Molecular Basis of Inheritance MCQs

DNA’s double‑helix structure was deduced from X‑ray diffraction data (Rosalind Franklin) and chemical clues about base pairing.

Watson applied Chargaff’s rules and the complementary pairing concept to build a model where two strands wind around each other.

Crick contributed the antiparallel orientation and the specific base‑pairing (A‑T, G‑C) that explained the helical geometry.

Thus the double helix model was proposed by Watson and Crick.

A nucleotide is built from three components: a nitrogenous base, a five‑carbon sugar (ribose or deoxyribose), and a phosphate group attached to the sugar. The phosphate group links successive nucleotides together via phosphodiester bonds to form DNA or RNA. The other options (hydroxyl group, fatty acid, amino acid) are not part of the basic nucleotide structure. Therefore the correct answer is A phosphate group.

Loosely packed chromatin allows RNA polymerase access for transcription.

Such regions are less condensed and stain lightly, indicating active gene expression.

The term for this transcriptionally active, open chromatin is euchromatin. Euchromatin

Heat denatures proteins, so it would destroy the transforming factor.

DNase degrades DNA and abolishes transformation, showing DNA is the genetic material.

Protease removes proteins and RNase removes RNA, but transformation still occurs because the active component is DNA.

Thus protease and RNase treatment did NOT stop transformation.

Okazaki fragments are synthesized as short DNA pieces on the lagging strand.

DNA polymerase III extends each fragment but leaves a nick between them.

DNA ligase catalyzes the formation of phosphodiester bonds, sealing the nicks and joining the fragments.

Hence the enzyme that joins Okazaki fragments is DNA ligase.

The nucleosome core particle consists of a protein core around which DNA wraps.

The protein core is an octamer formed by two copies each of H2A, H2B, H3 and H4.

Thus the total number of histone molecules in the octamer is 2 × 4 = 8.

DNA contains phosphorus in its backbone, so labeling phage DNA with ^32P makes it radioactive.

During infection the phage injects its DNA into the bacterial cell while the protein coat remains outside.

When the mixture is centrifuged, the radioactivity is found in the pellet (bacterial cells) only for ^32P.

Thus the radioactivity entering the bacterial cell was mainly 32P from DNA.

During DNA replication the lagging strand is synthesized discontinuously.

RNA primers are laid down and DNA polymerase extends them, creating short DNA pieces.

These short DNA pieces are later joined by DNA ligase to form a continuous strand.

Thus the short DNA fragments on the lagging strand are called Okazaki fragments.

Sigma factor binds to the core RNA polymerase to form holoenzyme that recognizes promoter sequences. This complex is required to initiate transcription at the start site. It is not involved in termination, elongation or splicing. Hence the sigma factor is associated with Initiation.

Base pairing (A‑T and G‑C) involves attraction between nitrogenous bases on opposite strands.

These attractions are hydrogen bonds: two for A‑T, three for G‑C.

The sugar‑phosphate backbone is linked by phosphodiester bonds, not between strands.

Thus the strands are held together mainly by hydrogen bonds between complementary bases.

Adenine and guanine are nitrogen‑containing bases that have a double‑ring structure.

Bases with two fused rings are classified as purines, whereas single‑ring bases are pyrimidines.

Since both adenine and guanine possess the double‑ring purine skeleton, they belong to the purines.

Chromatin is first packaged into nucleosomes by histone proteins.

Higher‑order folding of these nucleosome fibers needs extra proteins that bind to histones and DNA to stabilize loops and coils.

These extra proteins are not histones themselves and are termed non‑histone chromosomal (NHC) proteins.

Thus the correct answer is Non‑histone chromosomal (NHC) proteins.

The tobacco mosaic virus contains a single‑stranded nucleic acid that directs its replication.

Viruses use either DNA or RNA as their genome; TMV lacks DNA and any protein genome.

Its genome is a linear, positive‑sense RNA that serves directly as mRNA for protein synthesis.

Thus the genetic material of TMV is RNA.

During DNA synthesis the two parental strands separate, creating a Y‑shaped region where new strands are synthesized. This branched structure is termed the replication fork. The correct answer is Replication fork.

Adenine (A) pairs with thymine (T) through hydrogen bonding in the DNA double helix.

The base‑pairing rule states that A and T are joined by two hydrogen bonds, while G‑C have three.

Thus the number of hydrogen bonds between adenine and thymine is two. Two

Deoxyribose sugar units are joined to phosphate groups forming a repeating sugar‑phosphate chain.

Each linkage between the 3′‑OH of one sugar and the 5′‑phosphate of the next is a phosphodiester bond.

Thus the DNA backbone consists of alternating sugar and phosphate residues.

Each human diploid cell contains 46 chromosomes ≈ 3 × 10⁹ base pairs per haploid set.

Length of DNA = number of base pairs × 0.34 nm per base pair ≈ 6 × 10⁹ × 0.34 nm ≈ 2 m.

Including packing and the second set gives about 2.2 m of DNA per cell.

Thus the length is approximately 2.2 metres.

DNA has a deoxyribose sugar lacking the 2′‑OH group present in RNA, so it is less prone to hydrolysis.

The double‑helical structure further protects the bases from chemical damage.

Thus DNA’s greater chemical stability makes it a more reliable carrier of genetic information. It is more stable.

DNA polymerase adds nucleotides to the growing strand by forming phosphodiester bonds.

Each addition uses the high‑energy phosphate bond of an incoming deoxyribonucleoside triphosphate, releasing pyrophosphate.

The energy released from breaking this bond drives polymerisation.

Hence the required energy source is deoxyribonucleoside triphosphates.

Hydrogen bonds form between complementary nitrogenous bases in DNA.

Guanine (G) and cytosine (C) have three complementary sites for hydrogen bonding.

Each site forms one hydrogen bond, giving a total of three bonds.