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AS 2

Biological molecules

AS Level · Topic 2 · Papers 1, 2 and 3

🎯What you need to be able to do

  • Describe and carry out the Benedict’s, iodine, emulsion and biuret tests, and interpret the results.
  • Carry out a semi-quantitative Benedict’s test and a test for non-reducing sugars using acid hydrolysis.
  • Draw the ring forms of α- and β-glucose and describe the formation and breakage of a glycosidic bond.
  • Relate the structures of starch, glycogen and cellulose to their functions.
  • Describe triglycerides and phospholipids, and relate triglyceride structure to function.
  • Explain the four levels of protein structure and the bonds that hold them, using haemoglobin and collagen.
  • Relate hydrogen bonding in water to solvent action, high specific heat capacity and latent heat of vaporisation.

📚The biology

The four food tests

  • Reducing sugars — Benedict’s. Add Benedict’s solution, heat in a water bath. Blue → green → yellow → orange → brick-red precipitate as concentration rises. Glucose, fructose and maltose are reducing sugars; sucrose is not.
  • Starch — iodine. Iodine in potassium iodide solution: orange-brown → blue-black. No heating.
  • Lipids — emulsion test. Dissolve the sample in ethanol, then pour into water. A white emulsion means lipid. The ethanol step is not optional — lipid must dissolve first, and adding water straight to the sample gives nothing.
  • Proteins — biuret. Add biuret reagent (or sodium/potassium hydroxide then dilute copper sulfate): blue → purple/lilac. No heating. The colour comes from copper ions complexing with peptide bonds, so it detects peptide bonds, not amino acids.

Making Benedict’s semi-quantitative

Benedict’s is qualitative on its own. To estimate concentration you must standardise everything except the sugar: same volume of sample, same volume of Benedict’s, same water-bath temperature, same heating time. Then either

  • time how long it takes for the first colour change — a shorter time means a more concentrated sample; or
  • compare the final colour against a set of standards made from known concentrations.

The more precise version filters off the precipitate and reads the remaining blue colour in a colorimeter, which converts a judgement of colour into a number and removes the observer.

Testing for non-reducing sugars

A negative Benedict’s result does not mean “no sugar”. To test for a non-reducing sugar such as sucrose:

  1. Run Benedict’s on a fresh sample first and confirm it is negative — otherwise you cannot tell later whether the sugar was there all along.
  2. Boil a second sample with dilute hydrochloric acid. This hydrolyses the glycosidic bond, releasing the monosaccharides.
  3. Neutralise with sodium hydrogencarbonate — Benedict’s needs alkaline conditions and will not work in acid.
  4. Run Benedict’s again. A positive result now means a non-reducing sugar was present.
Skipping the control or the neutralisation. Without the first negative test, a positive second test proves nothing. Without neutralising, the second test stays blue and you wrongly conclude there is no sugar at all. Both steps carry marks in their own right.

Carbohydrates

Monomers join by condensation (water released) to form a glycosidic bond, and are separated by hydrolysis (water used) — the same reaction that the non-reducing sugar test drives with acid. The difference between the two glucose isomers sits in the position of one hydroxyl group on carbon 1, and it changes everything downstream:

  • Starch (α-glucose) — amylose is an unbranched 1,4-linked chain that coils into a compact helix; amylopectin is branched with 1,6 links. Compact, insoluble so it does not affect water potential, and the branch ends give many points for rapid hydrolysis. The plant storage carbohydrate.
  • Glycogen (α-glucose) — like amylopectin but more highly branched. The animal storage carbohydrate; the extra branching suits a higher metabolic rate.
  • Cellulose (β-glucose) — alternate residues are flipped 180°, so the chain is straight rather than helical. Straight chains lie parallel and hydrogen-bond to each other along their length, forming microfibrils of very high tensile strength. That is why plant cell walls resist turgor pressure without bursting.

Lipids

A triglyceride is glycerol plus three fatty acids joined by ester bonds, formed by condensation. Fatty acids are saturated (no C=C, straight chains that pack closely, solid at room temperature) or unsaturated (one or more C=C causing kinks, so they pack poorly and are liquid).

Triglycerides are non-polar and hydrophobic, which is exactly why they suit their functions: energy storage at more than twice the energy per gram of carbohydrate, stored anhydrously so no water is carried with them, thermal insulation, buoyancy, and a metabolic water source when oxidised.

A phospholipid replaces one fatty acid with a phosphate group. The result has a hydrophilic (polar) phosphate head and two hydrophobic (non-polar) fatty acid tails — the property that builds every membrane in Topic 4.

Proteins

Amino acids share a central carbon carrying an amino group, a carboxyl group, a hydrogen and a variable R group. Condensation between the amino group of one and the carboxyl of the next forms a peptide bond; hydrolysis breaks it.

  • Primary — the sequence of amino acids, held by peptide bonds. Everything else follows from it.
  • Secondary — α-helix or β-pleated sheet, held by hydrogen bonds between C=O and N–H groups of the backbone.
  • Tertiary — the overall 3-D fold, held by interactions between R groups: hydrophobic interactions, hydrogen bonds, ionic bonds, and covalent disulfide bonds.
  • Quaternary — two or more polypeptides, and often a prosthetic group, associating into one functional molecule.

Globular proteins fold with hydrophobic R groups inward and hydrophilic outward, so they are soluble and take physiological roles. Fibrous proteins have repetitive sequences, form long strands, are insoluble, and take structural roles.

Haemoglobin is the globular example: quaternary structure from two α-globin and two β-globin chains, each with a haem prosthetic group whose central Fe2+ ion binds one oxygen molecule. Four haems means four oxygens per molecule, and the interaction between the four subunits produces the cooperative binding you meet in Topic 8.

Collagen is the fibrous example: three polypeptides, each a left-handed helix, wound round each other into a triple helix held by hydrogen bonds. Every third residue is glycine — the smallest amino acid — because only glycine fits where the chains press together. Molecules lie staggered end to end and are cross-linked covalently into fibrils, giving very high tensile strength with some flexibility.

Water

Water is polar: oxygen pulls the shared electrons, leaving it slightly negative and the hydrogens slightly positive, so molecules hydrogen-bond to each other. Three consequences are examinable, and each must be linked to a biological role:

  • Solvent action — polar and ionic solutes dissolve, so metabolic reactions occur in solution and substances are transported dissolved in blood, xylem sap and phloem sap.
  • High specific heat capacity — much energy is needed to break the hydrogen bonds, so water warms and cools slowly. Aquatic habitats are thermally stable and organisms with a high water content resist temperature swings.
  • High latent heat of vaporisation — much energy is removed when water evaporates, so sweating and transpiration are efficient cooling.

Cohesion between water molecules also matters later — it is what holds the transpiration stream together in Topic 7.

✏️Worked example

A student is given solution X and told it contains one sugar. They carry out: Test 1 — Benedict’s on 2 cm³ of X, heated for 3 minutes: stays blue. Test 2 — 2 cm³ of X boiled with dilute HCl, cooled, sodium hydrogencarbonate added until fizzing stops, then Benedict’s heated for 3 minutes: orange precipitate. (a) Identify the sugar and justify it. (b) Explain the purpose of each step in Test 2. (c) The student now wants to compare X with three other solutions for concentration. Describe how.

(a) Test 1 negative rules out a reducing sugar — glucose, fructose and maltose would all have given a coloured precipitate. Test 2 positive shows that a reducing sugar appeared after hydrolysis, so X contained a non-reducing sugar, and at this level that means sucrose, which hydrolyses to glucose and fructose.

(b) Boiling with dilute HCl hydrolyses the glycosidic bond, releasing the two monosaccharides, both of which are reducing. Sodium hydrogencarbonate neutralises the acid, because Benedict’s reagent only works in alkaline conditions — left acidic, the tube stays blue and you would wrongly report no sugar. Test 1 is the control: it establishes that any reducing sugar detected in Test 2 was produced by the hydrolysis rather than present from the start.

(c) Make the test semi-quantitative by standardising every variable except the sugar: the same volume of each solution (2 cm³), the same volume of Benedict’s, the same water-bath temperature, and the same heating time of 3 minutes. Then either time how long each takes to reach its first colour change — the shortest time is the most concentrated — or compare final colours against standards prepared from known concentrations. For a numerical result, filter off the precipitate and read the absorbance of the remaining blue Benedict’s in a colorimeter: the more sugar, the less blue remains.

Check it. The logic of the pair of tests should read as a syllogism, and you can test it by asking what a different result would have meant. Positive in Test 1 would have meant a reducing sugar and Test 2 would then have been pointless. Negative in both would mean no sugar at all — or, more likely in a real lab, that you forgot to neutralise. Only the negative-then-positive pattern identifies a non-reducing sugar, which is why both tests must be reported together.
Reading the colour instead of the precipitate, and forgetting sucrose is a disaccharide of two reducing monosaccharides. Benedict’s results are judged on the precipitate that forms, not the shade of the solution above it — a green tinge with a red precipitate is a positive result. And in part (a), do not write that sucrose “becomes reducing”. It does not: it is hydrolysed into glucose and fructose, which were reducing all along but were locked into a glycosidic bond that tied up the reactive group.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Explain why cellulose has high tensile strength but starch does not, in terms of the monomer used.
Cellulose is built from β-glucose. Because the hydroxyl group on carbon 1 points the other way, alternate residues must be rotated 180° for the glycosidic bond to form. The result is a straight, unbranched chain. Straight chains lie parallel and form many hydrogen bonds between adjacent chains along their whole length, bundling into microfibrils that resist being pulled apart. Starch uses α-glucose, which gives a chain that coils into a helix (amylose) or branches (amylopectin); coiled and branched molecules cannot pack side by side, so no cross-chain bonding network forms. The trade-off is deliberate: starch is compact and easily hydrolysed for storage, cellulose is strong and hard to digest for structure.
2. A protein is heated to 70 °C and loses its function. State which levels of structure are affected and which is not, and name the bonds broken.
Heating breaks the weak interactions holding the tertiary structure — hydrogen bonds, ionic bonds and hydrophobic interactions between R groups — and also the hydrogen bonds of the secondary structure. Any quaternary structure is lost as the subunits separate. The primary structure is unaffected: peptide bonds are covalent and are not broken by this temperature. This is why denaturation changes shape without changing the amino acid sequence, and why a denatured enzyme has a disrupted active site but the same polypeptide. Note that disulfide bonds are also covalent and largely survive heating.
3. Describe how you would test a food sample for both lipid and protein, and state the positive result for each.
Lipid (emulsion test): grind the sample, add ethanol and shake to dissolve any lipid, allow the solid to settle, then pour the ethanol layer into a tube of distilled water. A white/milky emulsion indicates lipid, because the lipid comes out of solution as tiny droplets that scatter light. Protein (biuret test): add biuret reagent, or sodium hydroxide followed by a few drops of dilute copper sulfate, to the sample at room temperature. A colour change from blue to purple or lilac indicates protein. Neither test requires heating; only Benedict’s does. A control of distilled water treated identically strengthens both.
4. Relate the structure of haemoglobin to its function, including the role of iron.
Haemoglobin is a globular protein with a quaternary structure of four polypeptides — two α-globin and two β-globin chains. Hydrophobic R groups are folded inward and hydrophilic ones outward, making it soluble so it can be carried dissolved in the cytoplasm of red blood cells. Each chain holds a haem prosthetic group containing an Fe2+ ion, and each Fe2+ binds one oxygen molecule reversibly — reversibly is essential, since oxygen must be released at the tissues. Four haem groups mean up to four O₂ molecules per haemoglobin. Binding of the first oxygen changes the shape of the whole molecule so that the remaining sites bind more readily, which is cooperative binding and gives the sigmoid dissociation curve of Topic 8.
5. A student adds Benedict’s solution to a sample and heats it, obtaining a green colour. A second sample gives brick-red. What can and cannot be concluded?
Both samples contain a reducing sugar. The colour sequence blue → green → yellow → orange → brick-red follows increasing concentration, so the second sample contains more reducing sugar than the first. What cannot be concluded: (i) the identity of the sugar — glucose, fructose and maltose all give the same result; (ii) any numerical concentration, since the test is only semi-quantitative even when standardised, and colour judgement is subjective; (iii) that the first sample has little sugar in absolute terms, because volumes, heating time and temperature were not stated as controlled. To compare properly, standardise those variables and use a colorimeter.
6. Explain how the high specific heat capacity and high latent heat of vaporisation of water both arise from the same property, and give one biological consequence of each.
Both arise from hydrogen bonding between water molecules, which is possible because water is polar. High specific heat capacity: a large input of energy goes into weakening these hydrogen bonds rather than into increasing molecular kinetic energy, so the temperature rises slowly. Consequence — large bodies of water are thermally stable habitats, and organisms with high water content resist rapid temperature change, keeping enzymes near their optimum. High latent heat of vaporisation: a large input of energy is needed to break hydrogen bonds completely so that a molecule can escape as vapour. Consequence — evaporation removes a lot of heat for a small mass of water, making sweating in mammals and transpiration in plants efficient cooling mechanisms.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • RCSB Protein Data Bank — rotate real haemoglobin and collagen structures in 3-D; seeing the four subunits and the haem groups makes quaternary structure concrete
  • Any well-filmed A Level food tests practical — particularly the emulsion test, where the ethanol step is the one people get wrong
  • Nuffield Foundation practical biology — standardised protocols for the semi-quantitative Benedict’s test and colorimeter use