Chapter 18 of 18

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JEE Main Chemistry★ ~1 question (4 marks) in JEE Main, almost always a single-fact recall question on reducing sugars, protein structure, vitamin classification or DNA composition rather than a multi-step problem

Biomolecules

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JEE Main PYQsSolved previous-year questions

By the end of this chapter you'll be able to…

1Classify carbohydrates and decide reducing character by counting free anomeric carbons
2State which reaction proves which feature of glucose's open chain, and explain mutarotation and the cyclic hemiacetal
3Explain why glucose passes irreversible aldehyde tests and fails reversible ones, and why fructose reduces Fehling's through an enediol
4Give the units, linkages and reducing character of sucrose, maltose and lactose, and compute the rotation of invert sugar
5Explain amino acid properties from the zwitterion, use the isoelectric point to predict migration, and name the four levels of protein structure
6Distinguish denaturation from hydrolysis, relate enzyme specificity to tertiary structure, and apply Chargaff's rule to DNA composition and stability

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Why this chapter matters in JEE Main

Biomolecules are ordinary organic molecules doing ordinary organic chemistry. The only new thing is that they are big enough for their shape to matter, and almost every fact in the chapter follows from one of those two statements. Glucose's ring is a hemiacetal, the same reaction an aldehyde does with any alcohol; a peptide bond is an amide; DNA base pairing is hydrogen bonding; denaturation destroys weak interactions while the covalent skeleton survives untouched. The sharpest single result is that glucose passes three aldehyde tests and fails two, because only 0.02 per cent of it is open-chain at any moment: irreversible tests drain that trace and the ring keeps replacing it, while reversible tests need a standing concentration and find none.

Before you start — revise these

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Aldehyde and ketone chemistry, especially hemiacetal formation, from Organic Compounds Containing Oxygen

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Amide formation and the delocalisation of a nitrogen lone pair from Amines

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Hydrogen bonding and its effect on solubility and melting point from Chemical Bonding

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Optical activity and stereocentres from Basic Principles of Organic Chemistry

Biomolecules

Glucose is an aldohexose. Six carbons, a $CHO$ group at one end. Test it.

Test	Glucose
Tollens' reagent	Silver mirror
Fehling's solution	Red precipitate
Hydroxylamine	Forms an oxime
Schiff's reagent	Nothing
Sodium bisulphite	Nothing

Three aldehyde tests pass and two fail. What kind of aldehyde does that?

One that is 99.98 per cent not an aldehyde.

Dissolve pure $α$ -D-glucose, rotation $+ 11 1^{\circ}$ , and watch it fall. Dissolve pure $β$ -D-glucose, rotation $+ 19. 7^{\circ}$ , and watch it rise. Both settle at $+ 52. 7^{\circ}$ and stop. That is mutarotation, and it is direct evidence that the two forms interconvert through something in between.

The something is the open chain, and it is present at about 0.02 per cent. The other 99.98 per cent is a six-membered cyclic hemiacetal, formed when the C5 hydroxyl attacks the C1 aldehyde.

That is not a special biological reaction. It is the ordinary aldehyde-plus-alcohol addition from the previous chapter, made easy because both groups sit on the same molecule.

Which is the whole point of this chapter.

The two statements	What they cover
A biomolecule is an ordinary organic molecule with ordinary functional groups	Glucose's ring is a hemiacetal, a peptide bond is an amide, base pairing is hydrogen bonding
It is large enough for its shape to matter	Folding, denaturation, enzyme specificity, the double helix

Every fact here is one of those two things. Read the chapter through them and it stops being a list.

1. What Makes a Molecule a Biomolecule

Living cells build almost everything from four families: carbohydrates, proteins, nucleic acids and lipids. The JEE Main syllabus examines the first three, plus vitamins and a general introduction to hormones.

Three of the four families are polymers built by the same manoeuvre: two monomers join, and a molecule of water is expelled. Sugars join through a glycosidic linkage, amino acids through a peptide bond, nucleotides through a phosphodiester bridge.

Every one of those is a condensation, and every one is reversed by hydrolysis. That single fact tells you what happens to any of these molecules in acid, in base, or in the digestive tract.

The monomers decide the chemistry. The folding decides the function. A protein and a badly boiled protein have the same sequence and the same molar mass; only one of them works.

2. Carbohydrates: Classification

Carbohydrates were once written as hydrates of carbon, C $_{x}$ (H $_{2}$ O) $_{y}$ . The name survived even though the formula misleads, since the water is not present as water.

A carbohydrate is more usefully defined as a polyhydroxy aldehyde or ketone, or something that gives one on hydrolysis. That definition is the useful one because it names the functional groups you will actually reason with.

Basis	Categories
Carbonyl type	Aldose (aldehyde, e.g. glucose) or ketose (ketone, e.g. fructose)
Carbon count	Triose, tetrose, pentose, hexose
Hydrolysis	Monosaccharide (none), oligosaccharide (2 to 10 units), polysaccharide (many)
Tollens' or Fehling's	Reducing or non-reducing

The last row is the one JEE asks about most, and it has a single structural cause. A sugar reduces Tollens' reagent only if it has a free anomeric carbon somewhere, one that can open to a carbonyl.

Lock every anomeric carbon into a linkage and the sugar cannot open, so it cannot reduce anything. That is the whole of the reducing and non-reducing distinction.

3. Glucose: The Evidence for Its Open-Chain Structure

Glucose, C $_{6}$ H $_{12}$ O $_{6}$ , is the reference monosaccharide. Its structure was not assumed; it was deduced, and JEE asks which reaction proves which feature.

Observation	What it establishes
Prolonged heating with HI gives n-hexane	Six carbons in an unbranched chain
Forms an oxime with hydroxylamine and a cyanohydrin with HCN	A carbonyl group is present
Bromine water oxidises it to gluconic acid (six carbons)	The carbonyl is an aldehyde, not a ketone
Acetic anhydride gives a penta-acetate	Five hydroxyl groups
Nitric acid gives saccharic acid, a dicarboxylic acid	A primary alcohol sits at the far end from the CHO

Bromine water is the discriminating test. It oxidises aldehydes but not ketones, so it separates aldose from ketose.

Nitric acid is stronger and oxidises both ends. Getting a diacid with the same carbon count means the other terminal carbon must have been a CH $_{2}$ OH group.

Put those together and you get a straight six-carbon chain with CHO at one end, CH $_{2}$ OH at the other, and hydroxyl groups on the four carbons between.

4. The Cyclic Structure: Anomers and Mutarotation

The open chain then fails three tests, and the failures are the most examinable part of the topic.

Glucose does not give the Schiff's test. It does not form the hydrogensulphite addition product with sodium bisulphite. And its penta-acetate does not react with hydroxylamine at all.

A genuine free aldehyde would pass all three. Something is hiding the carbonyl.

The answer is that the C5 hydroxyl attacks the C1 aldehyde intramolecularly to give a six-membered cyclic hemiacetal. This is not a special biological reaction; it is the ordinary aldehyde-plus-alcohol addition, made easy because both groups are on the same molecule.

The six-membered form is called pyranose, after pyran. Its Haworth projection draws the ring flat with substituents above and below.

Anomers

Ring closure creates a new stereocentre at C1, called the anomeric carbon. The two products are diastereomers called anomers, and they are separable substances.

Anomer	Melting point	Specific rotation
$α$ -D-glucose	419 K	$+ 11 1^{\circ}$
$β$ -D-glucose	423 K	$+ 19. 7^{\circ}$

Dissolve either pure anomer in water and its rotation drifts, settling at $+ 52. 7^{\circ}$ from both directions. This slow change is mutarotation.

It happens because the ring opens and recloses through the trace open-chain form, so the two anomers interconvert until they reach their equilibrium mixture. The final value is fixed, which is what tells you it is an equilibrium and not a decomposition.

Resolving the apparent contradiction

If glucose is almost entirely cyclic, why does it still form an oxime and a cyanohydrin?

Because only about 0.02% exists as the open chain at any moment, but that fraction is continuously replenished. Oxime and cyanohydrin formation drain it irreversibly, and the ring keeps opening to restore the balance, so the reaction runs to completion.

Schiff's test and bisulphite addition are readily reversible and need a real standing concentration of free aldehyde. There is none, so they fail. The penta-acetate settles the matter: with C1 capped as an ester, the ring cannot open at all, and hydroxylamine finds nothing to react with.

Illustration 1

Sharpen that into a rule. Predict, for any glucose test, whether it will succeed on 0.02 per cent open chain.

Ask one question: does the test drain the aldehyde irreversibly, or does it need a standing concentration of it?

Test	Nature	On 0.02 per cent	Result
Oxime with $N H_{2} OH$	Irreversible, product is stable	Drains it; ring reopens to replace it	Passes
Tollens', Fehling's	Irreversible oxidation	Same	Passes
Schiff's reagent	Readily reversible	Needs real concentration	Fails
Bisulphite addition	Readily reversible	Needs real concentration	Fails
Any test on the penta-acetate	C1 capped, ring cannot open	Nothing to drain	Fails

A trace concentration is not a small amount of reagent. It is a tap that never runs dry, because every molecule consumed is replaced by another ring opening.

So an irreversible test eventually converts all the glucose, however little is open at any instant, and only the rate is affected. A reversible test reaches its own equilibrium against a concentration of $2 \times 1 0^{- 4}$ , which is far too small to give a visible result.

This is Le Chatelier applied to an analytical question, and it explains all five rows without a single new fact.

5. Fructose and the Disaccharides

Fructose is a ketohexose, with the carbonyl at C2. Its C5 hydroxyl reaches C2 to give a five-membered furanose ring, not a six-membered one.

Yet fructose reduces Tollens' and Fehling's reagents, which are meant to be aldehyde tests. The usual textbook line is that fructose is a reducing sugar and the matter is left there.

The honest statement is that it is not fructose that reduces the reagent. Both reagents are alkaline, and in base a ketose isomerises to the corresponding aldoses through an enediol intermediate. Glucose and mannose form, and those reduce the reagent.

So "fructose is a reducing sugar" is a statement about the conditions of the test, not about fructose's own carbonyl. Any $α$ -hydroxy ketone behaves the same way.

Illustration 2

Trace what actually happens in the Fehling's tube, and say what it predicts about propanone.

Fehling's and Tollens' are both alkaline. In base, the hydrogen on the carbon next to fructose's carbonyl is removed, and the resulting carbanion is stabilised by the neighbouring hydroxyl as an enediol, a species with an $OH$ on each carbon of a double bond.

$ketose ⇌ enediol ⇌ aldose$

The enediol can collapse either way. Collapsing towards C2 gives back fructose; collapsing towards C1 gives glucose and mannose, which are genuine aldoses and reduce the reagent at once.

Two predictions follow, and both are correct.

Propanone, an ordinary ketone with no adjacent hydroxyl, gives nothing with Fehling's, because it cannot form an enediol. The $α$ -hydroxyl is doing the work, not the ketone.

And the test destroys the sugar it is testing. Anything that isomerises a ketose to an aldose is not a mild reagent, which is why bromine water, run in neutral solution, remains the honest way to separate an aldose from a ketose.

The three named disaccharides

The syllabus names sucrose, maltose and lactose, and asks which monosaccharides each is built from.

Disaccharide	Units	Linkage	Reducing?
Sucrose	$α$ -D-glucose + $β$ -D-fructose	C1 to C2	No
Maltose	two $α$ -D-glucose	$α$ (1 to 4)	Yes
Lactose	$β$ -D-galactose + $β$ -D-glucose	$β$ (1 to 4)	Yes

Sucrose is the only non-reducing one, and the reason is structural rather than arbitrary. Its linkage joins the anomeric carbon of glucose to the anomeric carbon of fructose, so both are tied up and neither ring can open.

Maltose and lactose use the anomeric carbon of one unit and an ordinary C4 hydroxyl of the other. One anomeric carbon stays free, so that end opens, and the sugar reduces.

Illustration 3

Trehalose is a disaccharide that hydrolyses to give two molecules of D-glucose and nothing else. It is non-reducing. Deduce its linkage.

Two facts, and between them they leave one possibility.

Hydrolysis to two glucoses fixes the units. Non-reducing fixes the linkage, because a sugar reduces Tollens' reagent if and only if it has at least one free anomeric carbon.

Each glucose has exactly one anomeric carbon, at C1. A disaccharide has one glycosidic bond, which can tie up at most two positions. To leave zero free anomeric carbons, that single bond must use both of them.

$linkage is C1 to C1$

Compare the three named sugars and the rule does all the work.

Sugar	Bond uses	Free anomeric carbons	Reducing?
Sucrose	C1 and C2, both anomeric	0	No
Trehalose	C1 and C1, both anomeric	0	No
Maltose, lactose	one anomeric plus an ordinary C4	1	Yes

Note that the rule needed no memorising of which sugars are reducing. Count free anomeric carbons, and non-reducing means zero. Any polysaccharide is effectively non-reducing for the same reason: one free end in a chain of thousands is undetectable.

Invert sugar

Sucrose rotates plane-polarised light by $+ 66. 5^{\circ}$ . Hydrolyse it and the rotation turns negative, which is why the product is called invert sugar.

The arithmetic is worth doing. Hydrolysis gives equal moles of glucose at $+ 52. 5^{\circ}$ and fructose at $- 92. 4^{\circ}$ , and the mean of those is $- 19. 9^{\circ}$ . Fructose's large negative rotation simply outweighs glucose's positive one.

Illustration 4

Work it as a calculation, then say what changes about the sugar's reducing behaviour.

$[α]_{mixture} = \frac{( + 52.5 ) + ( - 92.4 )}{2} = \frac{- 39.9}{2} = - 19. 9^{\circ}$

The rotation has moved from $+ 66. 5^{\circ}$ to $- 19. 9^{\circ}$ , a swing of 86 degrees and a change of sign, which is why the product is called invert sugar and the enzyme that does it is called invertase.

The reducing behaviour changes just as sharply. Sucrose is non-reducing, because its glycosidic link ties up the anomeric carbon of glucose and the anomeric carbon of fructose, so neither ring can open. Hydrolysis breaks that one bond and frees both.

$one non-reducing disaccharide ⟶ two reducing monosaccharides$

Honey is mostly invert sugar, which is why it is sweeter than sucrose: fructose is the sweetest of the common sugars and hydrolysis releases it.

Note what the two observations have in common. Optical rotation and reducing power are both reporting on the same single bond, from opposite directions, and either one detects the hydrolysis.

Illustration 5

A disaccharide gives no precipitate with Tollens' reagent. On acid hydrolysis it yields two different hexoses, and the resulting mixture reduces Tollens' reagent immediately. Identify the disaccharide and explain both observations.

The disaccharide is sucrose.

The negative Tollens' test before hydrolysis means no free anomeric carbon exists anywhere in the molecule. In sucrose the glycosidic linkage joins C1 of glucose to C2 of fructose, and both of those are anomeric carbons, so neither ring can open.

Hydrolysis breaks that linkage and releases free glucose and free fructose. Glucose now has an anomeric carbon that can open to an aldehyde, so it reduces the reagent directly.

Fructose reduces it too, though indirectly: Tollens' reagent is alkaline, and in base fructose isomerises through an enediol to glucose and mannose, which are the species actually oxidised.

Illustration 6

A sucrose solution rotates plane-polarised light by $+ 66. 5^{\circ}$ . After hydrolysis the rotation is negative. Account for this, given that glucose has a specific rotation of $+ 52. 5^{\circ}$ and fructose $- 92. 4^{\circ}$ .

Hydrolysis of sucrose gives glucose and fructose in equal moles.

The mixture's specific rotation is the mean of the two, since the amounts are equal:

$(+ 52.5 - 92.4) /2 = - 19.95$ , so about $- 19. 9^{\circ}$ .

The sign has inverted from positive to negative, which is where the name invert sugar comes from. Note what has not happened: nothing has been inverted at any stereocentre. The change is purely arithmetic, because fructose's large negative rotation outweighs glucose's smaller positive one.

6. Amino Acids and the Peptide Bond

An $α$ -amino acid carries an amino group on the carbon next to the carboxyl group. About twenty occur in proteins, and ten of those are essential, meaning the body cannot make them and diet must supply them.

Amino acids melt very high, are appreciably water-soluble and are poor solutes in organic solvents. That is not how a small covalent molecule behaves; it is how a salt behaves.

The cause is that the carboxyl group protonates the amine internally, giving a doubly charged zwitterion with no net charge. An amino acid in the solid state is an internal salt, and its properties follow from that.

Because it carries both an acidic and a basic centre, an amino acid is amphoteric. At a particular pH the two charges balance exactly, the molecule has no net charge and will not migrate in an electric field. That pH is the isoelectric point.

Illustration 7

Three amino acids are placed at the centre of an electrophoresis gel buffered at pH 6.0: aspartic acid (pI 2.8), glycine (pI 6.0) and lysine (pI 9.7). Predict where each goes.

Compare each pI with the buffer pH, and one comparison settles each case.

Above its pI, an amino acid has lost more protons than it has gained and carries a net negative charge, so it moves to the anode.

Below its pI, it carries a net positive charge and moves to the cathode.

Amino acid	pI	Buffer at 6.0 is	Net charge	Migrates to
Aspartic acid	2.8	above its pI	Negative	Anode
Glycine	6.0	at its pI	Zero	Nowhere
Lysine	9.7	below its pI	Positive	Cathode

Three compounds separate into three places from one experiment, and the reasoning is the ordinary acid-base argument from the Equilibrium chapter applied to a molecule with two ionisable groups.

The pI values themselves are not arbitrary. Glycine's sits near neutrality because its only ionisable groups are the standard $COOH$ and $N H_{2}$ . Aspartic acid carries a second carboxyl, so more acid is needed to suppress the extra negative charge and its pI falls. Lysine carries a second amino group, so its pI rises. Count the extra acidic and basic groups on the side chain and the direction follows.

Two amino acids condense with loss of water to form a peptide bond, which is just an amide, $- CO - NH -$ . Chains are named as di-, tri- and polypeptides; beyond about a hundred residues, or a molar mass over 10000, we call it a protein.

Illustration 8

How many peptide bonds are present in a decapeptide? How many different tripeptides can be assembled from a pool of all twenty protein amino acids?

A decapeptide has ten residues joined in a line, so the number of links is one fewer.

$Peptide bonds = n - 1$

That gives nine peptide bonds.

For the tripeptides, each of the three positions can independently be any of the twenty amino acids, and order matters because a peptide has a direction.

$N = a^{n}$

With $a = 20$ and $n = 3$ , this gives $N = 8000$ distinct tripeptides.

The rapid growth is the point. Three positions already give eight thousand possibilities, which is why sequence carries so much information.

7. Proteins: Four Levels, and What Denaturation Destroys

The peptide bond is planar, and rotation about the C to N bond is restricted because the nitrogen lone pair delocalises into the carbonyl. That restriction is exactly why proteins fold into regular patterns rather than flopping at random.

Level	What it describes	Held together by
Primary	The sequence of amino acids	Covalent peptide bonds
Secondary	$α$ -helix and $β$ -pleated sheet	Hydrogen bonds between backbone C=O and N-H
Tertiary	The overall three-dimensional fold	H-bonds, disulphide bridges, ionic and van der Waals forces
Quaternary	Assembly of separate subunits	The same weak forces, between chains

The $α$ -helix is a right-handed coil held by hydrogen bonds within one chain. The $β$ -pleated sheet lies stretched with hydrogen bonds running between neighbouring stretches of chain.

Tertiary structure sorts proteins into two shapes with two functions. Fibrous proteins are long and insoluble and do structural work, like keratin and myosin. Globular proteins are folded into compact balls, are water-soluble and do chemical work, like insulin and haemoglobin.

Haemoglobin is the standard quaternary example, being four separate globular subunits held together in one working assembly.

Denaturation

Heat, strong acid, heavy metal ions or high salt will denature a protein: it loses its shape, loses its solubility and loses its biological activity. Coagulating egg white and curdling milk are the everyday cases.

The examinable point is what survives. Denaturation destroys secondary, tertiary and quaternary structure, all of which rest on weak interactions. It leaves the primary structure intact, because peptide bonds are covalent and heat alone will not break them.

So a denatured protein has the same sequence and the same molar mass as the working one. It simply no longer has the shape that made it work, and that is the clearest illustration in the chapter that shape is a chemical property.

Illustration 9

Boiling an egg and digesting an egg both destroy its biological activity. Distinguish the two by a single measurement.

Measure the molar mass.

	Boiling	Digestion
What is attacked	Hydrogen bonds, ionic and van der Waals forces, disulphide bridges	The peptide bonds themselves
Bond type broken	Weak, non-covalent	Covalent
Primary structure	Intact	Destroyed
Molar mass afterwards	Unchanged	Falls to that of amino acids
Reversible?	Sometimes, for mild denaturation	Never

Denaturation unfolds a chain that is still one chain. Hydrolysis cuts it into pieces.

That is why a denatured protein still runs as a single band of the original size on an appropriate analysis, while a digested one runs as a smear of fragments, and it is why the two words are not interchangeable even though both destroy function.

The everyday version is that egg white, once coagulated, cannot be uncoagulated by cooling, yet it remains nutritionally identical: your digestive enzymes get the same amino acids either way. Cooking changes the shape; digestion changes the molecules.

8. Enzymes

Enzymes are biological catalysts, and almost all of them are globular proteins. They obey ordinary catalysis rules: they lower the activation energy, leave the equilibrium constant untouched, and are recovered unchanged.

What sets them apart is scale and selectivity. Rate enhancements run to many powers of ten, and each enzyme typically handles one substrate and no other.

That specificity comes straight from tertiary structure. The fold creates an active site with a definite shape and a definite arrangement of polar and non-polar groups, and only a matching substrate can bind.

It also explains why enzymes are so fragile. Denature the protein and the active site is gone, which is why enzymes lose activity on heating or outside a narrow pH range.

Naming is usually the substrate plus -ase: maltase hydrolyses maltose, urease hydrolyses urea, lactase hydrolyses lactose.

9. Vitamins

Vitamins are organic compounds needed in small amounts that the body cannot synthesise in sufficient quantity, so diet must supply them. They are classified by solubility, and that single property predicts almost everything else.

Class	Members	Behaviour
Fat-soluble	A, D, E, K	Hydrocarbon-rich, stored in liver and adipose tissue, can accumulate to toxic levels
Water-soluble	B group, C	Many polar groups, excreted in urine, must be supplied regularly

Vitamin B $_{12}$ is the standard exception, being water-soluble yet stored in the liver.

The reasoning is ordinary solubility chemistry. A vitamin whose structure is mostly hydrocarbon dissolves in fat, so the body can store it; one carrying many hydroxyl and amino groups dissolves in water and is flushed out, so a daily supply is needed.

Vitamin	Deficiency disease
A	Night blindness, xerophthalmia
B $_{1}$	Beri-beri
B $_{2}$	Cheilosis
B $_{6}$	Convulsions
B $_{12}$	Pernicious anaemia
C	Scurvy
D	Rickets in children, osteomalacia in adults
E	Increased fragility of red blood cells
K	Increased blood clotting time

10. Nucleic Acids

Nucleic acids are polymers built from three parts: a nitrogen base, a pentose sugar and a phosphate group. Assembling them in order is the standard question.

A base joined to the C1 of the sugar gives a nucleoside, through a $β$ -N-glycosidic bond. Add a phosphate ester at the C5 hydroxyl and it becomes a nucleotide. Nucleotides then link through phosphodiester bridges from the C5 phosphate of one to the C3 hydroxyl of the next.

Feature	DNA	RNA
Sugar	2-deoxy-D-ribose	D-ribose
Purines	Adenine, guanine	Adenine, guanine
Pyrimidines	Cytosine, thymine	Cytosine, uracil
Strands	Double helix	Usually single

The two differences are one oxygen atom and one methyl group, and both matter. DNA's sugar lacks the C2 hydroxyl, which makes its backbone markedly more resistant to hydrolysis, and that is appropriate for a molecule that must store information for the life of the organism.

Base pairing

The two strands of DNA run antiparallel and are held together by hydrogen bonds between bases, always purine to pyrimidine.

Adenine pairs with thymine through two hydrogen bonds. Guanine pairs with cytosine through three.

Because pairing is strictly one-to-one, the amounts of A and T are equal in any double-stranded sample, and so are G and C.

$\frac{A}{T} = \frac{G}{C} = 1$

A consequence worth stating: DNA rich in G and C has more hydrogen bonds per base pair, so it takes more heat to separate the strands. Composition sets stability.

Illustration 10

Two bacterial DNA samples are heated until the strands separate. Sample X is 60 per cent G plus C, sample Y is 30 per cent. Predict which melts higher, and give the base composition of each.

Chargaff's rule fixes the composition from one number, because pairing is strictly one to one.

For sample X, G plus C is 60 per cent, and G equals C, so each is 30 per cent. The remaining 40 per cent is A plus T, so each is 20 per cent.

For sample Y, G and C are 15 per cent each and A and T are 35 per cent each.

Now count hydrogen bonds per hundred base pairs.

$X : 60 (3) + 40 (2) = 180 + 80 = 260 Y : 30 (3) + 70 (2) = 90 + 140 = 230$

Sample X has about 13 per cent more hydrogen bonding holding its strands together, so it melts at the higher temperature.

The effect is large enough to be a working measurement: melting temperature is a standard way of estimating GC content, and organisms living in hot springs have DNA noticeably richer in G and C than those living in cold water. A chemical composition is under selection because a hydrogen bond count is.

RNA comes in three working forms. Messenger RNA carries the sequence out of the nucleus, transfer RNA brings the correct amino acid, and ribosomal RNA forms part of the machine that joins them.

The two biological functions follow. DNA replicates itself, each strand acting as the template for its partner, and DNA directs protein synthesis through RNA.

Illustration 11

A sample of double-stranded DNA contains 22% adenine by base count. Find the percentage of each of the other three bases.

Adenine pairs only with thymine, so their amounts are equal.

Thymine is therefore also 22%, and together A and T account for 44%.

The remaining 56% must be guanine and cytosine, and those two are equal to each other as well.

Each is $56/2 = 28%$ , so guanine is 28% and cytosine is 28%.

A check on the reasoning: this DNA is comparatively rich in G and C, so its base pairs average nearly three hydrogen bonds rather than two, and it will need a higher temperature to separate the strands than an A-T-rich sample would.

11. Hormones

Hormones are chemical messengers secreted by endocrine glands directly into the bloodstream, which carries them to a distant target tissue. They act at very low concentration and are steadily destroyed, so a continuous supply is needed.

They fall into three chemical classes: steroids such as testosterone and estradiol, polypeptides such as insulin and glucagon, and amino acid derivatives such as adrenaline and thyroxine.

Insulin and glucagon act in opposite directions on blood glucose, insulin lowering it and glucagon raising it. Insufficient insulin gives diabetes mellitus. Thyroxine contains iodine, which is why dietary iodine deficiency produces thyroid disease.

Two distinctions are worth keeping straight. Hormones are made in the body, whereas vitamins must be taken in through diet. And enzymes catalyse reactions where they are produced, whereas hormones carry a signal to somewhere else.

A Note on Syllabus Emphasis

The official unit text names carbohydrate classification, glucose and fructose, and the constituent monosaccharides of sucrose, lactose and maltose. It does not separately name the polysaccharides.

Starch and cellulose are still worth a line, because they are the standard contrast. Both are glucose polymers; starch uses $α$ -linkages and is digestible, cellulose uses $β$ -linkages and is not. Human enzymes will hydrolyse one geometry and not the other, which is enzyme specificity in a single example.

Illustration 12

Cellulose and starch are both polymers of D-glucose and nothing else. Wood and bread therefore have the same empirical formula. Explain why one is food and the other is not, and why cattle manage both.

The difference is the geometry at one carbon.

	Starch	Cellulose
Linkage	$α (1 \to 4)$	$β (1 \to 4)$
Chain shape	Coils into a helix	Lies flat and straight
Chains pack?	Loosely	Tightly, hydrogen bonded into fibres
Human enzymes	Amylase hydrolyses it	Nothing hydrolyses it

An enzyme's active site is a shape, created by the tertiary fold of a protein. Human amylase is shaped to grip an $α$ -linkage, and a $β$ -linkage presents the same atoms in the wrong orientation, so it never binds and never reacts.

The failure is not thermodynamic. Cellulose hydrolysis is perfectly favourable, and hot acid does it easily. The barrier is entirely kinetic, and the only thing missing is a catalyst of the right shape.

Cattle do not make one either. They carry gut microorganisms that produce cellulase, and the animal digests the products. Termites do the same.

So one bond geometry decides which organisms can eat which polymer, and it does so purely by whether a protein happens to fold around it. That is enzyme specificity stated as concisely as the syllabus ever states it, and it is the clearest case in the chapter of shape being a chemical property.

Spend your preparation where the unit text points: reducing versus non-reducing sugars, the cyclic structure of glucose, protein structure levels and denaturation, and the DNA-versus-RNA comparison.

Summary

A biomolecule is an ordinary organic molecule with ordinary functional groups, big enough that its shape carries chemical meaning.

Carbohydrates are polyhydroxy aldehydes and ketones. A sugar reduces Tollens' reagent if and only if it retains a free anomeric carbon that can open to a carbonyl.

Glucose's open-chain structure is proved by HI, hydroxylamine, bromine water, acetic anhydride and nitric acid. It is disproved as the whole story by the failed Schiff's and bisulphite tests, which force the cyclic hemiacetal.

Ring closure creates the anomeric carbon, hence two anomers, hence mutarotation to $+ 52. 7^{\circ}$ from either starting anomer.

Sucrose is non-reducing because both anomeric carbons are consumed in its linkage; maltose and lactose keep one free and therefore reduce.

Amino acids exist as zwitterions, which is why they behave like salts, and the pH at which the charge balances is the isoelectric point. The peptide bond is an amide.

Protein structure has four levels. Denaturation destroys the upper three, all of which rest on weak interactions, and leaves the covalent primary sequence untouched.

Enzymes are globular proteins whose specificity comes from the shape of the active site, which is also why denaturation abolishes their activity.

Vitamins are classified by solubility, and solubility predicts storage: fat-soluble A, D, E and K accumulate, water-soluble B and C do not.

Nucleic acids run base to nucleoside to nucleotide to polymer. DNA differs from RNA in one hydroxyl and one methyl group, and pairs A with T through two hydrogen bonds and G with C through three.

Hormones are messengers made within the body, distinguishing them from vitamins, and acting away from their site of production, distinguishing them from enzymes.

Key formulas & results

Everything to memorise for the exam hall, in one card. Screenshot this for revision.

The organising principle

an ordinary organic molecule, large enough for its shape to matter

Glucose's ring is a hemiacetal, a peptide bond is an amide, base pairing is hydrogen bonding, denaturation is the loss of weak interactions. Nothing here is a new reaction type.

Definition of a carbohydrate

a **polyhydroxy aldehyde or ketone**, or something giving one on hydrolysis

More useful than $\mathrm{C_x(H_2O)_y}$, because it names the functional groups you actually reason with. Classify by carbonyl type, carbon count, hydrolysis behaviour and reducing character.

Reducing character

a sugar reduces Tollens' or Fehling's if and only if it has **at least one free anomeric carbon**

Sucrose and trehalose have zero, since their single glycosidic bond ties up both; maltose and lactose have one. Count free anomeric carbons and nothing needs memorising.

Proofs of glucose's open chain

HI gives hexane; $\mathrm{NH_2OH}$ gives an oxime; $\mathrm{Br_2}$ water gives gluconic acid; $\mathrm{(CH_3CO)_2O}$ gives a penta-acetate; $\mathrm{HNO_3}$ gives saccharic acid

Six unbranched carbons, a carbonyl, an aldehyde specifically, five hydroxyls, and a primary alcohol at the far end. Bromine water is the discriminating test, oxidising aldoses and not ketoses.

Mutarotation

$\alpha$ at $+111^\circ$ and $\beta$ at $+19.7^\circ$ both settle at $+52.7^\circ$

The two anomers interconvert through the 0.02 per cent open-chain form. Both ends converging on one fixed value is what identifies it as an equilibrium rather than a decomposition.

Irreversible against reversible tests

a trace concentration is a tap that never runs dry

Oxime, cyanohydrin, Tollens' and Fehling's drain the open chain irreversibly and go to completion. Schiff's and bisulphite are reversible and need a real standing concentration, so they fail. The penta-acetate caps C1 and fails everything.

Why fructose reduces Fehling's

in base, ketose $\rightleftharpoons$ **enediol** $\rightleftharpoons$ aldose

It is glucose and mannose that reduce the reagent, not fructose. Propanone has no $\alpha$-hydroxyl, cannot form an enediol, and gives nothing, which is the control that proves the mechanism.

The three disaccharides

sucrose $\alpha$-glucose + $\beta$-fructose C1-C2, **non-reducing**; maltose two glucose $\alpha(1\to4)$; lactose galactose + glucose $\beta(1\to4)$

Sucrose rotates $+66.5^\circ$ and hydrolyses to invert sugar at $[(+52.5)+(-92.4)]/2 = -19.9^\circ$, a change of sign that names the product.

Zwitterion and isoelectric point

$\mathrm{{}^-OOC{-}CHR{-}NH_3^+}$; at the pI the net charge is zero and it does not migrate

Explains high melting point, water solubility and poor solubility in benzene. Above its pI an amino acid is negative and moves to the anode; below it, positive and moves to the cathode.

Protein structure levels

primary **covalent**; secondary, tertiary and quaternary all **non-covalent**

Denaturation destroys the last three and leaves the first, so molar mass is unchanged. Hydrolysis breaks peptide bonds and molar mass collapses, which is the single measurement that tells them apart.

Starch against cellulose

$\alpha(1\to4)$ is digestible, $\beta(1\to4)$ is not

Same monomer, same empirical formula, one bond geometry. The barrier is kinetic, not thermodynamic: hot acid hydrolyses cellulose easily, and cattle rely on gut microbes for cellulase.

Nucleic acids and Chargaff

base + sugar = nucleoside; plus phosphate = nucleotide; joined by **phosphodiester** bridges; $A = T$ and $G = C$

A-T uses two hydrogen bonds and G-C three, so higher GC content means a higher melting temperature. DNA has deoxyribose and thymine; RNA has ribose and uracil.

Optical rotation of a mixture

α_{n e t} = \sum_{i} f_{i} α_{i} (f_{i} = mole fraction)

Rotations add with sign, weighted by how much of each species is present. Hydrolysing sucrose gives equimolar glucose ($+52.5^\circ$) and fructose ($-92^\circ$), so the sum turns negative and the sign of the rotation inverts — which is where the name invert sugar comes from.

JEE Main test series

Ready to be tested on this?

Biomolecules shows up in every JEE Main paper — sit a timed one and find out where you actually stand.

~1Est. Qs

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Chemistry-only sectionals & chapter tests
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Traps JEE Main sets — and how to dodge them

These are the exact option-traps and misreads that cost marks under negative marking.

WATCH OUT

✗ Treating glucose as a free aldehyde

✓ About 99.98 per cent of it is a cyclic hemiacetal. The three tests it passes are irreversible and drain the 0.02 per cent open chain, which the ring keeps replacing; Schiff's and bisulphite are reversible, need a standing concentration and find none. The penta-acetate, with C1 capped, fails everything, which settles the matter.

Why it happens: It is drawn as an open chain with a CHO group and it passes Tollens', Fehling's and the oxime test.

WATCH OUT

✗ Saying fructose is a reducing sugar because it has a carbonyl

✓ Its own carbonyl is a ketone and reduces nothing. Both reagents are alkaline, and in base a ketose isomerises through an enediol to glucose and mannose, which are the species that actually reduce the reagent. Propanone, lacking an

α

-hydroxyl, gives nothing at all, which is the control. The statement is about the conditions of the test, not about fructose.

Why it happens: It does reduce Fehling's and Tollens', and both are described as carbonyl tests.

WATCH OUT

✗ Memorising which disaccharides are reducing

✓ Count free anomeric carbons. Sucrose's single bond joins C1 of glucose to C2 of fructose, both anomeric, leaving zero free, so no ring can open. Maltose and lactose use one anomeric carbon and an ordinary C4 hydroxyl, leaving one free. The rule generalises: trehalose is non-reducing because its C1-C1 bond ties up both, and any polysaccharide is effectively non-reducing because one free end among thousands is undetectable.

Why it happens: Three named sugars with three answers looks like a list to learn.

WATCH OUT

✗ Confusing denaturation with hydrolysis

✓ Denaturation destroys the secondary, tertiary and quaternary structures, all held by weak non-covalent interactions, and leaves the primary structure and the molar mass unchanged. Hydrolysis breaks the covalent peptide bonds and cuts the chain into fragments. Boiling an egg and digesting one both stop it working, and only a molar mass measurement distinguishes them.

Why it happens: Both destroy biological activity and both are described as the protein being broken.

WATCH OUT

✗ Predicting amino acid migration from the pH alone

✓ Compare the pH with the amino acid's own pI. Above its pI it carries a net negative charge and moves to the anode; below it, a net positive charge and moves to the cathode; at it, nothing moves. At pH 6.0 aspartic acid (pI 2.8) goes to the anode, lysine (pI 9.7) to the cathode and glycine (pI 6.0) stays put, and one buffer separates all three.

Why it happens: It is tempting to call acidic solutions positive and alkaline ones negative without further thought.

WATCH OUT

✗ Forgetting that Chargaff's rule applies only to double-stranded DNA

✓ They follow from strict one-to-one base pairing between two strands, so they hold for double-stranded DNA and fail for single-stranded DNA and for RNA, which is usually single-stranded and carries uracil rather than thymine. When a question supplies one percentage and asks for the rest, check that it has said double-stranded, because that is the assumption doing all the arithmetic.

Why it happens:

A = T

and

G = C

are stated as though they were properties of DNA in general.

Exam-pattern practice

PYQ-style questions with full solutions. Work through them as a readiness check — mark yourself honestly and get your gap report at the end.

Readiness check

Are you exam-ready for Biomolecules?

12 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

12 questions~8 min worth ~4 marks in JEE Main exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Ordinary organic chemistry on molecules big enough for shape to matter; that is the whole chapter
•A sugar reduces if and only if it has at least one free anomeric carbon
•Glucose is 99.98 per cent cyclic hemiacetal; irreversible tests pass, reversible ones fail, the penta-acetate fails all
•Mutarotation: $+ 11 1^{\circ}$ and $+ 19. 7^{\circ}$ both settle at $+ 52. 7^{\circ}$ through the open chain
•Proofs: HI gives hexane, $B r_{2}$ water gives gluconic acid, $HN O_{3}$ gives saccharic acid, acetylation gives a penta-acetate
•Fructose reduces Fehling's only because base isomerises it to aldoses through an enediol
•Sucrose is the non-reducing one, C1 to C2, both anomeric; invert sugar rotates $- 19. 9^{\circ}$
•Amino acids are internal salts; at the pI net charge is zero and nothing migrates
•Denaturation destroys everything above primary and leaves molar mass unchanged; hydrolysis does not
•Enzyme specificity and fragility both come from the tertiary fold that makes the active site
•Fat-soluble A, D, E, K are stored and can accumulate; water-soluble B and C are excreted, with $B_{12}$ the exception
• $A = T$ and $G = C$ in double-stranded DNA; G-C has three hydrogen bonds so GC-rich DNA melts higher

JEE Main question blueprint

How this topic is asked, tier by tier — so you can prep to the pattern.

Typical weightage: ~1 question (4 marks) of the 100-mark Chemistry section

Question style	Marks each	Typical count	What it tests
Carbohydrate structure and reducing character	2	1	Which sugars reduce Fehling's and Tollens' and why fructose does despite being a ketose, mutarotation and anomers, the three named disaccharides and their linkages, and the sign change in invert sugar
Amino acids, proteins and enzymes	1	1	Zwitterions and the direction of migration at a given pH relative to the isoelectric point, counting possible peptides, the four levels of protein structure, and what denaturation destroys and what it leaves intact
Vitamins, nucleic acids and hormones	1	1	Fat- against water-soluble vitamins with their deficiency diseases, Chargaff's rules applied to base percentages, the sugar and base differences between DNA and RNA, and hormone classification

Exam-hall strategy

Battle-tested tips from mentors and toppers for this topic under the sectional clock.

For any reducing-sugar question, count free anomeric carbons rather than recalling which sugars are reducing. Zero means non-reducing, and that single rule covers sucrose, trehalose and every polysaccharide.
When a glucose test succeeds or fails unexpectedly, ask whether the test is reversible. Irreversible tests drain the 0.02 per cent open chain to completion; reversible ones need a standing concentration and find none.
Distinguish denaturation from hydrolysis by naming what happens to the primary structure and to the molar mass. Both destroy activity and only one changes the molecule.
For amino acid migration, compare the buffer pH with the pI rather than with 7. Above the pI means anion and anode; below means cation and cathode.
Chargaff's rule assumes double-stranded DNA. Check the question says so before using $A = T$ and $G = C$ , since RNA and single-stranded DNA obey neither.

Beyond the exam

Where this skill shows up in the job you're competing for — and in life.

Melting temperature is used routinely to estimate the GC …

Melting temperature is used routinely to estimate the GC content of a DNA sample, since G-C pairs carry three hydrogen bonds against A-T's two, and organisms living in hot springs have measurably GC-richer genomes than those in cold water

Honey is mostly invert sugar

Honey is mostly invert sugar, sucrose already hydrolysed by the bee's invertase, which is why it is sweeter than table sugar and why its optical rotation is negative where sucrose's is strongly positive

Lactose intolerance is enzyme specificity in a single exa…

Lactose intolerance is enzyme specificity in a single example: the $β (1 \to 4)$ linkage needs lactase, and adults who stop producing it cannot hydrolyse a disaccharide that is otherwise perfectly ordinary food

Where else this topic is tested

Prepare once, score in every exam that asks it.

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Questions aspirants ask

Pulled from the Q&A community and mentor sessions.

Because the open chain is continuously regenerated. Think of it as a tap rather than a reservoir. Every glucose molecule that opens and is captured by hydroxylamine is removed from the equilibrium, so by Le Chatelier the ring opens again to restore the balance, and the process repeats until essentially all the glucose has reacted. The tiny concentration affects only how fast this happens, not how far it goes. A reversible test behaves entirely differently: Schiff's reagent and bisulphite reach their own equilibrium against a free aldehyde concentration of about $2 \times 1 0^{- 4}$ , which is far too small to give any visible result. Whether a test succeeds on glucose is therefore a question about the test, not about the sugar.

It is the standard phrase and it will be accepted, but it is misleading and a good answer says why. Fructose's own carbonyl is a ketone and does not reduce anything. What happens in the tube is that both Tollens' and Fehling's are alkaline, and base removes the hydrogen next to the carbonyl to give an enediol, which can collapse to either the ketose or the corresponding aldoses. Glucose and mannose form and reduce the reagent. The control that proves this is propanone: an ordinary ketone with no $α$ -hydroxyl, it forms no enediol and gives nothing. So the phrase describes the outcome of a test carried out in base, and any $α$ -hydroxy ketone would behave the same way.

Because an enzyme's active site is a shape, and a shape is either matched or it is not. Starch and cellulose are both polymers of D-glucose with the same empirical formula, differing only in whether the link at C1 is $α$ or $β$ . That single inversion makes starch coil into a helix and cellulose lie flat and pack into hydrogen-bonded fibres, and it presents the bond to an enzyme in a completely different orientation. Human amylase grips the $α$ form and never binds the $β$ one. The failure is purely kinetic, since cellulose hydrolysis is thermodynamically favourable and hot acid achieves it easily. Cattle and termites solve it by hosting microorganisms that make cellulase.

The primary structure, and nothing above it. Primary structure is the sequence of amino acids held by peptide bonds, which are covalent amides and are not broken by heat, mild acid or heavy metal ions. Secondary structure, the alpha helix and beta sheet, rests on hydrogen bonds; tertiary and quaternary structure rest on hydrogen bonds, disulphide bridges, ionic attractions and van der Waals forces. All of those are weak enough to be disrupted, so the protein unfolds, becomes insoluble and stops working, while remaining one continuous chain of unchanged molar mass. That is why denaturation and hydrolysis are not interchangeable words, and why a denatured protein is nutritionally identical to a native one.

Where the syllabus points, which is narrower than the chapter looks. The unit text names carbohydrate classification, glucose and fructose, and the constituent monosaccharides of sucrose, lactose and maltose, so reducing against non-reducing sugars and the cyclic structure of glucose are the highest-yield topics by a clear margin. After those, protein structure levels with denaturation, and the DNA against RNA comparison with Chargaff arithmetic, are the reliable ones. Vitamins and hormones are asked as single recall facts and the deficiency table is worth a few minutes. Polysaccharides are not separately named, so starch against cellulose is worth one line as an illustration of enzyme specificity rather than a topic in itself.