CBSE Class 12 Chemistry Biomolecules Important Questions Set 05

Very Short Answer Questions (1 Mark)

Question 1. What is the importance of amino acids to us ?
Answer:
They are the building blocks of proteins which are essential for the growth and maintenance of life.

Teacher's Note:
1. Ten of the amino acids are essential amino acids; our body cannot make them, so they must come from the diet.

Question 2. How do amino acids form proteins ?
Answer:
α-Amino acids undergo condensation reaction between –NH₂ group of one acid and –COOH group of the other with elimination of H₂O molecule.

Teacher's Note:
1. The –CO–NH– bond so formed is the peptide bond; a long chain of such units is a polypeptide.

Question 3. In what respect would any two naturally occurring amino acids differ from one another ?
Answer:
All the naturally occurring α-amino acids differ from one another in the nature of the side chain groups R. For example, in alanine, the side chain is CH₃ group and in phenylalanine it is CH₂C₆H₅ group.

Teacher's Note:
1. All α-amino acids share the H₂N–CH(R)–COOH backbone; only R changes.

Question 4. What are zwitterions ? Give one example. (Assam Board 2012; Maharashtra Board 2012; HP Board 2013)
Or Write the zwitterion structure of glycine.
Or Name the species formed when an amino acid is dissolved in water. (CBSE Sample Paper 2019-2020)
Answer:
A zwitterion is a dipolar ion formed by neutralization of acidic and basic centres present within the molecule.
For example, all aminoacids and sulphanilic acid exist as zwitterions.
H₃N⁺–CH₂–COO⁻ (Glycine)
H₃N⁺–C₆H₄–SO₃⁻ (p-, Sulphanilic acid)

Teacher's Note:
1. In the zwitterion the proton moves from –COOH to –NH₂; this explains the high melting points and water solubility of amino acids.

Question 5. What is a peptide bond ? (Hr. Board 2011; CBSE 2011, 2012)
Answer:
The covalent bond –NH–CO– formed between –NH₂ group of one amino acid and –COOH group of the other with elimination of a molecule of H₂O is called a peptide bond.

Teacher's Note:
1. Chemically a peptide bond is an amide linkage; write it as –CO–NH–.

Question 6. What is isoelectric point ?
Answer:
The pH at which there is no net migration of the amino acid under the influence of an applied electric field is called isoelectric point. For example, the isoelectric point of glycine is 6.1, that of aspartic acid is 3.0 and that of lysine is 9.7.

Teacher's Note:
1. At the isoelectric point the amino acid is present mainly as the zwitterion and its solubility is minimum.

Question 7. Give one example each for fibrous protein and globular protein. (CBSE 2016)
Answer:
(i) Fibrous protein : Keratin in skin, hair, nails and wool.
(ii) Globular protein : Haemoglobin which transports oxygen from the blood to muscles.

Teacher's Note:
1. Fibrous proteins (keratin, myosin, collagen) are insoluble in water; globular proteins (insulin, albumins, haemoglobin) are usually water soluble.

Question 8. What happens when protein is denatured ? (CBSE 2010, 2018)
Answer:
When a protein in the native form is subjected to change of temperature, pH, etc. H-bonds are broken and the biological activity of the protein is lost. Such proteins are called denatured proteins. During denaturation, the secondary and tertiary structures are destroyed but primary structure remains intact.
The most common example of denaturation is the coagulation of white of an egg on boiling. When soluble globular protein albumin is converted into insoluble fibrous proteins. Another example is curdling of milk which is caused due to formation of lactic acid (from lactose present in milk) by the bacteria present in milk. Yet another example is the formation of cheese from milk when it is heated with an acid such as lemon juice or tartaric acid. During this denaturation, the globular milk protein becomes fibrous.

Teacher's Note:
1. Key exam point: denaturation leaves the primary structure (sequence of amino acids, peptide bonds) intact.

Question 9. Give the biological functions of the following proteins :
(i) Haemoglobin (ii) Collagen (Karnataka Board 2012)
Answer:
(i) Haemoglobin present in blood transports oxygen from lungs to all the tissues of the body and CO₂ from the tissues to the lungs for exhale.
(ii) Collagen is a fibrous protein present in connecting tissues of tendons, cartilages and bones.

Teacher's Note:
1. Collagen is the most abundant protein in the human body and gives strength to connective tissue.

Question 10. Give one example each of essential and non-essential amino acids.
Answer:
Essential amino acid – valine or phenylalanine
Non-essential amino acid — glycine or alanine.

Teacher's Note:
1. Essential amino acids cannot be synthesised in the body and must be taken in food; non-essential ones are made by the body itself.

Question 11. What are the ultimate products of digestion of proteins ?
Answer:
α-Amino acids.

Teacher's Note:
1. Proteins pass through proteoses, peptones and polypeptides before finally giving α-amino acids.

Question 12. What is a prosthetic group ?
Answer:
A prosthetic group is a non-protein portion obtained by hydrolysis of conjugated proteins. The main function of the prosthetic group is to control the biological functions of proteins.

Teacher's Note:
1. Example: haem is the prosthetic group of haemoglobin (a conjugated protein).

Question 13. Write the name of the linkage joining two amino acids. (CBSE 2013)
Answer:
Peptide bond, —CO—NH—.

Teacher's Note:
1. Writing the structure –CO–NH– along with the name makes the answer complete.

Question 14. How many peptide linkages are present in a tripeptide ? (CBSE Sample Paper 2018-19)
Answer:
A dipeptide contains one peptide bond, therefore, a tripeptide will contain two peptide bonds.

Teacher's Note:
1. General rule: a peptide made of n amino acids has (n − 1) peptide bonds.

Question 15. What is a polypeptide ? Give one example.
Answer:
Polypeptides are formed when several molecules of α-amino acids are joined together by peptide bonds. For example, oxytocin is a nanopeptide, i.e., contains 9 α-amino acid units while insulin contains 51 α-amino acid units.

Teacher's Note:
1. A polypeptide with more than about 100 amino acid residues (molecular mass above 10,000 u) is usually called a protein.

Question 16. What type of bonding occurs in globular proteins ?
Answer:
van der Waals interactions, disulphide bridges, dipolar interactions (ionic) and hydrogen bonding.

Teacher's Note:
1. These are the forces that fold the chain into its tertiary (globular) shape.

Question 17. What is the secondary structure of proteins ?
Answer:
Conformation which a polypeptide chain assumes as a result of H-bonding is called the secondary structure of protein. There are two types of secondary structures : α-helix and β-pleated sheet structure. For example, α-keratin in hair, nail, wool, etc. has α-helix structure while silk protein fibroin has β-pleated sheet structure.

Teacher's Note:
1. Primary structure = sequence of amino acids; secondary structure = shape due to H-bonding between >C=O and >N–H of peptide groups.

Question 18. What forces are responsible for stability of α-helix ?
Answer:
α-Helix structure is stabilized by intramolecular H-bonding between N—H of one α-amino acid residue and C = O of the fourth amino acid residue within the polypeptide chain.

Teacher's Note:
1. In an α-helix the H-bonds run within the same chain, roughly parallel to the axis of the helix.

Short Answer Questions (2 Marks)

Question 19. What are ketoses ? Explain the difference between α-D-(−)-fructose and β-D-(−)-fructose.
Answer:
Monosaccharides which contain a keto group (>C=O) are called ketoses, e.g., fructose.
Fructose exists in two cyclic (furanose) forms, α-D-(−)-fructofuranose and β-D-(−)-fructofuranose, obtained by addition of the –OH at C-5 to the >C=O group. They differ only in the configuration at C-2 (anomeric carbon) : in the α-form the –OH at C-2 is below the ring and in the β-form it is above the ring (Haworth structures).

Teacher's Note:
1. For fructose the anomeric carbon is C-2, not C-1, because the carbonyl group is at C-2.

Question 20. Draw the structure of lactose. (Hr. Board 2013)
Answer:
Lactose (milk sugar) is composed of β-D-galactose and β-D-glucose. The linkage is between C-1 of galactose and C-4 of glucose (C₁–C₄ β-glycosidic linkage).

Teacher's Note:
1. Galactose differs from glucose only at C-4 (OH up); the free C-1 of the glucose unit makes lactose a reducing sugar.

Question 21. Explain the following giving one example :
(i) Invert sugar (CBSE 2011) (ii) Glycosidic linkage (CBSE 2009, 2011; Hr. Board 2012)
Answer:
(i) Invert sugar : Sucrose is dextrorotatory but on hydrolysis it gives an equimolar mixture of D-(+)-glucose (+52.5°) and D-(−)-fructose (−92.4°) which is laevorotatory. This mixture is called invert sugar, since the sign of rotation changes from dextro (+) to laevo (−).
C₁₂H₂₂O₁₁ (Sucrose) + H₂O \( \xrightarrow{\text{H}^+ \text{ or Invertase}} \) C₆H₁₂O₆ (D-Glucose) + C₆H₁₂O₆ (D-Fructose)
(ii) Glycosidic linkage : The linkage between two monosaccharide units through an oxygen atom, formed by loss of a molecule of water, is called glycosidic linkage. For example, in maltose two α-D-glucose units are joined by a C₁–C₄ glycosidic linkage.

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

Question 22. Draw the structure of amylopectin. (Hr. Board 2011)
Answer:
Amylopectin is a branched chain polymer of α-D-glucose units. The main chain is formed by C₁–C₄ α-glycosidic linkages and branching occurs by C₁–C₆ α-glycosidic linkages.

Teacher's Note:
1. The branch point is the C₁–C₆ linkage, where the CH₂OH of a chain unit becomes CH₂–O–.

Question 23. Draw the structure of amylose. (Hr. Board 2013)
Answer:
Amylose is a long unbranched chain of 200–1000 α-D-(+)-glucose units held by C₁–C₄ α-glycosidic linkages.

Teacher's Note:
1. All links in amylose are α(1→4); there are no C₁–C₆ branches.

Question 24. Give one structural difference between amylose and amylopectin.
Answer:
Amylose is a linear (unbranched) polymer of α-D-glucose units joined only by C₁–C₄ glycosidic linkages, whereas amylopectin is a branched polymer in which the chain is formed by C₁–C₄ glycosidic linkages and branching occurs by C₁–C₆ glycosidic linkages.

Teacher's Note:
1. As a result, amylose is water soluble while amylopectin is insoluble.

Question 25. Amylopectin and glycogen both have branched structures. How do they differ from one another ?
Answer:
Both amylopectin and glycogen are branched polymers of α-D-glucose having C₁–C₄ linkages in the chains and C₁–C₆ linkages at the branch points. However, glycogen is more highly branched than amylopectin; in glycogen the branches occur after every 8–12 glucose units whereas in amylopectin after about 20–25 glucose units. Amylopectin is a component of plant starch while glycogen is stored in animals (liver, muscles and brain).

Teacher's Note:
1. Glycogen is called animal starch; the extra branching gives many chain ends for rapid release of glucose.

Question 26. Mention two structural differences between amylopectin and cellulose.
Answer:
(i) Amylopectin is made up of α-D-glucose units whereas cellulose is made up of β-D-glucose units.
(ii) Amylopectin is a branched chain polymer (C₁–C₄ linkages in the chain and C₁–C₆ linkages at branch points) whereas cellulose is a straight chain polymer with only C₁–C₄ β-glycosidic linkages.

Teacher's Note:
1. The two points to remember are the type of glucose (α or β) and branched versus straight chain.

Question 27. What are α-amino acids ? How are they related to proteins ? Give the structures of two α-amino acids.
Answer:
Amino acids which contain the amino group (–NH₂) on the α-carbon atom, i.e., the carbon next to the carboxyl group, are called α-amino acids. Their general formula is R–CH(NH₂)–COOH.
Proteins are condensation polymers of α-amino acids; α-amino acids are joined together by peptide (–CO–NH–) linkages to form proteins, and proteins on complete hydrolysis give α-amino acids.
Examples : Glycine, H₂N–CH₂–COOH ; Alanine, CH₃–CH(NH₂)–COOH.

Teacher's Note:
1. Except glycine (R = H), all naturally occurring α-amino acids are optically active and have the L-configuration.

Question 28. Illustrate the formation of a peptide bond by taking a suitable example. (Manipur Board 2011)
Answer:
When the –COOH group of one α-amino acid condenses with the –NH₂ group of another α-amino acid with the loss of a molecule of water, a peptide bond (–CO–NH–) is formed. For example, glycine and alanine combine to form the dipeptide glycylalanine :
H₂N–CH₂–COOH (Glycine) + H₂N–CH(CH₃)–COOH (Alanine) \( \xrightarrow{-\text{H}_2\text{O}} \) H₂N–CH₂–CO–NH–CH(CH₃)–COOH (Glycylalanine, Gly-Ala)

Teacher's Note:
1. The amino acid whose –COOH is used is named first (glycyl-), the one with the free –COOH is named last (-alanine).

Question 29. What is peptide linkage? How is dipeptide formed. (Maharashtra Board 2012)
Answer:
The amide linkage (–CO–NH–) formed between the –COOH group of one α-amino acid and the –NH₂ group of another α-amino acid with the loss of a molecule of water is called peptide linkage.
A dipeptide is formed when two α-amino acids combine through one peptide linkage. For example,
H₂N–CH₂–COOH (Glycine) + H₂N–CH(CH₃)–COOH (Alanine) \( \xrightarrow{-\text{H}_2\text{O}} \) H₂N–CH₂–CO–NH–CH(CH₃)–COOH (Glycylalanine, Gly-Ala)

Teacher's Note:
1. A dipeptide still has a free –NH₂ and a free –COOH group, so the chain can grow further at both ends.

Question 30. What is a peptide bond? Cite one example of a molecule in which peptide bond is present. (West Bengal Board 2012)
Answer:
The amide bond (–CO–NH–) formed between the –COOH group of one α-amino acid and the –NH₂ group of another α-amino acid with the elimination of a molecule of water is called a peptide bond.
Example : Peptide bonds are present in all peptides and proteins, e.g., glycylalanine, H₂N–CH₂–CO–NH–CH(CH₃)–COOH, and insulin.

Teacher's Note:
1. Any dipeptide, polypeptide or protein is a correct example; write the –CO–NH– group clearly.

Short Answer Questions (3 Marks)

Question 31. Indicating linkage involved draw the structure of sucrose. Also name the linkage. (Hr. Board 2011, 2013)
Answer:
Sucrose is formed by the condensation of α-D-glucose and β-D-fructose. The two units are linked through the C-1 of α-D-glucose and C-2 of β-D-fructose. The linkage is a glycosidic linkage (C₁–C₂ glycosidic linkage).

Teacher's Note:
1. Since both reducing groups are involved in the linkage, sucrose is a non-reducing sugar.

Question 32. Which polysaccharides make up starch ? What is the difference between them?
Answer:
Starch is a polymer of α-glucose and consists of two components — amylose and amylopectin.
Amylose is the water-soluble component which constitutes about 15–20% of starch. It is a long unbranched chain of 200–1000 α-D-(+)-glucose units held together by C₁–C₄ glycosidic linkages.
Amylopectin is insoluble in water and constitutes about 80–85% of starch. 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.

Teacher's Note:
1. Amylose gives a deep blue colour with iodine; amylopectin gives a red-violet colour.

Question 33. Write the structure of cellulose ? How does it differ from amylose ?
Answer:
Cellulose is a straight chain polysaccharide composed only of β-D-glucose units which are joined by glycosidic linkages between C-1 of one glucose unit and C-4 of the next glucose unit.

Difference : Amylose is made up of α-D-glucose units joined by C₁–C₄ α-glycosidic linkages, whereas cellulose is made up of β-D-glucose units joined by C₁–C₄ β-glycosidic linkages. Amylose is water soluble and is digested by humans; cellulose is insoluble and cannot be digested by humans.

Teacher's Note:
1. Humans lack the enzyme to break β-glycosidic linkages, which is why cellulose is not digested by us though cattle can digest it.

Question 34. What are carbohydrates ? Explain their important functions.
Answer:
Carbohydrates are optically active polyhydroxy aldehydes or ketones or the compounds which produce such units on hydrolysis.
Important functions :
(i) Carbohydrates are the major source of energy for plants and animals. Glucose is oxidised in the body to provide energy.
(ii) Starch in plants and glycogen in animals act as stored food (reserve carbohydrates).
(iii) Cellulose forms the cell walls of plants and bacteria, and is used to make wood, paper, cotton fibre, etc.
(iv) Honey (a mixture of glucose and fructose) is used as an instant source of energy and in Ayurvedic medicines.
(v) Ribose and deoxyribose are present in nucleic acids (RNA and DNA).
(vi) Carbohydrates are used as raw materials in many industries such as textile, paper, lacquer and brewery.

Teacher's Note:
1. When asked for functions, mention at least one energy role, one storage role and one structural role.

Question 35. What are carbohydrates ? How do they act as source of energy ?
Answer:
Carbohydrates are optically active polyhydroxy aldehydes or ketones or the compounds which produce such units on hydrolysis, e.g., glucose, fructose, sucrose, starch.
Carbohydrates such as starch and sucrose in our food are hydrolysed during digestion to glucose. Glucose is absorbed into the blood and is oxidised in the cells through a series of enzyme-catalysed reactions to CO₂ and H₂O, releasing a large amount of energy :
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O ; ΔH = −2880 kJ mol⁻¹
Excess glucose is stored as glycogen in the liver and muscles and is hydrolysed back to glucose when the body needs energy. Similarly, plants store energy as starch.

Teacher's Note:
1. The energy released in respiration is stored in the cells as ATP.

Question 36. What are proteins ? Name the compounds obtained on partial and complete hydrolysis of proteins. (Hr. Board 2011)
Answer:
Proteins are complex nitrogenous organic compounds of high molecular mass (more than 10,000 u) which are polymers of α-amino acids joined together by peptide linkages. They are essential for the growth and maintenance of life.
On partial hydrolysis, proteins give proteoses, peptones and polypeptides (peptides). On complete hydrolysis, they give α-amino acids :
Proteins → Proteoses → Peptones → Polypeptides → Simple peptides → α-Amino acids

Teacher's Note:
1. The end products of complete hydrolysis (or digestion) of proteins are always α-amino acids.

Question 37. Amino acids may be acidic, alkaline or neutral. How does this happen ? What are essential and non-essential amino acids ? Name one of each type. (CBSE 2010)
Answer:
Amino acids are classified as acidic, basic or neutral depending upon the relative number of amino and carboxyl groups in their molecule. If the number of amino and carboxyl groups is equal, the amino acid is neutral (e.g., glycine, alanine). If the number of amino groups is more than the carboxyl groups, it is basic (e.g., lysine). If the number of carboxyl groups is more than the amino groups, it is acidic (e.g., aspartic acid, glutamic acid).
Amino acids which cannot be synthesised in the body and must be obtained through diet are called essential amino acids, e.g., valine. Amino acids which can be synthesised in the body are called non-essential amino acids, e.g., glycine.

Teacher's Note:
1. Lysine has an extra –NH₂ in its side chain (basic), aspartic acid an extra –COOH (acidic).

Question 38. What are amino acids ? Describe their zwitterion structure. What do you know about the isoelectric point of amino acids ?
Answer:
Compounds which contain both an amino group (–NH₂) and a carboxyl group (–COOH) are called amino acids. α-Amino acids have the general formula R–CH(NH₂)–COOH.
Zwitterion structure : In aqueous solution, the carboxyl group loses a proton and the amino group accepts it, forming a dipolar ion called a zwitterion :
R–CH(NH₂)–COOH ⇌ H₃N⁺–CH(R)–COO⁻ (Zwitterion)
Because of the dipolar salt-like character, amino acids are crystalline solids with high melting points, soluble in water and amphoteric in nature (the zwitterion reacts with both acids and bases).
Isoelectric point : The pH at which the amino acid exists mainly as the zwitterion and shows no net migration towards either electrode in an electric field is called its isoelectric point. For example, the isoelectric point of glycine is 6.1. At the isoelectric point, the amino acid has the least solubility in water.

Teacher's Note:
1. In acid solution the amino acid exists as the cation (moves to cathode), in alkaline solution as the anion (moves to anode).

Question 39. What is meant by the following :
(i) Peptide linkage (Hr. Board 2011, 2012; CBSE 2008, 2009, 2011, 2012; West Bengal Board 2012)
(ii) Glycosidic linkage. (CBSE 2009, 2011; Hr. Board 2012)
Answer:
(i) Peptide linkage : The amide linkage (–CO–NH–) formed between the –COOH group of one α-amino acid and the –NH₂ group of another α-amino acid with the loss of a molecule of water is called peptide linkage. For example,
H₂N–CH₂–COOH (Glycine) + H₂N–CH(CH₃)–COOH (Alanine) \( \xrightarrow{-\text{H}_2\text{O}} \) H₂N–CH₂–CO–NH–CH(CH₃)–COOH (Glycylalanine, Gly-Ala)
(ii) Glycosidic linkage : The linkage between two monosaccharide units through an oxygen atom, formed with the loss of a molecule of water, is called glycosidic linkage. For example, in maltose two α-D-glucose units are joined through a C₁–C₄ glycosidic linkage.

Teacher's Note:
1. Peptide links join amino acids in proteins; glycosidic links join monosaccharides in carbohydrates.

Long Answer Questions (5 Marks)

Question 40. What are α-amino acids? How do they build proteins ? Discuss briefly their chemical properties.
Answer:
Amino acids which contain the amino group (–NH₂) on the α-carbon atom with respect to the carboxyl group are called α-amino acids, R–CH(NH₂)–COOH. They differ in the side chain R, e.g., glycine (R = H), alanine (R = CH₃).
Formation of proteins : The –COOH group of one α-amino acid condenses with the –NH₂ group of another with the loss of a molecule of water to form a peptide bond (–CO–NH–). Thus,
H₂N–CH₂–COOH (Glycine) + H₂N–CH(CH₃)–COOH (Alanine) \( \xrightarrow{-\text{H}_2\text{O}} \) H₂N–CH₂–CO–NH–CH(CH₃)–COOH (Glycylalanine, Gly-Ala)
The dipeptide still has free –NH₂ and –COOH groups, so it can combine with more amino acids to form tri-, tetra- and polypeptides. Polypeptides containing more than 100 amino acid residues (molecular mass > 10,000 u) are called proteins.
Chemical properties :
(i) Zwitterion formation and amphoteric nature : In aqueous solution, amino acids exist as dipolar ions (zwitterions), H₃N⁺–CH(R)–COO⁻. Therefore, they react with both acids and bases : with acids the –COO⁻ accepts a proton to give the cation, and with bases the –NH₃⁺ loses a proton to give the anion.
R–CH(NH₃⁺)–COOH \( \xleftarrow{\text{H}^+} \) H₃N⁺–CH(R)–COO⁻ \( \xrightarrow{\text{OH}^-} \) R–CH(NH₂)–COO⁻
(ii) Isoelectric point : At a particular pH, called the isoelectric point, the amino acid exists as the zwitterion and does not migrate in an electric field.
(iii) Peptide bond formation : They condense with each other to form peptides and proteins (as shown above).
(iv) Due to their dipolar salt-like character, amino acids are colourless crystalline solids with high melting points, soluble in water. Except glycine, all naturally occurring α-amino acids are optically active (L-configuration).

Teacher's Note:
1. Zwitterion formation explains most of the “unusual” properties of amino acids: high m.p., water solubility and amphoteric behaviour.