Q.Given below is the sequence of coding strand of DNA in a transcription unit: 3'-AATGCAGCTATTAGG-5'. Write the sequence of

Let’s go through each statement one by one, checking it against what actually happens in molecular biology.

1. Statement I: “DNA polymerase has capability of catalysing the process of elongation of polypeptide chain”

   This is false. DNA polymerase is an enzyme that synthesizes DNA — it adds deoxyribonucleotides to a growing DNA strand during replication or repair. The elongation of a polypeptide chain (protein synthesis) is catalysed by the ribosome (specifically, the peptidyl transferase activity of the rRNA in the large subunit). DNA polymerase has nothing to do with proteins. So I is wrong.

2. Statement II: “In splicing, introns are removed and exons are joined”

   This is correct. In eukaryotic gene expression, the primary transcript (pre-mRNA or hnRNA) contains both introns (non-coding) and exons (coding). Splicing is the process where introns are cut out and exons are ligated together to form the mature mRNA. So II is right.

3. Statement III: “In capping, methyl guanosine triphosphate is added to the 5' end of hnRNA” …

Understanding the history of major discoveries in molecular biology, along with the scientists who made them, is fundamental. Many key concepts in genetics and molecular biology are named after or strongly associated with specific researchers and their groundbreaking experiments. This question tests your knowledge of these historical milestones.

Let's evaluate each option:

1. Option (A): Genetic nature of DNA $\rightarrow$ Frankel-Conrat

   • The "genetic nature of DNA" refers to the discovery that DNA is the hereditary material responsible for transmitting genetic information from one generation to the next. This was established through a series of experiments:
     • Frederick Griffith (1928): Demonstrated transformation in bacteria, suggesting a "transforming principle."
     • Avery, MacLeod, and McCarty (1944): Identified DNA as the transforming principle.
     • Hershey and Chase (1952): Provided definitive proof that DNA, not protein, is the genetic material in bacteriophages.
   • Heinz Fraenkel-Conrat (along with B. Singer) conducted experiments in 1956 with the Tobacco Mosaic Virus (TMV). They showed that in TMV, RNA, not DNA, is the genetic material. They separated the RNA and protein components of the virus, recombined them, and demonstrated that the type of progeny virus produced was determined by the RNA, not the protein coat.
   • Therefore, Frankel-Conrat's work established RNA as the genetic material in some viruses, not the general genetic nature of DNA. This option is incorrect.

2. Option (B): In vitro DNA synthesis $\rightarrow$ Hanning

   • "In vitro DNA synthesis" refers to the artificial synthesis of DNA in a test tube, outside of a living organism.
   • Arthur Kornberg was awarded the Nobel Prize in 1959 for his discovery of the mechanisms in the biological synthesis of deoxyribonucleic acid. He isolated and characterized DNA polymerase I, the enzyme responsible for synthesizing new DNA strands from a DNA template and nucleotide precursors in vitro.
   • Elisabeth Hanning (1904) is known for her pioneering work in plant tissue culture, specifically the in vitro culture of immature plant embryos (embryo culture). Her work was foundational for plant biotechnology.
   • Therefore, Hanning is not associated with in vitro DNA synthesis. This option is incorrect.

3. Option (C): Double helical model of RNA $\rightarrow$ Watson and Crick

   • James Watson and Francis Crick are famously known for proposing the double helical model of DNA in 1953, based on X-ray diffraction data from Rosalind Franklin and Maurice Wilkins, and chemical data from Erwin Chargaff. …

The concept here is gene mapping on a linear chromosome. Recombination frequency (RF) between two genes tells you how far apart they are — the higher the RF, the farther apart they are. On a linear chromosome, distances are additive: if you know the distances between all pairs, you can deduce the order by finding which arrangement makes the distances consistent.

We have six RF values:

• a–c: 5%
• b–c: 15%
• b–d: 9%
• a–b: 20%
• c–d: 24%
• a–d: 29%

The smallest RF is 5% between a and c, so a and c are the closest pair. That means they must be neighbours on the chromosome. Let’s place them as a–c or c–a for now.

Next, look at b. b is 15% from c and 20% from a. If a and c are 5 apart, then b could be on either side of the pair. If b is on the a side, then distance a–b should be small, but it’s 20 — that’s larger than a–c (5). If b is on the c side, then b–c should be smaller than a–b, which matches: b–c = 15, a–b = 20. So b is on the c side: order a–c–b.

Now check distances: a–c = 5, c–b = 15, so a–b should be 5 + 15 = 20 — exactly matches the given 20%. Good. …

Gene regulation in prokaryotes, famously illustrated by the lac operon in E. coli, relies on a coordinated set of DNA sequences and genes working together. Understanding what each component does makes matching straightforward.

The promoter is where transcription begins. RNA polymerase recognizes and binds to this site to initiate mRNA synthesis. Without a functional promoter, the downstream genes remain silent.

The operator sits adjacent to the promoter and acts as a molecular switch. When a repressor protein binds here, it physically blocks RNA polymerase from transcribing the structural genes. Think of it as a gate that can be closed by the repressor.

The regulator gene (often denoted lacI in the lac operon) is located upstream and codes for the repressor protein itself. This gene is transcribed independently and produces the molecule that can bind to the operator. …

The question asks you to identify the correct sequence of processes and the scientist associated with the central dogma of molecular biology. The diagram likely shows a flow from DNA to a functional product, with blanks for the two key processes and the name of the scientist who first articulated this concept.

Concept & Intuition:

The central dogma, famously stated by Francis Crick, describes the directional flow of genetic information: DNA is transcribed into RNA, which is then translated into protein. This is the fundamental framework for gene expression. The other options mix up processes (e.g., replication, extension) or attribute the dogma to the wrong scientist (Franklin, Watson, Chargaff — each famous for other contributions: Franklin for X-ray crystallography of DNA, Watson for the double-helix model, Chargaff for base-pairing rules).

Step-by-step reasoning:

1. Identify the first blank (A): The process that converts DNA into RNA is transcription. This rules out options that start with "Translation" (B and D), because translation uses RNA to make protein, not DNA to RNA.

2. Identify the second blank (B): After transcription, the RNA molecule is used to direct protein synthesis via translation. So the second process must be translation. This eliminates option (C), which has "Replication" (copying DNA) in the second blank. …

Concept & Intuition

The problem tests basic knowledge of nucleic acid structure and the central dogma in retroviruses.

• A polynucleotide chain has a 5′ end (free phosphate) and a 3′ end (free hydroxyl).
• The classic DNA double helix is right-handed with a pitch of 3.4 nm (one full turn per 10 base pairs).
• Retroviruses (like HIV) reverse the usual flow: they use reverse transcriptase to go from RNA to DNA, not DNA to RNA.

Step-by-step reasoning

1. A) Free phosphate moiety polymer

   • In a polynucleotide, the phosphate group is attached to the 5′ carbon of the sugar. The end with a free phosphate is the 5′ end.
   • So A matches II (5′ end of the polynucleotide chain).

2. B) Free OH group end

   • The other end of the chain has a free hydroxyl (–OH) on the 3′ carbon of the sugar. This is the 3′ end.
   • So B matches III (3′ end of the polynucleotide chain).

3. C) The right handed fashion coiling

   • The B‑form DNA helix is right‑handed and has a pitch of 3.4 nm (the distance for one complete turn).
   • So C matches IV (Helix has 3.4 nm pitch).

4. D) The flow of information of gene in Retroviruses …

The key here is to recall the classic B‑form DNA structure, which is the most common form under physiological conditions. The double helix has a pitch (the length of one complete turn) of 34 Å, and the rise per base pair is 3.4 Å. Dividing the pitch by the rise gives the number of base pairs per turn:

So one full turn contains 10 base pairs, not 12. The assertion (A) is therefore false.

The reason (R) states that the pitch is 34 Å — this is a well‑established fact for B‑DNA, so (R) is true. However, even if (R) were correct, it does not explain a false statement.

Let’s go step by step:

1. Check the Assertion (A):

   “In DNA double helix, one full turn of helical strand contains twelve base pairs.”

   • In B‑DNA, the number of base pairs per turn is 10. In A‑DNA it is about 11, and in Z‑DNA it is 12, but the question refers to the standard double helix (B‑form).
   • Therefore, (A) is false.

2. Check the Reason (R):

   “The pitch of DNA helix is 34 Å.”

   • The pitch of B‑DNA is indeed 34 Å (the distance along the helix axis for one complete turn).
   • So (R) is true.

3. Determine the relationship: …

The structure of DNA relies on complementary base pairing between nitrogenous bases on opposite strands. Two purines (adenine and guanine, with double-ring structures) pair with two pyrimidines (thymine and cytosine, with single-ring structures). The number of hydrogen bonds between paired bases is fixed by their chemical structure: smaller bases form fewer bonds, larger complementary pairs form more.

The key is recognizing which base pairs with which, and how many hydrogen bonds stabilize each pair.

1. Identify the pairing pattern from hydrogen bond count.

   The problem states that A pairs with C via 2 hydrogen bonds, and B pairs with D via 3 hydrogen bonds. In DNA, only two pairing combinations exist:

   • Adenine (A) ↔ Thymine (T): 2 hydrogen bonds
   • Guanine (G) ↔ Cytosine (C): 3 hydrogen bonds

2. Match the 2-hydrogen-bond pair.

   Since 'A' pairs with 'C' through 2 hydrogen bonds, this must be the adenine–thymine pair. Therefore:

   • 'A' = Adenine
   • 'C' = Thymine

3. Match the 3-hydrogen-bond pair. …

The question asks what forms the F-actin of the thin filament. This is a classic structure-function question from muscle biology. The thin filament is a complex assembly, but its backbone is a double helix of actin. That actin exists in two forms: globular actin (G-actin), the monomer, and filamentous actin (F-actin), the polymer. The F-actin strand is literally made by G-actin subunits joining end-to-end.

Let’s walk through the options to see why only one fits.

1. G-actin is the monomeric form of actin. In muscle cells, these globular subunits polymerize to form the long, helical F-actin filament. So F-actin is formed by G-actin — this is the direct answer.

   Tip

   Think of G-actin as bricks and F-actin as the wall. The wall is made of bricks, not the other way around.

2. Meromyosin is a fragment of the myosin heavy chain (the thick filament protein). It has nothing to do with building actin filaments. It’s involved in the cross-bridge cycle, not thin filament structure. …