Match List-I (type of linkages) with List-II (Biomolecules)
|
List-I Type of linkage |
List-II Biomolecule |
|
(A) Glycosidic linkage |
(I) Tertiary structure of proteins |
|
(B) Peptide Linkage |
(II) Disaccharide |
|
(C) Phosphodiester linkage |
(III) Primary structure of proteins |
|
(D) Disulfide linkage |
(IV) Nucleotide |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
Each biomolecule is built by a typical link between its units: sugars by glycosidic links, amino acids by peptide links, nucleotides by phosphodiester links, and folded protein chains are held together in part by disulphide bridges.
(A) Glycosidic linkage — This matches (II) Disaccharide. Two monosaccharide units join through an oxygen atom with loss of water; in sucrose, for example, C1 of α-D-glucose is linked to C2 of β-D-fructose.
(B) Peptide Linkage — This matches (III) Primary structure of proteins. The –CO–NH– amide bonds fix the sequence of amino acids in the chain, which is the primary structure.
(C) Phosphodiester linkage — This matches (IV) Nucleotide. Nucleotides are joined by phosphodiester links between C5′ of one pentose sugar and C3′ of the next, forming the nucleic acid chain.
(D) Disulfide linkage — This matches (I) Tertiary structure of proteins. –S–S– bridges between cysteine residues, together with hydrogen bonds and other forces, hold the folded three-dimensional shape of the chain.
Shortcut: (A)-(II) appears in Options (2) and (3), and (C)-(IV) then leaves only Option (3).