The chapter “Molecular Basis of Inheritance” explains how genetic information is stored, copied, repaired, and expressed at the molecular level. Concepts like DNA replication, mutations, gene regulation, and recombination directly form the foundation for many board and competitive-exam questions because they connect DNA structure and enzyme mechanisms to inheritance patterns and evolutionary change.
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Q1. A linear eukaryotic chromosome of length has replication origins spaced every . Replication initiates simultaneously at all origins and proceeds bidirectionally; replication fork speed is . Approximately how long (in minutes) will it take to replicate the entire chromosome after initiation?
Q2. Consider a double-stranded linear DNA fragment of length with positions numbered to along the 5'→3' direction of the top strand. A forward primer has sequence identical to the top strand from position to (therefore it anneals to the bottom strand), and a reverse primer is complementary to the top strand from position to (therefore it anneals to the top strand). If PCR is performed using these primers, what is the expected length of the amplified product (in bp)?
Q3. A circular plasmid of total size has two EcoRI sites that partition the plasmid into arcs of and . A single BamHI site lies from the nearer EcoRI site measured along the arc. After digestion with both EcoRI and BamHI, which set of fragment sizes (in kb) will result?
Q4. Statement I: Cytosine methylation at CpG dinucleotides increases the frequency of C→T transitions at those sites.
Statement II: Spontaneous deamination of 5‑methylcytosine yields thymine, producing a G·T mismatch that is comparatively less efficiently recognized and repaired by base excision repair than a G·U mismatch.
Both statements are true but Statement II does not explain Statement I.
Statement I is true and Statement II is false.
Both statements are true and Statement II correctly explains Statement I.
Statement I is false and Statement II is true.
Q5. In a mammalian cell a gene is transcriptionally silent because its promoter CpG island is heavily methylated and the surrounding nucleosomes have deacetylated histone tails. Which treatment is most likely to produce robust reactivation of transcription from this promoter?
Treat cells with a DNA methyltransferase inhibitor (e.g., 5‑aza‑2'‑deoxycytidine) alone
Treat cells with a DNA demethylating agent (e.g., 5‑aza‑2'‑deoxycytidine) followed by (or together with) a histone deacetylase inhibitor (e.g., trichostatin A)
Treat cells with a histone deacetylase inhibitor (e.g., trichostatin A) alone
Treat cells with an inhibitor of histone methyltransferases (e.g., chaetocin) alone
Q6. In E. coli the circular chromosome of length replicates bidirectionally from oriC. If each replication fork moves at , approximately how long (in minutes) will one complete round of replication take?
Approximately
Approximately
Approximately
Approximately
Q7. In an E. coli strain the lac operator carries an mutation (repressor cannot bind) and the lac promoter has a mutation that reduces RNA polymerase binding to of wild‑type. Which statement best describes β‑galactosidase expression in this strain compared with wild‑type bacteria induced with lactose?
β‑galactosidase is expressed constitutively at about of the wild‑type lactose‑induced level (i.e., ~10% both with and without lactose).
No β‑galactosidase is produced because the weak promoter prevents transcription.
Expression occurs only in the presence of lactose and reaches wild‑type induced levels.
Expression is constitutive and equals wild‑type induced levels because the mutation overrides promoter weakness.
Q8. Three linked genes , and show recombination frequencies –, –, and –. Which linear gene order on the chromosome is most consistent with these data?
Q9. In a three‑point test cross with progeny the recombination frequencies are – and –. Only double‑crossover individual is observed. Using coefficient of coincidence and interference , what is the interference for the – interval?
Q10. A protein‑coding gene has an open reading frame of length (300 amino acids). In one allele two mutations occur: a deletion of nucleotides beginning within the codon and, farther downstream, an insertion of nucleotides immediately after the codon. Assuming no premature stop codon is created between the two mutations, which statement most accurately describes the resulting polypeptide?
No change in the amino acid sequence anywhere because the net nucleotide change is a multiple of three.
The frameshift persists to the normal termination codon, so the entire C‑terminal sequence is altered relative to wild type.
The reading frame is restored after the second mutation because nt (a multiple of ); amino acids N‑terminal to the first mutation are unchanged except at the affected codon, the stretch between the two mutations is translated in an altered frame, and the sequence downstream of the insertion resumes the original frame so the C‑terminal matches wild type but the polypeptide is shorter by one amino acid.
The protein will be truncated immediately after the first mutation regardless of the downstream insertion.