Chapter 10: Biomolecules Long Questions | chemistry Class 12 CBSE

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Chapter 10: Biomolecules Long Questions | chemistry Class 12 CBSE | ByteNotes.in

Long Answer

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Question 1

Long Form

Compare the structures of starch, cellulose, and glycogen with respect to their monomer units, types of glycosidic linkages, degree of branching, and biological roles.

Starch is a polymer of alpha-D-glucose with two components — amylose (unbranched, C1–C4 linkage) and amylopectin (branched, C1–C4 chain with C1–C6 branches); it serves as the main food storage polysaccharide in plants.
Cellulose is a straight-chain polysaccharide of beta-D-glucose units joined by C1–C4 glycosidic linkage; it is the most abundant organic substance in the plant kingdom and forms the structural cell wall.
Glycogen is the carbohydrate storage form in animals; it resembles amylopectin in structure but is more highly branched, and is present in liver, muscles, and brain.
The alpha-glycosidic linkage in starch and glycogen allows enzymatic digestion and serves energy storage; the beta-linkage in cellulose makes it resistant to digestion by most animals, giving structural rigidity.
Thus, the type of glycosidic linkage (alpha vs. beta) and degree of branching determine both the physical properties (solubility, texture) and the biological role (storage vs. structural) of these polysaccharides.

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Question 2

Long Form

Describe the four levels of protein structure. Which levels are destroyed during denaturation and why does the protein lose its biological activity?

Primary structure is the specific sequence of amino acids in a polypeptide chain linked by peptide bonds; any change in this sequence creates a different protein.
Secondary structure is the shape of the polypeptide chain — either alpha-helix (hydrogen bonds between C=O and –NH– within the chain twisting into a right-handed screw) or beta-pleated sheet (chains laid side by side with intermolecular hydrogen bonds).
Tertiary structure is the overall folding of the polypeptide chain, giving rise to fibrous or globular shapes, stabilised by hydrogen bonds, disulphide linkages, van der Waals forces, and electrostatic forces.
Quaternary structure describes the spatial arrangement of two or more polypeptide subunits relative to each other, as in haemoglobin.
Denaturation destroys secondary and tertiary structures (by disrupting hydrogen bonds) while the primary structure remains intact; since the specific 3D shape is essential for biological activity, its loss causes the protein to become non-functional.

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Long Answer

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Question 3

Long Form

Analyse the evidence used to establish the open-chain structure of glucose, pointing out which observations led to identification of specific functional groups.

The molecular formula C6H12O6 and formation of n-hexane on heating with HI established that all six carbons are in a straight chain.
Reaction with hydroxylamine to form an oxime and addition of HCN to give cyanohydrin confirmed the presence of a carbonyl (>C=O) group.
Oxidation with mild bromine water to gluconic acid (a 6-carbon monocarboxylic acid) established that the carbonyl is an aldehyde, not a ketone.
Acetylation with acetic anhydride yielding glucose pentaacetate confirmed the presence of five –OH groups, each on a different carbon atom.
Oxidation with nitric acid producing saccharic acid (a dicarboxylic acid) from both glucose and gluconic acid confirmed the presence of a primary alcoholic –OH at the terminal carbon.

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Question 4

Long Form

Explain why the open-chain structure of glucose is insufficient to account for all its properties, and describe how the cyclic hemiacetal structure resolves these anomalies.

Despite having a –CHO group, glucose does not give Schiff's test and does not form a hydrogen sulphite addition product with NaHSO3, which the open-chain structure cannot explain.
The pentaacetate of glucose does not react with hydroxylamine, indicating the absence of a free –CHO group, contradicting the open-chain structure.
Glucose exists in two crystalline forms — alpha (m.p. 419 K, from concentrated solution at 303 K) and beta (m.p. 423 K, from hot saturated solution at 371 K) — which cannot be explained by a single open-chain form.
The cyclic hemiacetal structure forms when the –OH at C-5 adds to the –CHO group, creating a six-membered pyranose ring and eliminating the free aldehyde.
The two cyclic forms (alpha and beta anomers) differ only in the configuration of –OH at C-1 (the anomeric carbon) and exist in equilibrium with the open-chain form, explaining both the absence of aldehyde reactions and the existence of two crystalline forms.

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Question 5

Long Form

Compare sucrose, maltose, and lactose as disaccharides with respect to their component units, glycosidic linkage positions, and reducing/non-reducing character.

Sucrose is formed from alpha-D-glucose (C1) and beta-D-fructose (C2) via a C1–C2 glycosidic linkage; since both reducing groups are used in bonding, sucrose is a non-reducing sugar.
Maltose consists of two alpha-D-glucose units joined by a C1–C4 glycosidic bond; a free aldehyde group can be generated at C1 of the second glucose unit, making maltose a reducing sugar.
Lactose (milk sugar) is composed of beta-D-galactose and beta-D-glucose linked by a C1–C4 glycosidic bond; a free aldehyde group may be produced at C1 of the glucose unit, so lactose is also a reducing sugar.
Hydrolysis of sucrose with dilute acid or enzyme gives an equimolar mixture of D-(+)-glucose and D-(–)-fructose; since fructose is more laevorotatory (–92.4°) than glucose is dextrorotatory (+52.5°), the product mixture is laevorotatory and called invert sugar.
All three disaccharides are formed by glycosidic linkages with loss of water, but the specific carbon atoms involved and whether reducing groups are free or bonded determine their chemical reactivity.

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Question 6

Long Form

Classify amino acids based on the nature of their R groups (acidic, basic, neutral) and explain why amino acids behave as salts with high melting points despite being organic compounds.

Neutral amino acids have equal numbers of –NH2 and –COOH groups (e.g., glycine, alanine); acidic amino acids have more –COOH groups than –NH2 (e.g., glutamic acid, aspartic acid); basic amino acids have more –NH2 than –COOH (e.g., lysine, arginine).
In aqueous solution, the –COOH group loses a proton and the –NH2 group gains a proton, forming a zwitter ion (dipolar ion) that is electrically neutral overall but carries both positive and negative charges.
This internal salt formation means that amino acids are essentially ionic compounds in the solid state, which is why they have high melting points comparable to inorganic salts.
They are highly soluble in water because the charged zwitter ionic form interacts strongly with the polar water molecules through electrostatic and hydrogen-bonding interactions.
This amphoteric nature — reacting with both acids (via –NH3+) and bases (via –COO–) — further reflects the salt-like character of amino acids.

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Question 7

Long Form

Analyse the chemical composition and structural differences between DNA and RNA, including sugar, bases, strand structure, and biological roles.

DNA contains beta-D-2-deoxyribose as the sugar, while RNA contains beta-D-ribose; the absence of an –OH at the 2' carbon in deoxyribose makes DNA chemically more stable.
DNA contains four bases: A, G, C, and thymine (T); RNA contains A, G, C, and uracil (U) — thymine is replaced by uracil in RNA.
DNA has a double-strand helix structure (Watson-Crick model) with hydrogen bonds between complementary base pairs (A–T, G–C); RNA is single-stranded and sometimes folds back on itself.
DNA is the chemical basis of heredity, maintains species identity, undergoes self-duplication during cell division, and contains the coded message for protein synthesis.
RNA carries out the actual protein synthesis: mRNA carries the genetic message to ribosomes, rRNA forms the ribosome, and tRNA brings amino acids to the ribosome during translation.

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Medium

Question 8

Long Form

Describe the classification of vitamins with examples and explain why water-soluble vitamins must be regularly consumed while fat-soluble vitamins need not be taken as frequently.

Vitamins are classified as fat-soluble (A, D, E, K) — soluble in fats and oils, insoluble in water — and water-soluble (B group: B1, B2, B6, B12; and C) — soluble in water.
Fat-soluble vitamins are stored in liver and adipose (fat-storing) tissues, so they accumulate in the body and do not need to be consumed every day.
Water-soluble vitamins are readily excreted in the urine and cannot be stored in the body (except Vitamin B12), so they must be supplied regularly in the diet to avoid deficiency.
Deficiency of fat-soluble vitamins leads to diseases such as night blindness (A), rickets/osteomalacia (D), increased RBC fragility (E), and prolonged blood clotting time (K).
Deficiency of water-soluble vitamins causes beriberi (B1), cheilosis (B2), convulsions (B6), pernicious anaemia (B12), and scurvy (C); since the body cannot store them, continuous dietary supply is critical.

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Question 9

Long Form

Evaluate the role of hormones in maintaining homeostasis in the body, citing specific examples of steroid and polypeptide hormones and their regulatory functions.

Hormones are chemical messengers produced by endocrine glands and transported in the blood to target sites, where they regulate biological activities and maintain the balance of body functions.
Insulin (polypeptide) lowers blood glucose by responding to its rise, while glucagon raises blood glucose; together they maintain blood glucose homeostasis within a narrow range.
Glucocorticoids (steroid hormones from adrenal cortex) control carbohydrate metabolism and modulate inflammatory responses; mineralocorticoids regulate water and salt excretion by the kidney.
Thyroxine (amino acid derivative from thyroid gland) regulates metabolic rate; its deficiency causes hypothyroidism (lethargy, obesity) and its excess causes hyperthyroidism.
Testosterone and estradiol (steroid sex hormones) are responsible for secondary male and female characteristics respectively, while progesterone prepares the uterus for implantation of a fertilised egg.

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Long Answer

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Question 10

Long Form

Describe the structure of a nucleotide and explain how nucleotides are linked together in a nucleic acid chain, including the type of bond and the positions of atoms involved.

A nucleotide consists of three components: a nitrogenous base (purine or pyrimidine), a pentose sugar (ribose in RNA, deoxyribose in DNA), and a phosphate group attached at the 5' position of the sugar.
The base attaches to the 1' position of the sugar forming a nucleoside; the phosphate group is then esterified at the 5' carbon of the nucleoside to form the nucleotide.
Nucleotides are joined together in a polynucleotide chain by phosphodiester linkages: the 3'-OH of one sugar is linked to the 5'-phosphate of the next sugar through a phosphate bridge.
This 3'–5' phosphodiester bond forms the backbone of the nucleic acid, which consists of alternating sugar and phosphate groups, with the bases projecting outward.
In DNA, two such strands are wound around each other in a double helix, held by hydrogen bonds between complementary base pairs (A–T, G–C); in RNA, the single strand can fold back on itself through base pairing.

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Chapter 10: Biomolecules | Long Questions

Compare the structures of starch, cellulose, and glycogen with respect to their monomer units, types of glycosidic linkages, degree of branching, and biological roles.

Describe the four levels of protein structure. Which levels are destroyed during denaturation and why does the protein lose its biological activity?

Analyse the evidence used to establish the open-chain structure of glucose, pointing out which observations led to identification of specific functional groups.

Explain why the open-chain structure of glucose is insufficient to account for all its properties, and describe how the cyclic hemiacetal structure resolves these anomalies.

Compare sucrose, maltose, and lactose as disaccharides with respect to their component units, glycosidic linkage positions, and reducing/non-reducing character.

Classify amino acids based on the nature of their R groups (acidic, basic, neutral) and explain why amino acids behave as salts with high melting points despite being organic compounds.

Analyse the chemical composition and structural differences between DNA and RNA, including sugar, bases, strand structure, and biological roles.

Describe the classification of vitamins with examples and explain why water-soluble vitamins must be regularly consumed while fat-soluble vitamins need not be taken as frequently.

Evaluate the role of hormones in maintaining homeostasis in the body, citing specific examples of steroid and polypeptide hormones and their regulatory functions.

Describe the structure of a nucleotide and explain how nucleotides are linked together in a nucleic acid chain, including the type of bond and the positions of atoms involved.

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