Molecular Basis of Inheritance

In human genome, Chromosome 1 has the highest number of genes, i.e., 2968.

The unequivocal proof that DNA is the genetic material came from the experiment of Alfred Hershey and Martha Chase.

Euchromatin are lightly stained region with loosely packed chromatin fibre.

Frederick Griffith performed series of experiments by selecting the different strains of Streptococcus pneumoniae.

Heterochromatin are darkly stained region with tightly packed chromatin fibre.

The process of copying genetic information from one strand of the DNA into RNA is known as transcription. It occurs in the cytoplasm with the help of transcripting enzyme.

Transport of pre-mRNA to cytoplasm prior to splicing is a part of transcription.

The primary transcript is converted into functional mRNA after post transcriptional processing involves 3 steps as follows-

$\bullet$ Modification of $5^{\prime}$ end by capping,

$\bullet$ Tailing,

$\bullet$ Splicing.

Base pairing of two complementary RNA is not on event of post-transcription. Hence, statements B, C, D are post-transcriptional modification events in eukaryotic cell.

In RNA world, RNA was the first genetic material as there are enough evidences to suggest that essential life processes (such as metabolism, translation, splicing, etc) evolved around RNA. RNA used to act as a genetic material as well as catalyst (there are some important biochemical reaction in living systems that are catalysed by RNA catalysts not by protein enzymes) so, statement I is correct statement II is also correct as DNA being double stranded and having complementary strands further resists changes by evolving a process of repair.

Both transfer RNAs and ribosomal RNA interact with mRNA.

RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence.

The idea that each amino acid is specified by a triplet of nucleotides was first proposed by George Gamow (Option C).

The key player in prokaryotic transcription termination is the ρ (rho) factor.

• ρ–dependent termination

– Rho is an RNA-helicase that binds a rut site on the nascent mRNA.

– It moves along the transcript, catches up with RNA polymerase and unwinds the RNA–DNA hybrid, releasing the mRNA.

By contrast:

• σ (sigma) is needed for initiation, not termination.

• α and γ are subunits of RNA polymerase (α for assembly/stability, γ isn’t a termination factor).

Answer: Option A (ρ)

The correct answer is:

Option D
ori site

Explanation:

The specific DNA sequence in a chromosome that is responsible for initiating replication is known as the "origin of replication," typically abbreviated as "ori." The ori site is a crucial element in DNA replication because it is the point where the replication machinery assembles and begins the process of copying the DNA. In prokaryotes like E. coli, there is usually a single ori site, whereas eukaryotes have multiple ori sites along their chromosomes to ensure timely and efficient replication.

Let's summarize each option:

Option A - Recognition sequence: This refers to a specific sequence of nucleotides that is recognized by proteins, such as enzymes, which can bind to it and perform a function. While these sequences are important, they are not specifically involved in initiating DNA replication.

Option B - Cloning site: A cloning site refers to a location within a plasmid or other vector where DNA can be inserted for cloning purposes. This is not related to the initiation of DNA replication.

Option C - Restriction site: A restriction site is a specific sequence of nucleotides recognized and cut by a restriction enzyme. These sites are important in molecular biology for cutting and manipulating DNA but are not involved in the initiation of DNA replication.

Option D - ori site: The origin of replication (ori site) is indeed the specific DNA sequence where replication begins. This site is essential for the replication of DNA and marks where the replication machinery assembles to start the process of copying DNA.

Let's analyze the given statements regarding RNA polymerase in prokaryotes:

Statement I: In prokaryotes, RNA polymerase is capable of catalysing the process of elongation during transcription.

In prokaryotes, RNA polymerase is indeed responsible for transcribing RNA from a DNA template. It performs various functions during transcription, including the elongation phase, where it synthesizes RNA by adding nucleotides to the growing RNA strand. Hence, Statement I is true.

Statement II: RNA polymerase associates transiently with the ‘Rho’ factor to initiate transcription.

This statement is incorrect. In prokaryotes, RNA polymerase associates with the sigma factor ($\sigma$) to initiate transcription. The sigma factor helps the RNA polymerase to recognize the promoter regions of the DNA. The Rho factor ($\rho$), on the other hand, is involved in the termination of transcription, not the initiation. Thus, Statement II is false.

Considering the above explanations, the correct answer is:

Option A: Statement I is true but Statement II is false

The correct answer is Option C: Both Statement I and Statement II are correct.

Let's break down why:

Statement I: In eukaryotes there are three RNA polymerases in the nucleus in addition to the RNA polymerase found in the organelles.

This statement is correct. Eukaryotic cells possess three main nuclear RNA polymerases:

• RNA polymerase I: This polymerase is responsible for transcribing ribosomal RNA (rRNA), which is essential for protein synthesis.
• RNA polymerase II: This polymerase transcribes messenger RNA (mRNA), which carries genetic information from DNA to ribosomes for protein production. It also transcribes small nuclear RNAs (snRNAs) involved in splicing.
• RNA polymerase III: This polymerase transcribes transfer RNA (tRNA), which brings amino acids to the ribosome during translation, and other small RNAs like 5S rRNA.

In addition to these nuclear RNA polymerases, eukaryotes also have RNA polymerases in their organelles, like mitochondria and chloroplasts. These organellar RNA polymerases have different structures and functions compared to the nuclear RNA polymerases.

Statement II: All the three RNA polymerases in eukaryotic nucleus have different roles.

This statement is also correct. As explained above, each of the three nuclear RNA polymerases has a specific function:

• RNA polymerase I transcribes rRNA.
• RNA polymerase II transcribes mRNA and snRNAs.
• RNA polymerase III transcribes tRNA and other small RNAs.

Therefore, all three RNA polymerases in the eukaryotic nucleus have distinct roles in the process of gene expression. Their specialized functions ensure the proper synthesis of different types of RNA molecules essential for cellular processes.

Let's examine both statements to determine their validity.

Statement I: In the lac operon, the z gene codes for beta-galactosidase which is primarily responsible for the hydrolysis of lactose into galactose and glucose.

Explanation: This statement is correct. The lac operon in E. coli consists of three structural genes: lacZ, lacY, and lacA. The lacZ gene encodes for the enzyme beta-galactosidase, which catalyzes the hydrolysis of lactose into glucose and galactose.

Statement II: In addition to lactose, glucose or galactose can also induce lac operon.

Explanation: This statement is incorrect. The lac operon is specifically induced by lactose or allolactose (an isomer of lactose), which act as inducers by deactivating the lac repressor. Glucose, on the other hand, represses the lac operon through catabolite repression, a mechanism that ensures the operon is not expressed when glucose is available. Galactose does not serve as an inducer for the lac operon.

Based on the explanations above, we can conclude that:

Option A: Statement I is true but Statement II is false

Thus, the correct answer is:

Option A

The correct answer is option (A) because Statement I is correct which states that RNA interference takes place in all eukaryotic organisms as method of cellular defense and Statement II is incorrect which states that RNAi involved the silencing of a specific mRNA due to a complementary ssRNA molecule that binds and prevents translation of mRNA.

In statement II, it should be complementary dsRNA molecule that is involved in silencing of a specific mRNA.

Options (B), (C) and (D) are incorrect as they represent wrong answer.

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.

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 correct statement regarding the process of replication in E.coli is found in Option D: "The DNA dependent DNA polymerase catalyses polymerization in 5' $\rightarrow$ 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' $\rightarrow$ 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 $3^{\prime} \rightarrow 5^{\prime}$, 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 $5^{\prime} \rightarrow 3^{\prime}$ 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.

To correctly match List I with List II concerning medical conditions and their associated chromosomes, it's essential to have some understanding of genetics and chromosome abnormalities related with each condition. Here's the correct matching based on that information:

• Down's syndrome: This genetic disorder is characterized by an extra copy (trisomy) of the 21$^\mathrm{st}$ chromosome, which makes it associated with 21$^\mathrm{st}$ chromosome (III).
• $\alpha$-thalassemia: This condition is related to a mutation or deletion in the alpha globin genes located on the 16$^\mathrm{th}$ chromosome (IV).
• $\beta$-thalassemia: This is a blood disorder caused due to mutations in the beta globin gene found on the 11$^\mathrm{th}$ chromosome (I).
• Klinefelter's syndrome: This syndrome arises from the presence of an extra X chromosome in males (usually XXY), so it is related to abnormality in the 'X' chromosome (II).

Given these matches:

• A - III
• B - IV
• C - I
• D - II

The correct option based on the details above is:

Option C: A-III, B-IV, 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 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

The correct answer is :

Option B : Chromosome 1

While the Human Genome Project, which aimed to sequence all the genes in the human genome, was officially completed in 2003, the sequencing and refinement of some parts of the genome continued past that date. Chromosome 1, the largest human chromosome, was the last to be fully sequenced, with this task not being completed until May 2006. This marked a significant milestone in human genetics.

The component that binds to the operator region of an operon and prevents RNA polymerase from transcribing the operon is the Repressor protein. Repressor proteins are produced by a regulator gene and can bind to an operator site to block transcription of downstream structural genes.

Therefore, the correct answer is Option C : Repressor protein.

Option C : Both Statement I and Statement II are true

Explanation :

Statement I is correct. Transcription is the process of copying genetic information from one strand of the DNA into RNA. The resulting RNA molecule is complementary to the DNA strand from which it was transcribed.

Statement II is also correct. A transcription unit in DNA is indeed defined by three main regions: the promoter, the structural gene, and a terminator.

1. 1. The promoter is a DNA sequence that defines where transcription of a gene by RNA polymerase begins.
2. 2. The structural gene contains the actual genetic information that is being transcribed.
3. 3. The terminator is a sequence of nucleotides that signals the end of transcription.

So, both the statements are true.

The experiment with ³²P and ³⁵S labelled phages for demonstrating that DNA is the genetic material was conducted by A. D. Hershey and M. J. Chase. So the correct answer is Option B.

The correct answer is :

Option C : Both Statement I and Statement II are true

Here's why :

Statement I : RNA is indeed generally less stable than DNA and mutates at a faster rate. This is due to the presence of a hydroxyl group on the 2' carbon of RNA, which makes it more susceptible to hydrolysis and other chemical reactions. This increased rate of mutation can be beneficial for RNA viruses, for example, as it allows them to rapidly adapt to new environments or hosts.

Statement II : RNA, particularly mRNA (messenger RNA), can directly code for the synthesis of proteins. In the process of gene expression, DNA is first transcribed into RNA, which is then translated into protein. Hence, changes in the RNA sequence can directly affect the resultant protein and thus easily express different characters.

Sickle cell anemia is caused by a single point mutation in the gene for the beta globin chain of hemoglobin.

(A) is correct because the mutation occurs in the gene for beta globin.

(B) is incorrect because, in sickle cell anemia, there is a change in the beta globin chain where valine is substituted for glutamic acid, not lysine.

(C) is correct because sickle cell anemia is caused by a single point mutation, which is a change in a single nucleotide in the DNA.

(D) is incorrect because the statement does not accurately describe the genetic mutation. In the normal gene, the DNA code is GAG, which codes for glutamic acid, but in sickle cell anemia, this is changed to GTG, which codes for valine. This does not involve replacing uracil (U) with adenine (A) as the statement suggests.

So, the correct answer is Option D : (A) and (C) only.

The correct answer is :

Option B : Lactose

Explanation :

In bacterial genetics, the lac operon is an operon (a functioning unit of genomic DNA containing a cluster of genes under the control of a single regulatory signal or promoter) required for the transport and metabolism of lactose in E. coli and some other enteric bacteria.

When lactose is present, it acts as an inducer for the lac operon. This means that lactose, when present, triggers the activation of the lac operon, allowing the bacteria to produce the necessary enzymes (beta-galactosidase, lactose permease, and thiogalactoside transacetylase) to metabolize lactose. The lac operon is a classic example of a gene system subject to both positive and negative control.

Option B : (A) and (B) only

Explanation : In the Hershey and Chase experiments, they used bacteriophages (viruses that infect bacteria) and grew one batch in the presence of radioactive phosphorus (P³²) and another batch in the presence of radioactive sulfur (S³⁵).

Phosphorus is a component of DNA but not of protein. Therefore, any virus particles grown in the presence of radioactive phosphorus would have radioactive DNA, which is statement (A).

Sulfur is found in proteins (in the amino acids methionine and cysteine) but not in DNA. Therefore, any virus particles grown in the presence of radioactive sulfur would have radioactive protein, which is statement (B).

Statements (C), (D), and (E) are incorrect because they do not align with the components of DNA and proteins, and where sulfur and phosphorus are found in these molecules.

The Hershey and Chase experiments were key in establishing that DNA, not protein, is the genetic material. When these radioactive viruses infected bacteria, they found that the radioactive DNA (from phosphorus) was transferred to the bacteria, not the radioactive protein (from sulfur).

The correct answer is Option D : (C) and (D) only.

The genetic code is unambiguous and specific, meaning each codon specifies only one amino acid. Also, the genetic code is nearly universal as the same codons specify the same amino acids in all organisms with a few minor exceptions. The code is not palindromic, and UGA does not act as an initiator codon; it is a stop codon in most organisms.

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 A : 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

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.

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.

In lac operon, z gene codes for $\beta$-galactosidase.

Transacetylase, permease and repressor protein are coded by genes 'a', 'y' and 'i' respectively.

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

In Iac operon,

The i gene codes for repressor protein.

The z gene codes for $\beta$-galactosidase.

The y gene codes for permease and the a gene codes for transacetylase.

Genetic material of –

$\bullet$ Bacteriophage $\phi$ $\times$ 174 contains 5386 nucleotides

$\bullet$ Bacteriophage lambda contains 48502 base pairs

$\bullet$ Escherichia coli contains 4.6 $\times$ 10⁶ base pairs

$\bullet$ Haploid content of human DNA contains 3.3 $\times$ 10⁹ base pairs

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'.

According to Chargaff’s rule, amount of Adenine (A) is equal to thymine and amount of cytosine (c) will be equal to that of Guanine.

So,

A = T = 30% + 30% = 60%

C = G = 20% + 20% = 40%

So, T and G will be 30% and 20% 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.

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

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.

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.

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.

Number of base pairs $\times$ distance between 2 consecutive base pairs = Length of DNA molecule

x . 0.34 $\times$ 10^$-$9 m = 1.1 m

x = ${{1.1} \over {0.3 \times {{10}^{ - 9}}}}$

= 3.6 $\times$ 10⁹

$\simeq$ 3.3 $\times$ 10⁹ bp

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.

From 10 parent E.coli cells

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