Q.(A) Define the following term :

Showing the 12 most recent of 20 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.A small segment of a polypeptide gave on complete hydrolysis 3 molecules of alanine, 2 molecules of glycine and 3 molecules of cysteine. What is the number of peptide linkages in the segment of the polypeptide? (A) 7 (B) 8 (C) 6 (D) 5

   ›Reveal solutionSolution

   The number of peptide bonds in a polypeptide is always one less than the number of amino acid residues. Here, total residues = 3 + 2 + 3 = 8, so peptide bonds = 8 − 1 = 7. The correct option is (A).

   Concept & Intuition

   A polypeptide is a chain of amino acids linked by peptide bonds. Each peptide bond forms between the carboxyl group of one amino acid and the amino group of the next. If you have $n$ amino acids in a chain, you need exactly $n-1$ peptide bonds to connect them — like linking $n$ beads on a string requires $n-1$ knots. Hydrolysis breaks all these bonds, releasing the individual amino acids. So counting the total number of amino acids released tells you the original chain length, and subtracting one gives the number of peptide bonds.

   Step-by-step reasoning

   1. Identify the total number of amino acid residues.

      Complete hydrolysis of the polypeptide yields:

      • 3 molecules of alanine
      • 2 molecules of glycine
      • 3 molecules of cysteine Total = $3 + 2 + 3 = 8$ amino acid molecules.

   2. Relate residues to peptide bonds. …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.The secondary structure of protein consists of: (A) A long chain of amino acids linked with each other in a specific sequence (B) A folding of the polypeptide chains which exists as fibrous and globular (C) A long polypeptide chain which exists $\alpha$-helix and $\beta$-pleated sheet structure (D) Two or more polypeptide chains which exist as sub-units

   ›Reveal solutionSolution

   The secondary structure of a protein refers to the local, regular folding patterns of the polypeptide backbone, specifically the α-helix and β-pleated sheet, not the sequence, overall shape, or subunit assembly. The correct option is (C).

   ---

   The key here is to recall the hierarchy of protein structure: primary, secondary, tertiary, and quaternary. Each level describes a different aspect of the protein’s architecture. The secondary structure is the local spatial arrangement of the polypeptide backbone, stabilized mainly by hydrogen bonds between the carbonyl oxygen and amide hydrogen of amino acids that are close in sequence. It does not involve the side chains (R-groups) or the overall 3D shape of the entire chain.

   Let’s examine each option:

   1. Option (A) describes “a long chain of amino acids linked in a specific sequence.” That is the primary structure — the linear order of amino acids held together by peptide bonds. This is not secondary structure.

   2. Option (B) says “a folding of the polypeptide chains which exists as fibrous and globular.” This refers to the tertiary structure (the overall 3D shape of a single polypeptide chain) or even the classification of proteins based on shape. Fibrous and globular are categories of tertiary/quaternary structure, not secondary. …

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3. COMEDK 2025Set 2025-E1 markMCQ
   Q.Two statements, one Assertion and the other Reason are given. Choose the right option. Assertion : Insulin is called a protein whereas Glycyl alanine is not called a protein Reason : A polypeptide with amino acid residue less than 100 can also be called as a protein if it has a well-defined conformation of a protein. (A) Assertion is correct but Reason is incorrect (B) Both Assertion and Reason are incorrect (C) Both Assertion and Reason are correct (D) Assertion is incorrect but Reason is correct

   ›Reveal solutionSolution

   Both statements are correct: insulin (51 residues, well-defined conformation) qualifies as a protein even though it has fewer than 100 residues, whereas glycyl-alanine (a dipeptide) does not.

   Assertion: Insulin is a protein; glycyl-alanine (a dipeptide) is not. This is true — insulin, though only 51 amino-acid residues, has a definite three-dimensional structure and biological function, so it is classed as a protein, while a simple dipeptide is not. …

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4. COMEDK 2025Set 2025-M1 markMCQ
   Q.Choose the incorrect statement. (A) The hormone Glucocorticoid controls the level of excretion of water and salts by the kidney. (B) Vitamins $A$ and $K$ are fat soluble and are stored in the liver of human beings. (C) Denaturation of protein is due to loss of both the secondary and tertiary structures of the protein. (D) Complete hydrolysis of RNA gives Nitrogen containing bases, a pentose sugar and phosphoric acid.

   ›Reveal solutionSolution

   The question asks for the incorrect statement. Glucocorticoids regulate metabolism and inflammation, not water/salt excretion (that’s mineralocorticoids like aldosterone). So option (A) is false; the others are true.

   Concept & Intuition

   This is a biology fact-checking problem. You need to recall specific functions of hormones, properties of vitamins, protein denaturation, and RNA hydrolysis. The trick is to spot the mismatch between a hormone’s actual role and the statement. Glucocorticoids (e.g., cortisol) are often confused with mineralocorticoids (e.g., aldosterone) because both come from the adrenal cortex. The key distinction: glucocorticoids handle stress, metabolism, and immune suppression; mineralocorticoids handle electrolyte and water balance.

   Step-by-step reasoning

   1. Evaluate statement (A):

      Glucocorticoids (like cortisol) primarily regulate glucose metabolism, suppress inflammation, and help the body respond to stress. They do not control water and salt excretion by the kidney. That job belongs to mineralocorticoids (e.g., aldosterone), which promote sodium retention and potassium excretion. Therefore, statement (A) is incorrect.

   2. Evaluate statement (B):

      Vitamins A, D, E, and K are fat-soluble. Vitamin A is stored in the liver (as retinol esters), and vitamin K is also stored in the liver (though in smaller amounts). This statement is correct.

   3. Evaluate statement (C): …

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5. KCET 2024Set B-21 markMCQ
   Q.$\alpha$–D–(+)–glucose and $\beta$–D–(+)–glucose are: (A) Enantiomers (B) Conformers (C) Epimers (D) Anomers

   ›Reveal solutionSolution

   α- and β-D-glucose differ in configuration at exactly one carbon — C-1, the anomeric carbon created on ring closure — so by definition they are anomers.

   1. Where the α/β difference comes from

   Open-chain D-glucose is an aldohexose. Its C-5 –OH attacks the C-1 aldehyde carbon intramolecularly, forming a six-membered cyclic hemiacetal (the pyranose ring):

   $\mathrm{CHO}\ \text{(C-1, planar, sp}^2) \longrightarrow \overset{\mathrm{OH}}{\underset{\mathrm{H}}{\mathrm{C}}}\text{(C-1, sp}^3\text{, a NEW stereocentre)}$

   Because the carbonyl carbon is planar, the ring oxygen can close from either face, so the new –OH at C-1 ends up either:

   • down (trans to the $\mathrm{CH_2OH}$; on the same side as the C-2 OH in Fischer projection) ⇒ α-D-(+)-glucose, or
   • up ⇒ β-D-(+)-glucose.

   The carbon whose configuration is fixed by this ring closure is called the anomeric carbon, and the two products are anomers. In solution they interconvert through the open-chain form — the phenomenon of mutarotation ($[\alpha]_D$ drifting from $+111^\circ$ for pure α and $+19^\circ$ for pure β to the equilibrium $+52.7^\circ$).

   2. Eliminate the other terms precisely

   1. (A) Enantiomers — non-superimposable mirror images: they must differ at every stereocentre. α- and β-D-glucose share the same configuration at C-2, C-3, C-4 and C-5, so they are not mirror images (they are diastereomers). ✗ …

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6. COMEDK 2024Set 2024-A1 markMCQ
   Q.Among the following peptides, identify the pair where the name and its structure are correctly matched. (A) (B) (C) (D)

   ›Reveal solutionSolution

   The key is to match the side‑chain sequence in the condensed structure with the standard three‑letter amino‑acid abbreviations (N‑terminal to C‑terminal). Only option (C) correctly names the tripeptide Ala‑Gly‑Phe.

   Concept & Intuition

   Peptide names are written from the N‑terminal (free amino end) to the C‑terminal (free carboxyl end). Each three‑letter code corresponds to a specific side chain:

   • Gly (glycine): side chain = H (i.e., the α‑carbon is –CH₂–).
   • Ala (alanine): side chain = –CH₃.
   • Phe (phenylalanine): side chain = –CH₂C₆H₅ (benzyl).

   To check correctness, read the structure from left (N‑terminus) to right (C‑terminus) and list the side chains in order. Then compare that sequence to the name.

   ---

   Step‑by‑step reasoning

   1. Option (A):

      Structure: H₂N–CH₂–C(=O)–NH–CH(–CH₂SH)–COOH

      • First residue: α‑carbon has no side chain (just –CH₂–) → Gly.
      • Second residue: α‑carbon has –CH₂SH (cysteine side chain) → Cys, not Ala. Name given: “Gly‑Ala” → mismatch (should be Gly‑Cys). ✗ Incorrect.

   2. Option (B):

      Structure: H₂N–CH₂–C(=O)–NH–CH(–CH₃)–COOH

      • First residue: –CH₂– → Gly.
      • Second residue: –CH₃ → Ala. Name given: “Ala‑Gly” → order reversed (N‑terminal is Gly, not Ala). ✗ Incorrect.

   3. Option (C):

      Structure: H₂N–CH(–CH₃)–C(=O)–NH–CH₂–C(=O)–NH–CH(–CH₂C₆H₅)–COOH

      • First residue: –CH₃ → Ala.
      • Second residue: –CH₂– (no side chain) → Gly.
      • Third residue: –CH₂C₆H₅ → Phe. Name given: “Ala‑Gly‑Phe” → matches exactly. ✓ Correct.

   4. Option (D): …

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7. COMEDK 2024Set 2024-A1 markMCQ
   Q.Which of the following statement is not true about glucose? (A) Glucose does not give Schiff's reagent test (B) Glucose on reaction with $\mathrm{HI}$ and red $\mathrm{P}$ at $373 \mathrm{~K}$ gives a mixture of cyclohexane and 1-iodohexane (C) Both $\alpha$-D-glucose and $\beta$-D-glucose undergo mutarotation in aqueous solutions (D) Glucose reacts with hydroxylamine to form an oxime

   ›Reveal solutionSolution

   The question asks which statement about glucose is false. By checking each option against known glucose chemistry, we find that option (B) is incorrect because the reaction with HI and red P at 373 K yields n-hexane, not a mixture of cyclohexane and 1-iodohexane.

   Concept & Intuition

   Glucose is an aldohexose — a six-carbon sugar with an aldehyde group (in open-chain form) and multiple hydroxyl groups. Its reactions depend on these functional groups: the aldehyde gives typical carbonyl reactions (oxime formation, no Schiff’s test because it’s usually cyclic), the hydroxyls can be reduced, and the anomeric carbon allows mutarotation. To spot the false statement, we need to recall the specific outcome of each reaction, especially the drastic reduction with HI and red phosphorus, which removes all oxygen atoms.

   Step-by-step reasoning

   1. Option (A): Glucose does not give Schiff’s reagent test

      Schiff’s reagent tests for aldehydes, but glucose exists predominantly in cyclic hemiacetal form, where the aldehyde group is masked. The equilibrium concentration of free aldehyde is too low to give a positive test under normal conditions. Thus, this statement is true.

   2. Option (B): Glucose on reaction with HI and red P at 373 K gives a mixture of cyclohexane and 1-iodohexane

      HI and red phosphorus are a powerful reducing agent that removes all hydroxyl groups (as water) and reduces any carbonyl to methylene. For glucose (C₆H₁₂O₆), the complete reduction yields straight-chain n-hexane (C₆H₁₄). No cyclohexane forms because the carbon skeleton remains unbranched and no ring closure occurs under these conditions. Also, 1-iodohexane would be an intermediate, but excess HI and red P drive reduction all the way to the alkane. So the product is n-hexane, not a mixture of cyclohexane and 1-iodohexane. This statement is false.

   3. Option (C): Both α-D-glucose and β-D-glucose undergo mutarotation in aqueous solutions …

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8. COMEDK 2024Set 2024-E1 markMCQ
   Q.Given are 4 statements related to the chemical properties of Glucose. Identify the two incorrect statements from the following. A. It reacts with $\mathrm{Br}_2$ (aq) to form Saccharic acid. B. Reacts with Acetic anhydride to form Glucose tetraacetate. C. Reacts with Hydroxylamine to give Glucose oxime. D. It reacts with ammoniacal $\mathrm{AgNO}_3$ to form ammonium salt of Gluconic acid with deposition of silver. (A) A & B (B) C & D (C) B & D (D) A & C

   ›Reveal solutionSolution

   Statements A (Br$_2$ water → saccharic acid) and B (acetic anhydride → tetraacetate) are the incorrect ones; C and D correctly describe glucose chemistry.

   A. Br$_2$(aq) → Saccharic acid — INCORRECT. Bromine water is a mild oxidising agent; it oxidises only the –CHO group of glucose to –COOH, giving gluconic acid. Saccharic (glucaric) acid, a dicarboxylic acid, requires the stronger oxidant conc. HNO$_3$.

   B. Acetic anhydride → Glucose tetraacetate — INCORRECT. Glucose has five –OH groups, all acetylated, giving glucose pentaacetate, not tetraacetate. …

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9. COMEDK 2024Set 2024-M1 markMCQ
   Q.Identify the 2 chemical tests which is not answered by Glucose having an open chain structure [A] Reaction with Schiff's reagent and with Sodium bisulphite [B] Reaction with $\mathrm{HCN}$ and with $\mathrm{HI}$ [C] Reaction with $\mathrm{HNO}_3$ and with Acetic anhydride [D] Reaction with aqueous Bromine and with Hydroxylamine (A) [D] (B) [B] (C) [C] (D) [A]

   ›Reveal solutionSolution

   Glucose’s open-chain aldehyde structure explains most of its reactions, but two tests—reaction with Schiff’s reagent and with sodium bisulphite—are not given by the open-chain form because in solution glucose exists predominantly as a cyclic hemiacetal, which lacks a free aldehyde group.

   The key concept here is mutarotation and the equilibrium between open-chain and cyclic forms of glucose. In water, glucose is almost entirely (over 99%) in its cyclic pyranose form. Only a tiny fraction exists as the free aldehyde. Many aldehyde-specific tests require a high enough concentration of the free aldehyde to give a visible positive result. Some tests are so sensitive that even the trace amount of open-chain form suffices; others are not.

   Let’s examine each pair of tests.

   1. Reaction with Schiff’s reagent and with Sodium bisulphite

      • Schiff’s reagent (fuchsine decolorized by SO₂) gives a magenta color with aldehydes. However, glucose gives a very slow or no color change because the equilibrium concentration of the free aldehyde is too low to react quickly.
      • Sodium bisulphite (NaHSO₃) adds to aldehydes to form a crystalline bisulphite addition product. Glucose does not form such a solid adduct under normal conditions, again because the open-chain form is too scarce.
      • Conclusion: These two tests are not answered by the open-chain structure in practice.

   2. Reaction with HCN and with HI

      • HCN adds to the carbonyl group of the open-chain form to give a cyanohydrin. This reaction does occur (though slowly) because the equilibrium shifts as the open-chain form is consumed.
      • HI reduces the aldehyde group (and also the alcohol groups) under drastic conditions, but the aldehyde is still the reactive site.
      • Conclusion: Both are possible via the open-chain form.

   3. Reaction with HNO₃ and with Acetic anhydride

      • HNO₃ oxidizes the aldehyde group to a carboxylic acid (giving gluconic acid) and also the terminal CH₂OH to COOH (giving glucaric acid). This clearly involves the open-chain aldehyde.
      • Acetic anhydride reacts with all –OH groups (including the hemiacetal OH) to form an acetate. This does not require the open-chain form; it reacts with the cyclic form as well. So this test is not exclusive to the open-chain structure, but the question asks which tests are not answered by the open-chain structure — meaning which tests fail if only the open-chain form is considered. Acetic anhydride works fine, so it’s not a “not answered” case.
      • Conclusion: HNO₃ works via the open-chain; acetic anhydride works anyway.

   4. Reaction with aqueous Bromine and with Hydroxylamine …

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10. COMEDK 2024Set 2024-M1 markMCQ
    Q.Given below are 4 statements about Insulin. Which of these statement/(s) is/are correct? [A] Insulin is a globular protein consisting of 51 Amino acids. [B] Insulin is constituted of 3 polypeptide chains linked together. [C] Insulin is constituted of 2 polypeptide chains linked together by disulphide bonds. [D] The 3 polypeptide chains in Insulin are linked together by Hydrogen bonding. (A) [D] (B) [A] & [C] (C) [B] & [D] (D) [B]

    ›Reveal solutionSolution

    Insulin is a globular protein with 51 amino acids arranged in two polypeptide chains (A and B) linked by disulphide bonds, so only statements [A] and [C] are correct.

    Concept and Intuition

    Insulin is a classic example of a small, well-studied protein. Its structure is fundamental in biochemistry: it is synthesized as a single chain (proinsulin) that is later cleaved to remove a connecting peptide (C-peptide), leaving two chains — the A chain (21 amino acids) and the B chain (30 amino acids). These two chains are held together by disulphide bonds (covalent bonds between cysteine residues), not by hydrogen bonds between separate chains. Hydrogen bonds are important for the protein’s three-dimensional folding, but they do not link the two chains together. Knowing this, we can evaluate each statement.

    Step-by-step reasoning

    1. Statement [A]: “Insulin is a globular protein consisting of 51 Amino acids.”

       • Insulin is indeed a globular protein (soluble, roughly spherical shape). The total number of amino acids in the active form is 21 (A chain) + 30 (B chain) = 51.
       • This statement is correct.

    2. Statement [B]: “Insulin is constituted of 3 polypeptide chains linked together.”

       • Mature insulin has only two chains (A and B). The third chain (C-peptide) is present only in the precursor proinsulin and is removed during processing.
       • This statement is incorrect.

    3. Statement [C]: “Insulin is constituted of 2 polypeptide chains linked together by disulphide bonds.”

       • Exactly right: the A and B chains are connected by two interchain disulphide bonds (between A7–B7 and A20–B19). There is also one intrachain disulphide bond within the A chain (A6–A11).
       • This statement is correct.

    4. Statement [D]: “The 3 polypeptide chains in Insulin are linked together by Hydrogen bonding.” …

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11. KCET 2023Set B-41 markMCQ
    Q.Match List-I and List-II with respect to proteins and their functions and select the correct option. List-I | List-II

    1. Collagen | p. Fights infectious agents
    2. Trypsin | q. Hormone
    3. Insulin | r. Enzyme
    4. Antibody | s. Intercellular ground substance (A) 1-s, 2-p, 3-r, 4-p (B) 1-q, 2-r, 3-q, 4-s (C) 1-s, 2-q, 3-r, 4-p (D) 1-s, 2-r, 3-q, 4-p

    ›Reveal solutionSolution

    Assign each protein its biological role: collagen → ground substance, trypsin → enzyme, insulin → hormone, antibody → defence.

    Concept — Proteins and their functions (Biomolecules). NCERT gives a table of proteins with their functions; each protein here has exactly one role.

    1. Collagen — the most abundant protein in the animal world; it is the fibrous protein of the intercellular ground substance of connective tissue. → s

    2. Trypsin — a proteolytic enzyme secreted (as trypsinogen) by the pancreas; it hydrolyses proteins in the small intestine. → r

    3. Insulin — a peptide hormone secreted by the β-cells of the islets of Langerhans; it lowers blood glucose. → q

    4. Antibody — an immunoglobulin that fights infectious agents (antigens/pathogens). → p …

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12. COMEDK 2023Set 2023-E1 markMCQ
    Q.Structures of 3 Monosaccharides are given below. Two of them are Anomers. Identify the two Anomers. (A) II is the anomer of both I and II (B) I and II are Anomers. (C) II and III are Anomers. (D) I and III are Anomers

    ›Reveal solutionSolution

    [I] and [III] differ only at the anomeric carbon C1 (identical at C2–C4), so they are anomers; [II] also differs at C2 and is not an anomer of either.

    Definition: anomers are a special pair of epimers (cyclic monosaccharides) that differ in configuration only at the anomeric carbon (C1, the former carbonyl carbon).

    Compare the structures at C1–C4 (left/right substituents):

    • [I]: C1 = H/OH, C2 = H/OH, C3 = HO/H, C4 = H/OH
    • [III]: C1 = HO/H, C2 = H/OH, C3 = HO/H, C4 = H/OH
    • [II]: C1 = HO/H, C2 = HO/H, C3 = HO/H, C4 = H/OH …

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