Chapter 10: Biomolecules

Solubility in water depends on the ability of a molecule to interact with water molecules, mainly through hydrogen bonding.

Glucose and sucrose are carbohydrates that contain a large number of polar hydroxyl ($$-\mathrm{OH}$$) groups. These $$-\mathrm{OH}$$ groups form extensive hydrogen bonds with water molecules, and the energy released in forming these bonds is enough to pull the solute molecules apart. Hence glucose and sucrose dissolve readily in water.

Cyclohexane and benzene are non-polar hydrocarbons. They have no polar group capable of forming hydrogen bonds with water; they can offer only weak van der Waals (London) forces, which cannot compensate for the strong hydrogen bonding among water molecules. Therefore they remain insoluble in water, although they dissolve readily in non-polar organic solvents.

Lactose (milk sugar) is a disaccharide. It is built up from one molecule of $$\beta$$-$$\mathrm{D}$$-galactose and one molecule of $$\mathrm{D}$$-glucose, joined together by a glycosidic linkage between C-1 of galactose and C-4 of glucose.

On hydrolysis (in the presence of dilute acid or the enzyme lactase), this glycosidic linkage is broken and a molecule of water is added:

$$\mathrm{Lactose} + \mathrm{H_2O} \xrightarrow{\mathrm{H^+}} \mathrm{D\text{-}Galactose} + \mathrm{D\text{-}Glucose}$$

Thus the expected products are one molecule of $$\mathrm{D}$$-galactose and one molecule of $$\mathrm{D}$$-glucose.

In aqueous solution glucose does not exist exclusively as the open-chain aldehyde. The $$-\mathrm{CHO}$$ group at C-1 reacts with the $$-\mathrm{OH}$$ group at C-5 to form a cyclic hemiacetal (the pyranose ring). In this cyclic form C-1 carries a new $$-\mathrm{OH}$$ group, called the anomeric hydroxyl.

When glucose is treated with acetic anhydride, all five $$-\mathrm{OH}$$ groups of the cyclic form get acetylated, including the anomeric $$-\mathrm{OH}$$ at C-1, giving glucose pentaacetate.

Once the anomeric $$-\mathrm{OH}$$ has been acetylated, the ring can no longer open up to regenerate the open-chain aldehyde form. Consequently the pentaacetate does not react with hydroxylamine ($$\mathrm{NH_2OH}$$); that is, it shows no reaction characteristic of a free aldehyde ($$-\mathrm{CHO}$$) group.

An amino acid contains both an acidic carboxyl group ($$-\mathrm{COOH}$$) and a basic amino group ($$-\mathrm{NH_2}$$) in the same molecule. The $$-\mathrm{COOH}$$ group transfers a proton to the $$-\mathrm{NH_2}$$ group, so in the solid state and in aqueous solution amino acids exist as dipolar ions called zwitterions:

$$\mathrm{H_2N\text{-}CHR\text{-}COOH} \longrightarrow \overset{+}{\mathrm{H_3N}}\text{-}\mathrm{CHR}\text{-}\mathrm{COO^-}$$

Because of this zwitterionic (ionic, salt-like) character, amino acid molecules are held together by strong electrostatic forces of attraction. A large amount of thermal energy is needed to break these forces, so amino acids have high melting points. The charged ends also interact strongly with polar water molecules, which makes amino acids readily soluble in water.

Halo acids ($$\mathrm{X\text{-}CHR\text{-}COOH}$$) do not form zwitterions. Their molecules are held together only by the weaker dipole-dipole and van der Waals forces, so they have comparatively lower melting points and lower solubility in water.

The egg (chiefly the egg white, albumin) is largely made up of globular protein dissolved in water.

On boiling, the protein gets denatured. Heat disrupts the hydrogen bonds and other weak interactions that maintain the protein's secondary and tertiary structure, so the coiled globular protein molecules uncoil and then coagulate into a solid mass.

During this coagulation the water that was present is absorbed (adsorbed and trapped) within the network of the coagulated protein. The water does not escape; it is held inside the solidified protein matrix, which is why a boiled egg appears dry.

On the basis of their solubility, vitamins are classified into fat-soluble vitamins (A, D, E, K) and water-soluble vitamins (the B-group and C).

Vitamin C (ascorbic acid) is a water-soluble vitamin. Water-soluble vitamins cannot be stored in the body because they are readily dissolved in water, and any excess is regularly excreted from the body through urine.

Since it cannot be stored, vitamin C must be supplied regularly through the diet.

A nucleotide is made up of three units: a nitrogenous base, a pentose sugar and a phosphate group. In DNA the sugar is 2-deoxy-$$\mathrm{D}$$-ribose.

On complete hydrolysis, a DNA nucleotide containing thymine breaks down into its three constituents:

• the nitrogenous base — thymine
• the pentose sugar — 2-deoxy-$$\mathrm{D}$$-ribose
• phosphoric acid ($$\mathrm{H_3PO_4}$$)

In DNA the two strands are complementary: adenine always pairs with thymine and guanine always pairs with cytosine. Because of this base pairing, on hydrolysis of DNA the amount of adenine equals that of thymine and the amount of guanine equals that of cytosine (Chargaff's rule).

When RNA is hydrolysed, no such fixed relationship is found among the quantities of the different bases. This indicates that there is no regular complementary base pairing in RNA.

Hence RNA does not possess a double-helix structure made of two complementary strands; it normally exists as a single-stranded molecule.

A carbohydrate that cannot be hydrolysed further to give a simpler unit of polyhydroxy aldehyde or ketone is called a monosaccharide.

Monosaccharides are the simplest carbohydrates and act as the building units of the larger carbohydrates (oligosaccharides and polysaccharides). About 20 monosaccharides are known to occur in nature. Common examples are glucose, fructose, ribose and galactose.

Carbohydrates which reduce Fehling's solution (giving a red-brown precipitate of $$\mathrm{Cu_2O}$$) and Tollens' reagent (giving a silver mirror) are called reducing sugars.

A sugar behaves as a reducing sugar if it has a free aldehydic or ketonic group, i.e. a free hemiacetal (anomeric) $$-\mathrm{OH}$$ group that can open up to give such a group. All monosaccharides (whether aldoses or ketoses) are reducing sugars. Among the disaccharides, maltose and lactose are reducing sugars, whereas sucrose is a non-reducing sugar.

Two main functions of carbohydrates in plants are:

• Storage of energy: Carbohydrates such as starch act as the reserve food material of the plant. Starch is stored in seeds, roots and other parts and is broken down to release energy whenever the plant needs it.
• Structural support: The carbohydrate cellulose is the chief structural material of the plant cell wall. It provides rigidity and mechanical strength to the plant body.

A monosaccharide cannot be hydrolysed into a simpler carbohydrate, whereas a disaccharide gives two monosaccharide units on hydrolysis.

Monosaccharides: ribose, 2-deoxyribose, galactose and fructose — none of these can be hydrolysed further.

Disaccharides: maltose and lactose — each gives two monosaccharide units on hydrolysis (maltose gives two glucose units; lactose gives glucose and galactose).

When two monosaccharide units combine, the reaction takes place between the hydroxyl groups of the two units with the elimination of a molecule of water. The two units then remain joined to each other through an oxygen atom.

This oxygen bridge that links two monosaccharide units is called the glycosidic linkage. For example, in sucrose a glycosidic linkage joins a glucose unit and a fructose unit, and in maltose a $$\mathrm{C_1\text{-}C_4}$$ glycosidic linkage joins two glucose units.

Glycogen is the carbohydrate in which excess glucose is stored in the animal body; for this reason it is also called animal starch. It is present mainly in the liver and in the muscles, and is broken down to glucose whenever the body needs energy.

Difference from starch:

• Starch is the reserve carbohydrate of plants, whereas glycogen is the reserve carbohydrate of animals (it also occurs in fungi).
• Starch is a mixture of two components — amylose (about 15–20%, an unbranched chain) and amylopectin (about 80–85%, a branched chain). Glycogen has a structure similar to amylopectin, but it is even more highly (more extensively) branched than amylopectin.

Step 1 ‒ Recall the monomer present in both biopolymers
Starch and cellulose are polysaccharides obtained on polymerising the same monosaccharide, namely $$\mathrm{D\text{-}glucose}$$.

Step 2 ‒ Identify the anomeric form of the repeating unit

• In starch every glucose residue enters the chain in its α-anomeric form, i.e. $$\alpha\text{-} \mathrm{D\!\text{-}glucopyranose}$$.
• In cellulose every glucose residue enters the chain in its β-anomeric form, i.e. $$\beta\text{-} \mathrm{D\!\text{-}glucopyranose}$$.

Step 3 ‒ Specify the type of glycosidic linkage

• Starch: The anomeric carbon 1 of an $$\alpha$$-glucose is linked to carbon 4 of the next residue giving an $$\alpha(1\,\rightarrow\,4)$$ glycosidic bond (with additional $$\alpha(1\,\rightarrow\,6)$$ bonds at branch points in amylopectin).
• Cellulose: The anomeric carbon 1 of a $$\beta$$-glucose is linked to carbon 4 of the next residue giving a straight $$\beta(1\,\rightarrow\,4)$$ glycosidic bond. There are no branches.

Step 4 ‒ Conclude the basic structural difference
The essential structural distinction, therefore, lies in the configuration about the anomeric carbon and, consequently, in the nature of the glycosidic linkage:

• Starch → chain of $$\alpha$$-D-glucose units joined mainly by $$\alpha(1\,\rightarrow\,4)$$ linkages.
• Cellulose → chain of $$\beta$$-D-glucose units joined exclusively by $$\beta(1\,\rightarrow\,4)$$ linkages.

Thus, although both are polymers of glucose, α-linking (starch) versus β-linking (cellulose) is the basic structural difference.

Objective: List the experimental facts shown by $$\mathrm{D}$$-glucose that contradict the simple open-chain aldehydic structure

$$\mathrm{CHO\!-\!(CHOH)_4\!-\!CH_2OH}$$

1. Expected behaviour of an open-chain aldehyde

An aldehyde group ($$-\mathrm{CHO}$$) is normally expected to:

• add $$\mathrm{NaHSO_3}$$ to give a crystalline bisulphite (hydrogensulphite) addition product,
• combine with ammonia ($$\mathrm{NH_3}$$) to give an aldehyde–ammonia addition compound,
• restore the magenta colour of decolourised Schiff's reagent.

When any of these characteristic tests fail, the carbonyl group must be "locked up" (not free) in the reacting species.

2. Reactions actually observed for $$\mathrm{D}$$-glucose that do not fit the open-chain picture

1. No bisulphite addition product. Glucose does not form a bisulphite (hydrogensulphite) addition product with $$\mathrm{NaHSO_3}$$, although a free $$-\mathrm{CHO}$$ would do so readily.

2. No aldehyde–ammonia addition compound. Glucose does not combine with ammonia ($$\mathrm{NH_3}$$) to give the characteristic aldehyde–ammonia adduct that simple aldehydes (e.g. acetaldehyde) produce readily.

3. No Schiff's test. A fresh glucose solution fails to restore the magenta colour of decolourised Schiff's reagent, again inconsistent with a free $$-\mathrm{CHO}$$ group.

4. Glucose pentaacetate does not react with $$\mathrm{NH_2OH}$$. Treatment of glucose with excess acetic anhydride gives the pentaacetate. If a free $$-\mathrm{CHO}$$ were still present in the pentaacetate it would form an oxime with hydroxylamine, but the pentaacetate gives no such reaction, showing that the carbonyl carbon is already tied up before acetylation.

5. Existence of two crystalline forms (α and β) and mutarotation. Glucose can be isolated in two crystalline forms: α-$$\mathrm{D}$$-glucose (m.p. 419 K; $$[\alpha]_\text{D}^{20}=+111^{\circ}$$) and β-$$\mathrm{D}$$-glucose (m.p. 423 K; $$[\alpha]_\text{D}^{20}=+19.2^{\circ}$$). In aqueous solution each form slowly attains the same equilibrium rotation of $$+52.5^{\circ}$$ — a behaviour called mutarotation that the open-chain structure cannot account for.

6. Formation of methyl α- and β-glucosides. Glucose reacts with methanol in the presence of dry $$\mathrm{HCl}$$ to give two isomeric methyl glucosides (acetal-like products). These glucosides do not reduce Fehling's solution and do not react with $$\mathrm{NH_2OH}$$, indicating that the carbonyl group is no longer free.

3. Conclusion

All the above anomalies are explained at once if glucose exists predominantly as a six-membered intramolecular hemiacetal (pyranose ring), in which the $$-\mathrm{CHO}$$ group at C-1 has reacted with the $$-\mathrm{OH}$$ at C-5; hence the open-chain formula alone is inadequate.

Human proteins are made up of 20 common α-amino acids.

For each amino acid we must ask: can the human metabolic machinery build it from simpler molecules?

1. Essential (indispensable) amino acids
   The human body cannot synthesise these in sufficient amount, so they have to be supplied ready-made in the food we eat.
2. Non-essential (dispensable) amino acids
   The body can synthesise these from other metabolites, so an external supply is not compulsory.

Examples required in the question:

*Histidine is essential only for growing children.

Step 1 – Recall the hierarchy of protein structure
The primary structure of a protein is the linear sequence of $$\alpha$$-amino acids linked through peptide ($$\mathrm{CONH}$$) bonds. Spatial folding of this chain, produced mainly by regular intra-chain hydrogen bonding, gives rise to the secondary structure.

Step 2 – Identify how peptide groups can hydrogen-bond regularly
The carbonyl oxygen $$\mathrm{C=O}$$ of one peptide bond can form a hydrogen bond with the $$\mathrm{N–H}$$ hydrogen of another peptide bond. When such $$\mathrm{C=O\;\cdots H–N}$$ interactions repeat in a uniform pattern, only a few stable conformations are geometrically possible.

Step 3 – List the conformations that satisfy these conditions

• $$\alpha$$-helix (“alpha helix”) — the polypeptide backbone coils into a right-handed spiral. Every $$n$$th $$\mathrm{C=O}$$ group hydrogen-bonds to the $$\mathrm{N–H}$$ of the $$(n+4)$$th residue, stabilising the helix.
• $$\beta$$-pleated sheet (“beta sheet”) — stretches of the chain lie side-by-side, forming a sheet in which adjacent strands are held together by hydrogen bonds. The strands may run in the same (parallel) or opposite (antiparallel) directions.

Hence, the two most common secondary structures observed in proteins are the $$\alpha$$-helix and the $$\beta$$-pleated sheet.

The $$\alpha$$-helix is a right-handed coiled (helical) secondary structure adopted by many polypeptide chains.

Inside the helix every peptide unit ($$-\mathrm{CO}-\mathrm{NH}-$$) is oriented so that:

• the carbonyl oxygen of the $$i^{\text{th}}$$ amino-acid residue points along the axis of the helix, while
• the $$\mathrm{N-H}$$ group of the $$(i+4)^{\text{th}}$$ residue points in the opposite direction.

This geometry allows a regular pattern of intramolecular hydrogen bonds to form:

$$\mathrm{C{=}O_{(i)} \;\cdots\; H{-}N_{(i+4)}}$$

Because each turn of the helix contains 3.6 amino-acid residues, every $$\mathrm{C{=}O}$$ group (except those at the ends) engages in one such bond, and the same is true for each $$\mathrm{N-H}$$ group. The cumulative effect of these multiple hydrogen bonds holds the coiled backbone firmly in place and stabilises the entire $$\alpha$$-helix.

Hence, the stabilising interaction is intramolecular hydrogen bonding between the carbonyl oxygen and the amide hydrogen of peptide bonds within the same polypeptide chain.

The primary structure of every protein is a long polypeptide chain. Depending on how this chain folds up in space, two broad classes of proteins are recognised — fibrous and globular.

Thus, fibrous proteins are tough, water-insoluble structural materials, whereas globular proteins are compact, water-soluble molecules responsible for nearly all biochemical activities in living systems.

Step 1 : Structure of an α-amino acid

An α-amino acid may be written as
$$\mathrm{H_2N\! -\! CHR\! -\! COOH}$$
where $$\mathrm{R}$$ is any side chain.

Step 2 : Internal acid–base reaction (zwitterion formation)

The –COOH group is acidic while the –NH₂ group is basic. In aqueous solution, a proton is transferred from the carboxyl to the amino group:

$$\mathrm{H_2N\! -\! CHR\! -\! COOH \;\rightleftharpoons\;{}^{-}OOC\! -\! CHR\! -\! NH_3^{+}}$$

The species on the right contains both a positively charged $$\mathrm{NH_3^{+}}$$ (basic centre) and a negatively charged $$\mathrm{COO^{-}}$$ (acidic centre); it is called a zwitterion.

Step 3 : Behaviour in an acidic medium

When extra $$\mathrm{H^{+}}$$ is present, the zwitterion accepts a proton at the carboxylate end, showing basic character:

$$\mathrm{^{-}OOC\! -\! CHR\! -\! NH_3^{+} + H^{+} \;\longrightarrow\; HOOC\! -\! CHR\! -\! NH_3^{+}}$$

Step 4 : Behaviour in an alkaline medium

When hydroxide ions are present, the zwitterion donates a proton from the ammonium end, showing acidic character:

$$\mathrm{^{-}OOC\! -\! CHR\! -\! NH_3^{+} + OH^{-} \;\longrightarrow\;{}^{-}OOC\! -\! CHR\! -\! NH_2 + H_2O}$$

Conclusion

Because an amino acid can accept a proton in acidic surroundings and donate a proton in basic surroundings, it behaves both as an acid and as a base. This dual (amphoteric) behaviour arises from its existence as a zwitterion containing the acidic $$\mathrm{COO^{-}}$$ and basic $$\mathrm{NH_3^{+}}$$ groups simultaneously.

The term enzyme was coined by Wilhelm Kühne (1878) from the Greek words en (within) and zyme (yeast) while studying fermentation. In modern biochemical language enzymes are defined with the help of their composition as well as their role in a reaction.

1  Composition

• Nearly all enzymes are globular proteins (a very small number are catalytically active RNA molecules called ribozymes).
• Being true proteins, they are formed from α-amino acids linked through peptide bonds and therefore are macromolecules of very high molar mass (10 kDa – >1 MDa).

2  Function

• Enzymes act as biological catalysts; i.e. they speed up the rate of a chemical reaction but emerge unchanged at the end of the reaction.
• They work by lowering the activation energy barrier, thereby permitting the reaction to proceed rapidly at the mild temperatures (≈ 300 K) and neutral pH that prevail in living cells.
• Like all catalysts, they do not affect the position of equilibrium and are required only in minute amounts.
• Each enzyme is highly specific for its substrate (or closely related group of substrates) and for the reaction it catalyses. For example, $$\mathrm{urease}$$ catalyses exclusively the hydrolysis of $$\mathrm{(NH_2)_2CO}$$ to $$\mathrm{NH_3}$$ and $$\mathrm{CO_2}$$.

3  Complete definition

Putting composition and function together, the definition generally quoted in textbooks is:

“Enzymes are proteinaceous biological catalysts that are produced by living cells and are capable of dramatically increasing the rates of specific biochemical reactions without themselves undergoing permanent change.”

Because of this property, enzymes control and regulate every metabolic pathway (digestion, respiration, DNA replication, etc.) in all forms of life.

Background — levels of protein structure

• Primary structure: the linear sequence of $$\alpha$$-amino acids linked through peptide (–CONH–) bonds.
  Peptide linkages are covalent and therefore very strong.
• Secondary structure: regular folding of the backbone, chiefly the $$\alpha$$-helix or $$\beta$$-pleated sheet, stabilised by intermolecular or intramolecular hydrogen bonds between $$\mathrm{C=O}$$ and $$\mathrm{N–H}$$ groups of the peptide chain.
• Tertiary structure: the overall three-dimensional folding of one polypeptide chain. It is maintained by comparatively weak, non-covalent interactions — hydrogen bonds, van der Waals forces, ionic (salt-bridge) interactions, hydrophobic interactions — and sometimes by covalent disulphide (–S–S–) bridges.
• Quaternary structure (if present): spatial arrangement of two or more polypeptide sub-units held together by the same weak interactions as the tertiary structure.

Denaturation — what happens?

1. When a native protein is subjected to heat, ultraviolet radiation, strong acids or bases, urea, organic solvents, heavy-metal ions, the weak forces (hydrogen bonds, hydrophobic and ionic interactions, van der Waals forces) are broken.
2. Because those interactions are responsible for the specific secondary, tertiary and quaternary arrangements, their disruption unfolds or uncoils the protein chain.
3. The primary structure remains intact because the covalent peptide bonds are not cleaved under the mild conditions that cause denaturation.

Observable consequences

• The globular, soluble form often changes into an insoluble, fibrous precipitate (e.g. coagulation of egg albumin on boiling).
• The protein loses its specific biological activity: enzymes lose catalytic power, antibodies lose antigen-binding ability, haemoglobin loses oxygen-carrying capacity, etc.
• The change is generally irreversible, although in a few cases gentle removal of the denaturing agent can allow renaturation.

Conclusion

Denaturation destroys the higher-order (secondary, tertiary, quaternary) structures of a protein without breaking its primary peptide sequence. Consequently the protein loses its native shape, becomes insoluble or coagulated, and its biological activity is abolished.

Meaning of vitamins: Vitamins are essential organic micronutrients that the human body either cannot synthesise at all or cannot synthesise in adequate amounts; hence they must be obtained through the diet.

Basis of classification: Because vitamins have very diverse chemical structures, they are classified on the basis of their solubility. This single property also correlates with the way each group is stored in and excreted from the body. On this basis vitamins are placed in two classes:

Vitamin needed for blood coagulation: The vitamin responsible for the normal coagulation (clotting) of blood is vitamin K.

Background about vitamins: Vitamins are organic compounds that are required in very small quantities but are indispensable for normal growth, maintenance of health and proper metabolic functioning of the body. Their prolonged absence in the diet gives rise to characteristic pathological conditions called deficiency diseases. The human body either cannot synthesise them at all or cannot make them in adequate amounts; therefore they must be supplied regularly in the diet.

The question asks about two vitamins of different solubility classes — vitamin A (fat-soluble) and vitamin C (water-soluble). For each we discuss (i) why it is essential — its main biological role and the symptoms produced by its deficiency — and (ii) the dietary sources that provide an adequate supply.

1. Vitamin A (retinol, $$\beta$$-carotene)

Why essential?

• It is required for the synthesis of visual purple (rhodopsin), the pigment of the rod cells of the retina, so it is indispensable for normal vision, especially night vision.
• It maintains the normal structure and functioning of the epithelial tissues that line the skin, eyes, respiratory tract and alimentary canal, preventing their dryness and keratinisation.
• It is needed for the proper growth of bones and teeth and acts as an antioxidant.

Deficiency of vitamin A causes xerophthalmia (hardening of the cornea), night-blindness, dryness of skin, retarded growth and lowered resistance to infection.

Important dietary sources:

• Animal sources rich in preformed retinol — fish-liver oil, liver, kidney, egg-yolk, butter, whole milk and cheese.
• Plant sources rich in the provitamin $$\beta$$-carotene (which the body converts into retinol) — carrots, spinach and other green leafy vegetables, and yellow/orange fruits such as mango, papaya and tomato.

2. Vitamin C (ascorbic acid)

Why essential?

• It is needed for the synthesis of collagen, the structural protein of connective tissues; hence it maintains the integrity of skin, cartilage, gums, tendons and blood-vessel walls.
• It promotes the healing of wounds and the repair of fractures.
• It facilitates the absorption of dietary iron by reducing $$\mathrm{Fe^{3+}}$$ ions to the $$\mathrm{Fe^{2+}}$$ state in the intestine.
• It functions as a powerful antioxidant and is a coenzyme in several enzymatic hydroxylation reactions.

Deficiency of vitamin C causes scurvy, characterised by spongy/bleeding gums, loose teeth, anaemia, delayed wound healing and general weakness.

Important dietary sources:

• Citrus fruits — lemon, orange, lime, grapefruit.
• Other fresh fruits and berries — amla (Indian gooseberry, one of the richest sources), guava, strawberry, pineapple.
• Fresh vegetables — tomato, green chilli, cabbage, cauliflower and green leafy vegetables such as spinach.
• Sprouted pulses and potato (especially when eaten with the skin) also contribute moderate amounts.

Conclusion: Vitamin A maintains normal vision and healthy epithelial tissues, while vitamin C maintains sound connective tissues and prevents scurvy. A balanced diet that includes dairy products and coloured fruits/vegetables for vitamin A, and plenty of fresh citrus fruits and vegetables for vitamin C, is therefore essential for good health.

Definition
Nucleic acids are long, unbranched, naturally occurring polymers of nucleotides; each nucleotide consists of a nitrogenous base (a purine or a pyrimidine), a pentose sugar (either $$\mathrm{D\!\text{-}ribose}$$ or $$\mathrm{D\!\text{-}2\!\text{-}deoxyribose}$$) and a phosphate group. By repeated 3′→5′ phosphodiester linkages the nucleotides form very long chains called polynucleotides. The two principal types are deoxyribonucleic acid (DNA) and ribonucleic acid (RNA).

Two key biological functions

1. Storage and transmission of genetic information: DNA stores the complete hereditary information of an organism and passes it unchanged (except for occasional mutation) from cell to cell and from one generation to the next during reproduction.
2. Directing and executing protein synthesis: RNA, copied from DNA, interprets that genetic code and guides the step-by-step assembly of specific amino-acid sequences into proteins (mRNA carries the code, tRNA brings amino acids, rRNA forms ribosomes). Thus nucleic acids ultimately control the structure and functioning of all cellular proteins.

In a nucleic-acid chain (DNA or RNA) each repeating unit contains three distinct parts:

• a nitrogen-containing heterocyclic base (a purine or a pyrimidine),
• a pentose sugar ($$\beta$$-$$\mathrm{D}$$-ribose in RNA or $$\beta$$-$$\mathrm{D}$$-2'-deoxyribose in DNA), and
• one or more phosphate groups.

The terms nucleoside and nucleotide refer to different combinations of these parts.

In short:

• A nucleoside is simply "base + sugar" and contains no phosphate.
• A nucleotide is "nucleoside + phosphate"; it is the phosphate-esterified form of a nucleoside and is the true monomer that is linked together to build DNA and RNA chains.

DNA (deoxyribonucleic acid) is a double-stranded helical polymer of nucleotides. Each nucleotide consists of a nitrogenous base, a pentose sugar (2'-deoxyribose) and a phosphate group. Neighbouring nucleotides in a strand are joined by phosphodiester bonds between the 3'-OH of one sugar and the 5'-phosphate of the next, so each strand has a definite direction — written from its free 5'-phosphate end to its free 3'-OH end.

When two polynucleotide strands come together to form the DNA double helix, the two strands do not carry the same sequence of bases. Instead, the bases on the two strands face each other and pair according to the strict base-pairing rules proposed by Watson and Crick:

• Adenine (A), a purine, pairs with thymine (T), a pyrimidine, through two hydrogen bonds: $$\mathrm{A \cdots T}$$.
• Guanine (G), a purine, pairs with cytosine (C), a pyrimidine, through three hydrogen bonds: $$\mathrm{G \cdots C}$$.

Only these two pairings (A–T and G–C) are geometrically and energetically compatible with the uniform width of the DNA double helix (≈ 20 Å) while keeping the sugar–phosphate backbones on the outside and the stacked bases on the inside. Hence the sequence on one strand uniquely determines the sequence on the opposite strand, yet the two sequences are clearly different. For example:

The two strands therefore:

1. Run in antiparallel directions (5'→3' on one strand faces 3'→5' on the other).
2. Carry different base sequences, so they are not identical.
3. Obey strict A–T and G–C pairing all along their length, so every base on one strand is matched uniquely by one and only one base on the other strand. Thus the two strands are complementary.

Because of this complementarity, given the sequence of one strand, the sequence of the other can be written at once by replacing every A with T, every T with A, every G with C and every C with G. This property is the fundamental basis for DNA replication and transcription in living cells.

The differences between deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) can be grouped under two headings — structural and functional. A comparison is summarised below.

Thus DNA and RNA differ in their pentose sugar, in one base (thymine vs uracil), in typical strandedness, in size, in cellular distribution and in chemical stability; and they perform different biological functions — DNA stores hereditary information, whereas RNA decodes and utilises that information to synthesise proteins (and in some viruses itself serves as the genetic material).

The nucleic acid present in cells occurs in two chemical forms — DNA and RNA. While DNA is largely restricted to the nucleus (in eukaryotes), RNA is synthesised from DNA and performs a variety of tasks in the cytoplasm as well as in the nucleus.

On the basis of both sedimentation coefficients obtained during ultracentrifugation and their distinct biological roles, RNA occurring in a typical cell is grouped into three main categories:

• Messenger RNA (mRNA)
• Ribosomal RNA (rRNA)
• Transfer RNA (tRNA)

Although modern molecular-biology texts additionally describe several small regulatory RNAs (snRNA, miRNA, siRNA, etc.), the NCERT treatment at class-XII level confines itself to the above three classical varieties. A brief account of each is given below so that the distinction between them is clear.

1. Messenger RNA (mRNA)
   It is transcribed from the DNA template strand by the enzyme RNA-polymerase. Its nucleotide sequence is complementary to the DNA template and it carries the genetic information in the form of codons (triplets of bases). During translation it serves as the template along which ribosomes move to assemble the amino-acid sequence of a specific polypeptide.

2. Ribosomal RNA (rRNA)
   This RNA, in association with proteins, forms the structural and catalytic framework of ribosomes (the cellular “work-benches” for protein synthesis). Depending on organism and ribosome type, several species of rRNA occur (e.g. 23 S, 16 S in prokaryotes; 28 S, 18 S in eukaryotes, where the “S” stands for the Svedberg sedimentation constant). rRNA also possesses peptidyl-transferase activity that catalyses peptide-bond formation.

3. Transfer RNA (tRNA)
   Sometimes called adapter RNA. It is the smallest of the three (≈ 73–93 nucleotides long) and exhibits a characteristic clover-leaf secondary structure. Each tRNA carries a specific amino acid at its 3′ end and possesses an anticodon loop that recognises the complementary codon on mRNA, thereby translating the nucleotide language into the amino-acid language.

Thus, the different types of RNA found in a cell are messenger RNA, ribosomal RNA and transfer RNA.