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Biomolecules: Class 12 Chemistry Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 12 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

A nutritionist is preparing a diet chart for patients recovering from illness. She includes two carbohydrates — one from boiled rice (starch) and one from fresh fruit juice (fructose). She also includes dietary proteins to help repair damaged tissues.

A nutritionist is preparing a diet chart for patients recovering from illness. She includes two carbohydrates — one from boiled rice (starch) and one from fresh fruit juice (fructose). She also includes dietary proteins to help repair damaged tissues.

Based on the above scenario, answer the following questions:
(a) The patient's doctor says both starch and fructose are carbohydrates, but only fructose is a reducing sugar. Give ONE reason why fructose is a reducing sugar, whereas starch is NOT. (2 marks)
(b) Amino acids in dietary proteins help repair body tissues. Why are amino acids described as amphoteric in nature? (1 mark)
(c) Starch is a polysaccharide. Name the two components of starch and identify the type of glycosidic linkage present in each component. (1 mark)

Show answer
(a) Fructose is a reducing sugar because, although it is a ketohexose, it has a free hemiacetal / hemiketal (potential free aldehyde) group available in its open-chain form — due to tautomerism (isomerisation in alkaline conditions), the ketone group converts to an aldehyde group, which can reduce Fehling's solution and Tollens' reagent, giving a positive test.

Starch, on the other hand, is a polysaccharide in which glucose units are joined by glycosidic linkages. The anomeric C-1 of almost every glucose unit is involved in the glycosidic bond, so there is effectively NO free hemiacetal (-CHO) group available. Hence starch does NOT reduce Fehling's solution / Tollens' reagent and is a non-reducing sugar.

∴ Fructose is a reducing sugar (free hemiacetal/ketal group present); starch is non-reducing (no free hemiacetal — all anomeric carbons locked in glycosidic bonds).

(b) Amino acids contain BOTH an acidic group (–COOH) and a basic group (–NH₂) in the same molecule.
— In the presence of a base (alkali), the –COOH group donates a proton → amino acid acts as an ACID.
— In the presence of an acid, the –NH₂ group accepts a proton → amino acid acts as a BASE.
Because an amino acid can behave both as an acid and as a base depending on the medium, it is said to be AMPHOTERIC in nature.

(c) Starch has two components:

(i) Amylose — linear (unbranched) polymer of α-D-glucose units joined by α-1,4-glycosidic linkages.

(ii) Amylopectin — branched polymer of α-D-glucose units; the main chain has α-1,4-glycosidic linkages and the branch points have α-1,6-glycosidic linkages.
Q2Case-based4 marks

A food scientist is analysing two carbohydrate samples, P and Q, isolated from plant sources. Sample P is extracted from the storage tissue of a potato and gives a deep blue-black colour with iodine solution. Sample Q is extracted from the cell walls of sugarcane fibre and does NOT give a colour with iodine solution. Both P and Q yield only glucose on complete acid hydrolysis. The scientist also notes that Sample Q cannot be digested by humans, while Sample P can.

A food scientist is analysing two carbohydrate samples, P and Q, isolated from plant sources. Sample P is extracted from the storage tissue of a potato and gives a deep blue-black colour with iodine solution. Sample Q is extracted from the cell walls of sugarcane fibre and does NOT give a colour with iodine solution. Both P and Q yield only glucose on complete acid hydrolysis. The scientist also notes that Sample Q cannot be digested by humans, while Sample P can.

On the basis of this information, answer the following:
(a) Identify Sample P and Sample Q. Give one structural reason why iodine gives a blue-black colour only with Sample P and not with Sample Q. (2 marks)
(b) Name the type of glycosidic linkage present in Sample Q and state why humans cannot digest it. (1 mark)
(c) The scientist heats an aqueous solution of Sample Q with Fehling's solution. Predict and justify the observation. (1 mark)

Show answer
(a) Sample P is Starch and Sample Q is Cellulose. (½ + ½ = 1 mark)

Structural reason for iodine colour: Starch (specifically its amylose component) has a helical coiled structure. Iodine molecules (I₂) fit inside this helix and form a blue-black coloured inclusion complex. Cellulose, on the other hand, has a long, unbranched, flat linear chain structure that does NOT form a helix and therefore cannot trap iodine molecules — hence no colour change is observed. (1 mark)

(b) Sample Q (Cellulose) contains β-1,4-glycosidic linkage — each glucose unit is linked to the next through a β-glycosidic bond between C-1 of one glucose and C-4 of the next. (½ mark)

Humans lack the enzyme β-glucosidase (cellulase), which is required to hydrolyse the β-1,4-glycosidic bonds. Therefore, cellulose cannot be digested by humans. (½ mark)

(c) Observation: No red/brick-red precipitate is formed — Fehling's solution remains blue. (½ mark)

Justification: Cellulose is a polysaccharide in which all anomeric (C-1) positions of glucose units are involved in β-1,4-glycosidic bonds. There is no free aldehyde (–CHO) group or free hemiacetal group available to reduce the Cu²⁺ ions in Fehling's solution. Hence cellulose is a non-reducing sugar/polysaccharide and gives a negative Fehling's test. (½ mark)
Q3Case-based4 marks

A biochemistry student is studying the structural and functional properties of two common disaccharides — sucrose and maltose — found in food samples. She observes the following:

(i) When sample A (sucrose solution) is treated with Fehling's solution, no red precipitate is formed.
(ii) When sample B (maltose solution) is treated with Fehling's solution, a brick-red precipitate forms.
(iii) On complete hydrolysis with dilute acid, both samples give reducing sugars.
(iv) She also notes that sucrose shows mutarotation only AFTER hydrolysis, while maltose shows mutarotation directly.

A biochemistry student is studying the structural and functional properties of two common disaccharides — sucrose and maltose — found in food samples. She observes the following:

(i) When sample A (sucrose solution) is treated with Fehling's solution, no red precipitate is formed.
(ii) When sample B (maltose solution) is treated with Fehling's solution, a brick-red precipitate forms.
(iii) On complete hydrolysis with dilute acid, both samples give reducing sugars.
(iv) She also notes that sucrose shows mutarotation only AFTER hydrolysis, while maltose shows mutarotation directly.

On the basis of the above observations, answer the following questions:

(a) Why does sucrose not reduce Fehling's solution, while maltose does? (2 marks)
(b) Identify the monomeric units obtained on complete hydrolysis of sucrose and name the type of glycosidic linkage present in it. (1 mark)
(c) Why does maltose show mutarotation directly but sucrose does not? (1 mark)

Show answer
CBSE Marking Scheme — 4 marks total

(a) Sucrose is a non-reducing sugar; maltose is a reducing sugar. (2 marks)

In sucrose, the glycosidic linkage is formed between the anomeric carbon (C-1) of α-D-glucose and the anomeric carbon (C-2) of β-D-fructose. Since both anomeric carbons are involved in glycosidic bond formation, no free aldehyde (–CHO) group or free hemiacetal group is available in sucrose. Therefore, sucrose cannot open its ring to generate a free aldehyde in solution and hence does NOT reduce Fehling's solution. (1 mark)

In maltose, the glycosidic linkage is formed between C-1 of one α-D-glucose unit and C-4 of the second glucose unit (α-1,4-glycosidic linkage). The second glucose unit retains a FREE anomeric carbon (C-1) that can open the ring and generate a free –CHO (aldehyde) group in solution. This free aldehyde reduces Cu²⁺ to Cu₂O (brick-red precipitate) in Fehling's solution. Hence maltose IS a reducing sugar. (1 mark)

∴ Sucrose = non-reducing (both anomeric carbons blocked); Maltose = reducing (one free anomeric carbon present).

(b) On complete hydrolysis of sucrose:

Sucrose → α-D-Glucose + β-D-Fructose

(Monomeric units: one molecule of α-D-glucose and one molecule of β-D-fructose.) (½ mark)

The glycosidic linkage in sucrose is an α,β-1,2-glycosidic linkage (C-1 of α-D-glucose linked to C-2 of β-D-fructose). (½ mark)

∴ Monomers = α-D-glucose + β-D-fructose; linkage = α,β-1,2-glycosidic bond.

(c) Mutarotation is the change in optical rotation of a sugar solution due to interconversion of α and β anomeric forms through the open-chain (free aldehyde/ketone) form. This interconversion requires a FREE anomeric carbon (free hemiacetal group). (½ mark)

In maltose, the second glucose unit has a free anomeric carbon (C-1 with free –OH), so the ring can open and re-close in either α or β configuration directly in solution — hence maltose shows mutarotation directly. (½ mark)

In sucrose, BOTH anomeric carbons are engaged in the glycosidic bond — there is NO free anomeric carbon. The ring cannot open without first breaking the glycosidic bond (i.e., hydrolysis). Therefore, sucrose shows NO mutarotation unless it is first hydrolysed.

∴ Maltose has a free hemiacetal (free anomeric C) → mutarotation directly. Sucrose has no free hemiacetal → mutarotation only after hydrolysis.
Q4Case-based4 marks

A school laboratory has four food samples labelled P, Q, R and S. The teacher tells the students the following about each sample:
• Sample P is the storage carbohydrate in plants; it gives a blue-black colour with iodine solution.
• Sample Q is the structural carbohydrate in plants; it is the most abundant organic compound on Earth and cannot be digested by humans.
• Sample R is a disaccharide that is the sweetest natural sugar; it is a non-reducing sugar.
• Sample S is a monosaccharide that is the primary fuel for cellular respiration and gives a positive Fehling's test.

A school laboratory has four food samples labelled P, Q, R and S. The teacher tells the students the following about each sample:
• Sample P is the storage carbohydrate in plants; it gives a blue-black colour with iodine solution.
• Sample Q is the structural carbohydrate in plants; it is the most abundant organic compound on Earth and cannot be digested by humans.
• Sample R is a disaccharide that is the sweetest natural sugar; it is a non-reducing sugar.
• Sample S is a monosaccharide that is the primary fuel for cellular respiration and gives a positive Fehling's test.

(a) Identify samples P and Q and state the type of glycosidic linkage present in each. (2 marks)
(b) Explain why sample R is a non-reducing sugar. (1 mark)
(c) Give ONE chemical reaction of sample S that confirms it contains an aldehyde group. (1 mark)

Show answer
(a) Sample P is Starch and Sample Q is Cellulose.

Starch is composed of two components — amylose and amylopectin. In amylose, glucose units are joined by α-1,4-glycosidic linkages (linear chain). In amylopectin, in addition to α-1,4-glycosidic linkages in the main chain, α-1,6-glycosidic linkages are present at the branch points.

In Cellulose, glucose units are joined by β-1,4-glycosidic linkages throughout the linear chain.

∴ Sample P (Starch): α-1,4-glycosidic linkage (and α-1,6- at branch points in amylopectin).
∴ Sample Q (Cellulose): β-1,4-glycosidic linkage.

(b) Sample R is Sucrose.

Sucrose is a non-reducing sugar because in its structure, the anomeric carbon (C-1) of glucose and the anomeric carbon (C-2) of fructose are both involved in the glycosidic bond formation. As a result, sucrose has no free aldehyde (–CHO) group or free hemiacetal group available to reduce Fehling's solution or Tollens' reagent.

∴ Since both anomeric carbons are locked in the glycosidic bond, there is no free reducing group — sucrose is a non-reducing sugar.

(c) Sample S is Glucose (an aldohexose).

Oxidation with Br₂ water (bromine water) confirms the presence of the aldehyde group:

Glucose + Br₂(aq) → Gluconic acid

Bromine water oxidises the –CHO group of glucose to a –COOH group, giving gluconic acid. The reddish-brown colour of bromine water is discharged (decolourised), confirming the presence of the aldehyde (–CHO) group.

(Alternatively: Glucose reacts with hydroxylamine (NH₂OH) to form glucose oxime, confirming the presence of the –CHO group.)
Q5Case-based4 marks

A nutritionist is studying the structural differences between two common dietary carbohydrates — starch and cellulose. Both are polysaccharides made entirely of glucose units, yet humans can digest starch but NOT cellulose.

A nutritionist is studying the structural differences between two common dietary carbohydrates — starch and cellulose. Both are polysaccharides made entirely of glucose units, yet humans can digest starch but NOT cellulose.

Based on the above context, answer the following questions:

(a) Name the type of glycosidic linkage present in (i) starch and (ii) cellulose. How does this difference explain why humans can digest starch but not cellulose? [2]

(b) The nutritionist also notes that starch itself has two components. Name them and state ONE structural difference between them. [1]

(c) A sample of glucose is dissolved in water and allowed to stand. After some time, the optical rotation of the solution changes and finally becomes constant at +52.5°. Name this phenomenon and identify the two forms of glucose responsible for it. [1]

Show answer
(a)
(i) Starch: α-1,4-glycosidic linkage (and α-1,6-glycosidic linkage in amylopectin at branch points).
(ii) Cellulose: β-1,4-glycosidic linkage.

Humans possess the enzyme α-glucosidase (amylase), which can hydrolyse only α-glycosidic linkages. Since cellulose contains β-1,4-glycosidic linkages, and humans lack the enzyme β-glucosidase required to hydrolyse these linkages, cellulose cannot be digested. Starch, having α-linkages, is readily hydrolysed and digested.

(b) The two components of starch are:
• Amylose — linear chain of glucose units joined by α-1,4-glycosidic linkages; water-soluble; forms a helical structure.
• Amylopectin — branched chain; backbone has α-1,4-glycosidic linkages and branch points have α-1,6-glycosidic linkages.

Structural difference: Amylose is unbranched (linear), whereas amylopectin is branched.

(c) The phenomenon is called Mutarotation.

It is the change in the specific rotation of a freshly prepared solution of glucose until it reaches a constant equilibrium value (+52.5°).

The two forms responsible are:
• α-D-glucose — specific rotation +112° (C-1 –OH is on the same side as C-6, i.e., below the ring in the Haworth projection).
• β-D-glucose — specific rotation +18.7° (C-1 –OH is opposite to C-6, i.e., above the ring in the Haworth projection).

In aqueous solution, both forms interconvert through the open-chain aldehyde form, establishing an equilibrium mixture that gives the constant rotation of +52.5°.
Q6Case-based4 marks

A food scientist is studying the nutritional composition of two common foods — cow's milk and table sugar (sucrose). Cow's milk contains lactose as its primary carbohydrate and casein as its major protein. When milk is heated at high temperature for a prolonged period, its taste and nutritional quality change noticeably. Sucrose, when hydrolysed completely, gives a mixture that rotates plane-polarised light in the opposite direction compared to sucrose itself.

A food scientist is studying the nutritional composition of two common foods — cow's milk and table sugar (sucrose). She observes the following:

• Cow's milk contains lactose as its primary carbohydrate and casein as its major protein.
• When milk is heated at high temperature for a prolonged period, its taste and nutritional quality change noticeably.
• Sucrose, when hydrolysed completely, gives a mixture that rotates plane-polarised light in the opposite direction compared to sucrose itself.

On the basis of the above information, answer the following questions:

(a) Lactose is a disaccharide. Name the two monosaccharide units obtained on complete hydrolysis of lactose and identify the type of glycosidic linkage present in it. (2 marks)

(b) The scientist observes that prolonged heating denatures casein. What is meant by denaturation of a protein? Which levels of protein structure are disrupted during denaturation? (1 mark)

(c) The hydrolysis of sucrose is called inversion. Why is this process given that name? Name the enzyme that brings about the hydrolysis of sucrose in biological systems. (1 mark)

Show answer
(a) On complete hydrolysis, lactose yields two monosaccharide units:
— β-D-Galactose and β-D-Glucose.
The two units are joined by a β-1,4-glycosidic linkage (C1 of galactose linked to C4 of glucose).
Lactose is a reducing sugar because the glucose unit retains a free anomeric —OH group (free hemiacetal), which is not involved in the glycosidic bond.
[Award 1 mark for correct monosaccharides; 1 mark for correct type of glycosidic linkage.]

(b) Denaturation of a protein is the process in which a protein loses its biological activity due to disruption of its secondary, tertiary and quaternary structures (the three-dimensional native conformation), while the primary structure (the sequence of amino acids / peptide bonds) remains intact.
During denaturation, the secondary structure (α-helix / β-sheet held by hydrogen bonds), the tertiary structure (3-D fold held by disulphide bridges, hydrogen bonds, ionic interactions and hydrophobic interactions) and, where applicable, the quaternary structure (subunit arrangement) are all disrupted. The primary structure is NOT affected.
[Award 1 mark for correct definition AND mention that 2°, 3°, 4° structures are disrupted while 1° is intact.]

(c) Sucrose is dextrorotatory and rotates plane-polarised light to the right (+66.5°). On hydrolysis it gives an equimolar mixture of glucose (+52.5°) and fructose (−92.4°). The net optical rotation of the product mixture is negative (levorotatory). Since the sign of optical rotation is inverted (from + to −) during hydrolysis, the process is called inversion and the product mixture is called invert sugar.
The enzyme that catalyses the hydrolysis of sucrose in biological systems is invertase (also called sucrase).
[Award ½ mark for correct reason (change in sign of rotation from + to −); ½ mark for naming the enzyme invertase / sucrase.]
Q7Case-based4 marks

A nutritionist is comparing two common carbohydrates — starch and cellulose — both made entirely of glucose units. She notes that humans can digest starch easily but cannot digest cellulose at all, even though both are polysaccharides of glucose. She also observes that starch gives a blue-black colour with iodine solution, while cellulose does not.

A nutritionist is comparing two common carbohydrates — starch and cellulose — both made entirely of glucose units. She notes that humans can digest starch easily but cannot digest cellulose at all, even though both are polysaccharides of glucose. She also observes that starch gives a blue-black colour with iodine solution, while cellulose does not.

Based on this information, answer the following questions:

(a) Identify the type of glycosidic linkage present in (i) starch (amylose component) and (ii) cellulose. How does this difference explain why humans can digest starch but not cellulose? [2 marks]

(b) Which component of starch is responsible for the blue-black colour with iodine solution, and why does cellulose not give this colour? [1 mark]

(c) Name the monosaccharide obtained on complete hydrolysis of both starch and cellulose. Write the product formed when this monosaccharide reacts with bromine water. [1 mark]

Show answer
(a) (i) Starch (amylose): α-1,4-glycosidic linkage — C1 of one α-D-glucose unit is linked to C4 of the next α-D-glucose unit.
(ii) Cellulose: β-1,4-glycosidic linkage — C1 of one β-D-glucose unit is linked to C4 of the next β-D-glucose unit.

Humans possess the enzyme α-amylase (and related α-glucosidases) which can hydrolyse α-1,4-glycosidic linkages present in starch. However, humans lack the enzyme β-glucosidase (cellulase), which is required to hydrolyse the β-1,4-glycosidic linkages in cellulose. Due to the absence of this enzyme, cellulose passes through the digestive tract undigested.
[Award 1 mark for correct identification of both linkages; 1 mark for the enzyme-based explanation.]

(b) The amylose component of starch is responsible for the blue-black colour with iodine solution. Amylose has a helical (coiled) structure, and iodine (I₂) molecules get trapped inside the helical cavity to form a blue-black coloured inclusion complex.

Cellulose has a long, unbranched, straight-chain structure (due to β-1,4-linkages) with extensive intermolecular hydrogen bonding; it does not form a helix. Therefore, iodine molecules cannot be accommodated within the structure, and no blue-black colour is produced.
[Award 1 mark for naming amylose AND giving the reason based on helical structure / absence of helix in cellulose.]

(c) The monosaccharide obtained on complete hydrolysis of both starch and cellulose is D-glucose.

Reaction of D-glucose with bromine water (Br₂/H₂O):

D-Glucose + Br₂/H₂O → D-Gluconic acid

Bromine water oxidises the aldehyde group (–CHO) of D-glucose to a carboxyl group (–COOH), forming D-gluconic acid. The bromine water is decolourised in this reaction.
[Award ½ mark for D-glucose; ½ mark for D-gluconic acid / correct product with or without the full equation.]

∴ Complete answers: (a) α-1,4 (starch/amylose), β-1,4 (cellulose); absence of cellulase in humans. (b) Amylose; helical structure traps I₂. (c) D-glucose; D-gluconic acid.
Q8Case-based4 marks

A food scientist is studying two samples of carbohydrate-rich foods. Sample X is obtained from potato starch and Sample Y is obtained from cotton fibres. She performs the following tests: (i) Both samples are treated with iodine solution — Sample X turns blue-black, Sample Y shows no colour change. (ii) Both samples give a brick-red precipitate with Fehling's test after complete hydrolysis. (iii) Humans can digest Sample X easily but cannot digest Sample Y.

A food scientist is studying two samples of carbohydrate-rich foods. Sample X is obtained from potato starch and Sample Y is obtained from cotton fibres. She performs the following tests:

(i) Both samples are treated with iodine solution — Sample X turns blue-black, but Sample Y shows no colour change.
(ii) Both samples are subjected to complete hydrolysis followed by Fehling's test — both give a brick-red precipitate.
(iii) She notes that humans can digest Sample X easily but cannot digest Sample Y.

Based on this information, answer the following questions:

(a) Identify the specific polysaccharide in Sample X responsible for the blue-black colour with iodine. Name the type of glycosidic linkage present in it. (2 marks)

(b) Identify the polysaccharide in Sample Y and state the type of glycosidic linkage that makes it indigestible to humans. (1 mark)

(c) Both samples give the same product on complete hydrolysis. Name this product and classify it. (1 mark)

Show answer
(a) The polysaccharide in Sample X responsible for the blue-black colour with iodine is amylose.

Amylose has a helical structure; iodine (I₂) molecules fit inside this helix to give the characteristic blue-black colour.

The type of glycosidic linkage present in amylose is α-1,4-glycosidic linkage (linear chain of glucose units joined by α-1,4 bonds).

∴ Polysaccharide = amylose; linkage = α-1,4-glycosidic linkage. (1 + 1 = 2 marks)

(b) The polysaccharide in Sample Y is cellulose.

Cellulose has β-1,4-glycosidic linkages between glucose units. Humans lack the enzyme β-glucosidase (cellulase) required to hydrolyse these β-linkages, making cellulose indigestible.

∴ Sample Y = cellulose; indigestible due to β-1,4-glycosidic linkage. (1 mark)

(c) Both starch (amylose) and cellulose are made entirely of glucose units. On complete hydrolysis, both yield glucose.

Glucose is a monosaccharide (hexose / aldohexose).

∴ Product of complete hydrolysis = D-glucose; classification = monosaccharide (aldohexose). (1 mark)
Q9MCQ1 mark

Which one of the following is a non-reducing sugar?
(A) Glucose
(B) Maltose
(C) Lactose
(D) Sucrose

Show answer
(D) Sucrose

Explanation: A reducing sugar must possess a free aldehydic (–CHO) group or a free hemiacetal group capable of being oxidised. In sucrose, the glycosidic linkage is formed between the anomeric carbon (C-1) of α-D-glucose and the anomeric carbon (C-2) of β-D-fructose; consequently, neither ring retains a free hemiacetal or hemiketal group. Because no free reducing group is available, sucrose cannot reduce Tollens' reagent or Fehling's solution and is therefore a non-reducing sugar. Glucose, maltose and lactose all retain a free hemiacetal group and hence are reducing sugars.
Q10MCQ1 mark

Name the type of glycosidic linkage present between the two glucose units in maltose.

Show answer
(A) α-1,4-glycosidic linkage

Explanation: Maltose is a disaccharide formed by the condensation of two α-D-glucose units. The C-1 (anomeric carbon) of one glucose unit is linked to the C-4 of the second glucose unit through an α-glycosidic bond, giving an α-1,4-glycosidic linkage. Because the anomeric –OH of the second glucose unit remains free, maltose is a reducing sugar.
Q11MCQ1 mark

Which of the following statements about the secondary structure of a protein is correct?

Show answer
(B)

Explanation: The secondary structure of a protein describes the spatial arrangement of the polypeptide backbone in regular, repeating patterns — chiefly the α-helix and the β-pleated sheet. Both of these patterns are stabilised by intramolecular (or intermolecular, in β-sheet) hydrogen bonds formed between the carbonyl (—C=O) and the imino (—N—H) groups of the peptide chain. Option (A) describes the primary structure (amino acid sequence held by peptide bonds). Option (C) describes the tertiary structure (overall 3D fold involving disulphide bridges, hydrophobic interactions and ionic bonds). Option (D) describes the quaternary structure (association of two or more polypeptide subunits).
Q12Short Answer1 mark

Assertion (A) : Sucrose is a non-reducing sugar.
Reason (R) : In sucrose, the anomeric carbon of glucose and the anomeric carbon of fructose are both involved in the glycosidic linkage.

Show answer
Option (A) — Both A and R are true, and R is the correct explanation of A.

Explanation: A reducing sugar must possess a free anomeric –OH group (free hemiacetal or hemiketal) that can open to give a free aldehyde or ketone group and thereby reduce Fehling's solution or Tollens' reagent. In sucrose, the C-1 anomeric carbon of the glucose unit and the C-2 anomeric carbon of the fructose unit are both used to form the α,β-1,2-glycosidic bond. Because neither anomeric carbon carries a free –OH, sucrose cannot open to a free aldehyde or ketone form and hence cannot act as a reducing agent — making it a non-reducing sugar. R correctly and completely explains A.
Q13MCQ1 mark

What are the monomeric units obtained on complete hydrolysis of starch?

Show answer
Option (C) — α-D-Glucose

Explanation: Starch is a polysaccharide made up of two components — amylose and amylopectin — both of which are polymers of α-D-glucose units joined by α-1,4-glycosidic linkages (and α-1,6-linkages at branch points in amylopectin). Complete hydrolysis of starch therefore yields only α-D-glucose as the monomer.
Q14Short Answer2 marks

Name the monosaccharide units present in sucrose and state the type of glycosidic linkage joining them.

Show answer
Sucrose is made up of one unit of α-D-glucose and one unit of β-D-fructose.

The two monosaccharide units are joined by an α,β-1,2-glycosidic linkage (i.e., C-1 of glucose is linked to C-2 of fructose).

Because this linkage involves the anomeric carbons of both monosaccharides, sucrose has no free aldehyde or ketone group and is therefore a non-reducing sugar.
Q15Short Answer2 marks

Write the chemical reactions involved when D-glucose is treated with the following reagents:
(a) Bromine water
(b) Acetic anhydride

Show answer
(a) D-glucose + Br₂/H₂O → D-gluconic acid

D-glucose is oxidised by bromine water (a mild oxidising agent) to give D-gluconic acid. This confirms the presence of an aldehyde (–CHO) group in glucose.

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

(1 mark)

(b) D-glucose + (CH₃CO)₂O → glucose pentaacetate

D-glucose reacts with excess acetic anhydride (acetylating agent) to give glucose pentaacetate, confirming the presence of five –OH groups in the molecule.

C₆H₇O(OH)₅ + 5(CH₃CO)₂O → C₆H₇O(OCOCH₃)₅ + 5CH₃COOH

(1 mark)
Q16Short Answer2 marks

Answer the following:
(a) Name the two strands of DNA are held together by which type of bonds? State the number of such bonds between Adenine–Thymine and Guanine–Cytosine base pairs.
(b) Give one difference between a nucleoside and a nucleotide.

Show answer
(a) The two strands of DNA are held together by hydrogen bonds between the complementary nitrogenous bases.
— Adenine (A) and Thymine (T) are linked by 2 hydrogen bonds.
— Guanine (G) and Cytosine (C) are linked by 3 hydrogen bonds.
(1 mark: correct bond type + both correct numbers)

(b) A nucleoside consists of a nitrogenous base linked to a pentose sugar (deoxyribose or ribose), whereas a nucleotide consists of a nitrogenous base, a pentose sugar, and a phosphate group.
∴ A nucleotide = nucleoside + phosphate group.
(1 mark: correct distinguishing feature stated)
Q17Short Answer2 marks

(a) Name the two strands of DNA that are held together by hydrogen bonds. How many hydrogen bonds are present between adenine–thymine (A–T) and guanine–cytosine (G–C) base pairs respectively?
(b) What type of glycosidic linkage is present in cellulose? Why can humans not digest cellulose?

Show answer
(a) The two strands of DNA are held together by hydrogen bonds between complementary nitrogenous bases. Adenine (A) pairs with thymine (T) through 2 hydrogen bonds, and guanine (G) pairs with cytosine (C) through 3 hydrogen bonds.

(b) Cellulose has β-1,4-glycosidic linkage between glucose units. Humans cannot digest cellulose because they lack the enzyme β-glucosidase (cellulase), which is required to hydrolyse the β-1,4-glycosidic bonds.
Q18Short Answer3 marks

(a) Give one chemical test to distinguish between glucose and sucrose.
(b) What structural feature of cellulose makes it unsuitable as a food source for humans?
(c) Name the vitamin whose deficiency causes rickets and state whether it is water-soluble or fat-soluble.

Show answer
(a) Tollens' reagent test (ammoniacal silver nitrate solution):
Glucose — being a reducing sugar, it reduces Tollens' reagent and gives a silver mirror on the inner wall of the test tube.
Sucrose — being a non-reducing sugar (no free aldehyde / hemiacetal group), it does NOT reduce Tollens' reagent and no silver mirror is formed.
∴ Glucose gives a positive Tollens' test; sucrose does not. [1 mark]

(b) Cellulose is a polysaccharide in which glucose units are linked by β-1,4-glycosidic linkages.
Humans lack the enzyme β-glucosidase (cellulase) needed to hydrolyse these β-1,4-glycosidic bonds, so cellulose cannot be digested and cannot serve as a food source. [1 mark]

(c) Vitamin D is responsible for preventing rickets (promotes absorption of Ca²⁺ and phosphate, essential for bone mineralisation).
Vitamin D is fat-soluble. [1 mark]
Q19Short Answer3 marks

Answer the following questions about the structure and reactions of glucose:
(a) What product is formed when D-glucose is treated with bromine water? Name the reaction and identify the functional group responsible for it.
(b) What is mutarotation? Give the optical rotation values of α-D-glucose and β-D-glucose.
(c) Why does glucose not give the Schiff's test despite having an aldehyde group?

Show answer
(a) When D-glucose is treated with bromine water, it is oxidised to gluconic acid (D-gluconic acid).

C₆H₁₂O₆ + Br₂(aq) → C₆H₁₂O₇
(D-glucose) (D-gluconic acid)

This is an oxidation reaction. The free aldehyde group (–CHO) present in the open-chain form of glucose is responsible for this reaction, as bromine water is a mild oxidising agent that selectively oxidises aldehydes.

(b) Mutarotation is the change in specific optical rotation of a freshly prepared solution of α- or β-form of glucose to an equilibrium value, due to interconversion of the two anomeric forms through the open-chain form in aqueous solution.

Optical rotation values:
• α-D-glucose: +112°
• β-D-glucose: +18.7°
At equilibrium, the specific rotation reaches +52.5°.

(c) In aqueous solution, glucose exists predominantly in the cyclic hemiacetal (pyranose) form, in which the –CHO group is involved in ring formation and is no longer free. Since the Schiff's test requires a free aldehyde group (–CHO) to react with Schiff's reagent (leucofuchsin), glucose does not give a positive Schiff's test. The ring structure masks the aldehyde group.
Q20Short Answer3 marks

Answer the following questions related to Biomolecules:
(a) What is the difference between a nucleoside and a nucleotide? Give one example of each.
(b) Name the type of glycosidic linkage present in cellulose and explain why humans cannot digest it.
(c) Glucose reacts with acetic anhydride to give a pentaacetate product. What does this observation indicate about the structure of glucose?

Show answer
(a) A nucleoside consists of a nitrogenous base linked to a pentose sugar (ribose or deoxyribose) by an N-glycosidic bond.
A nucleotide is a nucleoside esterified with a phosphate group at the 5′ position of the sugar.

Example of nucleoside: Adenosine (adenine + ribose)
Example of nucleotide: Adenosine monophosphate / AMP (adenine + ribose + phosphate)

(b) Cellulose contains β-1,4-glycosidic linkages — each glucose unit is joined to the next through the C-1 of one unit and the C-4 of the next in the β-configuration.

Humans cannot digest cellulose because the human digestive system lacks the enzyme β-glucosidase (cellulase), which is required to hydrolyse the β-1,4-glycosidic bonds. Hence cellulose passes through the gut undigested and acts as dietary fibre.

(c) Glucose reacts with acetic anhydride to form glucose pentaacetate, indicating that glucose contains five free hydroxyl (–OH) groups in its structure.

∴ This confirms that the open-chain form of glucose has five –OH groups (at C-1, C-2, C-3, C-4 and C-6) available for acetylation, consistent with its aldohexose structure.
Q21Short Answer3 marks

Answer the following questions on nucleic acids:
(a) Name the four nitrogenous bases present in DNA. Classify them as purines or pyrimidines.
(b) How does the base composition of RNA differ from that of DNA?
(c) A nucleotide isolated from DNA on complete hydrolysis gives three products. Name the three products obtained.

Show answer
(a) The four nitrogenous bases present in DNA are:

Purines: Adenine (A) and Guanine (G)
Pyrimidines: Cytosine (C) and Thymine (T)

(1 mark — all four names with correct classification)

(b) RNA contains Uracil (U) in place of Thymine (T). All other three bases — Adenine, Guanine and Cytosine — remain the same. Additionally, RNA contains ribose sugar instead of deoxyribose, and is generally single-stranded, whereas DNA is double-stranded.

(1 mark — stating Uracil replaces Thymine as the key difference)

(c) On complete hydrolysis, a nucleotide from DNA yields:
(i) A phosphoric acid molecule (H₃PO₄)
(ii) A pentose sugar — 2-deoxyribose
(iii) A nitrogenous base (a purine or pyrimidine, e.g., Adenine, Guanine, Cytosine or Thymine)

(1 mark — all three products correctly named)
Q22Short Answer3 marks

Answer the following:
(a) What is mutarotation? Name the phenomenon observed when α-D-glucose is dissolved in water and allowed to stand.
(b) Give one difference between amylose and amylopectin on the basis of their structure.
(c) Why is sucrose considered a non-reducing sugar?

Show answer
(a) Mutarotation is the change in the specific rotation of an optically active sugar when it is dissolved in water and allowed to stand until equilibrium is reached.
When α-D-glucose (+112°) is dissolved in water, it gradually interconverts with β-D-glucose (+18.7°) through the open-chain form, and the specific rotation of the solution changes until it reaches an equilibrium value of +52.5°. This phenomenon is called mutarotation.

(b)
Amylose: It is a linear polymer of α-D-glucose units joined by α-1,4-glycosidic linkages.
Amylopectin: It is a branched polymer of α-D-glucose units joined by α-1,4-glycosidic linkages in the main chain and α-1,6-glycosidic linkages at the branch points.

(c) In sucrose, glucose and fructose are linked through their anomeric carbons — C-1 of glucose and C-2 of fructose — via a glycosidic bond. As a result, sucrose has no free anomeric –OH group (no free hemiacetal or hemiketal group). Since it cannot be oxidised to give an aldehyde in solution, it cannot reduce Fehling's solution or Tollens' reagent.
∴ Sucrose is a non-reducing sugar.
Q23Case-based4 marks

Riya is a student who loves cooking. One day, she notices the following in her kitchen:
• Egg white, which is normally transparent, turns white and solid when boiled.
• A spoonful of honey (mainly glucose and fructose) gives a silver mirror when warmed with Tollen's reagent.
• Her mother uses sucrose (table sugar) to make jam, but warns that sucrose does not reduce Fehling's solution.
• Riya also learns from her biology textbook that DNA stores genetic information due to specific base pairing between complementary bases.

Riya is a student who loves cooking. One day, she notices the following in her kitchen:
• Egg white, which is normally transparent, turns white and solid when boiled.
• A spoonful of honey (mainly glucose and fructose) gives a silver mirror when warmed with Tollen's reagent.
• Her mother uses sucrose (table sugar) to make jam, but warns that sucrose does not reduce Fehling's solution.
• Riya also learns from her biology textbook that DNA stores genetic information due to specific base pairing between complementary bases.

Based on the above observations, answer the following questions:
(a) Name the phenomenon that occurs when egg white is boiled. Which bonds/interactions responsible for the secondary and tertiary structure of the protein are disrupted during this process? (2)
(b) Why does sucrose not reduce Tollen's reagent, while glucose does? (1)
(c) Which two bases in DNA are held together by THREE hydrogen bonds? (1)

Show answer
(a) The phenomenon is called DENATURATION of protein. (1 mark)

When egg white is boiled, the heat disrupts the following interactions that maintain the secondary and tertiary structure of the protein:
• Hydrogen bonds (responsible for α-helix and β-sheet secondary structure)
• Disulphide bonds, hydrophobic interactions, and ionic (electrostatic) bonds (responsible for tertiary structure)

As a result, the protein loses its native three-dimensional shape and biological activity, and the polypeptide chain gets unfolded/coagulated. The primary structure (sequence of amino acids held by peptide bonds) is NOT disrupted. (1 mark)

∴ The bonds/interactions disrupted are: hydrogen bonds, disulphide bonds, hydrophobic interactions, and ionic bonds.

(b) Sucrose is a NON-REDUCING sugar. In sucrose, the anomeric carbon (C-1) of glucose and the anomeric carbon (C-2) of fructose are both involved in the glycosidic linkage. Therefore, sucrose has NO free aldehyde group (–CHO) and no free hemiacetal group available to act as a reducing agent. Hence, it does not reduce Tollen's reagent (or Fehling's solution). (1 mark)

Glucose, on the other hand, has a free –CHO group (or exists in equilibrium with the open-chain form bearing –CHO), which acts as the reducing agent.

(c) In DNA, Guanine (G) and Cytosine (C) are held together by THREE hydrogen bonds. (1 mark)

(Note: Adenine–Thymine base pair is held by only TWO hydrogen bonds.)
Q24Case-based4 marks

A nutritionist is comparing the properties of two common carbohydrates: sucrose (table sugar) and lactose (milk sugar). When Fehling's solution is added to lactose solution and heated, a brick-red precipitate forms. The same test on sucrose solution gives no precipitate. Both sugars are then completely hydrolysed using dilute acid and their products are tested.

A nutritionist is comparing the properties of two common carbohydrates found in food: sucrose (table sugar) and lactose (milk sugar). She observes that when she adds a few drops of Fehling's solution to a lactose solution and heats it, a brick-red precipitate forms. However, when she performs the same test on a sucrose solution, no precipitate is observed. She then completely hydrolyses both sugars using dilute acid and tests the products.

(a) Identify whether lactose is a reducing sugar or non-reducing sugar. Give one reason for your answer. (2 marks)

(b) Name the monosaccharides obtained on complete hydrolysis of sucrose. (1 mark)

(c) Name the type of glycosidic linkage (α or β) present between the monosaccharide units in lactose. (1 mark)

Show answer
(a) Lactose is a reducing sugar.
Reason: In lactose, the anomeric carbon (C1) of glucose is free (not involved in the glycosidic bond), so it retains a free aldehyde group (or free hemiacetal). This free –CHO group reduces Cu²⁺ to Cu₂O, giving the brick-red precipitate with Fehling's solution.

∴ Lactose is a reducing sugar because it has a free hemiacetal (–CHO) group available to act as a reducing agent. (2 marks)

(b) On complete hydrolysis, sucrose gives:
• D-Glucose (from the glucopyranose unit)
• D-Fructose (from the fructofuranose unit)
∴ Sucrose yields one molecule of glucose and one molecule of fructose. (1 mark)

(c) In lactose, the glycosidic linkage between galactose and glucose is a β-1,4-glycosidic linkage (β-glycosidic bond). (1 mark)
Q25Case-based4 marks

Riya is a nutrition science student studying food labels. She notices that a sports drink contains 'sucrose', a breakfast cereal contains 'starch', and a probiotic yogurt contains 'lactose'. Her professor explains that these carbohydrates behave differently in Benedict's test and have different structural features.

Riya is a nutrition science student studying food labels. She notices that a sports drink contains 'sucrose', a breakfast cereal contains 'starch', and a probiotic yogurt contains 'lactose'. Her professor explains that these carbohydrates behave differently in Benedict's test and have different structural features.

Based on this context, answer the following questions:

(a) Riya observes that starch gives a blue-black colour with iodine solution but gives a negative Benedict's test. Give ONE structural reason why starch does not reduce Benedict's reagent. (1 mark)

(b) Her professor says lactose is a reducing sugar but sucrose is a non-reducing sugar. Identify the structural feature that makes sucrose a non-reducing sugar, whereas lactose is a reducing sugar. (1 mark)

(c) Riya learns that starch is made of two components. Name BOTH components and state the type of glycosidic linkage present in each. (2 marks)

Show answer
(a) Starch does not give a positive Benedict's test because all the anomeric –OH (reducing end) groups of the glucose units are involved in glycosidic linkages with other monosaccharide units throughout the polymer chain. Due to the absence of a free aldehyde group (–CHO) or a free hemiacetal/anomeric –OH group, starch cannot reduce Benedict's reagent and is therefore a non-reducing carbohydrate.

[Award 1 mark for: no free aldehyde / all anomeric –OH blocked in glycosidic bonds / no free hemiacetal group — any one correct point.]

(b) In sucrose, the glycosidic linkage is formed between the anomeric carbon (C-1) of α-D-glucose and the anomeric carbon (C-2) of β-D-fructose. Since BOTH anomeric carbons are locked in the glycosidic bond, there is no free hemiacetal or hemiketal group available. ∴ Sucrose cannot open to give a free –CHO and is a non-reducing sugar.

In lactose, the glycosidic bond is formed only at the anomeric carbon (C-1) of galactose; the anomeric carbon (C-1) of glucose remains FREE as a hemiacetal. ∴ This free anomeric –OH can open to give a free –CHO group, so lactose is a reducing sugar.

[Award 1 mark for: both anomeric carbons involved / no free anomeric –OH in sucrose — in contrast to lactose where C-1 of glucose is free.]

(c) The two components of starch are:

(i) Amylose — It is the linear, unbranched component of starch. It consists of glucose units linked by α-1,4-glycosidic linkages.

(ii) Amylopectin — It is the branched component of starch. The backbone consists of glucose units linked by α-1,4-glycosidic linkages, while the branch points are formed by α-1,6-glycosidic linkages.

[Award ½ mark for naming Amylose + ½ mark for its linkage (α-1,4); ½ mark for naming Amylopectin + ½ mark for both linkages (α-1,4 backbone and α-1,6 at branch points). Total = 2 marks.]
Q26Case-based4 marks

A biochemist is studying two patients with metabolic disorders. Patient X has a deficiency of an enzyme that breaks the β-1,4-glycosidic linkage in dietary polysaccharides, while Patient Y has a condition in which a storage polysaccharide accumulates abnormally in the liver due to defective branching enzyme activity.

A biochemist is studying two patients with metabolic disorders. Patient X has a deficiency of an enzyme that breaks the β-1,4-glycosidic linkage in dietary polysaccharides, while Patient Y has a condition in which a storage polysaccharide accumulates abnormally in the liver due to defective branching enzyme activity.

On the basis of the above information, answer the following questions:

(a) Name the polysaccharide that Patient X cannot digest. Identify the specific structural feature (type of linkage and monomeric unit) that makes this polysaccharide indigestible by humans under normal conditions. (2 marks)

(b) Name the storage polysaccharide that accumulates in Patient Y's liver. State ONE structural difference between the normal form of this polysaccharide and the linear storage polysaccharide found in plants. (1 mark)

(c) Patient Y is also found to have an abnormally high blood glucose level after meals. Suggest ONE biochemical reason for this observation in relation to the structural role of branching in the storage polysaccharide. (1 mark)

Show answer
CBSE Marking Scheme — 4 marks total

(a) [2 marks]

Patient X cannot digest Cellulose.

Structural feature:
Cellulose is a linear polysaccharide composed of D-glucose units linked by β-1,4-glycosidic linkages. Humans lack the enzyme β-glucosidase (cellulase) required to hydrolyse this β-1,4-glycosidic linkage. The β-configuration at C-1 of each glucose unit results in a straight, ribbon-like chain stabilised by intermolecular hydrogen bonds, making it structurally rigid and resistant to human digestive enzymes.

[Award 1 mark for correctly naming Cellulose AND identifying the monomer as D-glucose. Award 1 mark for correctly stating β-1,4-glycosidic linkage as the specific structural feature and connecting it to the absence of β-glucosidase / indigestibility in humans.]

(b) [1 mark]

The storage polysaccharide that accumulates in Patient Y's liver is Glycogen.

Structural difference between Glycogen and Amylose (the linear plant storage polysaccharide):
Glycogen is more extensively branched than amylose — glycogen has both α-1,4-glycosidic linkages along the main chain AND α-1,6-glycosidic linkages at branch points (occurring approximately every 8–10 glucose units), whereas amylose consists entirely of α-1,4-glycosidic linkages with no branching.

[Award 1 mark for naming Glycogen AND stating that it has α-1,6-glycosidic linkages at branch points / is more extensively branched compared to amylose / has branching every 8–10 units vs no branching in amylose. Either phrasing acceptable.]

(c) [1 mark]

Branching in glycogen creates a large number of free non-reducing (and reducing) terminal ends, allowing many glycogen phosphorylase enzymes to act simultaneously at multiple chain ends — this enables rapid mobilisation of glucose units from glycogen during periods of high glucose demand. In Patient Y, the defective branching enzyme produces glycogen with fewer branch points (abnormally long, less branched chains). This structurally abnormal glycogen is degraded much more slowly, so excess dietary glucose cannot be efficiently stored as normal glycogen and instead remains in the bloodstream, causing elevated blood glucose (hyperglycaemia) after meals.

[Award 1 mark for the correct biochemical reason: branching increases the number of terminal non-reducing ends available for simultaneous enzyme action → rapid glucose storage / mobilisation. Defective branching → fewer ends → slower glucose removal from blood → high blood glucose. Accept any correct equivalent reasoning linking reduced branching to impaired glucose storage and elevated blood glucose.]
Q27Case-based4 marks

A forensic scientist is analysing a biological sample recovered from a crime scene. The sample contains an unknown nucleic acid. The following observations are recorded:
(i) On complete hydrolysis, the sample yields adenine, guanine, cytosine, uracil, and a pentose sugar that gives a positive Bial's test (orcinol reagent).
(ii) The sample is single-stranded.
(iii) When the sample is treated with a specific enzyme, it produces nucleotides; each nucleotide on further hydrolysis gives a nitrogenous base, a sugar, and phosphoric acid.

A forensic scientist is analysing a biological sample recovered from a crime scene. The sample contains an unknown nucleic acid. The following observations are recorded:
(i) On complete hydrolysis, the sample yields adenine, guanine, cytosine, uracil, and a pentose sugar that gives a positive Bial's test (orcinol reagent).
(ii) The sample is single-stranded.
(iii) When the sample is treated with a specific enzyme, it produces nucleotides; each nucleotide on further hydrolysis gives a nitrogenous base, a sugar, and phosphoric acid.

Based on these observations, answer the following:
(a) Identify the nucleic acid present in the sample. Give TWO reasons from the observations to justify your identification. (2 marks)
(b) Name the type of bond that links the nucleotide units together in this nucleic acid. Which two functional groups of adjacent nucleotides participate in forming this bond? (1 mark)
(c) The scientist finds that in a related double-stranded nucleic acid from the same organism, the percentage of adenine is 28%. Calculate the percentage of each of the remaining three bases (guanine, cytosine, and the base that pairs with adenine in this double-stranded form). (1 mark)

Show answer
(a) The nucleic acid present in the sample is RNA (Ribonucleic Acid).

Justifications (any two, 1 mark each):

• Reason 1: The hydrolysis product includes uracil as a nitrogenous base. Uracil is exclusively present in RNA; DNA contains thymine in place of uracil. Since uracil is detected, the sample must be RNA.

• Reason 2: The pentose sugar gives a positive Bial's test (orcinol reagent), which is a characteristic test for ribose sugar. The sugar in RNA is ribose (has —OH at C-2), whereas DNA contains deoxyribose (has —H at C-2). A positive Bial's test confirms the presence of ribose, hence the nucleic acid is RNA.

• (Supporting, if stated) Observation (ii) confirms it is single-stranded, which is consistent with the typical structure of RNA.

∴ The nucleic acid is RNA.

(b) The nucleotide units in RNA are linked together by 3',5'-phosphodiester bonds.

The two functional groups that participate in forming this bond are:
— the 3'-hydroxyl group (—OH) of the sugar of one nucleotide, and
— the phosphate group (—OPO₃H—) attached to the 5'-carbon of the sugar of the next nucleotide.

Thus, one —OH group and one —H from the phosphate-OH condense to form the phosphodiester linkage (with loss of water).

(c) The double-stranded nucleic acid from the same organism is DNA.

In double-stranded DNA, by Chargaff's rule:
— Adenine pairs with thymine (A = T)
— Guanine pairs with cytosine (G = C)

Given: % A = 28%

∴ % T = % A = 28%

Remaining percentage = 100% − (28% + 28%) = 44%

Since % G = % C:

% G = % C = 44% ÷ 2 = 22%

∴ Percentage of bases:
— Adenine (A) = 28%
— Thymine (T) = 28% (the base that pairs with adenine in the double-stranded DNA)
— Guanine (G) = 22%
— Cytosine (C) = 22%
Q28Case-based4 marks

A food scientist is analysing two carbohydrate samples, X and Y, extracted from plant sources. Sample X gives a positive Fehling's test and forms an osazone with phenylhydrazine. On hydrolysis, it yields only glucose. Sample Y does not give Fehling's test, but on complete acid hydrolysis yields glucose and fructose in equimolar amounts. Sample Y also does not show mutarotation.

A food scientist is analysing two carbohydrate samples, X and Y, extracted from plant sources. She observes the following:

• Sample X gives a positive Fehling's test and forms an osazone with phenylhydrazine. On hydrolysis, it yields only glucose.
• Sample Y does not give Fehling's test, but on complete acid hydrolysis it yields glucose and fructose in equimolar amounts. Sample Y also does not show mutarotation.

Based on these observations, answer the following questions:

(a) Identify Sample X and Sample Y. Give one structural reason why Sample Y does not give Fehling's test. (2 marks)

(b) What type of glycosidic linkage is present between the two monosaccharide units in Sample Y? Name the two carbon atoms involved. (1 mark)

(c) The food scientist also notices that Sample X exists in two cyclic forms with specific optical rotations: one form shows [α] = +112° and the other shows [α] = +18.7°. Name this phenomenon and state the equilibrium optical rotation value reached in aqueous solution. (1 mark)

Show answer
(a) Sample X is maltose and Sample Y is sucrose.

Sample X (maltose): It gives a positive Fehling's test, indicating it is a reducing sugar with a free anomeric –CHO (hemiacetal) group available for oxidation. On hydrolysis it yields only glucose, confirming it is a disaccharide of two glucose units.

Sample Y (sucrose): It does not give Fehling's test because the glycosidic linkage in sucrose is formed between the anomeric carbon (C-1) of α-D-glucose and the anomeric carbon (C-2) of β-D-fructose. Both anomeric carbons are involved in bond formation, so there is no free hemiacetal or hemiketal group available to reduce Cu²⁺ in Fehling's solution. Hence sucrose is a non-reducing sugar.

(Value points: correct identification of X = maltose (½), Y = sucrose (½); structural reason — both anomeric carbons are locked in the glycosidic bond / no free hemiacetal/hemiketal group (1))

(b) In sucrose (Sample Y), the glycosidic linkage is an α,β-1,2-glycosidic linkage — formed between C-1 of α-D-glucose and C-2 of β-D-fructose.

(Value points: correct bond type α,β-1,2-glycosidic (½); both carbon positions C-1 and C-2 correctly named (½))

(c) The phenomenon is called mutarotation.

In aqueous solution, α-D-glucose ([α] = +112°) and β-D-glucose ([α] = +18.7°) interconvert through the open-chain form until an equilibrium mixture is reached.

∴ Equilibrium optical rotation = +52.5°

(Value points: correct term 'mutarotation' (½); equilibrium value +52.5° (½))
Q29Case-based4 marks

A biochemistry student is investigating two polysaccharides — Polysaccharide P and Polysaccharide Q — isolated from plant sources. Polysaccharide P gives a deep blue-black colour with iodine solution and yields only glucose on complete hydrolysis, consisting of a water-soluble linear fraction (P₁) and a water-insoluble branched fraction (P₂). Polysaccharide Q does not give a colour with iodine, yields only glucose on complete hydrolysis via β-1,4-glycosidic linkages, and cannot be digested by humans.

A biochemistry student is investigating two polysaccharides — Polysaccharide P and Polysaccharide Q — isolated from plant sources. The following observations are recorded:

• Polysaccharide P gives a deep blue-black colour with iodine solution and yields only glucose on complete hydrolysis. It contains two fractions: one fraction (P₁) is water-soluble and consists of a linear chain, while the other fraction (P₂) is water-insoluble and highly branched.

• Polysaccharide Q does not give a colour with iodine solution. It also yields only glucose on complete hydrolysis but its monomeric units are joined by β-1,4-glycosidic linkages. Humans cannot digest Polysaccharide Q.

On the basis of the above information, answer the following questions:

(a) Identify Polysaccharide P and its two fractions P₁ and P₂. State the type of glycosidic linkage present in fraction P₁. (2 marks)

(b) Identify Polysaccharide Q and give one reason why humans cannot digest it. (1 mark)

(c) Both Polysaccharide P and Polysaccharide Q are non-reducing in nature. Justify this statement. (1 mark)

Show answer
(a) Polysaccharide P is Starch.

Fraction P₁ is Amylose — it is the water-soluble, linear fraction of starch.
Fraction P₂ is Amylopectin — it is the water-insoluble, branched fraction of starch.

The type of glycosidic linkage present in fraction P₁ (Amylose) is α-1,4-glycosidic linkage, i.e., the C-1 of one glucose unit is linked to the C-4 of the next glucose unit through an α-glycosidic bond.

(b) Polysaccharide Q is Cellulose.

Humans cannot digest cellulose because the human digestive system lacks the enzyme β-glucosidase (cellulase), which is required to hydrolyse the β-1,4-glycosidic linkages present between the glucose units of cellulose.

(c) Both starch and cellulose are non-reducing polysaccharides because in their polymeric structures, all the anomeric –OH groups (C-1 –OH of glucose) of the individual glucose units are involved in forming glycosidic linkages between the monomer units. As a result, there is no free aldehyde group (–CHO) or free hemiacetal group available to reduce Fehling's solution or Tollens' reagent.

∴ Both Polysaccharide P (Starch) and Polysaccharide Q (Cellulose) are non-reducing polysaccharides.
Q30Case-based4 marks

A nutritionist is educating school students about carbohydrates in common foods — table sugar (sucrose), milk sugar (lactose), and animal starch (glycogen) — and their structural differences and reducing properties.

A nutritionist is educating a group of school students about carbohydrates present in common foods. She explains that table sugar (sucrose) is different from the sugar found in milk (lactose) and the sugar stored in our liver (glycogen). She then asks the students to answer the following questions based on their understanding of the structure and properties of these carbohydrates:

(a) Sucrose, on hydrolysis, gives two monosaccharides. Name both monosaccharides and identify the type of glycosidic linkage present in sucrose. (2 marks)

(b) Why is sucrose called a non-reducing sugar? (1 mark)

(c) Glycogen is called the 'animal starch'. State one structural feature that distinguishes glycogen from amylopectin (the branched component of starch). (1 mark)

Show answer
(a) On hydrolysis, sucrose gives:
— D-Glucose (an aldohexose)
— D-Fructose (a ketohexose)

The glycosidic linkage in sucrose is an α,β-1,2-glycosidic linkage — formed between C-1 of α-D-glucose and C-2 of β-D-fructose.

[1 mark for both correct monosaccharide names; 1 mark for correct glycosidic linkage (α,β-1,2)]

(b) In sucrose, the anomeric carbon (C-1) of glucose and the anomeric carbon (C-2) of fructose are both involved in the glycosidic bond. Therefore, sucrose has no free aldehyde (–CHO) group or free hemiacetal group available. Due to the absence of a free aldehyde / hemiacetal group, sucrose cannot reduce Tollens' reagent or Fehling's solution.
∴ Sucrose is a non-reducing sugar.

[1 mark for the reason: both anomeric carbons are involved in glycosidic bond / no free –CHO or hemiacetal group]

(c) Glycogen is more highly branched than amylopectin. In glycogen, branching occurs approximately every 8–10 glucose units (α-1,6-glycosidic linkage at branch points), whereas in amylopectin, branching occurs approximately every 24–30 glucose units.
∴ Glycogen is more extensively (more frequently) branched than amylopectin.

[1 mark for stating that glycogen is more highly/frequently branched than amylopectin / branching occurs every 8–10 units in glycogen vs. every 24–30 units in amylopectin]

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