This chapter is central to CBSE and competitive biology because it explains how genetic information is stored, copied, and expressed at the molecular level. Topics like DNA replication fidelity, mutation and repair, gene regulation in prokaryotes, and the logic of transcription/translation are repeatedly tested in both conceptual and numerical/formula-based questions.
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Minutes
15
Questions
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Marking
Q1. A eukaryotic gene's coding sequence (CDS) is 1233 nucleotides long and includes the stop codon at its 3' end. How many amino acids are present in the translated polypeptide?
411 amino acids
408 amino acids
410 amino acids
409 amino acids
Q2. Glycine is encoded by four codons . Using standard wobble rules in which inosine (I) at the 5'-position of the tRNA anticodon can pair with or , what is the minimum number of distinct tRNA species required to recognise all four glycine codons?
2
3
4
1
Q3. Assertion (A): Mutations in the promoter region of a eukaryotic gene always lead to a decrease in that gene's transcription level.
Reason (R): Changes in promoter sequence alter the binding affinity of transcription factors and the basal transcription machinery, which can either increase or decrease transcription initiation.
Both A and R are true, and R correctly explains A.
Both A and R are true, but R does not correctly explain A.
A is true, R is false.
A is false, R is true.
Q4. A PCR reaction starts with a single double‑stranded DNA template. Ideal amplification doubles product every cycle, giving copies after cycles. If the per‑cycle efficiency is only 90% (effective multiplication factor per cycle ), what is the approximate number of target molecules after 30 cycles?
Greater than
Approximately
Approximately
Approximately
Q5. In steady state, mRNA concentration is given by where is transcription rate and is degradation rate; half‑life . A mutation increases threefold but reduces mRNA half‑life to one‑half of the original value. What is the fold change in steady‑state mRNA level?
No change
Two‑fold increase
1.5‑fold increase
Six‑fold increase
Q6. A circular bacterial chromosome of length base pairs initiates replication at a single origin and proceeds bidirectionally. If each replication fork progresses at , approximately how long (in minutes) will it take to complete replication of the chromosome? (Assume replication is simultaneous in both directions and neglect pauses.)
Q7. In E. coli, Dam methylation transiently marks the parental strand at GATC sites after replication. Consider a replication error that produces a mismatch where the parental strand carries C (methylated) and the newly synthesized strand carries A (unmethylated). Which outcome is most likely due to the bacterial mismatch-repair system?
The methylated parental C will be excised and replaced, producing a permanent A–T pair.
The mismatch will be ignored until the new strand becomes methylated, producing a permanent A–T mutation after the next round of replication.
The mismatched A will be excised and replaced with C on the new strand, leaving an uncorrected C–C pair.
The repair machinery will recognise the parental (methylated) strand, remove the mismatched A from the unmethylated new strand and replace it with G, restoring the original C–G pair.
Q8. A bacterial cell is merodiploid: its chromosome carries a lac operon with an operator constitutive mutation (operator cannot bind repressor), while the same cell contains an F' plasmid encoding a wild-type lacI repressor. What is the expected expression behaviour of the chromosomal lac structural genes in absence and presence of lactose?
The chromosomal lac structural genes will be expressed constitutively (expressed both in absence and presence of lactose).
The plasmid-encoded LacI will repress the chromosomal operator in absence of lactose, so expression will be inducible.
The chromosomal operon will remain silent unless lactose is present because plasmid LacI complements the defect.
The chromosomal operon will be nonfunctional regardless of lactose due to creating a defective promoter.
Q9. DNA polymerase misincorporates a nucleotide at probability per base per replication. Proofreading corrects of such misincorporations. Post-replicative mismatch repair recognises the parental (methylated) strand and directs repair to the new strand of the time. What is the probability that a single misincorporation at that base will become a permanent mutation after these two correction steps?
Q10. A bacterium has a nonsense mutation (stop codon ) at codon 120 of an essential gene P, producing a truncated inactive enzyme. A second-site mutation in a tRNA gene alters its anticodon so that the tRNA now recognises and inserts tyrosine (a suppressor tRNA). Which statement most accurately predicts the likely consequences of this suppressor appearing in the population?
The suppressor is a true genotypic reversion restoring the original DNA sequence at the mutated site and will only affect gene P.
The suppressor tRNA allows read-through of the , producing full-length protein with tyrosine at position 120; gene P activity may be partially or fully restored depending on the substitution, but the suppressor acts extragenically and can cause read-through at other stops in different genes, potentially causing pleiotropic effects.
Suppression will be confined to gene P because altered tRNAs affect only the mRNA they are translated from, so other genes are unaffected.
The suppressor tRNA will correct frameshift mutations upstream of the stop codon as well, restoring proteins affected by frameshifts.
Q11. A eukaryotic gene's pre-mRNA contains exons totaling nucleotides and introns totaling nucleotides. After splicing, translation begins at the first AUG at the 5'-end of the first exon and terminates at a stop codon located immediately after the last exon; the stop codon is included within the -nt exon sequence. How many amino acids will the translated polypeptide contain (excluding the stop codon)?
amino acids
amino acids
amino acids
amino acids
Q12. The pre-mRNA of a eukaryotic gene has four exons with lengths nt, nt, nt and nt. Translation starts at the AUG at the 5'-end of and ends at the stop codon at the 3'-end of . Which alternative splicing event will produce a shorter polypeptide that lacks the protein domain encoded by exon but retains the original reading frame (i.e., causes no downstream frameshift)?
Skip exon only
Skip exons and together
Skip exon only
Skip exon only
Q13. In a eukaryotic chromosome replication origins are on average apart and each origin fires bidirectionally. If the average Okazaki fragment synthesized on the lagging strand is nucleotides long, approximately how many Okazaki fragments will be produced by a single replication fork before it meets the converging fork from the adjacent origin?
Q14. A point mutation in exon 4 of a human gene changes the codon –GAA– to –GAG– (both code for glutamic acid). However, analysis of mature mRNA shows exon 4 is completely skipped in transcripts from the mutant allele, producing a shorter mRNA. Which molecular explanation best accounts for exon 4 skipping despite the mutation being synonymous at the amino-acid level?
The synonymous change created a premature stop codon within exon 4
The mutation altered the branch point sequence located in the downstream intron
The nucleotide change disrupted an exonic splicing enhancer (ESE), preventing binding of splicing activators and causing exon skipping
The codon change altered the encoded amino acid, which caused the protein to destabilize and feedback to alter splicing
Q15. An open reading frame begins at an AUG start codon. Two independent mutations occur immediately downstream of the start codon: Mutation X inserts nucleotides at that position; Mutation Y inserts nucleotides at the same position. Which statement is most likely regarding the effect of these mutations on the translated polypeptide?
Both X and Y will cause frameshift mutations leading to premature termination
Mutation X will cause a frameshift likely producing an altered/truncated protein, whereas Mutation Y will insert two amino acids but preserve the downstream reading frame
Mutation X will be silent because small insertions do not affect translation; Mutation Y will always create a stop codon
Both insertions will be corrected by the spliceosome during mRNA processing because they are small insertions