Class 12 Chemistry Chapter 10: Biomolecules worksheet with answer key

Answer: (b) the –OH on C-5, the lowest asymmetric carbon, is on the right in the Fischer projection, as in (+)-glyceraldehyde

D gives the configuration relative to glyceraldehyde, judged at the lowest asymmetric carbon (C-5) with –CHO at the top; the sign of rotation is shown separately by (+). In D-glucose the –OH on C-3 is on the left, so not every –OH is on the right.

Answer: (c) a primary alcoholic (–CH₂OH) group

Gluconic acid already has COOH at C-1; a second COOH can appear at C-6 only if C-6 carries a primary alcoholic –CH₂OH group, which nitric acid oxidises.

Answer: (a) −19.95°

Hydrolysis of sucrose gives equal amounts of the two: (+52.5° − 92.4°)/2 = −19.95°. The sign is negative because fructose's laevorotation is larger. −39.9° forgets to take the mean; −72.45° adds the magnitudes.

Answer: (c) the –OH at C-5 adds to the keto group at C-2

Addition of the C-5 –OH to the >C=O at C-2 closes a ring of one oxygen and four carbons (C-2 to C-5), like furan. C-1 carries no carbonyl; a C-4 –OH would give a four-membered ring and a C-6 –OH a six-membered one.

Answer: (d) C-1

The two cyclic forms are anomers: they differ only in the position of the –OH at C-1, the anomeric carbon, which was the aldehyde carbon before cyclisation.

Adenine pairs with thymine and guanine with cytosine: complementary strand 3′–T–A–C–G–G–T–5′. Base pairs: A=T, T=A, A=T are three A–T pairs (3 × 2 = 6 hydrogen bonds); G≡C, C≡G, C≡G are three G–C pairs (3 × 3 = 9). Total 15 hydrogen bonds.

• complementary strand 3′–T–A–C–G–G–T–5′ (1)
• 15 hydrogen bonds, with the working (1)

(a) C₁₂H₂₂O₁₁ + H₂O —(dilute HCl or H₂SO₄, boiled in alcoholic solution)→ C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose). (b) (C₆H₁₀O₅)n + nH₂O —(dilute H₂SO₄, 393 K, 2–3 atm)→ nC₆H₁₂O₆ (glucose).

• (a) balanced equation from sucrose, acid named, glucose and fructose as products (1)
• (b) balanced equation from starch with dilute H₂SO₄, 393 K and pressure (1)

H₂N–CH(CH₃)–COOH + H₂N–CH₂–COOH → H₂N–CH(CH₃)–CO–NH–CH₂–COOH + H₂O. The peptide linkage is the –CO–NH– between the carbonyl carbon of alanine and the nitrogen of glycine; the free –NH₂ is on the alanine end and the free –COOH on the glycine end.

• correct equation with the water eliminated (1)
• –CO–NH– marked between alanine's C=O and glycine's N (1)

The ring has six members: five carbons and the oxygen between C-5 and C-1. C-5 carries CH₂OH up; the –OH at C-4 is down, at C-3 up, at C-2 down. C-1 is the anomeric carbon: in the β form its –OH is up; in α-D-(+)-glucopyranose it is down, all else the same. A ring of one oxygen and five carbons is named pyranose by analogy with pyran.

• ring O between C-5 and C-1; CH₂OH up at C-5, –OH down at C-4, up at C-3, down at C-2 (1)
• C-1 anomeric, –OH up in β, down in α (1)
• pyranose: six-membered ring like pyran (1)

(a) C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ (D-(+)-glucose) + C₆H₁₂O₆ (D-(–)-fructose): each mole hydrolysed gives one mole of each. With half the sucrose left, sucrose : glucose : fructose = 1 : 1 : 1. (b) Glucose rotates +52.5° and fructose –92.4°; the laevorotation of fructose is larger, so the mixture is laevorotatory. Sucrose itself is dextrorotatory, so the sign changes from (+) to (–) on hydrolysis: the product is invert sugar. (c) The glycosidic linkage joins C1 of α-D-glucose and C2 of β-D-fructose, using both reducing groups.

• (a) sucrose, D-(+)-glucose and D-(–)-fructose, 1 : 1 : 1 (1)
• (b) larger laevorotation of fructose; sign changes from + to – (1)
• (c) both reducing groups in the C1–C2 linkage (1)

(a) The sequence of bases in each person's DNA is unique to that person. (b) The sequence is the same in every cell of the body and cannot be altered by any known treatment, so hair-root DNA and blood DNA of one person match. (c) Any one: determining paternity; identifying a dead body by comparison with the DNA of parents or children; identifying racial groups.

• (a) unique base sequence for each person (1)
• (b) same in every cell, unalterable (1)
• (c) one valid use (1)