CBSE Class 12 Chemistry Biomolecules Important Questions Set 04

Very Short Answer Questions (1 Mark)

Question 1. Reaction of glucose that indicates that the carbonyl group is present as an aldehyde group in the open structure of glucose. Explain. (CBSE 2018 S)
Answer:
Glucose reduces Tollens’ reagent to give a shining silver mirror and itself gets oxidised to gluconic acid. It also reduces Fehling’s solution to give red ppt. of Cu₂O.

Teacher's Note:
1. Oxidation to an acid with the same number of carbon atoms is the proof of an aldehyde group; a ketone would need C–C bond cleavage.

Question 2. What are oligosaccharides ?
Answer:
Carbohydrates which on hydrolysis give 2-10 molecules of monosaccharides are called oligosaccharides. For example, sucrose, raffinose, stachyrose, etc.

Teacher's Note:
1. The range 2–10 is the key part of the definition; disaccharides are the most common oligosaccharides.

Question 3. What are disaccharides ? Give one example.
Answer:
Disaccharides on hydrolysis give two molecules of the same or different monosaccharides. For example, sucrose, maltose, lactose, etc.

Teacher's Note:
1. Learn the hydrolysis products too: sucrose gives glucose + fructose, maltose gives 2 glucose and lactose gives glucose + galactose.

Question 4. What are tetrasaccharides ? Give an example. (CBSE 2010)
Answer:
Tetrasaccharides on hydrolysis give four molecules of the same or different monosaccharides. For example stachyrose on hydrolysis gives one molecule of glucose, one molecule of fructose and two molecules of galactose.

Teacher's Note:
1. The prefix tells the number of monosaccharide units obtained on hydrolysis: di = 2, tri = 3, tetra = 4.

Question 5. What are polysaccharides ? Give one example. (CBSE 2010, 2018)
Answer:
Polysaccharides on hydrolysis give a large number of molecules of monosaccharides. For example, starch or cellulose.

Teacher's Note:
1. Polysaccharides are not sweet and are called non-sugars, unlike mono- and disaccharides.

Question 6. Name an aldohexose and a ketohexose and write their Fischer Projection formulae.
Answer:
Glucose is an aldohexose while fructose is a ketohexose. Their Fischer Projection formulae are :

Teacher's Note:
1. Fructose has the same configuration as glucose at C-3, C-4 and C-5; only the top two carbons differ (C=O at C-2 in fructose).

Question 7. What are reducing and non-reducing carbohydrates ? Give one example in each case.
Answer:
Carbohydrates which reduce Fehling’s solution to produce a red ppt. of Cu₂O or reduce Tollens’ reagent to produce shining silver mirror are called reducing carbohydrates. All monosaccharides and disaccharide except sucrose are reducing carbohydrates.
Carbohydrates which do not reduce these reagents are called non-reducing carbohydrates, e.g., sucrose.

Teacher's Note:
1. A sugar is reducing only if it has a free aldehydic or ketonic group (a free hemiacetal carbon); in sucrose both anomeric carbons are tied up in the glycosidic bond.

Question 8. What is Molisch test ? Explain.
Answer:
All carbohydrates, i.e., monosaccharides, oligosaccharides and polysaccharides are characterised by Molisch test. A 1% alcoholic solution of α-naphthol is called Molisch reagent. When Molisch reagent is added to an aqueous solution or suspension in water of a carbohydrate followed by conc. H₂SO₄ along the sides of the test tube, a violet ring is formed at the junction of the two layers.

Teacher's Note:
1. Molisch test is a general test for all carbohydrates; it does not distinguish one sugar from another.

Question 9. What are osazones ? How are they formed ?
Answer:
All the monosaccharides and reducing disaccharides such as maltose, lactose, etc. on treatment with excess of phenylhydrazone form crystalline compounds called osazones.
CHO–CHOH–(CHOH)₃–CH₂OH + 3C₆H₅NHNH₂ → CH(=NNHC₆H₅)–C(=NNHC₆H₅)–(CHOH)₃–CH₂OH + C₆H₅NH₂ + NH₃ + 2H₂O
All carbohydrates which contain the grouping –CHOH–CHO or –CO–CH₂OH give osazones. These are used for characterization of monosaccharides and reducing disaccharides. Sucrose, however, does not form an osazone since it is a non-reducing sugar.

Teacher's Note:
1. Three molecules of phenylhydrazine are used per sugar molecule; that is why glucose, fructose and mannose give the same osazone (only C-1 and C-2 are involved).

Question 10. What is mutarotation ? (Raj. Board 2011)
Answer:
The spontaneous change of specific rotation of an optically active substance with time is called mutarotation.
For example, when a freshly prepared solution of α-D-glucose is allowed to stand. Its specific rotation decreases from +111° to +52.5° with time and that of β-D-glucose increases from +19.2° to +52.5°.

Teacher's Note:
1. Both anomers end at the same value (+52.5°) because they interconvert through the open chain form and reach the same equilibrium mixture.

Question 11. What is invert sugar ? (CBSE 2010)
Answer:
An equimolar mixture of glucose and fructose, obtained by hydrolysis of sucrose in presence of an acid such as dil. HCl or the enzyme invertase or sucrase, is called invert sugar.

Teacher's Note:
1. The mixture is laevorotatory because fructose (−92.4°) rotates more strongly than glucose (+52.5°).

Question 12. What is inversion of sugar ?
Answer:
Sucrose is dextrorotatory but on hydrolysis it gives an equimolar mixture of D-(+)-glucose and D-(−)-fructose which is laevorotatory. This change of specific rotation from dextrorotation to laevorotation is called inversion of sugar.

Teacher's Note:
1. The word “inversion” refers to the change in the sign of rotation, not to any change of configuration.

Question 13. What is glycosidic linkage? (CBSE 2013)
Answer:
A linkage between two monosaccharide units through oxygen is called glycosidic linkage.

Teacher's Note:
1. Glycosidic linkage in carbohydrates plays the same role as the peptide linkage in proteins and the phosphodiester linkage in nucleic acids.

Question 14. Name the monosaccharide units in maltose. (Jharkhand Board 2011)
Answer:
Two α-D-glucose units.

Teacher's Note:
1. The two units are joined by a C₁–C₄ α-glycosidic linkage; the free C-1 of the second unit makes maltose a reducing sugar.

Question 15. Name the branch chain component of starch. (CBSE Sample Paper 2018-19)
Answer:
Amylopectin, it contains linear C₁–C₄ α-linkages and C₁–C₆ α-linkages which constitute the branch chains.

Teacher's Note:
1. The C₁–C₆ linkage is the one responsible for branching; glycogen has the same kind of structure but is more highly branched.

Question 16. How are carbohydrates stored in the animal body ? (Manipur Board 2011)
Answer:
Glucose is stored as glycogen in the liver and muscles of human body.

Teacher's Note:
1. Glycogen is called animal starch; plants store carbohydrate as starch.

Question 17. Name two carbohydrates which act as biofuels.
Answer:
Starch in plants and glycogen in animals.

Teacher's Note:
1. Both are storage polysaccharides that are broken down to glucose when the organism needs energy.

Short Answer Questions (2 Marks)

Teacher's Note:
1. Both are needed in very small amounts; the key difference is the source – hormones are made inside the body, vitamins mostly come from food.

Question 19. Carbohydrates are classified into mono-saccharides, oligosaccharides and polysaccharides.
(i) What is the basis of such classification?
(ii) Give an example of the oligosaccharide. (Kerala Board 2012)
Answer:
(i) The basis of this classification is the behaviour of carbohydrates on hydrolysis, i.e., the number of monosaccharide units obtained on hydrolysis : monosaccharides cannot be hydrolysed, oligosaccharides give 2–10 and polysaccharides give a large number of monosaccharide units.
(ii) Sucrose (a disaccharide) : C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ (Glucose) + C₆H₁₂O₆ (Fructose).

Teacher's Note:
1. Raffinose (trisaccharide) is another example; any disaccharide counts as an oligosaccharide.

Question 20. What are reducing sugars. Give an example each of a reducing sugar and a non-reducing sugar.
Answer:
Carbohydrates which reduce Fehling’s solution to red ppt. of Cu₂O and Tollens’ reagent to metallic silver are called reducing sugars. They contain a free aldehydic or ketonic group (free hemiacetal or hemiketal group). All monosaccharides and disaccharides except sucrose are reducing sugars.
Example : Reducing sugar – glucose (or maltose, lactose) ; Non-reducing sugar – sucrose.

Teacher's Note:
1. Sucrose is non-reducing because the reducing groups of both glucose (C-1) and fructose (C-2) are used up in the glycosidic bond.

Question 21. How will you obtain from glucose : (i) gluconic acid (ii) n-hexane? (Raj. Board 2012)
Answer:
(i) By oxidation of glucose with bromine water :
CHO–(CHOH)₄–CH₂OH \( \xrightarrow{\text{Br}_2\text{ water}} \) COOH–(CHOH)₄–CH₂OH (Gluconic acid)
(ii) By prolonged heating of glucose with HI :
CHO–(CHOH)₄–CH₂OH \( \xrightarrow{\text{HI},\ \Delta} \) CH₃CH₂CH₂CH₂CH₂CH₃ (n-Hexane)

Teacher's Note:
1. Name the reagent clearly in each step; “oxidation” alone is not enough because conc. HNO₃ would give saccharic acid instead.

Question 22. Give chemical reactions to show glucose :
(a) all the six carbon atoms are linked in a straight chain.
(b) one aldehyde group is present. (Raj. Board 2011; Karnataka Board 2012)
Answer:
(a) On prolonged heating with HI, glucose gives n-hexane, showing that all the six carbon atoms are linked in a straight chain.
CHO–(CHOH)₄–CH₂OH \( \xrightarrow{\text{HI},\ \Delta} \) CH₃CH₂CH₂CH₂CH₂CH₃
(b) Glucose on oxidation with a mild oxidising agent like bromine water gives gluconic acid having the same number of carbon atoms. This shows that the carbonyl group is present as an aldehyde group.
CHO–(CHOH)₄–CH₂OH \( \xrightarrow{\text{Br}_2\text{ water}} \) COOH–(CHOH)₄–CH₂OH

Teacher's Note:
1. Reaction with NH₂OH (oxime) or HCN only proves a carbonyl group; oxidation to a six-carbon acid proves that it is an aldehyde.

Question 23. Draw simple Fischer projections of D-glucose and L-glucose. Can these be enantiomers ?
Answer:

Yes, D-glucose and L-glucose are non-superimposable mirror images of each other (the configuration at every chiral carbon is reversed), hence these are enantiomers.

Teacher's Note:
1. D/L is decided only by the –OH on the lowest chiral carbon (C-5): on the right = D, on the left = L.

Question 24. Explain mutarotation taking D-glucose as an example.
Answer:
The spontaneous change in specific rotation of an optically active compound with time to an equilibrium value is called mutarotation.
When α-D-glucose (specific rotation +111°) is dissolved in water, its specific rotation gradually decreases and becomes constant at +52.5°. Similarly, the specific rotation of a freshly prepared solution of β-D-glucose (+19.2°) gradually increases to +52.5°. This is because in solution both the forms change into each other through the open chain form till an equilibrium is established :
α-D-Glucose (+111°) ⇌ Open chain form ⇌ β-D-Glucose (+19.2°)
The equilibrium mixture contains about 36% α-form, 64% β-form and only about 0.02% open chain form.

Teacher's Note:
1. Mutarotation is possible only for sugars with a free hemiacetal –OH; methyl glucosides and sucrose do not mutarotate.

Question 25. What is essentially the difference between α-glucose and β-glucose ? (CBSE 2011, 2012)
Answer:
α-Glucose and β-glucose differ only in the configuration of the hydroxyl group at C-1 (the anomeric carbon). In α-glucose, the –OH group at C-1 is on the same side as the ring oxygen in Fischer projection (below the ring in Haworth projection), while in β-glucose it is on the opposite side (above the ring in Haworth projection). Such isomers are called anomers.
They also differ in physical properties : α-glucose, m.p. 419 K, [α]_D = +111° ; β-glucose, m.p. 423 K, [α]_D = +19.2°.

Teacher's Note:
1. Anomers differ only at C-1; epimers (e.g., glucose and mannose) differ at any one other chiral carbon.

Question 26. Draw the structures of α-D-glucose and β-D-glucose? (J & K Board 2012)
Answer:
The Haworth (pyranose) structures are :

Teacher's Note:
1. In the Haworth projection, groups on the right in the Fischer projection are drawn below the ring; the α-anomer has the C-1 –OH down, trans to the CH₂OH group.

Question 27. What is meant by pyranose structure of glucose? (CBSE 2012)
Answer:
The six-membered cyclic structure of glucose containing five carbon atoms and one oxygen atom in the ring is called the pyranose structure, because it is similar to the six-membered heterocyclic compound pyran. It is formed when the –OH group at C-5 reacts with the –CHO group at C-1 to form a cyclic hemiacetal. Depending on the position of –OH at C-1, the two pyranose forms are α-D-glucopyranose and β-D-glucopyranose.

Teacher's Note:
1. Pyranose = six-membered ring (like pyran); furanose = five-membered ring (like furan), as in fructose.

Question 28. What is a monosaccharide? Write the Haworth’s structure of α-D-(+)-glucose. (Karnataka Board 2012)
Answer:
A carbohydrate which cannot be hydrolysed further to give simpler units of polyhydroxy aldehyde or ketone is called a monosaccharide, e.g., glucose, fructose, ribose.
Haworth structure of α-D-(+)-glucose (shown on the left) :

Teacher's Note:
1. In the α-form the anomeric –OH (C-1) is on the opposite side of the ring to the CH₂OH group.

Question 29. Represent the pyranose structures of α- and β-isomers of glucose in the form of Haworth structures.
Answer:

The two structures differ only in the orientation of the –H and –OH groups at C-1.

Teacher's Note:
1. Check your drawing: ring O at the back right, CH₂OH up on C-5, OH at C-2 down, C-3 up, C-4 down.

Question 30. How does fructose differ from glucose ? Draw their open chain structures.
Answer:
Glucose is an aldohexose (it contains an aldehyde group at C-1) while fructose is a ketohexose (it contains a keto group at C-2). Both have the molecular formula C₆H₁₂O₆ and the same configuration at C-3, C-4 and C-5. Glucose is dextrorotatory while fructose is laevorotatory. Glucose forms a six-membered (pyranose) ring while fructose forms a five-membered (furanose) ring.

Teacher's Note:
1. Since C-3 to C-6 are identical, glucose and fructose give the same osazone.

Short Answer Questions (3 Marks)

Question 31. (i) Fructose contains a keto group but still it reduces Tollens’ reagent. Explain.
(ii) Give the chemical name and sources of: (a) Vitamin C   (b) Vitamin B₁
Answer:
(i) Tollens’ reagent is alkaline. In the presence of base, fructose (an α-hydroxy ketone) undergoes rearrangement through an enediol intermediate to form glucose and mannose, which contain a free –CHO group. These reduce Tollens’ reagent to metallic silver.
(ii) (a) Vitamin C: ascorbic acid; sources – citrus fruits (lemon, orange), amla and green leafy vegetables.
(b) Vitamin B₁: thiamine; sources – yeast, milk, green vegetables and cereals.

Teacher's Note:
1. The same rearrangement is why fructose also reduces Fehling’s solution, even though ketones normally do not.

Question 32. Glucose does not give 2, 4-DNP test, Schiff's test or does not give sodium bisulphite adduct. Give reasons.
or Despite having an aldehyde group, glucose does not give 2, 4-DNP test? What does this indicate?
Answer:
Although the open chain structure of glucose does contain an aldehyde group, yet it does not give certain characteristic reactions of the –CHO group. For example, it neither reacts with 2, 4-dinitrophenylhydrazine to form a 2, 4-DNP derivative nor it gives Schiff's test nor it reacts with a concentrated solution of sodium bisulphite to give sodium bisulphite adduct.
The reason being that glucose actually exists in the cyclic hemiacetal form with only a small amount (< 0.05%) of the open chain form. Since the concentration of the open chain form is low and its reactions with 2, 4-DNP, NaHSO₃ and Schiff's reagent are reversible, therefore, formation of 2, 4-DNP derivative or NaHSO₃ adduct or observation of Schiff's test cannot disturb the equilibrium to regenerate more of the open chain form from the cyclic hemiacetal form and hence does not give these tests.

Teacher's Note:
1. The same three failed tests are the standard evidence for the cyclic structure of glucose – learn them as a set.

Question 33. Glucose and fructose give the same osazone? Explain.
Answer:
During osazone formation, the reaction occurs only at C₁ and C₂ while rest of the molecule remains intact. Since glucose and fructose differ from each other only in the arrangement of atoms at C₁ and C₂, therefore, they give the same osazone.
Glucose: CHO–CHOH–(CHOH)₃–CH₂OH   \( \xrightarrow[-C_6H_5NH_2,\ -NH_3,\ -2H_2O]{3\,C_6H_5NHNH_2} \)   CH=NNHC₆H₅–C(=NNHC₆H₅)–(CHOH)₃–CH₂OH (Glucosazone or Fructosazone)
Fructose: CH₂OH–CO–(CHOH)₃–CH₂OH   \( \xrightarrow[-C_6H_5NH_2,\ -NH_3,\ -2H_2O]{3\,C_6H_5NHNH_2} \)   the same osazone
The (CHOH)₃–CH₂OH portion (C₃ to C₆) is identical in glucose, fructose and the osazone.

Teacher's Note:
1. Three molecules of phenylhydrazine are used: one oxidises C-2 (or C-1 in fructose) and two form the hydrazone groups.

Question 34. Name the two components of starch. How do they differ from each other structurally?
Answer:
Starch consists of two components, (i) amylose and (ii) amylopectin. Although, both these components are made up of α-D-glucose units yet they differ in the manner in which the different α-D-glucose units are linked to one another.
Amylose is a linear condensation polymer of α-D-glucose in which C₁ of one glucose unit is attached to C₄ of the other through α-glycosidic linkage.
Amylopectin, on the other hand, is a highly branched polymer. It consists of a large number (several hundreds) of short chains each containing 20-25 glucose units which are joined together through α-glycosidic linkages involving C₁ of one glucose unit with C₄ of the other. The C₁ of terminal glucose unit in each chain is further linked to C₆ of some other glucose unit in the next chain through C₁–C₆ α-glycosidic linkage. This gives amylopectin a highly branched structure.

Teacher's Note:
1. Amylose is water soluble (15-20% of starch); amylopectin is insoluble (80-85%). Branching comes only from the C1–C6 links.

Teacher's Note:
1. Cysteine and methionine are the sulphur-containing amino acids; two cysteine –SH groups can join to form a disulphide bridge.

Question 36. Give reasons for the following : (i) Amino acids have high melting points and are soluble in water. (ii) What is meant by the secondary structure of proteins?
Answer:
(i) α-Amino acids contain both an acidic carboxyl group and a basic amino group. These two groups neutralize each other. As a result, α-amino acids largely exist as dipolar ions or zwitterions.
NH₂–CH(R)–COOH ⇄ H₃N⁺–CH(R)–COO⁻ (Zwitterion)
Due to dipole-dipole interactions, the molecules of α-amino acids are held together by strong forces of attraction and hence amino acids have high melting points. Further, due to strong interactions between the +ve and −ve poles of α-amino acids with polar water molecules, α-amino acids are moderately soluble in water.
(ii) Secondary structure of proteins refers to the conformation or the shape which the polypeptide chains assume as a result of H-bonding. There are two types of secondary structures of proteins. These are α-helix and β-pleated sheet structure.

Teacher's Note:
1. Amino acids behave like salts (ionic solids), not like ordinary organic acids or amines – that is the key to their high melting points.

Question 37. Towards which electrode would an α-amino acid migrate in an electric field at (i) pH < pI, (ii) pH > pI and (iii) pH = pI? Explain.
Answer:
The general formula of an amino acid is
H₃N⁺–CH(R)–COOH (I) \( \underset{H^+}{\overset{OH^-}{\rightleftharpoons}} \) H₃N⁺–CH(R)–COO⁻ (Zwitterion) \( \underset{H^+}{\overset{OH^-}{\rightleftharpoons}} \) H₂N–CH(R)–COO⁻ (II)
(i) When pH < pI, i.e., the solution is acidic, the cation (I) predominates and therefore, the amino acid migrates towards cathode.
(ii) When pH > pI, i.e., the solution is basic the anion (II) predominates, therefore, the acid migrates towards anode.
(iii) When pH = pI, the amino acid predominantly exists as a zwitterion. That is there is no net charge on the amino acid and hence it does not migrate.

Teacher's Note:
1. Cations go to the cathode and anions to the anode; decide the charge first from whether pH is below or above pI.

Question 38. Define enzymes? What is the most important reason for their specific action? (Uttarakhand Board 2012)
Answer:
Most of the reactions in living systems proceed with very high rates under physiological pH of 7.4 and temperature of 37°C (310 K) under one atmosphere pressure. All these reactions are catalysed by a group of biomolecules called enzymes. Thus, enzymes are biological catalysts. Chemically, they are globular (water soluble) proteins which have high molecular mass ranging from 15,000 to 1,000,000 g mol⁻¹.
Specificity. Enzymes are highly specific in their actions. This specificity is due to the presence of active sites of definite size and shape on their surfaces so that only specific substrates can fit into them. This specific binding leads to the formation of enzyme-substrate complex which accounts for the high specificity of enzymes.

Teacher's Note:
1. Use the lock-and-key picture: the active site is the lock and only the matching substrate is the key.

Question 39. Write the major classes in which the carbohydrates are divided depending upon whether these undergo hydrolysis and if so, on the number of products formed.
Answer:
Depending upon their behaviour on hydrolysis, carbohydrates are divided into three major classes :
(i) Monosaccharides. Carbohydrates which cannot be hydrolysed further to give simpler units of polyhydroxy aldehydes or ketones are called monosaccharides, e.g., glucose, fructose, ribose, etc.
(ii) Oligosaccharides. Carbohydrates which on hydrolysis give two to ten monosaccharide units are called oligosaccharides. These are further classified as disaccharides (e.g., sucrose, maltose, lactose), trisaccharides (e.g., raffinose), tetrasaccharides (e.g., stachyose), etc. depending upon the number of monosaccharide units obtained.
(iii) Polysaccharides. Carbohydrates which on hydrolysis give a large number of monosaccharide units are called polysaccharides, e.g., starch, cellulose, glycogen, etc.

Teacher's Note:
1. The basis of this classification is hydrolysis; the number of units obtained on hydrolysis decides the class.

Question 40. Write chemical equations for the reactions of glucose with
or What happens (write chemical equations) when D-Glucose is treated with the following reagents ?
(i) Acetic anhydride (CBSE 2019) (ii) Ammoniacal silver nitrate solution (iii) NH₂OH (iv) conc. HNO₃ (CBSE 2008, 2012) (v) HI (CBSE 2008, 2010 S) (vi) Bromine water (CBSE 2008, 2010 S, 2019) (vii) HCN. (CBSE 2019)
Answer:
(i) CHO–(CHOH)₄–CH₂OH + 5(CH₃CO)₂O \( \xrightarrow{\Delta} \) CHO–(CHOCOCH₃)₄–CH₂OCOCH₃ (Glucose pentaacetate) + 5CH₃COOH
(ii) CHO–(CHOH)₄–CH₂OH + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → COO⁻–(CHOH)₄–CH₂OH (Gluconate ion) + 2Ag↓ (Silver mirror) + 4NH₃ + 2H₂O
(iii) CHO–(CHOH)₄–CH₂OH + NH₂OH → CH(=NOH)–(CHOH)₄–CH₂OH (Glucose oxime) + H₂O
(iv) CHO–(CHOH)₄–CH₂OH \( \xrightarrow{\text{conc. HNO}_3,\ [O]} \) COOH–(CHOH)₄–COOH (Saccharic acid)
(v) CHO–(CHOH)₄–CH₂OH \( \xrightarrow{\text{HI},\ \Delta} \) CH₃–CH₂–CH₂–CH₂–CH₂–CH₃ (n-Hexane)
(vi) CHO–(CHOH)₄–CH₂OH \( \xrightarrow{\text{Br}_2\text{ water},\ [O]} \) COOH–(CHOH)₄–CH₂OH (Gluconic acid)
(vii) CHO–(CHOH)₄–CH₂OH + HCN → CH(OH)(CN)–(CHOH)₄–CH₂OH (Glucose cyanohydrin)

Teacher's Note:
1. Bromine water (mild) oxidises only –CHO to –COOH, while nitric acid (strong) oxidises both –CHO and the terminal –CH₂OH.

Long Answer Questions (5 Marks)

Question 41. What are carbohydrates ? How are they classified ? What are their important functions ?
Answer:
Carbohydrates are optically active polyhydroxy aldehydes or ketones or the compounds which produce such units on hydrolysis. Most of them have the general formula C_x(H₂O)_y.
Classification (on the basis of behaviour on hydrolysis) :
(i) Monosaccharides : cannot be hydrolysed further into simpler polyhydroxy aldehydes or ketones, e.g., glucose, fructose, ribose. They are further classified as aldoses (containing –CHO, e.g., glucose, an aldohexose) and ketoses (containing >C=O, e.g., fructose, a ketohexose), and as trioses, tetroses, pentoses, hexoses according to the number of carbon atoms.
(ii) Oligosaccharides : give two to ten monosaccharide units on hydrolysis, e.g., disaccharides (sucrose, maltose, lactose), trisaccharides (raffinose), tetrasaccharides (stachyose).
(iii) Polysaccharides : give a large number of monosaccharide units on hydrolysis, e.g., starch, cellulose, glycogen.
On the basis of taste and reducing character, they are also classified as sugars (sweet, crystalline, water-soluble : mono- and oligosaccharides) and non-sugars (polysaccharides), and as reducing sugars (all monosaccharides, maltose, lactose) and non-reducing sugars (sucrose).
Important functions :
(i) They are the major source of energy for plants and animals; glucose is oxidised in the body to give energy.
(ii) Starch (plants) and glycogen (animals) act as reserve food.
(iii) Cellulose forms the cell walls of plants; it is used in the form of wood, cotton and paper.
(iv) Ribose and deoxyribose are constituents of nucleic acids.
(v) Carbohydrates are raw materials for many industries such as textile, paper, lacquer and brewery; honey is used in Ayurvedic medicines.

Teacher's Note:
1. In a long answer, give the definition, all three classes with examples, and at least four functions.

Question 42. What are disaccharides ? Name two naturally occurring polysaccharides of α-D-glucose. Discuss briefly their structures.
Answer:
Carbohydrates which on hydrolysis give two molecules of the same or different monosaccharides are called disaccharides, e.g., sucrose (glucose + fructose), maltose (glucose + glucose) and lactose (galactose + glucose). The two monosaccharide units are joined through a glycosidic linkage.
Two naturally occurring polysaccharides of α-D-glucose are starch and glycogen.
Starch is the main storage polysaccharide of plants. It is a polymer of α-glucose and consists of two components :
(a) Amylose (15–20%) is water soluble. It is a long unbranched chain of 200–1000 α-D-(+)-glucose units held by C₁–C₄ glycosidic linkages.

(b) Amylopectin (80–85%) is insoluble in water. It is a branched chain polymer of α-D-glucose units in which the chain is formed by C₁–C₄ glycosidic linkages whereas branching occurs by C₁–C₆ glycosidic linkages.

Glycogen is the storage polysaccharide of animals (animal starch), present in liver, muscles and brain. Its structure is similar to amylopectin (C₁–C₄ linkages in the chain, C₁–C₆ linkages at branch points) but it is more highly branched. When the body needs glucose, enzymes break glycogen down to glucose.

Teacher's Note:
1. Cellulose is also a glucose polymer, but of β-D-glucose, so it is not a correct answer here.