Q.Differentiate between globular and fibrous proteins.

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. …

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. …

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. …

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): …

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): …

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.” …

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 …

Concept & Intuition

Solubility in water depends on how well a molecule can interact with water molecules through hydrogen bonding or ionic interactions. Water is polar, so "like dissolves like" — polar and charged substances tend to dissolve, while nonpolar, hydrophobic ones do not.

Let's look at each option:

1. Fibrous protein — These are structural proteins (like collagen, keratin, elastin) that form long, sheet-like or rope-like fibers. Their amino acid sequences are rich in hydrophobic residues, and they are tightly packed with extensive cross-linking. They are designed to be insoluble in water, providing mechanical strength to tissues. They do not form colloidal solutions in water; they remain as solid fibers.

2. Amylose — This is a linear polymer of glucose (a polysaccharide), a component of starch. Each glucose unit has many –OH groups, which form hydrogen bonds with water. Amylose is actually water-soluble (it forms a colloidal solution in hot water, and even in cold water it disperses to some extent). So it is not generally insoluble.

3. Vitamin-C (ascorbic acid) — This is a small, polar molecule with multiple –OH groups and a lactone ring. It is highly water-soluble. In fact, it is one of the most water-soluble vitamins. Definitely not insoluble.

4. Glycine — The simplest amino acid. It exists as a zwitterion in water ($\ce{^+H3N-CH2-COO-}$), making it highly polar and very soluble in water. It dissolves readily.

Step 1 — Recognise the curve. A plot of substrate concentration against enzyme activity that saturates at a maximum is the Michaelis–Menten hyperbola:

Step 2 — Read the axes.

• X-axis = substrate concentration $[S]$ (stated in the stem).
• Y-axis = the velocity (rate) of the reaction $v$; it rises steeply at low $[S]$ and asymptotes to the horizontal line marked $V_{max}$, the maximum velocity reached when every active site is occupied. …

Step 1 — The hypothesis being tested.

Oparin and Haldane proposed that the first organic molecules formed abiotically from simple inorganic molecules in a reducing (oxygen-free, hydrogen-rich) primitive atmosphere, with energy supplied by lightning and heat. Miller set out to test exactly this.

Step 2 — The apparatus and the gases.

In 1953 S. L. Miller built a closed flask containing:

• $\mathrm{CH_4}$ (methane) — the carbon source
• $\mathrm{NH_3}$ (ammonia) — the nitrogen source
• $\mathrm{H_2}$ (hydrogen) — makes the mixture strongly reducing
• water vapour ($\mathrm{H_2O}$) — from a flask of boiling water

Electric discharge (~800 °C, simulating lightning) was passed through the mixture.

Step 3 — Why the other options fail. …