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DNA/RNA constitution and biological functions: NEET MCQs with answers

50 questions on this topic of Biomolecules (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Which of the following pyrimidine bases present in nucleic acids is structurally named 5-methyluracil?

  1. Cytosine
  2. Thymine
  3. Uracil
  4. Adenine
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Answer: (B) Thymine

Thymine is chemically 5-methyluracil; it is a pyrimidine derivative with a methyl substitution at position 5 of the uracil ring. Cytosine is 4-amino-2-hydroxypyrimidine.

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2. Match List-I with List-II to choose the correct biological function or characteristic: List-I: (A) mRNA (B) tRNA (C) rRNA (D) DNA List-II: (I) Acts as a structural and catalytic component of ribosomes (II) Reservoir of genetic information in cells (III) Carries amino acids to the site of protein synthesis (IV) Carries genetic code from nucleus to ribosomes

  1. A-IV, B-III, C-I, D-II
  2. A-III, B-IV, C-I, D-II
  3. A-IV, B-I, C-III, D-II
  4. A-II, B-III, C-I, D-IV
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Answer: (A) A-IV, B-III, C-I, D-II

mRNA conveys genetic messages from DNA to the site of protein synthesis; tRNA adapts and transfers specific amino acids to growing polypeptide chains; rRNA forms the catalytic/structural core of ribosomes; and DNA serves as the central storehouse of genetic information.

hard

3. A linear trinucleotide sequence 5′-ApGpC-3′ is extracted from an RNA molecule. How many total phosphodiester linkages and free hydroxyl groups on the pentose sugars (excluding phosphate oxygens) are present in this molecule?

  1. Three phosphodiester linkages and three free pentose –OH groups
  2. Two phosphodiester linkages and four free pentose –OH groups
  3. Two phosphodiester linkages and three free pentose –OH groups
  4. Three phosphodiester linkages and two free pentose –OH groups
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Answer: (B) Two phosphodiester linkages and four free pentose –OH groups

In a linear trinucleotide of RNA, 3 nucleoside residues are joined by (3 - 1) = 2 phosphodiester linkages (between C3′ and C5′). Each of the 3 ribose sugars has an unlinked 2′-OH group (total 3), and the terminal 3′ sugar has an unesterified 3′-OH group (total 1). The 5′-end sugar has its C5′ either free or phosphorylated; considering the bare pentose hydroxyls not involved in internucleotide bonds, there are three 2′-OH and one 3′-OH, giving 4 free pentose –OH groups.

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4. The biological process by which a single strand of DNA serves as a template for the enzymatic synthesis of a complementary messenger RNA molecule is termed:

  1. Transcription
  2. Replication
  3. Translation
  4. Transversion
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Answer: (A) Transcription

Transcription is the biological process in which the genetic code from a DNA template strand is copied into a complementary single strand of messenger RNA (mRNA). Translation refers to protein synthesis directed by mRNA, while replication is DNA duplication.

easy

5. During which fundamental biological process does a cell synthesize an exact duplicate copy of its genetic material prior to cell division?

  1. Transcription
  2. Translation
  3. Mutation
  4. Replication
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Answer: (D) Replication

Replication is the enzyme-catalyzed process by which a single double-stranded DNA molecule duplicates to produce two identical DNA molecules, ensuring faithful transmission of genetic information during cell division. Transcription produces RNA from a DNA template, and translation synthesizes proteins from an mRNA template.

hard

6. A sample of double-stranded DNA contains 1200 nucleotides in total. If there are 180 thymine bases, what is the total number of hydrogen bonds holding the two strands together in this DNA fragment?

  1. 1380
  2. 1560
  3. 1740
  4. 1620
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Answer: (D) 1620

Total nucleotides = 1200, which means there are 600 base pairs. Thymine (T) = 180, so Adenine (A) = 180 (forming 180 A=T pairs). The remaining bases are G + C = 1200 - 360 = 840, meaning 420 G-C pairs. Each A=T pair has 2 H-bonds (180 × 2 = 360) and each G≡C pair has 3 H-bonds (420 × 3 = 1260). Total H-bonds = 360 + 1260 = 1620.

easy

7. Regarding the primary structure of RNA, which carbon position of the β-D-ribose unit bears a free hydroxyl group that is characteristically absent in the corresponding sugar unit of DNA?

  1. C1′
  2. C2′
  3. C3′
  4. C5′
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Answer: (B) C2′

RNA contains β-D-ribose which has a free –OH group at the C2′ position. In DNA, the sugar is β-D-2-deoxyribose, which lacks this hydroxyl group at C2′ (having two hydrogen atoms instead), contributing to the greater chemical stability of DNA.

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8. An analytical chemist isolated a pure nucleoside from yeast RNA. Alkaline hydrolysis does not break this unit down further into sugar and base. Which functional linkage links the components of this nucleoside?

  1. Phosphodiester linkage
  2. Ester linkage
  3. Peptide linkage
  4. C–N glycosidic linkage
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Answer: (D) C–N glycosidic linkage

A nucleoside consists solely of a nitrogenous base joined to a pentose sugar via an N-glycosidic (C1′–N) linkage. It lacks phosphate groups, meaning no phosphoester or phosphodiester bonds are present.

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9. Given below are two statements: Assertion (A): The genetic code is translated into proteins through the cooperative action of different RNA species. Reason (R): Protein synthesis occurs directly on the DNA template in eukaryotic cytoplasm without an RNA intermediate. Choose the correct option:

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (A) is true but (R) is false
  4. (A) is false but (R) is true
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Answer: (C) (A) is true but (R) is false

Assertion is true because mRNA, tRNA, and rRNA act collaboratively during the translation process. Reason is false because DNA remains inside the eukaryotic nucleus; it is first transcribed into mRNA, which moves to the cytoplasm where translation occurs on ribosomes.

easy

10. In a nucleoside, the nitrogenous base is attached to the pentose sugar via which specific linkage and at which position of the sugar?

  1. Phosphodiester linkage at C-5′ of pentose sugar
  2. α-N-glycosidic linkage at C-3′ of pentose sugar
  3. β-N-glycosidic linkage at C-1′ of pentose sugar
  4. O-glycosidic linkage at C-1′ of pentose sugar
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Answer: (C) β-N-glycosidic linkage at C-1′ of pentose sugar

A nucleoside is formed by the attachment of a nitrogenous base (purine or pyrimidine) to the C-1′ position of a pentose sugar (ribose or deoxyribose) through a β-N-glycosidic bond. Phosphodiester linkages connect adjacent nucleotides, not the base to the sugar.

easy

11. Given below are two statements: Statement I: DNA is the chemical basis of heredity and is regarded as the reserve of genetic information. Statement II: Protein synthesis in a cell is directed by various RNA molecules, but the message for the synthesis is present in DNA. In the light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (A) Both Statement I and Statement II are correct

According to NCERT Class 12 (Biomolecules), DNA is uniquely responsible for maintaining the identity of species over millions of years and serves as the reserve of genetic information. Although proteins are synthesized by various RNA molecules in the cell, the primary message for the synthesis of a particular protein is encoded in DNA.

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12. Which of the following chemical components remains entirely unchanged during the enzymatic cleavage of DNA into individual mononucleotide units?

  1. The hydrogen bonds between complementary bases
  2. The phosphodiester linkage
  3. The N-β-glycosidic bond
  4. The primary sequence of the intact chromosome
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Answer: (C) The N-β-glycosidic bond

Hydrolysis of DNA into mononucleotides cleaves the phosphodiester bonds linking adjacent nucleotides, but the N-β-glycosidic bond connecting the nitrogenous base to C-1′ of the β-D-2-deoxyribose unit remains intact within each mononucleotide unit. Hydrogen bonds and the continuous primary sequence are disrupted.

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