Biomolecules ISC Class 12 Chemistry: Notes and Questions

What Biomolecules is about

Weightage: Biomolecules is part of the Organic Chemistry group, which carries 31 of the 70 marks in the ISC Chemistry Paper 1 (Theory), shared with 4 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Chemistry syllabus, examination year 2027, distribution of marks

Key concepts

Classification of carbohydrates

Monosaccharides cannot be hydrolysed further: glucose (C₆H₁₂O₆, an aldohexose), fructose (a ketohexose), ribose (aldopentose). Aldose: contains -CHO; ketose: contains >C=O. Oligosaccharides give 2 to 10 monosaccharide units on hydrolysis: disaccharides (sucrose, maltose, lactose, C₁₂H₂₂O₁₁), trisaccharides (raffinose), tetrasaccharides (stachyose). Polysaccharides give many units: starch, cellulose, glycogen ((C₆H₁₀O₅)ₙ). Sugars: sweet, crystalline, water-soluble (mono- and oligosaccharides); polysaccharides are non-sugars. Reducing sugars have a free (or potentially free) aldehyde or keto group and reduce Tollens' reagent and Fehling's solution: all monosaccharides (including fructose, which isomerises in alkali), maltose and lactose. Non-reducing sugars: sucrose, whose two units are joined through both reducing carbons. Uses/importance: glucose is the body's main fuel; starch and glycogen store energy in plants and animals; cellulose builds plant cell walls and is used for paper, cotton and rayon; ribose and deoxyribose are parts of nucleic acids.

Worked example

Classify as aldose/ketose and reducing/non-reducing: glucose, fructose, sucrose, maltose.

Try it first, then show the solutionSolution

1. Glucose: aldohexose, reducing. Fructose: ketohexose, reducing (in alkaline Tollens'/Fehling's it isomerises to aldoses).
2. Sucrose: disaccharide, non-reducing (C1 of glucose and C2 of fructose both used in the glycosidic bond). Maltose: disaccharide, reducing (one free hemiacetal group).

Answer: Only sucrose is non-reducing

Open-chain structure of glucose: evidence

Molecular formula C₆H₁₂O₆. (1) Prolonged heating with HI (and red P) gives n-hexane: $C_6{H}_{12}{O}_6+14HI\to C_6{H}_{14}+6H_{2}O+7I_2$, so the six carbons form a straight chain. (2) With hydroxylamine it forms an oxime: $HOC{H}_2(CHOH{)}_{4}CHO+N{H}_{2}OH\to HOC{H}_2(CHOH{)}_{4}CHNOH+H_{2}O$, and it adds HCN: a carbonyl group is present. (3) Mild oxidation with bromine water gives gluconic acid with the same carbon count: $HOC{H}_2(CHOH{)}_{4}CHO+B{r}_2+H_{2}O\to HOC{H}_2(CHOH{)}_{4}COOH+2HBr$, so the carbonyl is an aldehyde at the chain end. (4) Acetic anhydride gives a pentaacetate: $C_6{H}_{12}{O}_6+5(C{H}_{3}CO{)}_{2}O\to C_6{H}_7{O}_6(COC{H}_3{)}_5+5C{H}_{3}COOH$, so there are five -OH groups, on different carbons (the compound is stable). (5) Nitric acid oxidises both ends to saccharic (glucaric) acid: $HOC{H}_2(CHOH{)}_{4}CHO+3[O]\to HOOC(CHOH{)}_{4}COOH+H_{2}O$, so one end is a primary alcohol (-CH₂OH). (6) With excess phenylhydrazine (3 molecules), glucose forms a yellow crystalline osazone (m.p. about 478 K): $C_6{H}_{12}{O}_6+3C_6{H}_{5}NHN{H}_2\to C_{18}{H}_{22}{N}_4{O}_4+C_6{H}_{5}N{H}_2+N{H}_3+2H_{2}O$; C1 and C2 are involved, so glucose, fructose and mannose give the same osazone. Conclusion: CHO-(CHOH)₄-CH₂OH. Test (bromine water): glucose decolourises bromine water (aldose); fructose does not (ketose).

Worked example

What does the reaction of glucose with bromine water show, and how does it distinguish glucose from fructose?

Try it first, then show the solutionSolution

1. Bromine water, a mild oxidant, converts -CHO into -COOH (gluconic acid) without breaking the chain: glucose has an aldehyde group.
2. Fructose, a ketose, is not oxidised by bromine water, so the orange-brown colour remains.

Answer: Glucose decolourises bromine water (aldehyde present); fructose does not

Cyclic structures, anomers and D-L configuration

Objections to the open chain: glucose does not give Schiff's test or form a NaHSO₃ adduct; glucose pentaacetate does not react with hydroxylamine (no free -CHO); two crystalline forms exist, α-D-glucose (m.p. 419 K, specific rotation +112°) and β-D-glucose (m.p. 423 K, +19°), and in water either changes to an equilibrium mixture of +52.7° (mutarotation). Cyclic structure: the C5 -OH adds to the C1 aldehyde to form a six-membered ring hemiacetal (pyranose, like pyran). C1 becomes a new asymmetric centre: in α the C1 -OH is on the same side as the ring oxygen bridge in the Fischer projection (below the ring in Haworth), in β on the opposite side (above the ring). α and β are anomers; C1 is the anomeric carbon. Fructose forms a five-membered ring (furanose) through the C5 -OH and the C2 keto group; it is a ketohexose and laevorotatory (D-(−)-fructose). D and L configuration: compare the -OH on the lowest asymmetric carbon (C5 in glucose) with glyceraldehyde: -OH on the right in the Fischer projection = D; on the left = L. D/L describes configuration, not the sign of rotation, which is shown as (+) or (−).

Worked example

Glucose is an aldehyde, yet it does not react with NaHSO₃ or Schiff's reagent. Explain.

Try it first, then show the solutionSolution

1. In solution glucose exists almost entirely as cyclic α- and β-pyranose forms (hemiacetals).
2. The concentration of the open-chain aldehyde form is extremely small, too small for these sensitive aldehyde tests, though it is enough for oxidising reagents that shift the equilibrium.

Answer: It exists mainly as a cyclic hemiacetal with no free -CHO

Disaccharides and polysaccharides

Monosaccharide units are joined by a glycosidic linkage (an oxide linkage formed between the anomeric carbon of one unit and an -OH of another, with loss of water). Sucrose (cane sugar): α-D-glucose C1 joined to β-D-fructose C2; non-reducing. Hydrolysis: $C_{12}{H}_{22}{O}_{11}+H_{2}O\to C_6{H}_{12}{O}_6+C_6{H}_{12}{O}_6$ (glucose + fructose). Sucrose is dextrorotatory (+66.5°) but the mixture is laevorotatory (fructose −92.4° outweighs glucose +52.5°): inversion of cane sugar, the product is invert sugar. Maltose (malt sugar): two α-D-glucose units, C1 to C4 (α-1,4-glycosidic link); reducing (free C1 of the second unit). Lactose (milk sugar): β-D-galactose C1 linked to C4 of D-glucose (β-1,4); reducing. Starch: polymer of α-glucose; amylose (15-20%, water-soluble, unbranched α-1,4 chains, blue with iodine) and amylopectin (80-85%, insoluble, branched: α-1,4 chains with α-1,6 branches every 20-25 units). Cellulose: unbranched chains of β-D-glucose (β-1,4 links); straight chains packed by hydrogen bonds into fibres; humans cannot digest it (no enzyme for β-links), but ruminants can with gut microbes. Glycogen: 'animal starch', stored in liver and muscles; like amylopectin but more highly branched.

Worked example

Maltose is a reducing sugar but sucrose is not. Explain, and name the products of hydrolysis of each.

Try it first, then show the solutionSolution

1. In maltose, C1 of one α-D-glucose is joined to C4 of the other; the second unit keeps a free hemiacetal C1 that can open to -CHO, so it reduces Tollens' and Fehling's reagents.
2. In sucrose, C1 of glucose is joined to C2 of fructose: both anomeric carbons are used in the glycosidic bond, so no free carbonyl can form.
3. Hydrolysis: maltose gives two molecules of D-glucose; sucrose gives D-glucose and D-fructose.

Answer: Maltose has a free anomeric carbon; sucrose has none. Maltose → 2 glucose; sucrose → glucose + fructose

Amino acids, zwitterions and peptides

α-Amino acids: R-CH(NH₂)-COOH; the -NH₂ is on the carbon next to -COOH. About 20 occur in proteins. Glycine (R = H) is the only one that is not optically active; all others have an asymmetric α-carbon and occur as L-forms. Classification: neutral (one -NH₂, one -COOH: glycine, alanine), acidic (extra -COOH: aspartic and glutamic acid), basic (extra -NH₂: lysine, arginine); essential amino acids cannot be made by the body and must come from food (valine, leucine, lysine, etc.); non-essential ones can be made (glycine, alanine). Zwitterion: in water the -COOH gives H⁺ to -NH₂, forming a dipolar ion: R-CH(NH₃⁺)-COO⁻. This explains their high melting points, solubility in water and insolubility in organic solvents, and their amphoteric behaviour (they react with both acids and bases). Isoelectric point: the pH at which the amino acid exists mainly as the zwitterion with no net charge and does not move in an electric field. Peptide bond: -CO-NH- formed between the -COOH of one amino acid and the -NH₂ of the next with loss of water: $H_{2}NC{H}_{2}COOH+H_{2}NCH(C{H}_3)COOH\to H_{2}NC{H}_{2}CONHCH(C{H}_3)COOH+H_{2}O$ (glycylalanine). Two amino acids give a dipeptide, ten or more a polypeptide; above about 10 000 Da (roughly 100 amino acids) it is a protein.

Worked example

How many peptide bonds are in a tripeptide, and how many different tripeptides (each using all three) can be formed from glycine, alanine and valine?

Try it first, then show the solutionSolution

1. Three amino acids are joined by 3 − 1 = 2 peptide bonds (2 water molecules lost).
2. The sequence matters (N-terminal to C-terminal): 3! = 6 different tripeptides.

Answer: 2 peptide bonds; 6 tripeptides

Proteins: classification, structure, denaturation, enzymes and hormones

By molecular shape: fibrous proteins have long thread-like chains held side by side by hydrogen and disulphide bonds; insoluble in water (keratin in hair and nails, myosin in muscle, collagen, silk fibroin). Globular proteins have chains folded into spheres; water-soluble (insulin, albumin, haemoglobin, enzymes). Levels of structure (definitions only): primary: the sequence of amino acids in the chain; a change of one amino acid can alter function (sickle-cell haemoglobin). Secondary: the shape of the chain due to hydrogen bonds between C=O and N-H of the backbone: α-helix (right-handed coil) or β-pleated sheet. Tertiary: overall folding of the chain into fibrous or globular shape, held by hydrogen bonds, disulphide links, ionic and van der Waals forces. Quaternary: arrangement of two or more polypeptide subunits (haemoglobin has four). Denaturation: heat, change of pH, alcohol or heavy-metal ions disrupt hydrogen bonds and other interactions, destroying secondary and tertiary structure and biological activity; the primary structure (peptide bonds) is unchanged. Examples: boiling an egg (albumin coagulates), curdling of milk by lactic acid or vinegar (casein). Enzymes (elementary): biological catalysts, mostly globular proteins, highly specific, working at body temperature and near-neutral pH; they lower activation energy (e.g. maltase hydrolyses maltose, invertase sucrose, pepsin proteins). Hormones (elementary): chemical messengers secreted by endocrine glands into blood: some are proteins/peptides (insulin, controls blood glucose), some steroids (oestrogen, testosterone), some amino acid derivatives (thyroxine, adrenaline).

Worked example

Egg white turns firm and opaque when an egg is boiled, and it does not become clear again on cooling. Explain.

Try it first, then show the solutionSolution

1. Heat breaks the hydrogen bonds and other weak forces that hold the globular protein albumin in its folded secondary and tertiary structure.
2. The unfolded chains tangle and coagulate into an insoluble mass: denaturation. The peptide bonds (primary structure) are unchanged, but the original folding cannot re-form on cooling.

Answer: Heat denatures (coagulates) albumin irreversibly

Vitamins

Vitamins are organic compounds needed in small amounts in the diet for normal growth and health; most cannot be made by the body. Fat-soluble vitamins (A, D, E, K): stored in liver and fatty tissue. Water-soluble vitamins (B group, C): not stored (except B₁₂) and excreted in urine, so they must be taken regularly. Deficiency diseases: Vitamin A: night blindness, xerophthalmia (hardening of the cornea). Vitamin B₁: beriberi (loss of appetite, retarded growth). Vitamin B₂: cheilosis (fissures at the corners of the mouth), digestive disorders. Vitamin B₆: convulsions. Vitamin B₁₂: pernicious anaemia. Vitamin C: scurvy (bleeding gums). Vitamin D: rickets in children (bone deformities), osteomalacia (soft bones, joint pain) in adults. Vitamin E: increased fragility of red blood cells, muscular weakness. Vitamin K: increased blood-clotting time. Sources (examples): A: carrots, milk, fish-liver oil; B group: cereals, milk, meat, yeast; C: citrus fruits, amla, green vegetables; D: sunlight on skin, fish, egg yolk; E: vegetable oils, wheat germ; K: green leafy vegetables. Chemical names and structures are not required.

Worked example

A child has bow legs and an adult complains of joint and bone pain, both living indoors with little sunlight. Which vitamin is lacking, and is it fat- or water-soluble?

Try it first, then show the solutionSolution

1. Rickets (children) and osteomalacia (adults) are caused by vitamin D deficiency, which reduces absorption of calcium and phosphate.
2. Vitamin D is fat-soluble and is made in the skin in sunlight.

Answer: Vitamin D (fat-soluble)

Nucleic acids: DNA and RNA

Nucleic acids are long polymers (polynucleotides). Nucleoside = pentose sugar + nitrogenous base; nucleotide = nucleoside + phosphate. Nucleotides are joined by phosphodiester links between C5' of one sugar and C3' of the next. Bases: purines (two rings): adenine (A), guanine (G); pyrimidines (one ring): cytosine (C), thymine (T, DNA only), uracil (U, RNA only). DNA: sugar β-D-2-deoxyribose; bases A, G, C, T; double helix (Watson and Crick, 1953): two antiparallel strands coiled round a common axis, sugar-phosphate backbones outside, bases inside; A pairs with T (two hydrogen bonds), G with C (three hydrogen bonds), so A = T and G = C in amount (complementary strands). RNA: sugar β-D-ribose; bases A, G, C, U; usually single-stranded; three types: mRNA, rRNA, tRNA. Differences: sugar (deoxyribose vs ribose); base (T vs U); structure (double helix vs single strand); function (DNA stores genetic information and replicates; RNA carries out protein synthesis); location (DNA mainly in the nucleus, RNA mostly in the cytoplasm); DNA is much larger. No chemical structure of RNA is required.

Worked example

A sample of double-stranded DNA contains 20% adenine. Find the percentages of thymine, guanine and cytosine.

Try it first, then show the solutionSolution

1. Base pairing: A = T, so thymine = 20%.
2. A + T = 40%, so G + C = 60%; G = C = 30% each.

Answer: T 20%, G 30%, C 30%

Chemistry in real life: Biomolecules

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Carbohydrates: Glucose and Disaccharides
• Proteins: Amino Acids, Peptide Bond and Structure
• Vitamins and Deficiency Diseases

Biomolecules formula sheet

Glucose with HI (straight chain)

$C_6{H}_{12}{O}_6+14HI\to C_6{H}_{14}+6H_{2}O+7I_2$

n-Hexane formed.

Glucose oxime

$HOC{H}_2(CHOH{)}_{4}CHO+N{H}_{2}OH\to HOC{H}_2(CHOH{)}_{4}CHNOH+H_{2}O$

Carbonyl group present.

Bromine water oxidation

$HOC{H}_2(CHOH{)}_{4}CHO+B{r}_2+H_{2}O\to HOC{H}_2(CHOH{)}_{4}COOH+2HBr$

Gluconic acid; aldehyde group; test vs fructose.

Glucose pentaacetate

$C_6{H}_{12}{O}_6+5(C{H}_{3}CO{)}_{2}O\to C_6{H}_7{O}_6(COC{H}_3{)}_5+5C{H}_{3}COOH$

Five -OH groups.

Nitric acid oxidation

$HOC{H}_2(CHOH{)}_{4}CHO+3[O]\to HOOC(CHOH{)}_{4}COOH+H_{2}O$

Saccharic acid; terminal -CH₂OH.

Osazone formation

$C_6{H}_{12}{O}_6+3C_6{H}_{5}NHN{H}_2\to C_{18}{H}_{22}{N}_4{O}_4+C_6{H}_{5}N{H}_2+N{H}_3+2H_{2}O$

Three phenylhydrazine molecules; C1 and C2 involved.

Hydrolysis of sucrose

$C_{12}{H}_{22}{O}_{11}+H_{2}O\to C_6{H}_{12}{O}_6+C_6{H}_{12}{O}_6$

Glucose + fructose (invert sugar).

Peptide bond formation

$H_{2}NC{H}_{2}COOH+H_{2}NCH(C{H}_3)COOH\to H_{2}NC{H}_{2}CONHCH(C{H}_3)COOH+H_{2}O$

Glycylalanine (Gly-Ala).

Zwitterion

$RCH(N{H}_2)COOH\rightleftharpoons RCH(N{H}_3^{+})CO{O}^{-}$

Exists at the isoelectric point.

Base pairing in DNA

$A=T(2H\text{-bonds}),G\equiv C(3H\text{-bonds})$

Chargaff: %A = %T, %G = %C.

Biomolecules: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Biomolecules

The question patterns to expect, and the method that works for each.

Tricks to remember Biomolecules

Short memory aids. Say each one aloud once, then use it on a question the same day.

• All Doctors Eat Kale

  Fat-soluble vitamins: A, D, E, K. Everything else (B, C) is water-soluble.

  Use it when: Classifying vitamins.

• Pure As Gold

  PURines: Adenine and Guanine (two rings). Pyrimidines: C, T, U ('CUT the Py').

  Use it when: Bases in nucleic acids.

• Sucrose is selfish

  Sucrose uses both reducing ends in its bond, so it reduces nothing.

  Use it when: Reducing vs non-reducing sugars.

• Bromine tells aldose from ketose

  Glucose decolourises bromine water; fructose does not.

  Use it when: Glucose vs fructose.

• Primary = sequence

  1° sequence, 2° helix/sheet, 3° folding, 4° subunits.

  Use it when: Protein structure definitions.

Biomolecules: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Biomolecules

The slips examiners see most often, and what to write instead.

• ✗Mistake: Naming the wrong vitamin

  CISCE examiners' comments, 2024 and 2025: for rickets and joint pain, candidates answered vitamin A or B instead of vitamin D; for bleeding gums, they wrote vitamin B or more than one vitamin.

  Wrong working

  Rickets: vitamin A. Bleeding gums: vitamins B and C.

  Right working

  Rickets: vitamin D (fat-soluble). Bleeding gums: vitamin C (water-soluble).

  ✓Do this: Learn a one-line table: vitamin, solubility, deficiency disease. Give one answer only.

  Usually costs: 1 mark.

• ✗Mistake: Not using scientific terms

  CISCE examiners' comments, 2024 and 2025: candidates wrote that milk 'bursts' or 'cracks' instead of curdles and did not name denaturation; wrote 'amino acid' instead of zwitterion; and wrote hydrogen bond instead of glycosidic linkage for the bond joining monosaccharides.

  Wrong working

  The milk cracks; the bond is a hydrogen bond.

  Right working

  The milk curdles (denaturation of protein); monosaccharides are joined by a glycosidic linkage.

  ✓Do this: Use the named term the syllabus uses.

  Usually costs: 1 mark each.

• ✗Mistake: Calling sucrose a reducing sugar

  CISCE examiners' comments, 2024: candidates named the products of hydrolysis of sucrose but called sucrose a reducing sugar.

  Wrong working

  Sucrose gives glucose and fructose, so it is reducing.

  Right working

  Sucrose itself is non-reducing (both anomeric carbons bonded); its hydrolysis products are reducing.

  ✓Do this: Separate the parent sugar from its hydrolysis products.

  Usually costs: 1 mark.

• ✗Mistake: Confusing D/L with (+)/(−)

  The syllabus asks for D-L configuration; D-fructose is laevorotatory, so D cannot mean dextrorotatory.

  Wrong working

  D-sugars rotate light to the right.

  Right working

  D/L is configuration relative to glyceraldehyde; (+)/(−) is the measured rotation.

  ✓Do this: Always write both, e.g. D-(−)-fructose.

  Usually costs: 1 mark.

Practice questions on Biomolecules

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   Which is a ketohexose?
   (a) glucose
   (b) fructose
   (c) ribose
   (d) galactose

   Hint

   Keto at C2.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   The linkage in cellulose is:
   (a) α-1,4
   (b) β-1,4
   (c) α-1,6
   (d) peptide

   Hint

   β-glucose.

   Answer

   (b)

3. Q3 · 1 mark · Very short answerBoard level

   Fill in the blanks: Deficiency of vitamin ______ causes pernicious anaemia, and of vitamin ______ causes beriberi.

   Hint

   B group.

   Answer

   B₁₂; B₁

4. Q4 · 2 marks · Short answerHarder board level

   What is mutarotation? A freshly made solution of α-D-glucose ([α] = +112°) changes to +52.7° at equilibrium. Taking β-D-glucose as +19° and ignoring the open-chain form, estimate the percentage of the β-anomer at equilibrium.

   Hint

   Let x be the fraction of α: 112x + 19(1 − x) = 52.7.

   Answer

   Mutarotation is the spontaneous change in specific rotation of a freshly prepared solution of an anomer to an equilibrium value as α and β interconvert through the open chain. 112x + 19(1 − x) = 52.7 gives 93x = 33.7, x = 0.362: about 36% α and 64% β.

5. Q5 · 1 mark · Very short answerBoard level

   How many molecules of phenylhydrazine react with one glucose molecule, and what product forms?

   Hint

   Osazone.

   Answer

   Three; glucosazone (a yellow crystalline osazone).

6. Q6 · 2 marks · Short answerHarder board level

   Why is glycine optically inactive while alanine is optically active?

   Hint

   Asymmetric carbon.

   Answer

   In glycine the α-carbon carries two H atoms, so it is not asymmetric. In alanine it carries H, CH₃, NH₂ and COOH (four different groups): it is chiral.

7. Q7 · 2 marks · Short answerBoard level

   Distinguish between nucleoside and nucleotide with one example each.

   Hint

   Phosphate.

   Answer

   Nucleoside = sugar + base (adenosine). Nucleotide = sugar + base + phosphate (adenosine monophosphate).

8. Q8 · 3 marks · Short answerHarder board level

   Compare amylose, amylopectin and glycogen.

   Hint

   Branching.

   Answer

   Amylose: unbranched α-1,4 glucose chains, water-soluble, 15-20% of starch, blue with iodine. Amylopectin: α-1,4 chains with α-1,6 branches every 20-25 units, insoluble, 80-85% of starch. Glycogen: animal starch with the same links but more highly branched, stored in liver and muscle.

9. Q9 · 2 marks · Short answerHarder board level

   In a DNA sample guanine is 35% of the bases. Calculate the % of adenine, and explain the rule used.

   Hint

   G = C.

   Answer

   G = C = 35%, so A + T = 30%, A = 15%. Complementary pairing (A with T, G with C) makes these amounts equal in double-stranded DNA.

10. Q10 · 2 marks · Short answerBoard level

    Give two reasons why enzymes are called biological catalysts and one example.

    Hint

    Activation energy.

    Answer

    They speed up biochemical reactions by lowering activation energy, are not used up, and are highly specific and active at body temperature; e.g. invertase hydrolyses sucrose.

11. Q11 · 5 marks · Long answerHarder board level

    (a) Draw the Haworth structure of β-D-glucopyranose. [2]
    (b) Explain why glucose and fructose give the same osazone. [1]
    (c) What are essential amino acids? Name two. [2]

    Hint

    C1 -OH up for β.

    Answer

    (a) Six-membered ring (5 C + O), CH₂OH on C5 above the ring, -OH at C1 above (β), C2 below, C3 above, C4 below. (b) Osazone formation involves only C1 and C2; glucose and fructose have identical configurations at C3-C6. (c) Amino acids that the body cannot make and must be obtained from food: e.g. valine, leucine (lysine, threonine).

12. Q12 · 2 marks · Short answerHOTS / competency

    A protein loses its enzyme activity when heated to 333 K, but its amino acid sequence is unchanged. Explain.

    Hint

    Which bonds break?

    Answer

    Heating breaks hydrogen bonds and other weak interactions that maintain the secondary and tertiary structure, so the active site loses its shape (denaturation). Peptide bonds are not broken, so the primary structure stays.

13. Q13 · 2 marks · Short answerHOTS / competency

    Glucose pentaacetate does not react with hydroxylamine, but glucose does. What does this show about glucose?

    Hint

    Hemiacetal.

    Answer

    In the pentaacetate the C1 -OH (hemiacetal) is acetylated, so the ring cannot open to give a free -CHO. Glucose's ring can open slightly, so it reacts. Hence glucose exists as a cyclic hemiacetal in equilibrium with a little open-chain form.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Define carbohydrates.Show answer

   Optically active polyhydroxy aldehydes or ketones, or compounds that give them on hydrolysis.

2. 2.Products of hydrolysis of lactose?Show answer

   D-galactose and D-glucose.

3. 3.What are anomers?Show answer

   Cyclic sugars differing only in configuration at the anomeric carbon (C1 of glucose): α and β.

4. 4.What is a peptide bond?Show answer

   The -CO-NH- amide link formed between -COOH of one amino acid and -NH₂ of another.

5. 5.What is denaturation?Show answer

   Loss of secondary and tertiary structure and biological activity of a protein on heating or pH change.

6. 6.Vitamin for blood clotting?Show answer

   Vitamin K.

7. 7.Pyrimidine bases in DNA?Show answer

   Cytosine and thymine.

8. 8.Sugar in RNA?Show answer

   β-D-ribose.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you classify any sugar as aldose/ketose and reducing/non-reducing?
• ☐Can you write what HI, NH₂OH, Br₂ water, (CH₃CO)₂O, HNO₃ and phenylhydrazine show about glucose?
• ☐Can you draw α- and β-glucopyranose?
• ☐Can you explain inversion of cane sugar?
• ☐Can you describe the glycosidic links in maltose, lactose, starch and cellulose?
• ☐Can you define zwitterion, isoelectric point and the four levels of protein structure?
• ☐Can you match each vitamin to its deficiency disease and solubility?
• ☐Can you give five differences between DNA and RNA?

Biomolecules: frequently asked questions