Q.Nucleic acids are the polymers of _______.

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Which of the following is/are present both in DNA and RNA? [FIGURE] (four structures labelled I-IV: I = thymine, a pyrimidinedione ring with a 5-methyl ($H_3C$) substituent and NH/C=O groups; II = a furanose sugar ring ($HOCH_2$ at C5, ring O, OH at the anomeric carbon, HO and OH substituents on the ring) resembling ribose; III = cytosine, a pyrimidine ring with an $NH_2$ substituent and a C=O; IV = uracil, a pyrimidinedione ring similar to I but without the methyl group) (only = only) (A) I, II (B) III only (C) II only (D) I, IV

   ›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

   1. Identify structure I: a pyrimidine ring with a 5-methyl group and two carbonyls (2,4-dioxo) — this is thymine, found only in DNA.
   2. 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.
   3. Identify structure III: a pyrimidine ring with an $NH_2$ group and one carbonyl — this is cytosine, present in BOTH DNA and RNA. …

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2. 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

   1. Identify ring type: six-membered, single ring $\Rightarrow$ pyrimidine (rules out adenine, a fused bicyclic purine). …

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3. 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

   1. (A) Adenine, Uracil — Adenine is a genuine DNA purine, but Uracil is an RNA base, not found in DNA. Rejected.
   2. (B) Cytosine, Guanine — the order is reversed: Cytosine is the pyrimidine and Guanine is the purine, but the question asks purine-then-pyrimidine. Rejected.
   3. (C) Thymine, Uracil — both entries are pyrimidines (no purine at all), and Uracil again doesn't belong to DNA. Rejected. …

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4. 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

   1. Purine bases (common to both DNA and RNA): adenine (A), guanine (G).
   2. Pyrimidine bases in DNA: cytosine (C) and thymine (T).
   3. Pyrimidine bases in RNA: cytosine (C) and uracil (U) — uracil replaces thymine. …

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5. 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

   1. 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.
   2. 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. …

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6. 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≡C, A≡T, G≡C, T≡A) (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≡C, A=T, C≡G, 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≡T, A=C, T≡G, 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≡T, U=G, A≡T)

   ›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:

   1. 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.
   2. 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

   1. 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.

   2. 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. …

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7. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
   Q.Which of the following are present in DNA? A. [FIGURE] (six-membered pyrimidine-base ring with an $\mathrm{NH_2}$ substituent, one ring N, and a C=O adjacent to an N–H — cytosine-like structure) B. [FIGURE] (six-membered pyrimidine-base ring with two C=O groups and two N–H — uracil-like structure) C. [FIGURE] (five-membered furanose sugar ring with $\mathrm{HOCH_2}$, OH and H substituents including an anomeric OH — a ribose-like sugar) D. [FIGURE] (fused five/six-membered purine-base ring system with an $\mathrm{NH_2}$ substituent, three ring N atoms, and an N–H — adenine-like structure) (A) A, D (B) A, C, D (C) B, C (D) B, C, D

   ›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.

   1. 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.

   2. 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.

   3. Match to the answer choices. …

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8. 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

   1. A nucleotide = base + sugar + phosphate (the full monomer unit of DNA/RNA).
   2. Removing the phosphate group leaves base + sugar only — this simpler unit is called a nucleoside.
   3. So the correct composition of a nucleoside is Sugar–Base, matching option (B).

   Common Mistakes …

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9. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
   Q.Nucleotides in RNA are joined by which of the following phosphodiester linkage? (A) $5^1, 3^1$ (B) $5^1, 2^1$ (C) $3^1, 2^1$ (D) $3^1, 3^1$

   ›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

   1. 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."
   2. This linkage is described as the 3',5'-phosphodiester bond (equivalently 5',3').
   3. Options B, C, D describe linkages between other positions (2', 3' combinations) that are not the actual biological linkage. …

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10. 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

    1. 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.
    2. The base-sugar (glycosidic) bond forms at the sugar's anomeric centre, which is C-1' for ribose/deoxyribose.
    3. Therefore the base is attached at C-1', matching option (A).

    Common Mistakes …

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11. 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

    1. 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.

    2. 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.

    3. 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. …

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12. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.Which of the following bases are present both in DNA and RNA? [FIGURE] (A: adenine - a purine bicyclic ring system with an $\mathrm{NH_2}$ substituent) [FIGURE] (B: uracil - a pyrimidine ring with two C=O groups and an NH, no amino or methyl substituent) [FIGURE] (C: cytosine - a pyrimidine ring with an $\mathrm{NH_2}$ substituent and one C=O group) [FIGURE] (D: thymine - a pyrimidine ring with a $\mathrm{CH_3}$ substituent and two C=O groups) (A) C, D (B) B, C (C) A, B (D) A, C

    ›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

    1. Identify each structure: A = adenine (purine, bicyclic, $\mathrm{NH_2}$ substituent) — found in both DNA and RNA.
    2. B = uracil (pyrimidine-2,4-dione, no methyl) — found only in RNA.
    3. C = cytosine (4-amino pyrimidin-2-one) — found in both DNA and RNA.
    4. D = thymine (5-methyl uracil) — found only in DNA. …

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