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 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.

The correct answer is Option 3 i.e. The RNA primer is removed by the combined action of RNase H, which degrades RNA in RNA-DNA hybrids, and FEN1 (Flap Endonuclease 1), which removes remaining RNA nucleotides or DNA flaps created during the process. DNA polymerase δ then fills the gap with DNA nucleotides, and DNA ligase I seals the nick.

Concept:

In eukaryotic DNA replication, RNA primers are necessary for initiating the synthesis of DNA strands, as DNA polymerases can only add nucleotides to an existing strand of DNA or RNA. Once an RNA primer has been utilized to initiate DNA synthesis, the replication machinery must remove it and replace it with DNA to ensure the newly synthesized DNA strand is complete and continuous. This process involves several key steps and enzymes:

• RNase H Activity: RNase H is an enzyme that recognizes RNA-DNA hybrids, which are structures formed by the RNA primer annealed to the newly synthesized DNA strand. RNase H selectively degrades the RNA component of these hybrids, removing most of the RNA primer but often leaving behind a few RNA nucleotides at the 5' end of the Okazaki fragment in the lagging strand.
• FEN1 (Flap Endonuclease 1): After RNase H action, the remaining RNA nucleotides (or the RNA-DNA flap that may result if DNA displacement synthesis occurs beyond the end of the RNA primer) need to be removed to prepare for gap filling. FEN1 precisely trims these remnants, ensuring that only DNA nucleotides are present at the primer site. FEN1's action is vital for maintaining genome stability by ensuring that the process does not generate mutations or single-stranded DNA breaks.
• DNA Polymerase δ Filling the Gap: Once the RNA primer is completely removed, the gap that remains must be filled in with DNA. DNA polymerase δ is responsible for extending the existing 3' end of the DNA near the gap, synthesizing DNA in a 5' to 3' direction. It adds nucleotides complementary to the template strand until it reaches the site where the RNA primer was located, thus filling the gap left by primer removal.
• DNA Ligase I Sealing the Nick: After DNA polymerase δ fills the gap with DNA nucleotides, there is still a discontinuity in the sugar-phosphate backbone, commonly referred to as a nick. DNA ligase I is an enzyme that catalyzes the final step of the repair process by forming a phosphodiester bond between the adjacent 3' OH and 5' phosphate groups of the DNA. This action effectively "seals" the nick, rendering the DNA strand continuous and intact.

This multistep process of primer removal and gap filling is critical for the fidelity of DNA replication. It ensures that the newly synthesized DNA is a precise copy of the template strand, which is essential for maintaining genetic stability and preventing mutations that could lead to various genetic disorders and diseases.

Explanation:

"DNA ligase directly recognizes and removes the RNA primers, subsequently filling the gaps with DNA nucleotides without the need for any DNA polymerase."

• DNA ligase does not recognize nor remove RNA primers. Its role is to join DNA strands together by forming a phosphodiester bond between the 3' hydroxyl end of one nucleotide and the 5' phosphate end of another. It does not have enzymatic activity to remove RNA primers or synthesize DNA, making this option incorrect. Primer removal and DNA synthesis are tasks performed by different enzymes before DNA ligase acts.

"RNA primers are removed by DNA polymerase α, which possesses both polymerase and 5' to 3' exonuclease activities, enabling it to replace the RNA with DNA nucleotides directly."

• While DNA polymerase α does initiate DNA synthesis by adding a short stretch of DNA nucleotides to the RNA primer on the lagging strand, it does not have exonuclease activity to remove primers. DNA polymerase α's main role is to synthesize the RNA-DNA hybrid primer used for DNA replication initiation. The removal of RNA primers and the subsequent replacement with DNA nucleotides are carried out by the functionalities of RNase H, FEN1, and DNA polymerase δ (not α).

"Telomerase enzyme recognizes and binds to the RNA primers at the ends of eukaryotic chromosomes, removing them and filling the resultant gaps using its RNA component as a template for DNA synthesis."

• Telomerase has a very specialized function unrelated to the routine removal of RNA primers during DNA replication. Its primary role is to extend telomeres, the repetitive nucleotide sequences at the ends of linear chromosomes, to prevent their shortening during cell division.
• Telomerase uses its intrinsic RNA as a template for adding telomeric DNA sequence repeats to the ends of chromosomes and does not participate in the removal of RNA primers within the internal regions of the chromosome being replicated. Thus, this option incorrectly assigns a role to telomerase that it does not perform.

Conclusion:

Therefore, the correct answer is Option 3

The correct answer is Option 3 i.e. Pyrimidine dimers

Explanation:

• Nucleotide excision repair is a mechanism that cells use to repair bulky DNA lesions, such as pyrimidine dimers, which result from ultraviolet (UV) light exposure. This process involves the recognition and removal of a short single-stranded DNA segment containing the damage, followed by DNA synthesis using the undamaged strand as a template and ligation to restore the integrity of the DNA.
• Pyrimidine dimers are a form of DNA damage resulting primarily from exposure to ultraviolet (UV) light. They are characterized by the covalent bonding of two adjacent pyrimidine bases within a DNA strand. Pyrimidines are one of the two classes of bases found in DNA, with cytosine (C) and thymine (T) being the pyrimidines. When DNA is exposed to UV light, particularly UV-B (280–315 nm), it can excite the electrons in the pyrimidine bases, causing them to form these abnormal covalent bonds.

Key PointsPyrimidine Dimers
Pyrimidine dimers are formed primarily as a result of exposure to ultraviolet (UV) light. This exposure leads to the covalent bonding of two adjacent pyrimidine bases (cytosine or thymine) on the same DNA strand. The most common types of these dimers are thymine dimers. These dimers create a bulge or distortion in the DNA double helix that can interfere with DNA replication and transcription.

Nucleotide Excision Repair Process
The NER process involves several steps and can be generally divided into two pathways: global genome NER (GG-NER) and transcription-coupled NER (TC-NER). Despite the differences in how they are initiated, both pathways follow a similar series of steps to repair DNA damage:

• Damage Detection: In GG-NER, damage recognition proteins scan the DNA for distortions typical of bulky lesions. In TC-NER, the process is initiated when RNA polymerase stalls at a lesion during transcription.
• DNA Unwinding: After damage is detected, several proteins, including helicases, are recruited to the site to unwind the DNA helix around the lesion, providing access to the damaged site.
• Damage Verification: Additional factors are recruited to verify the presence of damage. This step ensures that the repair machinery is precisely targeted to actual DNA lesions.
• Excision: A multiprotein complex excises a short, single-stranded DNA segment around the lesion, typically removing 24-32 nucleotides in humans, including the damaged bases.
• DNA Synthesis: The gap left by the excised segment is filled in by DNA polymerases, which synthesize new DNA using the undamaged strand as a template.
• Ligation: Finally, the remaining nick in the DNA backbone is sealed by DNA ligase, restoring the integrity of the DNA molecule.

NER is essential for preventing mutations that can lead to various pathological conditions, including skin cancers like xeroderma pigmentosum, a condition characterized by extreme sensitivity to UV light and a high predisposition to skin cancers, resulting from mutations in genes involved in the NER pathway.

Other DNA Damage:-
Other DNA repair mechanisms are responsible for addressing different types of DNA damage:-

• DNA Mismatch Repair (MMR) fixes base mismatches and insertion-deletion loops.
• Homologous Recombination (HR) and Non-Homologous End Joining (NHEJ) are two major pathways for repairing double-strand breaks.
• Base Excision Repair (BER) deals with small, non-helix-distorting base lesions.

Each of these mechanisms ensures DNA's stability and integrity, critical for maintaining cellular function and preventing diseases.

The correct answer is Mut S

Explanation:

• Mut S is a key protein involved in the mismatch repair (MMR) pathway of DNA. It recognizes and binds to mismatched base pairs (errors that occur during DNA replication) and initiates the repair process. A mutation in Mut S would inhibit the recognition of these mismatches, leading to an increase in the mutation rate.

Other Options:

• Mut H: This protein is also part of the MMR system but has a different role. It is involved in the strand discrimination step, identifying the newly synthesized strand that contains the error. Although important, a mutation in Mut H affects a later step in the repair process after mismatch recognition.
• Uvr D: UvrD is a helicase involved in nucleotide excision repair (NER), a different DNA repair pathway that removes bulky lesions like thymine dimers rather than mismatched base pairs. While critical for maintaining DNA integrity, mutations in UvrD affect NER, not the mismatch recognition in MMR.
• Mut L: Similar to Mut H, Mut L is another protein in the mismatch repair pathway. It works in conjunction with Mut S to initiate repair after a mismatch is recognized. Mut L's role is crucial in the recruitment of additional factors needed for repair, but it is not directly involved in the recognition of mismatched base pairs.

Summary:
Mut S's primary function is directly associated with the recognition of mismatched base pairs during DNA repair. A mutation in this protein would directly inhibit this initial and crucial step of the mismatch repair pathway, leading to increased mutation rates and potentially contributing to genetic diseases and cancer development. Other proteins mentioned have roles in either later steps of the MMR pathway or are part of different DNA repair mechanisms, making them less directly involved in the initial recognition of mismatched base pairs.