41 free MCQs on Molecular Basis of Inheritance with worked answers and explanations. DNA structure, replication, transcription, translation, the genetic code, and gene regulation (lac operon). The single most important chapter for NEET molecular biology.
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Helicase unwinds the double helix at the replication fork; one new strand (leading) is synthesised continuously toward the fork, while the other (lagging) must be made in short, separate Okazaki fragments because DNA polymerase only adds nucleotides 5'→3'.
Q1.
Who proposed the double helix model of DNA?
Answer: B. Watson and Crick
Why: James Watson and Francis Crick proposed the double helix structure of DNA in 1953, based on X-ray diffraction data from Rosalind Franklin.
Q2.
In DNA, adenine forms hydrogen bonds with which base?
Answer: C. Thymine
Why: Adenine pairs with Thymine through 2 hydrogen bonds; Guanine pairs with Cytosine through 3 hydrogen bonds.
Q3.
The two strands of the DNA double helix are arranged:
Answer: B. Antiparallel to each other
Why: The two polynucleotide strands of DNA run in opposite directions, one 5 prime to 3 prime and the other 3 prime to 5 prime, hence antiparallel.
Q4.
Histone proteins are rich in which type of amino acids?
Answer: B. Basic amino acids
Why: Histones are rich in basic amino acids like lysine and arginine, giving them a net positive charge that lets them bind the negatively charged DNA.
Q5.
DNA replication is described as semiconservative because:
Answer: B. Each daughter molecule retains one parental strand and gains one new strand
Why: In semiconservative replication, each of the two daughter DNA molecules consists of one old (parental) strand and one newly synthesised strand.
Q6.
Which experiment provided direct evidence for semiconservative DNA replication?
Answer: C. Meselson-Stahl experiment
Why: Meselson and Stahl used 15N and 14N labelled E. coli DNA and density gradient centrifugation to confirm semiconservative replication.
Q7.
Which enzyme unwinds the DNA double helix at the replication fork?
Answer: B. Helicase
Why: Helicase unwinds the parental DNA double helix at the replication fork, separating the two strands for replication.
Q8.
Which enzyme joins Okazaki fragments to form a continuous DNA strand?
Answer: C. DNA ligase
Why: DNA ligase seals the nicks between adjacent Okazaki fragments, joining them into a continuous lagging strand.
Q9.
According to the central dogma of molecular biology, genetic information flows:
Answer: C. DNA to RNA to Protein
Why: The central dogma proposed by Francis Crick states that genetic information flows from DNA to RNA to protein.
Q10.
Which enzyme catalyses transcription?
Answer: B. RNA polymerase
Why: RNA polymerase catalyses the synthesis of RNA using a DNA template during transcription.
Q11.
The DNA strand that is actually copied during transcription is called the:
Answer: B. Template strand
Why: Only the template strand is used by RNA polymerase to synthesise a complementary RNA transcript.
Q12.
Which of the following is the universal start codon?
Answer: C. AUG
Why: AUG, which codes for methionine, is the universal start codon for translation.
Q13.
How many stop codons exist in the genetic code?
Answer: C. 3
Why: There are three stop codons: UAA, UAG, and UGA. They do not code for any amino acid.
Q14.
Which molecule acts as the adaptor between mRNA codons and amino acids during translation?
Answer: B. tRNA
Why: Francis Crick proposed that tRNA acts as an adaptor molecule, reading the mRNA codon and carrying the corresponding amino acid.
Q15.
In the lac operon, which gene codes for beta-galactosidase?
Answer: B. The z gene
Why: The z gene of the lac operon codes for beta-galactosidase, which hydrolyses lactose into glucose and galactose.
Q16.
DNA fingerprinting relies mainly on variation in which type of DNA sequence?
Answer: B. Satellite DNA (VNTRs)
Why: DNA fingerprinting exploits polymorphism in repetitive satellite DNA sequences called VNTRs (Variable Number of Tandem Repeats), which vary between individuals.
Q17.
Why is a primer required to start DNA replication?
Answer: B. DNA polymerase can only add nucleotides to an existing 3 prime OH end
Why: DNA polymerase cannot initiate synthesis on a bare template; it can only extend an existing strand with a free 3 prime OH, which the RNA primer provides.
Q18.
Which enzyme synthesises the short RNA primer needed for DNA replication?
Answer: B. Primase
Why: Primase synthesises a short RNA primer complementary to the template, providing the free 3 prime end that DNA polymerase extends.
Q19.
On the lagging strand, DNA synthesis proceeds:
Answer: B. Discontinuously, away from the replication fork, in short fragments
Why: Because DNA polymerase synthesises only 5 prime to 3 prime, the lagging strand is made discontinuously in short Okazaki fragments moving away from the fork.
Q20.
Which RNA polymerase in eukaryotes transcribes the precursor of mRNA (hnRNA)?
Answer: B. RNA polymerase II
Why: RNA polymerase II transcribes the heterogeneous nuclear RNA (hnRNA), the precursor that is processed into mature mRNA.
Q21.
Splicing during mRNA processing refers to:
Answer: C. Removal of introns and joining of exons
Why: Splicing is the removal of non-coding intron sequences from hnRNA and the joining of coding exons to form mature mRNA.
Q22.
The genetic code is described as degenerate because:
Answer: B. Most amino acids are specified by more than one codon
Why: Degeneracy means that most of the 20 amino acids are coded by more than one codon, for example leucine and serine each have 6 codons.
Q23.
The wobble hypothesis explains how:
Answer: B. A single tRNA can recognise more than one codon for the same amino acid
Why: The wobble hypothesis states that the pairing at the third codon position can be flexible, letting one tRNA recognise multiple synonymous codons.
Q24.
In the lac operon, what is the role of the inducer (allolactose)?
Answer: B. It binds to the repressor protein, preventing it from binding the operator
Why: Allolactose (the inducer) binds the repressor protein, changing its shape so it can no longer bind the operator; this allows transcription of the structural genes.
Q25.
In the absence of lactose, the lac operon remains switched off because:
Answer: B. The repressor protein binds the operator and blocks transcription
Why: Without an inducer, the active repressor protein binds to the operator region, physically blocking RNA polymerase from transcribing the z, y, and a genes.
Q26.
The Human Genome Project revealed that the human genome contains roughly:
Answer: A. Fewer than 30,000 to 25,000 genes
Why: Contrary to earlier predictions, the HGP found the human genome contains only about 30,000 to 25,000 protein-coding genes or fewer.
Q27.
What proportion of the human genome consists of non-coding sequences?
Answer: C. Over 90%
Why: More than 90% of the human genome is made up of non-coding DNA sequences, once referred to as junk DNA.
Q28.
Which scientist is closely associated with developing DNA fingerprinting techniques in India?
Answer: B. Lalji Singh
Why: Dr. Lalji Singh made major contributions to developing DNA fingerprinting technology in India.
Q29.
Which feature of the genetic code means that a given codon always specifies the same amino acid?
Answer: C. Unambiguous nature
Why: The code is unambiguous: a specific codon always codes for the same amino acid in a given organism, without alternative meanings.
Q30.
DNA polymerase synthesises new DNA strands in which direction relative to the template?
Answer: A. It adds nucleotides only in the 5 prime to 3 prime direction on the new strand
Why: DNA polymerase always extends a new strand in the 5 prime to 3 prime direction, regardless of which template strand is being copied; this is why one strand is synthesised continuously and the other discontinuously.
Q31.
In the Meselson-Stahl experiment, after one round of replication in 14N medium, bacteria initially grown in 15N medium produced DNA that was:
Answer: C. Hybrid density (15N-14N)
Why: After one generation, all DNA molecules showed an intermediate (hybrid) density, consistent with each molecule having one heavy (15N) parental strand and one light (14N) new strand, confirming semiconservative replication.
Q32.
Which statement about RNA polymerase in bacteria is correct?
Answer: B. A single RNA polymerase, with the help of factors like sigma and rho, catalyses transcription of all RNA types
Why: In bacteria, unlike eukaryotes, a single RNA polymerase catalyses transcription of mRNA, tRNA, and rRNA; sigma factor helps with initiation and rho factor helps with termination.
Q33.
Why does eukaryotic mRNA require extensive processing before translation, while bacterial mRNA generally does not?
Answer: B. Bacterial genes do not have introns and transcription/translation are not separated by a nuclear membrane
Why: Bacterial (prokaryotic) genes typically lack introns and translation can begin even before transcription is complete since there is no nuclear membrane separating the processes, so extensive processing like splicing is not required as in eukaryotes.
Q34.
If a region of the coding strand of DNA reads 5 prime-ATGCCG-3 prime, what is the sequence of the corresponding template strand and mRNA?
Answer: A. Template: 3 prime-TACGGC-5 prime; mRNA: 5 prime-AUGCCG-3 prime
Why: The template strand is complementary and antiparallel to the coding strand (3 prime-TACGGC-5 prime), and mRNA has the same sequence as the coding strand except T is replaced by U, giving 5 prime-AUGCCG-3 prime.
Q35.
Which of the following best explains why far fewer than 61 distinct tRNAs are sufficient to translate all sense codons?
Answer: B. The wobble hypothesis allows flexible pairing at the third codon position, letting one tRNA read multiple synonymous codons
Why: Because of wobble pairing, a single tRNA species can recognise more than one codon differing only at the third (wobble) position, reducing the total number of tRNAs needed.
Q36.
A mutation that converts the lac operon's operator sequence so the repressor can never bind would most likely result in:
Answer: B. Constitutive (continuous) expression of the z, y, and a genes regardless of lactose presence
Why: If the repressor cannot bind the mutated operator, it can no longer block RNA polymerase, so the structural genes would be transcribed constitutively, irrespective of whether lactose (inducer) is present.
Q37.
Which statement correctly distinguishes euchromatin from heterochromatin?
Answer: B. Euchromatin is loosely packed, stains light, and is transcriptionally active; heterochromatin is densely packed, stains dark, and is transcriptionally inactive
Why: Euchromatin is loosely packed chromatin that stains light and is generally transcriptionally active, while heterochromatin is tightly packed, stains dark, and is largely transcriptionally inactive.
Q38.
What is the key structural distinction between the coding strand and the template strand of a gene during transcription?
Answer: A. The coding strand has the same base sequence as the mRNA (with T instead of U) and is not transcribed; the template strand is read by RNA polymerase
Why: The coding strand has a sequence equivalent to the mRNA (except T replaced by U) but is not directly copied; RNA polymerase reads the complementary template strand to synthesise the mRNA.
Q39.
Why is DNA replication confined largely to the S-phase of the cell cycle rather than continuing throughout interphase?
Answer: A. DNA polymerase is only active during S-phase due to availability of dNTPs and coordinated licensing of replication origins
Why: Replication is tightly regulated and licensed to occur only once per cell cycle, during S-phase, coordinated with the availability of replication machinery and resources, alongside duplication of other organelles like centrioles.
Q40.
During DNA fingerprinting, why is a probe complementary to VNTR sequences used in the hybridisation step?
Answer: B. VNTR regions vary in repeat number between individuals, so the probe binds to fragments whose size differs from person to person, producing a unique pattern
Why: Since the number of tandem repeats at VNTR loci varies between individuals, restriction digestion produces fragments of different lengths in different people; a labelled VNTR probe hybridises to these fragments, revealing a banding pattern unique to each individual.
Q41.
Which of the following correctly orders the size relationship of repeating units along the chromatin fibre, from smallest to largest packaging level?
Answer: B. DNA double helix, nucleosome, chromatin fibre (beads on string and higher coiling), chromosome
Why: Packaging proceeds from the naked DNA double helix wrapping around histones to form nucleosomes (beads on a string), which further coil into chromatin fibres, which condense further to form a visible chromosome.