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Case-Based Questions on DNA Replication | Prokaryotic & Eukaryotic DNA Replication Quiz
Enhance your understanding of DNA replication with these case-based multiple-choice questions (MCQs) designed to test your knowledge of prokaryotic and eukaryotic replication mechanisms. These scenario-driven MCQs challenge you to apply molecular biology concepts to real-world and clinical cases, making them perfect for medical students, researchers, and biology enthusiasts. Dive in and see how well you grasp the intricate process of DNA replication!
1. A patient with an inherited disorder affecting DNA replication has a mutation in an enzyme that is essential for synthesizing short RNA primers on the lagging strand. Laboratory studies show incomplete Okazaki fragments in their cells.
Which enzyme is most likely defective in this patient?
A) Ligase
B) DNA Polymerase III
C) Topoisomerase
D) Primase
E) Helicase
Correct Answer: D) Primase
Primase is responsible for synthesizing short RNA primers on the lagging strand to provide a starting point for DNA polymerase. Without primase, Okazaki fragments cannot be initiated, leading to incomplete DNA replication.
Why Are the Other Options Incorrect?
A) Ligase – Joins Okazaki fragments but does not initiate them.
B) DNA Polymerase III – Extends the primers but cannot start synthesis without RNA primers.
C) Topoisomerase – Reduces supercoiling, ensuring smooth replication fork progression, but does not contribute to primer synthesis.
E) Helicase – Unwinds DNA but does not synthesize primers.
2. A new chemotherapy drug is found to halt cancer cell division by preventing the unwinding of DNA at the replication fork, thereby inhibiting replication.
Which enzyme is the drug most likely targeting?
A) DNA Polymerase
B) Helicase
C) Ligase
D) Topoisomerase
E) Primase
Correct Answer: B) Helicase
Helicase unwinds the double-stranded DNA at the replication fork, allowing replication to proceed. Inhibiting helicase prevents DNA strands from separating, effectively stopping replication.
Why Are the Other Options Incorrect?
A) DNA Polymerase – Synthesizes new DNA but does not separate strands.
C) Ligase – Joins DNA fragments but does not affect the unwinding process.
D) Topoisomerase – Relieves supercoiling but does not separate strands directly.
E) Primase – Creates primers but does not influence DNA unwinding.
3. A researcher is studying DNA replication in bacterial cells using isotopically labeled nitrogen (15N and 14N). After growing the bacteria in 15N medium, she transfers them to 14N medium and allows replication for two generations. When she performs density gradient centrifugation, she observes two distinct DNA bands: one of intermediate density and one of lighter density.
Which conclusion can be drawn from these observations?
A) DNA replication is conservative, with one molecule retaining both original strands and another consisting of two newly synthesized strands.
B) DNA replication is semi-conservative, with each DNA molecule containing one parental strand and one newly synthesized strand.
C) DNA replication is dispersive, with each DNA molecule containing interspersed parental and newly synthesized segments.
D) DNA replication requires continuous synthesis on both strands, leading to a uniform DNA density.
E) The presence of two bands indicates that DNA replication occurs through random recombination of old and new strands.
Correct Answer: B) DNA replication is semi-conservative, with each DNA molecule containing one parental strand and one newly synthesized strand.
Reasoning: In the first generation, DNA molecules exhibit intermediate density because each consists of one heavy (15N) strand and one light (14N) strand. In the second generation, half the molecules remain intermediate (15N-14N), and half are lighter (14N-14N), confirming the semi-conservative model.
Why the Other Options Are Incorrect?
A) Conservative replication would have resulted in two distinct bands from the first generation: one fully heavy and one fully light, which was not observed.
C) Dispersive replication would result in a single band of gradually decreasing density, rather than two distinct bands.
D) Continuous synthesis occurs only on the leading strand, while the lagging strand is synthesized discontinuously as Okazaki fragments.
E) DNA strands do not randomly recombine during replication but follow a precise template-based mechanism.
4. A patient with a rare genetic disorder presents with symptoms of increased DNA strand breaks and genomic instability. Laboratory analysis reveals a mutation in an enzyme responsible for joining short DNA fragments during replication, particularly on the lagging strand.
Which enzyme is most likely defective in this patient?
A) Helicase
B) Ligase
C) DNA Polymerase III
D) Topoisomerase
E) Primase
Correct Answer: B) Ligase
Reasoning: DNA ligase is responsible for sealing the nicks between Okazaki fragments by forming phosphodiester bonds. A defect in ligase leads to incomplete lagging strand synthesis and increased strand breaks.
Why Are the Other Options Incorrect?
A) Helicase unwinds DNA but does not join Okazaki fragments.
C) DNA Polymerase III extends DNA strands but does not seal nicks.
D) Topoisomerase relieves supercoiling but does not process Okazaki fragments.
E) Primase synthesizes RNA primers but does not join DNA fragments.
5. A chemotherapy drug selectively targets rapidly dividing cancer cells by inhibiting an enzyme essential for synthesizing short RNA primers required for Okazaki fragment formation.
Which process is directly disrupted by this drug?
A) Initiation of replication
B) Elongation of the leading strand
C) Formation of the lagging strand
D) Supercoiling resolution
E) DNA ligation
Correct Answer: C) Formation of the lagging strand
Reasoning: Okazaki fragments are formed on the lagging strand, requiring RNA primers synthesized by primase. Inhibiting primase prevents proper initiation of Okazaki fragments, disrupting lagging strand synthesis.
Why Are the Other Options Incorrect?
A) Initiation of replication involves helicase and origin recognition, not Okazaki fragment synthesis.
B) Leading strand elongation is continuous and does not rely on Okazaki fragments.
D) Supercoiling resolution is handled by topoisomerase, unrelated to Okazaki fragment formation.
E) DNA ligation occurs after Okazaki fragments are formed, but this drug affects an earlier step.
7. A 35-year-old patient is diagnosed with a rare form of colorectal cancer. Genetic analysis reveals a mutation in an enzyme responsible for proofreading newly synthesized DNA strands, leading to an increased mutation rate.
Which enzyme is most likely defective in this patient?
A) DNA Polymerase III
B) Primase
C) Helicase
D) Ligase
E) Topoisomerase
Correct Answer: A) DNA Polymerase III
Reasoning: DNA polymerases have 3′ → 5′ exonuclease activity, allowing them to proofread and correct errors. In prokaryotes, DNA Polymerase III is the primary enzyme for both DNA synthesis and proofreading. A defect in this function leads to an accumulation of mutations, increasing cancer risk.
Why Are the Other Options Incorrect?
B) Primase synthesizes RNA primers but does not proofread.
C) Helicase unwinds DNA but does not edit errors.
D) Ligase seals DNA fragments but does not correct mismatches.
E) Topoisomerase relieves supercoiling but does not proofread DNA.
8. A cancer drug selectively inhibits the proofreading ability of DNA polymerase, increasing the mutation rate in rapidly dividing cells and triggering apoptosis.
Which specific function of DNA polymerase is blocked by this drug?
A) 5′ → 3′ polymerase activity
B) 3′ → 5′ exonuclease activity
C) RNA primer synthesis
D) DNA ligation
E) Helicase-mediated DNA unwinding
Correct Answer: B) 3′ → 5′ exonuclease activity
Reasoning: Proofreading occurs via 3′ → 5′ exonuclease activity, which removes incorrect nucleotides during replication. Inhibiting this function leads to an accumulation of errors, making cancer cells more vulnerable to damage and apoptosis.
Why Are the Other Options Incorrect?
A) 5′ → 3′ polymerase activity is essential for DNA synthesis but does not proofread.
C) RNA primer synthesis is performed by primase, not DNA polymerase.
D) DNA ligation seals nicks but does not remove errors.
E) Helicase unwinds DNA but does not participate in proofreading.
9. A cancer drug selectively inhibits the proofreading ability of DNA polymerase, increasing the mutation rate in rapidly dividing cells and triggering apoptosis.
Which specific function of DNA polymerase is blocked by this drug?
A) 5′ → 3′ polymerase activity
B) 3′ → 5′ exonuclease activity
C) RNA primer synthesis
D) DNA ligation
E) Helicase-mediated DNA unwinding
Correct Answer: B) 3′ → 5′ exonuclease activity
Reasoning: Proofreading occurs via 3′ → 5′ exonuclease activity, which removes incorrect nucleotides during replication. Inhibiting this function leads to an accumulation of errors, making cancer cells more vulnerable to damage and apoptosis.
Why Are the Other Options Incorrect?
A) 5′ → 3′ polymerase activity is essential for DNA synthesis but does not proofread.
C) RNA primer synthesis is performed by primase, not DNA polymerase.
D) DNA ligation seals nicks but does not remove errors.
E) Helicase unwinds DNA but does not participate in proofreading.
10. A researcher is studying a bacterial strain with a mutation in the origin of replication (OriC). Despite the presence of all necessary replication enzymes, the cells fail to initiate DNA replication and remain in a non-dividing state.
Which process is directly disrupted in these bacterial cells?
A) Unwinding of DNA by helicase
B) Binding of DnaA proteins to initiate replication
C) Elongation of the leading strand by DNA Polymerase III
D) Removal of RNA primers from Okazaki fragments
E) Termination of replication at termination sequences
Correct Answer: B) Binding of DnaA proteins to initiate replication
Reasoning: In prokaryotes, replication begins at a single origin of replication (OriC), where DnaA proteins bind to initiate strand separation. A mutation in OriC prevents DnaA from binding, blocking replication initiation and causing cell cycle arrest.
Why Are the Other Options Incorrect?
A) Helicase function is required after initiation; failure to initiate replication means helicase is not even recruited.
C) DNA Polymerase III is involved in elongation, not initiation.
D) RNA primers are synthesized and removed later in replication, not during initiation.
E) Termination occurs at specific sequences but does not affect initiation.
11. A genetic disorder is found to cause uncontrolled re-replication of DNA in human cells, leading to genomic instability and increased cancer risk. Further analysis reveals a defect in a protein complex that ensures each origin of replication is used only once per cell cycle.
Which regulatory mechanism is most likely impaired in this condition?
A) Helicase activation
B) DNA polymerase proofreading
C) Origin licensing by pre-replication complexes
D) Lagging strand synthesis
E) RNA primer removal
Correct Answer: C) Origin licensing by pre-replication complexes
Reasoning: In eukaryotic cells, each origin of replication must be licensed before S-phase to ensure it is used only once per cycle. The pre-replication complex (Pre-RC), including ORC, Cdc6, and Cdt1, prevents re-replication. A defect in this mechanism results in multiple initiations per cell cycle, leading to DNA damage and instability.
Why Are the Other Options Incorrect?
A) Helicase activation is necessary for unwinding DNA but does not regulate replication licensing.
B) Proofreading corrects errors in replication but does not prevent re-replication.
D) Lagging strand synthesis occurs after initiation and does not regulate replication frequency.
E) RNA primer removal is essential for replication completion but does not control origin usage.
12. A researcher observes that in human cells, some regions of the genome replicate early in S-phase, while others replicate late. Further investigation reveals that genes actively transcribed during interphase tend to replicate earlier than silent or heterochromatin-rich regions.
What primarily determines the timing of replication initiation in eukaryotic cells?
A) DNA polymerase speed
B) Chromatin structure and gene activity
C) The presence of multiple origins of replication
D) The number of replication forks per chromosome
E) The availability of nucleotides
Correct Answer: B) Chromatin structure and gene activity
Reasoning: In eukaryotic cells, actively transcribed (euchromatin) regions replicate early, while heterochromatin and silent regions replicate late in S-phase. This timing ensures that essential genes are duplicated first and non-essential or repetitive sequences are copied later.
Why Are the Other Options Incorrect?
A) DNA polymerase speed is relatively constant and does not dictate replication timing.
C) Multiple origins help speed up replication but do not determine which regions start first.
D) Replication forks arise from origins but do not regulate timing.
E) Nucleotide availability affects replication efficiency but not its timing.
13. A scientist compares DNA replication in human cells and bacterial cells. He observes that bacterial cells have a single origin of replication, whereas human cells have multiple origins along each chromosome.
What is the primary reason eukaryotic cells require multiple origins of replication?
A) Eukaryotic DNA polymerases are slower than bacterial polymerases
B) Eukaryotic genomes are much larger than bacterial genomes
C) Bacterial replication does not require RNA primers
D) Eukaryotic DNA has fewer replication forks
E) Bacteria replicate their DNA only once per cell cycle
Correct Answer: B) Eukaryotic genomes are much larger than bacterial genomes
Reasoning: Eukaryotic genomes are significantly larger than bacterial genomes. To ensure complete and timely DNA replication before cell division, multiple origins of replication are needed, allowing simultaneous replication at multiple sites.
Why Are the Other Options Incorrect?
A) DNA polymerase speed varies but is not the main reason for multiple origins.
C) Both eukaryotic and bacterial replication require RNA primers.
D) Eukaryotic replication has more replication forks due to multiple origins.
E) Bacteria also regulate replication per cell cycle, but this does not explain the need for multiple origins in eukaryotes.
14. A patient presents with a genetic disorder causing defective lagging strand synthesis. Genetic analysis identifies a mutation in an enzyme complex responsible for synthesizing RNA primers to initiate DNA replication. Cells from this patient exhibit incomplete Okazaki fragments and replication stalling.
Which enzyme complex is most likely affected?
A) DNA Polymerase α-Primase
B) DNA Polymerase δ
C) DNA Ligase
D) Helicase
E) Topoisomerase
Correct Answer: A) DNA Polymerase α-Primase
Reasoning: In eukaryotic cells, the DNA Polymerase α-Primase complex is responsible for synthesizing RNA primers on both the leading and lagging strands. A mutation in this complex prevents proper primer formation, leading to incomplete DNA synthesis and stalled replication forks.
Why Are the Other Options Incorrect?
B) DNA Polymerase δ extends DNA strands but does not synthesize primers.
C) DNA Ligase seals Okazaki fragments but does not initiate them.
D) Helicase unwinds DNA but does not synthesize primers.
E) Topoisomerase relieves supercoiling but does not participate in primer synthesis.
15. A new chemotherapy drug selectively inhibits the primase function of the DNA Polymerase α-Primase complex in rapidly dividing cancer cells. Treated cells show inability to initiate DNA replication, leading to replication failure and cell death.
Which phase of the cell cycle is most affected by this drug?
A) G1 phase
B) S phase
C) G2 phase
D) M phase
E) G0 phase
Correct Answer: B) S phase
Reasoning: DNA replication occurs in the S-phase of the cell cycle. Since the DNA Polymerase α-Primase complex is required for primer synthesis, inhibiting this function prevents replication from proceeding, leading to cell cycle arrest and apoptosis in cancer cells.
Why Are the Other Options Incorrect?
A) G1 phase precedes DNA replication and is not directly affected.
C) G2 phase involves preparation for mitosis but does not require primer synthesis.
D) M phase is mitosis, occurring after DNA replication is completed.
E) G0 phase is a resting state where cells are not actively replicating DNA.
16. A researcher is studying eukaryotic DNA replication and observes that two different DNA polymerases are responsible for synthesizing the leading and lagging strands. When a specific polymerase is inhibited, lagging strand synthesis is significantly impaired, while the leading strand remains relatively unaffected.
Which DNA polymerase is most likely inhibited?
A) DNA Polymerase α
B) DNA Polymerase β
C) DNA Polymerase δ
D) DNA Polymerase ε
E) DNA Polymerase γ
Correct Answer: C) DNA Polymerase δ
Reasoning: DNA Polymerase δ is primarily responsible for lagging strand synthesis in eukaryotic cells. It extends the Okazaki fragments after priming by DNA Polymerase α-Primase. Inhibition of Polymerase δ disrupts lagging strand elongation, leading to incomplete replication.
Why Are the Other Options Incorrect?
A) DNA Polymerase α synthesizes RNA-DNA primers but does not extend the lagging strand.
B) DNA Polymerase β is involved in DNA repair, not replication.
D) DNA Polymerase ε is primarily involved in leading strand synthesis.
E) DNA Polymerase γ is responsible for mitochondrial DNA replication, not nuclear DNA replication.
17. A patient presents with symptoms of a progressive neurological disorder linked to defective mitochondrial DNA replication. Genetic testing reveals a mutation in a DNA polymerase that is exclusively responsible for mitochondrial genome replication.
Which DNA polymerase is most likely defective?
A) DNA Polymerase α
B) DNA Polymerase β
C) DNA Polymerase δ
D) DNA Polymerase ε
E) DNA Polymerase γ
Correct Answer: E) DNA Polymerase γ
Reasoning: DNA Polymerase γ is the only polymerase responsible for mitochondrial DNA replication. Mutations in this polymerase lead to mitochondrial dysfunction, causing neurological and muscular disorders such as progressive external ophthalmoplegia (PEO).
Why Are the Other Options Incorrect?
A) DNA Polymerase α is involved in nuclear DNA primer synthesis, not mitochondrial replication.
B) DNA Polymerase β is involved in base excision repair, not replication.
C) DNA Polymerase δ synthesizes the lagging strand in nuclear DNA replication.
D) DNA Polymerase ε synthesizes the leading strand in nuclear DNA replication.
18. A patient with a rare genetic disorder exhibits genomic instability and frequent DNA strand breaks. Further analysis reveals that RNA primers are not properly removed from Okazaki fragments, leading to incomplete DNA replication.
Which enzyme complex is primarily responsible for removing primers in eukaryotic cells?
A) DNA Polymerase α
B) DNA Ligase
C) RNase H and FEN1
D) DNA Polymerase γ
E) Topoisomerase
Correct Answer: C) RNase H and FEN1
Reasoning: In eukaryotic cells, RNase H degrades most of the RNA primer, while FEN1 (Flap Endonuclease 1) removes the remaining RNA-DNA junction. A defect in these enzymes prevents proper primer removal, leading to replication errors and DNA strand breaks.
Why Are the Other Options Incorrect?
A) DNA Polymerase α synthesizes primers but does not remove them.
B) DNA Ligase seals Okazaki fragments but does not remove primers.
D) DNA Polymerase γ is responsible for mitochondrial DNA replication, not nuclear primer removal.
E) Topoisomerase relieves supercoiling but does not process RNA primers.
19. A new chemotherapy drug is designed to inhibit the enzyme responsible for removing the RNA portion of the primer during DNA replication. Treated cancer cells show persistent RNA segments within replicated DNA, leading to replication stress and cell death.
Which enzyme is most directly affected by this drug?
A) DNA Polymerase ε
B) RNase H
C) DNA Ligase
D) Topoisomerase
E) Helicase
Correct Answer: B) RNase H
Reasoning: RNase H is the key enzyme responsible for removing RNA primers from Okazaki fragments in eukaryotic cells. Inhibiting this enzyme leaves RNA segments embedded in newly synthesized DNA, causing replication stress and genome instability, which can selectively kill rapidly dividing cancer cells.
Why Are the Other Options Incorrect?
A) DNA Polymerase ε is involved in leading strand synthesis, not primer removal.
C) DNA Ligase seals DNA fragments but does not remove RNA primers.
D) Topoisomerase relieves supercoiling but does not degrade RNA primers.
E) Helicase unwinds DNA but does not process primers.
20. A patient presents with a neurodevelopmental disorder associated with excessive DNA supercoiling and strand breakage. Genetic analysis reveals a mutation in an enzyme responsible for reducing torsional strain ahead of the replication fork in eukaryotic cells.
Which enzyme is most likely defective?
A) Helicase
B) DNA Polymerase δ
C) Topoisomerase I
D) Ligase
E) RNase H
Correct Answer: C) Topoisomerase I
Reasoning: Topoisomerase I prevents excessive supercoiling by making single-strand cuts in DNA, allowing it to unwind smoothly during replication. A defect in this enzyme leads to accumulated supercoiling, causing DNA damage and replication stress.
Why Are the Other Options Incorrect?
A) Helicase unwinds DNA but does not relieve supercoiling.
B) DNA Polymerase δ synthesizes DNA but does not alter DNA topology.
D) Ligase seals DNA nicks but does not relieve torsional stress.
E) RNase H removes RNA primers but does not affect DNA supercoiling.
21. A chemotherapy drug is designed to inhibit an enzyme that makes transient double-stranded DNA breaks to prevent excessive supercoiling during replication. Treated cancer cells accumulate DNA damage and chromosomal fragmentation, leading to apoptosis.
Which enzyme is most directly inhibited by this drug?
A) DNA Polymerase α
B) Topoisomerase II
C) Helicase
D) Primase
E) DNA Ligase
Correct Answer: B) Topoisomerase II
Reasoning: Topoisomerase II prevents DNA tangling by making double-stranded cuts to relieve excessive supercoiling. Inhibiting this enzyme leads to chromosomal breaks and replication failure, which selectively kills rapidly dividing cancer cells.
Why Are the Other Options Incorrect?
A) DNA Polymerase α synthesizes primers but does not regulate DNA supercoiling.
C) Helicase unwinds DNA but does not prevent tangling.
D) Primase synthesizes RNA primers but does not affect DNA topology.
E) DNA Ligase seals nicks but does not resolve supercoiling issues.
22. A 45-year-old patient presents with signs of bone marrow failure and premature aging, including skin atrophy and early graying of hair. Genetic testing reveals a mutation in the TERT (telomerase reverse transcriptase) gene, leading to a loss of telomerase activity.
What is the most likely cellular consequence of this mutation?
A) Increased DNA replication errors
B) Shortening of chromosome ends with each cell division
C) Increased synthesis of Okazaki fragments
D) Failure to remove RNA primers from lagging strands
E) Continuous elongation of telomeres
Correct Answer: B) Shortening of chromosome ends with each cell division
Reasoning: Telomerase adds repetitive nucleotide sequences to the ends of chromosomes (telomeres) to prevent progressive shortening during DNA replication. A loss of telomerase activity leads to telomere shortening, triggering cellular senescence and premature aging disorders like dyskeratosis congenita and aplastic anemia.
Why Are the Other Options Incorrect?
A) DNA replication errors are corrected by DNA polymerases and mismatch repair mechanisms, not telomerase.
C) Okazaki fragments are formed during lagging strand synthesis, but telomerase does not regulate their production.
D) RNA primer removal is carried out by RNase H and FEN1, not telomerase.
E) Continuous telomere elongation occurs in cancer cells, but telomerase deficiency leads to telomere shortening, not elongation.
23. A pharmaceutical company is developing a drug that inhibits telomerase activity in cancer cells. After treatment, cancer cells undergo progressive cell cycle arrest and apoptosis due to critically shortened telomeres.
Which type of cells would be least affected by this drug?
A) Germ cells
B) Stem cells
C) Cancer cells
D) Differentiated somatic cells
E) Rapidly dividing bone marrow cells
Correct Answer: D) Differentiated somatic cells
Reasoning: Most differentiated somatic cells naturally lack telomerase activity and rely on finite telomere length. Therefore, an anti-telomerase drug would primarily impact cancer cells, stem cells, and germ cells, which actively maintain their telomeres using telomerase.
Why Are the Other Options Incorrect?
A) Germ cells rely on telomerase to maintain telomere length for reproductive viability.
B) Stem cells use telomerase to sustain self-renewal and long-term proliferation.
C) Cancer cells depend on telomerase to maintain unlimited replication potential, making them a major drug target.
E) Rapidly dividing bone marrow cells express telomerase to support hematopoiesis and would be affected by the drug.
24. A 37-year-old patient is diagnosed with Lynch syndrome, a hereditary condition associated with a high risk of colorectal cancer. Genetic analysis reveals a mutation in a gene involved in mismatch repair (MMR), a key post-replication repair mechanism.
Which type of DNA damage is most likely to accumulate in this patient’s cells?
A) Double-strand breaks
B) Thymine dimers
C) Mismatched base pairs
D) Depurination events
E) Telomere shortening
Correct Answer: C) Mismatched base pairs
Reasoning: Post-replication mismatch repair (MMR) corrects mismatched bases that escape the proofreading activity of DNA polymerase. Mutations in MMR genes (e.g., MLH1, MSH2) lead to microsatellite instability (MSI) and an increased risk of cancer, especially colorectal cancer in Lynch syndrome.
Why Are the Other Options Incorrect?
A) Double-strand breaks are repaired by homologous recombination (HR) or non-homologous end joining (NHEJ), not MMR.
B) Thymine dimers result from UV damage and are repaired by nucleotide excision repair (NER).
D) Depurination involves loss of purine bases and is repaired by base excision repair (BER), not MMR.
E) Telomere shortening is due to loss of telomerase activity, unrelated to post-replication repair.
25. A patient with a developmental disorder presents with global gene expression abnormalities. Genetic analysis reveals a mutation in a protein responsible for modifying histones, leading to altered chromatin structure and dysregulated transcription.
Which of the following is most likely affected in this patient?
A) Nucleosome formation
B) DNA replication fork progression
C) Okazaki fragment ligation
D) RNA primer synthesis
E) DNA mismatch repair
Correct Answer: A) Nucleosome formation
Reasoning: DNA organization in eukaryotic cells is achieved by wrapping DNA around histone proteins, forming nucleosomes. Mutations affecting histone modifications (e.g., acetylation, methylation) can alter chromatin structure, leading to changes in gene expression and developmental disorders like Rett syndrome or Rubinstein-Taybi syndrome.
Why Are the Other Options Incorrect?
B) DNA replication fork progression is regulated by helicase and topoisomerase, not histones.
C) Okazaki fragment ligation is performed by DNA ligase and is unrelated to nucleosome formation.
D) RNA primer synthesis is carried out by primase, which does not regulate chromatin structure.
E) DNA mismatch repair corrects replication errors but does not influence chromatin organization.
