Write the structures of fragments produced on complete hydrolysis of DNA. How are they linked in DNA molecule?

NCERT Exemplar · Q53

Q. Write the structures of fragments produced on complete hydrolysis of DNA. How are they linked in DNA molecule? Draw a diagram to show pairing of nucleotide bases in double helix of DNA.

Andhra Pradesh BieapLong· 5mImportance★★★★★

78% · 86/110 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Nucleic Acid Components

Nucleic Acid Components: The Alphabet of Life

Imagine you want to write a book that contains all the instructions for building and running a living organism. You'd need an alphabet — a set of letters that can be combined in endless ways to form words, sentences, and chapters. In living cells, that alphabet is made of nucleic acids (DNA and RNA), and each "letter" is called a nucleotide.

The Big Picture: What Are Nucleic Acids?

Nucleic acids are long, chain-like molecules that store and transmit genetic information. DNA holds the master blueprint; RNA helps execute it. But both are built from the same basic building block: the nucleotide.

Note

Think of a nucleotide as a single Lego brick. DNA and RNA are long chains of these bricks, each brick slightly different from the next.

The Three Parts of a Nucleotide

Every nucleotide has three components, like a three-part key:

A phosphate group — a small, negatively charged group ( $\text{PO}_4^{3-}$ ). It acts like the "handle" that links nucleotides together.
A sugar — either ribose (in RNA) or deoxyribose (in DNA). This is the "body" of the brick.
A nitrogenous base — a ring-shaped molecule containing nitrogen. This is the "colored part" that carries the actual information.

The sugar and base together form a nucleoside. When you add the phosphate, you get a nucleotide.

Nucleoside = Sugar + Base Nucleotide = Sugar + Base + Phosphate

The Two Families of Bases

The bases come in two structural types:

Purines (double-ring structures): Adenine (A) and Guanine (G)
Pyrimidines (single-ring structures): Cytosine (C), Thymine (T) (only in DNA), and Uracil (U) (only in RNA)

Tip

A mnemonic: Purines are Pure as All Gold (A and G). Pyrimidines are CUT (C, U, T).

DNA vs. RNA: The Key Differences

Feature	DNA	RNA
Sugar	Deoxyribose (missing one oxygen)	Ribose (has that oxygen)
Bases	A, G, C, T	A, G, C, U
Structure	Double-stranded helix	Usually single-stranded
Function	Stores genetic information	Carries and executes instructions

Watch out

A common mistake: thinking "nucleoside" and "nucleotide" are the same. Remember: nucleotide has the phosphate; nucleoside does not. ATP (adenosine triphosphate) is a nucleotide — it's the energy currency of the cell.

Why This Matters

The sequence of bases along a DNA strand spells out the genetic code. A change in even one base (a mutation) can alter a protein, sometimes with dramatic consequences — like sickle cell anemia, where a single base change in the hemoglobin gene causes red blood cells to deform. …

Why this formula?

Nucleic Acid Components: Understanding the "Why" Behind the Key Relationships

Let’s start with the big picture: Nucleic acids (DNA and RNA) are polymers made of nucleotides. Each nucleotide has three parts: a nitrogenous base, a pentose sugar, and a phosphate group. The key formulas and relationships in this topic arise from how these parts are linked and how they behave chemically.

1. The Basic Composition Formula

What it says:

A nucleotide = Base + Sugar + Phosphate

Why this holds:

Chemical necessity: The sugar (ribose in RNA, deoxyribose in DNA) has a 5-carbon ring. The base attaches to the 1' carbon (via a glycosidic bond), and the phosphate attaches to the 5' carbon (via an ester bond).
Directionality: This creates a 5' → 3' linkage in the polymer. The phosphate of one nucleotide bonds to the 3' carbon of the next sugar.
Reasoning: Without the phosphate, you have a nucleoside (base + sugar). Adding the phosphate makes it a nucleotide — the monomer that can polymerize.

Key takeaway: The formula isn’t arbitrary — it reflects the specific carbon positions on the sugar that allow for stable, directional chain formation.

2. Chargaff’s Rules (for DNA)

What it says:

In double-stranded DNA:

$[A] = [T]$
$[G] = [C]$
$[A] + [G] = [T] + [C]$

Why this holds:

Base pairing: Adenine (A) forms two hydrogen bonds with Thymine (T). Guanine (G) forms three hydrogen bonds with Cytosine (C).
Structural constraint: The DNA double helix has a constant width (20 Å). A purine (A or G) always pairs with a pyrimidine (T or C) — otherwise the helix would bulge or narrow.
Derivation: If every A on one strand must pair with a T on the opposite strand, then the number of A equals number of T in the whole molecule. Same for G and C.
Consequence: The sum of purines equals sum of pyrimidines ( $A+G = T+C$ ).

Why it’s not just a rule: It’s a geometric and energetic necessity — hydrogen bonding and helix stability force this equality.

3. The Phosphodiester Bond Energy Formula

What it says:

Formation of a phosphodiester bond requires ~30 kJ/mol of energy (from ATP).

Why this holds:

Mechanism: The 3' hydroxyl of one nucleotide attacks the α-phosphate of a nucleotide triphosphate (e.g., ATP). This releases pyrophosphate (PPi).
Energy source: The hydrolysis of PPi to two inorganic phosphates ( $PPi \rightarrow 2P_i$ ) is highly exergonic (ΔG ≈ -30 kJ/mol). This drives the reaction forward.
Reasoning: The bond itself is a covalent ester linkage — strong but not spontaneously formed. The energy comes from breaking a high-energy phosphate bond in the triphosphate.

Key insight: The formula isn’t about the bond’s strength — it’s about the thermodynamic cost of making it in a cell.

4. The Melting Temperature ( $T_m$ ) Formula

What it says:

$T_m$ (in °C) ≈ $4(G+C) + 2(A+T)$ for short oligonucleotides.

Why this holds:

Hydrogen bonds: G-C pairs have 3 H-bonds, A-T pairs have 2 H-bonds. More H-bonds = more energy needed to separate strands.
Stacking interactions: G-C base pairs also have stronger π-stacking (aromatic ring overlap) than A-T.
Derivation: The formula is empirical — it comes from measuring $T_m$ for many sequences. The coefficients (4 and 2) reflect the relative stability contributed by each base pair.
Limitation: For long DNA, this simple formula fails because nearest-neighbor interactions matter more.

Why it works: It’s a linear approximation of the free energy needed to break all base pairs, weighted by H-bond count.

5. The Central Dogma (Information Flow)

What it says:

DNA → RNA → Protein

Why this holds: …

Start your 14-day free trial to unlock the full solution →

The key idea is that DNA is a polynucleotide; complete hydrolysis breaks all phosphodiester and N-glycosidic bonds, yielding three distinct components.

Reasoning:

Hydrolysis of the backbone: The phosphodiester bonds between the $3'$ ‑OH of one sugar and the $5'$ ‑phosphate of the next are broken, releasing free phosphoric acid ( $H_3PO_4$ ) and nucleosides.
Hydrolysis of nucleosides: The N‑glycosidic bond linking the nitrogenous base to the $1'$ ‑carbon of deoxyribose is cleaved, separating the base from the sugar.
Final fragments: The three products are:
- Phosphoric acid ( $H_3PO_4$ )
- Deoxyribose sugar ( $C_5H_{10}O_4$ )
- Nitrogenous bases — Adenine (A), Guanine (G), Cytosine (C), Thymine (T)

Linking in DNA: Nucleotides are linked by $3'$ – $5'$ phosphodiester bonds — the phosphate group bridges the $3'$ ‑OH of one deoxyribose to the $5'$ ‑OH of the next, forming a sugar‑phosphate backbone.

Base pairing in the double helix:

Adenine pairs with Thymine via two hydrogen bonds.
Guanine pairs with Cytosine via three hydrogen bonds.
The strands are antiparallel ( $5' \to 3'$ and $3' \to 5'$ ).

     5'                   3'
      |                   | …

Start your 14-day free trial to unlock the full solution →

Complete hydrolysis of DNA breaks it down into three components: a phosphate group, a deoxyribose sugar, and nitrogenous bases (adenine, guanine, cytosine, thymine). In the DNA molecule, these are linked by phosphodiester bonds between sugars and phosphates, and hydrogen bonds between complementary base pairs (A–T and G–C). The double helix is stabilised by these bonds, with base pairing shown in a diagram.

Concept and Intuition

DNA is a long polymer made of repeating units called nucleotides. Each nucleotide has three parts: a phosphate group, a deoxyribose sugar (with five carbons), and a nitrogenous base. When DNA is completely hydrolysed — meaning all chemical bonds are broken — it falls apart into these individual building blocks. But in the intact DNA molecule, these pieces are linked in a specific way: the sugar of one nucleotide connects to the phosphate of the next via a phosphodiester bond, forming a sugar-phosphate backbone. The bases stick out from this backbone and pair up with bases on the opposite strand through hydrogen bonds, following the rule that adenine (A) pairs with thymine (T), and guanine (G) pairs with cytosine (C). This pairing is what gives the DNA double helix its famous structure.

Watch out

A common mistake is to confuse the products of complete hydrolysis with those of partial hydrolysis. Complete hydrolysis yields individual components (phosphate, sugar, base), not nucleotides or dinucleotides. Partial hydrolysis would give smaller fragments like nucleotides or oligonucleotides.

Step-by-Step Solution

1. Identify the products of complete hydrolysis of DNA

Complete hydrolysis breaks all covalent bonds in the DNA polymer. This means:

The phosphodiester bonds between sugars and phosphates are broken.
The N-glycosidic bonds between sugars and bases are also broken.

So the final fragments are:

Phosphoric acid ( $H_3PO_4$ ) — the phosphate group.
Deoxyribose sugar — a pentose sugar with the formula $C_5H_{10}O_4$ .
Nitrogenous bases — these are of two types:
- Purines: Adenine (A) and Guanine (G).
- Pyrimidines: Cytosine (C) and Thymine (T).

Thus, the complete hydrolysis of DNA yields a mixture of phosphate, deoxyribose, and the four bases (A, G, C, T).

\text{DNA} \xrightarrow{H_2O, H^+} \text{Phosphoric acid} + \text{Deoxyribose} + \text{Adenine} + \text{Guanine} + \text{Cytosine} + \text{Thymine}

2. How are these components linked in the DNA molecule?

In the intact DNA molecule, the linkages are as follows:

Between sugar and phosphate: The phosphate group forms an ester bond with the $5'$ carbon of one deoxyribose sugar and another ester bond with the $3'$ carbon of the next deoxyribose sugar. This creates a phosphodiester bond ( $-O-PO_2-O-$ ) that links nucleotides together in a chain. This forms the sugar-phosphate backbone.
Between sugar and base: The nitrogenous base is attached to the $1'$ carbon of deoxyribose via an N-glycosidic bond (a bond between the anomeric carbon of the sugar and a nitrogen atom of the base).
Between two strands: The two strands of DNA are held together by hydrogen bonds between complementary bases. Adenine (A) pairs with Thymine (T) via two hydrogen bonds, and Guanine (G) pairs with Cytosine (C) via three hydrogen bonds. This is called complementary base pairing.

Tip

Remember the base pairing rule with a mnemonic: Apple Tree (A–T, 2 bonds) and Golf Cart (G–C, 3 bonds). The number of hydrogen bonds is important for stability — G–C pairs are stronger because they have three bonds. …

Start your 14-day free trial to unlock the full solution →

Concept: Structure of DNA and its Hydrolysis Products

DNA (Deoxyribonucleic Acid) is a polymer of nucleotides. Complete hydrolysis breaks all the bonds in the DNA backbone, yielding its simplest building blocks.

Method: Sequential Hydrolysis Analysis

Step 1: Identify the primary structure of DNA

DNA is a polynucleotide chain.
Each nucleotide has three components:
- A nitrogenous base (Adenine, Guanine, Cytosine, Thymine)
- A deoxyribose sugar (a pentose sugar)
- A phosphate group

Step 2: Understand what "complete hydrolysis" means

Complete hydrolysis breaks all covalent bonds between nucleotides.
This yields the monomeric units — not larger fragments.

Step 3: Write the fragments produced

Complete hydrolysis of DNA gives three types of molecules:

Nitrogenous bases:
- Purines: Adenine (A) and Guanine (G)
- Pyrimidines: Cytosine (C) and Thymine (T)
Deoxyribose sugar:
- $\text{C}_5\text{H}_{10}\text{O}_4$ (a pentose sugar)
Phosphoric acid:
- $\text{H}_3\text{PO}_4$

Key result: Complete hydrolysis → Bases + Sugar + Phosphate (no nucleotides remain).

How Are They Linked in the DNA Molecule?

The linkage is through phosphodiester bonds:

A phosphodiester bond forms between the 3' carbon of one deoxyribose sugar and the 5' carbon of the next deoxyribose sugar, via a phosphate group.
This creates a sugar-phosphate backbone with bases projecting inward.

Bonding summary:

N-glycosidic bond: Between base and deoxyribose (C1' of sugar to N1 of pyrimidine or N9 of purine)
Phosphodiester bond: Between adjacent sugars via phosphate

Diagram: Base Pairing in DNA Double Helix

     5'                3'
      |                |
      A = = = = = = = T
      |                |
      G = = = = = = = C
      |                |
      C = = = = = = = G
      |                | …

Start your 14-day free trial to unlock the full solution →

Here are the common mistakes students make on this DNA hydrolysis and structure question, along with how to avoid each.

1. Confusing "Complete Hydrolysis" with "Partial Hydrolysis"

The Mistake:

Students often list nucleotides (sugar + phosphate + base) as the final products.

Complete hydrolysis breaks all chemical bonds in the backbone, yielding three separate components.

How to Avoid:

Remember the hierarchy:

DNA → (complete hydrolysis) → Phosphoric acid + Deoxyribose sugar + Nitrogenous bases (A, T, G, C)
Partial hydrolysis would give nucleotides or nucleosides.

Correct answer:

Fragments = Phosphate group, Deoxyribose sugar, and Nitrogenous bases (Adenine, Thymine, Guanine, Cytosine).

2. Forgetting the Specific Sugar Name

The Mistake:

Writing "ribose" instead of deoxyribose.

This is a classic exam trap — RNA has ribose, DNA has deoxyribose.

How to Avoid:

Link the name to the structure:

Deoxyribose lacks one –OH group at the 2′ carbon (compared to ribose).
Write it explicitly: 2-deoxy-D-ribose.

3. Incorrect Linkage Description (Phosphodiester Bond)

The Mistake:

Saying "phosphate links sugar to base" or "bases are linked by hydrogen bonds" when describing the backbone linkage.

How to Avoid:

Be precise about which atoms are involved:

The linkage is a 3′–5′ phosphodiester bond.
It connects the 3′ carbon of one deoxyribose to the 5′ carbon of the next deoxyribose, via a phosphate group.

Correct phrasing:

"Nucleotides are linked by 3′–5′ phosphodiester bonds between the sugar of one nucleotide and the phosphate of the next."

4. Drawing the Base Pairing Diagram Incorrectly

The Mistake:

Showing A–G or T–C pairs.
Drawing two hydrogen bonds for G–C (it has three).
Forgetting the antiparallel orientation of the two strands.

How to Avoid:

Memorise the Chargaff’s rule pairs:

A = T (2 hydrogen bonds)
G ≡ C (3 hydrogen bonds)

In your diagram:

Draw one strand 5′ → 3′ and the other 3′ → 5′.
Show dashed lines for H-bonds (2 between A–T, 3 between G–C).
Label the sugar-phosphate backbone as two outer ribbons.

Quick diagram (text representation):

5'   3'
|    |
S—P—S—P—S—P
|    |    |
A===T   G≡C
|    |    |
P—S—P—S—P—S
|    |    |
3'   5'

(In your exam, draw a clear, labelled diagram with arrows showing direction.)

5. Omitting the "Antiparallel" Nature

The Mistake: …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 17 on this concept.

AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$H_3C$
›Reveal solutionSolution
Of the four structures, only cytosine (III) is a base found in both DNA and RNA; thymine is DNA-only, uracil is RNA-only, and the drawn sugar (with a free 2′-OH) is ribose — RNA-specific, since DNA's sugar (deoxyribose) lacks that 2′-OH.

Concept and Intuition

DNA and RNA share three of the five standard nitrogenous bases — adenine, guanine, and cytosine — but differ in the fourth pyrimidine: DNA uses thymine (5-methyluracil), RNA uses uracil. They also differ in their sugar: DNA has 2′-deoxyribose (no OH at C2′), RNA has ribose (OH at C2′). So any question asking "common to both" hinges on recognising exactly these two DNA/RNA differences.

Step-by-Step Solution

Identify structure I: a pyrimidine ring with a 5-methyl group and two carbonyls (2,4-dioxo) — this is thymine, found only in DNA.

Identify structure II: a furanose (5-membered) sugar ring with $HOCH_2-$ at C5, and OH groups shown on the ring including near the ring oxygen — matches ribose (has the 2′-OH). Since DNA's sugar (deoxyribose) lacks the 2′-OH, this specific sugar (ribose) belongs to RNA only, not DNA.

Identify structure III: a pyrimidine ring with an $NH_2$ group and one carbonyl — this is cytosine, present in BOTH DNA and RNA. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$Q. The structure of the nitrogen containing heterocyclic base given below represents [FIGURE] (a six-membered ring: a carbon bearing a C=O group at the top is bonded to an adjacent NH; that NH is bonded to a carbon which carries another C=O group; that carbon is bonded to an N-H; the N-H is bonded back to the first C=O carbon via a C=C double bond, completing a pyrimidine-2,4-dione ring) (A) Adenine (B) Thymine (C) Uracil (D) Cytosine$
›Reveal solutionSolution
The figure is the bare pyrimidine-2,4-dione ring (two ring $\mathrm{N-H}$ , two ring $\mathrm{C=O}$ , one ring $\mathrm{C=C}$ ) with no extra substituent — that is uracil, not thymine (which carries a 5- $\mathrm{CH_3}$ group) or the purine bases adenine/cytosine (cytosine is also a pyrimidine, but is an amino-oxo tautomer, not a dione).

Concept and Intuition

Nucleic-acid bases split into purines (adenine, guanine — fused bicyclic) and pyrimidines (cytosine, thymine, uracil — single six-membered ring). Among the pyrimidines: cytosine has one ring $\mathrm{C=O}$ and one ring $\mathrm{C-NH_2}$ (amino group, not a second carbonyl); thymine and uracil both have TWO ring carbonyls (a 2,4-dione) flanked by two ring N-H's, differing only in that thymine additionally carries a methyl group at C5. Since the figure shows a bare 2,4-dione ring with a plain $\mathrm{C=C}$ and no extra substituent drawn, it must be uracil (RNA-specific; thymine is DNA-specific and always drawn with its 5-methyl group).

Step-by-Step Solution

Identify ring type: six-membered, single ring $\Rightarrow$ pyrimidine (rules out adenine, a fused bicyclic purine). …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$Q. Identify the set containing purine and pyrimidine base of DNA respectively. (A) Adenine, Uracil (B) Cytosine, Guanine (C) Thymine, Uracil (D) Adenine, Cytosine$
›Reveal solutionSolution
Tests knowing which bases are purines vs. pyrimidines, and which set belongs to DNA (as opposed to RNA).

Concept and Intuition

Purines (two fused rings) in nucleic acids are Adenine and Guanine — common to both DNA and RNA. Pyrimidines (single ring) are Cytosine (common to both), Thymine (DNA only), and Uracil (RNA only, replacing thymine). A correct "purine, pyrimidine of DNA" pair must therefore avoid Uracil altogether.

Step-by-Step Solution

(A) Adenine, Uracil — Adenine is a genuine DNA purine, but Uracil is an RNA base, not found in DNA. Rejected.

(B) Cytosine, Guanine — the order is reversed: Cytosine is the pyrimidine and Guanine is the purine, but the question asks purine-then-pyrimidine. Rejected.

(C) Thymine, Uracil — both entries are pyrimidines (no purine at all), and Uracil again doesn't belong to DNA. Rejected. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$Q. The pyrimidine bases found in RNA are (A) Thymine & Uracil (B) Thymine & Cytosine (C) Adenine & Guanine (D) Uracil & Cytosine$
›Reveal solutionSolution
RNA's pyrimidine bases are uracil and cytosine — RNA replaces DNA's thymine with uracil.

Concept and Intuition

Nucleic acid bases are classified as purines (two fused rings: adenine, guanine) or pyrimidines (single ring: cytosine, thymine, uracil). DNA and RNA share adenine, guanine and cytosine, but differ in their fourth base: DNA uses thymine, while RNA uses uracil instead (uracil lacks the methyl group that thymine has).

Step-by-Step Solution

Purine bases (common to both DNA and RNA): adenine (A), guanine (G).

Pyrimidine bases in DNA: cytosine (C) and thymine (T).

Pyrimidine bases in RNA: cytosine (C) and uracil (U) — uracil replaces thymine. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$Q. Which of the following represents nucleoside of RNA ? (A) [FIGURE] (a sugar ring bearing base T (thymine), with hydroxyl groups drawn at both ring-carbon positions below the ring — i.e. two -OH/HO groups present) (B) [FIGURE] (a sugar ring bearing base G (guanine), with only a single hydroxyl group drawn below the ring) (C) [FIGURE] (a sugar ring bearing base C (cytosine), with only a single hydroxyl group drawn below the ring) (D) [FIGURE] (a sugar ring bearing base U (uracil), with hydroxyl groups drawn at both ring-carbon positions below the ring — i.e. two -OH/HO groups present)$
›Reveal solutionSolution
An RNA nucleoside must combine an RNA-specific base (A, G, C or U — never T) with ribose sugar (two ring -OH groups, at C2' and C3'); only the Uracil + two-OH structure satisfies both.

Concept and Intuition

A nucleoside = a nitrogenous base joined to a pentose sugar by an N-glycosidic bond, with no phosphate group. DNA uses 2'-deoxyribose, which has lost the oxygen at C2', so on the ring it shows only ONE hydroxyl (at C3'). RNA uses ribose, which retains the C2'-OH as well as the C3'-OH, so its ring shows TWO hydroxyls. Separately, RNA and DNA differ in one base: RNA uses Uracil in place of DNA's Thymine; the other three bases (Adenine, Guanine, Cytosine) are common to both. So a genuine RNA nucleoside must (a) show two ring -OH's (ribose) AND (b) carry a base that is RNA-specific or RNA-compatible, never Thymine.

Step-by-Step Solution

Sugar check: Structures with base T and base U both show two ring hydroxyls $\Rightarrow$ these have ribose sugar. Structures with base G and base C show only one ring hydroxyl $\Rightarrow$ these have deoxyribose sugar, i.e. they are DNA nucleosides (deoxyguanosine, deoxycytidine), even though G and C themselves appear in both nucleic acids.

Base check: Thymine (T) is exclusively a DNA base — it never occurs in RNA, regardless of what sugar is drawn attached to it, so the "T + ribose" structure does not correspond to any real RNA nucleoside. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
$Q. Which of the following represents double stranded helix structure of DNA? (A) [FIGURE] (a double-helix diagram: both strand ends at the top are labelled 5' and both strand ends at the bottom are labelled 3', i.e. the two strands are drawn running in the SAME direction (parallel, not antiparallel); the rungs between them, top to bottom, are labelled G\equivC, A\equivT, G\equivC, T\equivA) (B) [FIGURE] (a double-helix diagram: the top ends are labelled 5' (left strand) and 3' (right strand); the bottom ends are labelled 3' (left strand) and 5' (right strand), i.e. the two strands run in opposite directions (antiparallel); the rungs, top to bottom, are labelled G\equivC, A=T, C\equivG, T=A) (C) [FIGURE] (a double-helix diagram: both strand ends at the top are labelled 5' and both strand ends at the bottom are labelled 3' (parallel, not antiparallel); the rungs, top to bottom, are labelled G\equivT, A=C, T\equivG, A=C) (D) [FIGURE] (a double-helix diagram: the top ends are labelled 5' (left strand) and 3' (right strand); the bottom ends are labelled 3' (left strand) and 5' (right strand), i.e. antiparallel strands; the rungs, top to bottom, are labelled G=U, A\equivT, U=G, A\equivT)$
›Reveal solutionSolution
The correct double‑stranded DNA helix must have antiparallel strands (one 5′→3′, the other 3′→5′) and standard base pairs (A‑T and G‑C, with correct hydrogen‑bond counts). Only option (B) satisfies both conditions.

Why this approach works

DNA’s double helix is not just any twisted ladder — it has two non‑negotiable features:

Antiparallel orientation – The two sugar‑phosphate backbones run in opposite directions. One strand goes 5′→3′ downward, the other goes 3′→5′ downward. This is essential for the geometry of base pairing and for enzymes like DNA polymerase.

Complementary base pairing – Adenine (A) pairs only with thymine (T) via two hydrogen bonds (A=T), and guanine (G) pairs only with cytosine (C) via three hydrogen bonds (G≡C). Any other pairing (e.g., G‑T, A‑C, or using uracil U) is not found in DNA.

The question tests whether you can spot both features simultaneously in the given diagrams.

Step‑by‑step reasoning

Check strand direction (parallel vs. antiparallel)

In a correct DNA helix, if the left strand’s top end is labelled 5′, its bottom end must be 3′. The right strand must have the opposite: top 3′, bottom 5′.

Options (A) and (C) show both top ends as 5′ and both bottom ends as 3′ → parallel strands. This is wrong.

Options (B) and (D) show top: 5′ (left) and 3′ (right); bottom: 3′ (left) and 5′ (right) → antiparallel strands. This is correct for direction.

Check base‑pair composition

(A) has G≡C, A≡T, G≡C, T≡A. The pairs are standard (A‑T, G‑C), but the strands are parallel → invalid because of direction.

(B) has G≡C, A=T, C≡G, T=A. All pairs are standard (note: A=T uses two lines, G≡C uses three lines — correct). Strands are antiparallel → valid. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$\mathrm{NH_2}$
›Reveal solutionSolution
DNA contains the bases adenine, guanine, cytosine, and thymine, plus the sugar deoxyribose. Here, structure A is cytosine, D is adenine, C is ribose (not deoxyribose), and B is uracil (found in RNA, not DNA). So only A and D are present in DNA, making option (A) correct.

Identify each structure by its chemical features.

Structure A is a six-membered pyrimidine ring with one carbonyl (C=O), one ring N–H, and an exocyclic amino group (NH₂). This matches cytosine, one of the four DNA bases.

Structure B is a six-membered pyrimidine ring with two carbonyls and two ring N–H groups, but no methyl group. This is uracil, which is found in RNA, not DNA. (DNA uses thymine, which has a methyl group at the position where uracil has a hydrogen.)

Structure C is a five-membered furanose ring with an oxygen in the ring, a CH₂OH group, and OH groups on every ring carbon except the one bonded to the ring oxygen. That is ribose, the sugar in RNA. DNA uses deoxyribose, which lacks an OH on the 2′ carbon (the carbon adjacent to the anomeric carbon). Since all non-anomeric carbons here have an OH, this is ribose, not deoxyribose.

Structure D is a fused purine ring system with an NH₂ group and three ring nitrogens (two in the six-membered ring, two in the five-membered ring, one of which has an N–H). This is adenine, a DNA base.

Determine which are present in DNA.

DNA contains the bases adenine (D), guanine, cytosine (A), and thymine. It does not contain uracil (B) or ribose (C).

Therefore, only A and D are components of DNA.

Match to the answer choices. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$Q. Which of the following represents simplified version of nucleoside? (A) Base- sugar- phosphate (B) Sugar- base (C) Sugar- Phosphate (D) Base- Phosphate$
›Reveal solutionSolution
A nucleoside is simply a sugar joined to a nitrogenous base (no phosphate); adding phosphate to a nucleoside gives a nucleotide.

Concept and Intuition

Nucleic acid building blocks have a hierarchy: base + sugar = nucleoside; base + sugar + phosphate = nucleotide. The question asks for the SIMPLER of the two (without phosphate), which is the nucleoside.

Step-by-Step Solution

A nucleotide = base + sugar + phosphate (the full monomer unit of DNA/RNA).

Removing the phosphate group leaves base + sugar only — this simpler unit is called a nucleoside.

So the correct composition of a nucleoside is Sugar–Base, matching option (B).

Common Mistakes …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
$5^1, 3^1$
›Reveal solutionSolution
Nucleic acid backbones (both DNA and RNA) are held together by 3',5'-phosphodiester linkages between adjacent sugar residues.

Concept and Intuition

Each nucleotide consists of a sugar (ribose in RNA), a nitrogenous base, and a phosphate group. The polynucleotide chain is formed when the phosphate group bridges the 3'-hydroxyl of one sugar to the 5'-hydroxyl of the next sugar, forming a phosphodiester bond (two ester linkages from one phosphate — one to each sugar). This defines the chain's directionality (5' end to 3' end).

Step-by-Step Solution

Each phosphate in the backbone forms an ester bond with the 5'-OH of the upcoming sugar and another ester bond with the 3'-OH of the previous sugar — hence "phosphodiester."

This linkage is described as the 3',5'-phosphodiester bond (equivalently 5',3').

Options B, C, D describe linkages between other positions (2', 3' combinations) that are not the actual biological linkage. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
$Q. In nucleoside, the base is attached to which position of sugar molecule? (A) C - 1 (B) C - 2 (C) C - 3 (D) C - 5$
›Reveal solutionSolution
A nucleoside forms when a nitrogenous base bonds to the sugar's anomeric carbon; this glycosidic linkage is always at C-1' of the ribose/deoxyribose.

Concept and Intuition

A nucleoside = base + sugar (no phosphate). The bond connecting the base to the sugar is an N-glycosidic bond, analogous to the O-glycosidic bonds that link sugars to each other, formed specifically between the sugar's anomeric carbon (C-1', the carbon that was the carbonyl carbon in the open-chain form) and a nitrogen of the purine or pyrimidine base. The phosphate group, by contrast, attaches separately at C-5' (or sometimes C-3') to make a nucleotide.

Step-by-Step Solution

Recall the sugar numbering convention in nucleic acid chemistry: carbons of the sugar are primed (1', 2', 3', 4', 5') to distinguish them from the base's own numbering.

The base-sugar (glycosidic) bond forms at the sugar's anomeric centre, which is C-1' for ribose/deoxyribose.

Therefore the base is attached at C-1', matching option (A).

Common Mistakes …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
$Q. Given below are two statements Assertion (A): Hydrolysis of DNA does not form equal number of A and T; G and C Reason (R): In DNA adenine forms hydrogen bonds with thymine and cytosine forms hydrogen bonds with guanine The correct answer is (A) Both A and R are correct and R is the correct explanation of A (B) Both A and R are correct but R is not the correct explanation of A (C) A is correct but R is incorrect (D) A is incorrect but R is correct$
›Reveal solutionSolution
The assertion is false because hydrolysis of DNA does yield equal numbers of A and T, G and C due to base pairing; the reason is true but does not explain a false assertion. The correct option is (D).

Concept and Intuition

The question tests your understanding of Chargaff’s rules and the nature of DNA hydrolysis.

Hydrolysis of DNA breaks it into its constituent nucleotides (or bases, sugars, phosphates). Because DNA is a double helix where A always pairs with T and G always pairs with C, the total number of A residues equals the total number of T residues, and G equals C. So hydrolysis does produce equal amounts of A and T, and G and C.

The Reason correctly states the base-pairing rules (A–T, G–C), which is a true fact about DNA structure.

The Assertion claims the opposite — that hydrolysis does not give equal numbers — which is false. Thus, the Assertion is wrong, the Reason is right, and the Reason cannot explain a false statement.

Step-by-step reasoning

Understand the Assertion (A)

“Hydrolysis of DNA does not form equal number of A and T; G and C.”

Hydrolysis breaks the phosphodiester bonds, releasing free nucleotides.

In a double-stranded DNA molecule, the number of adenine (A) residues equals the number of thymine (T) residues because every A on one strand pairs with a T on the opposite strand. Similarly, guanine (G) equals cytosine (C).

Therefore, upon complete hydrolysis, the molar amounts of A and T are equal, and likewise for G and C.

Conclusion: The Assertion is incorrect.

Examine the Reason (R)

“In DNA adenine forms hydrogen bonds with thymine and cytosine forms hydrogen bonds with guanine.”

This is a fundamental fact of DNA structure (Watson–Crick base pairing).

A pairs with T via two hydrogen bonds; G pairs with C via three hydrogen bonds.

Conclusion: The Reason is correct.

Check the logical connection

The Reason correctly describes base pairing, which explains why A = T and G = C in double-stranded DNA.

However, the Assertion claims the opposite (that they are not equal). Since the Assertion is false, the Reason cannot be a correct explanation of it. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
$\mathrm{NH_2}$
›Reveal solutionSolution
Tests which nitrogenous bases are shared between DNA and RNA. Answer: adenine and cytosine (option D).

Concept and Intuition

DNA and RNA share three of their four bases — the two purines adenine (A) and guanine (G), and the pyrimidine cytosine (C). The fourth base differs: DNA uses thymine (T, a methylated pyrimidine), while RNA uses uracil (U, the non-methylated pyrimidine) in its place. This substitution (methyl group present/absent) is a classic structural distinguishing feature.

Step-by-Step Solution

Identify each structure: A = adenine (purine, bicyclic, $\mathrm{NH_2}$ substituent) — found in both DNA and RNA.

B = uracil (pyrimidine-2,4-dione, no methyl) — found only in RNA.

C = cytosine (4-amino pyrimidin-2-one) — found in both DNA and RNA.

D = thymine (5-methyl uracil) — found only in DNA. …
Start your 14-day free trial to see the full solution →

←Previous questionQ52 — Explain the terms primary and secondary structure of proteins. What is…→Next questionQ54 — Carbohydrates are classified on the basis of their behaviour on hydrol…

Q.Write the structures of fragments produced on complete hydrolysis of DNA. How are they linked in DNA molecule? Draw a diagram to show pairing of nucleotide bases in double helix of DNA.

Concept and Intuition

Step-by-Step Solution

🎓Unlock everything free for 14 days

Unlock everything free for 14 days