COMPETITIVE EXAM MCQs SERIES of LIFE SCIENCES for CSIR-UGC NET/JRF, SLET, GATE, and other entrance tests: FUNDAMENTAL PROCESSES – DNA Replicon and Recombination.
Syllabus Outline
- Unit of replication and concept of replicon.
- Homologous Recombination: Role in meiosis, gene targeting and DNA repair
- Site-Specific Recombination: Phage integration/excision and recombinase systems
- Illegitimate and Non-homologous Recombination: Transposition and chromosomal rearrangements.
This quiz contains concept-based, most frequently asked 25 MCQs of “FUNDAMENTAL PROCESSES – DNA Replicon and Recombination”. Each question has a single correct/most appropriate answer.
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1. The high fidelity of E. coli DNA replication is primarily achieved through three distinct steps. Which of the following lists these steps in the correct chronological order of their action during a single base incorporation event?
A) Base selection → 3′ → 5′ Proofreading → Mismatch Repair
B) 3′ → 5′ Proofreading → Base selection → Mismatch Repair
C) Mismatch Repair → Base selection → 3′ → 5′ Proofreading
D) Base selection → Mismatch Repair → 5′ → 3′ Proofreading
2. E. coli DNA Polymerase I and Polymerase III differ in their physiological roles and enzymatic properties. Which of the following statements correctly differentiates them?
A) Pol III is a monomeric protein responsible for primer removal, while Pol I is a multimeric complex responsible for genomic replication.
B) Pol I possesses 5′ → 3′ exonuclease activity used for nick translation, whereas Pol III lacks this activity and is the primary replicative enzyme.
C) Pol III has a higher processivity but a lower rate of polymerisation compared to Pol I.
D) Pol I requires an RNA primer to initiate synthesis, while Pol III can initiate DNA synthesis de novo.
3. Which of the following describes the mechanism of Tus-mediated termination of replication in E. coli?
A) Tus acts as a topoisomerase to decatenate the two daughter circular chromosomes.
B) Tus binds to specific Ter sites and acts as a directional “trap” by blocking the DnaB helicase.
C) Tus catalyses the addition of telomeric repeats to the circular chromosome to prevent shortening.
D) Tus is a component of the primosome that signals the polymerase to dissociate from the DNA.
4. Which of the following best describes the telomerase mechanism to solve the “end-replication problem” in linear eukaryotic chromosomes?
A) Telomerase is a DNA-dependent DNA polymerase that fills in the gaps left by primer removal.
B) Telomerase uses the sister chromatid as a template to perform homologous recombination at the chromosome ends.
C) Telomerase recruits a specialised DNA Ligase to circularise the linear chromosome before replication.
D) Telomerase is a ribonucleoprotein that uses its own RNA component as a template to extend the 3′ end of the parental DNA.
5. Which of the following is the correct function of the Proliferating Cell Nuclear Antigen (PCNA) protein in eukaryotic DNA replication?
A) It serves as a helicase to unwind the DNA double helix.
B) It binds to the origin of replication to initiate the “licensing” process.
C) It is an endonuclease that removes the RNA primers from Okazaki fragments.
D) It acts as a sliding clamp that tethers DNA polymerases delta and epsilon to the DNA template.
6. Which of the following correctly describes the role of the sigma subunit in E. coli RNA polymerase function?
A) It is the processivity factor for DNA Polymerase III.
B) It is required for the specific recognition of promoter sequences during transcription initiation.
C) It acts as a primase to synthesise RNA primers for DNA replication.
D) It stabilises single-stranded DNA at the replication fork.
7. During meiotic prophase I in eukaryotes, homologous chromosomes pair and undergo crossover. This process is initiated by which of the following?
A) Spontaneous melting of the DNA double helix at GC-rich regions.
B) The activity of DNA Polymerase alpha at recombination hotspots.
C) Programmed double-strand breaks introduced by the Spo11 protein.
D) Non-homologous end joining triggered by oxidative stress.
8. Unequal crossover between repetitive DNA sequences on homologous chromosomes can result in which of the following?
A) A DNA break on one chromosome with no change in gene dosage.
B) The deletion of a segment on one chromosome and a corresponding duplication on the other.
C) The inversion of a segment on both homologous chromosomes
D) The formation of a Robertsonian translocation between non-homologous chromosomes.
9. Which statement regarding meiotic recombination in Saccharomyces cerevisiae is TRUE?
A) It uses the sister chromatid as a template more frequently than the homologous chromosome.
B) It uses both homologous chromosomes and sister chromatid depends on abiotic stress to ensure genetic variation and conservation of genes.
C) It predominantly uses the homologous chromosome as a template to ensure genetic variation.
D) It occurs only at the centromeric regions to stabilise the spindle attachment.
10. Which of the following molecular events explains gene conversion through homologous recombination?
A) The physical exchange of large segments of chromosomes without any DNA synthesis.
B) The mismatch repair of heteroduplex DNA formed during strand invasion, where one allele is “corrected” to match the other.
C) The mutation of a gene triggered by the presence of RecA.
D) The insertion of a transposon into the middle of a coding sequence.
11. Why is the repair of double-strand DNA breaks by homologous recombination restricted to the S and G₂ phases of the eukaryotic cell cycle?
A) Homologous recombination requires a sister chromatid as a template, which is available only after DNA replication has occurred.
B) DNA polymerases are available for double-strand break repair only after DNA replication is completed.
C) RecA, required for homologous recombination, is synthesised during the S and G₂
D) The proofreading activity of DNA polymerase is activated only after DNA replication has occurred.
12. Which of the following amino acid residues is used by the Cre recombinase to form a covalent intermediate with the DNA backbone during the reaction?
A) Serine
B) Tyrosine
C) Aspartic Acid
D) Lysine
13. In Salmonella, the “phase variation” of flagellar proteins (H1 and H2) is controlled by which mechanism?
A) The random mutation of the flagellin genes.
B) The transcriptional repression of the H2 gene by the Lac operon.
C) The degradation of the H1 protein by the proteasome.
D) The reversible inversion of a DNA segment containing a promoter, mediated by a site-specific recombinase.
14. Which of the following is a key difference between Tyrosine and Serine recombinases?
A) Tyrosine recombinases require ATP, whereas Serine recombinases do not.
B) Tyrosine recombinases cleave all four strands simultaneously, while Serine recombinases cleave them one at a time.
C) Tyrosine recombinases involve a Holliday junction intermediate, while Serine recombinases cleave all four strands at once and rotate the complex.
D) Serine recombinases are only found in eukaryotes.
15. Which of the following statements correctly describes a Class I transposable element?
A) It moves by a “cut-and-paste” mechanism using a transposase.
B) It moves via an RNA intermediate that is reverse-transcribed into cDNA before integration.
C) It is only found in prokaryotes like Bacillus subtilis.
D) It requires the RecBCD complex to find a homologous site for insertion.
16. What is the molecular cause of Target Site Duplication?
A) The transposase enzyme performs a staggered cut at the target site.
B) The transposon is replicated twice by DNA Polymerase alpha before it is inserted.
C) The target DNA already contains a duplication that “attracts” the transposon.
D) The Target Site Duplication is an artefact of the PCR used to detect the transposon.
17. Non-Homologous End Joining is the primary pathway for repairing double-strand breaks in which phase of the human cell cycle?
A) S phase
B) G2 phase
C) G1 phase
D) Late M phase
18. Microhomology-Mediated End Joining differs from Non-Homologous End Joining in which of the following ways?
A) Microhomology-Mediated End Joining is error-free, whereas Non-Homologous End Joining is mutagenic.
B) Microhomology-Mediated End Joining is independent of Ku proteins and DNA Ligase IV
C) Microhomology-Mediated End Joining only occurs in bacteria.
D) Microhomology-Mediated End Joining uses the sister chromatid as a template.
19. A bacterial strain extremely resistant to UV radiation is found to possess multiple copies of a gene encoding a 3′ → 5′ exonuclease. How might this contribute to UV resistance?
A) It allows the cell to perform faster translesion synthesis.
B) It improves the fidelity of DNA replication
C) It is a component of the telomerase complex, reducing the burden of mutations caused by UV-induced lesions.
D) It is required for the proper integration of prophages, reducing the burden of mutations caused by UV-induced lesions.
20. A circular plasmid containing a single EcoRI site is replicated in vitro. If the reaction is interrupted and the DNA is analysed by gel electrophoresis, which of the following patterns would indicate the presence of theta-type replication intermediates?
A) A single sharp band corresponding to the linear plasmid.
B) A series of bands with increasing mass that migrate faster than the supercoiled plasmid.
C) A smeared pattern of branched molecules that migrate slower than the relaxed circular plasmid.
D) No bands, as replication intermediates are degraded simultaneously.
21. A pedigree shows a trait that appears in every generation, affecting both males and females equally. When an affected individual mates with a non-affected individual, approximately 50% of the offspring are affected. This pattern of inheritance is:
A) X-linked recessive
B) Autosomal dominant
C) Mitochondrial
D) Autosomal recessive
22. Assertion (A): In eukaryotes, the lagging strand is synthesised discontinuously as Okazaki fragments, while the leading strand is synthesised continuously.
Reason (R): DNA polymerases can only catalyse the addition of nucleotides to the 3′-OH end of a primer, meaning synthesis can only occur in the 5′ → 3′ direction.
A) Both (A) and (R) are true, and (R) is the correct reason for (A).
B) Both (A) and (R) are true, and (R) is not the correct reason for (A).
C) (A) is true, but (R) is false.
D) (A) is false, but (R) is true.
23. Assertion (A): In E. coli, the RecA protein is essential for both homologous recombination and the induction of the SOS response.
Reason (R): RecA binds to single-stranded DNA to form a nucleoprotein filament that promotes strand exchange and stimulates the autoproteolysis of the LexA repressor.
A) Both (A) and (R) are true, and (R) is the correct reason for (A).
B) Both (A) and (R) are true, and (R) is not the correct reason for (A).
C) (A) is true, but (R) is false.
D) (A) is false, but (R) is true.
24. Assertion (A): Recombination is an energetic process that requires ATP hydrolysis for the strand exchange.
Reason (R): RecA uses ATP hydrolysis for the orderly assembly and disassembly of the nucleoprotein filament.
A) Both (A) and (R) are true, and (R) is the correct reason for (A).
B) Both (A) and (R) are true, and (R) is not the correct reason for (A).
C) (A) is true, but (R) is false.
D) (A) is false, but (R) is true.
25. Assertion (A): Transposons are often suppressed in the somatic cells of eukaryotes by small RNA pathways.
Reason (R): Unregulated transposition in the somatic cells would lead to an accumulation of deleterious mutations.
A) Both (A) and (R) are true, and (R) is the correct reason for (A).
B) Both (A) and (R) are true, and (R) is not the correct reason for (A).
C) (A) is true, but (R) is false.
D) Both (A) and (R) are false.
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References
- Nelson, David L. & Cox, Michael M. (2021). Lehninger Principles of Biochemistry, W. H. Freeman, 8th Edition.
- Cooper, G. M. (2022). The Cell: A Molecular Approach, Sinauer Associates, 9th Edition
- Kumar, P., & Mina, U. (2025). Life Sciences: Fundamentals and Practice – Part I & II, Pathfinder Academy, 9th Edition
- Verma, P. S., & Agarwal, V. K. (2022). Cell Biology, Genetics, Molecular Biology, Evolution and Ecology, S. Chand Publishing, 1st Edition
- Alberts, B., Johnson, A., Lewis, J., Morgan, D., Raff, M., Roberts, K., & Walter, P. (2014). Molecular Biology of the Cell, Garland Science, 6th Edition.
- Gupta, P.K. (2022). Cell and Molecular Biology, Rastogi Publications, 5th Edition.
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