31 Years NEET Previous Year Questions: Molecular Basis of Inheritance - 1 - NEET MCQ

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.

The two chains of DNA are coiled in a right-handed helicle fashion. The pitch of the helix is 3.4 nm and has a diameter of 2 nm. In the B-form of DNA, the distance between a bp in a helix is approximately 0.34 nm which means there are about 10 nucleotides present in a complete turn.
The total length of DNA molecule in nm = 1.1 × 10⁹ nm.
The approximate number of base pairs = 1.1 × 10⁹/ 0.34
= 3.3 × 10⁹ 

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

When the small subunit of ribosome encounters an mRNA, the process of translation of the mRNA to protein begins. This process is followed by the binding of bigger/larger subunit.
t-RNA is activated by the addition of amino acid prior to the attachment or ribosome, in the first phase.

• 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

Polymorphism in DNA sequence is the basis of genetic mapping of human genome as well as of DNA finger printing.

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. 

• Heterochromatin is transcriptionally inactive. A typical nucleosome contains 200 bp of DNA helix.
• Euchromatin is the loosely packed chromatin region.
• The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome. Histones are rich in basic amino acid residues lysine and arginine.

• As the product of 'i' gene binds with the operator region and blocks the transcription and translation of z, y and a genes.
• It's product is prevented from binding to the operator by attaching it with the inducer. As the inducer can not no more capable of binding with the repressor, thus, in all the cases, operator always gets attached with the repressor thereby preventing the transcription and transmission of z, y and a.
• Even in the presence of lactose, transcription and translation of z, y and a would not occur.

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.

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).

Hence, the correct option is D.
NCERT Reference: Page no. 111 of topic “6.5.3 Types of RNA and the process of Transcription” of chapter 6.

RNA polymerase holoenzyme binds to the promoter, unwinds DNA (open complex) and forms phosphodiester links between the initiating nucleotides. DNA polymerase, DNA ligase & DNA helicase are involved in the process of replication and not transcription.

Hence, the correct option is B.
NCERT Reference: Page no. 109 and 110 of topic “6.5.3 Types of RNA and the process of Transcription” of chapter 6.

As genetic code is nearly universal means almost all organisms (from a virus, bacteria to a tree or human being) will have amino acids coded by the same kind of codons as given in checkerboard. So, this property is utilised to produce human insulin using bacteria.

Hence, the correct option is D.
NCERT Reference: Page no. 111 of topic “6.6 GENETIC CODE” of chapter 6.
NOTE: This question is based on the basic understanding of the topic genetic code.

Semiconservative replication of DNA was proved by the work of Matthew Meselson and Franklin Stahl (1958) using bacterium Escherichia coli. 

Hence, the correct option is B.
NCERT Reference: Page no. 104 of topic “6.4 REPLICATION” of chapter 6.

• The process in which the DNA is replicated and two copies are synthesized is called DNA replication.
• It involves the formation of a replication fork. The two strands of the DNA helix separate and the new strands are formed on the original strands, known as the template strands.
• The strand which is synthesized in 3'-5' direction is the leading strand and the strand which is synthesized in the opposite direction in 5'-3' is the lagging strand which contains the Okazaki fragments as its growth is discontinuous. This strand is synthesized away from the replication fork. 

Hence, the correct option is C.
NCERT Reference: Page no. 106 of topic “6.4 REPLICATION” of chapter 6.