperactive Dam methyltransferase would eliminate this crucial distinction. Normally, there's a brief period after replication where only the template strand is methylated while the new strand remains unmethylated. This hemimethylated state tells the MMR machinery which strand is "correct" (the methylated template) and which potentially contains errors (the unmethylated new strand). If Dam methyltransferase instantly methylates the new strand, both strands become fully methylated immediately, leaving the MMR system with no way to determine which base in a mismatch is correct. Answer B incorrectly assumes faster methylation improves repair efficiency, but rapid methylation actually disrupts the targeting mechanism. Answer C misunderstands the consequence—the system wouldn't become hyperactive against correct pairs, but rather unable to function properly at all. Answer D incorrectly suggests that nucleotide excision repair would compensate, but these systems target different types of DNA damage and don't functionally substitute for each other. Remember: MMR effectiveness depends on asymmetric methylation patterns that create directional information. Without this asymmetry, the repair machinery loses its ability to make informed corrections.

Question 2

1. Nucleotide excision repair would recognize the 8-oxoG:A pair as a bulky lesion and remove a short oligonucleotide containing the A.
2. The mismatch repair system would recognize the A as incorrect and replace it with a C, but leave the 8-oxoG in place.
3. The 8-oxoG:A mismatch would cause a replication fork to collapse, leading to repair by homologous recombination.
4. A specialized DNA glycosylase (MUTYH in humans) from the BER pathway would recognize and remove the adenine that is mispaired with 8-oxoG. (correct answer)

Explanation: When you encounter questions about DNA repair mechanisms, focus on which specific repair pathway handles each type of damage and how different systems recognize their target lesions. 8-oxoguanine (8-oxoG) is a common oxidative DNA lesion that's normally repaired by base excision repair (BER). However, if it escapes initial repair and pairs with adenine during replication, the cell has a backup mechanism. The specialized DNA glycosylase MUTYH specifically recognizes 8-oxoG:A mismatches and removes the incorrectly incorporated adenine, allowing the BER pathway to complete repair by inserting the correct cytosine opposite 8-oxoG. Option A is incorrect because nucleotide excision repair (NER) targets bulky lesions that distort the DNA helix, like UV-induced thymine dimers. The 8-oxoG:A mismatch doesn't create sufficient helix distortion to trigger NER. Option B misrepresents how mismatch repair (MMR) works. While MMR does recognize base mismatches, it cannot distinguish between 8-oxoG and normal guanine, so it wouldn't specifically target this oxidative lesion. MMR would treat this as a standard G:A mismatch. Option C is wrong because 8-oxoG:A mismatches don't typically cause replication fork collapse. These lesions are relatively minor and don't block DNA polymerase progression enough to trigger homologous recombination. The key study tip: Remember that cells often have specialized backup repair mechanisms. MUTYH represents the cell's "second chance" to fix 8-oxoG lesions that escaped initial BER, demonstrating how multiple repair pathways can target the same type of damage at different stages.

Question 3

1. The mismatch will be corrected, and the original G:C base pair will be restored in all daughter cells.
2. The G-T mismatch will be excised entirely, leading to a deletion at that position in the chromosome.
3. The T on the daughter strand will be correctly excised and replaced with a C, restoring the proper G:C base pair.
4. The G on the parental strand will be excised and replaced with an A, resulting in a permanent G:C to A:T transition mutation. (correct answer)

Explanation: When you encounter questions about DNA mismatch repair (MMR), focus on understanding the strand identification mechanism and the consequences of repair errors. Normal MMR relies on Dam methylation patterns to distinguish the newly synthesized (unmethylated) daughter strand from the methylated parental strand, ensuring errors are corrected on the daughter strand. Without functional Dam methyltransferase, the MMR system loses this crucial guidance system. In this scenario, a replication error created a G-T mismatch where T was incorrectly incorporated opposite the template G. The MMR system mistakenly targets the parental strand containing G for repair instead of the daughter strand containing the erroneous T. When the G is excised from the parental strand, DNA polymerase fills the gap using the daughter strand as template. Since the daughter strand contains T, an A will be incorporated opposite it, creating an A-T base pair. After the next replication round, one daughter cell will have the original G-C pair, but the other will have A-T, establishing a permanent G:C to A:T transition mutation. Answer choice A is wrong because the error isn't corrected—it becomes fixed as a mutation. Choice B incorrectly suggests deletion rather than base substitution. Choice C describes normal MMR function where the daughter strand error is corrected, but that's not what happens when strand identification fails. Remember: MMR errors don't just fail to fix mistakes—they can actively convert temporary mismatches into permanent mutations by "correcting" the wrong strand. Dam methylation is essential for proper strand discrimination.

Question 4

1. G1 phase, because this is when the genome is scanned for any pre-existing damage prior to replication.
2. S phase, because MMR's primary function is to correct errors made by DNA polymerase during DNA synthesis. (correct answer)
3. G2 phase, because the presence of an intact sister chromatid is required to serve as a template for repair.
4. M phase, because chromosome condensation reveals structural errors in the DNA that are then targeted by MMR.

Explanation: Mismatch repair (MMR) corrects errors of replication, such as misincorporated bases and small insertions/deletions. These errors are generated during DNA synthesis, which occurs in the S phase of the cell cycle. Furthermore, the strand discrimination mechanisms that allow MMR to identify the newly synthesized (and erroneous) strand are also linked to the replication process itself (e.g., transient nicks in the lagging strand). Therefore, MMR function is temporally and mechanistically coupled to S phase.

Question 5

1. Initial recognition of the pyrimidine dimer lesion within the DNA helix.
2. Unwinding the DNA around the lesion to create a stable repair bubble.
3. Making an incision in the damaged DNA strand 3' to the lesion. (correct answer)
4. Synthesizing a new DNA strand using the undamaged strand as a template.

Explanation: The XPG protein is a structure-specific endonuclease. In the context of NER, it makes the 3' incision relative to the DNA lesion. The 5' incision is made by the XPF-ERCC1 complex. Distractor A refers to XPC or CSA/B. Distractor B refers to the helicases XPB and XPD. Distractor D refers to DNA polymerase δ/ε.

Question 6

1. An A-C mismatch occurring in the middle of an Okazaki fragment on the lagging strand.
2. A small deletion loop formed near the 5' end of a lagging strand Okazaki fragment.
3. A G-T mismatch occurring on the continuously synthesized leading strand, far from the origin of replication. (correct answer)
4. A base misincorporation that occurs immediately adjacent to the RNA primer on a newly synthesized Okazaki fragment.

Explanation: The premise is that nicks serve as the signal to identify the new strand. The lagging strand is synthesized discontinuously as Okazaki fragments, and thus is transiently rich in nicks before they are sealed by DNA ligase. The leading strand, however, is synthesized continuously. Therefore, a mismatch on the leading strand, especially one far from the origin, would lack a nearby nick to direct the MMR machinery. This would make it difficult for the system to distinguish the new strand from the template strand, leading to inefficient or incorrect repair.

Question 7

1. uracil is a bulky lesion that significantly distorts the DNA double helix, blocking replication.
2. uracil can arise from the spontaneous deamination of cytosine, and if unrepaired, it would be read as thymine during replication. (correct answer)
3. DNA polymerase frequently misincorporates dUTP from the nucleotide pool, and this represents the only source of uracil in DNA.
4. uracil's presence in a G:U pair is recognized by the mismatch repair system, which creates a double-strand break.

Explanation: The chemical instability of cytosine leads to its spontaneous deamination to form uracil. Because uracil has the same base-pairing properties as thymine (pairing with adenine), if it is not removed before replication, a C:G base pair will be converted to a T:A base pair in one of the daughter duplexes. This is a common C to T transition mutation. The uracil-DNA glycosylase specifically removes U from DNA to prevent this outcome. Uracil is not bulky and does not significantly distort the helix, making it a poor substrate for other repair systems.

Question 8

1. DNA synthesis → Excision by endonucleases → DNA unwinding by helicase → Ligation.
2. DNA unwinding by helicase → Excision by endonucleases → DNA synthesis → Ligation. (correct answer)
3. Excision by endonucleases → DNA unwinding by helicase → DNA synthesis → Ligation.
4. DNA unwinding by helicase → DNA synthesis → Excision by endonucleases → Ligation.

Explanation: The logical and experimentally determined order of events in NER after damage recognition is as follows: first, the DNA around the lesion must be opened and unwound by helicases (XPB/XPD in TFIIH). This creates a bubble that allows access for the endonucleases (XPF/XPG), which then excise the damaged oligonucleotide. Next, a DNA polymerase fills in the resulting gap. Finally, DNA ligase seals the nick to complete the repair.

Question 9

1. The DNA would still contain the U:G mispair, which would persist until the next round of replication.
2. The DNA would contain an abasic site with an intact phosphodiester backbone at that position. (correct answer)
3. The DNA would have a single-nucleotide gap with a 3'-OH and a 5'-phosphate, ready for DNA polymerase.
4. The DNA would be cleaved on both sides of the original uracil, releasing a short oligonucleotide.

Explanation: Base excision repair is a sequential process. The first step is the removal of the damaged base (uracil) by a glycosylase. This creates an abasic (AP) site but leaves the sugar-phosphate backbone intact. The second step is the cleavage of the backbone at the AP site by AP endonuclease. If AP endonuclease is non-functional, the pathway is blocked after the first step, resulting in the accumulation of AP sites with an intact backbone.

Question 10

1. Formation of thymine dimers induced by ultraviolet radiation.
2. Incorporation of uracil instead of thymine opposite adenine.
3. Slippage of DNA polymerase during replication of short tandem repeats. (correct answer)
4. Double-strand breaks caused by exposure to ionizing radiation.

Explanation: Lynch syndrome is caused by inherited mutations in mismatch repair (MMR) genes. Microsatellites are regions of repetitive DNA (e.g., ...CACACACA...). These regions are prone to errors during replication because DNA polymerase can 'slip,' creating a small loop of inserted or deleted bases on the new strand. The MMR system is responsible for recognizing and repairing these insertion/deletion loops. When MMR is defective, these errors accumulate, causing the length of microsatellites to change, a phenomenon called microsatellite instability.

Question 11

1. Both uracil and benzo[a]pyrene adducts will persist, leading to a mix of transition and transversion mutations.
2. Uracil will be efficiently removed, but the benzo[a]pyrene adducts will persist, leading primarily to replication fork stalling and chromosomal breaks. (correct answer)
3. Benzo[a]pyrene adducts will be slowly removed by base excision repair, but uracil will persist, leading predominantly to C:G to T:A transitions.
4. The mismatch repair pathway will be upregulated to compensate for the loss of NER, but it will be unable to recognize the uracil bases.

Explanation: This question requires differentiating the lesions handled by Base Excision Repair (BER) and Nucleotide Excision Repair (NER). Deamination of cytosine to uracil is repaired by BER, initiated by uracil-DNA glycosylase. This pathway is functional. Bulky adducts like those from benzo[a]pyrene are repaired by NER. With NER inhibited, these lesions will persist. Bulky lesions are potent blockers of DNA replication, leading to stalled replication forks, which can collapse into double-strand breaks.

Question 12

1. NER involves the removal of a single damaged base, whereas MMR excises a patch of several nucleotides.
2. MMR functions almost exclusively to correct errors made during replication, whereas NER primarily repairs DNA lesions caused by exogenous mutagens. (correct answer)
3. NER requires a DNA polymerase and ligase for the final steps, whereas MMR can directly correct the mismatch without DNA synthesis.
4. MMR can only occur on the lagging strand during replication, while NER can occur on either the leading or lagging strand.

Explanation: This question highlights the fundamental difference in the biological roles of the two pathways. Mismatch repair (MMR) is intrinsically linked to DNA replication, where its job is to correct misincorporated bases and small insertions/deletions, acting as a 'proofreader' for the replication machinery. Nucleotide excision repair (NER) is a general damage repair pathway that deals with lesions, often caused by environmental mutagens like UV light or chemical carcinogens, that distort the DNA helix.

Question 13

1. AP endonuclease (APE1).
2. DNA Polymerase β.
3. DNA ligase III.
4. Flap endonuclease 1 (FEN1). (correct answer)

Explanation: When you encounter questions about DNA repair pathways, focus on the key differences between short-patch and long-patch base excision repair (BER). Both pathways start similarly but diverge based on whether the 5' deoxyribose phosphate (dRP) residue can be easily removed. In short-patch BER, AP endonuclease (APE1) creates a nick 5' to the abasic site, DNA polymerase β fills in a single nucleotide while removing the dRP group with its lyase activity, and DNA ligase III seals the nick. However, when the dRP residue is chemically modified or difficult to remove, the pathway switches to long-patch BER. Long-patch BER involves DNA polymerase β (or δ/ε) adding 2-10 nucleotides, which displaces the problematic dRP-containing strand as a "flap" structure. This is where flap endonuclease 1 (FEN1) becomes essential—it specifically cleaves these displaced DNA flaps, allowing the repair to proceed. Without FEN1, the flap cannot be removed and repair stalls. Looking at the wrong answers: (A) APE1 is required in both pathways to create the initial nick. (B) DNA polymerase β participates in both pathways for gap filling. (C) DNA ligase III functions in both pathways to seal the final nick. Only (D) FEN1 is uniquely required for long-patch BER because short-patch repair doesn't generate flap structures. Remember this pattern: when DNA repair creates displaced flaps or overhangs, FEN1 is usually involved to clean them up. This principle applies beyond BER to other repair mechanisms as well.

Question 14

1. A DNA polymerase to fill the resulting single-nucleotide gap.
2. A DNA ligase to seal the final nick in the phosphodiester backbone.
3. A damage recognition protein that binds to a non-standard base.
4. ATP-dependent helicase activity to unwind the DNA around the lesion. (correct answer)

Explanation: When comparing DNA repair mechanisms, focus on the scope and complexity of the damage being repaired. Base excision repair (BER) handles single damaged bases, while nucleotide excision repair (NER) tackles bulky lesions that distort the DNA helix structure. The correct answer is D because NER requires ATP-dependent helicase activity to unwind DNA around bulky lesions like thymine dimers or chemical adducts. These large distortions create significant structural changes that must be unwound for proper access by repair enzymes. In contrast, short-patch BER deals with single base modifications (like uracil or 8-oxoguanine) that don't severely distort the helix, so helicases aren't needed. Option A is wrong because both pathways require DNA polymerase - BER needs it to fill the single-nucleotide gap after base removal, while NER needs it to fill the longer gap after excision. Option B is incorrect since both pathways require DNA ligase to seal the final nick and restore the continuous phosphodiester backbone. Option C is wrong because both pathways begin with damage recognition proteins - BER uses glycosylases that recognize specific damaged bases, while NER uses proteins that detect helix distortions. Remember this key distinction: BER is like precision surgery on individual bases, while NER is like major reconstruction around bulky damage. The structural complexity of NER lesions demands additional machinery (helicases) that simple base modifications in BER don't require. Focus on the relationship between lesion size, structural distortion, and repair complexity.

Question 15

1. It creates a nick in the phosphodiester backbone, signaling for DNA polymerase to bind at the damaged site.
2. It removes the entire damaged nucleotide, including its associated sugar and phosphate group from the DNA strand.
3. It generates an abasic (AP) site by cleaving the N-glycosidic bond between the damaged base and its deoxyribose sugar. (correct answer)
4. It directly reverses the chemical modification on the base without removing any part of the DNA strand.

Explanation: The defining role of a DNA glycosylase is to recognize a specific type of damaged or incorrect base and catalyze the hydrolysis of the N-glycosidic bond that links the base to the deoxyribose sugar of the DNA backbone. This action removes the base, leaving behind an 'abasic' or 'apurinic/apyrimidinic' (AP) site, while the sugar-phosphate backbone remains intact. The nick is created in the next step by AP endonuclease.

Question 16

1. DNA synthesis → Excision by endonucleases → DNA unwinding by helicase → Ligation.
2. DNA unwinding by helicase → Excision by endonucleases → DNA synthesis → Ligation. (correct answer)
3. Excision by endonucleases → DNA unwinding by helicase → DNA synthesis → Ligation.
4. DNA unwinding by helicase → DNA synthesis → Excision by endonucleases → Ligation.

Explanation: The logical and experimentally determined order of events in NER after damage recognition is as follows: first, the DNA around the lesion must be opened and unwound by helicases (XPB/XPD in TFIIH). This creates a bubble that allows access for the endonucleases (XPF/XPG), which then excise the damaged oligonucleotide. Next, a DNA polymerase fills in the resulting gap. Finally, DNA ligase seals the nick to complete the repair.

Question 17

1. Formation of thymine dimers induced by ultraviolet radiation.
2. Incorporation of uracil instead of thymine opposite adenine.
3. Slippage of DNA polymerase during replication of short tandem repeats. (correct answer)
4. Double-strand breaks caused by exposure to ionizing radiation.

Explanation: Lynch syndrome is caused by inherited mutations in mismatch repair (MMR) genes. Microsatellites are regions of repetitive DNA (e.g., ...CACACACA...). These regions are prone to errors during replication because DNA polymerase can 'slip,' creating a small loop of inserted or deleted bases on the new strand. The MMR system is responsible for recognizing and repairing these insertion/deletion loops. When MMR is defective, these errors accumulate, causing the length of microsatellites to change, a phenomenon called microsatellite instability.

Question 18

1. NER involves the removal of a single damaged base, whereas MMR excises a patch of several nucleotides.
2. MMR functions almost exclusively to correct errors made during replication, whereas NER primarily repairs DNA lesions caused by exogenous mutagens. (correct answer)
3. NER requires a DNA polymerase and ligase for the final steps, whereas MMR can directly correct the mismatch without DNA synthesis.
4. MMR can only occur on the lagging strand during replication, while NER can occur on either the leading or lagging strand.

Explanation: This question highlights the fundamental difference in the biological roles of the two pathways. Mismatch repair (MMR) is intrinsically linked to DNA replication, where its job is to correct misincorporated bases and small insertions/deletions, acting as a 'proofreader' for the replication machinery. Nucleotide excision repair (NER) is a general damage repair pathway that deals with lesions, often caused by environmental mutagens like UV light or chemical carcinogens, that distort the DNA helix.

Question 19

1. presence of a stalled DNA polymerase at the site of a bulky DNA lesion.
2. direct recognition of a helix-distorting lesion by the XPC-RAD23B protein complex.
3. physical blockage and stalling of an elongating RNA polymerase II complex on the template strand. (correct answer)
4. detection of hemimethylation patterns near the promoter of the actively transcribed gene.

Explanation: The defining feature of TC-NER is its link to transcription. When a transcribing RNA polymerase encounters a bulky lesion on the template DNA strand, it stalls. This stalled complex acts as a damage signal, recruiting TC-NER specific factors (like CSA and CSB) to the site, which then assemble the rest of the NER machinery. In contrast, GG-NER is initiated by the XPC complex recognizing helix distortions anywhere in the genome, independent of transcription.

Question 20

1. The DNA would still contain the U:G mispair, which would persist until the next round of replication.
2. The DNA would contain an abasic site with an intact phosphodiester backbone at that position. (correct answer)
3. The DNA would have a single-nucleotide gap with a 3'-OH and a 5'-phosphate, ready for DNA polymerase.
4. The DNA would be cleaved on both sides of the original uracil, releasing a short oligonucleotide.

Explanation: Base excision repair is a sequential process. The first step is the removal of the damaged base (uracil) by a glycosylase. This creates an abasic (AP) site but leaves the sugar-phosphate backbone intact. The second step is the cleavage of the backbone at the AP site by AP endonuclease. If AP endonuclease is non-functional, the pathway is blocked after the first step, resulting in the accumulation of AP sites with an intact backbone.