NEET Molecular Basis of Inheritance Previous Year Questions with Solutions

The lac operon consists of one regulatory gene (the i gene, from the term inhibitor) and three structural genes (z, y, and a). The i gene codes for the repressor protein of the lac operon. It is synthesized constitutively (all the time) at a basal level.

• • Constitutive expression means continuous, non-regulated transcription regardless of lactose presence.

Lactose (or allolactose) acts as an inducer by binding to the repressor tetramer, causing conformational change and inactivating it. The structural genes code for enzymes: z = β-galactosidase, y = permease, a = transacetylase.

• • Negative regulation: the repressor bound to operator blocks RNA polymerase; inducer lifts repression.

In eukaryotes, there are at least three RNA polymerases in the nucleus (in addition to the RNA polymerase found in the organelles). There is a clear division of labour among them.

• • Eukaryotic RNA polymerase II also synthesizes microRNAs (miRNAs).

The primary transcript (hnRNA) synthesized by RNA polymerase II contains both exons (coding) and introns (non-coding). It undergoes post-transcriptional processing: 5' capping with methylguanosine triphosphate, splicing out of introns by spliceosomes, and 3' polyadenylation (tailing).

• • DNA polymerases synthesize DNA, not RNA.

Analyze the canonical principles of the RNA World hypothesis from NCERT Chapter 6:

• • Prebiotic evolution centered around RNA performing both informational storage and enzymic catalysis.
• • The reactive 2'-OH group on ribose makes RNA labile and prone to alkaline hydrolysis.
• • Statement I is verbatim accurate from NCERT and completely TRUE.

Analyze the structural adaptations that favored DNA as the permanent repository of genetic information:

• • Double-stranded DNA with complementary base-pairing provides redundancy: if one strand is damaged, the complementary strand directs high-fidelity enzymatic repair.
• • Statement II is verbatim accurate from NCERT and completely TRUE.

Both statements are true and accurate:

• • Matches Option (1) perfectly.

The complete sequencing of the human genome revealed extraordinary insights into the genetic architecture of humans:

• • The human genome contains 3164.7 million nucleotide base pairs (3.164 Gb).
• • The total number of genes is estimated at 30,000, much lower than previous estimates of 80,000 to 140,000 genes.
• • Almost all (99.9%) nucleotide bases are exactly the same in all people.

Genes are not evenly distributed across human chromosomes:

• • Chromosome 1 (the largest human autosome) contains the highest number of genes: exactly 2,968 genes.
• • Chromosome 1 was the last chromosome to be completely sequenced (completed in May 2006).
• • Chromosome Y contains the fewest genes: only 231 genes.

Chromosome 1 holds the record for harboring the highest number of annotated protein-coding genes in the human genome.

During protein translation, multiple RNA species exhibit essential, highly coordinated physical and functional interactions:

• • tRNA (transfer RNA) possesses an anticodon loop that specifically base-pairs with complementary codons on mRNA.
• • rRNA (ribosomal RNA) forms the structural and catalytic framework of ribosomes; the smaller subunit binds mRNA, and rRNA sequences (like 16S rRNA in prokaryotes or 18S in eukaryotes) interact with mRNA to position it correctly for translation.
• • Therefore, asserting that tRNA and rRNA "do not interact with mRNA" is factually and conceptually INCORRECT.

RNA interference (RNAi) is a conserved post-transcriptional gene silencing mechanism:

• • NCERT explicitly states: "RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence."
• • It involves silencing of a specific mRNA due to a complementary double-stranded RNA (dsRNA) molecule that binds to and prevents translation (or degrades) the target mRNA.
• • Therefore, Statement II is completely CORRECT.

Statement I is incorrect, but Statement II is correct. This corresponds to Option (4).

In eukaryotes, RNA Polymerase II transcribes heterogeneous nuclear RNA (hnRNA), which contains both coding exons and non-coding introns and is non-functional until post-transcriptional processing:

• • Splicing (Event B): The non-coding introns are excised by spliceosomes and the coding exons are ligated together in a specific, defined sequence. This occurs entirely within the nucleoplasm.
• • Capping (Event C): An unusual nucleotide, 7-methylguanosine triphosphate (m⁷G), is added to the 5'-end of the primary transcript. This cap protects the mRNA from exonucleases and assists in ribosome binding during translation.
• • Tailing (Event D): Polyadenylation involves the template-independent enzymatic addition of 200–300 adenylate residues (poly-A tail) to the 3'-end of hnRNA.

Analyzing incorrect statements:

• • Statement A claims "Transport of pre-mRNA to cytoplasm prior to splicing". Eukaryotic nuclear pore complexes enforce strict quality control: unspliced pre-mRNAs are retained within the nucleus until splicing and processing are complete. Only mature, fully spliced mRNA is exported to the cytoplasm.
• • Statement E claims "Base pairing of two complementary RNAs". This is the mechanism of RNA interference (RNAi) or antisense regulation, not a regular step in eukaryotic mRNA maturation.

The genuine post-transcriptional modification events are B (splicing), C (capping), and D (tailing). Hence, B, C, D only are correct.

Foundational experiments in the discovery and proof of the genetic material:

• • Frederick Griffith (1928): Conducted experiments with *Streptococcus pneumoniae* (bacteria responsible for pneumonia). Discovered that non-virulent R strain was transformed into virulent S strain by heat-killed S bacteria (Transforming Principle). Hence, C matches I.
• • Alfred Hershey and Martha Chase (1952): Provided the unequivocal, definitive proof that DNA is the genetic material. Worked with bacteriophage T2 and *E. coli*, labeling viral DNA with ³²P and viral protein capsid with ³⁵S. Hence, A matches IV.

Cytological differentiation of interphase chromatin:

• • Euchromatin: In a typical nucleus, some regions of chromatin are loosely packed and stain light with basic dyes. Euchromatin is transcriptionally active. Hence, B matches III.
• • Heterochromatin: The chromatin regions that are more densely packed and stain dark are called heterochromatin. Heterochromatin is transcriptionally inactive. Hence, D matches II.

Combining the verified pairings:

• • A -> IV (DNA as genetic material confirmation)
• • B -> III (Loosely packed and light-stained)
• • C -> I (Streptococcus pneumoniae)
• • D -> II (Densely packed and dark-stained)

The unambiguous combination is A-IV, B-III, C-I, D-II, matching Option (3).

Human diploid cells contain approximately 2.2 meters of DNA ($6.6 \times 10^9$ bp) that must be condensed to fit inside a nucleus of ~10 $\mu$m diameter:

• • DNA possesses a net negative electrical charge because of the phosphate groups in its phosphodiester backbone.
• • Packaging is facilitated by specialized basic proteins called histones.

Why histones carry a high positive charge density:

• • Histone proteins are enriched with the basic amino acids Lysine and Arginine.
• • At physiological pH (7.4), the amino side chains of lysine ($epsilon$-amino group) and arginine (guanidino group) are protonated ($-\text{NH}_3^+$ and $-\text{C(NH}_2)_2^+$), giving histones a strong positive charge.
• • Electrostatic attraction between the positively charged basic histone octamer and negatively charged acidic DNA forms the nucleosome core particle.

Histones are specifically enriched with the basic amino acids Lysine & Arginine, corresponding to Option (1).

After the discovery of the double helix structure of DNA, deciphering how a sequence of 4 nucleotide bases (A, U, G, C) directs the synthesis of proteins containing 20 distinct amino acids emerged as the central challenge:

• • If a codon were a singlet ($4^1 = 4$), only 4 amino acids could be specified.
• • If a codon were a doublet ($4^2 = 16$), only 16 amino acids could be specified, which is insufficient for 20 amino acids.

The physical and mathematical argument:

• • George Gamow (a theoretical physicist and cosmologist) proposed that the genetic code must be a triplet of nucleotides.
• • A triplet code generates $4^3 = 64$ possible codons, providing ample capacity to specify all 20 amino acids with degeneracy.
• • Subsequent biochemical experiments by Har Gobind Khorana, Marshall Nirenberg, and Severo Ochoa experimentally verified Gamow's triplet hypothesis.

Contributions of the other scientists listed:

• • Francis Crick: Proposed the Central Dogma, adaptor hypothesis (tRNA), and comma-less reading frame.
• • Jacques Monod (with François Jacob): Elucidated the operon model (*lac* operon) of transcriptional gene regulation.
• • Franklin Stahl (with Matthew Meselson): Proved the semi-conservative replication of DNA using $^{15}\text{N}$ isotope labeling.

George Gamow is the scientist who proposed that the genetic code must be made up of three nucleotides.

Bacterial RNA polymerase core enzyme (alpha_2 eta eta' omega) is catalytically competent for chain elongation but cannot recognize promoters or terminate faithfully on its own:

• • To ensure stage-specific regulation, the core enzyme associates reversibly with auxiliary protein factors.
• • Initiation: The core enzyme binds the sigma ($\sigma$) factor to form the holoenzyme, enabling sequence-specific promoter recognition at the $-10$ (Pribnow box) and $-35$ consensus sequences.

How RNA synthesis is brought to an end:

• • Elongation proceeds until the polymerase reaches a terminator sequence.
• • In rho-dependent termination, the termination factor rho ($\rho$) (an ATP-dependent hexameric RNA helicase) binds to a C-rich *rut* site on the nascent RNA transcript.
• • Rho migrates along the RNA transcript toward the stalled RNA polymerase.
• • Upon contacting the transcription elongation complex, Rho unwinds the RNA-DNA hybrid within the transcription bubble, releasing the newly synthesized nascent RNA chain and dissociating RNA polymerase from the template DNA.

The factor important for the termination of transcription is $\rho$ (rho), corresponding to Option (3).

The lac operon consists of three structural genes: z, y, and a. Each encodes a specific polypeptide essential for lactose metabolism in E. coli.

The z gene codes for beta-galactosidase, which cleaves lactose into galactose and glucose. The y gene codes for permease, a transmembrane protein that increases cell permeability to beta-galactosides, allowing lactose uptake. The a gene encodes a transacetylase.

Transport of lactose across the bacterial cell membrane into the cytoplasm is strictly mediated by lactose permease.

A transcription unit in DNA is defined primarily by three regions: a Promoter, the Structural gene, and a Terminator.

By convention, positions of sequence elements are defined with respect to the coding strand (5' to 3'). The promoter is situated towards the 5'-end (upstream). It provides the binding site for RNA polymerase. The structural gene lies between the promoter and terminator. The terminator is situated towards the 3'-end (downstream) and defines the end of transcription.

Reading from upstream to downstream yields: Promoter -> Structural gene -> Terminator.

Frederick Griffith conducted transformation experiments with Streptococcus pneumoniae, demonstrating transfer of genetic virulence (A -> III). Francois Jacob (geneticist) and Jacques Monod (biochemist) elucidated the lac operon mechanism in E. coli (B -> IV).

Har Gobind Khorana developed chemical synthesis of RNA molecules with defined combinations of bases to crack the genetic code (C -> I). Matthew Meselson and Franklin Stahl performed equilibrium density gradient centrifugation with heavy isotope 15N in E. coli, proving semi-conservative DNA replication (D -> II).

A-III, B-IV, C-I, D-II matches Option (4).

In living cells, such as E. coli, the main enzyme is DNA-dependent DNA polymerase. It uses a DNA template to catalyse the polymerisation of deoxynucleotides with high speed and exceptional fidelity.

The DNA-dependent DNA polymerases catalyse polymerisation only in one direction, that is, $5' \rightarrow 3'$. Nucleotides can only be added to a pre-existing free $3'\text{-OH}$ group on the elongating strand.

Consequently, at the replication fork, one strand (template with polarity $3' \rightarrow 5'$) experiences continuous synthesis, whereas the other strand (template with polarity $5' \rightarrow 3'$) experiences discontinuous synthesis.

DNA-dependent RNA polymerase catalyses the polymerisation in only one direction, that is, $5' \rightarrow 3'$. The strand that has the polarity $3' \rightarrow 5'$ acts as a template, and is referred to as the template strand.

The RNA transcript is complementary to the DNA template strand with Adenine ($A$) pairing with Uracil ($U$), and Cytosine ($C$) pairing with Guanine ($G$): $$\text{DNA Template:}\quad 3'\;\text{TACATGGCAAATATCCATTCA}\;5'$$ $$\text{RNA Transcript:}\quad 5'\;\text{AUGUACCGUUUAUAGGUAAGU}\;3'$$

Option (2) contains Thymine ($T$), which cannot exist in an RNA transcript. Option (1) has incorrect terminal bases. Option (3) is exact.

In eukaryotes, there are at least three RNA polymerases in the nucleus. RNA polymerase I transcribes rRNAs (28S, 18S, and 5.8S). RNA polymerase II transcribes precursor of mRNA (hnRNA). RNA polymerase III is responsible for transcription of tRNA, 5S rRNA, and snRNAs (small nuclear RNAs) (A -> IV).

- In bacteria, RNA polymerase associates transiently with termination factor ($\rho$, rho factor) to terminate transcription (B -> III). - Splicing involves removal of introns and joining of exons, catalyzed by the spliceosome formed of small nuclear ribonucleoproteins (snRNPs) (C -> I). - The TATA box is a conserved hexanucleotide sequence found within the core promoter region of eukaryotic genes where transcription factors bind (D -> II).

A-IV, B-III, C-I, D-II corresponds to Option (2).

Unlike prokaryotes where a single RNA polymerase transcribes all classes of RNA, eukaryotes possess at least three distinct RNA polymerases in the nucleus:

NCERT states verbatim: "The RNA polymerase I transcribes rRNAs (28S, 18S, and 5.8S), whereas the RNA polymerase III is responsible for transcription of tRNA, 5srRNA, and snRNAs (small nuclear RNAs). The RNA polymerase II transcribes precursor of mRNA, the heterogeneous nuclear RNA (hnRNA)."

Notice that while 28S, 18S, and 5.8S rRNAs are transcribed by RNA Pol I as a single polycistronic pre-rRNA in the nucleolus, 5S rRNA is uniquely transcribed outside the nucleolus by RNA Pol III.

The search for the genetic material followed three landmark historical stages:

NCERT states verbatim: "The unequivocal proof that DNA is the genetic material came from the experiments of Alfred Hershey and Martha Chase (1952). They worked with viruses that infect bacteria called bacteriophages."

Bacteria infected with viruses grown on radioactive phosphorus ($^{32}\text{P}$) were radioactive because DNA entered the host cells. Bacteria infected with viruses grown on radioactive sulfur ($^{35}\text{S}$) were not radioactive, confirming that protein did not enter bacteria.

The Human Genome Project utilized two contrasting sequencing paradigms:

NCERT states verbatim: "The methodologies involved two major approaches. One approach focused on identifying all the genes that are expressed as RNA (referred to as Expressed Sequence Tags (ESTs))."

ESTs are generated by reverse-transcribing cellular messenger RNAs into complementary DNAs (cDNAs) and sequencing short fragments, completely bypassing non-coding intergenic and intronic genomic DNA.

Ribosomes are ribonucleoprotein complexes consisting of structural ribosomal RNAs ($rRNAs$) and a diverse set of ribosomal proteins ($r$-proteins):

NCERT states verbatim in the section on Translation: "The ribosome consists of structural RNAs and about 80 different proteins. In its inactive state, it exists as two subunits; a large subunit and a small subunit."

These ~80 proteins stabilize the complex tertiary and quaternary folded architecture of the ribosomal RNAs and facilitate translation factor binding, while the peptidyl transferase catalytic activity itself is mediated by the 23S rRNA ribozyme.

RNA possesses a 2′-hydroxyl group on each ribose sugar ring, making the phosphodiester backbone catalytic and chemically labile, while uracil is less stable than thymine:

NCERT states verbatim under "RNA World / Properties of Genetic Material": - "RNA being unstable, mutates at a faster rate." - "Consequently, viruses having RNA genome and having shorter life span mutate and evolve faster."

Examples include influenza virus, HIV, and SARS-CoV-2, whose error-prone RNA-dependent RNA polymerases lack 3′ to 5′ proofreading, leading to rapid antigenic drift and antigenic shifts.

The lactose operon of $\textit{E. coli}$ consists of one regulatory gene and three polycistronic structural genes:

NCERT states verbatim: "The z gene codes for beta-galactosidase... The y gene codes for permease... The a gene encodes a transacetylase... The i gene codes for the repressor of the lac operon."

Matching C-IV and D-I immediately narrows to Option (3), completely ruling out Options 1, 2, and 4.

Every phosphodiester linkage in the DNA backbone carries a net negative charge at physiological pH due to ionization of the phosphate group ($PO_4^{3-}$):

In prokaryotes (e.g., $\textit{E. coli}$), lacking histones, the negatively charged DNA is held in large loops by positively charged non-histone polyamine proteins in a region termed the nucleoid. Statement I states "positively charged DNA is held with negatively charged proteins", which is a complete polarity inversion and unambiguously FALSE.

In eukaryotes, basic proteins rich in lysine and arginine residues (histones) form an octamer bearing positive charges on their side chains. Approximately 200 bp of negatively charged DNA wraps around this core:

The DNA strand with $3' \to 5'$ polarity acts as the template for RNA polymerase. The other strand with $5' \to 3'$ polarity has the sequence identical to mRNA (except $T$ instead of $U$) and is termed the coding strand:

NCERT states verbatim under Transcription Unit: "The other strand which has the polarity (5'->3') and the sequence same as RNA (except thymine at the place of uracil), is displaced during transcription. Strangely, this strand (which does not code for anything) is referred to as coding strand."

Option (1) has reversed $3' \to 5'$ polarity. Options (2) and (3) contain RNA Uracil residues and complementary antisense sequences. Only Option (4) has the correct 5′ to 3′ polarity with DNA Thymine residues.

Francis Crick formulated the fundamental paradigm of information transfer in molecular biology:

- (a): DNA replication (autocatalytic function). - (b): Transcription (synthesis of RNA from DNA template). - (c): Translation (synthesis of polypeptide chain specified by mRNA codons). - (d): Protein (the structural or catalytic end-product).

In prokaryotes (bacteria), a single DNA-dependent RNA polymerase catalyzes the transcription of all three types of RNA ($mRNA, tRNA, rRNA$). Although the core enzyme is capable of elongation, it requires specific accessory protein factors for initiation and termination:

- Option (1): Capping involves adding an unusual nucleotide—methyl guanosine triphosphate—to the $5'-end$ of $hnRNA$ (polyadenylation/tailing occurs at the $3'-end$). - Option (3): The strand that acts as a template during transcription is the template strand ($3' \to 5'$), whereas the coding strand ($5' \to 3'$) has the same sequence as the newly synthesized RNA (except thymine is replaced by uracil). - Option (4): The split-gene arrangement (interrupted by non-coding introns) is a diagnostic feature of eukaryotes; prokaryotes have continuous coding cistrons.

Human genomic DNA contains regions that do not code for proteins but consist of tandemly repeated nucleotide sequences:

Verbatim from the textbook: *"DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called as repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times."*

During density gradient centrifugation, bulk genomic DNA forms a major peak, while repetitive DNA separates out as distinct smaller peaks termed satellite DNA. The underlying DNA sequence category is repetitive DNA.

Unlike bacteria where a single RNA polymerase transcribes all RNAs, eukaryotic nuclei contain three distinct RNA polymerases showing complete division of labor:

Question asks for the specific role of RNA polymerase III. Option (2) explicitly lists "tRNA, 5s rRNA and snRNA", which is identical to the canonical NCERT text.

Erwin Chargaff established that in double-stranded DNA (dsDNA): 1. The molar ratio of Adenine (A) to Thymine (T) is always 1: $[A] = [T]$. 2. The molar ratio of Guanine (G) to Cytosine (C) is always 1: $[G] = [C]$. 3. The total percentage of all four nitrogenous bases equals 100%:

Given that Adenine makes up $30\%$ of the DNA: - Because Adenine strictly pairs with Thymine via two hydrogen bonds ($A = T$): $$\%T = \%A = 30\%$$ - The sum of A and T is: $$\%A + \%T = 30\% + 30\% = 60\%$$ - Therefore, the remaining fraction of the molecule comprises Guanine and Cytosine: $$\%G + \%C = 100\% - 60\% = 40\%$$ - Because Guanine pairs with Cytosine via three hydrogen bonds ($G \equiv C$): $$\%G = \%C = \frac{40\%}{2} = 20\%$$