CBSE Class 12 Chemistry Biomolecules MCQs Set 12

Answer: (a) adenine and thymine; guanine and cytosine

Explanation:
1. In DNA, adenine pairs with thymine (two hydrogen bonds) and guanine pairs with cytosine (three hydrogen bonds).

Teacher's Note:
1. A purine always pairs with a pyrimidine; uracil never appears in DNA, so options with uracil can be ruled out at once.

Answer: (b) beri-beri

Explanation:
1. Deficiency of vitamin B₁ (thiamine) causes beri-beri (loss of appetite, retarded growth).

Teacher's Note:
1. Pairs to remember: B₂ – cheilosis, B₆ – convulsions, E – sterility (loss of fertility).

Answer: (c) TACGAA

Explanation:
1. Replace each base by its partner: A → T, T → A, G → C, C → G.
2. ATGCTT → TACGAA

Teacher's Note:
1. Write the complement letter by letter under the given strand – it is quick and avoids skipping a base.

Answer: (d) The sugar component in RNA is ribose and sugar component in DNA is 2′-deoxyribose

Explanation:
1. RNA contains β-D-ribose, while DNA contains β-D-2-deoxyribose (no –OH at C2′).

Teacher's Note:
1. The names give it away: Ribo-nucleic acid has ribose and Deoxyribo-nucleic acid has deoxyribose.

Answer: (a) vitamin C

Explanation:
1. Vitamins B group and C are water soluble; vitamins A, D, E and K are fat soluble.

Teacher's Note:
1. Memory aid: fat soluble vitamins are A, D, E, K – everything else (B and C) is water soluble.

Answer: (c) C3′ and C5′ respectively of sugar molecule

Explanation:
1. In a polynucleotide chain, the phosphate group links the C3′–OH of one sugar to the C5′–OH of the next sugar (3′,5′-phosphodiester linkage).
2. The base is attached at C1′ of the sugar.

Teacher's Note:
1. Do not mix up the two positions: base at C1′, phosphodiester linkage between C3′ and C5′.

Answer: (c) >C=O and –OH

Explanation:
1. Carbohydrates are polyhydroxy aldehydes or ketones, so they contain a carbonyl group (>C=O, as aldehyde or ketone) and hydroxyl (–OH) groups.

Teacher's Note:
1. The >C=O covers both aldoses (–CHO) and ketoses (>C=O), which is why it is the best general answer.

Answer: (b) sucrose

Explanation:
1. Sucrose is dextrorotatory, but on hydrolysis it gives an equimolar mixture of glucose (+52.5°) and fructose (−92.4°) which is laevorotatory.
2. This change in the sign of rotation is called inversion, so sucrose is called invert sugar.

Teacher's Note:
1. Strictly, the laevorotatory glucose–fructose mixture formed on hydrolysis is the invert sugar; questions often name sucrose as the sugar that is inverted.

Answer: (a) maltose

Explanation:
1. Maltose on hydrolysis gives two molecules of α-D-glucose, so it is a disaccharide.
2. Cellulose is a polysaccharide, maltase is an enzyme and glucose is a monosaccharide.

Teacher's Note:
1. Watch the spelling trap: maltose (sugar, ends in -ose) vs maltase (enzyme, ends in -ase).

Answer: (b) amylose and amylopectin

Explanation:
1. Starch is a polymer of α-glucose and consists of two components: amylose (15–20%, water soluble, linear) and amylopectin (80–85%, water insoluble, branched).

Teacher's Note:
1. Amylopectin is the major part of starch; amylose is the minor, water soluble part.

Answer: (a) Vitamin B₁₂

Explanation:
1. Vitamin B₁₂ (cyanocobalamin) contains cobalt, a transition metal.
2. Chlorophyll contains magnesium, which is not a transition metal; glycogen and RNA contain no metal.

Teacher's Note:
1. Metal-containing biomolecules to remember: Co in vitamin B₁₂, Fe in haemoglobin, Mg in chlorophyll.

Answer: (c) Galactose

Explanation:
1. Glycosidic linkage joins two monosaccharide units through an oxygen atom.
2. Maltose and sucrose (disaccharides) and amylose (polysaccharide) have glycosidic linkages; galactose is a monosaccharide and has none.

Teacher's Note:
1. Only di-, oligo- and polysaccharides have glycosidic linkages – a single monosaccharide unit has nothing to link to.

Answer: (b) albumin

Explanation:
1. Myosin (muscles), collagen (tendons) and fibroin (silk) are fibrous proteins.
2. Albumin is a globular protein.

Teacher's Note:
1. Fibrous proteins are thread-like and water insoluble; globular proteins such as albumin and insulin are spherical and water soluble.

Answer: (d) A–s, B–r, C–q, D–p

Explanation:
1. Lactose (A) is a disaccharide which on hydrolysis gives β-D-galactose and β-D-glucose (q, s); starch (B) is a polysaccharide (r); sucrose (C) is a disaccharide (q); fructose (D) is a ketohexose (p).
2. The option that fits is A–s, B–r, C–q, D–p.

Teacher's Note:
1. When a Column I entry fits two Column II entries (lactose fits q and s), use the entries that fit only one to pick the option.

Answer: (b) A–q, B–p, C–s, D–r

Explanation:
1. Keratin (A) is a protein with β-pleated structure (p, q); haemoglobin (B) is a (globular) protein (p); riboflavin (C) is vitamin B₂, water soluble (s); glycine (D) is an α-amino acid (r).
2. Since haemoglobin can only be (p), keratin must be (q): A–q, B–p, C–s, D–r.

Teacher's Note:
1. Riboflavin is vitamin B₂ – all B vitamins are water soluble.

Answer: (a) Ribose

Explanation:
1. Ribose, C₅H₁₀O₅, contains five carbon atoms; glucose, fructose and galactose are hexoses.

Teacher's Note:
1. Count the carbon atoms: pent- = 5 (ribose, deoxyribose), hex- = 6 (glucose, fructose, galactose).

Answer: (d) Existence of alpha and beta forms of glucose

Explanation:
1. Open chain structure cannot explain existence of α- and β-forms of glucose.

Teacher's Note:
1. The α- and β-forms (anomers) arise only from the cyclic hemiacetal structure, which creates a new chiral centre at C-1.

Answer: (c) gluconic acid

Explanation:
1. Glucose on reaction (i.e. oxidation) with Br₂ water gives gluconic acid.

Teacher's Note:
1. Bromine water is a mild oxidant: only –CHO → –COOH (gluconic acid). Conc. HNO₃ oxidises both ends to give saccharic acid.

Answer: (d) NaHSO₃

Explanation:
1. Because glucose exists mainly in the cyclic hemiacetal form, it does not form the bisulphite addition product with NaHSO₃.

Teacher's Note:
1. Remember the three ‘missing’ aldehyde tests of glucose: no 2,4-DNP derivative, no Schiff's test, no NaHSO₃ adduct.

Answer: (a) glucosazone

Explanation:
1. With excess (3 molecules of) phenylhydrazine, C-1 and C-2 of glucose react to form glucosazone.

Teacher's Note:
1. One molecule of phenylhydrazine gives a phenylhydrazone; excess gives the osazone – read the word ‘excess’ carefully.

Answer: (a) five

Explanation:
1. Number of chiral carbons in β-D(+)-glucose is five (C-1 to C-5 of the pyranose ring).

Teacher's Note:
1. Open-chain glucose has 4 chiral carbons; ring closure makes C-1 chiral too, giving 5 in the cyclic form.

Answer: (d) configuration at the penultimate chiral carbon

Explanation:
1. D indicates that the –OH group on the lowest (penultimate) chiral carbon is on the right, as in D-glyceraldehyde.

Teacher's Note:
1. D/L tells configuration, not the direction of rotation; (+)/(−) tells the sign of rotation.

Answer: (c) Sucrose

Explanation:
1. In sucrose, the reducing groups of glucose (C-1) and fructose (C-2) are both used in the glycosidic bond, so it is non-reducing.

Teacher's Note:
1. A disaccharide is reducing only if one anomeric carbon remains free; sucrose has none free.

Answer: (a) Sucrose

Explanation:
1. Sucrose on hydrolysis gives glucose which is an aldose and fructose which is a ketose.

Teacher's Note:
1. Maltose gives glucose + glucose and lactose gives galactose + glucose – all aldoses.

Answer: (c) lactose

Explanation:
1. Lactose (milk sugar) is the disaccharide present in milk.

Teacher's Note:
1. Cellulose is a polysaccharide, so it cannot be the answer to a disaccharide question.

Answer: (c) glucose and galactose

Explanation:
1. Lactose is made of β-D-galactose and β-D-glucose; hydrolysis gives glucose and galactose.

Teacher's Note:
1. Match each disaccharide to its units: sucrose = glucose + fructose, maltose = glucose + glucose, lactose = galactose + glucose.

Answer: (d) inversion

Explanation:
1. Hydrolysis of sucrose (cane sugar) changes the sign of rotation from dextro (+66.5°) to laevo (overall −20°); hence it is called inversion of cane sugar.

Teacher's Note:
1. The product mixture (invert sugar) is laevorotatory because fructose (−92.4°) rotates more than glucose (+52.5°).

Answer: (c) A is true, but R is false.

Explanation:
1. Assertion is correct: sucrose is dextrorotatory and its hydrolysis product is laevorotatory.
2. Reason is wrong: hydrolysis gives equal (equimolar) amounts of glucose and fructose; the sign changes because laevorotation of fructose (−92.4°) is greater than dextrorotation of glucose (+52.5°).

Teacher's Note:
1. The amounts are equal; it is the larger magnitude of fructose's rotation that makes the mixture laevorotatory.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Tollens' reagent is alkaline. In the basic medium fructose rearranges (through an enediol) to glucose and mannose, which contain –CHO groups and reduce Tollens' reagent.
2. So both statements are true and the reason explains the assertion.

Teacher's Note:
1. Fructose is also an α-hydroxy ketone; such ketones are oxidised by Tollens' and Fehling's reagents because of this base-catalysed rearrangement.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Depending upon whether the OCH₃ group on the anomeric carbon of the hemiacetal form lies on the right or left side in the Fischer projection, glucose exists in two isomeric methyl glycosides.

Teacher's Note:
1. The two methyl glucosides are α and β – they differ only in configuration at C-1.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Glucose and fructose differ in configuration only at C₁ and C₂ and since during osazone formation configuration at C₁ and C₂ are destroyed, therefore, they form the same osazone.

Teacher's Note:
1. Osazone formation involves only C-1 and C-2; any sugars identical from C-3 to C-6 (glucose, fructose, mannose) give the same osazone.

Answer: (c) A is true, but R is false.

Explanation:
1. In sucrose glucose exists in the pyranose form while fructose exists in the furanose form.

Teacher's Note:
1. Sucrose = α-D-glucopyranose + β-D-fructofuranose linked C1–C2.

Answer: (d) A is false, but R is true.

Explanation:
1. Glycogen is a highly branched polymer.

Teacher's Note:
1. Of the glucose polymers only amylose and cellulose are linear; amylopectin and glycogen are branched.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Under basic conditions of the reagent, fructose undergoes Lobry de Bruyn van Eikenstein rearrangement to form glucose and mannose both of which contain an –CHO group and hence reduce Tollens' reagent.

Teacher's Note:
1. Being a ketohexose does not explain reduction of Tollens' reagent; the base-catalysed rearrangement to aldoses does.

Answer: (d) A is false, but R is true.

Explanation:
1. Glucose and fructose are two different monosaccharides (an aldose and a ketose); anomers differ only in configuration at the anomeric carbon of the same sugar.

Teacher's Note:
1. α- and β-D-glucose are anomers; glucose and fructose are functional isomers.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Formation of an oxime and a cyanohydrin are reactions of the carbonyl group, which is present in the open chain structure of glucose.

Teacher's Note:
1. These reactions only show a carbonyl group; Tollens'/bromine water oxidation shows that it is an aldehyde group.

Answer: (c) A is true, but R is false.

Explanation:
1. Phenylalanine cannot be synthesized by the body but has to be supplied in food.

Teacher's Note:
1. The definition of an essential amino acid is exactly that it is not made in the body.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. In the zwitterionic structure, the acidic group of α-amino acids is –NH₃⁺ instead of –COOH group and basic group is –COO⁻ instead of NH₂.
2. Since –NH₃⁺ is a much weaker acid than –COOH and –COO⁻ is a much weaker base than –NH₂, the K_a and K_b values are very low.

Teacher's Note:
1. Zwitterion formation explains several properties of amino acids together: high melting points, water solubility and low K_a/K_b.