NCERT Solutions for Class 12 Chemistry Chapter 10: Biomolecules

Concept. A substance dissolves in water only if it can interact strongly with water molecules — chiefly by forming hydrogen bonds — so that the energy released on solvation compensates for breaking the water–water hydrogen bonds. Polar molecules rich in –OH (or other H-bonding) groups dissolve; non-polar molecules do not.

Explanation. Glucose and sucrose carry many –OH (hydroxyl) groups (glucose has five, sucrose eight). Each –OH can both donate and accept hydrogen bonds to water, so these sugars are extensively hydrogen-bonded to water and dissolve readily. Cyclohexane and benzene are non-polar hydrocarbons with no –OH or other polar group; they cannot form hydrogen bonds with water, and the weak water–hydrocarbon attraction cannot make up for the water–water hydrogen bonds that would have to be broken. Hence they are insoluble in water (though soluble in non-polar solvents). The six-membered ring shape is irrelevant — what matters is the presence or absence of H-bonding groups.

Conclusion: Glucose and sucrose dissolve because their many –OH groups form hydrogen bonds with water, whereas non-polar cyclohexane and benzene cannot hydrogen-bond with water and are therefore insoluble.

Concept. Lactose (“milk sugar”) is a disaccharide: two monosaccharide units joined by a glycosidic linkage. Hydrolysis (acid or the enzyme lactase) breaks this glycosidic C–O–C bond, adding a molecule of water and giving the two constituent monosaccharides.

Explanation. Lactose is made of one unit of β-D-galactose and one unit of D-glucose, joined by a β-1,4-glycosidic linkage between C1 of galactose and C4 of glucose. On hydrolysis it gives equal amounts of D-galactose and D-glucose. (People who lack the enzyme lactase cannot hydrolyse lactose — “lactose intolerance”.)

Conclusion: Hydrolysis of lactose gives one molecule of D-galactose and one molecule of D-glucose.

Concept. The open-chain form of glucose has a free –CHO (aldehyde) group, which would react with reagents such as hydroxylamine (to give an oxime). But glucose exists overwhelmingly in the cyclic hemiacetal form, in which the C1 aldehyde has reacted with the C5 –OH; in the ring there is no free aldehyde group, only the anomeric –OH at C1.

Explanation. When glucose is acetylated with acetic anhydride, the reaction acetylates the cyclic form: all five –OH groups (including the anomeric –OH at C1) are converted to –OCOCH₃ (acetate) groups, giving glucose pentaacetate. Because the C1 oxygen is now locked as an acetate within the ring, the ring cannot open back to the open-chain aldehyde. Therefore glucose pentaacetate does not react with hydroxylamine (it forms no oxime) — showing the absence of a free aldehyde group. This is one of the key pieces of evidence that glucose has a cyclic structure.

Conclusion: In the pentaacetate the anomeric –OH at C1 is acetylated, locking the cyclic (hemiacetal) form so it cannot revert to the open-chain aldehyde; the aldehyde group is therefore absent and the pentaacetate does not react with hydroxylamine.

Concept. An amino acid has both an acidic (–COOH) and a basic (–NH₂) group on the same molecule. These react internally: the –COOH transfers its proton to the –NH₂, so the amino acid exists as a zwitterion (dipolar ion), ⁺H₃N–CHR–COO⁻. A halo acid (e.g. ClCH₂COOH) has no basic group, so it stays as an ordinary molecule.

Explanation. Because an amino acid exists as a zwitterion, it is effectively an ionic (salt-like) substance: the crystal is held together by strong electrostatic (ionic) forces between the ⁺NH₃ and COO⁻ groups of neighbouring molecules. Breaking this ionic lattice needs a lot of energy, so amino acids have high melting points (they usually melt with decomposition above 200 °C). The charged ⁺NH₃ and COO⁻ groups also interact strongly with the polar water molecules (ion–dipole attraction), so amino acids are very soluble in water. A halo acid, being a normal molecular compound held only by weaker dipole–dipole and hydrogen-bonding forces, has a much lower melting point and lower water solubility.

Conclusion: Amino acids exist as zwitterions (⁺H₃N–CHR–COO⁻), so their crystals are held by strong ionic forces (high melting points) and interact strongly with water (high solubility), unlike the simple molecular halo acids.

Concept. The white of an egg is largely the soluble globular protein albumin dissolved in water. On heating, the protein undergoes denaturation — the hydrogen bonds and other weak interactions that held its secondary and tertiary (folded) structure are broken, and the unfolded polypeptide chains tangle and coagulate into a solid network.

Explanation. When the egg is boiled, the denatured protein chains coagulate and form a semi-solid, three-dimensional network; the water molecules that were formerly dispersed around the soluble protein become trapped (absorbed/entangled) within this coagulated protein network. The water does not escape as vapour (the egg is sealed in its shell) — it is simply held inside the solid mass of coagulated protein. That is why a boiled egg feels dry and firm rather than watery, even though no water has actually been removed.

Conclusion: On boiling, the egg’s albumin is denatured and coagulates into a solid protein network, within which the water present in the egg becomes trapped (absorbed) — it is not expelled.

Concept. Vitamins are classified by their solubility into fat-soluble (A, D, E, K), which are stored in the liver and fatty (adipose) tissues, and water-soluble (the B-group and C), which are not stored because they dissolve in the body’s aqueous fluids and are readily excreted.

Explanation. Vitamin C (ascorbic acid) is water-soluble. Any excess dissolves in the body fluids (blood and tissue fluid) and is readily excreted in the urine rather than being deposited in a storage tissue. Because it is continuously lost and not stored, vitamin C must be supplied regularly in the diet (its deficiency causes scurvy). By contrast, the fat-soluble vitamins can be stored in fat and so need not be taken daily.

Conclusion: Vitamin C is water-soluble, so any excess is readily excreted in the urine rather than stored in the body; it must therefore be supplied regularly in the diet.

Concept. A nucleotide is made of three parts joined together: a nitrogenous base, a pentose sugar and a phosphate group. In DNA the sugar is 2-deoxy-D-ribose. Hydrolysis breaks the bonds joining these three components, releasing them separately.

Explanation. On complete hydrolysis, a thymine-containing DNA nucleotide gives three products: (1) the base thymine (a pyrimidine), (2) the sugar 2-deoxy-D-ribose (the DNA sugar, which lacks the –OH at C2), and (3) phosphoric acid (H₃PO₄). (If only the phosphate ester bond is broken, the product is the nucleoside thymidine plus phosphoric acid; full hydrolysis also splits the base from the sugar.)

Conclusion: Hydrolysis of a thymine-containing DNA nucleotide gives thymine, 2-deoxy-D-ribose and phosphoric acid.

Concept. In DNA, the two strands are held together by complementary base pairing (A pairs with T, G pairs with C), so the amount of adenine equals that of thymine and the amount of guanine equals that of cytosine (the Chargaff rule, A = T and G = C). This fixed relationship is a direct consequence of the double-stranded structure.

Explanation. When RNA is hydrolysed, the quantities of its bases (A, G, C, U) show no such fixed relationship — adenine need not equal uracil, nor guanine equal cytosine. This tells us that RNA is not a regularly base-paired double helix: in other words, RNA is normally a single-stranded molecule. Because there is no complementary strand forcing A to pair with U and G with C throughout, the base composition is not constrained, and the amounts of the four bases are unrelated. (RNA can fold back on itself locally, but it is not a continuously paired double strand like DNA.)

Conclusion: The lack of any fixed relationship among the base amounts shows that RNA is single-stranded (not a regularly base-paired double helix like DNA).

Concept and definition. Carbohydrates are classified by whether they can be broken down into simpler sugars. Monosaccharides are the simplest carbohydrates — polyhydroxy aldehydes or ketones that cannot be hydrolysed further into smaller sugar units. They are the building blocks from which all larger carbohydrates (disaccharides and polysaccharides) are made.

Explanation and examples. A monosaccharide has the general formula roughly Cₙ(H₂O)ₙ and contains one carbonyl group (an aldehyde → an aldose, or a ketone → a ketose) together with several –OH groups. They are further named by the number of carbon atoms: triose (3C), tetrose (4C), pentose (5C), hexose (6C). Common examples are glucose and galactose (aldohexoses), fructose (a ketohexose) and ribose (an aldopentose). About 20 monosaccharides occur in nature; glucose is the most important.

Conclusion: Monosaccharides are the simplest carbohydrates — polyhydroxy aldehydes or ketones (e.g. glucose, fructose, ribose) — that cannot be hydrolysed to smaller sugar molecules.

Concept and definition. A reducing sugar is a carbohydrate that reduces Tollens’ reagent (gives a silver mirror) and/or Fehling’s solution (gives a red precipitate of Cu₂O). It can do so because it has a free aldehyde or ketone group, or a free anomeric hydroxyl (hemiacetal/hemiketal –OH) that can open to a free carbonyl group, which is then oxidised.

Explanation and examples. All monosaccharides (whether aldoses like glucose, or ketoses like fructose) are reducing sugars. Among disaccharides, maltose and lactose are reducing (they still have a free anomeric –OH that can open to a carbonyl), whereas sucrose is non-reducing because both of its anomeric carbons are tied up in the glycosidic linkage, leaving no free carbonyl group. (Fructose reduces Fehling’s/Tollens’ because in the basic medium it isomerises to aldoses.)

Conclusion: Reducing sugars are carbohydrates with a free aldehyde/ketone or anomeric –OH group that reduce Tollens’ and Fehling’s reagents — all monosaccharides and the disaccharides maltose and lactose (but not sucrose).

Concept. Carbohydrates, the first products of photosynthesis, serve plants in two broad ways — as an energy store and as structural material.

Explanation. Both functions are carried out by polysaccharides built from glucose but differing in their glycosidic linkages: starch (α-linkages, coiled, easily hydrolysed → an energy reserve) and cellulose (β-linkages, straight chains, strong and indigestible → a structural material).

Conclusion: In plants carbohydrates (1) store energy, as starch, and (2) provide structural support, as cellulose in the cell walls.

Concept. A monosaccharide cannot be hydrolysed to a simpler sugar; a disaccharide is made of two monosaccharide units joined by a glycosidic linkage and gives two monosaccharides on hydrolysis.

Explanation. Ribose, 2-deoxyribose, galactose and fructose each exist as a single sugar unit that cannot be hydrolysed further → monosaccharides. Maltose and lactose each hydrolyse to give two monosaccharide molecules → disaccharides.

Conclusion: Monosaccharides — ribose, 2-deoxyribose, galactose, fructose; disaccharides — maltose and lactose.

Concept and definition. A glycosidic linkage is the ether-type (C–O–C) bond that joins two monosaccharide units in a disaccharide or polysaccharide. It is formed between the anomeric carbon (C1) of one sugar and a hydroxyl group of the other sugar, with the loss of a molecule of water (a condensation reaction).

Explanation and example. For example, in sucrose the glycosidic linkage joins C1 of α-D-glucose to C2 of β-D-fructose; in maltose it joins C1 of one α-D-glucose to C4 of another (an α-1,4 linkage). Because this bond is an acetal (formed from the hemiacetal –OH), it can be broken by hydrolysis (acid or enzymes) to regenerate the two monosaccharides.

Conclusion: A glycosidic linkage is the ether (C–O–C) bond, formed with loss of water, that joins the anomeric carbon of one monosaccharide to a hydroxyl group of another in di- and polysaccharides.

Concept. Glycogen is the storage polysaccharide of animals (also called “animal starch”): a polymer of α-D-glucose units stored in the liver and muscles, and hydrolysed to glucose when the body needs energy. Structurally it resembles the amylopectin of starch but is much more highly branched.

Explanation. Both are polymers of α-D-glucose joined by α-1,4 linkages (with α-1,6 at the branch points), and both are energy reserves. The difference is that glycogen consists only of highly branched molecules, whereas starch is a mixture of the linear amylose and the moderately branched amylopectin; glycogen is also found in animals, starch in plants. The extensive branching of glycogen allows glucose to be released rapidly when energy is needed.

Conclusion: Glycogen is the highly branched storage polysaccharide of animals (“animal starch”); it differs from starch in being found in animals, in consisting only of highly branched chains (no linear amylose), and in being more extensively branched than amylopectin.

Concept. A disaccharide is hydrolysed (by acid or a specific enzyme) at its glycosidic linkage into its two constituent monosaccharides.

Explanation. Sucrose (α-D-glucose joined to β-D-fructose) hydrolyses to an equimolar mixture of glucose and fructose; because fructose is strongly laevorotatory, the overall rotation changes sign (dextrorotatory sucrose → laevorotatory mixture), so the product is called invert sugar. Lactose (β-D-galactose joined to D-glucose) hydrolyses to glucose and galactose.

Conclusion: (i) Sucrose gives D-glucose + D-fructose (invert sugar); (ii) lactose gives D-glucose + D-galactose.

Concept. Both starch and cellulose are polysaccharides made entirely of D-glucose units, but they differ in the configuration of the glycosidic linkage (α vs β), which in turn decides their shape, properties and whether humans can digest them.

Explanation. The basic structural difference is the type of glycosidic linkage: starch is built from α-D-glucose with α-1,4 linkages, giving coiled, branched molecules, whereas cellulose is built from β-D-glucose with β-1,4 linkages, giving long straight chains that pack together into strong fibres. This single difference (α vs β) explains why starch is a digestible energy store and cellulose a tough, indigestible structural material.

Conclusion: Starch consists of α-D-glucose units joined by α-glycosidic (α-1,4) linkages, while cellulose consists of β-D-glucose units joined by β-glycosidic (β-1,4) linkages — this difference in linkage gives their different shapes, digestibility and roles.

Concept. These three reactions are part of the evidence for the open-chain structure of glucose (an aldohexose, CH₂OH–(CHOH)₄–CHO). HI is a strong reducing/reducing-plus-deoxygenating agent that strips off the oxygens; bromine water is a mild oxidant specific for the aldehyde group; nitric acid is a strong oxidant that oxidises both ends of the chain.

Explanation. (i) Prolonged heating with HI reduces glucose all the way to n-hexane, proving that its six carbon atoms form a straight chain. (ii) Bromine water, a mild oxidising agent, oxidises only the aldehyde group to a –COOH, giving the six-carbon monocarboxylic acid gluconic acid — proving the presence of a –CHO group. (iii) Concentrated nitric acid, a stronger oxidant, oxidises both the aldehyde (–CHO) and the terminal primary alcohol (–CH₂OH) to –COOH groups, giving the dicarboxylic saccharic acid — proving glucose has a primary –CH₂OH at the end opposite the aldehyde.

Conclusion: (i) HI gives n-hexane; (ii) bromine water gives D-gluconic acid (oxidises –CHO only); (iii) nitric acid gives D-saccharic acid (oxidises both –CHO and –CH₂OH).

Concept. Although the open-chain (aldohexose) structure explains most of glucose’s reactions, several observations cannot be explained by a free –CHO group. These are the facts that led chemists to propose the cyclic hemiacetal (pyranose) structure, in which the C5 –OH has added to the C1 aldehyde, creating the anomeric carbon.

Explanation. All four observations are explained by the cyclic structure: glucose exists mainly as a six-membered ring in which the aldehyde is “hidden” as a hemiacetal, so it does not behave as a free aldehyde (facts 1 and 2); ring closure creates a new stereocentre at C1, giving two anomers (fact 3) that interconvert through the small amount of open-chain form, producing mutarotation (fact 4).

Conclusion: The reactions not explained by the open chain are: glucose’s failure to give the Schiff’s/NaHSO₃ aldehyde tests, the pentaacetate forming no oxime, the existence of two anomers (α- and β-D-glucose), and mutarotation — all accounted for by the cyclic hemiacetal structure.

Concept. The amino acids needed by the body are divided by whether the body can make them. Essential amino acids cannot be synthesised in the body (or not in sufficient amounts) and must be supplied in the diet. Non-essential amino acids can be synthesised by the body itself, so they need not be obtained from food.

Explanation. A healthy diet must include the essential amino acids because their absence impairs protein synthesis, growth and health; the non-essential ones are built up by the body from other molecules as needed. (Of the about 20 amino acids in proteins, roughly 10 are essential for humans.)

Conclusion: Essential amino acids cannot be made by the body and must come from the diet (e.g. valine, leucine); non-essential amino acids can be synthesised by the body (e.g. glycine, alanine).

(i) Peptide linkage. A peptide linkage is the amide bond (–CO–NH–) that joins two amino acids: it is formed between the –COOH group of one amino acid and the –NH₂ group of the next, with the elimination of a molecule of water. A chain of many amino acids joined by peptide bonds is a polypeptide (protein).

(ii) Primary structure. The primary structure of a protein is the specific sequence (order) in which the amino acids are linked to one another in the polypeptide chain. Each protein has its own unique amino-acid sequence; any change in the order gives a different protein.

(iii) Denaturation. Denaturation is the loss of the biological activity of a protein due to the disruption of its secondary and tertiary structures by heat, change in pH, or heavy-metal ions and other chemicals. The hydrogen bonds and other weak interactions are broken, the protein unfolds and coagulates; the primary structure (the sequence) remains intact, but the natural shape and function are lost (e.g. boiling an egg, curdling of milk).

Conclusion: (i) A peptide linkage is the –CO–NH– amide bond between two amino acids (formed with loss of water); (ii) the primary structure is the sequence of amino acids in the chain; (iii) denaturation is the disruption of a protein’s secondary and tertiary structure (loss of function), the primary sequence remaining intact.

Concept. The secondary structure of a protein is the regular, local folding of the polypeptide backbone, held together by hydrogen bonds between the >C=O and >N–H groups of the peptide bonds. There are two common types.

Explanation. Both arrangements maximise hydrogen bonding between the backbone peptide groups. In the α-helix a single chain twists into a coil stabilised by H-bonds running along the helix (as in wool keratin); in the β-pleated sheet several extended chains line up side by side and are held by H-bonds between adjacent chains (as in silk fibroin).

Conclusion: The two common secondary structures are the α-helix (a coiled chain held by intramolecular hydrogen bonds) and the β-pleated sheet (extended chains held side-by-side by inter-chain hydrogen bonds).

Concept and answer. The α-helix is stabilised by hydrogen bonds. Specifically, these are intramolecular hydrogen bonds between the carbonyl oxygen (>C=O) of one peptide bond and the amide hydrogen (>N–H) of another peptide bond further along the same polypeptide chain (the C=O of one residue bonds to the N–H of the fourth residue ahead).

Explanation. As the chain coils, every backbone C=O group lines up with an N–H group one turn away, and the resulting regular array of hydrogen bonds running roughly parallel to the helix axis holds the coil rigidly in shape. Because these H-bonds are intramolecular (within one chain), the α-helix is a property of a single polypeptide chain (in contrast to the β-sheet, whose H-bonds are between different chains).

Conclusion: The α-helix is stabilised by intramolecular hydrogen bonds between the >C=O and >N–H groups of peptide bonds in the same polypeptide chain.

Concept. Proteins are classified by their shape (tertiary structure) into two broad types — globular and fibrous — which differ in the way the polypeptide chains are folded and hence in their solubility and function.

Explanation. In globular proteins the chain coils and folds upon itself into a compact ball, so the molecule is soluble and mobile — suited to functional roles such as catalysis (enzymes) and transport. In fibrous proteins the chains run parallel and are cross-linked into tough, insoluble fibres — suited to structural and protective roles.

Conclusion: Globular proteins have compact, folded, water-soluble molecules with functional roles (e.g. insulin, enzymes); fibrous proteins have long, parallel, water-insoluble chains with structural roles (e.g. keratin, collagen).

Concept. A substance is amphoteric if it can react both as an acid (donate a proton) and as a base (accept a proton). An amino acid has both an acidic group (–COOH) and a basic group (–NH₂), so it can do both.

Explanation via the zwitterion. In the solid state and in aqueous solution, an amino acid exists as a zwitterion (dipolar ion), ⁺H₃N–CHR–COO⁻, formed by internal transfer of the –COOH proton to the –NH₂ group. This dipolar ion has both a positive (⁺NH₃) and a negative (COO⁻) centre, so it can respond to both acids and bases:

Conclusion of the mechanism. Because the zwitterion contains both an acidic (⁺NH₃) and a basic (COO⁻) group, the amino acid reacts with acids (the carboxylate accepts H⁺) and with bases (the ammonium donates H⁺) — this dual behaviour is its amphoteric character. At the isoelectric point (a particular pH) the molecule exists entirely as the neutral zwitterion with no net charge.

Conclusion: Amino acids exist as zwitterions (⁺H₃N–CHR–COO⁻), which contain both a basic –COO⁻ group (accepts H⁺ from acids) and an acidic ⁺NH₃ group (donates H⁺ to bases); this dual reactivity is their amphoteric behaviour.

Concept and definition. Enzymes are biological catalysts — almost all of them globular proteins — that catalyse the biochemical reactions occurring in living organisms. They speed up reactions enormously (by lowering the activation energy) while remaining unchanged themselves, and they are highly specific, each enzyme usually catalysing only one particular reaction of one particular substrate.

Explanation. Enzymes are extremely efficient (a tiny amount transforms a large amount of substrate) and work best at an optimum temperature (around body temperature) and an optimum pH. They are usually named after the substrate or the reaction they catalyse, with the suffix -ase — for example maltase hydrolyses maltose to glucose, urease hydrolyses urea, and oxidases catalyse oxidation. Their specificity is often explained by the “lock-and-key” fit between the enzyme’s active site and the substrate.

Conclusion: Enzymes are biocatalysts — globular proteins that are highly specific and efficient — which catalyse the biochemical reactions in living organisms (e.g. maltase hydrolyses maltose to glucose).

Concept. A native protein owes its biological activity to its precisely folded three-dimensional shape, which is held by hydrogen bonds, disulphide bridges, ionic and van der Waals interactions. Denaturation is the disruption of this folding by heat, a change in pH, or chemicals/heavy-metal ions.

Effect on structure. During denaturation the weak forces that maintain the secondary and tertiary structures are broken: the hydrogen bonds of the α-helix/β-sheet and the interactions holding the tertiary fold are disrupted, so the globular protein uncoils/unfolds and the chains become disordered and often coagulate. Crucially, the primary structure (the sequence of amino acids joined by peptide bonds) is NOT affected — the covalent peptide backbone remains intact. Because the natural shape is lost, the protein loses its biological activity. Familiar examples are the coagulation of egg white (albumin) on boiling and the curdling of milk.

Conclusion: Denaturation disrupts the secondary and tertiary structures of a protein (uncoiling and coagulating it, with loss of biological activity) while the primary structure — the amino-acid sequence — remains unchanged.

Concept. Vitamins are organic compounds needed in small amounts for normal growth and health, which the body (mostly) cannot synthesise. They are classified by their solubility into two groups.

Vitamin for blood coagulation. The vitamin responsible for the coagulation (clotting) of blood is vitamin K (a fat-soluble vitamin); its deficiency leads to delayed clotting and excessive bleeding.

Conclusion: Vitamins are classified as fat-soluble (A, D, E, K — stored in the body) and water-soluble (B-group and C — not stored); vitamin K is responsible for the coagulation of blood.

Concept. Each vitamin is essential because its absence causes a specific deficiency disease; the body cannot synthesise them (in adequate amounts), so they must be supplied in the diet.

Explanation. Vitamin A is essential because it is required for the proper functioning of the eyes (and for growth and healthy epithelial tissue); its lack causes night blindness and xerophthalmia. Vitamin C is essential for maintaining healthy connective tissue and preventing scurvy; because it is water-soluble it cannot be stored and must be taken regularly.

Conclusion: Vitamin A is essential for healthy vision and skin (its lack causes night blindness/xerophthalmia; sources: fish-liver oil, carrots, butter, milk), and vitamin C is essential to prevent scurvy and maintain connective tissue (sources: citrus fruits, amla, green vegetables).

Concept and definition. Nucleic acids are the biomolecules that carry the hereditary (genetic) information of living cells. They are long-chain polymers (polynucleotides) of nucleotides, each nucleotide being made of a nitrogenous base, a pentose sugar and a phosphate group. The two kinds are DNA (deoxyribonucleic acid) and RNA (ribonucleic acid), found mainly in the nucleus of the cell.

Conclusion: Nucleic acids (DNA and RNA) are polymers of nucleotides that carry genetic information; their two main functions are the replication and transmission of hereditary information, and the control of protein synthesis in the cell.

Concept. Both are the structural units of nucleic acids, built from a nitrogenous base, a pentose sugar and (in one of them) a phosphate group. The difference is simply the presence or absence of the phosphate group.

Explanation. A nucleoside is formed when a nitrogenous base is joined to the C1 of a pentose sugar (a β-N-glycosidic bond). A nucleotide is formed when a phosphate group is esterified to the C5 –OH of the sugar of a nucleoside. The nucleotides are the monomers that join through phosphodiester linkages to build DNA and RNA.

Conclusion: A nucleoside is made of a base and a sugar only, whereas a nucleotide is a nucleoside plus a phosphate group (base + sugar + phosphate).

Concept. DNA is a double helix of two polynucleotide strands held together by hydrogen bonds between the nitrogenous bases. The pairing is strictly selective: a purine always pairs with a specific pyrimidine — adenine (A) pairs with thymine (T) by two hydrogen bonds, and guanine (G) pairs with cytosine (C) by three hydrogen bonds. This selective pairing is what “complementary” means.

Explanation. Because of the fixed base-pairing rule, the sequence of bases on one strand determines the sequence on the other: wherever one strand has A, the other must have T; wherever one has G, the other must have C. So the two strands are not identical (they do not have the same sequence) but complementary (each base on one strand is matched by its partner base on the other). For example, if one strand reads 5′–A–G–C–T–3′, its partner reads 3′–T–C–G–A–5′. This is why, in any DNA, the amount of adenine equals that of thymine and the amount of guanine equals that of cytosine (Chargaff’s rule, A = T and G = C). Complementarity is also the basis of accurate DNA replication: each strand serves as a template for building a new complementary partner.

Conclusion: The two DNA strands are held by specific base pairing (A–T and G–C), so each base on one strand is matched by its complementary partner on the other; the strands are therefore complementary (one determines the other) rather than identical.

Concept. DNA and RNA are both nucleic acids built from nucleotides, but they differ in their sugar, one of their bases, the number of strands and their biological role.

Explanation. Structurally, DNA has the sugar deoxyribose, the base thymine and two strands wound in a double helix, whereas RNA has ribose, uracil (in place of thymine) and is normally a single strand. Functionally, DNA is the store of genetic information and duplicates itself in replication (the chemical basis of heredity), while RNA uses that information to synthesise proteins.

Conclusion: DNA has deoxyribose, thymine and two strands and stores/transmits genetic information; RNA has ribose, uracil and a single strand and carries out protein synthesis.

Concept. RNA (ribonucleic acid) carries out the synthesis of proteins using the information copied from DNA. There are three main types of RNA in the cell, each with a distinct role in protein synthesis.

Explanation. Together these three cooperate to build proteins: mRNA brings the coded instructions from DNA, tRNA ferries the amino acids, and rRNA (in the ribosome) provides the site and machinery where the amino acids are joined in the order specified by the mRNA.

Conclusion: The three types of RNA in the cell are messenger RNA (mRNA), ribosomal RNA (rRNA) and transfer RNA (tRNA), which together carry out protein synthesis.