DNA replication, repair and recombination MCQ Quiz in தமிழ் - Objective Question with Answer for DNA replication, repair and recombination - இலவச PDF ஐப் பதிவிறக்கவும்

The correct answer is Histone modification is not required for mitotic chromosome condensation; it is mainly required for epigenetic control of gene repression in interphase.

Concept:

• DNA packaging in chromosomes is a highly complex process that ensures that the entire genome is compactly packed within the nucleus while still allowing accessibility for transcription, replication, and repair.
• Histones, condensins, and other chromatin-associated proteins play critical roles in this process by facilitating the structural organization of chromatin.
• Histone modifications and additional factors like histone H1 and condensins are required for different levels of chromatin organization, particularly during mitosis when chromosomes condense for segregation.

Explanation:

Histone modification is not required for mitotic chromosome condensation; it is mainly required for epigenetic control of gene repression in interphase:

• This statement is incorrect because histone modifications are crucial for mitotic chromosome condensation as well as epigenetic regulation.
• During mitosis, histone modifications such as phosphorylation of histone H3 at serine 10 (H3S10ph) are essential for chromatin compaction and segregation.
• These modifications facilitate the recruitment of condensins and other chromatin remodeling factors required for higher-order chromatin organization.
• While histone modifications also play a role in epigenetic regulation during interphase (e.g., methylation and acetylation for gene repression or activation), it is incorrect to say they are not required for mitotic chromosome condensation.

Other Options:

• Condensin I creates loops of nucleosomal chromatin for packaging in mitosis:
  • This statement is correct. Condensin I is a protein complex critical for mitotic chromosome condensation. It works by creating and stabilizing loops of nucleosomal chromatin, facilitating the compaction and organization of chromosomes during mitosis.
• Histone H1 is required for higher order packaging of mammalian chromosomes:
  • This statement is correct. Histone H1, also known as the linker histone, is responsible for stabilizing the 30 nm chromatin fiber structure, a higher-order organization of chromatin.
  • Histone H1 binds to the DNA between nucleosomes (linker DNA) and helps in the compaction of chromatin into more condensed forms required for mitosis.
• Histones form hydrogen bonds with the sugar-phosphate backbone of DNA:
  • This statement is correct. Histones interact with DNA through hydrogen bonds, primarily with the sugar-phosphate backbone.

The correct answer is A (iv), B (ii), C (iii), D (i)

Explanation:

A. DnaB (iv. MCM complex):

• DnaB is the helicase in E. coli, responsible for unwinding the DNA helix at the replication fork.
• Its functional orthologue in eukaryotes is the MCM (Mini-Chromosome Maintenance) complex, which performs the same role of helicase activity during replication initiation and elongation.

B. DnaC (ii. Cdc6):

• DnaC is a loader protein in E. coli that helps load the DnaB helicase onto the DNA strand.
• Its eukaryotic orthologue is Cdc6, which plays a similar role in loading the MCM complex onto the origin of replication during the initiation of replication.

C. β clamp (iii. PCNA):

• The β clamp in E. coli is a sliding clamp that enhances the processivity of DNA polymerase by holding it onto the DNA strand.
• Its eukaryotic counterpart is PCNA (Proliferating Cell Nuclear Antigen), which performs the same function during replication.

D. DnaG (i. Polα/primase):

• DnaG is the primase in E. coli, responsible for synthesizing short RNA primers required for DNA polymerase to begin synthesis.
• Its functional orthologue in eukaryotes is the Polα/primase complex, which also synthesizes RNA primers during replication initiation.

The correct answer is Type A > Type B

Concept:

• The MMR system is a crucial DNA repair mechanism that corrects errors introduced during DNA replication, such as base mismatches and small insertion-deletion loops. It discriminates between the newly synthesized strand and the parental strand using the methylation status of GATC sequences. The parental strand is methylated, while the newly synthesized strand is temporarily unmethylated, allowing the repair system to identify and correct errors in the daughter strand.
• Methylation at GATC sequences is a critical marker for the MMR system. If the methylation pattern is disrupted, the system cannot correctly identify the newly synthesized strand, leading to an accumulation of mutations.
• There are two types of mutants based on methylation and its impact on the MMR system:
  • Type A (Hypermethylation Mutant): In this mutant, GATC sequences are methylated immediately after DNA synthesis, leaving no window for the MMR system to distinguish between the parental and daughter strands.
  • Type B (Non-methylation Mutant): In this mutant, GATC sequences are never methylated, leaving both strands indistinguishable in terms of methylation status.

Explanation:

Why Type A has a greater impact on MMR:

• In Type A mutants, the immediate methylation of GATC sequences eliminates the time window during which the MMR system can recognize and repair mismatches based on methylation differences. As a result, the system fails to correct replication errors, leading to a significant accumulation of spontaneous mutations.
• In the absence of a functional MMR system, these errors persist and propagate, increasing the mutation rate in the genome.

Why Type B has a lesser impact:

• In Type B mutants, GATC sequences are never methylated, which also impairs the ability of the MMR system to distinguish between the parental and daughter strands. However, unlike Type A, there may be alternative repair mechanisms or compensatory pathways that could partially mitigate the effects, leading to a slightly reduced impact compared to Type A mutants.

Conclusion:

• The correct answer is Type A > Type B because the hypermethylation in Type A mutants completely disrupts the function of the MMR system, leading to a higher accumulation of spontaneous mutations compared to Type B mutants.

The correct answer is B only 

Concept:

• Topoisomerase I: Relieves supercoiling by creating transient single-strand breaks in the DNA, which changes the linking number by ±1.
• Topoisomerase II: Relieves supercoiling by creating transient double-strand breaks in the DNA, which changes the linking number by ±2.

Inhibitors:

• Novobiocin: Specifically inhibits Topoisomerase II by blocking the ATPase activity required for its function, thus preventing it from creating double-strand breaks and passing one segment of DNA through another.
• ICRF-193: Another specific inhibitor of Topoisomerase II, which stabilizes the closed-clamp form of the enzyme and prevents strand passage and decatenation activities, thereby inhibiting its function.

Explanation:

• Novobiocin is an inhibitor of Topoisomerase II activity, which means it prevents Topoisomerase II from altering the linking number of DNA.
• When Novobiocin inhibits Topoisomerase II, the enzyme cannot perform its double-strand break-and-reseal function. Thus, there would be no change in the linking number due to Topoisomerase II activity in the presence of Novobiocin.

Option A: In the presence of Novobiocin, Topoisomerase I will lead to change in the linking number by ±2.

• Incorrect. Topoisomerase I changes the linking number by ±1, not ±2.

Option B: In the presence of ICRF-193, Topoisomerase I will lead to change in the linking number by ±1.

• Correct. ICRF-193 specifically inhibits Topoisomerase II, not Topoisomerase I, so it wouldn’t affect Topoisomerase I activity. However, Topoisomerase I changes the linking number by ±1 regardless.

Option C: In the presence of Novobiocin, Topoisomerase II will lead to change in the linking number by ±2.

• Incorrect. Since Novobiocin inhibits Topoisomerase II, it would prevent any change in the linking number by Topoisomerase II. Thus, no change in linking number would occur due to Topoisomerase II activity when Novobiocin is present.

Option D: In the presence of ICRF-193, Topoisomerase II will lead to change in the linking number by ±1.

• Incorrect. Topoisomerase II changes the linking number by ±2 under normal conditions. However, since ICRF-193 inhibits Topoisomerase II, it would prevent any change in the linking number from Topoisomerase II.

The correct answer is A, B and C only.

Explanation:

A double-stranded DNA break (DSB) is one of the most deleterious forms of DNA damage. If not repaired properly, DSBs could lead to mutations, deletions, translocations, and amplifications in the genome.

A. Repair occurs primarily through the use of a sister chromatid as a template.

• This statement is true. During homologous recombination, especially in the S and G2 phases of the cell cycle when a sister chromatid is available, the cell often uses the sister chromatid as a template for repair. This is because the sister chromatid is an identical copy, ensuring high-fidelity repair.

B. Non-homologous end joining (NHEJ) is the dominant pathway for repairing DSBs in G1 phase.

• This statement is also true. NHEJ is the predominant DSB repair mechanism in the G1 phase when no sister chromatid is available. NHEJ directly ligates the broken ends of DNA without needing a homologous template, making it the primary pathway of repair in the absence of a sister chromatid.

C. The presence of RAD51 protein is essential for strand invasion during repair.

• This statement is true. RAD51 plays a crucial role in the homologous recombination repair process by facilitating strand invasion, where a single strand of DNA searches for homology with the sister chromatid.

D. Homologous recombination leads to the formation of chimeric chromosomes.

• This statement is incorrect. Homologous recombination typically leads to precise repair using a homologous sequence as a template, thereby maintaining genomic integrity. Chimeric chromosomes (consisting of mixed genetic material from different sources) are more characteristic of inappropriate recombination events or unequal crossovers between non-identical sequences, but not a regular outcome of typical homologous recombination processes.

The correct answer is It inhibits the initiation of replication.

Explanation:

To understand the mode of action of chemical D, let's analyze the experimental data:

Analysis:

1. 0 minute: At the beginning, there is 0% hybrid DNA, as the culture originally grown in ¹⁴N medium is shifted to ¹⁵N medium containing the chemical D.
2. 15 minutes to 60 minutes: The proportion of hybrid DNA remains constant at 50% throughout these time points. This plateau at 50% hybrid DNA suggests that no new rounds of replication are initiated after the initial shift.

From this analysis, it is clear that the presence of chemical D does not prevent the elongation or the termination of already initiated replication because if it did, we would expect to see changes (possibly an increase or decrease) in the hybrid DNA proportion over the experimental period. Instead, the data indicate that DNA replication is halted entirely right after the medium shift, implying that no new replication forks are initiated.

The most likely mode of action is that chemical D inhibits the initiation of replication. This prevents any new DNA strands from starting to form, resulting in a constant 50% hybrid DNA because existing replication forks complete their rounds but no new replication forks are formed.

Other possibilities:

• It inhibits the elongation phase of replication: If this were true, we would observe a buildup of incomplete DNA strands and the hybrid DNA percentage would likely change over time.
• It inhibits the termination of replication: This would likely result in stalled replication forks and an increasing proportion of incomplete hybrid DNA, which is not observed.
• It competes with dNTPs for incorporation into newly synthesized DNA: This would likely cause errors or stalls during elongation, changing the proportion of hybrid DNA over time.

Therefore, the consistent 50% hybrid DNA observed suggests the correct answer is inhibition at the initiation stage of replication.

The correct answer is A, B and C

Explanation:

A. The genome of multicellular animals contains many potential origins of replication.

• This statement is correct. Eukaryotic genomes have multiple origins of replication to ensure that the entire genome is replicated efficiently during the S phase of the cell cycle.

B. During early development, when embryos are undergoing rapid cell divisions, origin sites are uniformly activated.

• In rapidly dividing embryonic cells, replication origins are uniformly activated to ensure the complete and rapid duplication of the genome. This uniform activation is essential because there is limited time for genome replication during early embryonic development.

C. "Pulse-chase" technique is used to label sites of DNA replication.

• This statement is correct. The pulse-chase technique involves briefly exposing cells to a labeled nucleotide (pulse) followed by an excess of unlabeled nucleotide (chase). This technique can be used to study the dynamics of DNA replication and label replication sites.

D. The rate of elongation of different DNA chains during genome replication varies drastically.

• False: The rate of elongation during replication is generally uniform for a given organism and cell type. Variations in replication rates across the genome are minimal under normal conditions, as the replication machinery operates efficiently to maintain fidelity and speed.

The correct answer is MCM helicases get activated in the S phase of the cell cycle.

Explanation:

In eukaryotic cells, DNA replication is tightly regulated and restricted to the S phase of the cell cycle to ensure that the genome is replicated only once per cycle.

• MCM (Mini-Chromosome Maintenance) helicases are loaded onto DNA at the origins of replication during the G1 phase of the cell cycle, but they remain inactive.
• The activation of MCM helicases, which are responsible for unwinding the DNA to initiate replication, occurs only in the S phase.
• This activation is triggered by specific kinases (such as CDKs and DDKs) that phosphorylate the MCM complex, allowing replication to begin.
• This regulation ensures that replication occurs only during the S phase and is tightly controlled to prevent re-replication of the genome.

Other Options:

• DNA polymerase is present only in the S phase of the cell cycle: DNA polymerase is always present, but it becomes active in the S phase due to the activation of the replication machinery.
• Origin recognition complex (ORC) recognizes origin only in the S phase: The ORC binds to replication origins in the G1 phase, not just in the S phase. It remains bound throughout the cell cycle but does not trigger replication until the S phase.
• MCM helicases get activated in the G1 phase of the cell cycle: MCM helicases are loaded in the G1 phase but are only activated in the S phase, making this statement incorrect.

The correct answer is base excision repair

Explanation:

DNA glycosylases are enzymes that play a crucial role in the DNA repair mechanism. These enzymes initiate the process of base excision repair (BER) by recognizing and removing damaged or inappropriate bases from DNA. This is the first step in correcting small base lesions that do not significantly distort the DNA helix structure.

• DNA replication involves synthesizing a new strand of DNA based on the template strand. While DNA repair mechanisms are vital for maintaining the integrity of the DNA that is replicated, DNA glycosylases specifically are not enzymes that directly participate in DNA replication.
• Negative supercoiling of DNA refers to the underwinding of the DNA helix, a process primarily managed by topoisomerases. This condition facilitates processes like replication and transcription by making the DNA more accessible, but it is not directly related to the function of DNA glycosylases.
• The SOS response is a bacterial DNA repair system activated by extensive DNA damage. It involves a coordinated cellular response to damage that includes error-prone repair mechanisms to save the cell. Although some DNA repair enzymes are upregulated during the SOS response, DNA glycosylases primarily function in the base excision repair pathway, which operates both independently of and in concert with SOS in different contexts.
• Base excision repair (BER) is the correct  for the function of DNA glycosylases. In BER, these enzymes first identify and remove damaged or inappropriate bases, creating an abasic site. Subsequent enzymes then cut the DNA backbone at this site, remove the abasic sugar, and fill in the gap with the correct base(s), ultimately restoring the DNA to its undamaged state.

Conclusion:

Therefore, DNA glycosylases are DNA repair enzymes involved in Base excision repair

The correct answer is Option 2

Explanation:

• PCR (Polymerase Chain Reaction) is a technique used to amplify specific segments of DNA. It consists of three main steps: denaturation, annealing, and extension.
• During the denaturation step, the reaction mixture is heated to a high temperature (usually around 94-98°C) to denature the DNA, causing the double-stranded DNA to separate into two single strands.
• This high-temperature denaturation step effectively prevents the formation of supercoils because it separates the DNA strands.
• Supercoiling typically occurs in double-stranded DNA due to the unwinding of the helix; however, since the DNA strands are separated during PCR, the supercoiling issue that might necessitate topoisomerase in other contexts does not arise here.

The other options do not accurately describe why topoisomerase is not needed in PCR:

• Taq polymerase does not have innate topoisomerase activity. Its primary functions are to withstand high temperatures used in PCR and to synthesize new DNA strands by adding nucleotides to a DNA template.
• The reaction buffer in PCR is not intended to denature DNA through its pH. Instead, its role is to maintain an optimal pH for the activity of the Taq polymerase and other components of the reaction. DNA denaturation is achieved through the application of heat.
• The 5´→ 3´ exonuclease activity of Taq polymerase relates to its ability to remove nucleotides from the ends of DNA strands and is not directly related to the prevention of supercoiling during PCR.