CBSE Class 12 Chemistry Biomolecules Important Questions Set 01

Very Short Answer Questions (1 Mark)

Question 1. Why are carbohydrates generally optically active?
Answer:
Carbohydrates have chiral or asymmetric carbon atom.

Teacher's Note:
1. A chiral carbon carries four different groups; glucose has four such carbons in its open-chain form.

Question 2. What happens when glucose is treated with bromine water? [CBSE (F) 2010]
Answer:
When glucose is treated with bromine water it forms gluconic acid.
HOCH₂–(CHOH)₄–CHO \( \xrightarrow{\text{Br}_2\text{/H}_2\text{O}} \) HOCH₂–(CHOH)₄–COOH (Gluconic acid)

Teacher's Note:
1. Bromine water is a mild oxidant: only –CHO is oxidised to –COOH; this proves the carbonyl group is an aldehyde.

Question 3. What happens when glucose reacts with nitric acid? [CBSE (F) 2010]
Answer:
Glucose gets oxidised to saccharic acid.
HOCH₂–(CHOH)₄–CHO \( \xrightarrow{\text{HNO}_3} \) HOOC–(CHOH)₄–COOH (Saccharic acid)

Teacher's Note:
1. HNO₃ oxidises BOTH ends (–CHO and –CH₂OH) to give the dicarboxylic saccharic (glucaric) acid; Br₂ water oxidises only –CHO.

Question 4. Write a reaction which shows that all the carbon atoms in glucose are linked in a straight chain. [CBSE (AI) 2012]
Answer:
On prolonged heating with HI, glucose gives n-hexane.
OHC–(CHOH)₄–CH₂OH \( \xrightarrow[\Delta]{\text{HI}} \) CH₃CH₂CH₂CH₂CH₂CH₃ (n-Hexane)

Teacher's Note:
1. Formation of n-hexane (not a branched hexane) is the proof of an unbranched six-carbon chain.

Question 5. Which component of starch is a branched polymer of α-glucose and insoluble in water? [CBSE Delhi 2014]
Answer:
Amylopectin

Teacher's Note:
1. Amylopectin (80–85% of starch) is branched through C1–C6 linkages.

Question 6. Which of the two components of starch is water soluble? [CBSE Delhi 2014]
Answer:
Amylose is water soluble whereas amylopectin is water insoluble component.

Teacher's Note:
1. Amylose is the unbranched C1–C4 linked chain and gives a blue colour with iodine.

Question 7. Name two α-amino acids which form a dipeptide which is 100 times more sweet than cane sugar. [NCERT Exemplar]
Answer:
Aspartic acid and phenylalanine.

Teacher's Note:
1. The dipeptide (as its methyl ester) is aspartame, an artificial sweetener.

Question 8. What type of bonding helps in stabilising the α-helix structure of proteins? [CBSE Delhi 2013]
Answer:
The α-helix structure of proteins is stabilised by intramolecular H-bonding between C=O of one amino acid residue and the N–H of the fourth amino acid residue in the chain.

Teacher's Note:
1. In an α-helix the hydrogen bonds are within the same chain; in a β-pleated sheet they are between neighbouring chains.

Question 9. How do you explain the presence of five –OH groups in glucose molecule? [NCERT Exemplar] [HOTS]
Answer:
Glucose gives pentaacetate derivative on acetylation with acetic anhydride. This confirms the presence of five –OH groups.
OHC–(CHOH)₄–CH₂OH \( \xrightarrow{(\text{CH}_3\text{CO})_2\text{O}} \) OHC–(CH–O–CO–CH₃)₄–CH₂–O–CO–CH₃ (Glucose pentaacetate)

Teacher's Note:
1. Five –OH groups on five different carbons: the pentaacetate is stable because no carbon carries two –OH groups.

Question 10. What are biocatalysts? Give an example. [CBSE (F) 2014]
Answer:
Enzymes are termed as biocatalysts as they catalyse numerous reactions that occur in the bodies of animals and plants to maintain life process, e.g., invertase, pepsin, urease.

Teacher's Note:
1. Enzymes are mostly globular proteins and are highly specific – each one acts on a particular substrate.

Question 11. What are three types of RNA molecules which perform different functions? [CBSE Delhi 2013]
Answer:
There are three types of RNAs:
(i) Ribosomal RNA (rRNA)
(ii) Messenger RNA (mRNA)
(iii) Transfer RNA (tRNA)

Teacher's Note:
1. mRNA carries the code, tRNA brings amino acids, and rRNA forms the ribosome where proteins are made.

Question 12. Name the components of starch. [CBSE Delhi 2013]
Answer:
There are two components of starch, i.e., amylose and amylopectin.

Teacher's Note:
1. Amylose is about 15–20% of starch and is unbranched; amylopectin (80–85%) is branched.

Question 13. Write the products of hydrolysis of maltose? [CBSE OD 2014]
Answer:
2 molecules of Glucose.

Teacher's Note:
1. Maltose is made of two α-D-glucose units joined by a C1–C4 glycosidic link, so both products are glucose.

Question 14. What are the products of hydrolysis of sucrose? [CBSE Delhi 2014]
Answer:
Fructose and Glucose are the products of hydrolysis of sucrose.

Teacher's Note:
1. The equimolar mixture formed is called invert sugar because the sign of rotation changes from (+) to (−).

Question 15. What are the products of hydrolysis of lactose? [CBSE OD 2013; OD 2014]
Answer:
It gives β-D-galactose and β-D-glucose.

Teacher's Note:
1. Lactose (milk sugar) is the one common disaccharide that contains galactose – do not write two glucose units.

Teacher's Note:
1. Most enzyme names tell you the substrate: malt-ase acts on maltose, ure-ase on urea; invertase is named after the invert sugar it produces.

Question 17. Write the linkage joining amino acids? [CBSE OD 2013]
Answer:
Peptide bond.

Teacher's Note:
1. A peptide bond is an amide link, –CO–NH–, formed between the –COOH of one amino acid and the –NH₂ of the next with loss of water.

Short Answer Questions (2 Marks)

Question 18. Write the reactions involved when D-glucose is treated with the following reagents:
(i) HCN
(ii) Br₂ water [CBSE (F) 2013]
Answer:
(i) OHC–(CHOH)₄–CH₂OH \( \xrightarrow{\text{HCN}} \) NC–CH(OH)–(CHOH)₄–CH₂OH (Glucose cyanohydrin)
(ii) OHC–(CHOH)₄–CH₂OH \( \xrightarrow{\text{Br}_2\text{ water}} \) HOOC–(CHOH)₄–CH₂OH (Gluconic acid)

Teacher's Note:
1. HCN adds across C=O (proves a carbonyl group); Br₂ water oxidises –CHO to –COOH (proves it is an aldehyde).

Question 19. How do you explain the presence of an aldehydic group in a glucose molecule? [NCERT Exemplar]
Answer:
Glucose reacts with hydroxylamine to form a monoxime and adds one molecule of hydrogen cyanide to give cyanohydrin so it contains a carbonyl group which can be an aldehyde or a ketone. On mild oxidation with bromine water, glucose gives gluconic acid which is a six carbon carboxylic acid. This indicates that carbonyl group present in glucose is an aldehydic group.

Teacher's Note:
1. Write the argument in two steps: first prove a C=O group, then prove it is –CHO by oxidation without loss of carbon.

Question 20. Label the glucose and fructose units in the following disaccharide and identify anomeric carbon atoms in these units. Is the sugar reducing in nature? Explain. [NCERT Exemplar] [HOTS]

Answer:
C-1 of glucose unit and C-2 of fructose unit are anomeric carbon atoms in the given disaccharide. The disaccharide is non-reducing sugar because –OH groups attached to anomeric carbon atoms are involved in the formation of glycosidic bond.

Teacher's Note:
1. The six-membered ring is the glucose (pyranose) unit and the five-membered ring is the fructose (furanose) unit; the anomeric carbon is the one bonded to the ring O and to the glycosidic O.

Question 21. Amino acids behave like salts rather than simple amines or carboxylic acids. Explain. [NCERT Exemplar]
Answer:
In aqueous solution, the carboxyl group loses a proton and amino group accepts a proton to form a zwitter ion.
R–CH(NH₂)–COOH ⇄ R–CH(N⁺H₃)–COO⁻ (Zwitter ion)

Teacher's Note:
1. Because of the dipolar salt-like structure, amino acids are crystalline, water-soluble and have high melting points, like salts.

Question 22. Explain what is meant by the following:
(i) peptide linkage
(ii) pyranose structure of glucose. [CBSE (AI) 2011; (F) 2011]
Answer:
(i) A peptide linkage is an amide (–CO–NH–) linkage formed between –COOH group of one α-amino acid and –NH₂ group of other α-amino acid by loss of a water molecule.
(ii) The six-membered cyclic structure of glucose is called pyranose structure (α- or β-), in analogy with heterocyclic compound pyran.

Teacher's Note:
1. Pyranose = six-membered ring with five C atoms and one O atom, like pyran; fructose instead forms a five-membered furanose ring.

Question 23. Describe what you understand by primary structure and secondary structure of proteins. [CBSE Delhi 2011; (F) 2011]
Answer:
Primary structure: The specific sequence in which the various α-amino acids present in a protein are linked to one another is called its primary structure. Any change in the primary structure creates a different protein.
Secondary structure: The conformation which the polypeptide chain assumes as a result of hydrogen bonding is known as secondary structure. The two types of secondary structures are α-helix and β-pleated sheet structures.
In α-helix structure, the polypeptide chain forms all the possible hydrogen bonds by twisting into a right-handed screw (helix) with the –NH group of each amino acid residue hydrogen bonded to the >C=O groups of an adjacent turn of the helix. In β-pleated structure, all peptide chains are stretched out to nearly maximum extension and then laid side by side and are held together by hydrogen bonds.

Teacher's Note:
1. Primary structure is held by covalent peptide bonds, secondary by hydrogen bonds – mention the type of bond for full marks.

Question 24. (i) What type of linkage is present in nucleic acids?
(ii) Give one example each for fibrous protein and globular protein. [CBSE Central 2016]
Answer:
(i) Phosphodiester linkage.
(ii) Fibrous protein: Myosin, keratin, collagen, etc.
Globular protein: Insulin, haemoglobin, etc.

Teacher's Note:
1. Do not mix up the linkages: glycosidic in carbohydrates, peptide in proteins, phosphodiester in nucleic acids.

Question 25. Name the bases present in RNA. Which one of these is not present in DNA? [CBSE Delhi 2011, 2020 (56/3/3)]
Answer:
The bases present in RNA are adenine (A), guanine (G), cytosine (C) and uracil (U). Uracil is not present in DNA.

Teacher's Note:
1. DNA has thymine in place of uracil; A, G and C are common to both.

Question 26. If one strand of a DNA has the sequence –ATGCTTCA–, what is the sequence of the bases in the complementary strand?
Answer:
As we know that in DNA molecule, adenine (A) always pairs with thymine (T) and cytosine (C) always pairs with guanine (G). Thus,
Sequence of bases in one strand: A T G C T T C A
Sequence of bases in the complementary strand: T A C G A A G T

Teacher's Note:
1. Write the partner base under each base, one by one; then count – both strands must have the same number of bases.

Question 27. Write any two reactions of glucose which cannot be explained by the open chain structure of glucose molecule. [CBSE Delhi 2012]
Answer:
Reactions of glucose that cannot be explained by open chain structure:
(i) Glucose gives the characteristic reaction of alcohols and carbonyl group (aldehydes and ketones). However, it does not form addition compound with ammonia and sodium bisulphite and does not respond to 2,4-DNP test and Schiff’s test.
(ii) The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free –CHO group.

Teacher's Note:
1. All these failures point the same way: the –CHO group is tied up in a cyclic hemiacetal, which is why the ring structure was proposed.
2. Also remember the existence of two anomers (α and β) as another fact the open chain cannot explain.

Question 28. What is meant by:
(i) Peptide linkage,
(ii) Biocatalysts. [CBSE Delhi 2012]
Answer:
(i) Peptide linkage: Proteins are complex nitrogenous polymeric substances present in all forms of living matter. These are obtained by the condensation polymerisation of α-amino acids through the formation of peptide bonds.
H₂N–CH(R)–COOH + H–NH–CH(R)–COOH \( \xrightarrow{-H_2O} \) H₂N–CH(R)–CO–NH–CH(R)–COOH (–CO–NH– = peptide linkage)
(ii) Biocatalyst: Enzymes are complex nitrogenous substances (proteins) having molecular weight of 10,000 or even more. Enzymes increase the rate of biochemical reactions by providing alternative path of lower activation energy. Therefore, enzymes are termed as biochemical catalyst and phenomenon is known as biochemical catalysis.

Teacher's Note:
1. When asked to explain the peptide linkage, always show the –CO–NH– group in the equation; the drawing usually carries the mark.
2. Enzymes do not change the equilibrium or ΔG of a reaction – they only lower the activation energy.

Teacher's Note:
1. The sugar difference is the safest ‘main structural difference’: DNA’s 2-deoxyribose lacks the –OH at C-2′.

Short Answer Questions (3 Marks)

Question 30. Define the following terms:
(i) Glycosidic linkage
(ii) Invert sugar
(iii) Oligosaccharides [CBSE (AI) 2014] [CBSE 2020 (56/3/2)]
Answer:
(i) The linkage between two monosaccharides through oxygen atom in an oligosaccharide or a polysaccharide is known as glycosidic linkage.
(ii) Sucrose is dextrorotatory (+66.5°) but after 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 and the mixture obtained is called invert sugar.
(iii) Carbohydrates which on hydrolysis give two to ten molecules of monosaccharides are called oligosaccharides, e.g., sucrose.

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

Question 31. (i) Which of the following biomolecules is insoluble in water? Justify.
Insulin, Haemoglobin, Keratin.
(ii) Draw the Haworth structure for α-D-glucopyranose.
(iii) Write chemical reaction to show that glucose contains aldehyde as carbonyl group. [CBSE Sample Paper 2015]
Answer:
(i) Keratin, being a fibrous protein, is insoluble in water.
(ii)

(iii) Glucose reduces Tollens’ reagent to metallic silver.
OHC–(CHOH)₄–CH₂OH + 2[Ag(NH₃)₂]⁺ (Tollens’ reagent) + 3OH⁻ → ⁻OOC–(CHOH)₄–CH₂OH (Gluconate ion) + 2Ag↓ (Silver mirror) + 4NH₃ + 2H₂O

Teacher's Note:
1. Insulin and haemoglobin are globular (water-soluble); keratin of hair and nails is fibrous and insoluble.
2. In the Haworth structure of α-D-glucopyranose, the C1 –OH points down and the CH₂OH at C5 points up.

Question 32. (i) Give one structural difference between amylose and amylopectin.
(ii) Name the protein and its shape present in oxygen carrier in human body.
(iii) Name two fat storing tissues in human body. [CBSE Sample Paper 2014]
Answer:
(i) Amylose is a long unbranched chain polymer of α-D-(+)-glucose.
Amylopectin is a branched chain polymer of α-D-glucose.
(ii) Haemoglobin; it is a globular protein and its shape is spherical.
(iii) Liver and adipose tissue.

Teacher's Note:
1. Amylose has only C1–C4 links; amylopectin also has C1–C6 links at the branch points.

Question 33. (i) A non-reducing disaccharide ‘A’ on hydrolysis with dilute acid gives an equimolar mixture of D-(+)-glucose and D-(−)-fructose.
A + H₂O \( \xrightarrow{\text{HCl}} \) C₆H₁₂O₆ + C₆H₁₂O₆
[α]_D = +66.5°    +52.5°    −92.4°
Identify A. What is the mixture of D-(+)-glucose and D-(−)-fructose known as? Name the linkage that holds the two units in the disaccharide.
(ii) α-Amino acids have relatively higher melting points than the corresponding halo acids. Explain. [CBSE Sample Paper 2016] [HOTS]
Answer:
(i) A = C₁₂H₂₂O₁₁ (sucrose).
Invert sugar.
Glycosidic linkage.
(ii) α-Amino acids act as zwitter ions (H₃N⁺–CHR–COO⁻) or dipolar ions. Due to this dipolar salt-like structure, they have strong dipole-dipole interactions. Therefore, their melting points are higher than the corresponding halo acids which do not exist as zwitter ions.

Teacher's Note:
1. The clue ‘non-reducing’ + ‘glucose and fructose’ points straight to sucrose.

Question 34. (i) Draw the pyranose structure of glucose.
(ii) What type of linkage is present in proteins?
(iii) Give one example each for water-soluble vitamins and fat-soluble vitamins. [CBSE (F) 2016]
Answer:
(i)

(ii) Peptide linkage (–CO–NH–)
(iii) Vitamin C and B group vitamins are water soluble vitamins.
Vitamins A, D, E and K are fat soluble vitamins.

Teacher's Note:
1. α- and β-glucopyranose differ only in the position of –OH at C1 (down in α, up in β).

Question 35. (i) Write the product obtained when D-glucose reacts with H₂N–OH.
(ii) Amino acids show amphoteric behaviour. Why?
(iii) Why cannot vitamin C be stored in our body? [CBSE Patna 2015]
Answer:
(i) OHC–(CHOH)₄–CH₂OH + NH₂OH → HON=CH–(CHOH)₄–CH₂OH (Glucose oxime) + H₂O
(ii) In aqueous solution, amino acids exist as a zwitter ion. In zwitter ionic form, amino acids show amphoteric behaviour as they react both with acids and bases.
H₃N⁺–CH(R)–COOH (I) \( \underset{\text{H}^+}{\overset{\text{OH}^-}{\rightleftharpoons}} \) H₃N⁺–CH(R)–COO⁻ (Zwitter ion) \( \underset{\text{H}^+}{\overset{\text{OH}^-}{\rightleftharpoons}} \) H₂N–CH(R)–COO⁻ (II)
In acidic medium COO⁻ ion of the zwitter ion accepts a proton to form the cation (I) while in basic medium N⁺H₃ loses a proton to form the anion (II).
(iii) As vitamin C is water soluble, therefore, it is readily excreted in urine and hence cannot be stored in the body.

Teacher's Note:
1. In acid an amino acid is a cation, in base it is an anion – check the net charge on each form you write.

Question 36. Give three reactions of glucose which cannot be explained by its chain structure. [CBSE SQP 2018-2019]
Answer:
(i) Glucose does not give 2,4-DNP test.
(ii) Glucose does not give Schiff’s test.
(iii) The pentaacetate of glucose does not react with hydroxylamine.
(iv) Glucose does not form the hydrogensulphite addition product with NaHSO₃.
(Any three points)

Teacher's Note:
1. Each point is a test for a free –CHO group that glucose fails, because in solution almost all of it is in the cyclic hemiacetal form.

Question 37. How will you distinguish 1° and 2° hydroxyl groups present in glucose? Explain with reaction. [NCERT Exemplar]
Answer:
Glucose on treatment with acetic anhydride in presence of pyridine or a few drops of conc. H₂SO₄. It forms penta-acetyl derivative indicating the presence of 5-OH groups. Out of which, one –OH group is primary (1°) alcoholic and four (C₂, C₃, C₄ and C₅) –OH groups are secondary (2°) alcoholic groups.
OHC–(CHOH)₄–CH₂OH + 5(CH₃CO)₂O \( \xrightarrow[\text{Acetylation}]{\text{Pyridine}} \) OHC–(CH–O–CO–CH₃)₄–CH₂–O–CO–CH₃ (Glucose pentaacetate)
Glucose (or gluconic acid) on oxidation with HNO₃ gives saccharic acid (a dicarboxylic acid) indicating that one of the primary (1°) alcoholic group is oxidized to –COOH group but secondary (2°) hydroxyl groups undergo oxidation only in drastic conditions.
OHC–(CHOH)₄–CH₂OH (Glucose) \( \xrightarrow[\text{HNO}_3]{\text{Oxidation}} \) HOOC–(CHOH)₄–COOH (Saccharic acid, dicarboxylic acid)
HOOC–(CHOH)₄–CH₂OH (Gluconic acid, monocarboxylic acid) \( \xrightarrow[\text{HNO}_3]{\text{Oxidation}} \) HOOC–(CHOH)₄–COOH (Saccharic acid)

Teacher's Note:
1. Acetylation only counts the –OH groups; it is the HNO₃ oxidation of the terminal CH₂OH to –COOH that proves one of them is primary.
2. Do not mix the oxidants: bromine water stops at gluconic acid (only –CHO oxidised), nitric acid goes on to saccharic acid.

Question 38. Define the following as related to proteins:
(i) Peptide linkage, (ii) Primary structure, (iii) Denaturation [CBSE OD 2015; OD 2014]
Answer:
(i) Peptide linkage: Proteins are complex nitrogenous polymeric substances present in all forms of living matter. These are obtained by the condensation polymerisation of α-amino acids through the formation of peptide bonds.
H₂N–CH(R)–COOH + H–NH–CH(R)–COOH \( \xrightarrow{-H_2O} \) H₂N–CH(R)–CO–NH–CH(R)–COOH (–CO–NH– = peptide linkage)
(ii) Primary structure: each polypeptide in a protein molecule having amino acids which are linked with each other in a specific sequence.
(iii) Denaturation: When a protein is subjected to physical change like change in temperature or chemical change like change in pH, protein loses its biological activity.

Teacher's Note:
1. During denaturation the 2° and 3° structures are destroyed but the primary structure (the peptide bonds) remains intact.
2. Coagulation of egg white on boiling is the standard example of denaturation.

Long Answer Questions (5 Marks)

Question 39. (i) (a) What is the difference between native protein and denatured protein?
(b) Which one of the following is a disaccharide?
Glucose, Lactose, Amylose, Fructose
(c) Write the name of the vitamin responsible for the coagulation of blood. [CBSE 2019 (56/4/3)]
(ii) Define the following terms:
(a) Native protein [CBSE 2020 (56/3/2)]
(b) Nucleotide [CBSE 2019 (56/2/2)]
Answer:
(i) (a) Protein found in a biological system with unique three-dimensional structure and biological activity is called native protein.
When a protein in its native form is subjected to change such as change in temperature, change in pH, its 2° and 3° structures are destroyed and it loses its biological activity. The protein thus formed is called denatured protein.
(b) Lactose
(c) Vitamin K.
(ii) (a) Protein found in biological system with unique three dimensional structure and biological activity is called native protein.
(b) A unit formed by the combination of nitrogenous base, pentose sugar and phosphate.

Teacher's Note:
1. Coagulation of egg white on boiling is the standard example of denaturation; the primary structure remains intact.
2. Amylose is a polysaccharide and glucose and fructose are monosaccharides, so lactose is the only disaccharide.