Q.In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called:

Understanding the fundamental processes of gene expression – transcription and translation – along with the concept of an operon, is crucial for answering this question. Each option describes a key component or process in molecular biology. We need to evaluate if the description accurately matches the term.

1. Analyze Option (A): Repressor $\rightarrow$ Protein coded by regulatory gene of operon

   • In an operon system (like the lac operon in bacteria), a regulatory gene (often denoted as the i gene) is responsible for synthesizing a repressor protein.
   • This repressor protein then binds to the operator region of the operon, thereby regulating the transcription of the structural genes.
   • Therefore, this statement is correct.

2. Analyze Option (B): Operon $\rightarrow$ Promotor, Structural genes, operator

   • An operon is a functional unit of DNA containing a cluster of genes under the control of a single promoter.
   • It typically consists of:
     • A promoter region: where RNA polymerase binds to initiate transcription.
     • An operator region: a segment of DNA to which a repressor protein binds, regulating gene expression.
     • Structural genes: genes that code for proteins involved in a particular metabolic pathway.
   • These three components are indeed the core parts of an operon.
   • Therefore, this statement is correct.

3. Analyze Option (C): Translation $\rightarrow$ Polymerisation of amino acids and polypeptide chain formation

   • Translation is the biological process where messenger RNA (mRNA) is decoded by a ribosome to produce a specific amino acid chain, or polypeptide.
   • This process involves the sequential addition of amino acids, linked by peptide bonds, to form a growing polypeptide chain. This is precisely the "polymerisation of amino acids and polypeptide chain formation."
   • Therefore, this statement is correct.

4. Analyze Option (D): Transcription $\rightarrow$ Attachment of amino acid to a specific t-RNA …

Transketolase always moves a two-carbon "ketol" fragment ($-CO-CH_2OH$) from a ketose sugar to an aldose sugar, converting the ketose donor into an aldose 2 carbons shorter, and the aldose acceptor into a ketose 2 carbons longer.

The given equation, $2A + 2\ G\text{-}3P \xrightarrow{\text{Transketolase}} 2\ \text{xylulose-5-P} + 2B$, halves to:

Glyceraldehyde-3-phosphate (G–3P) is a 3-carbon aldose. As the acceptor, it gains 2 carbons to become the 5-carbon ketose xylulose-5-phosphate ($3+2=5$) — consistent with the given product.

So A must be the donor ketose, losing 2 carbons to become the aldose B. In the pentose phosphate pathway, the transketolase reaction that uses glyceraldehyde-3-phosphate as the acceptor and produces xylulose-5-phosphate is:

$\text{Fructose-6-phosphate (6C ketose)} + G\text{-}3P \rightarrow \text{Xylulose-5-P (5C ketose)} + \text{Erythrose-4-phosphate (4C aldose)}.$ …

The concept first: why the fork looks lop-sided

Rule 1 — DNA polymerase has only one direction. The enzyme adds a nucleotide by attacking the incoming dNTP with the free $3'$-OH of the growing chain. It has no chemistry for adding to a $5'$ end. Therefore:

and, since the new strand is antiparallel to its template, the enzyme reads its template $3' \rightarrow 5'$.

Rule 2 — the two parental strands are antiparallel. In the duplex, one strand runs $5'\!\rightarrow\!3'$ left-to-right and the other $3'\!\rightarrow\!5'$ left-to-right. So when helicase opens the fork, the two exposed templates point in opposite directions relative to the moving fork.

The consequence. Consider the direction the fork travels (into the unopened duplex):

• On the template whose $3'$ end lies toward the fork, reading $3'\!\rightarrow\!5'$ means moving in the same direction as the fork. The polymerase can therefore chase the fork continuously $\Rightarrow$ this is the leading strand, synthesised in one unbroken piece, its arrow pointing towards the fork.
• On the other template the polarity is reversed. Reading $3'\!\rightarrow\!5'$ now means moving away from the fork. The polymerase must wait for a stretch of template to be exposed, then back-track and synthesise a short piece away from the fork; repeat. $\Rightarrow$ this is the lagging strand, made as Okazaki fragments later joined by DNA ligase, with arrows pointing away from the fork.

So the signature of a correct fork diagram is asymmetry: one arrow in, one arrow out.

Step-by-step: audit the four figures

A useful two-part checklist for each option — (i) do the arrows show one strand made towards and one away from the fork? (ii) are those directions consistent with the printed $3'$/$5'$ polarity labels? …

The lac operon is a classic example of gene regulation in prokaryotes, specifically in E. coli. It controls the expression of genes necessary for the metabolism of lactose. Understanding the function of each component — the promoter, operator, regulator gene, and structural genes — is crucial to grasp how gene expression is turned on or off in response to environmental cues.

Here's how the components match their functions:

1. A. Operator site:

   The operator site is a specific DNA sequence located immediately downstream of the promoter and upstream of the structural genes. Its primary role is to act as a switch for transcription. It serves as the binding site for the repressor protein. When the repressor protein binds to the operator, it physically obstructs the movement of RNA polymerase, preventing it from transcribing the structural genes. This effectively turns off gene expression.

   Therefore, A matches with II (Binding site for repressor).

2. B. Promoter site:

   The promoter site is a DNA sequence located upstream of the operator and structural genes. It is the recognition and binding site for RNA polymerase. RNA polymerase must bind to the promoter to initiate the transcription of the genes that follow. Without a functional promoter, transcription cannot begin.

   Therefore, B matches with I (Binding site for RNA polymerase).

3. C. Regulator gene (or i gene): …

The concept first — what a transcription unit is made of

A transcription unit has three components:

1. Promoter — the DNA sequence, located upstream (towards the 5' end of the coding strand), where RNA polymerase binds. It defines the template strand.
2. The structural gene — the stretch that is actually copied.
3. Terminator — located downstream (towards the 3' end of the coding strand), where transcription stops.

And the two strands have fixed names:

• Template (antisense/minus) strand — the one the polymerase reads. Since RNA is built $5' \to 3'$ and DNA–RNA pairing is antiparallel, this strand must have $3' \to 5'$ polarity in the direction of transcription.
• Coding (sense/plus) strand — the other one. It has $5' \to 3'$ polarity and its base sequence is identical to the mRNA (except T instead of U). It is not copied at all; it is called "coding" only because it reads like the message.

Step-by-step, using the figure

1. Find the direction of transcription. The bent arrow, which marks the start point and points rightwards, sits immediately to the right of box A. RNA polymerase always begins at the promoter, so:

2. The far end must therefore be the stop signal: $\mathbf{B = Terminator}$ …

The concept first

This is a straight history-of-biology recall item. The safe strategy is to anchor the two or three you are certain about, then let the options eliminate one another.

Step-by-step

1. A) Buchner → IV (Zymase). In 1897 Eduard Buchner ground yeast cells and showed that the resulting cell-free extract could still ferment sugar to alcohol. He named the active principle zymase. This was a landmark: it proved fermentation is a chemical process catalysed by a substance, not a mystical property of living cells. Nobel Prize, 1907.
2. B) Sumner → I (Urease). In 1926 James B. Sumner crystallised urease from jack bean — the first enzyme ever obtained in pure crystalline form — and demonstrated that it was a protein. This settled a long-running dispute about the chemical nature of enzymes. Nobel Prize, 1946.
3. C) Knoll and Ruska → V (Electron microscope). Max Knoll and Ernst Ruska built the first transmission electron microscope in 1931–32, breaking the resolution limit imposed on light microscopy by the wavelength of visible light. Nearly everything we know about cell ultrastructure — ER, ribosomes, membranes — flowed from this instrument. Ruska received the Nobel Prize in 1986. …

The concept first

Biomolecules split cleanly along three axes, and this question tests all three at once: the formula (List-II sometimes), the hydrolysis product/example (List-II), and the chemical class or status (List-III). Anchor on the item you know with certainty and let the rest follow.

Step-by-step

1. A) Deoxyribose. Ribose is $C_5H_{10}O_5$. "Deoxy" means one oxygen fewer at C-2, so deoxyribose is $C_5H_{10}O_4$ $\Rightarrow$ List-II (ii). It is a pentose sugar, i.e. a carbohydrate $\Rightarrow$ List-III (IV). A → ii, IV.

2. B) Lipids. A triglyceride hydrolyses as

So lipids are built from glycerol $\Rightarrow$ (iv) and fatty acids $\Rightarrow$ (I). B → iv, I.

3. C) Polymeric substances. Gums (i) are polysaccharide polymers, and cellulose (III) is the archetypal plant polymer — a $\beta$-1,4 linked glucose homopolymer. C → i, III. …