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Biology Set-1 MCQ Online Practice Test | Class 12 | Quiz Tweek

Biology Set-1

Practice Important MCQs for Molecular Basis of Inheritance — Biology, Class 12.

“Molecular Basis of Inheritance” explains how genetic information is stored, copied, and interpreted at the molecular level—linking DNA structure, replication, transcription, translation, and DNA repair to real experimental evidence. This makes the chapter highly scoring for CBSE boards and competitive exams because most questions are concept-based (with numericals and reasoning), directly testing understanding of semiconservative replication, gene expression control, mutations, and DNA damage-repair mechanisms.

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Q1. A double‑stranded DNA molecule of length $N=1000\ \mathrm{bp}$ has GC content $60\%$ . Using the approximate formula for melting temperature for long DNA
$T_m \approx 81.5 + 0.41(\%\mathrm{GC}) - \dfrac{675}{N}$ (in $^\circ\mathrm{C}$ ), what is the expected $T_m$ (to one decimal place)?

(A)

$97.8\;^\circ\mathrm{C}$

(B)

$82.1\;^\circ\mathrm{C}$

(C)

$105.4\;^\circ\mathrm{C}$

(D)

$110.0\;^\circ\mathrm{C}$

Q2. E. coli are grown for many generations in medium containing $^{15}\mathrm{N}$ and then transferred to medium containing $^{14}\mathrm{N}$ . DNA extracted after one round of replication in $^{14}\mathrm{N}$ forms a single band of intermediate density in a CsCl gradient; DNA extracted after a second round shows two bands: one intermediate and one light. Which replication model do these observations best support and why?

(A)

Semiconservative replication — after the first replication each duplex contains one old ( $^{15}\mathrm{N}$ ) and one new ( $^{14}\mathrm{N}$ ) strand (hybrid); after the second round half the duplexes are hybrid and half are entirely light.

(B)

Conservative replication — the original heavy duplex would remain intact and a fully light duplex would be produced, giving heavy and light bands even after the first replication.

(C)

Dispersive replication — strands would be mixtures of old and new segments and should remain of intermediate density even after multiple rounds.

(D)

A mixed model where some molecules replicate semiconservatively and some conservatively, explaining the observed two bands after the second round.

Q3. A circular plasmid vector of size $5\ \text{kb}$ contains a unique EcoRI site. A $2\ \text{kb}$ DNA fragment with EcoRI‑compatible ends is ligated into this EcoRI site to yield a recombinant circular plasmid ( $7\ \text{kb}$ ). If the recombinant plasmid is digested with EcoRI and run on an agarose gel, which band pattern is expected?

(A)

Single linear band at $7\ \text{kb}$

(B)

Three bands of $5\ \text{kb}$ , $2\ \text{kb}$ and an uncut recombinant plasmid band at about $7\ \text{kb}$ (if digestion is incomplete)

(C)

Single band at $2\ \text{kb}$ (insert only)

(D)

Two bands of sizes $5\ \text{kb}$ and $2\ \text{kb}$ corresponding to vector backbone and insert

Q4. Assertion (A): E. coli cells lacking Dam methylase exhibit elevated spontaneous mutation rates. Reason (R): The mismatch‑repair system distinguishes the parental strand from the newly synthesized strand by methylation of adenines in GATC sites; without Dam methylation the system cannot identify which strand to correct. Choose the correct option.

(A)

Both A and R are true but R does not explain A.

(B)

Both A and R are true and R correctly explains A.

(C)

A is true but R is false.

(D)

A is false but R is true.

Q5. A eukaryotic gene has two exons and one intron. Exon‑1 = $120\ \mathrm{bp}$ , intron = $90\ \mathrm{bp}$ , exon‑2 = $300\ \mathrm{bp}$ . All exon lengths are multiples of 3 and the coding sequence starts at the first nucleotide of exon‑1 and continues in‑frame into exon‑2 after splicing. A point mutation in the 5' splice donor causes retention of the intron in the mature mRNA; the retained intron contains an in‑frame stop codon located $60\ \mathrm{bp}$ from the exon1–intron junction. What is the most likely length (in amino acids) of the polypeptide produced from this mutant mRNA?

(A)

$60\ \text{amino acids}$

(B)

$140\ \text{amino acids}$

(C)

$40\ \text{amino acids}$

(D)

$100\ \text{amino acids}$

Q6. A double‑stranded DNA oligonucleotide of 20 bp has 60% GC content. Using the empirical formula $T_m = 2\times(\#A+\#T) + 4\times(\#G+\#C)$ (in °C), estimate its melting temperature $T_m$ .

(A)

56 °C

(B)

72 °C

(C)

64 °C

(D)

60 °C

Q7. The distance between two adjacent replication origins on a eukaryotic chromosome is $120\ \text{kb}$ . Replication forks move at a speed of $3\times10^3\ \text{nt min}^{-1}$ . If both adjacent origins fire simultaneously, what is the minimum time to replicate the entire 120 kb region? If one origin fails to fire and the neighbouring origin alone replicates the whole 120 kb, how long will it take? (Give answers in minutes.)

(A)

$20\ \text{min}$ ; $40\ \text{min}$

(B)

$40\ \text{min}$ ; $80\ \text{min}$

(C)

$10\ \text{min}$ ; $20\ \text{min}$

(D)

$30\ \text{min}$ ; $60\ \text{min}$

Q8. Two statements about base excision repair (BER) are given below.
P: "Cells deficient in DNA glycosylase accumulate fewer abasic (AP) sites than wild‑type cells after oxidative DNA damage."
Q: "DNA glycosylase initiates BER by removing damaged bases to create AP sites; AP endonuclease then cleaves the sugar–phosphate backbone producing a single‑strand break that is processed further."
Which option correctly describes P and Q?

(A)

P and Q are true but Q does not explain P.

(B)

P is true, Q is false.

(C)

P is false, Q is true.

(D)

P and Q are true and Q explains P.

Q9. A bacterium carries a nonsense mutation that converts a sense codon at position 50 of an essential gene into the stop codon UAG, yielding a truncated nonfunctional protein. A suppressor mutation alters the anticodon of a tRNA‑Phe so it now recognizes UAG and inserts Phe at this position. Which outcome is most likely?

(A)

Full restoration of the wild‑type protein sequence and normal function with no additional effects.

(B)

Insertion of Phe at the premature stop can produce a full‑length polypeptide (possibly with altered activity), but the suppressor tRNA may also cause read‑through of normal UAG stop codons in other genes, producing aberrant elongated proteins.

(C)

The suppressor tRNA will act only on the mutated mRNA and will not affect translation of other genes.

(D)

The suppressor tRNA will equally suppress all three stop codons (UAG, UAA, UGA).

Q10. A covalently closed circular plasmid has $N=4200$ base pairs of B‑form DNA. For relaxed B‑DNA $L_{k_0}=\dfrac{N}{10.5}$ . If topoisomerase introduces 10 negative supercoils so that $\Delta Lk=-10$ , calculate the superhelical density $\sigma=\dfrac{\Delta Lk}{L_{k_0}}$ . Which value is closest?

(A)

$-0.025$

(B)

$-0.010$

(C)

$-0.0025$

(D)

$-0.050$

Q11. A double‑stranded DNA molecule has $10{,}000$ base pairs and a GC content of $40\%$ . How many hydrogen bonds stabilize the two strands in the entire molecule?

(A)

$20{,}000\ \text{H-bonds}$

(B)

$24{,}000\ \text{H-bonds}$

(C)

$26{,}000\ \text{H-bonds}$

(D)

$22{,}000\ \text{H-bonds}$

Q12. A contiguous chromosomal region of length $270\ \text{kb}$ contains three replication origins that fire simultaneously. Replication forks progress bidirectionally from each origin at $2{,}000\ \text{nt min}^{-1}$ . What is the minimum time required (in minutes) to replicate the entire $270\ \text{kb}$ region after S‑phase begins?

(A)

$45\ \text{min}$

(B)

$67.5\ \text{min}$

(C)

$11.25\ \text{min}$

(D)

$22.5\ \text{min}$

Q13. A eukaryotic gene's coding sequence comprises five exons with lengths $150\ \text{bp},\ 102\ \text{bp},\ 99\ \text{bp},\ 123\ \text{bp}$ and $36\ \text{bp}$ respectively (all exonic sequences are translated in‑frame; translation starts in exon 1). If an alternatively spliced mRNA lacks exons 2 and 4, what will be the length of the translated polypeptide (in amino acids)?

(A)

$95\ \text{aa}$

(B)

$125\ \text{aa}$

(C)

$170\ \text{aa}$

(D)

$140\ \text{aa}$

Q14. Assertion (A): A premature termination codon (PTC) introduced by a nonsense mutation more than $55\ \text{nt}$ upstream of the final exon–exon junction usually triggers nonsense‑mediated decay (NMD), resulting in rapid degradation of the mRNA.
Reason (R): During splicing the exon‑junction complex (EJC) is deposited $\sim 20\!-\!24\ \text{nt}$ upstream of exon–exon junctions; if a stop codon is encountered with an EJC remaining downstream, the termination is recognized as premature and recruits the NMD machinery.

(A)

Both A and R are false.

(B)

A is true but R is false.

(C)

Both A and R are true and R explains A.

(D)

Both A and R are true but R does not explain A.

Q15. A bacterial population whose DNA is uniformly labeled with $^{15}\text{N}$ is transferred to medium containing $^{14}\text{N}$ and allowed to replicate for three generations. After three generations, what fraction of the DNA duplexes will be hybrid (one $^{15}\text{N}$ strand + one $^{14}\text{N}$ strand) and what fraction of all nitrogen atoms in DNA will be $^{15}\text{N}$ ? Choose the correct pair.

(A)

hybrid $50\%$ ; $^{15}\text{N}$ atoms $25\%$

(B)

hybrid $25\%$ ; $^{15}\text{N}$ atoms $12.5\%$

(C)

hybrid $12.5\%$ ; $^{15}\text{N}$ atoms $6.25\%$

(D)

hybrid $75\%$ ; $^{15}\text{N}$ atoms $37.5\%$

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