Biomolecules

What to remember

• Carbohydrates are polyhydroxy aldehydes or ketones (or compounds that give them on hydrolysis). Glucose has an open-chain form (evidence: HI, NH₂OH, HCN, Br₂ water, HNO₃, acetic anhydride) and a cyclic hemiacetal form (α and β anomers) that shows mutarotation.
• Sucrose (non-reducing, α-D-glucose C1–β-D-fructose C2), maltose and lactose (reducing, C1–C4 links) are the disaccharides. Amylose and amylopectin (α-glucose), glycogen (α-glucose, more branched) and cellulose (β-glucose, unbranched) are the polysaccharides.
• Proteins are polymers of α-amino acids joined by peptide bonds (–CO–NH–). They have four levels of structure; denaturation destroys the secondary and tertiary structure but not the primary structure.
• DNA (deoxyribose; A, G, C, T) and RNA (ribose; A, G, C, U) are polynucleotides linked by 3′–5′ phosphodiester bonds. Vitamin deficiency diseases and hormone types (steroid, polypeptide, amino-acid derivative) are direct-recall topics.

Carbohydrates: classification

Carbohydrates are optically active polyhydroxy aldehydes or ketones, or substances that give such units on hydrolysis. Many are sweet and are called sugars, but starch and cellulose are not sweet.

• A monosaccharide with an aldehyde group (–CHO) is an aldose (glucose, galactose). One with a ketone group (>C=O) is a ketose (fructose).
• A name also shows the carbon number: glucose is an aldohexose, ribose is an aldopentose, fructose is a ketohexose.
• Reducing sugars reduce Tollens' reagent and Fehling's solution because they have a free aldehyde or ketone (hemiacetal) group. Glucose, fructose, maltose and lactose are reducing. Sucrose is non-reducing because both anomeric carbons are used in the glycosidic link.

Glucose

Preparation

• From sucrose: C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose), with dilute HCl or H₂SO₄ in alcohol.
• From starch: (C₆H₁₀O₅)ₙ + nH₂O → nC₆H₁₂O₆, by boiling with dilute H₂SO₄ at 393 K under 2–3 atm pressure. This is the commercial method.

Open-chain structure. Glucose is an aldohexose, HOCH₂–(CHOH)₄–CHO, with the molecular formula C₆H₁₂O₆. It is the D-(+) isomer.

Limitations of the open-chain structure

• Glucose does not give Schiff's test and does not form the NaHSO₃ addition product.
• Glucose pentaacetate does not react with NH₂OH, so a free –CHO group is absent.
• Glucose shows mutarotation.

Cyclic structure. The –OH on C5 adds to the C1 aldehyde group and forms a six-membered ring (pyranose) that is a cyclic hemiacetal. C1 becomes a new chiral centre and is called the anomeric carbon. The two forms are anomers.

• Haworth projection in words: a flat six-membered ring with five carbons and one oxygen; the CH₂OH group at C5 stands above the plane for the D series.
• Mutarotation: a freshly made solution of either anomer slowly changes its specific rotation to the same equilibrium value, +52.5°. The open-chain form is the intermediate: α ⇌ open chain ⇌ β.

Worked example 1. Find the percentage of α-anomer at equilibrium. Let the fraction of α be x. Then 111x + 19.2(1 – x) = 52.5. So 91.8x = 33.3 and x = 0.363. The mixture is about 36% α and 64% β.

Fructose

Fructose, C₆H₁₂O₆, is a ketohexose with the C=O at C2. It is laevorotatory, with specific rotation about –92.4°, and is the sweetest sugar. Its ring form is a furanose: the –OH on C5 adds to the C2 ketone group and gives a five-membered ring (four carbons and one oxygen). Glucose and fructose are functional isomers of each other.

Disaccharides

Two monosaccharides join through an oxygen bridge called the glycosidic linkage, with loss of one H₂O.

Invert sugar. Sucrose is dextrorotatory (+66.5°). On hydrolysis it gives an equimolar mixture of glucose (dextro, +52.5°) and fructose (laevo, –92.4°). The laevorotation of fructose is larger, so the mixture is laevorotatory. The change of sign of rotation is called inversion, and the enzyme invertase brings it about.

Polysaccharides

Importance: carbohydrates are a main source of energy. Cellulose is used for paper, cloth and wood. Honey and fruits contain glucose and fructose. Starch is stored as food in plants and glycogen in animals. Humans cannot digest cellulose because they lack the enzyme for β-links.

D and L configuration

D-Glyceraldehyde and L-glyceraldehyde are the reference compounds. In the Fischer projection, a sugar is D if the –OH on the chiral carbon farthest from the carbonyl group (C5 in glucose) is on the right, and L if it is on the left. D and L show configuration only. The signs (+) and (–) (or d and l) show the direction of rotation of plane polarised light, and they are not linked to D and L. Natural glucose is D-(+); natural fructose is D-(–). Open-chain glucose has four chiral carbons (C2 to C5), so 2⁴ = 16 stereoisomers are possible.

Proteins

Amino acids contain an amino group (–NH₂) and a carboxyl group (–COOH). In α-amino acids both are on the same carbon, RCH(NH₂)COOH. About twenty occur in proteins.

• Essential amino acids cannot be made in the body and must come from food: valine, leucine, isoleucine, arginine, lysine, threonine, methionine, phenylalanine, tryptophan and histidine.
• Non-essential amino acids can be made in the body: glycine, alanine, glutamic acid, aspartic acid, serine, proline and others.
• Zwitterion: the –COOH gives H⁺ to the –NH₂, so the molecule exists as ⁺H₃N–CHR–COO⁻ in the solid and in water. This explains the high melting points and solubility in water.
• Amino acids are amphoteric. In acid solution they are cations; in base they are anions.
• Isoelectric point (pI): the pH at which the amino acid has no net charge, does not move in an electric field and has its least solubility.

Peptide bond. The –CO–NH– link is formed between the –COOH of one amino acid and the –NH₂ of the next, with loss of one H₂O. A chain of n amino acids has n – 1 peptide bonds. Three different amino acids can form 3! = 6 different tripeptides.

Levels of protein structure

Fibrous and globular proteins

Denaturation. Heat, change of pH, UV light, salts or chemicals break the hydrogen bonds and disturb the secondary and tertiary structures. The primary structure is not changed. The protein loses biological activity. Examples: coagulation of egg white on boiling, curdling of milk. Denaturation turns an α-helix into a random coil.

Enzymes

Enzymes are biological catalysts, and almost all are globular proteins. They are specific for a substrate, act at an active site and work best at an optimum temperature (about 298–310 K) and pH. Names usually end in -ase: maltase converts maltose to glucose, invertase converts sucrose to glucose and fructose, and urease converts urea to ammonia and carbon dioxide. Inhibitors reduce enzyme activity.

Vitamins

Vitamins are organic compounds needed in small amounts that the body cannot make. Fat-soluble vitamins (A, D, E, K) are stored in liver and fat tissue. Water-soluble vitamins (B group and C) are not stored and must be supplied regularly.

Nucleic acids

A nucleoside is a base joined to a sugar (at C1′ of the sugar). A nucleotide is a nucleoside with a phosphate group on C5′ of the sugar. In a nucleic acid, nucleotides are joined by phosphodiester linkages between C3′ of one sugar and C5′ of the next.

• Double helix: two strands run in opposite directions (antiparallel) and coil around a common axis. They are held by hydrogen bonds between base pairs: A=T (two H-bonds) and G≡C (three H-bonds). So the amount of A equals T, and G equals C.
• DNA carries the hereditary message. It replicates itself and directs protein synthesis through RNA.

Hormones

Hormones are chemical messengers made by ductless (endocrine) glands and carried by blood to target organs.

Insulin lowers blood glucose; glucagon raises it. Deficiency of iodine, needed for thyroxine, causes goitre. Deficiency of insulin causes diabetes.

Worked example 2. A protein chain has 101 amino acid residues. Peptide bonds = 101 – 1 = 100, and 100 molecules of water are lost in forming it.

NEET traps

• 1. Sucrose is non-reducing; maltose and lactose are reducing.
• 2. The linkage in sucrose is C1–C2, but in maltose and lactose it is C1–C4.
• 3. α-Glucose and β-glucose differ only at C1 (they are anomers), not in the D/L series.
• 4. Starch has α-links (digestible); cellulose has β-links (not digested by humans).
• 5. Amylose is unbranched and water soluble; amylopectin is branched and insoluble.
• 6. Glycogen is like amylopectin but more branched; it is stored in animals, not plants.
• 7. Br₂ water gives gluconic acid (only –CHO oxidised); HNO₃ gives saccharic acid (both ends oxidised).
• 8. HI heating of glucose gives n-hexane, which shows a straight chain, not a carbonyl group.
• 9. D and L have no link with + and –: D-fructose is laevorotatory.
• 10. Water-soluble vitamins (B, C) are not stored; fat-soluble ones (A, D, E, K) are stored.

One-liners

• 1. Glucose is an aldohexose; fructose is a ketohexose.
• 2. Glucose is dextrorotatory (+52.5°); fructose is laevorotatory (–92.4°).
• 3. Anomeric carbon of glucose is C1; of fructose it is C2.
• 4. Mutarotation ends at +52.5° for glucose.
• 5. Glucose forms a pyranose ring; fructose forms a furanose ring.
• 6. Invert sugar is an equimolar mixture of glucose and fructose, laevorotatory.
• 7. Lactose gives galactose and glucose on hydrolysis.
• 8. Starch gives a blue colour with iodine.
• 9. Zwitterion is ⁺H₃N–CHR–COO⁻.
• 10. At the isoelectric point an amino acid has least solubility.
• 11. The peptide bond is an amide bond (–CO–NH–).
• 12. α-Helix has intra-chain hydrogen bonds; β-sheet has inter-chain hydrogen bonds.
• 13. Haemoglobin is an example of quaternary structure.
• 14. Vitamin C deficiency causes scurvy; vitamin D deficiency causes rickets.
• 15. DNA has thymine; RNA has uracil. A pairs with T by two H-bonds, G with C by three.

Practice questions

1. Prolonged heating of glucose with HI gives

   1. hexanoic acid
   2. 2-methylpentane
   3. hexan-1-ol
   4. n-hexane
   Answer

   D. n-hexane

   HI reduces glucose completely to n-hexane, showing that all six carbon atoms are in a straight chain.

2. Glucose forms a pentaacetate with acetic anhydride. This shows that glucose has

   1. four –OH groups
   2. five –CHO groups
   3. five –OH groups
   4. six –OH groups
   Answer

   C. five –OH groups

   One acetyl group is added for each –OH group; five acetyl groups mean five –OH groups.

3. Mild oxidation of glucose with bromine water gives

   1. gluconic acid
   2. n-hexane
   3. saccharic acid
   4. glucose pentaacetate
   Answer

   A. gluconic acid

   Br₂ water oxidises only the –CHO group to –COOH, giving gluconic acid. Saccharic acid needs the stronger HNO₃.

4. Which of the following sugars is non-reducing?

   1. Maltose
   2. Sucrose
   3. Lactose
   4. Glucose
   Answer

   B. Sucrose

   In sucrose the C1 of glucose and C2 of fructose are joined, so no free aldehyde or ketone group remains.

5. The glycosidic linkage in maltose is between

   1. C1 of one α-D-glucose and C4 of another α-D-glucose
   2. C1 of one α-D-glucose and C6 of another α-D-glucose
   3. C1 of α-D-glucose and C2 of β-D-fructose
   4. C1 of β-D-galactose and C4 of β-D-glucose
   Answer

   A. C1 of one α-D-glucose and C4 of another α-D-glucose

   Maltose is made of two α-D-glucose units joined by a C1–C4 link.

6. Lactose on hydrolysis gives

   1. α-D-glucose and β-D-fructose
   2. two molecules of α-D-glucose
   3. β-D-galactose and β-D-glucose
   4. two molecules of β-D-galactose
   Answer

   C. β-D-galactose and β-D-glucose

   Lactose (milk sugar) is a galactose–glucose disaccharide.

7. Amylopectin differs from amylose because amylopectin

   1. is made of β-D-glucose
   2. is branched, with C1–C6 links at the branch points
   3. gives no colour with iodine
   4. is soluble in water and unbranched
   Answer

   B. is branched, with C1–C6 links at the branch points

   Amylose is unbranched (C1–C4 only); amylopectin has additional C1–C6 links and is branched and insoluble.

8. The structure of an α-amino acid in the zwitterion form is

   1. H₂N–CHR–COOH
   2. H₂N–CHR–COO⁻
   3. ⁺H₃N–CHR–COOH
   4. ⁺H₃N–CHR–COO⁻
   Answer

   D. ⁺H₃N–CHR–COO⁻

   The –COOH transfers H⁺ to the –NH₂ group, giving ⁺H₃N–CHR–COO⁻ with no net charge.

9. Which of the following is an essential amino acid?

   1. Glycine
   2. Lysine
   3. Glutamic acid
   4. Alanine
   Answer

   B. Lysine

   Lysine cannot be made in the body. Glycine, alanine and glutamic acid are non-essential.

10. The α-helix of a protein is stabilised by

    1. hydrogen bonds between different polypeptide chains
    2. disulphide bridges only
    3. peptide bonds between adjacent turns
    4. hydrogen bonds between C=O and N–H groups of the same chain
    Answer

    D. hydrogen bonds between C=O and N–H groups of the same chain

    α-Helix has intra-chain hydrogen bonds; the β-pleated sheet has hydrogen bonds between neighbouring chains.

11. Pernicious anaemia is caused by deficiency of vitamin

    1. B₁₂
    2. B₁
    3. B₆
    4. K
    Answer

    A. B₁₂

    Deficiency of vitamin B₁₂ causes pernicious anaemia; B₁ causes beri-beri, B₆ causes convulsions, K increases clotting time.

12. A nucleotide is made of

    1. a base and a sugar only
    2. a sugar and a phosphate group only
    3. a base, a sugar and a phosphate group
    4. a base and a phosphate group only
    Answer

    C. a base, a sugar and a phosphate group

    A nucleoside is base + sugar; adding a phosphate group on C5′ gives a nucleotide.

13. Which base is present in RNA but not in DNA?

    1. Uracil
    2. Adenine
    3. Guanine
    4. Thymine
    Answer

    A. Uracil

    RNA has A, G, C and U; DNA has A, G, C and T.

14. The number of hydrogen bonds between guanine and cytosine in DNA is

    1. one
    2. two
    3. four
    4. three
    Answer

    D. three

    G pairs with C by three hydrogen bonds; A pairs with T by two.

15. Which of the following hormones is a steroid?

    1. Thyroxine
    2. Glucagon
    3. Testosterone
    4. Insulin
    Answer

    C. Testosterone

    Testosterone is a steroid; insulin and glucagon are polypeptides and thyroxine is an amino-acid derivative.

16. The anomeric carbon of D-fructose in its ring form is

    1. C5
    2. C2
    3. C3
    4. C1
    Answer

    B. C2

    Fructose is a 2-ketose; its ring closes at C2, which becomes the anomeric carbon.

17. Given below are two statements: Statement I: Sucrose is a non-reducing sugar. Statement II: In sucrose the C1 of α-glucose is linked to the C2 of β-fructose, so both reducing groups are involved in the glycosidic bond. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Both statements are correct and II explains I.

18. Given below are two statements: Statement I: Glucose pentaacetate reacts with hydroxylamine to form an oxime. Statement II: Glucose does not give Schiff's test. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    The pentaacetate has no free –CHO group, so it gives no oxime; Statement I is incorrect. Glucose does not give Schiff's test, so II is correct.

19. Given below are two statements: Statement I: Glycogen is more highly branched than amylopectin. Statement II: Glycogen is the reserve carbohydrate of plants. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Glycogen is animal starch and is more branched; plants store starch, so II is incorrect.

20. Given below are two statements: Statement I: Amylose is a branched polymer that is insoluble in water. Statement II: Cellulose is made of α-D-glucose units. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    Amylose is unbranched and water soluble; cellulose is made of β-D-glucose units. Both are incorrect.

21. Which of the following statements are correct? 1. Vitamins A, D, E and K are fat soluble. 2. Vitamin C is stored in the liver. 3. Deficiency of vitamin K increases the blood clotting time.

    1. 1 and 2 only
    2. 1 and 3 only
    3. 2 and 3 only
    4. 1, 2 and 3
    Answer

    B. 1 and 3 only

    Vitamin C is water soluble and is not stored; statements 1 and 3 are correct.

22. Given below are two statements: Statement I: Vitamins of the B group and vitamin C are water soluble. Statement II: Water-soluble vitamins are stored in the body for a long time. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    They are water soluble but are not stored, so they must be supplied regularly.

23. Which of the following statements are correct? 1. In DNA, adenine pairs with thymine by two hydrogen bonds. 2. In DNA, guanine pairs with cytosine by two hydrogen bonds. 3. The two strands of DNA run in opposite directions.

    1. 1 and 2 only
    2. 2 and 3 only
    3. 1, 2 and 3
    4. 1 and 3 only
    Answer

    D. 1 and 3 only

    G pairs with C by three hydrogen bonds, so statement 2 is incorrect.

24. Given below are two statements: Statement I: In an α-helix the hydrogen bonds join different polypeptide chains. Statement II: In a β-pleated sheet the hydrogen bonds are within the same chain. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    The positions are swapped: α-helix has intra-chain and β-sheet has inter-chain hydrogen bonds.

25. Given below are two statements: Statement I: Insulin is a steroid hormone. Statement II: Thyroxine is a hormone that is a derivative of an amino acid. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Insulin is a polypeptide hormone, so I is incorrect; thyroxine is an amino-acid derivative, so II is correct.

26. Assertion (A): Glucose does not give Schiff's test. Reason (R): In solution glucose exists mainly in the cyclic hemiacetal form, with very little free –CHO group.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    The free aldehyde group is almost absent in the ring form, so the test is negative.

27. Assertion (A): Sucrose is a non-reducing sugar. Reason (R): Sucrose has a free aldehyde group at C1 of glucose.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    Sucrose has no free aldehyde or ketone group because both anomeric carbons are in the glycosidic link.

28. Assertion (A): Amino acids have high melting points and are soluble in water. Reason (R): Amino acids exist as zwitterions.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    The dipolar ionic (salt-like) form explains both properties.

29. Assertion (A): Starch gives a blue colour with iodine. Reason (R): Starch is a polymer of glucose.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Starch is a glucose polymer, but that fact does not explain the blue colour with iodine.

30. Assertion (A): On boiling an egg, the primary structure of its albumin is destroyed. Reason (R): Heat disrupts the hydrogen bonds that hold the secondary and tertiary structure.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    Boiling denatures albumin by breaking hydrogen bonds but the peptide bonds (primary structure) stay intact.

31. Assertion (A): Haemoglobin shows quaternary structure. Reason (R): Haemoglobin is made of a single polypeptide chain.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    Haemoglobin has four polypeptide subunits, which is why it shows quaternary structure.

32. Match List I with List II. List I: a. Sucrose b. Maltose c. Lactose d. Cellulose List II: i. β-D-galactose and β-D-glucose ii. α-D-glucose and β-D-fructose iii. two α-D-glucose units iv. β-D-glucose units only

    1. a-i, b-iii, c-ii, d-iv
    2. a-ii, b-iii, c-iv, d-i
    3. a-iii, b-ii, c-i, d-iv
    4. a-ii, b-iii, c-i, d-iv
    Answer

    D. a-ii, b-iii, c-i, d-iv

    Sucrose = glucose + fructose; maltose = two α-glucose; lactose = galactose + glucose; cellulose = β-glucose units.

33. Match List I with List II. List I: a. Vitamin A b. Vitamin B₁ c. Vitamin C d. Vitamin K List II: i. Scurvy ii. Night blindness iii. Increased blood clotting time iv. Beri-beri

    1. a-iv, b-ii, c-i, d-iii
    2. a-ii, b-iv, c-i, d-iii
    3. a-ii, b-iv, c-iii, d-i
    4. a-ii, b-i, c-iv, d-iii
    Answer

    B. a-ii, b-iv, c-i, d-iii

    A: night blindness; B₁: beri-beri; C: scurvy; K: increased clotting time.

34. Match List I with List II. List I: a. HI b. NH₂OH c. Br₂ water d. Acetic anhydride List II: i. Gluconic acid ii. n-Hexane iii. Oxime iv. Pentaacetate

    1. a-ii, b-iii, c-i, d-iv
    2. a-ii, b-iii, c-iv, d-i
    3. a-i, b-iii, c-ii, d-iv
    4. a-iii, b-ii, c-i, d-iv
    Answer

    A. a-ii, b-iii, c-i, d-iv

    HI gives n-hexane; NH₂OH gives an oxime; Br₂ water gives gluconic acid; acetic anhydride gives the pentaacetate.

35. Match List I with List II. List I: a. Primary structure b. α-Helix c. Tertiary structure d. Quaternary structure List II: i. Disulphide bridges and overall folding ii. Sequence joined by peptide bonds iii. Hydrogen bonds within one chain iv. Assembly of several subunits

    1. a-iii, b-ii, c-i, d-iv
    2. a-i, b-iii, c-ii, d-iv
    3. a-ii, b-iii, c-i, d-iv
    4. a-ii, b-iii, c-iv, d-i
    Answer

    C. a-ii, b-iii, c-i, d-iv

    Primary: peptide-bonded sequence; α-helix: intra-chain H-bonds; tertiary: folding with S–S bridges; quaternary: subunits.

36. Match List I with List II. List I: a. Insulin b. Glucagon c. Thyroxine d. Testosterone List II: i. Lowers blood glucose ii. Raises blood glucose iii. Iodine is needed for it iv. Steroid sex hormone

    1. a-ii, b-i, c-iii, d-iv
    2. a-i, b-ii, c-iv, d-iii
    3. a-i, b-ii, c-iii, d-iv
    4. a-iii, b-ii, c-i, d-iv
    Answer

    C. a-i, b-ii, c-iii, d-iv

    Insulin lowers and glucagon raises blood glucose; thyroxine contains iodine; testosterone is a steroid.

37. Match List I with List II. List I: a. Nucleoside b. Nucleotide c. A base found only in DNA d. A base found only in RNA List II: i. Uracil ii. Thymine iii. Base joined to a sugar iv. Base, sugar and phosphate

    1. a-iii, b-iv, c-i, d-ii
    2. a-iv, b-iii, c-ii, d-i
    3. a-ii, b-iv, c-iii, d-i
    4. a-iii, b-iv, c-ii, d-i
    Answer

    D. a-iii, b-iv, c-ii, d-i

    Nucleoside = base + sugar; nucleotide = base + sugar + phosphate; thymine only in DNA; uracil only in RNA.

38. Insulin has two polypeptide chains, one of 21 and one of 30 amino acids. The number of peptide bonds in insulin (ignoring the disulphide bridges) is

    1. 49
    2. 48
    3. 51
    4. 50
    Answer

    A. 49

    Chain A has 20 bonds and chain B has 29 bonds: 20 + 29 = 49. (Also 51 residues – 2 chains = 49.)

39. Pure α-D-glucose has specific rotation +111° and pure β-D-glucose has +19.2°. A solution at equilibrium shows +52.5°. The percentage of α-form at equilibrium is about

    1. 64%
    2. 36%
    3. 48%
    4. 52%
    Answer

    B. 36%

    111x + 19.2(1 – x) = 52.5 gives x = 33.3/91.8 = 0.36, so about 36% α (and 64% β).

40. A solution of sucrose is dextrorotatory. After acid hydrolysis the solution turns plane polarised light to the left. The reason is that

    1. glucose becomes laevorotatory on hydrolysis
    2. the fructose formed has a larger laevo rotation than the dextro rotation of glucose
    3. fructose is dextrorotatory
    4. sucrose is converted into galactose
    Answer

    B. the fructose formed has a larger laevo rotation than the dextro rotation of glucose

    Equimolar glucose (+52.5°) and fructose (–92.4°) give a net laevorotation: invert sugar.

41. A sugar does not reduce Fehling's solution but gives glucose and fructose on hydrolysis with dilute acid. The sugar is

    1. lactose
    2. maltose
    3. sucrose
    4. starch
    Answer

    C. sucrose

    Only sucrose is non-reducing and gives glucose plus fructose; maltose and lactose reduce Fehling's solution.

42. The number of different tripeptides that can be made from three different amino acids, using each once in a chain, is

    1. 3
    2. 6
    3. 9
    4. 8
    Answer

    B. 6

    Arrangements = 3! = 3 × 2 × 1 = 6.

43. A person has a delayed blood clotting time after a long period of poor diet. The vitamin that is most likely lacking is

    1. K
    2. C
    3. B₁
    4. D
    Answer

    A. K

    Deficiency of vitamin K increases the blood clotting time.

44. A DNA sample has 30% adenine. The percentage of cytosine in it is

    1. 10%
    2. 40%
    3. 30%
    4. 20%
    Answer

    D. 20%

    A = T = 30%, so G + C = 40%, and G = C = 20%.