DNA replication, repair and recombination MCQ Quiz - Objective Question with Answer for DNA replication, repair and recombination - Download Free PDF

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 Glycosylase, AP endonuclease 1, DNA Polδ/ε, Flap endonuclease 1, DNA Ligase 1

Explanation:

• Base excision repair (BER) is a critical DNA repair pathway that corrects small base lesions caused by oxidation, alkylation, deamination, or spontaneous hydrolysis.
• The BER pathway is divided into two sub-pathways:
  • Short-patch BER: Repairs a single nucleotide.
  • Long-patch BER: Repairs a stretch of 2-10 nucleotides.
• Long-patch BER involves the removal of a damaged base followed by the synthesis of a new DNA strand and ligation to restore the DNA helix.
• The correct order of enzyme actions during long-patch BER is essential to ensure accurate repair and maintenance of genomic stability.
  • Step 1: Glycosylase: This enzyme recognizes and removes the damaged or modified base by cleaving the N-glycosidic bond, leaving behind an abasic site (AP site).
  • Step 2: AP Endonuclease 1: AP endonuclease 1 cleaves the phosphodiester bond at the AP site, creating a nick in the DNA backbone adjacent to the site.
  • Step 3: DNA Polymerase δ/ε: DNA polymerase δ or ε begins synthesizing a new DNA strand, displacing the damaged strand and creating a flap structure.
  • Step 4: Flap Endonuclease 1: Flap endonuclease 1 (FEN1) removes the displaced DNA flap, ensuring the newly synthesized strand is properly aligned.
  • Step 5: DNA Ligase 1: DNA ligase 1 seals the remaining nick in the DNA backbone, completing the repair process and restoring the DNA structure.

Fig: A simplified scheme illustrates the Short Patch and Long Patch base excision repair (BER) pathways in eukaryotes.

The correct answer is option 2.

In News

• The Sample Registration System (SRS) 2021 Report released by the Registrar-General of India provided key demographic insights on fertility rates and the aging population across Indian States.

Key Points

• Statement 1: The Total Fertility Rate (TFR) is defined as the average number of children born to a woman over her reproductive lifespan. Hence, Statement 1 is correct.
• Statement 2: The SRS 2021 Report states that India’s TFR stands at 2.0, which is below the replacement level fertility rate of 2.1. Hence, Statement 2 is correct.
• Statement 3: Kerala (14.4%), Tamil Nadu (12.9%), and Himachal Pradesh (12.3%) are reported to have the highest percentage of population aged 60 and above. Hence, Statement 3 is correct.
• Statement 4: Bihar (6.9%), Assam (7%), and Delhi (7.1%) are reported to have the lowest, not the highest, percentage of elderly population. Hence, Statement 4 is incorrect.

Additional Information

• Bihar has the highest TFR at 3.0, while Delhi and West Bengal have the lowest at 1.4.
• The report notes a demographic shift with a rising working-age population (15-59 years).
• The mean age at marriage for females has increased from 19.3 years in 1990 to 22.5 years in 2021.
• A high-powered committee was announced in the Union Budget 2024 to address demographic challenges.

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 Option 3 i.e.Cdc45

Concept:

• DNA replication in eukaryotes could be divided into three major parts:
  • Initiation
  • Elongation
  • Termination.
• DNA replication initiation could be divided into:
  •  pre-replicative complex
  •  initiation complex.
• Pre-replicative complex majorly consists of
  • ORC (origin recognition complex)+ Cdc6 + Cdt + MCM complex (mini-chromosome maintenance complex)
• Initiation complex consists of 
  • Cdc45 + MCM 10 + GINS + DDK and CDK kinase + Dpb11, Sld3, Sld2 protein complex.

Explanation:

• All the proteins given in the options belong to eukaryotic cells and so we must consider only eukaryotic DNA replication here.

Option 1: ORC - INCORRECT

• DNA replication is initiated from the origin of replication, having specific sequences to initiate replication.
• The ORC is a hexameric DNA binding complex that binds with the origin of replication followed by the recruitment of Cdc6 protein followed by Cdt1.
• ORC dephosphorylates and becomes inactivated before the elongation process.
• Hence, this option is incorrect. 

Option 2: Geminin - INCORRECT

• It binds to cdt1 to prevent the re-initiation of DNA replication and hence it works as a regulator/inhibitor rather than an initiator of DNA replication.
• It is an inhibitor of Cdt1.

Option 3: Cdc45 - CORRECT

• Cdc refers to the cell division control proteins that are involved in various steps of DNA replication process.
• Cdc45 remains with MCM complex and GINS to work as a helicase.
• Thus, it helps in the initiation of DNA replication as well as advancement of the replication fork.

​Option 4: Cdc6 - INCORRECT

• It helps in the assembling of the pre-replicative complexes and interacts with the ORC.
• Cdc6 degrades before initiation of the replication fork.
• The concentration of both cdc6 and cdt1 declines before DNA elongation starts.

​Hence, the correct option is option 3.

The correct order in which the proteins are recruited during DNA double-strand break (DSB) repair in prokaryotes is RecBCD, Ssb, RecA, DNA Pol III, DNA Ligase.

Explanation:

The process of repairing double-strand breaks in prokaryotes, particularly in E. coli, involves several key steps and specific proteins:

1. RecBCD Complex:

   • The first step in DSB repair involves the RecBCD complex, which recognizes and binds to the double-strand break. This complex unwinds the DNA and processes the ends to create a single-stranded DNA overhang necessary for the next steps in repair.

2. Ssb (Single-Strand Binding Protein):

   • After RecBCD processes the DNA, Ssb proteins bind to the single-stranded DNA (ssDNA) to protect it from degradation and to prevent the formation of secondary structures.

3. RecA:

   • Once the ssDNA is stabilized by Ssb, RecA is recruited to the ssDNA. RecA plays a critical role in homologous recombination, facilitating the search for homology and the strand invasion process necessary for repair.

4. DNA Pol III:

   • After the strand invasion, DNA Polymerase III (DNA Pol III) synthesizes new DNA to fill in the gaps left after the repair process.

5. DNA Ligase:

   • Finally, DNA Ligase seals the nicks in the DNA backbone, completing the repair process.

​

Thus, the correct order of recruitment is:

1. RecBCD
2. Ssb
3. RecA
4. DNA Pol III
5. DNA Ligase

Therefore, the correct answer is  RecBCD, Ssb, RecA, DNA Pol III, DNA Ligase.

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 Option 4 i.e. B and C

Concept:

• DNA polymerases are enzymes that catalyze the synthesis of DNA in a template-dependent manner. 

DNA polymerases in prokaryotes:

1. DNA polymerase I - It is mainly involved in repair. It is involved in the excision with 3'-5' and 5'-3' exonuclease activity and processing of the Okazaki fragment.
2. DNA polymerase II - it has a 3'-5' exonuclease that participates in DNA repair. It is also considered to be the backup for pol III as it can interact with holoenzyme protein complex and have a high level of processivity
3. DNA polymerase III - It is the primary enzyme that is involved i the replication in E.coli. 
4. DNA polymerase IV - It is an error-prone DNA polymerase that is involved in non-targeted mutagenesis. In SOS induction Pol IV production is increased to 10 fold.
5. DNA polymerase V - It is the Y family of DNA polymerase that is involved in SOS response and translation synthesis DA repair mechanisms. 

Explanation:

Statement A: INCORRECT

• 5' to 3' exonuclease removes nucleotides at the 5' end, it removed ribonucleotides of the primers.
• If this activity is compromised then only RNA primer will not be removed. 
• Hence, this is an incorrect statement. 

Statement B: CORRECT

• 3' to 5' exonuclease activity of the DNA polymerase plays a central role in genetic stability. If this activity is compromised then high errors will be introduced in the DNA and the cell will be under stress response. 
• Hence, this is a correct statement.

Statement C: INCORRECT

• If polymerase frequently dissociated them it most likely results in the formation of short fragments of the DNA. 
• Hence, this is a correct statement. 

Statement D: INCORRECT

• DNA helicases is responsible for unwinding of the DNA during DNA replication.
• Hence, this is a correct statement. 

Hence, the correct answer is option 4.

The correct answer is Tube 1 will have more T3 DNA and tube 2 will have more T7 DNA.

Explanation:

DNA polymerase PolA has:

• High fidelity (accuracy in adding nucleotides).
• Low processivity (the enzyme falls off the template frequently, leading to less DNA synthesis).

DNA polymerase PolB has:

• Low fidelity (less accurate in adding nucleotides).
• High processivity (the enzyme stays bound to the template longer and synthesizes more DNA).

The experiment involves:

• Tube 1: PolA synthesizing T7 DNA for 20 minutes, followed by T3 DNA synthesis for 40 minutes.
• Tube 2: PolB synthesizing T7 DNA for 20 minutes, followed by T3 DNA synthesis for 40 minutes.

Factors to Consider:

• Processivity: Higher processivity means the enzyme can synthesize more DNA in a given time.
• Fidelity: High fidelity ensures accurate DNA replication, but the question is focused on the quantity of DNA synthesized, so fidelity is less relevant here.

In each tube:

• T7 DNA (X μM) is provided for 20 minutes.
• T3 DNA (10X μM) is provided for 40 minutes.

Tube 1 (PolA):

• PolA has low processivity, meaning it will not synthesize much DNA in the 20-minute window for T7 DNA.
• T3 DNA is provided for 40 minutes, and since PolA continues to function, more T3 DNA will be synthesized (despite the enzyme's low processivity) because of the longer time frame and larger amount of DNA substrate available.

Tube 2 (PolB):

• PolB has high processivity, so it will synthesize more T7 DNA in the 20 minutes compared to PolA.
• However, due to its high processivity, it will continue synthesizing a large amount of T3 DNA as well, but the amount of T7 DNA synthesized will likely still be greater than in Tube 1.

Conclusion:

Therefore, the correct answer is Tube 1 will have more T3 DNA, and Tube 2 will have more T7 DNA.

The correct answer is Option 3 i.e. C and D only

Concept:-

• Replication occurs in three major steps: the opening of the double helix and separation of the DNA strands, the priming of the template strand, and the assembly of the new DNA segment.

Explanation:

Statement A: INCORRECT

• Camptothecin selectively poisons topoisomerase I by trapping topoisomerase I cleavage complexes, which correspond to enzyme-linked DNA breaks. 
• Camptothecin induces replication-dependent DNA lesions, and arrests cells in the S and G2 phase of the cell cycle.

• By creating DNA double-strand breaks (DSBs) localized in replicating DNA, camptothecin promotes S-phase cytotoxicity.

Statement B:- INCORRECT

• The dynamic control of DNA replication initiation is mediated by a two-step mechanism:
  • formation of the pre-replicative complex (pre-RC) comprised of ORC1–6 complex, Cdc6, Cdt1 and MCM2–7 complex in late mitosis and early G1.
  • activation of MCM2–7 complexes to initiate origin firing and DNA replication during S phase.
• Prevention of reinitiation of DNA replication during the same cell cycle is mediated by regulating the loading of the core replicative helicase Mcm2-7 and not ORC. 
• Therefore, statement B is also incorrect. 

Statement C:- CORRECT

• In DNA polymerase, the nucleotide-binding pocket cannot accommodate a 2' -OH on the incoming nucleotide.
• This space is occupied by two amino acids that make van der Waals contact with the sugar ring.
• Changing these amino acids to other amino acids with smaller side chains (e.g., by changing glutamate to an alanine) results in a DNA polymerase with significantly reduced discrimination between dNTPs and rNTPs.
• Therefore, statement C is correct.

Statement D:- CORRECT

• Without topoisomerase, DNA will not unwind, and the supercoiling strain will rise to the point where DNA may fragment.
• Therefore, statement D is correct.

Hence, option 3 (C and D) is correct.

The correct answer is Option 4 i.e. A and D

Concept:-

• The first DNA-based genome to be sequenced is the single-stranded DNA virus known as Bacteriophage phi x 174, which infects Escherichia coli. Its DNA contains 5386 nucleotides in total.
• When the G protein attaches to lipopolysaccharides on the bacterial host cell surface, the infection starts.
• The projected N-terminal transmembrane domain helix is most likely how the H protein (also known as the DNA Pilot Protein) guides the viral genome through the bacterial membrane of E. coli bacteria.
• Once within the host bacteria, a negative sense DNA intermediate is used to continue the replication of the [+] ssDNA genome.
• This occurs because the secondary structure created by the phage genome's supercoiling attracts a primosome protein complex.
• This moves once around the genome and creates an [-]ssDNA from the original, healthy positive genome.
• This is used to generate [+]ssDNA genomes that are packaged into viruses using a rolling circle process.
• This is the method by which a virus-encoded A protein cleaves the double-stranded supercoiled genome on the positive strand, luring a bacterial DNA polymerase (DNAP) to the point of cleavage. DNAP creates positive sense DNA by using the negative strand as a template.

Explanation:

Statement A - CORRECT

• Supercoiled ssDNA is reproduced during phiX174 bacteriophage reproduction to create dsRF (replication form) of DNA, which produces ssDNA (+), which produces more RF forms and about 500 or more virus particles, which are discharged by cell lyses.

Statement B - INCORRECT

•  A negative sense DNA intermediate is used to continue the replication of the [+] ssDNA genome.

Statement C - INCORRECT

• DNAP uses the negative strand as a template to produce positive sense DNA.

Statement D - CORRECT

• The process a virus-encoded file is A bacterial DNA polymerase (DNAP) is drawn to the site of cleavage by a protein that splits the double-stranded supercoiled genome on the positive strand.
• DNAP uses the negative strand as a template to produce positive sense DNA.
• Within the host bacteria, a negative sense DNA intermediate is used to continue the replication of the [+] ssDNA genome.

Therefore, option 4 (A and D) is correct.

The correct answer is Option 1

Concept:

• Direct repair, base excision repair (BER), nucleotide excision repair (NER), mismatch repair (MMR), homologous recombination (HR) and non-homologous end joining (NHEJ) are a few types of repair mechanisms functional throughout the life of the organism. 

Direct repair - 

• It is the simplest and most efficient repair mechanism.  
• In the direct repair mechanism, the damaged nucleotide in the DNA is eliminated by chemical reversion with the help of specific enzymes.
• This process does not require a nucleotide template, breaking of phosphodiester backbone and addition of new nucleotides. 
• Only a few types of DNA damage are repaired by this method.
• Pyrimidine dimers and O6-methylguanine are two examples of damage repaired by direct repair mechanisms.
• Pyrimidine dimers occur as a result of exposure to UV light and are repaired by a light-dependent direct system called photoreactivation.
• O6-methylguanine occurs as a result of exposure to alkylating agents. It is modified by the addition of ethyl or methyl group at the O6 position in the purine ring.

Base excision repair - 

• Base excision repair mechanism involves the removal of the damaged base and replacement of the damaged base with a new nucleotide.
• In the base excision repair mechanism, firstly the damaged base is removed by the action of the glycosylase enzyme, resulting in the generation of AP-site (Apurinic and Apyrimidinic site).
• AP-Endinucleaase cleaves the sugar-phosphate backbone at 5' of the AP site to produce a single-stranded break in the DNA.
• DNA glycosylases are lesion specific and there are multiple such enzymes with different specificities in the cell. 
• The next step is the addition of the new nucleotide in its place by a repair DNA polymerase and DNA ligase enzyme using undamaged strands as a template. 

Explanation:

• In the direct repair mechanism, the damaged base is modified by chemical reversion with the help of enzymes.
• For example, in the case of pyrimidine dimers, DNA photolyase captures the light energy and uses it to break the bond that links the adjacent pyrimidines.
• So, in direct repair, the damaged base is modified/altered not replaced.
• While in base excision, DNA glycosylase removes the damaged base.
• The new nucleotide is added by DNA polymerase and DNA ligase enzyme using the undamaged strand as a template. 
• So, in base excision repair damaged base is removed and replaced.

Hence, the correct answer is option 1. 

Additional Information

• Even though DNA repair systems quickly focus on the damaged DNA for repair, some lesions still exist and prevent the cell's replicator from duplicating the genome.
• Cells need specialised polymerases called translesion polymerase, to traverse the damage in order to prevent the negative effects of a halted replication fork.
• The cell is given more time to fix the damage during this procedure, known as "translesion DNA synthesis" (TLS) before the genome is duplicated.
• The translesion polymerase has low fidelity and it is associated with a higher risk of mutagenesis and carcinogenesis.

The correct answer is Option 3 i.e.Cdc45

Concept:

• DNA replication in eukaryotes could be divided into three major parts:
  • Initiation
  • Elongation
  • Termination.
• DNA replication initiation could be divided into:
  •  pre-replicative complex
  •  initiation complex.
• Pre-replicative complex majorly consists of
  • ORC (origin recognition complex)+ Cdc6 + Cdt + MCM complex (mini-chromosome maintenance complex)
• Initiation complex consists of 
  • Cdc45 + MCM 10 + GINS + DDK and CDK kinase + Dpb11, Sld3, Sld2 protein complex.

Explanation:

• All the proteins given in the options belong to eukaryotic cells and so we must consider only eukaryotic DNA replication here.

Option 1: ORC - INCORRECT

• DNA replication is initiated from the origin of replication, having specific sequences to initiate replication.
• The ORC is a hexameric DNA binding complex that binds with the origin of replication followed by the recruitment of Cdc6 protein followed by Cdt1.
• ORC dephosphorylates and becomes inactivated before the elongation process.
• Hence, this option is incorrect. 

Option 2: Geminin - INCORRECT

• It binds to cdt1 to prevent the re-initiation of DNA replication and hence it works as a regulator/inhibitor rather than an initiator of DNA replication.
• It is an inhibitor of Cdt1.

Option 3: Cdc45 - CORRECT

• Cdc refers to the cell division control proteins that are involved in various steps of DNA replication process.
• Cdc45 remains with MCM complex and GINS to work as a helicase.
• Thus, it helps in the initiation of DNA replication as well as advancement of the replication fork.

​Option 4: Cdc6 - INCORRECT

• It helps in the assembling of the pre-replicative complexes and interacts with the ORC.
• Cdc6 degrades before initiation of the replication fork.
• The concentration of both cdc6 and cdt1 declines before DNA elongation starts.

​Hence, the correct option is option 3.