AQA A Level Biology: Biological Molecules — Revision Notes

Condensed for the final weeks. For the full explanation, use the Biological Molecules study guide.

Condensation and hydrolysis

Condensation joins two molecules, forming a bond and releasing one water molecule. Hydrolysis is the reverse — it breaks a bond using a water molecule.

Monomer	Polymer	Bond
Monosaccharide	Polysaccharide	Glycosidic
Amino acid	Polypeptide	Peptide
Nucleotide	Nucleic acid	Phosphodiester

Lipids are not polymers — triglycerides are one glycerol plus three fatty acids joined by ester bonds, and there is no repeating monomer unit.

Carbohydrates

Polysaccharide	Made of	Structure	Suits its function because
Starch (amylose)	α-glucose, 1,4	Unbranched helix	Compact, insoluble — storage without osmotic effect
Starch (amylopectin)	α-glucose, 1,4 + 1,6	Branched	Many ends for rapid hydrolysis
Glycogen	α-glucose	Highly branched	Even faster release — suits high animal metabolic rate
Cellulose	β-glucose	Straight chains, alternate molecules inverted	Hydrogen bonds between chains form microfibrils of great tensile strength

The α/β distinction is the whole answer to cellulose questions. Because β-glucose has the OH group above carbon 1, alternate molecules must flip 180° to bond, producing straight chains rather than a helix.

Lipids

Triglyceride = glycerol + 3 fatty acids, ester bonds. Saturated fatty acids have no C=C; unsaturated have one or more, which produce kinks that stop close packing — hence oils are liquid at room temperature.

Phospholipids replace one fatty acid with a phosphate group, giving a hydrophilic head and hydrophobic tails. In water they form a bilayer, tails inward — the basis of every membrane.

Proteins

Level	Description	Bonds
Primary	Sequence of amino acids	Peptide
Secondary	α-helix or β-pleated sheet	Hydrogen
Tertiary	Overall 3-D shape	Hydrogen, ionic, disulfide, hydrophobic
Quaternary	Two or more polypeptides	Same as tertiary

Primary structure determines all the others, because the sequence of R groups determines which bonds can form where.

Disulfide bridges are tertiary, not secondary, even though they are bonds between amino acids like hydrogen bonds are. Secondary structure comes only from hydrogen bonds between nearby amino acids in a regular, repeating pattern (the alpha helix or beta sheet); disulfide bridges instead form between R groups that may be far apart in the primary sequence, contributing to the molecule’s overall 3-D shape — which is exactly what makes them a tertiary-level feature.

Worked example. A polypeptide is 250 amino acids long. The number of bonds in the chain is one fewer than the number of monomers, so 249 water molecules are released during its synthesis by condensation, and 249 water molecules are used during its complete hydrolysis. It is the number of bonds, not the number of monomers, that the calculation turns on — a mistake worth watching for.

• Fibrous (collagen, keratin) — long, insoluble, structural, repetitive sequence.
• Globular (haemoglobin, enzymes) — compact, soluble, hydrophilic groups outward.

Enzymes

Enzymes lower the activation energy by forming an enzyme–substrate complex.

Induced fit rather than lock-and-key: the active site is not a rigid complementary shape but changes shape as the substrate binds, straining the substrate’s bonds and so lowering activation energy. Say “changes shape to become complementary” — that phrase carries the mark.

Factor	Effect
Temperature	Rate rises to an optimum, then falls sharply as hydrogen and ionic bonds break, changing the tertiary structure — denaturation
pH	Similar; extreme pH disrupts ionic and hydrogen bonds in the active site
Substrate concentration	Rises, then plateaus when all active sites are occupied — enzyme concentration is now limiting
Competitive inhibitor	Similar shape to substrate, binds the active site; effect reduced by more substrate
Non-competitive inhibitor	Binds elsewhere, changing the active site’s shape; not overcome by more substrate

Denaturation is not “the enzyme is killed” — enzymes are not alive. It is the loss of tertiary structure so the active site is no longer complementary.

Food tests

Test	Reagent	Positive result
Starch	Iodine in KI	Yellow-brown → blue-black
Reducing sugar	Benedict’s, heat	Blue → brick red
Non-reducing sugar	Boil with HCl, neutralise, then Benedict’s	Brick red
Protein	Biuret	Blue → purple
Lipid	Ethanol then water	White emulsion

The non-reducing sugar test must be preceded by a negative Benedict’s test, and the acid must be neutralised before adding Benedict’s, or the result is invalid.

Exam traps

• Calling lipids polymers.
• Describing enzymes with lock-and-key when induced fit is required.
• Saying enzymes are “killed” or “destroyed” by heat.
• Forgetting to neutralise after acid hydrolysis in the non-reducing sugar test.
• Explaining cellulose strength without mentioning inverted alternate β-glucose molecules.
• Saying more substrate overcomes a non-competitive inhibitor.

Self-test

1. Define condensation and name the bond formed between two monosaccharides.
2. Why is cellulose strong, in structural terms?
3. Explain induced fit and how it lowers activation energy.
4. Why does the rate plateau at high substrate concentration?
5. Describe the non-reducing sugar test in full.

Answers: 1. A reaction joining two molecules with the release of a water molecule; a glycosidic bond. 2. β-glucose requires alternate molecules to be inverted, producing straight unbranched chains that hydrogen-bond to each other to form microfibrils with high tensile strength. 3. The active site is not initially complementary but changes shape as the substrate binds, straining the substrate’s bonds so less energy is needed to break them. 4. All active sites are occupied, so enzyme concentration becomes the limiting factor. 5. Confirm a negative Benedict’s test first; boil with dilute HCl to hydrolyse, neutralise with sodium hydrogencarbonate, then heat with Benedict’s — brick red indicates a non-reducing sugar was present.