Biomolecules

Given: Glucose/sucrose vs. cyclohexane/benzene — all are six-membered ring compounds, yet solubility in water differs.

Concept: 'Like dissolves like' — polar solutes dissolve in polar solvents (water), non-polar solutes do not.

Explanation:
• Glucose and sucrose contain a large number of –OH (hydroxyl) groups in their structures. These –OH groups form hydrogen bonds with water molecules (H-bonding between O–H of solute and O–H of water). This strong interaction makes them readily soluble in water.
• Cyclohexane and benzene are non-polar hydrocarbons. They have no –OH or any other polar group capable of forming hydrogen bonds with water. Hence they are insoluble in water.

Conclusion: The presence of multiple –OH groups in glucose and sucrose enables extensive hydrogen bonding with water, making them water-soluble, whereas the non-polar nature of cyclohexane and benzene prevents any such interaction.

Given: Lactose undergoes hydrolysis.

Concept: Lactose is a disaccharide formed by a glycosidic linkage between one molecule of D-galactose and one molecule of D-glucose (β-1,4-glycosidic bond). On hydrolysis, the glycosidic bond is broken.

Reaction:
$$\text{Lactose} \xrightarrow{\text{H}_2\text{O}/\text{H}^+\text{ or enzyme}} \text{D-Galactose} + \text{D-Glucose}$$

Answer: The expected products of hydrolysis of lactose are D-galactose and D-glucose (one molecule each).

Given: D-glucose forms a pentaacetate (reacts with 5 molecules of acetic anhydride).

Concept: If D-glucose existed only in the open-chain form, it would have 4 –OH groups and 1 –CHO group. Acetic anhydride acetylates free –OH groups; it does not react with –CHO groups. So the open-chain form should give a pentaacetate with the –CHO group still intact.

Explanation:
• In reality, D-glucose exists predominantly in the cyclic (pyranose) hemiacetal form. In the cyclic form, the –CHO group reacts intramolecularly with the –OH at C-5 to form a hemiacetal, generating a new –OH group at C-1 (anomeric carbon).
• This cyclic form now has 5 free –OH groups (at C-1, C-2, C-3, C-4, and C-6) and no free –CHO group.
• Acetic anhydride acetylates all 5 –OH groups to give glucose pentaacetate.
• Since all 5 acetyl groups have replaced –OH groups and no –CHO is present, the pentaacetate shows no aldehyde group.

Conclusion: The absence of –CHO in the pentaacetate of D-glucose confirms that glucose exists in the cyclic hemiacetal form in which the aldehyde carbon is involved in ring formation, converting –CHO into a –OH (at C-1), which then gets acetylated.

Given: Amino acids have higher melting points and greater water solubility compared to corresponding halo acids (e.g., halo acetic acids).

Concept: Amino acids exist as zwitter ions (internal salts), whereas halo acids exist as covalent molecules.

Explanation:
• In amino acids, the –NH₂ group accepts a proton from the –COOH group to form a zwitter ion (dipolar ion): $\text{H}_3\text{N}^+$–CHR–$\text{COO}^-$.
• Because amino acids exist as ionic species (zwitter ions), they behave like salts. Ionic compounds have strong electrostatic (ionic) interactions between oppositely charged groups, requiring more energy to break — hence higher melting points.
• Halo acids (e.g., ClCH₂COOH) are covalent molecules with weaker intermolecular forces, so they have lower melting points.
• The ionic nature of zwitter ions also makes amino acids highly soluble in water (a polar solvent), since water solvates the charged groups effectively. Halo acids are less polar and hence less soluble.

Conclusion: The zwitter ionic (salt-like) nature of amino acids is responsible for their higher melting points and greater water solubility compared to halo acids.

Given: An egg is boiled; the egg white (albumin protein) coagulates.

Concept: Denaturation of proteins — on heating, the secondary and tertiary structures of proteins are disrupted.

Explanation:
• Egg white contains the protein albumin dissolved/dispersed in water. In the native state, the protein chains are folded in specific three-dimensional structures stabilised by hydrogen bonds, disulfide bonds, etc.
• On boiling, the thermal energy disrupts these bonds. The protein chains unfold and get denatured.
• The denatured (unfolded) protein chains expose their hydrophobic groups and aggregate together, trapping water molecules within the coagulated protein network.
• The water does not escape; it gets entrapped (bound) within the coagulated/denatured protein mass.

Conclusion: The water present in the egg gets trapped (entrapped) within the coagulated denatured protein network after boiling. It is held inside the solid coagulated egg white.

Given: Vitamin C (ascorbic acid) cannot be stored in the body.

Concept: Classification of vitamins based on solubility — fat-soluble (A, D, E, K) and water-soluble (B group and C).

Explanation:
• Vitamin C is a water-soluble vitamin. It dissolves readily in water (body fluids).
• Water-soluble vitamins are not stored in the body because any excess amount is readily excreted through urine by the kidneys.
• Unlike fat-soluble vitamins (A, D, E, K) which can be stored in adipose (fatty) tissue and liver, vitamin C has no storage depot in the body.

Conclusion: Since vitamin C is water-soluble, excess amounts are continuously excreted in urine and cannot be stored in the body. Therefore, it must be supplied regularly through diet.

Given: A nucleotide from DNA containing the base thymine is hydrolysed.

Concept: A nucleotide consists of three components: (i) a nitrogenous base, (ii) a pentose sugar, and (iii) a phosphate group. In DNA, the sugar is 2-deoxyribose.

Hydrolysis:
On complete hydrolysis, the nucleotide (thymidine monophosphate / deoxythymidine monophosphate) breaks down into:
1. Thymine (nitrogenous base — a pyrimidine)
2. 2-Deoxyribose (the pentose sugar present in DNA)
3. Phosphoric acid ($\text{H}_3\text{PO}_4$)

Answer: The products of hydrolysis are:
(i) Thymine
(ii) 2-Deoxyribose (deoxyribose sugar)
(iii) Phosphoric acid

Given: On hydrolysis of RNA, the four bases (adenine, guanine, cytosine, uracil) are obtained in no fixed ratio — i.e., [A] ≠ [U] and [G] ≠ [C].

Concept: In DNA (double-stranded), Chargaff's rule holds: A = T and G = C because of complementary base pairing between the two strands.

Explanation:
• In DNA, because of the double-helical complementary structure, adenine always pairs with thymine (A = T) and guanine always pairs with cytosine (G = C). This gives a fixed 1:1 ratio between complementary bases.
• In RNA, there is no such fixed ratio between the bases. This means RNA does not have a regular double-stranded complementary structure.
• The absence of a fixed ratio suggests that RNA is a single-stranded molecule. The bases are not required to be complementary to each other, so their quantities are not equal.

Conclusion: The fact that no fixed relationship exists among the quantities of different bases in RNA hydrolysate suggests that RNA is a single-stranded molecule (unlike the double-stranded complementary structure of DNA).

Definition: Monosaccharides are the simplest carbohydrates that cannot be hydrolysed further into smaller carbohydrate units.

Key features:
• They are the building blocks (monomeric units) of all carbohydrates.
• They are polyhydroxy aldehydes (aldoses) or polyhydroxy ketones (ketoses).
• They are classified on the basis of the number of carbon atoms: trioses (3C), tetroses (4C), pentoses (5C), hexoses (6C), heptoses (7C).

Examples: Glucose ($\text{C}_6\text{H}_{12}\text{O}_6$), fructose ($\text{C}_6\text{H}_{12}\text{O}_6$), ribose ($\text{C}_5\text{H}_{10}\text{O}_5$), 2-deoxyribose ($\text{C}_5\text{H}_{10}\text{O}_4$).

Definition: Reducing sugars are carbohydrates that can reduce mild oxidising agents such as Fehling's solution or Tollens' reagent.

Structural basis:
• Reducing sugars contain a free aldehyde (–CHO) group or a free ketone (–C=O) group (or a potential free –CHO in the hemiacetal form) that can act as a reducing agent.
• All monosaccharides (both aldoses and ketoses) are reducing sugars.
• Among disaccharides, maltose and lactose are reducing sugars (they have a free anomeric –OH / hemiacetal group), whereas sucrose is a non-reducing sugar (both anomeric carbons are involved in the glycosidic bond).

Examples: Glucose, fructose, maltose, lactose.

Two main functions of carbohydrates in plants:

1. Energy storage: Carbohydrates serve as the primary energy reserve in plants. Starch is the main storage polysaccharide in plants (stored in seeds, tubers, roots). It is hydrolysed to glucose when energy is needed.
$$\text{Starch} \xrightarrow{\text{hydrolysis}} \text{Glucose} \xrightarrow{\text{oxidation}} \text{Energy (ATP)}$$

2. Structural support: Cellulose, a polysaccharide made of D-glucose units linked by β-1,4-glycosidic bonds, forms the cell wall of plant cells. It provides rigidity, mechanical strength, and structural framework to the plant cell and plant body.

Classification:

Monosaccharides (cannot be hydrolysed further into simpler sugars):
1. Ribose ($\text{C}_5\text{H}_{10}\text{O}_5$) — an aldopentose
2. 2-Deoxyribose ($\text{C}_5\text{H}_{10}\text{O}_4$) — a deoxy aldopentose
3. Galactose ($\text{C}_6\text{H}_{12}\text{O}_6$) — an aldohexose
4. Fructose ($\text{C}_6\text{H}_{12}\text{O}_6$) — a ketohexose

Disaccharides (on hydrolysis give two monosaccharide units):
1. Maltose — hydrolysis gives 2 molecules of D-glucose
2. Lactose — hydrolysis gives D-glucose + D-galactose

Definition: A glycosidic linkage (or glycosidic bond) is the bond formed between two monosaccharide units through the loss of a water molecule. It is an acetal (or ketal) type of linkage formed between the anomeric carbon (C-1 of aldose or C-2 of ketose) of one monosaccharide and the –OH group of another monosaccharide.

Formation:
$$\text{Monosaccharide}_1\text{–OH} + \text{HO–Monosaccharide}_2 \rightarrow \text{Monosaccharide}_1\text{–O–Monosaccharide}_2 + \text{H}_2\text{O}$$

The –C–O–C– linkage formed is called the glycosidic linkage.

Example: In maltose, two glucose units are joined by an α-1,4-glycosidic linkage (C-1 of one glucose to C-4 of another glucose). In sucrose, glucose and fructose are joined by an α,β-1,2-glycosidic linkage.

Glycogen:
Glycogen is the storage polysaccharide of animals (and fungi). It is also called 'animal starch'. It is a polymer of D-glucose units linked by α-1,4-glycosidic bonds in the main chain and α-1,6-glycosidic bonds at branch points. It is stored mainly in the liver and muscles.

Differences between Glycogen and Starch:

| Property | Glycogen | Starch |
|---|---|---|
| Occurrence | Found in animals (liver, muscles) | Found in plants (seeds, tubers) |
| Structure | Highly branched; branching occurs every 8–10 glucose units | Amylose (unbranched, α-1,4 links) + Amylopectin (branched, α-1,4 and α-1,6 links); less branched than glycogen |
| Branching frequency | Very high (branch every 8–10 units) | Amylopectin branches every 24–30 units |
| Molecular mass | Very high | High |
| Colour with iodine | Gives red-brown colour | Gives blue-black colour (amylose) |

Both glycogen and starch are made of D-glucose units with α-glycosidic linkages.

(i) Hydrolysis of Sucrose:
Sucrose is a disaccharide formed by α-D-glucose and β-D-fructose joined by an α,β-1,2-glycosidic bond.
$$\text{Sucrose} \xrightarrow{\text{H}^+/\text{H}_2\text{O or sucrase}} \text{D-Glucose} + \text{D-Fructose}$$
Products: D-Glucose and D-Fructose (one molecule each).

(ii) Hydrolysis of Lactose:
Lactose is a disaccharide formed by β-D-galactose and D-glucose joined by a β-1,4-glycosidic bond.
$$\text{Lactose} \xrightarrow{\text{H}^+/\text{H}_2\text{O or lactase}} \text{D-Galactose} + \text{D-Glucose}$$
Products: D-Galactose and D-Glucose (one molecule each).

Starch:
• Starch is a polymer of D-glucose in which the glucose units are joined by α-glycosidic linkages.
• It consists of two components: Amylose (linear, α-1,4-glycosidic bonds) and Amylopectin (branched, α-1,4 in chain and α-1,6 at branch points).
• The α-linkage gives starch a helical (coiled) structure.
• Starch is digestible by humans (enzyme amylase acts on α-linkages).

Cellulose:
• Cellulose is a polymer of D-glucose in which the glucose units are joined by β-glycosidic linkages (specifically β-1,4-glycosidic bonds).
• It is a linear, unbranched polymer.
• The β-linkage gives cellulose a straight, ribbon-like structure that allows chains to align and form strong hydrogen bonds, giving mechanical strength.
• Cellulose is not digestible by humans (we lack the enzyme cellulase to break β-linkages).

Key difference: Starch has α-1,4 (and α-1,6) glycosidic linkages, while cellulose has β-1,4 glycosidic linkages.

(i) D-Glucose + HI:
D-Glucose on prolonged treatment with HI (hydroiodic acid) undergoes reduction. All the –OH groups are replaced and the carbonyl group is also reduced. The product is n-hexane.
$$\text{D-Glucose} \xrightarrow{\text{HI, } \Delta} \text{n-Hexane (CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3\text{)}$$
This confirms that all six carbon atoms of glucose are in a straight chain.

(ii) D-Glucose + Bromine water:
Bromine water is a mild oxidising agent. It oxidises only the aldehyde group (–CHO) to a carboxyl group (–COOH), without affecting the –OH groups.
$$\text{D-Glucose (CHO group)} \xrightarrow{\text{Br}_2/\text{H}_2\text{O}} \text{D-Gluconic acid (COOH group)}$$
Product: D-Gluconic acid.
This confirms the presence of an aldehyde group in glucose (distinguishes aldoses from ketoses, as ketoses do not decolourise bromine water).

(iii) D-Glucose + HNO₃:
HNO₃ (dilute nitric acid) is a stronger oxidising agent. It oxidises both the terminal groups — the aldehyde group (–CHO) at C-1 and the primary alcohol group (–CH₂OH) at C-6 — to carboxyl groups (–COOH).
$$\text{D-Glucose} \xrightarrow{\text{dil. HNO}_3} \text{D-Saccharic acid (Glucaric acid)}$$
Product: D-Saccharic acid (glucaric acid) — a dicarboxylic acid.
This confirms the presence of a primary alcohol group at C-6 in addition to the aldehyde at C-1.

The following reactions/observations of D-glucose cannot be explained by its open-chain (aldehyde) structure:

1. Existence of two anomers (α and β forms):
D-Glucose exists in two crystalline forms — α-D-glucose (m.p. 419 K) and β-D-glucose (m.p. 423 K). An open-chain aldehyde can give only one form. The existence of two forms is explained only by the cyclic hemiacetal structure with an anomeric carbon (C-1).

2. Mutarotation:
When either α-D-glucose or β-D-glucose is dissolved in water, the optical rotation changes and reaches an equilibrium value of +52.7°. This interconversion (mutarotation) cannot be explained by a fixed open-chain structure; it is explained by the ring-opening and ring-closing equilibrium between α and β cyclic forms.

3. Reaction with acetic anhydride to give pentaacetate:
D-Glucose reacts with acetic anhydride to give a pentaacetate (5 acetyl groups). The open-chain structure has only 4 –OH groups and 1 –CHO group. Acetic anhydride does not acetylate –CHO. The pentaacetate is explained by the cyclic form which has 5 –OH groups (including the anomeric –OH at C-1) and no free –CHO.

4. Failure to give Schiff's test and no reaction with NaHSO₃:
D-Glucose does not give Schiff's test (with fuchsin-aldehyde reagent) and does not form a crystalline addition product with NaHSO₃, as expected of a free aldehyde. This is because in the cyclic form, the –CHO is converted to a hemiacetal and is not freely available.

5. Existence of two methyl glucosides:
When D-glucose reacts with methanol in the presence of dry HCl, it forms two methyl glucosides (α-methyl glucoside and β-methyl glucoside). A simple aldehyde would give only one acetal. The two glucosides correspond to the two anomeric forms of cyclic glucose.

Essential Amino Acids:
Amino acids that cannot be synthesised by the human body and must be supplied through diet are called essential amino acids. There are 10 essential amino acids.

Examples:
1. Valine
2. Leucine
(Others: Isoleucine, Lysine, Methionine, Phenylalanine, Threonine, Tryptophan, Histidine, Arginine)

Non-Essential Amino Acids:
Amino acids that can be synthesised by the human body itself and do not need to be supplied through diet are called non-essential amino acids.

Examples:
1. Glycine
2. Alanine
(Others: Serine, Aspartic acid, Glutamic acid, etc.)

Summary:
| Type | Definition | Examples |
|---|---|---|
| Essential | Cannot be synthesised in body; must be taken in diet | Valine, Leucine |
| Non-essential | Can be synthesised in body | Glycine, Alanine |

(i) Peptide Linkage:
A peptide linkage (peptide bond) is the amide bond formed between the carboxyl group (–COOH) of one amino acid and the amino group (–NH₂) of another amino acid with the elimination of a water molecule.
$$\text{–NH}_2 + \text{HOOC–} \rightarrow \text{–NH–CO–} + \text{H}_2\text{O}$$
The –CO–NH– group is called the peptide linkage. It is the bond that links amino acid residues in a protein chain.

(ii) Primary Structure:
The primary structure of a protein refers to the specific sequence of amino acids in the polypeptide chain. It describes the number, type, and order (sequence) of amino acid residues linked by peptide bonds from the N-terminal (free –NH₂ end) to the C-terminal (free –COOH end). The primary structure is unique for each protein and determines its higher-order structure and function.

(iii) Denaturation:
Denaturation is the process by which a protein loses its native three-dimensional structure (secondary, tertiary, and quaternary structures) without breaking the peptide bonds (primary structure remains intact). It is caused by changes in pH, temperature, addition of organic solvents, heavy metal salts, etc.

During denaturation:
• Hydrogen bonds, disulfide bonds, and other non-covalent interactions maintaining the 3D structure are disrupted.
• The protein unfolds and loses its biological activity.
• Example: Coagulation of egg white on boiling; curdling of milk.

The secondary structure of a protein refers to the regular, repeating local folding patterns of the polypeptide backbone, stabilised mainly by hydrogen bonds between the –C=O and –N–H groups of the peptide backbone.

The two most common types of secondary structures are:

1. α-Helix:
• The polypeptide chain coils into a right-handed helical structure.
• It is stabilised by intramolecular hydrogen bonds formed between the –C=O group of one amino acid residue and the –N–H group of the amino acid four residues ahead in the chain.
• There are 3.6 amino acid residues per turn of the helix.
• Example: α-keratin (hair, wool, nails).

2. β-Pleated Sheet (β-sheet):
• The polypeptide chains are extended and lie side by side (parallel or antiparallel).
• It is stabilised by intermolecular hydrogen bonds between –C=O and –N–H groups of adjacent polypeptide chains (or different segments of the same chain).
• The structure resembles a pleated (folded) sheet.
• Example: Silk fibroin (β-keratin).

The α-helix structure of proteins is stabilised by intramolecular hydrogen bonds.

Specifically:
• Hydrogen bonds are formed between the carbonyl oxygen (–C=O) of one amino acid residue and the imino hydrogen (–N–H) of the amino acid residue that is four residues ahead in the same polypeptide chain.
$$\text{–C=O} \cdots \text{H–N–}$$
• These hydrogen bonds run roughly parallel to the helix axis.
• The regular repetition of these hydrogen bonds along the entire length of the helix stabilises the coiled structure.

Note: The side chains (R groups) of amino acids project outward from the helix and do not participate in the hydrogen bonding that stabilises the helix backbone.

| Property | Globular Proteins | Fibrous Proteins |
|---|---|---|
| Shape | Spherical or ellipsoidal (compact, rounded) | Long, thread-like or fibre-like (elongated) |
| Structure | Polypeptide chains are folded into a compact, spherical shape | Polypeptide chains are arranged parallel along one axis, forming fibres |
| Solubility | Generally soluble in water and dilute salt solutions | Generally insoluble in water |
| Function | Dynamic functions — act as enzymes, hormones, antibodies, transport proteins | Structural/mechanical functions — provide strength and support |
| Secondary structure | Contain α-helix, β-sheet, and random coil regions | Predominantly α-helix (α-keratin) or β-sheet (silk fibroin) |
| Examples | Haemoglobin, insulin, albumin, enzymes | Keratin (hair, nails), collagen (tendons), myosin (muscles), silk fibroin |

Amino acids contain two functional groups:
• An acidic group: –COOH (carboxyl group)
• A basic group: –NH₂ (amino group)

Because of the presence of both an acidic and a basic group in the same molecule, amino acids can act as both acids and bases — i.e., they are amphoteric.

As a base (accepting a proton): The –NH₂ group accepts H⁺ from an acid:
$$\text{H}_2\text{N–CHR–COOH} + \text{H}^+ \rightarrow \text{H}_3\text{N}^+\text{–CHR–COOH}$$

As an acid (donating a proton): The –COOH group donates H⁺ to a base:
$$\text{H}_2\text{N–CHR–COOH} + \text{OH}^- \rightarrow \text{H}_2\text{N–CHR–COO}^- + \text{H}_2\text{O}$$

Zwitter ion formation: In aqueous solution, the –COOH group donates a proton to the –NH₂ group within the same molecule, forming a dipolar ion (zwitter ion):
$$\text{H}_2\text{N–CHR–COOH} \rightleftharpoons \text{H}_3\text{N}^+\text{–CHR–COO}^-$$

This internal salt (zwitter ion) can donate a proton (act as acid) or accept a proton (act as base), confirming the amphoteric nature of amino acids.

Definition: Enzymes are biological catalysts (biocatalysts) that speed up the rate of biochemical reactions in living organisms without themselves being consumed in the reaction.

Key features:
• Almost all enzymes are globular proteins in nature (a few RNA molecules called ribozymes also have catalytic activity).
• They are highly specific — each enzyme catalyses only one particular reaction or acts on one particular type of substrate (lock and key specificity).
• They are required only in small quantities.
• They function by lowering the activation energy of the reaction.
• They are sensitive to temperature and pH — they work best at an optimum temperature and pH.
• They are generally named after the substrate or the reaction they catalyse, with the suffix -ase (e.g., maltase, sucrase, urease).

Example: Maltase catalyses the hydrolysis of maltose to glucose:
$$\text{Maltose} \xrightarrow{\text{Maltase}} 2\,\text{Glucose}$$

Effect of denaturation on protein structure:

When a protein is denatured (by heat, change in pH, organic solvents, heavy metal salts, etc.):

1. Secondary structure is destroyed: The hydrogen bonds stabilising the α-helix or β-pleated sheet are broken. The regular folding pattern is lost.

2. Tertiary structure is destroyed: The non-covalent interactions (hydrogen bonds, hydrophobic interactions, van der Waals forces) and disulfide bonds (in some cases) that maintain the three-dimensional folded shape are disrupted. The protein unfolds.

3. Quaternary structure is destroyed (if present): The association between multiple polypeptide subunits is disrupted.

4. Primary structure remains intact: The peptide bonds (covalent bonds) linking amino acids in the polypeptide chain are NOT broken during denaturation. The sequence of amino acids is unchanged.

5. Loss of biological activity: Since the specific 3D shape (active site) is lost, the protein can no longer perform its biological function.

Example: Boiling of egg white causes coagulation — the albumin protein denatures and precipitates. Curdling of milk is another example.

Classification of Vitamins:

Vitamins are classified into two groups based on their solubility:

1. Fat-soluble vitamins:
• Soluble in fats and oils; insoluble in water.
• Can be stored in the body (liver and adipose tissue).
• Vitamins: A, D, E, and K

2. Water-soluble vitamins:
• Soluble in water; cannot be stored in the body (excess is excreted in urine).
• Must be supplied regularly through diet.
• Vitamins: B group (B₁, B₂, B₆, B₁₂, etc.) and C

Vitamin responsible for coagulation (clotting) of blood:
Vitamin K is responsible for the coagulation (clotting) of blood. It is essential for the synthesis of prothrombin and other clotting factors in the liver. Deficiency of Vitamin K leads to increased bleeding time.

Vitamin A:

Why essential:
• Vitamin A (retinol) is essential for normal vision, especially in dim light (night vision). It is a component of the visual pigment rhodopsin.
• It is necessary for the growth and maintenance of epithelial tissues (skin, mucous membranes).
• It is important for normal bone and tooth development.
• Deficiency causes: Night blindness (nyctalopia), xerophthalmia (dryness of eyes), and increased susceptibility to infections.

Important sources: Fish liver oil, liver, butter, milk, eggs, carrots, green leafy vegetables (as β-carotene, a precursor).

Vitamin C:

Why essential:
• Vitamin C (ascorbic acid) is essential for the synthesis of collagen (a structural protein in connective tissue, bones, and blood vessels).
• It acts as an antioxidant, protecting cells from oxidative damage.
• It is necessary for the proper functioning of the immune system.
• It enhances the absorption of iron from food.
• Deficiency causes: Scurvy — characterised by bleeding gums, swollen joints, slow wound healing, and anaemia.

Important sources: Citrus fruits (lemon, orange, amla/Indian gooseberry), tomatoes, green leafy vegetables, guava, capsicum.

Nucleic Acids:
Nucleic acids are high molecular weight, naturally occurring biopolymers (polynucleotides) found in the nuclei of all living cells. They are polymers of nucleotides linked by 3',5'-phosphodiester bonds.

There are two types:
1. DNA (Deoxyribonucleic acid) — contains 2-deoxyribose sugar
2. RNA (Ribonucleic acid) — contains ribose sugar

Two Important Functions of Nucleic Acids:

1. Storage and transmission of genetic information (Heredity):
DNA is the chemical basis of heredity. It carries the genetic information (coded in the sequence of bases) that is passed from parents to offspring during reproduction. The sequence of bases in DNA encodes the information for the synthesis of all proteins in the cell.

2. Protein synthesis:
RNA molecules (mRNA, tRNA, rRNA) are directly involved in the synthesis of proteins (translation). mRNA carries the genetic message from DNA to the ribosomes; tRNA brings the appropriate amino acids; rRNA forms part of the ribosome structure. Thus, nucleic acids control the synthesis of proteins that determine the structure and function of the organism.

| Feature | Nucleoside | Nucleotide |
|---|---|---|
| Composition | Nitrogenous base + Pentose sugar | Nitrogenous base + Pentose sugar + Phosphate group |
| Phosphate group | Absent | Present (one or more phosphate groups) |
| Formation | Base linked to sugar by N-glycosidic bond | Nucleoside esterified with phosphoric acid at 5'-OH of sugar |
| Relationship | Nucleoside is a component of nucleotide | Nucleotide = Nucleoside + Phosphate |
| Example | Adenosine (adenine + ribose), Thymidine (thymine + deoxyribose) | Adenosine monophosphate/AMP (adenosine + phosphate), Thymidine monophosphate/TMP |

In short:
$$\text{Nucleoside} = \text{Base} + \text{Sugar}$$
$$\text{Nucleotide} = \text{Base} + \text{Sugar} + \text{Phosphate}$$
$$\text{Nucleotide} = \text{Nucleoside} + \text{Phosphate}$$

Given: DNA is a double-stranded molecule.

Explanation:

In the double helix structure of DNA (Watson-Crick model):
• The two strands run antiparallel to each other (one strand runs 5'→3' and the other runs 3'→5').
• The two strands are held together by hydrogen bonds between specific pairs of nitrogenous bases — this is called complementary base pairing:
- Adenine (A) always pairs with Thymine (T) — connected by 2 hydrogen bonds
- Guanine (G) always pairs with Cytosine (C) — connected by 3 hydrogen bonds

This means:
• If one strand has the sequence 5'-A-T-G-C-3', the other strand must have 3'-T-A-C-G-5'.
• The two strands are NOT identical (they have different base sequences).
• But they are COMPLEMENTARY — the base sequence of one strand determines the base sequence of the other strand.

This complementarity is the basis of:
1. DNA replication (each strand serves as a template for a new complementary strand)
2. Transcription (DNA strand serves as template for mRNA synthesis)

Chargaff's rule confirms this: In any DNA, [A] = [T] and [G] = [C].

Structural Differences:

| Feature | DNA | RNA |
|---|---|---|
| Sugar | 2-Deoxyribose (lacks –OH at C-2) | Ribose (has –OH at C-2) |
| Bases | Adenine, Guanine, Cytosine, Thymine | Adenine, Guanine, Cytosine, Uracil |
| Strands | Double-stranded (double helix) | Single-stranded |
| Base pairing | A–T (2 H-bonds), G–C (3 H-bonds) | Partial/local base pairing possible |
| Stability | More stable (no 2'-OH) | Less stable (2'-OH makes it susceptible to hydrolysis) |
| Location | Mainly in nucleus (also mitochondria, chloroplasts) | Nucleus and cytoplasm |
| Chargaff's rule | [A]=[T], [G]=[C] | No fixed ratio between bases |

Functional Differences:

| Function | DNA | RNA |
|---|---|---|
| Primary role | Storage and transmission of genetic information (heredity) | Protein synthesis (translation) |
| Template | Serves as template for its own replication and for RNA synthesis (transcription) | mRNA serves as template for protein synthesis; tRNA brings amino acids; rRNA forms ribosomes |
| Permanence | Permanent genetic material | Temporary (mRNA is degraded after use) |

There are three main types of RNA found in the cell, all involved in protein synthesis:

1. Messenger RNA (mRNA):
• It is synthesised in the nucleus using DNA as a template (transcription).
• It carries the genetic message (codons — triplet base sequences) from DNA to the ribosomes in the cytoplasm.
• It serves as the template for protein synthesis (translation).
• It is the largest RNA but present in the smallest amount.

2. Transfer RNA (tRNA) — also called Soluble RNA (sRNA):
• It is the smallest RNA molecule.
• It acts as an adaptor molecule — it reads the codons on mRNA and brings the specific amino acid to the ribosome during translation.
• Each tRNA is specific for one amino acid.
• It has a cloverleaf secondary structure with an anticodon loop (complementary to mRNA codon) and an amino acid attachment site at the 3'-end.

3. Ribosomal RNA (rRNA):
• It is the most abundant RNA in the cell.
• It is a structural and functional component of ribosomes (along with ribosomal proteins).
• Ribosomes are the sites of protein synthesis.
• rRNA plays a catalytic role in peptide bond formation (ribozyme activity).

Summary:
| Type | Function | Size |
|---|---|---|
| mRNA | Carries genetic code from DNA to ribosome | Largest |
| tRNA | Brings amino acids to ribosome | Smallest |
| rRNA | Forms ribosome structure; catalyses peptide bond formation | Most abundant |