Biomolecules | ISC Class 12 Chemistry Notes

This note covers carbohydrate classification, glucose and fructose structures and reactions, sugar configurations, glycosidic linkages, polysaccharides, amino acids, protein structure and denaturation, enzymes, hormones, vitamins, and the composition and functions of nucleic acids.

What are carbohydrates, and how are they classified?

Definition: Carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds that yield such units on hydrolysis. Polyhydroxy means containing several hydroxyl groups, written as -OH; hydrolysis means bond cleavage through reaction with water.

Optical activity is the ability to rotate the plane of plane-polarised light, whose vibrations are confined to one plane. An aldehyde contains the terminal -CHO group; a ketone contains a carbonyl group, >C=O, between carbon-containing groups.

In these formulae, C denotes carbon, H hydrogen and O oxygen. A line denotes a bond, and a double line denotes a double bond. Subscript numerals count atoms or repeated groups. Carbohydrates are primarily produced by plants; glucose, cane sugar and starch are examples.

Why is the old hydrate formula insufficient?

Most carbohydrates fit Cₓ(H₂O)ᵧ, where x and y denote the numbers of carbon atoms and water units in this compositional expression. Glucose, C₆H₁₂O₆, fits it. However, acetic acid also fits this pattern but is not a carbohydrate, while rhamnose, C₆H₁₂O₅, is a carbohydrate that does not.

Disaccharides, trisaccharides and tetrasaccharides yield two, three and four monosaccharide units respectively. Sucrose is a disaccharide; raffinose and stachyose are examples of a trisaccharide and a tetrasaccharide respectively. Polysaccharides are not sweet and are also called non-sugars.

How do functional groups distinguish monosaccharides?

An aldose contains an aldehyde group; a ketose contains a ketonic group. A hexose contains six carbon atoms. Glucose is an aldohexose and fructose a ketohexose. These descriptions combine functional-group identity with carbon count.

Reducing sugars reduce Fehling’s solution or Tollens’ reagent, chemical reagents used to test reducing behaviour. All monosaccharides, whether aldoses or ketoses, are reducing sugars. Among disaccharides, maltose and lactose are reducing, whereas sucrose is non-reducing.

How do reactions establish the open-chain structure of glucose?

Glucose, also called dextrose, occurs in sweet fruits and honey and is present in large amounts in ripe grapes. Its molecular formula is C₆H₁₂O₆. Its open-chain condensed structure is CHO-(CHOH)₄-CH₂OH, containing an aldehyde end and a primary alcohol end.

A primary alcohol group, -CH₂OH, has its hydroxyl-bearing carbon attached to one other carbon. A condensed structure groups attached atoms together; (CHOH)₄ represents four successive carbon atoms, each carrying one hydrogen and one hydroxyl group.

Derivation: How is the open-chain structure of glucose established?

Begin with the molecular formula $\mathrm{C_6H_{12}O_6}$, then combine the reaction evidence to determine the carbon skeleton and the functional groups.

1. Prolonged heating with hydrogen iodide, $\mathrm{HI}$, produces n-hexane, the unbranched six-carbon hydrocarbon. This places all six carbon atoms in a straight chain.
2. Hydroxylamine, $\mathrm{NH_2OH}$, produces an oxime, while hydrogen cyanide produces a cyanohydrin. Both reactions establish a carbonyl group. Here N denotes nitrogen.
3. Bromine water, bromine dissolved in water, mildly oxidises glucose to the six-carbon carboxylic acid gluconic acid. The carbonyl group is therefore aldehydic, giving an aldehyde end to the chain.
4. Acetic anhydride produces glucose pentaacetate, establishing five hydroxyl groups. The stability of this product supports placing these groups on different carbon atoms.
5. Nitric acid, $\mathrm{HNO_3}$, oxidises both glucose and gluconic acid to the dicarboxylic acid saccharic acid. This establishes a primary alcohol group at the other end of the chain.

Result: Combining the straight six-carbon chain, the aldehyde group and the five hydroxyl groups gives the open-chain structure $\mathrm{CHO{-}(CHOH)_4{-}CH_2OH}$. The four middle carbon atoms each carry a hydroxyl group, and the terminal primary alcohol supplies the fifth.

Acetylation replaces the hydrogen of a hydroxyl group with an acetyl group, CH₃CO-. The stability of glucose pentaacetate supports placement of the five hydroxyl groups on different carbon atoms. Gluconic acid also yields saccharic acid on oxidation with nitric acid.

How is the bromine-water reaction written?

In the following equation, Br denotes bromine, Br₂ is molecular bromine, H₂O is water, HBr is hydrogen bromide, and → means “forms”. A coefficient before a formula gives the relative number of molecules reacting or formed.

CHO-(CHOH)₄-CH₂OH + Br₂ + H₂O → COOH-(CHOH)₄-CH₂OH + 2HBr

The -CHO group becomes -COOH, a carboxyl group; the terminal -CH₂OH remains unchanged. Thus mild oxidation gives gluconic acid. With nitric acid, oxidation at both ends gives HOOC-(CHOH)₄-COOH, saccharic acid. Reagent identity is therefore essential when predicting the product.

Glucose also adds hydrogen cyanide, HCN, to form a cyanohydrin, a compound with hydroxyl and cyanide groups attached to the former carbonyl carbon. This provides further evidence for the carbonyl group but does not by itself distinguish an aldehyde from a ketone.

What do D and L mean, and why does glucose form rings?

Configuration describes the spatial arrangement of groups in a molecule. D and L indicate configuration relative to glyceraldehyde, a three-carbon aldehyde with a hydroxyl group on its middle carbon. An asymmetric carbon has four different attached groups.

In a Fischer projection, a conventional flat representation of configuration, place the most oxidised carbon at the top. Compare the lowest asymmetric carbon with glyceraldehyde. Its hydroxyl group lies on the right for the D series and on the left for the L series.

Does D mean a positive optical rotation?

No. D and L do not specify the direction of optical rotation. The sign (+) means dextrorotatory, rotating plane-polarised light to the right; (−) means laevorotatory, rotating it to the left. Glucose is D-(+)-glucose, while fructose is D-(−)-fructose.

In the open-chain Fischer projection of D-glucose, the hydroxyl groups at carbon positions 2, 3, 4 and 5 lie right, left, right and right respectively. Here C-1, C-2 and similar labels mean carbon position 1, carbon position 2, and so on.

Derivation: Why does glucose require a cyclic hemiacetal structure?

1. The open-chain structure contains an aldehyde group, yet glucose does not give Schiff’s test, an aldehyde colour test, or the addition product with sodium hydrogensulphite. The open-chain structure alone therefore does not explain its behaviour.
2. Glucose pentaacetate does not react with hydroxylamine, indicating absence of a free aldehyde group. This provides further evidence that the aldehyde group is not freely available.
3. Glucose exists in two crystalline forms called α and β, pronounced alpha and beta. The structure must also account for these two forms.
4. Addition of the hydroxyl group at C-5 to the aldehyde group at C-1 forms a six-membered cyclic hemiacetal. This accounts for the absence of a free aldehyde group.
5. The resulting cyclic forms differ in hydroxyl-group configuration at C-1, explaining the α and β forms. C-1, formerly the aldehyde carbon, is the anomeric carbon.

Result: Glucose forms two cyclic hemiacetal anomers, α and β, which exist in equilibrium with the open-chain structure and can interconvert through it. Its six-membered ring contains five carbon atoms and one oxygen atom and is called a pyranose ring.

A hemiacetal carbon bears both a hydroxyl group and an ether-like oxygen linkage. The α and β forms are anomers, differing only in the hydroxyl-group configuration at C-1.

Draw two six-membered Haworth rings, conventional ring representations, with oxygen at the upper right and C-1 at the right. Place the C-1 hydroxyl below the ring for α-D-glucopyranose and above for β-D-glucopyranose; retain the other groups in matching positions.

How do fructose structure and reactions compare with glucose?

Fructose is a natural monosaccharide found in fruits, honey and vegetables. Hydrolysis of sucrose produces fructose together with glucose. Pure fructose is used as a sweetener. It has the same molecular formula as glucose, C₆H₁₂O₆, but a different functional group.

Its open-chain condensed structure is CH₂OH-CO-(CHOH)₃-CH₂OH. Here -CO- denotes the ketonic carbonyl group at C-2. The six carbon atoms form a straight chain. In its Fischer projection, the hydroxyl groups at C-3, C-4 and C-5 lie left, right and right respectively.

What ring does fructose form?

The hydroxyl group at C-5 adds to the ketonic group at C-2. This gives a five-membered furanose ring containing four carbon atoms and one oxygen atom. The cyclic product is a hemiketal, the ketone counterpart of a hemiacetal, and C-2 becomes the anomeric carbon.

The α and β forms differ in configuration at this carbon. D-(−)-fructose illustrates why configuration and optical rotation must be stated separately: its D configuration coexists with a negative rotation.

Which reactions reveal the functional groups?

For the distinguishing test: fructose + bromine water → no reaction under the mild test conditions. Glucose decolourises bromine water as it forms gluconic acid. Fructose nevertheless reduces Fehling’s and Tollens’ reagents; these alkaline conditions allow conversion into aldoses.

What does phenylhydrazine show?

Phenylhydrazine, C₆H₅NHNH₂, reacts at a sugar’s carbonyl group to form a phenylhydrazone. With excess reagent on warming, glucose and fructose give the same yellow crystalline osazone, a derivative with adjacent carbon-nitrogen double-bond groups at C-1 and C-2.

Osazone formation alters those first two carbon positions. Glucose and fructose retain the same configuration at their remaining asymmetric carbons, explaining the common derivative. This test therefore does not distinguish them; the mild bromine-water test does.

How do glycosidic linkages determine disaccharide properties?

Definition: A glycosidic linkage joins two monosaccharide units through an oxygen atom, with loss of a water molecule during bond formation. Hydrolysis with dilute acids or enzymes releases the component monosaccharides.

Enzymes are biological catalysts, substances that speed biological reactions. Whether a disaccharide is reducing depends on the availability of a reducing group. When both component reducing groups participate in the linkage, the disaccharide is non-reducing.

What products form on hydrolysis?

Sucrose gives equal amounts in moles of glucose and fructose. A mole is a unit for amount of substance; equal mole amounts contain equal numbers of molecules. Maltose gives two glucose molecules per molecule hydrolysed. Lactose, the sugar in milk, gives galactose and glucose.

Sucrose + water → glucose + fructose

Maltose + water → 2 glucose

Lactose + water → galactose + glucose

Why is hydrolysed sucrose called invert sugar?

Sucrose is dextrorotatory. Its hydrolysis gives dextrorotatory glucose and laevorotatory fructose. Fructose’s negative rotation exceeds glucose’s positive rotation in magnitude, so the mixture is laevorotatory. The change from positive to negative rotation is inversion, and the product is invert sugar.

For sucrose, join the oxygen bridge between C-1 of the glucose ring and C-2 of the fructose ring. For maltose, join C-1 of one glucose ring to C-4 of the second. Label each carbon position and the glycosidic oxygen.

A description of the linkage should therefore identify the component sugars, the carbon positions joined and the configuration of the linking group. Merely saying “two sugar rings” does not explain why sucrose and maltose behave differently towards reducing-sugar reagents.

How do starch, cellulose and glycogen differ?

Polysaccharides contain many monosaccharide units joined by glycosidic linkages. They mainly serve as food-storage or structural materials. A polymer is a large molecule built from repeating smaller units, called monomers. All three polysaccharides here are built from glucose, but their linkages and branching differ.

What are the two components of starch?

Starch is the main storage polysaccharide of plants and an important dietary source for humans. Cereals, roots, tubers and some vegetables contain much starch. Its components are amylose and amylopectin, both formed from α-D-glucose.

The percentage sign means “per hundred”. The component proportions are approximate. Branching means that additional chains extend from positions along a main chain; it does not mean that each glucose molecule itself has a branched carbon skeleton.

How do structure and role compare?

Cellulose is a predominant constituent of plant cell walls. It consists of straight chains of β-D-glucose, linked from C-1 of one unit to C-4 of the next. Its β linkages distinguish it from the α-glucose polymers in starch.

Glycogen, also called animal starch, stores carbohydrate in animals. It occurs in liver, muscles and brain. Its structure resembles amylopectin but is rather more highly branched. When glucose is needed, enzymes break glycogen down. Glycogen also occurs in yeast and fungi.

Carbohydrates provide energy and storage materials, while cellulose supplies structural material and raw materials for textiles and paper. Cotton fibre and wood contain cellulose. The sugars ribose and deoxyribose are constituents of nucleic acids, molecules involved in heredity and protein synthesis.

What are amino acids, zwitterions and the isoelectric point?

Amino acids contain an amino group, -NH₂, and a carboxyl group, -COOH. In an α-amino acid, the amino group is attached to the carbon immediately next to the carboxyl group. Proteins yield α-amino acids on hydrolysis.

The general structure is H₂N-CH(R)-COOH, where R is the side chain, the variable attached group. For glycine, R is hydrogen; for alanine it is -CH₃, a methyl group. Different side chains help distinguish the amino acids.

How are amino acids classified?

Amino acids are usually colourless crystalline solids. They are water soluble and have high melting points, behaving like salts. In water, the carboxyl group can lose a proton, a positively charged hydrogen ion, while the amino group can accept one.

This forms ⁺H₃N-CH(R)-COO⁻, a zwitterion: an ion carrying both positive and negative charges but neutral overall. Superscript + and − indicate electrical charges. Its two charged groups explain why “net neutral” does not mean “without charges”.

How do acidity and charge relate?

Amino acids are amphoteric, meaning they react with both acids and bases. The isoelectric point is the pH at which an amino acid has zero average net charge and shows no net migration in an electric field. Here pH is a measure of acidity or alkalinity.

Glycine is optically inactive because its α-carbon has two hydrogen atoms. The other naturally occurring α-amino acids are optically active. Most naturally occurring amino acids have L configuration, represented with the amino group on the left in the conventional projection.

How are proteins linked, folded and denatured?

Proteins are polymers of α-amino acids and contribute to growth, maintenance, structure and biological function. Their amino-acid units are linked by peptide bonds, the amide linkage -CO-NH- formed between a carboxyl group and an amino group.

How does a peptide bond form?

The carboxyl group of glycine can combine with the amino group of alanine, eliminating water and forming the dipeptide glycylalanine. A dipeptide contains two amino-acid units. A tripeptide contains three units joined by two peptide bonds; longer chains are polypeptides.

H₂N-CH₂-COOH + H₂N-CH(CH₃)-COOH → H₂N-CH₂-CO-NH-CH(CH₃)-COOH + H₂O

An amino-acid residue is the part of an amino acid retained in the chain after linkage formation. The distinction between a polypeptide and a protein is not very sharp. Insulin, with 51 amino-acid residues, is a protein despite its relatively short chain.

What are the four levels of protein structure?

An α-helix is a coiled secondary structure; a β-pleated sheet consists of extended chains arranged side by side. Hydrogen bonds are attractions involving hydrogen attached to an electronegative atom and another electronegative atom. They help stabilise these folded structures.

Fibrous proteins have fibre-like arrangements and are generally insoluble in water; keratin and myosin are examples. Globular proteins have coiled, approximately spherical shapes and are usually soluble in water; insulin and albumins are examples.

What changes during denaturation?

A native protein has its characteristic three-dimensional structure and biological activity. Denaturation is the loss of this activity when changes such as heating or altered pH disturb its higher structure. Secondary and tertiary structures are destroyed, but the primary structure remains intact.

Coagulation of egg white on boiling and curdling of milk illustrate denaturation. In curdling, bacteria produce lactic acid. Denaturation therefore differs from hydrolysis of peptide bonds: an unfolded protein has not necessarily been split into its constituent amino acids.

How do enzymes and hormones control biological activity?

Enzymes are biocatalysts that accelerate reactions in living systems under mild conditions. Almost all enzymes are globular proteins. A substrate is the substance on which an enzyme acts. Enzymes are very specific for a particular reaction and a particular substrate.

How does enzyme action affect a reaction?

Enzymes reduce the activation energy, the energy barrier that reacting substances must overcome. They are needed only in small quantities. Maltase catalyses hydrolysis of maltose into glucose, showing how an enzyme’s name can reflect its substrate.

Other names reflect the reaction. Oxidoreductases catalyse oxidation of one substrate with simultaneous reduction of another. Oxidation and reduction involve loss and gain of electrons respectively. Maltase illustrates the use of the ending “-ase” in an enzyme name.

What makes a hormone different?

Hormones are intercellular messengers, carrying information between cells. Endocrine glands produce them and release them directly into the bloodstream, which transports them to their sites of action. Their role is communication and regulation of biological activities.

Hormones have different chemical natures. Insulin is a polypeptide hormone; epinephrine is an amino-acid derivative; estrogens and androgens are steroid hormones, compounds with a characteristic framework of four fused carbon rings. Thus the word hormone describes a function rather than a single chemical structure.

Insulin is released in response to a rapid rise in blood glucose and helps keep its level within a narrow limit. Glucagon tends to increase blood glucose. Together they regulate its level. Growth and sex hormones participate in growth and development.

Note: Retain the distinction between catalysis and signalling. Maltase speeds a chemical reaction on maltose; insulin acts as a hormone regulating blood glucose. Both involve proteins, but their described biological roles differ.

How are vitamins classified, and what does deficiency cause?

Vitamins are organic compounds required in small dietary amounts for specific biological functions, growth and health. Most cannot be synthesised in our bodies, although gut bacteria can produce some. Their chemical structures differ, so solubility provides a useful classification.

Which vitamins are fat soluble or water soluble?

Fat-soluble vitamins A, D, E and K dissolve in fats and oils but not in water. They are stored in the liver and adipose tissue, the tissue that stores fat. Water-soluble vitamins comprise the B group and vitamin C.

Water-soluble vitamins need regular dietary supply because they are readily excreted in urine and cannot be stored, except vitamin B₁₂. Vitamin letters and subscripts are names, not molecular formulae. The B group contains several vitamins with different deficiency effects.

Which sources and deficiency effects should be distinguished?

In the table, red blood cells are the blood cells containing haemoglobin, the oxygen-carrying protein. A deficiency effect identifies the consequence of inadequate vitamin supply; it should be paired with the correct vitamin rather than assigned to the whole group.

These effects connect vitamins with normal functions: vitamin A supports vision, vitamin D supports healthy bones, vitamin E helps maintain red blood cells and muscles, and vitamin K is needed for blood coagulation, the formation of a clot. Vitamin C prevents the deficiency condition scurvy.

The solubility grouping and the deficiency association answer different questions. For example, vitamins A and D belong to the same solubility class but their characteristic deficiency effects differ. Likewise, the water-soluble B vitamins should not all be equated with vitamin C.

What are nucleosides, nucleotides and nucleic acids?

Nucleic acids are long-chain polymers of nucleotides, also called polynucleotides. The two main types are deoxyribonucleic acid, abbreviated DNA, and ribonucleic acid, abbreviated RNA. They are associated with heredity, the transmission of characteristics between generations.

Complete hydrolysis of either type yields a pentose sugar, phosphoric acid and nitrogen-containing bases. A pentose is a five-carbon sugar. DNA contains 2-deoxyribose, while RNA contains ribose. These sugar identities are a central chemical distinction between the two polymers.

Which nitrogenous bases occur?

A nitrogenous base is a nitrogen-containing ring compound in a nucleic acid. Adenine and guanine are purines, with two fused rings. Cytosine, thymine and uracil are pyrimidines, with one ring. These two classes describe the base structures.

Thus both nucleic acids contain adenine, guanine and cytosine. The fourth base is thymine in DNA and uracil in RNA. The conventional abbreviations A, G, C, T and U stand for adenine, guanine, cytosine, thymine and uracil respectively.

How does a nucleotide differ from a nucleoside?

A nucleoside consists of a base attached to a sugar. A nucleotide contains the sugar, base and phosphate component. Phosphate is the group derived from phosphoric acid. A base alone is therefore neither a nucleoside nor a complete nucleotide.

A phosphodiester linkage joins successive nucleotide sugar units through phosphate. The resulting chain has a sugar-phosphate backbone with bases attached to the sugars. Its nucleotide sequence is its primary structure; changing that sequence changes the information carried along the chain.

This organisation separates two ideas: nucleotides are the repeating chemical units, while purines and pyrimidines classify their base components. The sugar and phosphate must be included when describing a complete nucleotide, even when attention is focused on the base sequence.

How do DNA and RNA structures explain their functions?

DNA consists of two nucleic-acid chains wound around one another in a double helix, a paired spiral arrangement. Hydrogen bonds between specific base pairs hold the chains together. Adenine pairs with thymine; cytosine pairs with guanine.

What does complementary pairing mean?

The strands are complementary: the base on one strand determines its partner on the other. Complementary does not mean identical. Where one strand has adenine, the partner has thymine; where it has guanine, the partner has cytosine.

The drawing shows two strands winding around one another. The interior base-pair labels show adenine opposite thymine and guanine opposite cytosine. Follow corresponding pairs across the helix to distinguish complementarity from identical base sequences.

See Fig. 10.7 in your NCERT textbook

How are heredity and protein synthesis linked?

DNA is the chemical basis of heredity and may be regarded as a reserve of genetic information. It can duplicate during cell division so that daughter cells receive DNA. The information for synthesis of a particular protein is present in DNA.

Protein synthesis is the production of proteins in cells. RNA molecules participate in carrying it out. The three types are messenger RNA, ribosomal RNA and transfer RNA, abbreviated mRNA, rRNA and tRNA respectively. They perform different functions in this process.

Comparing DNA and RNA therefore requires both structural and functional features. A sugar difference alone does not describe their roles, while saying that both concern proteins omits their distinct bases and strand arrangements. Keep composition, shape and biological role as separate comparison points.

Glossary

• Carbohydrate — An optically active polyhydroxy aldehyde or ketone, or a compound yielding such units on hydrolysis.
• Monosaccharide — A carbohydrate that cannot be hydrolysed into simpler polyhydroxy aldehyde or ketone units.
• Reducing sugar — A carbohydrate capable of reducing Fehling’s solution or Tollens’ reagent under the test conditions.
• Anomers — Cyclic sugar forms differing only in configuration at the carbon that was the carbonyl carbon.
• Glycosidic linkage — An oxygen bridge joining monosaccharide units after elimination of water during bond formation.
• Osazone — A sugar derivative formed with excess phenylhydrazine, involving reaction at two adjacent carbon positions.
• Zwitterion — A dipolar ion containing both positive and negative charges while remaining electrically neutral overall.
• Isoelectric point — The pH at which an amino acid has zero average net charge and no net electrical migration.
• Peptide linkage — The amide bond joining amino-acid residues through their carboxyl and amino functional groups.
• Denaturation — Loss of a protein’s biological activity following disruption of higher structure while primary structure remains intact.
• Enzyme — A biological catalyst that accelerates a specific reaction involving a particular substrate.
• Hormone — An intercellular messenger released by an endocrine gland and transported through the bloodstream to its site of action.
• Vitamin — An organic compound required in small dietary amounts for specific biological functions, growth and health.
• Nucleoside — A unit consisting of a nitrogenous base attached to a sugar, without the additional phosphate component.
• Nucleotide — A sugar-base-phosphate unit that serves as a building block of a nucleic acid chain.

Common errors and misconceptions

• Misconception: Every compound fitting the hydrate formula is a carbohydrate. Correct: Acetic acid fits that pattern but is not a carbohydrate; functional groups and hydrolysis behaviour matter.
• Misconception: D means dextrorotatory. Correct: D denotes relative configuration, while (+) or (−) denotes optical rotation. D-(−)-fructose illustrates the distinction.
• Misconception: Fructose is non-reducing because it is a ketose. Correct: It reduces Fehling’s and Tollens’ reagents but does not undergo the mild bromine-water oxidation used to distinguish it from glucose.
• Misconception: Every disaccharide is non-reducing. Correct: Sucrose is non-reducing; maltose and lactose retain the ability to produce a free reducing group.
• Misconception: A zwitterion has no charges. Correct: It contains positive and negative charges that balance overall; electrical neutrality does not remove its charged groups.
• Misconception: Denaturation destroys the amino-acid sequence. Correct: It disrupts secondary and tertiary structure while leaving primary structure intact.
• Misconception: All water-soluble vitamins cannot be stored. Correct: Vitamin B₁₂ is the stated exception; vitamin C and the other water-soluble vitamins require regular supply.
• Misconception: DNA’s two strands have identical bases facing each other. Correct: They have complementary pairs: adenine with thymine and guanine with cytosine.

Exam-style questions with model answers

Q1. Define a monosaccharide and give one example. [2 marks]

1. A monosaccharide is a carbohydrate that cannot be hydrolysed further into a simpler polyhydroxy aldehyde or ketone.
2. Glucose is an example; it is a six-carbon monosaccharide containing an aldehyde group in its open-chain structure.

Q2. D-(−)-fructose has its lowest asymmetric carbon’s hydroxyl group on the right in its conventional Fischer projection and rotates plane-polarised light to the left. Explain the two symbols D and (−). [2 marks]

1. D specifies configuration relative to D-glyceraldehyde, consistent with the stated hydroxyl position at the lowest asymmetric carbon.
2. (−) specifies laevorotation, the stated rotation of plane-polarised light to the left. Configuration does not determine the sign of rotation.

Q3. Glucose gives n-hexane on prolonged heating with hydrogen iodide, an oxime with hydroxylamine, and gluconic acid with bromine water. State one structural inference from each observation. [3 marks]

1. Formation of n-hexane indicates that glucose’s six carbon atoms are connected in an unbranched chain, matching the carbon skeleton of this hydrocarbon.
2. Formation of an oxime indicates a carbonyl group, since hydroxylamine reacts with that group to form the oxime derivative.
3. Oxidation to gluconic acid by mild bromine water identifies the carbonyl group as aldehydic, producing a carboxyl group at that end.

Q4. Sucrose links C-1 of glucose with C-2 of fructose, involving both reducing groups. Maltose links C-1 of one glucose to C-4 of another, leaving the second unit able to form an aldehyde group. Compare their reducing behaviour and state each sugar’s hydrolysis products. [4 marks]

1. Sucrose is non-reducing because both component reducing groups participate in the given glycosidic linkage and are unavailable as free reducing groups.
2. Maltose is reducing because the second glucose unit can produce the free aldehyde group described in the question.
3. Hydrolysis of one sucrose molecule gives one glucose molecule and one fructose molecule, releasing the two different component monosaccharides.
4. Hydrolysis of one maltose molecule gives two glucose molecules, since both of its component monosaccharide units are glucose.

Q5. Explain zwitterion formation in an α-amino acid, its net charge, its amphoteric behaviour and the meaning of its isoelectric point. [4 marks]

1. The carboxyl group can lose a proton while the amino group accepts one, giving the zwitterionic form ⁺H₃N-CH(R)-COO⁻, where R is the side chain.
2. The positive amino-group charge and negative carboxyl-group charge balance, so the zwitterion is neutral overall despite containing charged groups.
3. It behaves amphoterically: it can react with both acids and bases because its groups can accept or release protons.
4. Its isoelectric point is the pH at which average net charge is zero and there is no net migration in an electric field.

Q6. Define the four levels of protein structure. Then explain what denaturation changes and what remains intact, using boiling egg white as the example. [6 marks]

1. Primary structure is the specific sequence of amino-acid residues in a polypeptide chain. It identifies the order in which the units are linked.
2. Secondary structure is the regular folding of the chain into forms such as an α-helix or a β-pleated sheet, stabilised by hydrogen bonding.
3. Tertiary structure is the overall three-dimensional folding of the polypeptide chain, incorporating the further arrangement of its secondary structure.
4. Quaternary structure is the spatial arrangement of two or more polypeptide subunits relative to one another in a protein.
5. Denaturation disrupts secondary and tertiary structure and causes loss of biological activity. The coagulation of egg white on boiling is an example.
6. The primary structure remains intact during denaturation: disruption of folding does not itself mean cleavage of peptide bonds or loss of the amino-acid sequence.

Q7. Classify vitamins A, B₁, C and K by solubility and name one characteristic deficiency effect for each. Give one vitamin per answer point. [4 marks]

1. Vitamin A is fat soluble. Its deficiency causes night blindness; it belongs to the same solubility group as vitamins D, E and K.
2. Vitamin B₁ is water soluble. Its deficiency causes beri beri, associated with loss of appetite and retarded growth.
3. Vitamin C is water soluble. Its deficiency causes scurvy, characterised here by bleeding gums, and it requires regular dietary supply.
4. Vitamin K is fat soluble. Its deficiency increases blood-clotting time, reflecting its role in normal blood coagulation.

Q8. Distinguish DNA and RNA using sugar, distinctive base, strand structure and function. Add a fifth point explaining complementary base pairing in DNA. [5 marks]

1. DNA contains the pentose sugar 2-deoxyribose, whereas RNA contains ribose. The sugar is a constituent of each nucleotide rather than the nitrogenous base.
2. DNA contains thymine, whereas RNA contains uracil in its place. Adenine, guanine and cytosine occur in both types of nucleic acid.
3. DNA has two strands wound into a double helix. RNA is single stranded and sometimes folds back on itself.
4. DNA stores genetic information and carries the message for a particular protein. Different RNA molecules participate in carrying out protein synthesis in the cell.
5. DNA strands are complementary because adenine pairs specifically with thymine and guanine with cytosine through hydrogen bonding; facing bases are therefore paired, not identical.

Key takeaways

• Classify carbohydrates by hydrolysis behaviour, then distinguish aldoses from ketoses and reducing sugars from non-reducing sugars.
• Glucose reactions establish a straight six-carbon chain, an aldehyde group, five hydroxyl groups and a primary alcohol group.
• D and L describe configuration; optical rotation requires a separate positive or negative sign.
• Sucrose uses both reducing groups in its linkage, whereas maltose and lactose can generate a free aldehyde group.
• Starch and glycogen store carbohydrate, while cellulose forms structural material; glucose configuration and branching distinguish their structures.
• Amino acids form zwitterions and peptide bonds; protein sequence, folding and subunit arrangement describe different structural levels.
• Denaturation disrupts higher protein structure while preserving primary structure; enzymes catalyse reactions and hormones act as intercellular messengers.
• Vitamins A, D, E and K are fat soluble; the B group and vitamin C are water soluble.
• Nucleotides contain sugar, base and phosphate; DNA stores genetic information and RNA participates in protein synthesis.

Test yourself

Why does a hydrate-like molecular formula not establish that a compound is a carbohydrate?

Acetic acid fits that pattern but is not a carbohydrate, whereas rhamnose is a carbohydrate that does not fit it.

Which glucose reaction shows five hydroxyl groups?

Acetylation with acetic anhydride produces glucose pentaacetate, demonstrating the presence of five hydroxyl groups.

Which carbon is anomeric in glucose, and which in fructose?

C-1 is the anomeric carbon in glucose; C-2 is the anomeric carbon in fructose.

Can the osazone test distinguish glucose from fructose?

No. Both give the same osazone; mild bromine-water oxidation distinguishes glucose from fructose instead.

Which starch component is branched, and where do branches occur?

Amylopectin is branched. Its chains have C-1 to C-4 linkages, with C-1 to C-6 linkages at branch points.

Does zero net charge mean a zwitterion contains no charged groups?

No. It contains positively and negatively charged groups whose charges balance to give overall neutrality.

Which protein structural level survives denaturation?

The primary structure, meaning the amino-acid sequence, remains intact despite disruption of secondary and tertiary structures.

Which water-soluble vitamin is the exception to the non-storage statement?

Vitamin B₁₂ is the exception; the other water-soluble vitamins require regular dietary supply because they are not stored.

What must be added to a nucleoside to obtain a nucleotide?

A phosphate component must be added to the nucleoside, which already contains a sugar and a base.