This MCQ module is based on: NCERT Exercises and Solutions: Biomolecules
This assessment will be based on: NCERT Exercises and Solutions: Biomolecules
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Carbohydrates are optically active polyhydroxy aldehydes or ketones, or molecules which provide such units on hydrolysis. They are broadly classified into three groups — monosaccharides, disaccharides and polysaccharides. Glucose, the most important source of energy for mammals, is obtained by the digestion of starch. Monosaccharides are held together by glycosidic linkages to form disaccharides or polysaccharides.
Proteins are the polymers of about twenty different α-amino acids linked by peptide bonds. Ten amino acids are called essential amino acids because they cannot be synthesised by our body and must be provided through diet. Proteins perform various structural and dynamic functions in organisms. Proteins which contain only α-amino acids are called simple proteins. The secondary or tertiary structure gets disturbed on change of pH or temperature and the protein can no longer perform its function — this is denaturation. Enzymes are biocatalysts which speed up reactions in biosystems; they are very specific and selective in their action, and chemically the majority of enzymes are proteins.
Vitamins are accessory food factors required in the diet, classified as fat soluble (A, D, E and K) and water soluble (B group and C). Deficiency of vitamins leads to many diseases.
Nucleic acids are polymers of nucleotides, which in turn consist of a base, a pentose sugar and a phosphate moiety. They are responsible for the transfer of characters from parents to offspring. DNA contains 2-deoxyribose whereas RNA contains ribose. Both contain adenine, guanine and cytosine; the fourth base is thymine in DNA and uracil in RNA. DNA is double stranded while RNA is a single strand molecule. DNA is the chemical basis of heredity and carries the coded message for proteins to be synthesised in the cell. There are three types of RNA — mRNA, rRNA and tRNA — which actually carry out protein synthesis.
10.1 What are monosaccharides? A monosaccharide is a carbohydrate that cannot be hydrolysed further to give a simpler unit of polyhydroxy aldehyde or ketone. About 20 occur in nature; examples are glucose, fructose and ribose.
10.2 What are reducing sugars? Carbohydrates which reduce Fehling's solution and Tollens' reagent are called reducing sugars. All monosaccharides, whether aldose or ketose, are reducing sugars; among disaccharides, maltose and lactose are reducing while sucrose is not.
10.3 Two main functions of carbohydrates in plants. (i) Storage of energy — carbohydrates are stored as starch, the main storage polysaccharide of plants. (ii) Structural role — cellulose is the predominant constituent of the cell wall of plant cells.
10.4 Classify into monosaccharides and disaccharides. Monosaccharides: ribose, 2-deoxyribose, galactose, fructose. Disaccharides: maltose, lactose.
10.5 Glycosidic linkage. The linkage between two monosaccharide units through an oxygen atom, formed by the loss of a water molecule, is called a glycosidic linkage.
10.6 What is glycogen? How is it different from starch? Glycogen is the polysaccharide in which carbohydrates are stored in the animal body, present in liver, muscles and brain, and also found in yeast and fungi. It is called animal starch because its structure is similar to amylopectin, but it is rather more highly branched. Starch, by contrast, is the plant storage polysaccharide and is a mixture of unbranched amylose (15–20%) and branched amylopectin (80–85%).
10.7 Hydrolysis products. (i) Sucrose gives one molecule of D-(+)-glucose and one of D-(−)-fructose. (ii) Lactose gives one molecule of D-(+)-galactose and one of D-(+)-glucose.
10.8 Basic structural difference between starch and cellulose. Starch is a polymer of α-glucose (amylose with C1–C4 linkages, amylopectin with C1–C4 chains and C1–C6 branches). Cellulose is a straight chain polysaccharide composed only of β-D-glucose units joined by C1–C4 linkages. The α versus β configuration is the essential difference.
10.9 What happens when D-glucose is treated with: (i) HI — on prolonged heating it gives n-hexane, showing that all six carbons are in a straight chain. (ii) Bromine water — the aldehyde group is oxidised to give the six-carbon gluconic acid. (iii) HNO₃ — both the aldehyde and the primary alcoholic group are oxidised to give the dicarboxylic saccharic acid.
10.10 Reactions not explained by the open chain structure. (i) Despite having an aldehyde group, glucose does not give Schiff's test and does not form the hydrogensulphite addition product with NaHSO₃. (ii) The pentaacetate does not react with hydroxylamine, indicating absence of a 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).
10.11 Essential and non-essential amino acids. Essential amino acids cannot be synthesised in the body and must be obtained through diet — examples valine and leucine. Non-essential amino acids can be synthesised in the body — examples glycine and alanine.
10.12 Define as related to proteins. (i) Peptide linkage — the –CO–NH– bond, chemically an amide, formed between the –COOH group of one amino acid and the –NH₂ group of another with elimination of a water molecule. (ii) Primary structure — the specific sequence of amino acids in a polypeptide chain; any change in this sequence creates a different protein. (iii) Denaturation — loss of the unique three-dimensional structure and biological activity of a native protein on change of temperature or pH, in which the secondary and tertiary structures are destroyed but the primary structure remains intact.
10.13 Common types of secondary structure. The α-helix and the β-pleated sheet.
10.14 Bonding that stabilises the α-helix. Hydrogen bonding — specifically, the –NH group of each amino acid residue is hydrogen bonded to the >C=O group of an adjacent turn of the helix.
10.15 Globular versus fibrous proteins. In fibrous proteins the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds to give a fibre-like structure; they are generally insoluble in water — keratin and myosin. In globular proteins the chains coil around to give a spherical shape and are usually soluble in water — insulin and albumins.
10.16 Amphoteric behaviour of amino acids. An amino acid contains both an acidic –COOH and a basic –NH₂ group, so in aqueous solution an internal proton transfer gives the zwitter ion ⁻OOC–CHR–NH₃⁺. This dipolar ion can react with an acid — ⁻OOC–CHR–NH₃⁺ + H⁺ → HOOC–CHR–NH₃⁺ — and with a base — ⁻OOC–CHR–NH₃⁺ + OH⁻ → ⁻OOC–CHR–NH₂ + H₂O. Since it reacts with both, the amino acid is amphoteric.
10.17 What are enzymes? Enzymes are biocatalysts that enable the chemical reactions of living organisms to occur under very mild conditions. Almost all enzymes are globular proteins, they are very specific for a particular reaction and substrate, are required only in small quantities, and act by reducing the magnitude of the activation energy. Their names generally end in -ase.
10.18 Effect of denaturation on the structure of proteins. The hydrogen bonds are disturbed, so globules unfold and the helix uncoils. The secondary and tertiary structures are destroyed while the primary structure remains intact, and the protein loses its biological activity.
10.19 How are vitamins classified? Which vitamin is responsible for coagulation of blood? Vitamins are classified by solubility into fat soluble (A, D, E and K), stored in liver and adipose tissues, and water soluble (B group and C), which are excreted in urine and cannot be stored except B₁₂. The vitamin responsible for blood coagulation is vitamin K — its deficiency causes increased blood clotting time.
10.20 Why are vitamins A and C essential? Sources. Vitamin A is essential because its deficiency causes xerophthalmia (hardening of the cornea of the eye) and night blindness; sources are fish liver oil, carrots, butter and milk. Vitamin C is essential because its deficiency causes scurvy (bleeding gums); sources are citrus fruits, amla and green leafy vegetables. Vitamin C additionally cannot be stored in the body, so it must be supplied regularly.
10.21 What are nucleic acids? Two important functions. Nucleic acids are long chain polymers of nucleotides, also called polynucleotides, present in the nucleus of the cell and responsible for the transmission of hereditary characters. Functions: (i) DNA is the chemical basis of heredity and the reserve of genetic information, capable of self duplication during cell division so that identical strands pass to daughter cells; (ii) nucleic acids direct protein synthesis — proteins are synthesised by RNA molecules, while the message for a particular protein is present in DNA.
10.22 Difference between a nucleoside and a nucleotide. A nucleoside is formed by the attachment of a base to the 1′ position of the sugar. A nucleotide is formed when the nucleoside is linked to phosphoric acid at the 5′ position of the sugar. In short, nucleotide = nucleoside + phosphate.
10.23 The two strands of DNA are not identical but complementary. Explain. The two chains are held together by hydrogen bonds between specific pairs of bases: adenine pairs with thymine and cytosine pairs with guanine. Therefore wherever one strand carries A the other must carry T, and wherever one carries C the other must carry G. The sequences of the two strands are consequently different from each other — not identical — yet each fully determines the other, which is what the word complementary means. Example: if one strand reads A–T–G–C, the other reads T–A–C–G.
10.24 Structural and functional differences between DNA and RNA.
10.25 Types of RNA. There are three: messenger RNA (m-RNA), ribosomal RNA (r-RNA) and transfer RNA (t-RNA). They perform different functions in carrying out protein synthesis in the cell.
| Point of difference | DNA | RNA |
|---|---|---|
| Pentose sugar | β-D-2-deoxyribose | β-D-ribose |
| Fourth base | thymine (with A, G, C) | uracil (with A, G, C) |
| Strandedness | double strand helix | single stranded helix, sometimes folded back |
| Base pairing | complementary A–T and C–G, so base quantities are related | no such relationship among base quantities |
| Occurrence | mainly in the nucleus (chromosomes) | mainly in the cytoplasm; three types m-RNA, r-RNA, t-RNA |
| Function | chemical basis of heredity; reserve of genetic information; self duplicates during cell division | carries out protein synthesis in the cell |
Chapter 10 covers four families of molecules. This audit checks whether you can move between them rather than recalling each in isolation.
| Family | Monomer | Linkage | Lost on joining |
|---|---|---|---|
| Carbohydrates | monosaccharide | glycosidic (through O) | H₂O |
| Proteins | α-amino acid | peptide –CO–NH– | H₂O |
| Nucleic acids | nucleotide | phosphodiester 5′–3′ | H₂O |
Storage and information: starch and glycogen store energy; DNA stores genetic information; proteins carry out function. The common idea is condensation polymerisation — in every family the monomers join with the elimination of a water molecule, and hydrolysis reverses the process.
The deeper idea the chapter keeps returning to is that structure determines function. α-Glucose gives digestible starch while β-glucose gives indigestible cellulose. The sequence of amino acids fixes the fold, and the fold fixes the biological activity — lose the fold by denaturation and the activity goes even though the sequence survives. The specific pairing A–T and C–G makes the DNA strands complementary, and that complementarity is exactly what permits heredity. If you can state that principle and give one example from each family, you have the chapter.
1. The urine sample gives a positive Fehling's test. Name two classes of carbohydrate that could be responsible and one that could not. L2 Understand
2. The child's diet contains adequate total protein but lacks lysine and tryptophan. Why is this still a problem? L3 Apply
3. The diagnostic enzyme has been heated and no longer works, though its amino acid sequence is unchanged. Explain and state whether the activity can be restored. L4 Analyse
4. Fill in the blanks: Nucleotides are joined by a ______ linkage between the ______ and ______ carbon atoms of the pentose sugar, and the sequence of nucleotides is called the ______ structure of a nucleic acid. L1 Remember
5. Explain why DNA fingerprinting is more reliable for identification than conventional fingerprints, and list two of its applications. L5 Evaluate
For each pair choose: (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
Assertion (A): Vitamin K deficiency increases blood clotting time.
Reason (R): Vitamin K is a fat soluble vitamin obtained from green leafy vegetables.
Assertion (A): Sucrose does not reduce Fehling's solution.
Reason (R): In sucrose the reducing groups of glucose and fructose are both involved in the glycosidic linkage.
Assertion (A): RNA contains thymine as one of its four bases.
Reason (R): RNA has a single stranded helical structure.
Sit a full paper on what you have been studying, marked question by question.