Intext Questions
1. Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain.
Answer
Glucose has five –OH groups and sucrose eight; these form many hydrogen bonds with water molecules, so they dissolve readily. Cyclohexane and benzene are non-polar hydrocarbons with no groups able to hydrogen-bond with water, so they do not dissolve.
2. What are the expected products of hydrolysis of lactose?
Answer
Lactose (milk sugar) is a disaccharide of β-D-galactose and β-D-glucose. On hydrolysis (dil. acid or the enzyme lactase) it gives one molecule each of D-galactose and D-glucose: C₁₂H₂₂O₁₁ + H₂O
⇒ C₆H₁₂O₆ (D-galactose) + C₆H₁₂O₆ (D-glucose).
3. How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
Answer
Glucose pentaacetate does not react with hydroxylamine, so it has no free –CHO group. In the cyclic (pyranose) form of glucose, the aldehyde group has combined with the C-5 –OH to form a hemiacetal; this cyclic form can open into the open chain (–CHO) form only through the free –OH at C-1. In the pentaacetate the C-1 –OH is also acetylated, so the ring cannot open and no aldehyde group is generated. This supports the cyclic structure of glucose.
4. The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.
Answer
Amino acids exist as dipolar ions (zwitterions), H₃N⁺–CHR–COO⁻, because the acidic –COOH gives its proton to the basic –NH₂. Strong electrostatic attractions between these ions give them salt-like character, hence high melting points; their ionic nature also makes them highly soluble in water. Halo acids have no such ionic form and are held only by weaker dipole–dipole forces and hydrogen bonds.
5. Where does the water present in the egg go after boiling the egg?
Answer
On boiling, the globular egg-white proteins (albumin) are denatured and coagulate. The water is trapped (held by hydrogen bonding/absorbed) within the network of the coagulated, denatured protein, so the egg becomes solid without losing it.
6. Why cannot vitamin C be stored in our body?
Answer
Vitamin C is water-soluble. Excess of it is readily excreted in urine and is not stored in body tissues (unlike fat-soluble vitamins, which are stored in liver and adipose tissue). Hence it must be supplied regularly through diet.
7. What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Answer
Complete hydrolysis gives the three components: the base thymine, the sugar 2-deoxy-D-ribose and phosphoric acid (H₃PO₄).
8. When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Answer
In DNA, base pairing makes A = T and G = C in amount. The lack of any such relation in RNA shows that RNA has no complementary base pairing throughout; it is a single-stranded molecule (not a double helix).
Exercises
1. What are monosaccharides?
Answer
Carbohydrates that cannot be hydrolysed into simpler polyhydroxy aldehydes or ketones. Examples: glucose, fructose, ribose. About 20 occur in nature; they are classified by the number of C atoms and the carbonyl group (aldose or ketose).
2. What are reducing sugars?
Answer
Carbohydrates that reduce Fehling's solution and Tollens' reagent. They have a free aldehyde or ketone group (or a free hemiacetal/hemiketal carbon). All monosaccharides and disaccharides such as maltose and lactose are reducing sugars; sucrose is non-reducing.
3. Write two main functions of carbohydrates in plants.
Answer
(i) Structural material: cellulose forms the cell walls of plant cells (and wood, cotton fibre). (ii) Storage of food/energy: starch is the main reserve polysaccharide stored in seeds, tubers, etc.
4. Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Answer
Monosaccharides: ribose, 2-deoxyribose, galactose, fructose.
Disaccharides: maltose, lactose.
5. What do you understand by the term glycosidic linkage?
Answer
The oxide (ether-type) linkage –O– that joins two monosaccharide units through an oxygen atom, formed by loss of a water molecule between the hemiacetal –OH of one sugar and an –OH of the other. E.g. in sucrose, C-1 of α-D-glucose is joined to C-2 of β-D-fructose by a glycosidic linkage.
6. What is glycogen? How is it different from starch?
Answer
Glycogen is the storage polysaccharide of animals (found in liver, muscles, brain), also called animal starch; it is broken down to glucose when the body needs it. It is a polymer of α-D-glucose like amylopectin, but more highly branched.
Starch is the storage polysaccharide of plants and is a mixture of amylose (15–20%, linear, C1–C4 links, water soluble) and amylopectin (80–85%, branched, with C1–C6 branch points every 20–25 glucose units). Glycogen is a single, much more branched polymer, so the difference is in source and degree of branching.
7. What are the hydrolysis products of (i) sucrose and (ii) lactose?
Answer
(i) Sucrose + H₂O
⇒ D-(+)-glucose + D-(–)-fructose (equimolar; the mixture is laevorotatory, called invert sugar).
(ii) Lactose + H₂O
⇒ D-(+)-galactose + D-(+)-glucose.
8. What is the basic structural difference between starch and cellulose?
Answer
Starch is made of α-D-glucose units (amylose: C1–C4 α-glycosidic links in a chain; amylopectin: branched with C1–C6 links). Cellulose is an unbranched straight chain of β-D-glucose units joined by C1–C4 β-glycosidic linkages.
9. What happens when D-glucose is treated with the following reagents? (i) HI (ii) Bromine water (iii) HNO₃
Answer
(i) Prolonged heating with HI reduces it to n-hexane, showing six carbons in a straight chain: CHO–(CHOH)₄–CH₂OH —HI, heat
⇒ CH₃–(CH₂)₄–CH₃
(ii) Bromine water (mild oxidant) oxidises –CHO to –COOH, giving gluconic acid: CHO–(CHOH)₄–CH₂OH —Br₂ water
⇒ COOH–(CHOH)₄–CH₂OH
(iii) Nitric acid oxidises both –CHO and the primary alcohol –CH₂OH, giving saccharic (glucaric) acid: CHO–(CHOH)₄–CH₂OH —HNO₃
⇒ COOH–(CHOH)₄–COOH
10. Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
Answer
(i) Although it has an aldehyde group, glucose does not give Schiff's test and does not form the hydrogensulphite addition product with NaHSO₃.
(ii) Glucose pentaacetate does not react with hydroxylamine, showing no free –CHO group.
(iii) Glucose exists in two crystalline forms, α (m.p. 419 K, from concentrated solution at 303 K) and β (m.p. 423 K, from hot saturated solution at 371 K), and shows mutarotation. These are explained by the cyclic hemiacetal (pyranose) structure with C-1 as the anomeric carbon.
11. What are essential and non-essential amino acids? Give two examples of each type.
Answer
Essential amino acids cannot be synthesised by the body and must be taken in the diet. Examples: valine, leucine (also isoleucine, lysine, phenylalanine, threonine, methionine, tryptophan, histidine, arginine).
Non-essential amino acids can be synthesised in the body. Examples: glycine, alanine (also glutamic acid, aspartic acid, serine).
12. Define the following as related to proteins (i) Peptide linkage (ii) Primary structure (iii) Denaturation.
Answer
(i) Peptide linkage: the amide bond –CO–NH– formed between the –COOH group of one α-amino acid and the –NH₂ group of another with loss of water, e.g. glycine + alanine
⇒ glycylalanine (H₂N–CH₂–CO–NH–CH(CH₃)–COOH).
(ii) Primary structure: the specific sequence in which amino acids are linked in each polypeptide chain of a protein. Any change in this sequence gives a different protein.
(iii) Denaturation: loss of biological activity of a protein when its native secondary and tertiary structure is disturbed by heat, change in pH or chemicals; hydrogen bonds break, the helix uncoils and the globules unfold, but the primary structure stays intact. E.g. coagulation of egg white on boiling, curdling of milk.
13. What are the common types of secondary structure of proteins?
Answer
(i) α-Helix: the polypeptide chain coils into a right-handed spiral, held by hydrogen bonds between the –C=O of one amino acid residue and the –NH of the fourth residue along the chain.
(ii) β-Pleated sheet: polypeptide chains lie side by side, almost fully extended, and are joined by intermolecular hydrogen bonds, giving a pleated (folded-sheet) appearance.
14. What type of bonding helps in stabilising the α-helix structure of proteins?
Answer
Intramolecular hydrogen bonding between the C=O group of one peptide unit and the N–H group of the fourth peptide unit along the same chain (–C=O···H–N–).
15. Differentiate between globular and fibrous proteins.
Answer
Fibrous proteins: polypeptide chains run parallel and are held together by hydrogen and disulphide bonds, forming fibre-like structures; generally insoluble in water. Examples: keratin (hair, wool, silk), myosin (muscles).
Globular proteins: chains coil around to give a spherical shape; usually soluble in water. Examples: insulin, albumins.
16. How do you explain the amphoteric behaviour of amino acids?
Answer
In aqueous solution an amino acid exists as a zwitterion, H₃N⁺–CHR–COO⁻, containing both an acidic group (–NH₃⁺, which can give a proton) and a basic group (–COO⁻, which can accept a proton). With acid: H₃N⁺CHRCOO⁻ + H⁺
⇒ H₃N⁺CHRCOOH; with base: H₃N⁺CHRCOO⁻ + OH⁻
⇒ H₂NCHRCOO⁻ + H₂O. Thus it reacts with both acids and bases, i.e. it is amphoteric.
17. What are enzymes?
Answer
Enzymes are biocatalysts, mostly globular proteins, that catalyse biochemical reactions in living systems under mild conditions. They are very specific for a particular reaction and substrate, and lower the activation energy. E.g. maltase converts maltose into glucose; oxidoreductases catalyse oxidation–reduction of one substrate with another.
18. What is the effect of denaturation on the structure of proteins?
Answer
Denaturation disrupts the hydrogen bonds and other forces that hold the secondary and tertiary structures; the helices uncoil and the globules unfold, so the protein loses its specific 3-D shape and biological activity. The primary structure (amino acid sequence) is not affected.
19. How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Answer
On the basis of solubility: (i) fat-soluble vitamins – A, D, E and K (stored in liver and adipose tissue); (ii) water-soluble vitamins – B group and C (must be supplied regularly, as they are excreted in urine; vitamin B₁₂ is an exception that can be stored).
Vitamin K is responsible for coagulation of blood.
20. Why are vitamin A and vitamin C essential to us? Give their important sources.
Answer
Vitamin A: its deficiency causes xerophthalmia (hardening of the cornea) and night blindness. Sources: fish liver oil, carrots, butter and milk.
Vitamin C: its deficiency causes scurvy (bleeding gums). Sources: citrus fruits (amla, oranges, lemons), green leafy vegetables.
21. What are nucleic acids? Mention their two important functions.
Answer
Nucleic acids are long-chain polymers of nucleotides (polynucleotides) present in the nucleus of cells; they are of two types, DNA and RNA.
Functions: (i) DNA is the chemical basis of heredity; it carries genetic information and transmits inherited characters from one generation to the next (it can replicate). (ii) Nucleic acids (DNA and RNA) direct the synthesis of specific proteins in the cell; the message in DNA is carried and translated by RNA.
22. What is the difference between a nucleoside and a nucleotide?
Answer
Nucleoside = base + pentose sugar (base attached at C-1′ of the sugar). Nucleotide = base + pentose sugar + phosphate group (phosphoric acid esterified at the C-5′ position of the sugar of a nucleoside). Nucleotides are the monomers of nucleic acids.
23. The two strands in DNA are not identical but are complementary. Explain.
Answer
In the DNA double helix the two strands are held by hydrogen bonds between specific base pairs: adenine always pairs with thymine (two H-bonds) and guanine with cytosine (three H-bonds). So the base sequence of one strand fixes the sequence of the other (e.g. A-G-C-T on one strand faces T-C-G-A on the other). The strands are therefore not identical but complementary.
24. Write the important structural and functional differences between DNA and RNA.
Answer
Structural: (i) DNA contains 2-deoxy-D-ribose; RNA contains D-ribose. (ii) DNA has bases A, G, C and thymine; RNA has A, G, C and uracil. (iii) DNA is a double-stranded α-helix; RNA is single-stranded. (iv) DNA molecules are much larger (higher molecular mass).
Functional: (i) DNA is the chemical basis of heredity and carries genetic information; it can replicate itself. (ii) RNA does not usually replicate; its main role is protein synthesis, using the message coded in DNA.
25. What are the different types of RNA found in the cell?
Answer
(i) Messenger RNA (m-RNA) – carries the coded message from DNA for protein synthesis. (ii) Ribosomal RNA (r-RNA) – present in ribosomes, where proteins are synthesised. (iii) Transfer RNA (t-RNA) – brings specific amino acids to the ribosome for joining into the protein chain.
Important Extra Questions
What is meant by invert sugar?
Answer
Sucrose is dextrorotatory, but its hydrolysis gives an equimolar mixture of D-(+)-glucose and D-(–)-fructose, which is laevorotatory because fructose (–92.4°) rotates more than glucose (+52.5°). The sign of rotation is inverted, so the mixture is called invert sugar.
What are anomers?
Answer
Cyclic stereoisomers of a sugar that differ only in configuration at the hemiacetal (anomeric) carbon, C-1 for glucose; e.g. α-D-glucose and β-D-glucose.
Why is sucrose a non-reducing sugar?
Answer
Its glycosidic bond joins the reducing groups of both units (C-1 of glucose and C-2 of fructose), so no free hemiacetal/aldehyde or ketone group is available to reduce Tollens' or Fehling's reagent.
What is a zwitterion?
Answer
A dipolar ion carrying both a positive and a negative charge in the same molecule, e.g. H₃N⁺–CH₂–COO⁻ formed by glycine in aqueous solution.
Which bases are purines and which are pyrimidines?
Answer
Purines: adenine and guanine. Pyrimidines: cytosine, thymine (in DNA) and uracil (in RNA).