31 Years NEET Previous Year Questions: Molecular Basis of Inheritance

• Humans have 23 pairs of chromosomes (46 total), with 22 pairs being autosomes and 1 pair of sex chromosomes (X and Y).
• Chromosomes carry genes, which are the functional units of heredity made up of DNA. Each chromosome contains a unique set of genes that determine various biological functions.
• Chromosome 1 is the largest human chromosome and contains the highest number of genes compared to other chromosomes in the human genome.
• Chromosome 1 is the largest of all human chromosomes and contains approximately 2,968 genes, making it the chromosome with the highest number of genes.
• Chromosome Y has the fewest ~231 genes.

A. Alfred Hershey and Martha Chase - IV. DNA as genetic material confirmation: Hershey and Chase conducted the famous "blender experiment" which confirmed that DNA, and not protein, is the genetic material in bacteriophages. 

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. 

The experiment used the T2 bacteriophage, a type of virus that infects bacteria. The bacteriophage consists of a protein coat and DNA.

B. Euchromatin - III. Loosely packed and light-stained: Euchromatin is a form of chromatin that is loosely packed and appears light under a microscope, allowing active gene transcription.

C. Frederick Griffith - I. Streptococcus Pneumoniae: Griffith's experiments involved Streptococcus pneumoniae bacteria, leading to the discovery of the phenomenon of transformation.

• The "Transforming Principle" is a term used to describe the substance responsible for transformation in bacteria, which was first identified by Frederick Griffith in 1928.
• Griffith's experiments involved two strains of the bacterium Streptococcus pneumoniae, a virulent smooth strain (S) and a non-virulent rough strain (R).

D. Heterochromatin - II. Densely packed and dark-stained: Heterochromatin is tightly packed chromatin that appears dark under a microscope and is generally transcriptionally inactive.

• Option B (Removal of introns and joining of exons) — splicing: Introns (non-coding sequences) are removed and exons are joined to form a continuous coding sequence that can be translated into protein. This is a key post-transcriptional processing event occurring in the nucleus.
• Option C (Addition of methyl group at 5' end of hnRNA) — 5' capping: A 7-methylguanosine cap is added to the 5' end of the primary transcript (hnRNA). This cap protects the RNA from degradation and helps ribosome recognition during translation.
• Option D (Addition of adenine residues at 3' end of hnRNA) — polyadenylation: A poly(A) tail of adenine residues is added at the 3' end; this increases RNA stability and facilitates nuclear export.
• Option A is incorrect because splicing occurs in the nucleus; pre-mRNA is not exported before splicing. Option E (base pairing of two complementary RNAs) is not a standard post-transcriptional processing event — it describes processes like RNA interference rather than processing of hnRNA to mature mRNA.

Explanation (Assertion–Reason format):

• (i) Assertion: Statement I correctly summarises the RNA world hypothesis: RNA can both store genetic information and catalyse reactions (ribozymes), but is chemically more reactive and less stable than DNA.
• (ii) Reason: Statement II is also correct: DNA, with its deoxyribose backbone and double-stranded complementary structure, is chemically more stable and allows repair mechanisms that preserve genetic integrity over time.
• (iii) Justification: Together, these points support the hypothesis that early life relied on RNA for both information and catalysis and that DNA later evolved as a more stable repository for genetic information, enabling accurate inheritance and reduced mutational load.

• Statement I — incorrect: tRNA interacts directly with mRNA during translation by recognising codons via its anticodon loop and delivering the correct amino acid. rRNA is an essential structural and catalytic component of ribosomes; it aligns mRNA and catalyses peptide bond formation, so it also interacts functionally with mRNA.
• Statement II — correct: RNA interference (RNAi) is a conserved eukaryotic mechanism for gene regulation and defence. Small RNAs such as siRNA and miRNA guide protein complexes (e.g., RISC) to complementary mRNAs to degrade them or block their translation, contributing to antiviral defence and regulation of endogenous genes.

• The genetic code refers to the set of rules by which information encoded in DNA or RNA is translated into proteins, the functional molecules in cells.
• Proteins are composed of amino acids, and the sequence of amino acids is determined by the sequence of nucleotides in the genetic material.
• The concept of the genetic code being made up of three nucleotides, known as codons, was first proposed by George Gamow, a physicist.
• George Gamow proposed the idea of the triplet code in 1954. He reasoned that combinations of three bases (4^3 = 64 codons) would be sufficient to code for the 20 common amino acids.
• This triplet idea anticipated the eventual elucidation of the genetic code, which indeed uses three-nucleotide codons to specify amino acids.

Other Options:

• Francis Crick: Francis Crick is well known for co-discovery of the double helix and later contributions to deciphering the genetic code, but the specific triplet-code hypothesis was proposed by Gamow.
• Jacques Monod: Known for work on gene regulation (lac operon), not for proposing the triplet code.
• Franklin Stahl: Known for the Meselson–Stahl experiment on DNA replication, not the genetic code hypothesis.

• The nucleosome is the basic structural unit of chromatin in eukaryotic cells and helps package DNA efficiently.
• A nucleosome consists of DNA wound around a core of histone proteins.
• Histones are rich in the basic amino acid residues lysine and arginine. These positively charged residues bind to the negatively charged DNA backbone, facilitating tight packing.
• Histones form an octamer (two each of H2A, H2B, H3 and H4) around which DNA is wrapped, forming the nucleosome.
• A typical nucleosome protects ~200 bp of DNA, including the core ~147 bp wrapped directly around the histone octamer.

Transcription is the process by which a DNA sequence is copied into RNA. It is carried out by RNA polymerase and includes three main stages: initiation, elongation, and termination.

• Initiation: RNA polymerase binds to promoter; in prokaryotes sigma factor (σ) helps promoter recognition and then dissociates.
• Elongation: RNA polymerase synthesises RNA in 5' → 3' direction by adding ribonucleotides complementary to the DNA template.
• Termination: In many prokaryotic genes, termination requires the ρ (rho) factor, a helicase-like protein that recognises specific sequences on the nascent RNA and causes RNA polymerase to dissociate, releasing the transcript.

Other Options:

• α (Alpha): Alpha subunits are structural components of bacterial RNA polymerase and help enzyme assembly and promoter binding, but they are not termination factors.
• σ (Sigma): Sigma factors are required for initiation, not termination; they help RNA polymerase locate promoters and then dissociate.
• γ (Gamma): Gamma is not associated with bacterial transcription termination.

X-ray diffraction images (notably Franklin's Photo 51) provided evidence of DNA's helical structure and dimensions; Watson and Crick used this information to build the double helix model.

• The genetic code is a triplet code (three bases per codon).
• It is nearly universal across organisms.
• It is degenerate: several amino acids are encoded by more than one codon.
• However, the code does not have punctuation marks; translation proceeds codon by codon without extra separators. Start and stop codons signal initiation and termination but are not punctuation in the ordinary sense.

The correct answer is Option C: Permease.
Lactose, a disaccharide sugar composed of glucose and galactose, needs particular mechanisms to be transported into bacterial cells for metabolism. Among the options provided, permease is the protein responsible for this function. Specifically, in the case of bacterial cells such as Escherichia coli, the lactose permease enzyme plays a crucial role.
Lactose permease is encoded by the lacY gene, which is part of the lac operon. The lac operon is a famous example of gene regulation in bacteria. When lactose is present outside the cell, lactose permease facilitates its transport into the cell across the cell membrane. Once inside, lactose can be utilized as a source of energy and carbon.

Once lactose is inside the bacterial cell:
It is broken down by the enzyme β-galactosidase into glucose and galactose. This enzyme is encoded by the lacZ gene, which is a different component of the lac operon.

We can rule out the other options because:
Beta-galactosidase (Option A) is involved in the hydrolysis of lactose into glucose and galactose but not in its transport.
Acetylase (Option B) refers to enzymes involved in the addition of acetyl groups to substrates, unrelated to sugar transport.
Polymerase (Option D) are enzymes that synthesize RNA and DNA, thus also unrelated to transporting sugars like lactose.
Thus, to facilitate the transport of lactose across the cell membrane into the bacterial cell, lactose permease (Option C) is required, making it the correct choice.

The goal is to correctly match the terms in List I with the descriptions in List II. Let's analyze and match each term from List I with its appropriate partner in List II:

List I:
A. RNA polymerase III: This enzyme is primarily responsible for transcribing DNA to synthesize tRNA, 5S rRNA, and other small RNAs.
B. Termination of transcription: In prokaryotes, specific termination factors such as the Rho factor are involved in stopping transcription. In eukaryotes, different mechanisms and sequences are used.
C. Splicing of Exons: The process involving the removal of introns and joining of exons during mRNA processing. Small nuclear ribonucleoproteins (snRNPs) are critical components in this process.
D. TATA box: A DNA sequence within the promoter region, which is crucial for forming the transcription initiation complex.

List II:
I. snRNPs: Small nuclear ribonucleoproteins involved in mRNA splicing.
II. Promoter: A region of DNA that initiates transcription of a particular gene, typically containing sequences like the TATA box.
III. Rho factor: A protein essential for terminating transcription in prokaryotes.
IV. SnRNAs, tRNA: Molecules transcribed primarily by RNA polymerase III.

Matching these descriptions:
A (RNA polymerase III) matches with IV (SnRNAs, tRNA).
B (Termination of transcription) matches with III (Rho factor).
C (Splicing of Exons) matches with I (snRNPs).
D (TATA box) matches with II (Promoter).
Therefore, the correct matches according to options listed are: Option D: A-IV, B-III, C-I, D-II

DNA-dependent RNA polymerase is an enzyme responsible for transcribing DNA into RNA. During transcription, the RNA polymerase reads the template strand of DNA and synthesizes a complementary RNA strand. A key point in this process is that RNA polymerase builds RNA by replacing thymine (T) with uracil (U).

The provided DNA template is: 3'TACATGGCAAATATCCATTCA5'
To find the correct RNA sequence, we need to identify the complementary base for each base in the template strand while considering RNA bases. Remember, in RNA:
A (Adenine) pairs with U (Uracil)
T (Thymine) pairs with A (Adenine)
C (Cytosine) pairs with G (Guanine)
G (Guanine) pairs with C (Cytosine)

The complementary RNA sequence to the DNA template is generated as follows:
3'T --> 5'A
3'A --> 5'U
3'C --> 5'G
3'G --> 5'C
3'T --> 5'A
3'A --> 5'U
3'T --> 5'A
3'G --> 5'C
3'G --> 5'C
3'C --> 5'G
3'A --> 5'U
3'A --> 5'U
3'T --> 5'A
3'A --> 5'U
3'T --> 5'A
3'C --> 5'G
3'C --> 5'G
3'A --> 5'U
3'T --> 5'A
3'T --> 5'A
3'C --> 5'G'

This constructs the RNA sequence: 5'AUGUACCGUUUAUAGGUAAGU3'
Thus, Option A correctly represents the RNA sequence transcribed by DNA-dependent RNA polymerase from the given DNA template:
Option A: 5'AUGUACCGUUUAUAGGUAAGU3'

The correct statement regarding the process of replication in E.coli is found in Option D: "The DNA dependent DNA polymerase catalyses polymerization in 5' → 3' direction".
To elaborate, replication in E.coli involves the synthesis of new DNA strands from a DNA template. This synthesis is catalyzed by an enzyme known as DNA polymerase. DNA polymerases are key enzymes that add nucleotides to the growing DNA strand during replication. However, the key characteristic of these enzymes is their directionality. DNA polymerases can only add nucleotides to the 3' end of the DNA strand, thereby synthesizing new DNA in the direction of 5' → 3'.
This directional limitation is due to the structure of deoxyribonucleotide triphosphates (dNTPs) that are used as substrates by the enzyme. Each dNTP has a 5' phosphate group, a 3' hydroxyl group, and a nitrogenous base. The formation of DNA strands occurs via the formation of phosphodiester bonds between the 3' hydroxyl group of one nucleotide and the phosphate group at the 5' position of the next nucleotide. Therefore, nucleotides can only be added to the 3' end of the growing strand.
The other options are incorrect for the following reasons:
Option A incorrectly states the direction of synthesis as  which is impossible with the mechanics of DNA polymerases.
Option B refers to DNA dependent RNA polymerase. This enzyme is indeed involved in transcription (synthesizing RNA from DNA), not replication, and synthesizes RNA in the  direction.
Option C incorrectly states that DNA polymerase can synthesize in both directions, which contradicts the inherent directional nature of this enzyme
Therefore, understanding the precise enzyme mechanics and functionality helps in recognizing that Option D is the correct description of how DNA dependent DNA polymerase facilitates DNA replication in E.coli.

To solve this matching question, let's discuss each scientist(s) and their contribution:
A. Frederick Griffith: Known for discovering the "transforming principle," which showed that a substance from dead bacteria could genetically transform living bacteria. This process is called transformation. The correct match for Frederick Griffith is III. Transformation.
B. Francois Jacob & Jacques Monod: They are famous for their work on the lac operon, a set of genes involved in lactose metabolism in bacteria. Their study elucidated how genes are regulated and expressed in cells. The correct match for Francois Jacob and Jacques Monod is IV. Lac operon.
C. Har Gobind Khorana: Known for his research on the genetic code and its role in protein synthesis. Khorana was one of the scientists who elucidated how the sequence of nucleotides in nucleic acids is translated into protein sequences. The correct match for Har Gobind Khorana is I. Genetic code.
D. Meselson & Stahl: Famous for their experiment confirming the semi-conservative mechanism of DNA replication, where each new DNA molecule consists of one old strand and one newly synthesized strand. The correct match for Meselson and Stahl is II. Semi-conservative mode of DNA replication.

Comparing this information with the options provided:
Option A: A-III, B-II, C-I, D-IV (Incorrect: B does not match II)
Option B: A-III, B-IV, C-I, D-II (Correct: All matches are accurate)
Option C: A-II, B-III, C-IV, D-I (Incorrect: A, B, C, and D do not match correctly)
Option D: A-IV, B-II, C-II, D-III (Incorrect: A, C, and D do not match correctly)

Therefore, the correct answer is Option B.

A transcription unit in DNA is critical for the process of transcription, wherein a particular segment of DNA is copied into RNA (especially mRNA) by the enzyme RNA polymerase. This unit is composed of sequences that include both coding regions, which are directly transcribed into RNA, and regulatory regions, which ensure that transcription is initiated and terminated at the correct locations on the DNA.
The correct answer is: Option D: Promotor, Structural gene, Terminator
Here's a detailed explanation of each component:
Promoter: The promoter is a sequence in DNA that signals the RNA polymerase to start transcription. It is located at the upstream end (5' end) of the gene. Promoters are essential for transcription initiation and are typically found just before the genes they regulate.

Structural gene: This region of the transcription unit is actually expressed or translated into protein (or functional RNA), depending on the kind of gene. These genes contain the functional sequences that are copied during the transcription process.

Terminator: The terminator is found at the downstream end (3' end) of the transcription unit and includes sequences that signal the RNA polymerase enzyme to stop transcription. This ensures that the newly synthesized RNA contains only the necessary genetic message.

The other options contain components that do not accurately define the typical structure of a transcription unit:

Option A mixes regulatory proteins and DNA regions, which does not accurately represent the structural components of a transcription unit.
Option B includes "transposons" which are genetic elements that can move around within the genomes but are not typically part of the transcription unit.
Option C again refers to regulatory proteins (inducer and repressor) along with structural genes, confusing the functions of proteins and DNA regions.
Therefore, Option D correctly represents the standard components of a transcription unit in the context of gene transcription in DNA.

The question relates to the lac operon model in E. coli, which is a well-studied example of gene regulation. In this model :
- Gene 'z' codes for beta-galactosidase (breaks down lactose into glucose and galactose)
- Gene 'y' codes for permease (transports lactose into the cell)
- Gene 'a' codes for transacetylase (may help in lactose metabolism, but its exact role is unclear)
- Gene 'i' codes for the repressor protein (binds to the operator to prevent transcription)
So, matching the List I (genes) with List II (proteins they code for) :
- A (Gene 'a') matches with II (Transacetylase)
- B (Gene 'y') matches with III (Permease)
- C (Gene 'i') matches with IV (Repressor protein)
- D (Gene 'z') matches with I (Beta-galactosidase)
Therefore, the correct option is : Option A : A-II, B-III, C-IV, D-I

• In prokaryotes, the negatively charged DNA is held with some positively charged proteins in a region termed as nucleoid.
• In eukaryotes, the negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome.

Option (C) is the correct answer because in recombinant DNA technology the separated DNA fragments can be visualised only after staining the DNA with a substance known as ethidium bromide followed by exposure to U.V. radiation. You can see bright orange coloured bands of DNA in an ethidium bromide stained gel exposed to U.V. light.

• The first unequivocal proof that DNA is the genetic material came from the experiments conducted by Alfred Hershey and Martha Chase in 1952. They used bacteriophages (viruses that infect bacteria) and radioactively labeled the protein and DNA of the phage separately in two different sets of experiments. They found that it was the DNA, not the protein, of the phage that was injected into the bacteria and carried the genetic information necessary for the production of new phage particles.
• Avery, Macleoid and McCarty gave the biochemical characterisation of Transforming Principle.
• The transformation experiments by using Pneumococcus was conducted by Frederick Griffith.
• Wilkins and Franklin produced X-ray diffraction data of DNA.

So, the correct answer is : Option B : Alfred Hershey and Martha Chase.

• In eukaryotes there are three major types of RNA polymerases.
• RNA polymerase I transcribes : 5.8S, 18S, 28S rRNAs 
• RNA polymerase II transcribes : hnRNAs (precurssor of mRNA) 
• RNA polymerase III transcribes : tRNAs, ScRNA, 5S rRNA and snRNA

Expressed Sequence Tags (ESTs) are short sub-sequences of a cDNA sequence. They may identify expressed genes, so they are derived from mRNA which is transcribed from expressed genes. They serve as a kind of "tag" or marker for identifying the gene from which it was transcribed. Therefore, ESTs represent genes that are expressed as RNA. The other options listed do not accurately describe what ESTs are.

Sequencing the whole set of genome that contained all the coding and non-coding sequences and later assigning different regions in the sequence with functions is called sequence annotation.

• Option (b) is the correct answer as the source of the complementary RNA for RNAi could be mobile genetic elements (transposons) that replicate via an RNA intermediate.
• Option (c) is incorrect as autoradiography usually follows hybridisation.
• Option (a) is incorrect because polymerase chain reaction is used to make copies of the DNA sample and does not need transposons.
• Option (d) is incorrect because transposons are not required during gene sequencing.

From 10 parent E.coli cells

Therefore, after 60 minutes, 60 E.coli cells will have DNA totally free from ¹⁵N.

In lac operon, z gene codes for β-galactosidase.
Transacetylase, permease and repressor protein are coded by genes 'a', 'y' and 'i' respectively.

• In the process of translation, the mRNA codon pairs with the tRNA anticodon to bring the correct amino acid to the growing polypeptide chain. The pairing between the codon and anticodon follows the complementary base-pairing rules, with adenine (A) pairing with uracil (U), and guanine (G) pairing with cytosine (C).
• In this case, the mRNA codon is given as 5' UAC 3'. To determine the tRNA anticodon sequence, we must follow the base-pairing rules:
• The U (uracil) in the codon pairs with A (adenine) in the anticodon.
• The A (adenine) in the codon pairs with U (uracil) in the anticodon.
• The C (cytosine) in the codon pairs with G (guanine) in the anticodon.
• Therefore, the anticodon sequence on the tRNA would be 5' AUG 3'.

Genetic material of –

• Bacteriophage ∅ × 174 contains 5386 nucleotides
• Bacteriophage lambda contains 48502 base pairs
• Escherichia coli contains 4.6 × 10⁶ base pairs
• Haploid content of human DNA contains 3.3 × 10⁹ base pairs

The DNA-dependent DNA polymerases catalyse polymerisation only in one direction, that is 5' → 3'. This creates some additional complications at the replicating fork. Consequently, on one strand (the template with polarity 3' → 5'), the replication is continuous, while on the other (the template with polarity 5' → 3'), it is discontinuous. 

In Iac operon,

• The i gene codes for repressor protein.
• The z gene codes for β-galactosidase.
• The y gene codes for permease and the a gene codes for transacetylase.