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Biomolecules — Class 12 Chemistry Notes & Practice

Biomolecules — Class 12 Chemistry Notes & Practice

Take a breath. If the word biomolecules makes you picture a scary wall of ring structures, you are not alone — almost every Class 12 student feels that on day one. Here is the good news: this chapter is one of the most scoring units in the whole paper. It is worth 7 marks in the CBSE 2026–27 Chemistry syllabus (Unit 10), and almost every mark comes from clear definitions, neat comparisons and a handful of reactions you can genuinely memorise in a weekend. There is very little heavy calculation. If you can describe things carefully, you can score here.

Think of your body as a busy kitchen. Carbohydrates are the fuel on the stove, proteins are the utensils and workers, enzymes are the specialised chefs who only cook one dish each, vitamins are the tiny pinches of spice you cannot live without, nucleic acids are the recipe book locked in the safe, and hormones are the notes the manager slips under the door telling everyone to speed up or slow down. Every heading in this chapter is just one of those characters. Once you meet them properly, they stop being scary.

This page is a full biomolecules class 12 notes with solved examples guide. We build every idea from zero, then practise it. You will find a compact biomolecules class 12 formula and reaction list woven into the sections, seventeen worked model answers written the way an examiner wants to see them, three hand-drawn schematic diagrams, and a worksheet of biomolecules class 12 chemistry important questions with hidden answers so you can test yourself honestly. Work through it slowly. You do not need to finish it in one sitting.

Meet Your Tutor

Biomolecules become easier when structure, linkage and function stay in the same picture. I will help you compare carbohydrates, proteins, enzymes, vitamins and nucleic acids through small structural clues rather than isolated lists.

What You’ll Learn

Your Game Plan

Do not read this chapter front to back in one go. Follow this order instead — it is designed so each step makes the next one easier.

  1. Day 1 — Carbohydrates. Learn the classification tree first, then glucose, then the disaccharides and polysaccharides. This is roughly half the unit.
  2. Day 2 — Proteins. Amino acids → zwitterion → peptide bond → the four structural levels → denaturation. Each idea sits on the one before it.
  3. Day 3 — Enzymes and Vitamins. Short, definition-heavy and very examinable. Make a deficiency-disease table in your own handwriting.
  4. Day 4 — Nucleic acids and hormones. Learn DNA vs RNA as a table, not as paragraphs.
  5. Day 5 — Practice only. Close the notes, attempt the worksheet at the bottom of this page, then check. Rewrite any answer you got wrong, in full.
Exam Tip — know what is actually assessed
In the CBSE 2026–27 Class 12 Chemistry syllabus, Biomolecules is Unit 10 and carries 7 marks in the theory paper. Two neighbouring topics — Polymers and Chemistry in Everyday Life — are still in the syllabus but are assessed formatively only, i.e. through internal/class assessment, not in the summative board paper. Do not skip them entirely (they are easy and they build vocabulary), but when your revision time is tight, Biomolecules earns you board marks and they do not. There is also a practical link: characteristic tests of carbohydrates and proteins in foodstuffs appear in the Class 12 practical syllabus, so the Molisch, Fehling, Tollens, biuret, ninhydrin and xanthoproteic tests are worth knowing by name and by observation.

Study Notes

Carbohydrates — Classification and Structure

Carbohydratespolyhydroxy aldehydes or ketones Monosaccharidescannot be hydrolysed further Oligosaccharidesgive 2 to 10 units on hydrolysis Polysaccharidesmany units; not sweet, not soluble Glucose — aldohexose Fructose — ketohexose Ribose — aldopentose Sucrose = Glu + Fru Maltose = Glu + Glu Lactose = Glu + Gal Starch — plant store Glycogen — animal store Cellulose — plant fibre
Schematic classification of carbohydrates. Colour marks the branch: blue = monosaccharide, green = oligosaccharide, purple = polysaccharide.

Start with the name. Old chemists noticed that many sugars had the formula Cx(H2O)y — carbon plus what looked like water — so they called them carbohydrates, meaning “hydrates of carbon”. The name stuck even though it is not really accurate (rhamnose and deoxyribose break the pattern, and acetic acid fits the pattern without being a sugar). What actually defines a carbohydrate chemically is much tidier: a carbohydrate is a polyhydroxy aldehyde or a polyhydroxy ketone, or a compound that gives one on hydrolysis. That single sentence is worth writing on the inside cover of your notebook.

From there, classification is just a question of “how many sugar units do you get when you boil it with dilute acid?”. Get none extra (it will not break down) → monosaccharide. Get two to about ten → oligosaccharide, of which the two-unit ones, the disaccharides, are the only ones the board cares about. Get very many → polysaccharide. That is the entire tree in the diagram above.

There is a second, cross-cutting classification you must also know, and it is the one students most often confuse with the first. Based on reducing power, sugars are either reducing or non-reducing. A reducing sugar has a free aldehyde or free ketone group (a free hemiacetal carbon), so it can reduce Tollens’ reagent to a silver mirror and Fehling’s solution to a red-brown Cu2O precipitate. All monosaccharides are reducing. Among disaccharides, maltose and lactose are reducing; sucrose is not, because both of its anomeric carbons are tied up in the linkage. If you have already met these tests in the oxidation and reduction language used in Electrochemistry, the logic will feel familiar — the sugar is the reducing agent, the metal ion is reduced.

Naming, too, is systematic and free marks. Count the carbons: triose (3), tetrose (4), pentose (5), hexose (6). Look at the carbonyl: aldehyde → aldo-, ketone → keto-. Glue them together. Glucose has six carbons and an aldehyde, so it is an aldohexose. Fructose has six carbons and a ketone, so it is a ketohexose. Ribose has five carbons and an aldehyde: aldopentose. That is all there is to it.

Key Idea — two independent classifications
By hydrolysis: mono / oligo (di) / poly. By reducing power: reducing / non-reducing. A sugar has one label from each list. Maltose, for example, is a reducing disaccharide. Sucrose is a non-reducing disaccharide. Examiners love a question that quietly tests whether you can hold both ideas at once.
Example 1 — Classify and justify (2 marks)
Question: Classify the following as monosaccharide, disaccharide or polysaccharide, and state in each case whether it is reducing or non-reducing: (i) lactose (ii) starch.

Model answer:
(i) Lactose is a disaccharide (hydrolysis gives one glucose + one galactose). It is a reducing sugar because the anomeric carbon of its glucose unit is free, so an open-chain aldehyde form is available. (1 mark)
(ii) Starch is a polysaccharide (hydrolysis gives a very large number of α-D-glucose units). It is non-reducing for practical purposes, since the single free reducing end on a huge molecule is chemically negligible. (1 mark)

Examiner’s note: the marks are for the reason, not the label. Always write “because the anomeric carbon is free / not free”.
Example 2 — A genuine calculation (3 marks)
Question: Glucose has the molecular formula C6H12O6. Calculate (a) its molar mass and (b) the percentage by mass of carbon, hydrogen and oxygen in it. (Atomic masses: C = 12.011, H = 1.008, O = 15.999 u)

Solution:
(a) M = 6(12.011) + 12(1.008) + 6(15.999)
= 72.066 + 12.096 + 95.994 = 180.16 g mol–1 (1 mark)
(b) %C = 72.066 ÷ 180.16 × 100 = 40.00 %
%H = 12.096 ÷ 180.16 × 100 = 6.71 %
%O = 95.994 ÷ 180.16 × 100 = 53.28 % (2 marks)
Check: 40.00 + 6.71 + 53.28 = 99.99 ≈ 100 % ✓

Notice that the empirical formula is CH2O — literally “carbon + water”. That is exactly why the family got the name carbohydrate.

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Monosaccharides: Glucose and Fructose

Glucose is the single most important molecule in this chapter. It is the sugar your blood carries, the sugar plants make in photosynthesis, and the building block of starch, cellulose, glycogen, sucrose, maltose and lactose. Learn glucose properly and half the unit falls into place.

How it is prepared. Two routes are in the syllabus. From sucrose: boil cane sugar with dilute HCl or H2SO4 in alcohol and you get equal amounts of glucose and fructose — C12H22O11 + H2O → C6H12O6 (glucose) + C6H12O6 (fructose). Glucose is less soluble in alcohol, so it crystallises out first. From starch: hydrolyse starch with dilute H2SO4 at about 393 K under roughly 2–3 atm pressure — (C6H10O5)n + n H2O → n C6H12O6. Remember the conditions; they are frequently asked as a one-mark fill-in.

How chemists proved the open-chain structure. This is a lovely piece of detective work and it is very examinable, because each clue is a separate mark. Glucose adds one molecule of hydroxylamine to give a monoxime and one molecule of HCN to give a cyanohydrin → there is exactly one carbonyl group. It gives a silver mirror with Tollens’ reagent and a red precipitate with Fehling’s solution → that carbonyl is an aldehyde. Acetylation with acetic anhydride gives a pentaacetate → there are five –OH groups, and since the molecule is stable they must be on five different carbons. Prolonged heating with HI gives n-hexane → the six carbons are in a straight, unbranched chain. Mild oxidation with bromine water gives gluconic acid (six carbons, one –COOH); strong oxidation with nitric acid gives saccharic acid (six carbons, two –COOH, one at each end), confirming the aldehyde sits at C–1 and a primary alcohol at C–6.

Where the open chain fails. Three stubborn facts refuse to fit: glucose does not give the 2,4-DNP (Schiff’s) test, does not form a hydrogensulphite addition product with NaHSO3, and the pentaacetate does not react with hydroxylamine — all three suggest no free –CHO. Worse, glucose exists as two crystalline forms, α-D-glucose (m.p. 419 K, specific rotation +111°) and β-D-glucose (m.p. 423 K, +19.2°), whose solutions both drift to +52.5° on standing. That drift is called mutarotation. The resolution is the cyclic hemiacetal: the –OH on C–5 attacks the C–1 aldehyde, closing a six-membered pyranose ring and creating a new stereocentre at C–1 called the anomeric carbon. The two ring forms differing only at that carbon are anomers.

Fructose is glucose’s partner: same molecular formula C6H12O6, same molar mass, but a ketone at C–2 instead of an aldehyde at C–1. It cyclises to a five-membered furanose ring. Here is the fact that trips everyone up: fructose still reduces Tollens’ and Fehling’s reagents even though it is a ketose. Why? Because those reagents are alkaline, and in base fructose isomerises (via an enediol) into glucose and mannose, which then reduce the reagent. So “reducing sugar” does not mean “must be an aldose”.

Because glucose and fructose have identical molar masses, any technique that measures molar mass — freezing-point depression, osmotic pressure and the other colligative properties you practised in the Class 12 Chemistry chapter on Solutions — cannot tell them apart. Only chemistry can.

Diagram: Fischer and Haworth projections of α- and β-D-glucopyranose with full stereochemistry — to be added by illustrator.

Common Mistake — “D” does not mean dextrorotatory
The capital D in D-glucose is a configurational label: it says the –OH on the highest-numbered chiral carbon (C–5 in glucose) is on the right in the Fischer projection. It says nothing about which way the compound rotates plane-polarised light. Direction of rotation is shown by (+) or (–), or by the prefixes d/l. D-fructose, for instance, is laevorotatory. Writing “D means it rotates light to the right” costs you the mark every single time.
Example 3 — Counting stereoisomers (3 marks)
Question: (a) How many chiral carbons are present in the open-chain form of glucose? (b) How many optically active stereoisomers are therefore possible for an aldohexose? (c) Repeat for the open-chain form of fructose.

Solution:
(a) Open-chain glucose is OHC–CHOH–CHOH–CHOH–CHOH–CH2OH. C–1 is the aldehyde (not chiral) and C–6 is –CH2OH (two H atoms, not chiral). The four carbons C–2, C–3, C–4 and C–5 each carry four different groups, so there are 4 chiral carbons. (1 mark)
(b) Number of stereoisomers = 2n with n = 4, so 24 = 16 stereoisomers (8 D-forms and 8 L-forms). (1 mark)
(c) In open-chain fructose the carbonyl is at C–2, so C–1 (–CH2OH), C–2 (C=O) and C–6 (–CH2OH) are all non-chiral. The chiral carbons are C–3, C–4 and C–5, i.e. n = 3, giving 23 = 8 stereoisomers. (1 mark)

Watch out: once glucose cyclises, C–1 becomes chiral too (n = 5, 25 = 32 for the ring forms). Read the question — open chain or ring?
Example 4 — Explain the evidence (3 marks)
Question: Give chemical reasons for each of the following observations about glucose. (i) It gives a silver mirror with Tollens’ reagent. (ii) It does not react with sodium hydrogensulphite. (iii) Its solution shows mutarotation.

Model answer:
(i) The open-chain form of glucose carries a free –CHO group, which is oxidised to gluconic acid while Ag+ is reduced to metallic silver, depositing as a mirror. (1 mark)
(ii) In the crystalline and predominant solution state glucose exists as a cyclic hemiacetal, so no free –CHO is available for the nucleophilic addition of NaHSO3. (1 mark)
(iii) Freshly dissolved α-D-glucose (+111°) or β-D-glucose (+19.2°) slowly opens to the open-chain form and re-closes both ways, until an equilibrium mixture with specific rotation +52.5° is reached. This gradual change in optical rotation is mutarotation. (1 mark)

Examiner’s note: (i) and (ii) look contradictory. State plainly that only a small fraction exists open-chain — enough for the sensitive Tollens’ test, not enough for a bulk addition reaction.

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Disaccharides: Sucrose, Maltose and Lactose

Join two monosaccharides by squeezing out one molecule of water between their –OH groups and you get a disaccharide. The oxygen bridge left behind is called a glycosidic linkage. Three of them are in your syllabus, and the cleanest way to remember them is a single line each: sucrose = glucose + fructose, maltose = glucose + glucose, lactose = glucose + galactose. All three share the formula C12H22O11 and the molar mass 342.30 g mol–1.

Sucrose (table sugar) is the odd one out and the board knows it. Its linkage runs C–1 of α-glucose to C–2 of β-fructose, which are the anomeric carbons of both units. With no free anomeric carbon left, neither ring can open, so sucrose is non-reducing. It also shows the beautiful phenomenon of inversion: sucrose itself is dextrorotatory (+66.5°), but on hydrolysis it gives glucose (+52.5°) and fructose (–92.4°), and because fructose rotates further to the left than glucose does to the right, the mixture comes out laevorotatory (–39.9°). The sign of rotation literally flips, which is why the product is called invert sugar.

Maltose (from starch digestion) joins C–1 of one α-D-glucose to C–4 of the next: an α-1,4-glycosidic linkage. The second unit’s anomeric carbon is free, so maltose is reducing. Lactose (milk sugar) joins C–1 of β-D-galactose to C–4 of D-glucose: a β-1,4-glycosidic linkage. Again a free anomeric carbon on the glucose unit, so lactose is reducing too.

Feature α-Glycosidic linkage β-Glycosidic linkage
–OH at anomeric CPoints down (opposite side to the C–6 group)Points up (same side as the C–6 group)
Chain shape it producesCoiled, helical, easy to pack looselyStraight, flat ribbons that stack tightly
Typical examplesMaltose, starch (amylose, amylopectin), glycogenLactose, cellulose
Digested by humans?Yes — we have α-amylase and maltaseCellulose: no (no cellulase). Lactose: yes, if lactase is present
Biological roleEnergy storageStructural strength (and milk nutrition)
Exam Tip — one sentence that answers half the disaccharide questions
Memorise this and adapt it: “X is reducing/non-reducing because the anomeric carbon of its ____ unit is free / is involved in the glycosidic linkage, so the ring can / cannot open to give a free carbonyl group.” Fill in the blanks and you have a complete two-mark answer for sucrose, maltose or lactose.
Example 5 — Invert sugar mass balance (3 marks)
Question: 342.30 g of pure sucrose is completely hydrolysed. Write the equation and calculate the mass of each product. (M: sucrose 342.30, water 18.02, glucose 180.16, fructose 180.16 g mol–1)

Solution:
C12H22O11 + H2O → C6H12O6 (D-glucose) + C6H12O6 (D-fructose) (1 mark)
Moles of sucrose = 342.30 ÷ 342.30 = 1.00 mol, so 1.00 mol of each product is formed. (1 mark)
Mass of glucose = 1.00 × 180.16 = 180.16 g; mass of fructose = 180.16 g. (1 mark)
Mass check: 342.30 + 18.02 = 360.32 g in; 180.16 + 180.16 = 360.32 g out ✓ The water is not a spectator — its mass appears in the products.
Example 6 — Why the name “invert sugar”? (2 marks)
Question: Sucrose has a specific rotation of +66.5°. Its hydrolysis products glucose and fructose have +52.5° and –92.4° respectively. Explain, with a calculation, why the hydrolysis mixture is called invert sugar.

Solution: Hydrolysis gives glucose and fructose in a 1 : 1 mole ratio, so the observed rotation is the average:
(+52.5°) + (–92.4°) = –39.9° for the equimolar pair. (1 mark)
The starting sucrose was dextrorotatory (+66.5°) and the product mixture is laevorotatory (–39.9°). The sign of optical rotation has been inverted, so the product is called invert sugar and the process is called inversion. (1 mark)

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Polysaccharides: Starch, Cellulose and Glycogen

Now string hundreds or thousands of glucose units together. All three polysaccharides in your syllabus are made from the same monomer — glucose — and yet one is soft food, one is wood and cotton, and one is your body’s emergency fuel tank. The whole difference is in the type of linkage and the amount of branching. That is a genuinely beautiful chemistry lesson: same bricks, different mortar, completely different building.

Starch is the plants’ store cupboard, found in rice, wheat, potato and maize. It is a mixture of two fractions. Amylose is about 15–20 % of starch, is water-soluble, and is a long unbranched chain of 200–1000 α-D-glucose units linked C–1 to C–4. Amylopectin is the other 80–85 %, is water-insoluble, and is branched: mainly α-1,4 chains with an α-1,6 branch roughly every 20–25 units. Amylose is the fraction that gives the deep blue-black colour with iodine, because its helix traps I2 molecules inside.

Cellulose is the most abundant organic polymer on Earth — cotton is about 90 % cellulose, wood roughly half. It is a straight, unbranched chain of β-D-glucose units joined C–1 to C–4. Those β-linkages make flat ribbons that lie side by side and hydrogen-bond to one another, which is why cellulose is tough, insoluble and fibrous. Humans have no cellulase enzyme, so we cannot digest it — it passes through as dietary fibre. Cows and termites manage only because gut microbes do the job for them.

Glycogen is “animal starch”, stored in liver and muscle. Structurally it is like amylopectin but much more highly branched. That matters practically: more branches means more free chain ends, and enzymes chew glucose off the ends, so glycogen can be mobilised into blood glucose very quickly when you sprint or when blood sugar falls.

Point of comparison Starch Cellulose Glycogen
Monomer unitα-D-glucoseβ-D-glucoseα-D-glucose
Linkageα-1,4 (amylose); α-1,4 with α-1,6 branches (amylopectin)β-1,4 onlyα-1,4 with very frequent α-1,6 branches
BranchingAmylose none; amylopectin moderateNone — perfectly linearHighest of the three
OccurrenceRice, wheat, potato, maizeCell walls of plants; cotton, wood, paperLiver, muscle, brain of animals; also in yeast and fungi
FunctionEnergy reserve in plantsStructural supportRapid-release energy reserve in animals
Test with iodineDeep blue-blackNo colourRed-brown
Digestible by humansYesNoYes
Key Rule — the repeating unit is C6H10O5, not C6H12O6
Every time one glucose joins the chain, one molecule of water is lost. So the polysaccharide formula is written (C6H10O5)n, with a repeat-unit mass of 162.14 g mol–1 (that is 180.16 – 18.02). Use 162, not 180, in any degree-of-polymerisation calculation.
Example 7 — Degree of polymerisation (2 marks)
Question: A cellulose sample has an average molar mass of 1.62 × 105 g mol–1. Estimate the average number of glucose units per chain. (Repeat unit C6H10O5 = 162.14 g mol–1)

Solution:
Degree of polymerisation n = total molar mass ÷ repeat-unit mass
n = (1.62 × 105) ÷ 162.14 = 999.1 ≈ 1.0 × 103 glucose units (2 marks)

Common slip: dividing by 180.16 gives 899, which is wrong. Once a glucose is inside the chain it has already lost its molecule of water.
Example 8 — The classic 3-marker (3 marks)
Question: Starch and cellulose are both polymers of glucose, yet humans can digest starch but not cellulose. Explain, and state one further structural difference between them.

Model answer:
• Starch is built from α-D-glucose joined by α-1,4-glycosidic linkages, whereas cellulose is built from β-D-glucose joined by β-1,4-glycosidic linkages. (1 mark)
• Human digestive enzymes (α-amylase, maltase) are stereospecific and can only hydrolyse the α-linkage. We secrete no cellulase, so the β-linkage is not attacked and cellulose passes through undigested as roughage. (1 mark)
• Further difference: starch contains a branched fraction (amylopectin, with α-1,6 branch points) and gives a blue-black colour with iodine, while cellulose is strictly linear, forms tightly hydrogen-bonded fibres and gives no colour with iodine. (1 mark)

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Amino Acids: Structure, Zwitterion and Classification

Proteins are polymers, and their monomers are α-amino acids. The general formula is short enough to sketch in three seconds: a central carbon carrying –NH2, –COOH, –H and a side chain R, written R–CH(NH2)–COOH. “α” simply means the amino group sits on the carbon immediately next to the carboxyl group. Only the R group changes from one amino acid to the next, and there are about twenty R groups in nature. Twenty letters; every protein you own is a word spelled with them.

Classification by R group. If R is a plain hydrocarbon or otherwise neutral, the amino acid is neutral (glycine, alanine, valine). If R carries an extra –COOH, it is acidic (aspartic acid, glutamic acid). If R carries an extra basic nitrogen, it is basic (lysine, arginine). Count the acidic and basic groups in the whole molecule — whichever wins decides the label.

Essential versus non-essential. Ten of the twenty cannot be made by the human body at all and must come from the diet — these are the essential amino acids (valine, leucine, isoleucine, lysine, methionine, phenylalanine, threonine, tryptophan, histidine, arginine). The other ten the body synthesises for itself and they are called non-essential. Note carefully: “non-essential” does not mean unimportant, it means “you do not have to eat it”. That distinction is a favourite one-mark trap.

The zwitterion — the single most examined idea here. An amino acid contains an acid (–COOH) and a base (–NH2) in the same molecule, so the proton simply hops across internally. The result is the zwitterion (also called the dipolar or internal salt form): R–CH(NH3+)–COO–. Because the molecule is really a salt, amino acids behave like ionic solids — they are crystalline, have unexpectedly high melting points (they usually decompose above 470 K rather than melt cleanly), are soluble in water but poorly soluble in organic solvents, and are amphoteric: they neutralise both acids and bases. If you have seen how a substance can react with both an acid and a base in the Class 10 Science chapter on Acids, Bases and Salts, the amphoteric behaviour here is exactly the same idea, just carried by one molecule instead of two.

The pH at which the amino acid exists entirely as the zwitterion — net charge zero, so it does not migrate to either electrode in an electric field — is called the isoelectric point (pI). Below pI the molecule picks up a proton and becomes a cation, so it migrates to the cathode; above pI it loses a proton and becomes an anion, so it migrates to the anode.

One more structural point: except for glycine (R = H), the α-carbon of every amino acid carries four different groups, so every amino acid except glycine is chiral and optically active. Naturally occurring amino acids are almost all of the L-configuration.

Common Mistake — drawing the zwitterion with the wrong charges
The proton leaves the acid and lands on the base. So the carboxyl loses H+ and becomes –COO–; the amino gains H+ and becomes –NH3+. Students routinely write –COOH+ and –NH2–, which is chemical nonsense and scores zero. Also remember the net charge is zero — the ion is dipolar, not charged overall.
Example 9 — Zwitterion and amphoteric behaviour (3 marks)
Question: (a) Write the zwitterionic form of glycine. (b) Show how it reacts with HCl and with NaOH. (c) Give one physical property of amino acids that this structure explains.

Model answer:
(a) Glycine is H2N–CH2–COOH; its zwitterion is H3N+–CH2–COO–, net charge zero. (1 mark)
(b) With acid the carboxylate is protonated: H3N+–CH2–COO– + H+ → H3N+–CH2–COOH (a cation).
With base the ammonium loses a proton: H3N+–CH2–COO– + OH– → H2N–CH2–COO– + H2O (an anion). Hence glycine is amphoteric. (1 mark)
(c) Because the solid is really an internal salt held together by strong electrostatic forces, glycine is a high-melting crystalline solid (it decomposes near 505 K), soluble in water but almost insoluble in benzene or ether. (1 mark)
Example 10 — Molar mass and chirality (3 marks)
Question: For glycine (C2H5NO2) and alanine (C3H7NO2): (a) calculate the molar masses; (b) state which is optically active and why; (c) how many stereoisomers does the optically active one have?

Solution:
(a) Glycine: 2(12.011) + 5(1.008) + 14.007 + 2(15.999) = 24.022 + 5.040 + 14.007 + 31.998 = 75.07 g mol–1
Alanine: 3(12.011) + 7(1.008) + 14.007 + 2(15.999) = 36.033 + 7.056 + 14.007 + 31.998 = 89.09 g mol–1 (1 mark)
(b) In glycine the α-carbon carries –NH2, –COOH and two hydrogen atoms, so it is not chiral and glycine is optically inactive. In alanine the α-carbon carries –NH2, –COOH, –H and –CH3, four different groups, so it is chiral and alanine is optically active. (1 mark)
(c) One chiral centre, so n = 1 and the number of stereoisomers = 21 = 2, namely D-alanine and L-alanine (an enantiomeric pair). (1 mark)

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Proteins: Peptide Bond and Primary to Quaternary Structure

1. PrimarySequence of amino acidsjoined in a definite orderHeld by: peptide bonds(covalent, very strong) 2. SecondaryLocal shape of the chain:α-helix or β-pleated sheetHeld by: hydrogen bonds(C=O ··· H–N backbone) 3. TertiaryWhole chain folds into a3-D fibrous or globular shapeHeld by: –S–S–, ionic,H-bond, van der Waals 4. QuaternaryTwo or more folded chainsassembled into one unitHeld by: same weak forcesExample: haemoglobin (4)
Schematic flow of the four levels of protein structure. Each level is built on the one before it; only the primary level uses covalent peptide bonds.

Two amino acids can join by losing a molecule of water between the –COOH of one and the –NH2 of the other. The –CO–NH– link created is a peptide bond, and it is nothing exotic — it is simply an amide bond. Two units give a dipeptide, three a tripeptide, many a polypeptide. When a polypeptide reaches a molar mass of roughly 10 000 g mol–1 or more and takes up a definite biological shape, we call it a protein.

Proteins split into two families by shape and behaviour. Fibrous proteins have chains lying parallel and held by hydrogen and disulphide bonds; they are tough, insoluble in water and structural — keratin in hair and nails, myosin in muscle, collagen in tendons. Globular proteins have chains coiled into compact balls, are soluble in water, and do the chemistry — insulin, haemoglobin, albumin and all enzymes.

The four structural levels in the diagram are the classic five-mark question, so learn them as a story rather than a list. Primary structure is the order of the amino acids — the spelling of the word. Change one letter and you can change everything; sickle-cell haemoglobin differs from normal haemoglobin by a single amino acid. Secondary structure is how short stretches of that chain curl up locally, into a right-handed α-helix (hydrogen bonds run along the coil, as in wool keratin) or a β-pleated sheet (chains stretched side by side and hydrogen-bonded across, as in silk fibroin). Tertiary structure is the overall three-dimensional fold of the whole chain, stabilised by disulphide bridges, ionic (salt) bridges, hydrogen bonds, van der Waals forces and hydrophobic interactions. Quaternary structure appears only when two or more folded chains club together into one working unit — haemoglobin is four subunits, which is why it can carry four oxygen molecules.

Key Rule — only the primary level is covalent (mostly)
Peptide bonds (primary) are strong covalent bonds. Levels 2, 3 and 4 are held mainly by weak interactions — hydrogen bonds, ionic attractions, van der Waals and hydrophobic forces — with disulphide bridges as the one covalent exception at the tertiary level. That is precisely why gentle heat or a change in pH can wreck the shape of a protein without breaking the chain itself. Remember this sentence; it hands you the whole of the next section.
Example 11 — Peptide-bond and water-loss counting (3 marks)
Question: (a) How many peptide bonds and how many molecules of water are involved when 6 amino acid molecules condense into one linear hexapeptide? (b) Generalise for n residues. (c) Calculate the molar mass of the tripeptide glycylglycylglycine, given glycine = 75.07 and water = 18.02 g mol–1.

Solution:
(a) Six units in a straight line have 5 joins, and each join releases one water molecule. So 5 peptide bonds and 5 molecules of water. (1 mark)
(b) For n residues: peptide bonds = n – 1 and water molecules lost = n – 1. (1 mark)
(c) M = 3(75.07) – 2(18.02) = 225.21 – 36.04 = 189.17 g mol–1 (1 mark)

Sanity check: a tripeptide has n = 3, so 2 bonds and 2 waters lost — exactly the 2 × 18.02 we subtracted. If your subtraction count and your bond count ever disagree, one of them is wrong.
Example 12 — The five-mark structure question (5 marks)
Question: Describe the four levels of protein structure, naming in each case the type of bonding responsible, and give one example.

Model answer:
Primary structure — the specific linear sequence in which the amino acid residues are joined. Bonding: covalent peptide (–CO–NH–) bonds, plus disulphide links between chains. Example: the 51-residue sequence of insulin. (1 mark)
Secondary structure — the regular local conformation adopted by the backbone, either a right-handed α-helix or a β-pleated sheet. Bonding: intramolecular or intermolecular hydrogen bonds between the C=O of one residue and the N–H of another. Example: α-helix in keratin, β-sheet in silk fibroin. (1 mark)
Tertiary structure — the complete three-dimensional folding of a single polypeptide chain into a fibrous or globular form. Bonding: disulphide bridges, hydrogen bonds, ionic salt bridges, van der Waals and hydrophobic interactions. Example: the globular fold of myoglobin. (1 mark)
Quaternary structure — the arrangement of two or more separately folded subunits into one functional protein. Bonding: the same non-covalent forces acting between subunits. Example: haemoglobin, made of four subunits. (1 mark)
Overall point — only the primary level is fully covalent, so the higher levels can be destroyed (denaturation) while the primary sequence survives intact. (1 mark)

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Denaturation of Proteins

Crack an egg into a hot pan. The clear, runny white turns opaque, white and solid, and no amount of cooling will turn it back. You have just watched denaturation, and the kitchen is the best laboratory you have for this topic.

Definition to memorise word for word: denaturation is the process in which a protein loses its secondary, tertiary and quaternary structure — its biological activity along with it — while the primary structure remains intact, when it is subjected to physical or chemical stress such as heat, strong acid or alkali, heavy-metal salts, alcohol, urea or ultraviolet radiation.

The mechanism follows straight from the Key Rule you just read. The higher levels are held by weak forces; heat supplies enough energy to shake those hydrogen bonds and salt bridges apart, and acid or alkali changes the charges on the side chains so the ionic bridges no longer attract. The globular protein uncoils into a random tangle, exposes its water-hating interior, and clumps together — which is exactly why the coagulated protein looks cloudy and becomes insoluble. But the peptide bonds are covalent and far too strong to be affected, so the sequence survives.

Everyday examples worth quoting in the exam: boiling an egg (soluble globular albumin becomes insoluble solid), curdling of milk when lactic acid is produced, using alcohol as a disinfectant (it denatures the proteins of bacteria), and the medical use of an egg-white or milk drink as a first-aid antidote in heavy-metal poisoning — the metal ion binds the food protein rather than the patient’s own enzymes. In most laboratory cases denaturation is irreversible.

Common Mistake — denaturation is not hydrolysis
Denaturation unfolds a protein; hydrolysis chops it up. In denaturation not a single peptide bond breaks and the primary structure is untouched. In hydrolysis (hot acid or a protease enzyme) the peptide bonds themselves are cleaved and you end up with free amino acids. If your answer says “the protein breaks down into amino acids on boiling an egg”, you have described the wrong process.
Example 13 — Define and illustrate (3 marks)
Question: What is meant by denaturation of a protein? Which structural levels are affected and which is not? Give two everyday examples.

Model answer:
• Denaturation is the loss of the specific three-dimensional shape and hence the biological activity of a protein, caused by heat, change of pH, heavy-metal ions, alcohol or UV radiation. (1 mark)
• The secondary, tertiary and quaternary structures are destroyed because the weak hydrogen bonds, ionic salt bridges and hydrophobic interactions holding them are disrupted. The primary structure is not affected, since the covalent peptide bonds are not broken. (1 mark)
• Examples: (i) the coagulation of egg white on boiling, in which soluble albumin becomes an insoluble white solid; (ii) the curdling of milk by lactic acid produced by bacteria. (1 mark)

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Enzymes: Nature, Specificity and Mechanism

An enzyme is a biocatalyst — a substance produced by a living cell that speeds up a biochemical reaction without being used up. Chemically, almost all enzymes are globular proteins. They are the reason your body can do at 310 K and pH 7 what a chemistry laboratory would need high temperature and strong acid to achieve.

Naming. Take the substrate or the reaction and add -ase. Maltase hydrolyses maltose, urease hydrolyses urea, oxidase carries out oxidation, invertase inverts sucrose. If a question says “name the enzyme that hydrolyses lactose”, the answer is almost always sitting inside the question.

Specificity is the headline property. Enzymes are extraordinarily fussy. Urease hydrolyses urea and nothing else; maltase attacks the α-glycosidic bond of maltose but ignores the β-bond of cellulose. The standard picture is the lock-and-key model: the substrate fits a specially shaped cavity on the enzyme surface called the active site, much as one key fits one lock. Substrate binds to form an enzyme–substrate complex, the strained bonds react, and the products leave, freeing the enzyme to go round again. In modern language the fit is a little flexible (induced fit), but for CBSE the lock-and-key picture is what is wanted.

How they work, in one line: an enzyme lowers the activation energy of the reaction by providing an alternative path through the enzyme–substrate complex — it does not change the position of equilibrium or the enthalpy of reaction, only the speed. Enzymes are also astonishingly efficient, raising rates by factors of 106 to 1020, and they need only tiny amounts. Being proteins, they work best in a narrow band of temperature (around 310 K in humans) and pH (usually 5–7), and outside that band they denature and stop working — which is why a high fever is dangerous.

Diagram: realistic enzyme active-site and induced-fit rendering — to be added by illustrator.

Exam Tip — borrow the language of Chemical Kinetics
Marks are given for the phrase “lowers the activation energy by providing an alternative reaction pathway”. That is the same catalysis idea you met in physical chemistry, so if you can already explain a catalyst on an energy-profile diagram, you can already explain an enzyme. Add one sentence that enzymes differ from ordinary catalysts by being protein in nature, highly specific and active only in a narrow temperature and pH range, and the answer is complete.
Example 14 — Mechanism and specificity (3 marks)
Question: Explain the mechanism of enzyme action. Why is an enzyme said to be highly specific? Name the enzyme that converts sucrose into glucose and fructose.

Model answer:
• Mechanism: the substrate molecule binds to a cavity of complementary shape on the enzyme surface, the active site, forming an enzyme–substrate complex. Within this complex the bonds of the substrate are weakened, the reaction proceeds along a path of lower activation energy, and the products are released, regenerating the free enzyme. (2 marks)
• Specificity: the active site has a definite size, shape and arrangement of functional groups, so only a substrate that fits it exactly (the lock-and-key fit) can bind and react. Hence one enzyme generally catalyses only one reaction. (½ mark)
• The enzyme is invertase (also called sucrase). (½ mark)
Example 15 — A reasoning question (2 marks)
Question: A student boils a solution of the enzyme amylase for five minutes, cools it back to 310 K and then adds starch. No hydrolysis occurs. Explain.

Model answer: Amylase is a globular protein. Boiling supplies enough thermal energy to break the weak hydrogen bonds and ionic interactions that maintain its secondary and tertiary structure, so the enzyme is denatured and its active site loses its specific shape. (1 mark) Because starch can no longer bind to form an enzyme–substrate complex, catalysis cannot occur; and since denaturation is irreversible here, cooling the solution back to 310 K does not restore activity. (1 mark)

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Vitamins: Classification and Deficiency Diseases

Vitamins are organic compounds required in very small amounts in the diet, which the body cannot synthesise (or cannot make in sufficient quantity), and whose absence causes a specific deficiency disease. They are not fuel and they are not building material — think of them as the tiny screws that hold a large machine together. A few grams a year can be the difference between health and blindness.

Two honest exceptions are worth knowing because examiners like them: vitamin D can be made in the skin under sunlight, and vitamin K and some B vitamins are produced by intestinal bacteria. So “the body cannot make them at all” is a slight over-statement — write “cannot synthesise them in adequate amounts” instead.

The classification is based on solubility, and this single idea explains almost every difference between the two groups. Fat-soluble vitamins A, D, E and K dissolve in fats and oils, get stored in the liver and adipose tissue, need not be eaten daily — and, because they accumulate, can actually be toxic in excess. Water-soluble vitamins — the B group and C — are not stored (vitamin B12 being the exception), are excreted in urine, and must therefore be supplied regularly in the diet.

Property Fat-soluble vitamins Water-soluble vitamins
MembersA, D, E, KB-complex (B1, B2, B6, B12 and others) and C
Soluble inFats and oils (non-polar solvents)Water
Stored in body?Yes — in liver and adipose tissueNo (except B12, stored in the liver)
Excreted in urine?NoYes, regularly
Must be eaten daily?Not necessarilyYes, supplied regularly
Risk on overdoseCan accumulate and become toxicExcess is flushed out; toxicity rare

Now the deficiency table. Do not try to understand it — just drill it. Two marks in the paper are almost guaranteed and they are the easiest two marks in the unit.

Vitamin Chemical name Good dietary sources Deficiency disease
A (fat-soluble)RetinolFish liver oil, carrots, butter, milk, green leafy vegetablesXerophthalmia (hardening of the cornea) and night blindness
B1 (water-soluble)ThiamineYeast, milk, unpolished rice, green vegetablesBeri-beri (loss of appetite, weak muscles)
B2 (water-soluble)RiboflavinMilk, egg white, liver, kidneyCheilosis — cracking at the corners of the mouth, burning skin, digestive upset
B6 (water-soluble)PyridoxineYeast, milk, egg yolk, cereals, gramsConvulsions
B12 (water-soluble)CyanocobalaminMeat, fish, egg, curdPernicious anaemia
C (water-soluble)Ascorbic acidCitrus fruits, amla, tomatoes, green leafy vegetablesScurvy — bleeding gums, delayed wound healing
D (fat-soluble)CalciferolSunlight on skin, fish, egg yolk, cod liver oilRickets in children; osteomalacia in adults
E (fat-soluble)TocopherolVegetable oils such as wheat germ oil and sunflower oilIncreased fragility of red blood cells, muscular weakness
K (fat-soluble)PhylloquinoneGreen leafy vegetables; also made by gut bacteriaIncreased blood clotting time, excessive bleeding
Exam Tip — a memory hook that actually works
Fat-soluble = “A, D, E, K” — say it aloud as one word, adek. Everything not in adek is water-soluble. For diseases, pair them by picture: A → eyes, D → bones, K → blood clotting, C → gums, B1 → nerves and muscles, B12 → blood. Six pictures beat nine paragraphs on exam morning.
Example 16 — Name the vitamin (2 marks)
Question: Name the vitamin whose deficiency causes (i) rickets in children (ii) scurvy (iii) night blindness (iv) pernicious anaemia. State in each case whether it is fat-soluble or water-soluble.

Model answer:
(i) Rickets — vitamin D (calciferol), fat-soluble. (½)
(ii) Scurvy — vitamin C (ascorbic acid), water-soluble. (½)
(iii) Night blindness — vitamin A (retinol), fat-soluble. (½)
(iv) Pernicious anaemia — vitamin B12 (cyanocobalamin), water-soluble. (½)
Example 17 — Reason it out (2 marks)
Question: Why must vitamin C be included in the diet almost every day, whereas vitamin A need not be?

Model answer: Vitamin C is water-soluble. It is not stored in the body and any excess is continuously excreted in the urine, so the supply is quickly exhausted and must be replenished regularly through the diet. (1 mark) Vitamin A is fat-soluble. It dissolves in body fat and is stored in the liver and adipose tissue, so a reserve is available for weeks and daily intake is not essential — indeed, a large excess can accumulate and become toxic. (1 mark)

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Nucleic Acids: DNA and RNA Structure

DNA — double stranded RNA — single stranded A T G C T A C G A U G C Key differencesSingle polynucleotidechain, not a helix pairSugar = riboseUracil replaces thymineA – U : 2 H‑bonds Sugar–phosphate backbone (tan) · A–T held by 2 H‑bonds · G–C held by 3 H‑bonds Bases hang off one backbone; base pairing happens only on folding or with DNA
Schematic ladder (not a true double helix). Dashed lines are hydrogen bonds — count them: two for A–T and A–U, three for G–C.

Nucleic acids are the instruction manuals of life, and the chemistry is more orderly than the biology makes it sound. Build one from the bottom up in three steps and you will never be confused again.

Step 1 — the nucleoside. Take a pentose sugar and attach a nitrogen base to its C–1 by a β-glycosidic linkage. Sugar + base = nucleoside. Step 2 — the nucleotide. Now attach a phosphoric acid unit to the C–5 –OH of that sugar. Sugar + base + phosphate = nucleotide. Step 3 — the polymer. Link nucleotides through phosphodiester bridges running from C–3 of one sugar to C–5 of the next, and you have a polynucleotide, which is a nucleic acid.

The bases. There are two structural families. Purines have a fused two-ring system: adenine (A) and guanine (G). Pyrimidines have a single ring: cytosine (C), thymine (T) and uracil (U). DNA uses A, G, C and T. RNA uses A, G, C and U — uracil takes thymine’s place.

The Watson–Crick double helix. DNA is two polynucleotide strands wound round a common axis, running in opposite directions, with the bases turned inwards and paired by hydrogen bonds. The pairing is strict: adenine always pairs with thymine through two hydrogen bonds, and guanine always pairs with cytosine through three hydrogen bonds. Notice the design logic — a big purine always pairs with a small pyrimidine, so the ladder keeps a constant width all the way down. This complementarity is what makes replication possible: unzip the ladder and each half tells you exactly how to rebuild the other. That is, at the chemical level, the same faithful copying of information you studied as inheritance in the Class 10 Science chapter on Heredity.

What each does. DNA stores the genetic information and passes it to daughter cells; RNA carries it out. Three RNAs are worth naming: messenger RNA (m-RNA) carries the message from DNA to the ribosome, ribosomal RNA (r-RNA) forms the ribosome where the protein is assembled, and transfer RNA (t-RNA) ferries individual amino acids into place. The overall flow — DNA makes RNA makes protein — means the proteins section you read earlier and this section are really one story.

Point of comparison DNA RNA
Full nameDeoxyribonucleic acidRibonucleic acid
Pentose sugar2-Deoxy-D-ribose (no –OH at C–2)D-Ribose (–OH present at C–2)
Bases usedA, G, C and T (thymine)A, G, C and U (uracil)
Number of strandsTwo, wound into a double helixUsually one
Base pairingA–T (2 H-bonds), G–C (3 H-bonds)A–U (2 H-bonds), G–C (3 H-bonds), only where the chain folds back
LocationMainly the nucleusMainly the cytoplasm
FunctionStores and transmits hereditary information; self-replicatingCarries out protein synthesis; generally does not replicate itself
StabilityMore stable (no C–2 –OH, and double stranded)Less stable and more reactive
Key Rule — Chargaff’s base-pairing arithmetic
Because every A is opposite a T and every G is opposite a C in double-stranded DNA:
%A = %T and %G = %C, therefore %A + %G = %T + %C = 50 % and (A + T) + (G + C) = 100 %.
Any question that gives you one base percentage and asks for the rest is solved with these two lines and nothing else. Note that this arithmetic does not hold for single-stranded RNA.
Example 18 — Chargaff calculation (3 marks)
Question: A sample of double-stranded DNA contains 22 % adenine. Calculate the percentage of thymine, guanine and cytosine, and state how many hydrogen bonds hold a G–C pair together.

Solution:
Since A pairs only with T, %T = %A = 22 %. (1 mark)
%A + %T = 22 + 22 = 44 %, so %G + %C = 100 – 44 = 56 %. (1 mark)
Since G pairs only with C, %G = %C = 56 ÷ 2 = 28 % each.
A G–C pair is held by three hydrogen bonds (A–T has only two). (1 mark)
Check: 22 + 22 + 28 + 28 = 100 % ✓

Follow-on to try yourself: if the DNA were 30 % adenine, %G would come out as 20 %. Confirm it with the same two lines.
Example 19 — Definitions and differences (3 marks)
Question: (a) Distinguish between a nucleoside and a nucleotide. (b) Give any two structural differences between DNA and RNA. (c) Name the linkage that joins two nucleotides.

Model answer:
(a) A nucleoside is a pentose sugar joined to a nitrogen base at C–1 by a β-glycosidic linkage. A nucleotide is a nucleoside in which the C–5 hydroxyl of the sugar has additionally been esterified with phosphoric acid. In short, nucleotide = nucleoside + phosphate. (1 mark)
(b) (i) DNA contains 2-deoxy-D-ribose whereas RNA contains D-ribose. (ii) DNA contains the base thymine whereas RNA contains uracil in its place. (A third valid answer: DNA is double stranded, RNA is usually single stranded.) (1 mark)
(c) The phosphodiester linkage, running from C–3 of one sugar to C–5 of the next. (1 mark)

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Hormones: Types and Functions

If enzymes are the workers, hormones are the management memos. A hormone is a molecule produced in an endocrine (ductless) gland and released straight into the bloodstream, which carries it to a distant target organ where it triggers a response. They act as intercellular messengers, work at extremely low concentration, and are continuously destroyed and replaced — which is exactly why the body can turn a signal off as well as on.

Chemical classification — three families, and the board asks for these names:

  • Steroid hormones — built on the four-ring steroid skeleton. Examples: testosterone (male sex hormone), estradiol and other estrogens (female sex hormones), cortisone and the glucocorticoids.
  • Polypeptide and protein hormones — chains of amino acids. Examples: insulin, glucagon, oxytocin, vasopressin and growth hormone.
  • Amino-acid-derived hormones — small molecules made by modifying a single amino acid. Examples: adrenaline (epinephrine), noradrenaline and thyroxine.

Functions you should be able to quote. Insulin, from the pancreas, lowers blood glucose by promoting its uptake and storage as glycogen; its deficiency causes diabetes mellitus. Glucagon, also from the pancreas, does the opposite and raises blood glucose — the pair works like an accelerator and a brake, keeping the level roughly steady. Thyroxine, from the thyroid, contains iodine and controls the basal metabolic rate; too little iodine in the diet causes hypothyroidism and goitre, while an excess of thyroxine causes hyperthyroidism. Adrenaline and noradrenaline, from the adrenal medulla, are the stress hormones — they raise heart rate, blood pressure and blood glucose in the classic “fight or flight” response. Testosterone and estradiol control the development of male and female secondary sexual characters respectively.

Common Mistake — mixing up hormones, enzymes and vitamins
All three are needed in tiny amounts, so students blur them. Keep them apart with three questions. Is it made inside the body? Hormones yes, enzymes yes, vitamins no (they come from the diet). Is it a catalyst? Enzymes yes, hormones no — a hormone is a chemical message, not a catalyst. Is it consumed? Hormones are used up and must be replenished; enzymes are regenerated unchanged at the end of every cycle. Never write “insulin catalyses the breakdown of glucose”.
Example 20 — Classify and explain (3 marks)
Question: (a) What are hormones? (b) Classify insulin, testosterone and adrenaline according to their chemical nature. (c) State one biological function of insulin and one of thyroxine.

Model answer:
(a) Hormones are chemical messengers produced by endocrine glands and released directly into the bloodstream, which carries them to target tissues where they regulate a specific physiological process. They are effective at very low concentration and are continuously destroyed and resynthesised. (1 mark)
(b) Insulin — a polypeptide (protein) hormone. Testosterone — a steroid hormone. Adrenaline — an amino-acid-derived hormone. (1 mark)
(c) Insulin lowers the blood glucose level by promoting the uptake of glucose by cells and its storage as glycogen; its deficiency causes diabetes mellitus. Thyroxine controls the basal metabolic rate; its deficiency causes hypothyroidism and goitre. (1 mark)

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Quick Revision Table for Last-Minute Study

On the morning of the paper you will not have time to re-read this page. Read only this table. It is the whole unit compressed into the answers examiners actually award marks for.

Term The one-line answer that scores
CarbohydrateA polyhydroxy aldehyde or ketone, or a compound giving one on hydrolysis
Reducing sugarHas a free anomeric carbon, so it reduces Tollens’ and Fehling’s reagents (all monosaccharides, plus maltose and lactose)
MutarotationGradual change in specific rotation of a freshly prepared sugar solution until equilibrium (+52.5° for glucose) is reached
AnomersCyclic sugars differing in configuration only at the anomeric carbon, e.g. α- and β-D-glucose
Invert sugarThe equimolar glucose + fructose mixture from sucrose hydrolysis; rotation changes from +66.5° to –39.9°
Glycosidic linkageThe oxygen bridge joining two monosaccharide units, formed by loss of a water molecule
ZwitterionThe dipolar form of an amino acid, R–CH(NH3+)–COO–, net charge zero
Isoelectric pointThe pH at which the amino acid exists entirely as the zwitterion and does not migrate in an electric field
Peptide bondThe –CO–NH– (amide) link formed between two amino acids with loss of water
DenaturationLoss of secondary, tertiary and quaternary structure and of biological activity; primary structure survives
EnzymeA globular-protein biocatalyst that lowers activation energy and is highly substrate-specific
VitaminOrganic compound needed in trace amounts in the diet; A, D, E, K are fat-soluble, B group and C are water-soluble
NucleotideBase + pentose sugar + phosphate (a nucleoside is base + sugar only)
HormoneChemical messenger from an endocrine gland, carried in blood to a target organ; not a catalyst

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Practice Worksheet with Answers

Close the notes now. Attempt all ten questions on paper first, in full sentences, with the marks in mind — then open the answers. Marking your own work honestly is worth more than reading the chapter a second time. These are original questions written in the style of biomolecules class 12 chemistry important questions, covering the whole 7-mark unit.

Q1 (1 mark) — Why is sucrose described as a non-reducing sugar? Show Answer
In sucrose the glycosidic linkage is formed between C–1 of α-glucose and C–2 of β-fructose, which are the anomeric carbons of both units. With no free anomeric carbon, neither ring can open to give a free aldehyde or ketone group, so sucrose cannot reduce Tollens’ or Fehling’s reagent.
Q2 (2 marks) — Fructose is a ketose, yet it gives a positive Tollens’ test. Explain. Show Answer
Tollens’ reagent is alkaline. In basic medium fructose undergoes base-catalysed isomerisation through a common enediol intermediate and is converted into the aldoses glucose and mannose. (1 mark) These aldoses possess a free –CHO group, which reduces Ag+ to metallic silver, giving the silver mirror. So the positive test is due to the isomerised aldose, not to the ketone group itself. (1 mark)
Q3 (2 marks) — A pentapeptide is formed from five molecules of alanine (M = 89.09 g mol–1). How many peptide bonds are present, and what is the molar mass of the pentapeptide? (H2O = 18.02) Show Answer
Number of peptide bonds = n – 1 = 5 – 1 = 4, and 4 molecules of water are eliminated. (1 mark)
M = 5(89.09) – 4(18.02) = 445.45 – 72.08 = 373.37 g mol–1 (1 mark)
Q4 (2 marks) — A double-stranded DNA contains 18 % guanine. Find the percentages of cytosine, adenine and thymine. Show Answer
G pairs only with C, so %C = %G = 18 %. (1 mark)
%G + %C = 36 %, therefore %A + %T = 100 – 36 = 64 %, and since A pairs only with T, %A = %T = 32 %. (1 mark)
Check: 18 + 18 + 32 + 32 = 100 % ✓
Q5 (3 marks) — Give reasons: (i) amino acids have unusually high melting points; (ii) glycine is optically inactive; (iii) vitamin C must be taken daily. Show Answer
(i) In the solid state an amino acid exists as a zwitterion, R–CH(NH3+)–COO–, so the crystal is held by strong electrostatic (ionic) forces like a salt, and a great deal of energy is needed to melt it. (1 mark)
(ii) In glycine the α-carbon carries –NH2, –COOH and two identical hydrogen atoms, so it is not a chiral centre and glycine has no enantiomers. (1 mark)
(iii) Vitamin C is water-soluble, is not stored in the body and any excess is excreted in the urine, so the supply must be replenished regularly through the diet. (1 mark)
Q6 (3 marks) — 68.46 g of sucrose is completely hydrolysed. Calculate the moles of sucrose taken and the mass of glucose formed. (M: sucrose 342.30, glucose 180.16 g mol–1) Show Answer
C12H22O11 + H2O → C6H12O6 + C6H12O6 (1 mark)
Moles of sucrose = 68.46 ÷ 342.30 = 0.200 mol (1 mark)
1 mol sucrose gives 1 mol glucose, so moles of glucose = 0.200 mol and mass = 0.200 × 180.16 = 36.03 g. (An equal 36.03 g of fructose is also formed.) (1 mark)
Q7 (2 marks) — How many chiral carbons are there in the open-chain form of an aldopentose such as ribose, and how many stereoisomers are therefore possible? Show Answer
The open chain is OHC–CHOH–CHOH–CHOH–CH2OH. C–1 is the aldehyde and C–5 is –CH2OH, so the chiral carbons are C–2, C–3 and C–4, i.e. 3 chiral carbons. (1 mark)
Number of stereoisomers = 2n = 23 = 8. (1 mark)
Q8 (3 marks) — Amylopectin and glycogen are both branched α-glucose polymers. Give two differences and explain why the extra branching in glycogen is biologically useful. Show Answer
• Source: amylopectin is the branched fraction of plant starch (about 80–85 % of it), whereas glycogen is the storage polysaccharide of animals, found in liver and muscle. (1 mark)
• Extent of branching: amylopectin has an α-1,6 branch roughly every 20–25 glucose units, while glycogen is branched considerably more frequently; it also gives a red-brown colour with iodine rather than blue-black. (1 mark)
• Why it helps: more branch points means many more free non-reducing chain ends. Enzymes remove glucose units from the ends, so a highly branched molecule can be broken down at many places at once, releasing glucose into the blood very rapidly when energy is suddenly required. (1 mark)
Q9 (5 marks) — (a) What is denaturation? (b) Which structural levels are lost? (c) Why does an enzyme stop working when denatured? (d) Give two causes and (e) one everyday example. Show Answer
(a) Denaturation is the process in which a protein loses its native three-dimensional conformation, and with it its biological activity, on exposure to physical or chemical stress. (1 mark)
(b) The secondary, tertiary and quaternary structures are destroyed because the weak hydrogen bonds, ionic salt bridges, van der Waals and hydrophobic interactions holding them are disrupted; the primary structure remains intact since covalent peptide bonds are not broken. (1 mark)
(c) Enzymes are globular proteins whose catalytic power depends on the precise shape of the active site. Once the protein unfolds, the active site loses its complementary shape, the substrate can no longer bind to form the enzyme–substrate complex, and catalysis ceases. (1 mark)
(d) Any two of: heating above the optimum temperature; strong acid or alkali (change of pH); heavy-metal ions such as Pb2+ or Hg2+; alcohol; urea; ultraviolet radiation. (1 mark)
(e) Boiling an egg — soluble albumin in the egg white coagulates into an insoluble white solid; or the curdling of milk by lactic acid. (1 mark)
Q10 (5 marks) — (a) Distinguish DNA from RNA on any three counts. (b) State the base-pairing rule with hydrogen-bond counts. (c) Name the three types of RNA and the role of each. Show Answer
(a) Sugar: DNA contains 2-deoxy-D-ribose, RNA contains D-ribose. Base: DNA contains thymine, RNA contains uracil in its place. Strands: DNA is double stranded and helical, RNA is usually single stranded. (A fourth valid point: DNA is found mainly in the nucleus and is self-replicating, RNA is found mainly in the cytoplasm and is not.) (2 marks)
(b) Adenine pairs with thymine through two hydrogen bonds, and guanine pairs with cytosine through three hydrogen bonds. A purine always pairs with a pyrimidine, which keeps the width of the helix constant. In RNA, adenine pairs with uracil, again through two hydrogen bonds. (1 mark)
(c) m-RNA (messenger) carries the coded instruction from the DNA in the nucleus to the ribosome. r-RNA (ribosomal) is a structural and functional part of the ribosome, the site where protein is assembled. t-RNA (transfer) brings individual amino acids to the ribosome and places them in the order specified by the m-RNA. (2 marks)

Score yourself out of 28. Anything above 20 on a first honest attempt means this unit is already safe for you.

Exam Tip — the practical link is free marks too
The Class 12 practical syllabus includes characteristic tests of carbohydrates and proteins in foodstuffs. Learn the observation, not just the name: Molisch gives a violet ring (any carbohydrate); Fehling’s gives a red-brown precipitate of Cu2O and Tollens’ a silver mirror (reducing sugars); iodine gives blue-black (starch); biuret gives a violet colour and ninhydrin a blue-violet colour (proteins); xanthoproteic gives a yellow stain with nitric acid (aromatic amino acids). The same tests turn up in theory papers as one-markers.

One last thought before you close this page. Biomolecules is not a chapter you conquer in a single heroic night; it is a chapter you win by small, boring, daily improvement. Learn one more deficiency disease than you knew yesterday. Write one more model answer than you wrote yesterday. Aim for one more correct answer than yesterday — do that for two weeks and the 7 marks stop being a worry and start being a certainty. That is kaizen, and it works.

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