This chapter is central for both board and competitive exams because it connects the structure of DNA with replication, mutation, gene expression regulation, and experimental proof (e.g., Meselson–Stahl, PCR, lac operon). Concepts like base pairing, semiconservative replication, repair mechanisms, restriction mapping, and transcription control are frequently tested and form the foundation for higher-level genetics problems.
20
Minutes
15
Questions
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Marking
Q1. A double-stranded DNA molecule of base pairs contains adenine. Assuming Watson–Crick pairing (– pairs have hydrogen bonds and – pairs have hydrogen bonds), the total number of hydrogen bonds in the molecule is:
Q2. Consider the following double-stranded DNA fragment; only the sense (coding) strand is shown : -ATGCCATTAGGCTGAACGGATCCGTAATGC-185'\rightarrow3'18$ nucleotides? (Forward primers are given as matching the sense strand.)
Forward: -ATGCCATT- ; Reverse: -ACGGATCC-
Forward: -CATTAGGC- ; Reverse: -GATCCGTT-
Forward: -AGGCTGAA- ; Reverse: -GCATTACG-
Forward: -GGATCCGT- ; Reverse: -CGTAATGC-
Q3. In E. coli a lac operon mutant carries two mutations: (i) an operator mutation that prevents lac repressor binding (so the repressor cannot block transcription), and (ii) a mutation in the CAP‑binding site that prevents CAP–cAMP activation. Predict the level of expression of lac structural genes under these four environmental states: (1) glucose present, lactose absent; (2) glucose absent, lactose present; (3) glucose absent, lactose absent; (4) glucose present, lactose present.
Constitutive low-to-moderate expression in all four conditions (derepressed but no CAP activation)
High expression only when lactose is present and glucose is absent; negligible expression otherwise (wild‑type pattern)
Constitutive high expression in all four conditions (CAP not required)
Expression only when glucose is absent regardless of lactose presence (CAP‑dependent expression only)
Q4. Assertion (A): Telomerase is active in most somatic human cells.
Reason (R): DNA polymerases require an RNA primer and synthesize DNA only in the direction, producing the end‑replication problem on linear chromosomes.
Both A and R are true and R explains A
Both A and R are true but R does not explain A
A is true but R is false
A is false but R is true
Q5. A linear DNA molecule of length gives the following digestion pattern: enzyme → fragments of and ; enzyme → fragments of and ; double digest → fragments of , and . Which one of the following linear maps (positions measured from the left end at ) is consistent with these data?
at ; at
at ; at
at ; at
at ; at
Q6. A PCR reaction is set up with an initial number of double-stranded target molecules and is run for cycles with perfect efficiency (each cycle doubles the number of target molecules). Assuming no loss and perfect doubling each cycle, how many double-stranded target molecules will be present at the end?
Q7. A circular plasmid of length has a single recognition site for enzyme A and two recognition sites for enzyme B. Digestion results: A alone → one fragment of (linear), B alone → two fragments of and , and A+B (double digest) → three fragments of , and . Which of the following best describes the relative positions of the restriction sites?
The A site lies within the fragment produced by B and is from one B site (and from the other), producing fragments on double digest.
The A site lies within the B fragment and is from one B site and from the other.
The A site coincides with one of the B sites (they cut at the same nucleotide), so a double digest would not change fragment number relative to B alone.
The A site lies outside the interval between the two B sites; therefore a double digest should still yield only two fragments.
Q8. E. coli cells are grown for many generations in medium containing so that all genomic DNA is heavy (HH). They are then shifted to medium containing and allowed to replicate for generations. Assuming semiconservative replication and no selection, what fraction of the DNA molecules extracted after the fourth generation will be hybrid (one heavy strand + one light strand)?
Q9. Assertion (A): In a merodiploid bacterium where the chromosomal lac operon carries an operator-constitutive mutation () and a second copy on a plasmid supplies a wild-type repressor gene (), the chromosomal lac structural genes will be expressed constitutively even in the absence of lactose.
Reason (R): Operator sequences act in cis, so a functional repressor supplied in trans from a different DNA molecule cannot bind and repress expression at a mutated operator on the chromosome.
Both A and R are true but R is NOT the correct explanation for A.
Both A and R are true and R is the correct explanation for A.
A is true but R is false.
A is false but R is true.
Q10. During DNA replication in E. coli a G:T mismatch occurs such that the parental (original) strand contains G and the newly synthesized strand contains T. In wild-type cells Dam methylase methylates adenine in GATC and the mismatch repair (MMR) system uses this methylation to identify and remove the newly synthesized strand; in a Dam-deficient mutant there is no methylation and MMR cannot distinguish strands and instead excises one strand at random. Under these assumptions, what is the probability that after MMR the original (parental) base G is retained at that locus in the Dam-deficient strain?
(original base will never be retained)
(original base will always be retained)
Q11. A double-stranded DNA fragment contains 300 base pairs and has GC content of 40%. Using the relation , where and are numbers of GC and AT base pairs respectively, what is the total number of hydrogen bonds in the fragment?
600
720
660
480
Q12. Consider a eukaryotic diploid cell whose haploid genome size is bp. DNA polymerase incorporates wrong nucleotides at a rate per base. Proofreading reduces this error rate by 100-fold and mismatch-repair reduces the remaining errors by another 100-fold. Using (where is the diploid genome size and is the final error rate), estimate the expected number of replication errors per cell division across the entire diploid genome.
6
0.06
60
0.6
Q13. In a bacterium mRNA is synthesized at rate molecules min and degrades with first-order kinetics having half-life min. Using and steady state , calculate (i) the steady-state number of mRNA molecules, and (ii) if a mutation increases the half-life to min, the new steady-state number and the fold-change relative to the original. Choose the option that correctly states (original ; new ; fold-change).
-fold
-fold
-fold
-fold
Q14. Assertion (A): In a Meselson–Stahl type experiment starting from a single (heavy–heavy) DNA molecule, the absolute number of hybrid () DNA molecules remains constant in all generations after the first replication in light medium.
Reason (R): Each original heavy strand continues to serve as a template in every subsequent replication and therefore produces exactly one molecule per original heavy strand in each generation.
Both A and R are true and R is the correct explanation of A
Both A and R are true but R is not the correct explanation of A
A is true but R is false
A is false but R is true
Q15. A linear eukaryotic chromosome has length bp. Replication forks move at speed . Assume all origins fire synchronously at the start of S-phase, each origin gives bidirectional replication, and no additional origins fire later. To finish replication within S-phase duration min, the minimum number of origins required is given by . What is (nearest integer)?
100
200
400
50