Exam map of the chapter

Knowledge questions with answers

1. ★★★ What is the molecular formula of deoxyribose?
   C₅H₁₀O₄.
2. ★★ What functional group makes glucose an aldohexose?
   The aldehyde group (–CHO).
3. ★★★ Which functional group is present in fructose?
   The ketone group (>C=O).
4. ★★ What is the general molecular formula of a disaccharide mentioned in the source?
   C₁₂H₂₂O₁₁.
5. ★★★ What are the products of sucrose hydrolysis?
   Glucose and fructose.
6. ★★★ What type of bond joins two monosaccharides in a disaccharide?
   A glycosidic bond.
7. ★★ Which two monosaccharides form maltose?
   Two glucose molecules.
8. ★★ What is the linkage in maltose?
   An α-1,4 glycosidic linkage.
9. ★★ How many monosaccharide units are obtained by hydrolysing an oligosaccharide?
   Three to ten units.
10. ★★ Approximately how many isoprene units form natural rubber?
    About 3,000–4,000 isoprene units.
11. ★★★ What is a hexose?
    A monosaccharide containing six carbon atoms, with the general formula C₆H₁₂O₆.
12. ★★★ Define a reducing sugar and name one example from the source.
    A sugar capable of reducing agents; glucose is an example.
13. ★★★ Why is sucrose a non-reducing sugar although glucose and fructose are reducing sugars?
    During sucrose formation, the aldehyde group of glucose and the ketone group of fructose become involved in the glycosidic linkage, so their reducing ability is lost.
14. ★★★ What is a peptide bond?
    The C–N bond formed when the carboxyl group of one amino acid reacts with the amino group of another, releasing one molecule of water.
15. ★★★ What is a protein?
    A protein is a high-molecular-weight polymer of amino acids joined by peptide bonds.
16. ★★ What is meant by the amphoteric nature of proteins?
    Because their amino acids contain both amino (–NH₂) and carboxyl (–COOH) groups, proteins can show both basic and acidic properties.
17. ★★ What is a structural protein? Give one example.
    A protein that forms body structures; keratin is an example.
18. ★★ What is a conjugated protein?
    A protein combined with a non-protein component called a prosthetic group.
19. ★★ What is albumin?
    A protein that dissolves readily in water and dilute salt solutions and coagulates on heating.
20. ★ What are terpenes?
    Compounds of isoprene containing 10–40 carbon atoms, with the general formula (C₅H₈)ₙ.

Admission-style MCQ with answers

1. ★★★ Which molecular formula correctly represents deoxyribose?

   a) C₅H₁₀O₅	b) C₅H₁₀O₄
   c) C₆H₁₂O₆	d) C₇H₁₄O₇

   ✔ b) C₅H₁₀O₄ — Deoxyribose is a pentose sugar with the molecular formula C₅H₁₀O₄.
2. ★★★ Why is deoxyribose essential for DNA formation?

   a) It forms the nitrogenous bases of DNA	b) It forms the sugar component of DNA nucleotides
   c) It supplies the phosphate groups of DNA	d) It acts as the enzyme for DNA replication

   ✔ b) It forms the sugar component of DNA nucleotides — DNA nucleotides contain deoxyribose as their sugar component.
3. ★★ Which structural difference distinguishes deoxyribose from ribose?

   a) Deoxyribose has no oxygen atom at its second carbon	b) Deoxyribose has an extra hydroxyl group at its second carbon
   c) Deoxyribose contains six carbon atoms instead of five	d) Deoxyribose lacks an aldehyde functional group

   ✔ a) Deoxyribose has no oxygen atom at its second carbon — At the second carbon, deoxyribose has hydrogen instead of the hydroxyl group found in ribose.
4. ★★ A deoxyribose molecule becomes a deoxyribonucleotide when which group is added at its fifth carbon?

   a) Amino group	b) Phosphate group
   c) Carboxyl group	d) Ketone group

   ✔ b) Phosphate group — Addition of a phosphate group at carbon 5 produces a deoxyribonucleotide.
5. ★★★ Which set contains only hexose monosaccharides mentioned in the source?

   a) Ribose, deoxyribose and sedoheptulose	b) Glucose, fructose and galactose
   c) Glucose, ribose and sucrose	d) Fructose, sucrose and maltose

   ✔ b) Glucose, fructose and galactose — Glucose, fructose and galactose are six-carbon monosaccharides.
6. ★★★ Glucose is classified as an aldohexose because it contains—

   a) a ketone group and six carbon atoms	b) an aldehyde group and six carbon atoms
   c) an aldehyde group and five carbon atoms	d) a carboxyl group and six carbon atoms

   ✔ b) an aldehyde group and six carbon atoms — Glucose has six carbon atoms and an aldehyde group, so it is an aldohexose.
7. ★★★ Which statement about glucose is correct according to the source?

   a) It is a non-reducing sugar and is insoluble in water	b) It is a reducing sugar and is readily soluble in water
   c) It is a ketose and is less sweet than sucrose	d) It is stored extensively by plants as their main reserve carbohydrate

   ✔ b) It is a reducing sugar and is readily soluble in water — Glucose is described as a reducing sugar that dissolves readily in water.
8. ★★ A fruit sample contains a high concentration of a reducing aldohexose known as grape sugar. The sugar is most likely—

   a) Fructose	b) Glucose
   c) Sucrose	d) Maltose

   ✔ b) Glucose — Glucose is called grape sugar and is abundant in ripe grapes.
9. ★★★ Cyclization of glucose mainly results from interaction between which groups?

   a) The aldehyde group at C1 and the hydroxyl group at C5	b) The ketone group at C2 and the hydroxyl group at C6
   c) The carboxyl group at C1 and the amino group at C5	d) The phosphate group at C5 and the aldehyde group at C6

   ✔ a) The aldehyde group at C1 and the hydroxyl group at C5 — The aldehyde group at carbon 1 reacts with the hydroxyl group at carbon 5 to form a cyclic structure.
10. ★★ The α and β forms of glucose differ primarily in the—

    a) number of carbon atoms	b) position of the newly formed hydroxyl group at the anomeric carbon
    c) presence or absence of the aldehyde group in the open chain	d) molecular formula of the sugar

    ✔ b) position of the newly formed hydroxyl group at the anomeric carbon — The α and β forms differ in the orientation of the newly formed hydroxyl group at carbon 1.
11. ★★ In the open-chain Fischer representation described in the source, glucose is called D-glucose when the hydroxyl group attached to the terminal chiral carbon lies on the—

    a) left side	b) right side
    c) upper side	d) lower side

    ✔ b) right side — The source defines D-glucose by the hydroxyl group being on the right at the terminal carbon.
12. ★★ Which statement correctly describes dextrorotation?

    a) Rotation of plane-polarized light anticlockwise	b) Rotation of plane-polarized light clockwise
    c) Absorption of all visible light	d) Conversion of D-glucose into L-glucose

    ✔ b) Rotation of plane-polarized light clockwise — Dextrorotation means rotating plane-polarized light in the clockwise direction.
13. ★ According to the source, which statement is correct about D- and L-glucose?

    a) D-glucose is levorotatory and L-glucose is dextrorotatory	b) D-glucose is dextrorotatory and L-glucose is levorotatory
    c) Both are always dextrorotatory	d) Both have identical optical rotation

    ✔ b) D-glucose is dextrorotatory and L-glucose is levorotatory — The source associates D-glucose with dextrorotation and L-glucose with levorotation.
14. ★★★ Fructose is best classified as a—

    a) reducing ketohexose	b) non-reducing aldohexose
    c) reducing pentose	d) non-reducing disaccharide

    ✔ a) reducing ketohexose — Fructose has six carbon atoms, a ketone group and reducing-sugar properties.
15. ★★★ Glucose and fructose have the same molecular formula but different functional groups. Therefore, they are—

    a) polysaccharides	b) structural isomers
    c) nucleotides	d) disaccharides

    ✔ b) structural isomers — They share C₆H₁₂O₆ but differ in structure, making them structural isomers.
16. ★★ The name 'fruit sugar' is most directly associated with—

    a) Glucose	b) Fructose
    c) Galactose	d) Mannose

    ✔ b) Fructose — Fructose is found in many fruits and honey and is also called fruit sugar.
17. ★★ Which statement correctly describes the relationship between fructose and sucrose?

    a) Fructose is one of the two monosaccharide components of sucrose	b) Fructose is formed only by polymerisation of sucrose
    c) Fructose is a protein component of sucrose	d) Fructose and sucrose have identical molecular formulas

    ✔ a) Fructose is one of the two monosaccharide components of sucrose — Hydrolysis of sucrose produces glucose and fructose.
18. ★★★ Based on the relative sweetness values in the source, which is the sweetest?

    a) Sucrose	b) Glucose
    c) Fructose	d) Maltose

    ✔ c) Fructose — The listed relative sweetness values are sucrose 100, glucose 74, fructose 173 and maltose 32.
19. ★★ Which statement about mannose and galactose is correct?

    a) Both are pentoses with formula C₅H₁₀O₅	b) Both are hexose aldoses with formula C₆H₁₂O₆
    c) Both are ketoses with formula C₆H₁₂O₆	d) Both are disaccharides with formula C₁₂H₂₂O₁₁

    ✔ b) Both are hexose aldoses with formula C₆H₁₂O₆ — The source identifies both mannose and galactose as hexose aldoses with formula C₆H₁₂O₆.
20. ★ A monosaccharide with seven carbon atoms and formula C₇H₁₄O₇ is classified as a—

    a) pentose	b) hexose
    c) heptose	d) disaccharide

    ✔ c) heptose — Monosaccharides containing seven carbon atoms are heptoses.
21. ★★★ A disaccharide is formed when—

    a) one monosaccharide loses carbon dioxide	b) two monosaccharides join with removal of one water molecule
    c) two amino acids join with removal of one phosphate group	d) a polysaccharide gains one water molecule

    ✔ b) two monosaccharides join with removal of one water molecule — Condensation between two monosaccharides removes one water molecule and forms a disaccharide.
22. ★★★ Which molecular formula is characteristic of the disaccharides described in the source?

    a) C₅H₁₀O₄	b) C₆H₁₂O₆
    c) C₇H₁₄O₇	d) C₁₂H₂₂O₁₁

    ✔ d) C₁₂H₂₂O₁₁ — The general formula given for the listed disaccharides is C₁₂H₂₂O₁₁.
23. ★★★ Which pairing correctly represents hydrolysis of a disaccharide?

    a) Sucrose → glucose + fructose	b) Sucrose → glucose + galactose
    c) Maltose → glucose + fructose	d) Lactose → fructose + galactose

    ✔ a) Sucrose → glucose + fructose — Sucrose hydrolyses into one glucose molecule and one fructose molecule.
24. ★★★ The bond formed between two monosaccharide units in a disaccharide is called a—

    a) peptide bond	b) glycosidic bond
    c) phosphodiester bond	d) hydrogen bond

    ✔ b) glycosidic bond — The C–O–C linkage joining monosaccharides is the glycosidic bond.
25. ★★★ Why is sucrose a non-reducing sugar although both glucose and fructose are reducing sugars?

    a) Sucrose contains no oxygen atoms	b) The relevant aldehyde and ketone groups become unavailable during sucrose formation
    c) Sucrose contains only one carbon atom	d) Sucrose cannot dissolve in water

    ✔ b) The relevant aldehyde and ketone groups become unavailable during sucrose formation — Formation of sucrose involves the linked aldehyde and ketone groups, eliminating its reducing ability.
26. ★★ Sucrose is formed by removal of water from the hydroxyl groups attached to—

    a) C1 of α-D-glucose and C2 of β-D-fructose	b) C2 of α-D-glucose and C1 of β-D-fructose
    c) C5 of glucose and C6 of fructose	d) C1 of β-D-glucose and C5 of α-D-fructose

    ✔ a) C1 of α-D-glucose and C2 of β-D-fructose — The source specifies α-D-glucose C1 and β-D-fructose C2 as the participating positions.
27. ★★ When sucrose is hydrolysed by mild acid, the products are—

    a) equal amounts of glucose and fructose	b) two molecules of glucose only
    c) equal amounts of glucose and galactose	d) fructose and galactose only

    ✔ a) equal amounts of glucose and fructose — Mild-acid hydrolysis of sucrose yields equal amounts of glucose and fructose.
28. ★★★ Which combination correctly describes cellobiose?

    a) Two α-D-glucose units joined by an α-1,4 linkage; non-reducing	b) Two β-D-glucose units joined by a β-1,4 linkage; reducing
    c) Glucose and fructose joined by a 1,2 linkage; non-reducing	d) Two α-D-glucose units joined by an α-1,6 linkage; reducing

    ✔ b) Two β-D-glucose units joined by a β-1,4 linkage; reducing — Cellobiose consists of two β-D-glucose units joined by a β-1,4 linkage and is reducing.
29. ★★★ Which statement correctly distinguishes maltose from cellobiose?

    a) Maltose contains fructose, whereas cellobiose contains galactose	b) Maltose has an α-1,4 linkage, whereas cellobiose has a β-1,4 linkage
    c) Maltose is non-reducing, whereas cellobiose is reducing	d) Maltose is a polysaccharide, whereas cellobiose is a monosaccharide

    ✔ b) Maltose has an α-1,4 linkage, whereas cellobiose has a β-1,4 linkage — Maltose has two α-D-glucose units joined by an α-1,4 linkage, while cellobiose has a β-1,4 linkage.
30. ★★ A carbohydrate that yields 3 to 10 monosaccharide molecules on hydrolysis is called an—

    a) aldose	b) oligosaccharide
    c) amino sugar	d) inorganic phosphate

    ✔ b) oligosaccharide — Oligosaccharides are defined in the source as carbohydrates yielding 3 to 10 monosaccharides on hydrolysis.
31. ★★★ Which property of amino acids enables them to show both acidic and basic behaviour?

    a) They contain only a carboxyl group	b) They contain both amino and carboxyl groups
    c) They contain only an amino group	d) They contain a phosphate group

    ✔ b) They contain both amino and carboxyl groups — The amino group is basic and the carboxyl group is acidic, so amino acids are amphoteric.
32. ★★★ A peptide bond is formed between which groups of two amino acids?

    a) Two amino groups	b) Two carboxyl groups
    c) The carboxyl group of one and the amino group of another	d) The amino group of one and the phosphate group of another

    ✔ c) The carboxyl group of one and the amino group of another — The carboxyl group of one amino acid combines with the amino group of another, releasing water and forming a C–N bond.
33. ★★★ If three amino acids combine in a linear chain, the resulting product is called a—

    a) Dipeptide	b) Tripeptide
    c) Polysaccharide	d) Nucleoprotein

    ✔ b) Tripeptide — Two amino acids form a dipeptide, while three form a tripeptide.
34. ★★ Which function of amino acids is directly related to removal of nitrogenous waste from the body?

    a) Formation of urea	b) Formation of cellulose
    c) Formation of sucrose	d) Formation of rubber

    ✔ a) Formation of urea — Amino acids help in the synthesis of urea.
35. ★★★ Which statement best defines a protein?

    a) A polymer of monosaccharides joined by glycosidic bonds	b) A polymer of amino acids joined by peptide bonds
    c) A polymer of isoprene units joined by ester bonds	d) A compound formed only from fatty acids

    ✔ b) A polymer of amino acids joined by peptide bonds — Proteins are polypeptides formed by amino acids linked through peptide bonds.
36. ★★★ Which statement is correct regarding enzymes and proteins?

    a) All proteins are enzymes	b) All enzymes are proteins, but all proteins are not enzymes
    c) No enzyme is a protein	d) Only structural proteins act as enzymes

    ✔ b) All enzymes are proteins, but all proteins are not enzymes — The source states that all enzymes are proteins, but all proteins are not enzymes.
37. ★★★ Which protein is primarily associated with hair, nails, skin and feathers?

    a) Keratin	b) Insulin
    c) Hemoglobin	d) Caseinogen

    ✔ a) Keratin — Keratin is a structural protein found in hair, nails, skin, wool and feathers.
38. ★★ Which pair contains one structural protein and one functional protein, respectively?

    a) Keratin and insulin	b) Hemoglobin and collagen
    c) Myosin and keratin	d) Insulin and collagen

    ✔ a) Keratin and insulin — Keratin is structural, whereas insulin is a functional protein.
39. ★★ Which of the following is a globular protein?

    a) Keratin	b) Collagen
    c) Hemoglobin	d) Elastin

    ✔ c) Hemoglobin — Hemoglobin is listed as a globular protein; keratin, collagen and elastin are fibrous or structural proteins.
40. ★★★ Which characteristic most specifically identifies albumin?

    a) It is insoluble in water but soluble in dilute salt solution	b) It dissolves readily in water and forms a concentrated solution
    c) It is soluble only in 70–80% alcohol	d) It contains a metal as its prosthetic group

    ✔ b) It dissolves readily in water and forms a concentrated solution — Albumins are readily soluble in water and dilute salt solutions.
41. ★★ A protein is almost insoluble in water but soluble in dilute salt solution. It is most likely a—

    a) Globulin	b) Albumin
    c) Prolamin	d) Histone

    ✔ a) Globulin — Globulins are nearly insoluble in water but soluble in dilute salt solutions.
42. ★★ Which solubility pattern belongs to glutelins?

    a) Soluble in water and dilute salt solution	b) Insoluble in water and dilute salt solution but soluble in acid or dilute alkali
    c) Soluble only in 100% alcohol	d) Insoluble in all solvents

    ✔ b) Insoluble in water and dilute salt solution but soluble in acid or dilute alkali — Glutelins dissolve in acids or dilute alkalis but not in water or dilute salt solutions.
43. ★★ Which protein is correctly matched with its characteristic solubility?

    a) Zein—soluble in 70–80% alcohol	b) Albumin—soluble only in 70–80% alcohol
    c) Globulin—soluble only in strong acid	d) Histone—insoluble in water

    ✔ a) Zein—soluble in 70–80% alcohol — Prolamins such as zein are soluble in 70–80% alcohol.
44. ★★ Which distinction between histones and protamines is correct?

    a) Histones contain no sulfur, whereas protamines contain sulfur	b) Protamines are alkaline proteins and contain no sulfur
    c) Histones are insoluble in water, whereas protamines are insoluble in acids	d) Both are acidic proteins rich in glutamic acid

    ✔ b) Protamines are alkaline proteins and contain no sulfur — Protamines are alkaline simple proteins rich in basic amino acids and contain no sulfur.
45. ★★ A protein combined with a non-protein prosthetic group is called a—

    a) Simple protein	b) Conjugated protein
    c) Derived carbohydrate	d) Fibrous sugar

    ✔ b) Conjugated protein — Conjugated proteins contain a protein part and a non-protein prosthetic group.
46. ★★★ Hydrolysis of a nucleoprotein produces—

    a) A simple protein and a nucleic acid	b) A lipid and a simple sugar
    c) A metal and an amino acid	d) Only monosaccharides

    ✔ a) A simple protein and a nucleic acid — Nucleoproteins yield a simple protein and nucleic acid on hydrolysis.
47. ★★ A protein combined with carbohydrate, especially a monosaccharide, is known as—

    a) Lipoprotein	b) Glycoprotein
    c) Metalloprotein	d) Phosphoprotein

    ✔ b) Glycoprotein — Glycoproteins or mucoproteins contain protein linked with carbohydrate.
48. ★★★ Which combination forms the lipid component of the lipoproteins described in the source?

    a) Cholesterol and phospholipids	b) Glucose and fructose
    c) FAD and iron	d) DNA and RNA

    ✔ a) Cholesterol and phospholipids — The lipid portion of lipoproteins is formed from cholesterol and phospholipids.
49. ★★ Which protein is an example of a chromoprotein?

    a) Hemoglobin	b) Glutenin
    c) Ovalbumin	d) Prolamin

    ✔ a) Hemoglobin — Chromoproteins contain pigment; hemoglobin is listed as an example.
50. ★★ A protein containing Fe, Mn or Zn as an activating metal is classified as a—

    a) Metalloprotein	b) Glycoprotein
    c) Phosphoprotein	d) Flavoprotein

    ✔ a) Metalloprotein — Proteins or enzymes associated with metals such as Fe, Mn or Zn are metalloproteins.

Short and explain questions with answers

1. ★★★ Explain why deoxyribose is essential for the formation of DNA.
   Deoxyribose is a pentose sugar with the molecular formula C₅H₁₀O₄. At its 2nd carbon, an –OH group is replaced by a hydrogen atom, unlike ribose. A nitrogenous base attaches at its 1st carbon to form a deoxyribonucleoside. A phosphate group attaches at its 5th carbon to form a deoxyribonucleotide. These deoxyribonucleotides join together to form DNA.

   • States the formula and pentose nature of deoxyribose.
   • Explains the absence of –OH at the 2nd carbon and attachment of a base.
   • Relates deoxyribonucleotides to DNA formation.
2. ★★★ Describe the formation of α-D-glucose and β-D-glucose from open-chain glucose.
   In glucose, the aldehyde group at carbon 1 reacts with the hydroxyl group at carbon 5. This forms an intramolecular bond and produces a cyclic hemiacetal structure. A new –OH group is formed at carbon 1, called the anomeric carbon. If this –OH occupies the α position, α-D-glucose is formed. If it occupies the β position, β-D-glucose is formed.

   • Explains ring formation between carbon 1 and carbon 5.
   • Identifies carbon 1 as the anomeric carbon with a newly formed –OH group.
   • Distinguishes α and β forms by the position of the anomeric –OH group.
3. ★★ Explain the distinction between D-glucose and L-glucose, and relate it to optical rotation.
   D- and L-glucose are distinguished by the position of the –OH group on the terminal chiral carbon in the open-chain structure. In D-glucose, this –OH group is on the right; in L-glucose, it is on the left. D/L notation describes configuration, not necessarily the direction of optical rotation. The symbols (+) and (−) describe clockwise and anticlockwise rotation of plane-polarised light. Thus, configuration and optical rotation should not be treated as identical concepts.

   • Correctly identifies the structural basis of D and L forms.
   • Defines (+) and (−) optical rotation.
   • Distinguishes configurational nomenclature from optical rotation.
4. ★★★ Give reasons why glucose and fructose are structural isomers although both have the formula C₆H₁₂O₆.
   Both glucose and fructose contain six carbon atoms and have the molecular formula C₆H₁₂O₆. Glucose contains an aldehyde group and is therefore an aldohexose. Fructose contains a ketone group and is therefore a ketohexose. Their atoms are arranged differently, producing different structures and properties. Therefore, they are structural isomers.

   • States that both have the same molecular formula.
   • Identifies glucose as an aldohexose and fructose as a ketohexose.
   • Explains that different atom arrangements make them structural isomers.
5. ★★★ Describe the formation of a disaccharide and explain the role of a glycosidic bond.
   A disaccharide forms when two monosaccharide molecules join by a condensation reaction. One molecule of water is removed from hydroxyl groups of the two monosaccharides. The remaining molecules become connected through a C–O–C linkage. This linkage is called a glycosidic bond. For example, glucose and fructose combine to form sucrose.

   • Mentions the joining of two monosaccharides by condensation.
   • States that one water molecule is removed.
   • Defines the C–O–C linkage as a glycosidic bond and gives a valid example.
6. ★★★ Explain why sucrose is a non-reducing sugar, although glucose and fructose are reducing sugars.
   Sucrose is formed from glucose and fructose by a glycosidic linkage. The linkage involves the aldehyde-related carbon of glucose and the ketone-related carbon of fructose. Their reactive reducing groups become involved in the bond formation. As a result, sucrose does not retain a free reducing group. Therefore, sucrose has no reducing property, although its component monosaccharides are reducing sugars.

   • Identifies glucose and fructose as the component sugars of sucrose.
   • Explains involvement of their reactive carbonyl-related groups in the glycosidic bond.
   • Relates the absence of a free reducing group to the non-reducing nature of sucrose.
7. ★★ Describe the hydrolysis of sucrose and state the products formed.
   Sucrose undergoes hydrolysis in the presence of water and a mild acid. Its glycosidic bond is broken during the reaction. One molecule of sucrose produces one molecule of glucose and one molecule of fructose. The reaction may be represented as C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆. Thus, hydrolysis converts the disaccharide into two monosaccharides.

   • Mentions water and mild acid as conditions for hydrolysis.
   • Identifies glucose and fructose as the products.
   • Writes or correctly explains the balanced hydrolysis reaction.
8. ★★★ Compare cellobiose and maltose on the basis of their constituent units, linkage and source.
   Cellobiose consists of two β-D-glucose units joined by a β-1,4 glycosidic linkage. It is mainly produced by partial hydrolysis of cellulose and is a reducing sugar. Maltose consists of two α-D-glucose units joined by an α-1,4 linkage. It is generally formed by partial breakdown of starch and is also a reducing sugar. Both have the molecular formula C₁₂H₂₂O₁₁.

   • Correctly compares β-D-glucose/β-1,4 in cellobiose with α-D-glucose/α-1,4 in maltose.
   • States cellulose and starch as their respective sources.
   • Mentions their reducing nature and common molecular formula.
9. ★★ Explain the importance of amino acids in living organisms.
   Amino acids are the basic units used to form proteins. They contribute to the growth and structural development of cells and tissues. They also help in the synthesis of enzymes, hormones, vitamins and insulin. Amino acids participate in urea formation and help regulate body pH. They also contribute to the formation of haemoglobin and blood cells.

   • Identifies amino acids as units of protein formation.
   • Explains their roles in growth and synthesis of biological substances.
   • Mentions urea formation, pH regulation or haemoglobin formation.
10. ★★★ Describe how a peptide bond is formed between two amino acids.
    The carboxyl group (–COOH) of one amino acid reacts with the amino group (–NH₂) of another. The –OH from the carboxyl group and one hydrogen from the amino group combine to form water. The remaining groups become joined by a C–N bond. This bond is called a peptide bond. Two linked amino acids form a dipeptide, while many form a polypeptide.

    • Identifies the reacting carboxyl and amino groups.
    • Explains removal of one water molecule during condensation.
    • Defines the resulting C–N bond and relates it to dipeptides or polypeptides.
11. ★★★ Explain why proteins are called amphoteric substances.
    Proteins are made of amino acids, and amino acids contain both amino and carboxyl groups. The amino group can accept a proton and show basic behaviour. The carboxyl group can donate a proton and show acidic behaviour. Therefore, amino acids and proteins can show both acidic and basic properties. For this reason, they are described as amphoteric substances.

    • Mentions the presence of both –NH₂ and –COOH groups.
    • Explains the basic action of the amino group and acidic action of the carboxyl group.
    • Correctly concludes that proteins are amphoteric.
12. ★★ Describe the difference between structural proteins and functional proteins, with examples.
    Structural proteins form important physical parts of the body and tissues. Keratin occurs in hair, nails, skin, wool and feathers, while collagen occurs in bone and connective tissue. Functional proteins participate in metabolic and other biological activities. Examples include enzymes, hormones and haemoglobin. Thus, structural proteins mainly provide support, whereas functional proteins perform active biological roles.

    • Defines structural proteins and gives suitable examples.
    • Defines functional proteins and gives suitable examples.
    • Clearly contrasts support or formation with biological activity.

Fill in the blanks and True/False

Fill in the blanks

1. The molecular formula of deoxyribose is ______. [C₅H₁₀O₄]
2. A monosaccharide containing six carbon atoms is called a ______. [hexose]
3. Glucose is classified as an ______ because it contains an aldehyde group. [aldohexose]
4. Fructose contains a ______ group in its structure. [keto]
5. The hydrolysis of sucrose produces ______ and ______. [glucose and fructose]
6. The general molecular formula of a disaccharide is ______. [C₁₂H₂₂O₁₁]
7. The C–O–C linkage between two monosaccharides is called a ______ bond. [glycosidic]
8. Maltose is formed when two α-D-glucose molecules are joined by an ______ linkage. [α-1,4]
9. A peptide bond is a ______ bond formed between amino acids. [C–N]
10. Natural rubber is obtained mainly from the latex of ______ brasiliensis. [Hevea]

True / False

1. Deoxyribose contains an –OH group at its 2nd carbon atom. — False (At the 2nd carbon, deoxyribose has hydrogen instead of –OH.)
2. Glucose and fructose have the same molecular formula, C₆H₁₂O₆, but differ in structure. — True (They are structural isomers with the same molecular formula.)
3. Fructose is an aldohexose because it contains an aldehyde group. — False (Fructose is a ketohexose because it contains a keto group.)
4. Sucrose is a non-reducing sugar. — True (Its aldehyde and ketone groups lose their reducing capacity during formation.)
5. Cellobiose is formed by the linkage of two β-D-glucose molecules through a β-1,4 bond. — True (This is the linkage described for cellobiose.)
6. The formation of a peptide bond between two amino acids releases one molecule of water. — True (The amino and carboxyl groups combine with loss of water.)
7. All proteins are enzymes, although all enzymes are proteins. — False (The source states that all enzymes are proteins, but all proteins are not enzymes.)
8. Globulins are readily soluble in water and form concentrated solutions. — False (Globulins are nearly insoluble in water but soluble in dilute salt solutions.)
9. Approximately 3000–4000 isoprene units combine to form natural rubber. — True (The source gives this range for rubber formation.)
10. Sucrose has a melting point of 186° Celsius according to the source. — True (The stated melting point of sucrose is 186° Celsius.)

Exam strategy: shortcuts, traps and practice plan

Shortcuts and memory tricks

Common traps

Practice plan

1. First, make a one-page source map with these headings: nucleotides and sugars, monosaccharides, disaccharides, oligosaccharides, amino acids, proteins, protein classification, and terpenes/rubber. This prevents missing the less frequently revised topics.
2. Second, build comparison tables for ribose/deoxyribose; glucose/fructose; sucrose/maltose/lactose/cellobiose; and fibrous/globular proteins. Include formula, monomers, linkage, reducing property and examples.
3. Third, practise active recall without looking at notes: write the formulas C₅H₁₀O₅, C₅H₁₀O₄, C₆H₁₂O₆ and C₁₂H₂₂O₁₁, then reproduce the three key disaccharide hydrolysis equations and the peptide-bond reaction.
4. Fourth, revise amino-acid and protein applications separately. Test yourself on amino-acid functions, peptide-bond formation, structural versus functional proteins, solubility classes, conjugated proteins and examples such as haemoglobin, caseinogen, lipoprotein and flavoprotein.
5. Fifth, cover the final source topics with short flashcards: oligosaccharides contain 3–10 monosaccharide units, terpenes contain 10–40 carbon atoms and have the general formula (C₅H₈)n, and natural rubber contains about 3000–4000 isoprene units and is obtained mainly from Hevea brasiliensis.
6. Sixth, solve timed mixed questions in admission-test style. For every wrong answer, record whether the error was formula-based, linkage-based, classification-based, solubility-based or caused by confusing a similar term. Reattempt the error list after 24 hours.
7. Seventh, take a final full mock under realistic time pressure, then review every option rather than only the selected answer. Finish by memorizing the highest-yield contrasts: RNA/ribose versus DNA/deoxyribose, aldehyde versus ketone, reducing versus non-reducing, α-1,4 versus β-1,4, and fibrous versus globular.

Last-minute revision

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Formulas

Common mistakes