RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Chemistry · 11 questions · 26 marks
Every living cell runs on a handful of molecule families — sugars for energy and structure, amino acids strung into proteins, nucleotides encoding information, and hormones that carry signals between cells. This chapter builds the chemical vocabulary to describe carbohydrates, proteins, enzymes, vitamins, nucleic acids and hormones in terms of the same functional groups you have studied all year.
Sucrose is classified as a non-reducing sugar because:
Answer
Both the anomeric carbons of glucose and fructose are involved in the glycosidic bond, leaving no free reducing group. In sucrose the C-1 of glucose is linked to the C-2 of fructose, and these are exactly the carbons that would otherwise carry the free aldehyde/hemiacetal or hemiketal group capable of reducing Tollens' or Fehling's reagent. With both tied up in the glycosidic linkage, sucrose has no reducing end and gives a negative test, unlike maltose or lactose, which retain one free anomeric carbon.
The secondary structure of a protein, such as the α-helix, is primarily held together by:
Answer
Hydrogen bonds between the C=O and N–H groups of the backbone. In an α-helix the carbonyl oxygen of one peptide bond hydrogen-bonds to the N–H hydrogen four residues further along the chain, and this repeating pattern coils the backbone into a helix. Peptide bonds themselves define the primary structure (the covalent sequence); disulfide bridges and ionic/hydrophobic interactions are mainly responsible for tertiary structure.
In the DNA double helix, adenine always pairs with thymine and guanine always pairs with cytosine. This is best explained by:
Answer
Complementary hydrogen-bonding patterns and matching geometric size between each specific base pair. Adenine and thymine fit together and form exactly two hydrogen bonds in the correct geometric arrangement, while guanine and cytosine fit together and form three hydrogen bonds; a purine (A or G) pairs with a pyrimidine (T or C) in each case so that the two strands stay a constant, uniform width all along the helix, which would not happen with any other pairing.
A prolonged deficiency of vitamin C in the diet leads to:
Answer
Scurvy. Vitamin C (ascorbic acid) is essential for collagen synthesis, and its prolonged absence weakens connective tissue, causing bleeding gums, joint pain and slow wound healing — the classic symptoms of scurvy. Night blindness follows vitamin A deficiency, rickets follows vitamin D deficiency, and beriberi follows vitamin B1 (thiamine) deficiency.
Assertion (A): An amino acid at its isoelectric point does not move towards either electrode in an electric field. Reason (R): At the isoelectric point the amino acid exists as a zwitterion carrying both a positive and a negative charge that exactly cancel to give zero net charge.
Answer
Both A and R are true and R is the correct explanation of A. At the isoelectric pH, the –COOH has donated its proton to the –NH₂ group, giving the structure H₃N⁺–CHR–COO⁻. Because the positive charge on nitrogen and the negative charge on the carboxylate are numerically equal, the molecule carries zero net charge overall, so an applied electric field exerts no net force on it and it stays stationary — exactly the reasoning R provides for observation A.
Describe one chemical test that shows glucose has a free aldehyde group, and one chemical test that shows it has five hydroxyl groups.
Answer
Aldehyde group: glucose reduces Tollens' reagent to give a bright silver mirror (and reduces Fehling's solution to brick-red Cu₂O), which only an oxidisable carbonyl carbon such as an aldehyde can do; also, glucose reacts with hydroxylamine to form an oxime, glucose oxime, confirming the presence of a C=O group. Five hydroxyl groups: glucose reacts with excess acetic anhydride to form glucose pentaacetate, in which all five –OH groups are acetylated; since exactly five acetate groups are incorporated, glucose must contain exactly five hydroxyl groups.
Distinguish between the primary, secondary and tertiary structure of a protein, and state what denaturation destroys.
Answer
Primary structure is simply the linear sequence in which amino acids are joined to one another by peptide bonds, exactly analogous to the order of letters in a word; it is fixed by covalent bonds and is the most stable level of structure. Secondary structure describes a regular, repeating local shape such as the α-helix or β-pleated sheet, held together by hydrogen bonds between the backbone C=O and N–H groups at regular intervals along the chain. Tertiary structure is the overall three-dimensional folding of the entire polypeptide into a compact globular or fibrous shape, stabilised by disulfide bridges, hydrogen bonds between side chains, and hydrophobic interactions between non-polar residues that cluster away from water. Denaturation (by heat, extreme pH, or heavy-metal ions) disrupts the hydrogen bonds and other weak interactions holding the secondary and tertiary structure together, unfolding the protein and destroying its biological activity, while the primary sequence of peptide bonds itself is left intact.
Name the two components of a nucleotide besides the nitrogenous base, and state one structural difference between DNA and RNA.
Answer
A nucleotide is built from a nitrogenous base, a pentose sugar, and a phosphate group; the base and sugar alone (without phosphate) form a nucleoside. One structural difference: DNA contains 2-deoxyribose as its sugar (missing the 2′-OH), while RNA contains ribose (which has the 2′-OH); a second acceptable difference is that DNA uses thymine as one of its four bases while RNA replaces thymine with uracil.
(i) List, in order, the five pieces of experimental evidence used to establish that glucose is an open-chain aldohexose with five –OH groups. (ii) Explain, in terms of the cyclic hemiacetal structure, why glucose actually exists mostly as a six-membered ring and what an anomer is.
Answer
(i) 1. Elemental analysis and molecular mass show glucose has the formula C₆H₁₂O₆. 2. Glucose reacts with hydroxylamine to form a monoxime, confirming one carbonyl group. 3. Glucose is oxidised by bromine water to a monocarboxylic acid (gluconic acid), showing the carbonyl is specifically an aldehyde (a ketone would resist mild bromine-water oxidation). 4. Glucose reacts with acetic anhydride to give a pentaacetate, showing exactly five –OH groups are present. 5. On prolonged reduction with HI, glucose gives n-hexane, showing the six carbon atoms form a single straight, unbranched chain; and vigorous oxidation with HNO₃ gives a dicarboxylic (saccharic) acid, confirming a primary –OH at the other end of the chain in addition to the aldehyde. (ii) The open-chain structure has an aldehyde at C-1 and a hydroxyl at C-5, and these two groups are close enough in space that the C-5 –OH oxygen attacks the C-1 carbonyl carbon intramolecularly, forming a cyclic hemiacetal — a six-membered ring called the pyranose form, which is far more stable than the small equilibrium amount of open-chain aldehyde and is why glucose gives only a weak/slow Schiff's test despite being an aldehyde. This cyclisation converts C-1 into a new stereocentre, since it now bears an –OH, an H, the ring oxygen, and the rest of the chain, all different. The two possible spatial arrangements at this new C-1 centre are called anomers: α-D-glucose has the C-1 –OH on the opposite side to the reference –CH₂OH at C-5 (in the Haworth projection, below the ring), and β-D-glucose has it on the same side (above the ring); the two anomers interconvert in solution through the open-chain form, a phenomenon called mutarotation.
(i) Explain what an enzyme is and why enzymes are highly specific for their substrates. (ii) Describe how temperature and pH affect enzyme activity, and explain why boiling a solution destroys an enzyme's catalytic power. (iii) Give one example each of a fat-soluble and a water-soluble vitamin, with the corresponding deficiency disease.
Answer
(i) An enzyme is a globular protein that acts as a biological catalyst, speeding up a specific biochemical reaction by binding the substrate at a precisely shaped active site and stabilising the transition state, thereby lowering the activation energy without itself being consumed. Enzymes are highly specific because the active site has a particular three-dimensional shape, and only a substrate (or a small family of closely related substrates) whose shape and functional groups complement the active site — the classic 'lock and key' idea, refined by the 'induced fit' model in which the site adjusts slightly around the substrate — can bind productively; a molecule of the wrong shape or the wrong functional groups simply does not fit and is not converted. (ii) Each enzyme has an optimum temperature (often near human body temperature, about 37 °C, for human enzymes) at which activity is highest; below this, activity falls because molecular motion and collision frequency decrease, while above it activity rises briefly and then falls sharply as heat disrupts the weak interactions (hydrogen bonds, hydrophobic contacts) holding the tertiary structure together. Each enzyme also has an optimum pH, and moving away from it in either direction changes the ionisation state of key side chains in the active site, weakening substrate binding and catalysis. Boiling a solution supplies far more thermal energy than the enzyme's tertiary structure can withstand: the hydrogen bonds and hydrophobic interactions that hold the specific three-dimensional shape of the active site break down irreversibly, the protein denatures and unfolds, and once the active site's precise shape is lost the enzyme can no longer bind its substrate, so catalytic activity is permanently destroyed even though the primary sequence of amino acids remains unchanged. (iii) Fat-soluble: vitamin A (retinol); deficiency causes night blindness (and, if prolonged, xerophthalmia). Water-soluble: vitamin C (ascorbic acid); deficiency causes scurvy, with bleeding gums and slow wound healing.
A biochemistry student is given three white solids labelled X, Y, Z, each a carbohydrate: one is glucose, one is sucrose, one is starch. Test 1 (Tollens' reagent, warm): X gives a silver mirror, Y and Z do not. Test 2 (acid hydrolysis followed by Tollens' reagent on the hydrolysate): both Y and Z now give a silver mirror after hydrolysis. Test 3 (iodine solution): Z turns deep blue-black, Y shows no colour change. (a) Identify X, Y and Z with reasoning. (b) Write the hydrolysis equation for Y. (c) Explain why Z, unlike Y, gives a positive test with iodine. (d) Explain why X gives a positive Tollens' test directly without needing prior hydrolysis, unlike Y and Z.
Answer
(a) X reduces Tollens' reagent directly, so it is the monosaccharide, glucose, which has a free aldehyde/hemiacetal group available immediately. Y and Z are both negative before hydrolysis but positive afterwards, so both must be sugars whose reducing groups are tied up until hydrolysed; Z gives a blue-black colour with iodine, the classic starch–iodine complex, so Z is starch, and by elimination Y is sucrose. (b) C₁₂H₂₂O₁₁ (sucrose) + H₂O --H⁺/heat--> C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose). (c) Starch's amylose component coils into a helix, and iodine molecules slot inside this helical cavity to form a charge-transfer complex that strongly absorbs visible light, producing the blue-black colour. Sucrose is a small disaccharide with no such helical cavity, so it cannot trap iodine in this way and shows no colour change. (d) Glucose already possesses a free anomeric carbon (C-1) bearing an aldehyde/hemiacetal group in equilibrium with its open-chain aldehyde form, so Tollens' reagent can oxidise it immediately. Sucrose has both anomeric carbons locked into the glycosidic bond and starch is a large polysaccharide with only one reducing end per enormous chain (negligible concentration of free aldehyde), so both must first be hydrolysed by acid to release free monosaccharide units with unlocked anomeric carbons before a positive Tollens' test can be observed.
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