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This lesson covers the structure, properties and biological functions of lipids — specifically triglycerides and phospholipids — as required by the Edexcel A-Level Biology B specification (9BI0), Topic 1: Biological Molecules. You need to understand the formation of ester bonds, the difference between saturated and unsaturated fatty acids, and the relationship between the structure and function of these molecules.
Lipids are a diverse group of biological molecules that are insoluble in water but soluble in organic solvents such as ethanol and acetone. They contain the elements carbon (C), hydrogen (H) and oxygen (O), but they have a much higher proportion of hydrogen relative to oxygen compared with carbohydrates.
Lipids are not true polymers — they are not built from repeating monomer subunits in the same way as polysaccharides or proteins. However, triglycerides are formed from glycerol and fatty acids by condensation reactions.
The main types of lipid you need to know are:
A fatty acid consists of a carboxyl group (–COOH) attached to a long hydrocarbon chain (–R). The general formula is:
CH₃(CH₂)ₙCOOH
The hydrocarbon chain is typically 14–22 carbon atoms long. This long chain is non-polar and hydrophobic, while the carboxyl group is polar and hydrophilic.
| Feature | Saturated Fatty Acid | Unsaturated Fatty Acid |
|---|---|---|
| Carbon–carbon bonds | All single bonds (C–C) | One or more double bonds (C=C) |
| Hydrogen atoms | Maximum number — every carbon is "saturated" with hydrogen | Fewer hydrogen atoms — double bonds reduce the number of H atoms |
| Chain shape | Straight | Kinked/bent at the double bond(s) |
| Melting point | Higher — straight chains pack closely together, forming more van der Waals forces | Lower — kinks prevent close packing, fewer intermolecular forces |
| State at room temperature | Solid (fat) — e.g. butter, lard | Liquid (oil) — e.g. olive oil, sunflower oil |
| Source | Mainly animal fats | Mainly plant oils and fish oils |
Monounsaturated fatty acids contain one C=C double bond. Polyunsaturated fatty acids contain two or more C=C double bonds.
Key Definition: A saturated fatty acid has no carbon–carbon double bonds in its hydrocarbon chain. An unsaturated fatty acid has one or more carbon–carbon double bonds, which cause kinks in the chain.
A triglyceride is formed when one molecule of glycerol reacts with three fatty acid molecules in a series of condensation reactions.
Glycerol is a small molecule with three hydroxyl groups (–OH). Each –OH group on glycerol reacts with the –COOH group of a fatty acid:
The resulting molecule is a triglyceride (also called a triacylglycerol).
Key Definition: An ester bond is a covalent bond formed between a hydroxyl group (–OH) and a carboxyl group (–COOH) with the elimination of water. In triglycerides, ester bonds link fatty acids to glycerol.
The reverse reaction is hydrolysis: three water molecules are added to break the three ester bonds, releasing glycerol and three fatty acids.
This is catalysed by lipase enzymes during digestion.
The long hydrocarbon chains of the fatty acids are non-polar and hydrophobic. Although glycerol and the ester bonds are slightly polar, this region is small compared to the three long hydrophobic tails. Overall, the molecule is hydrophobic and does not dissolve in water.
| Function | Explanation |
|---|---|
| Energy storage | Triglycerides release more than twice the energy per gram (approximately 39 kJ g⁻¹) compared to carbohydrates (approximately 17 kJ g⁻¹), because they have a higher proportion of C–H bonds |
| Thermal insulation | Adipose tissue (containing stored triglycerides) beneath the skin reduces heat loss — important in marine mammals and Arctic animals |
| Buoyancy | Lipids are less dense than water; fat deposits help aquatic organisms float |
| Protection | Fat deposits cushion and protect internal organs (e.g. kidneys) |
| Metabolic water | When triglycerides are oxidised during respiration, they release water as a by-product — important for desert animals like the kangaroo rat |
| Electrical insulation | Myelin sheath (lipid-rich) insulates neurones and speeds up nerve impulse transmission |
| Waterproofing | Waxy lipids on leaf surfaces (cuticle) and animal skin/fur reduce water loss |
Exam Tip: In comparison questions, remember that fat produces more energy per gram than carbohydrate and more metabolic water. However, carbohydrate is a better short-term energy source because it is more easily hydrolysed and is soluble (can be transported in the blood).
Phospholipids are modified triglycerides in which one of the three fatty acid chains is replaced by a phosphate group (often with an additional polar molecule attached).
A phospholipid consists of:
This gives the phospholipid an amphipathic structure — it has both a hydrophilic region and a hydrophobic region.
| Part | Polarity | Interaction with Water |
|---|---|---|
| Phosphate head | Polar / hydrophilic | Attracted to water — faces the aqueous environment |
| Fatty acid tails | Non-polar / hydrophobic | Repelled by water — faces inward, away from the aqueous environment |
Key Definition: An amphipathic (or amphiphilic) molecule has both a hydrophilic (water-loving) region and a hydrophobic (water-hating) region. This property is fundamental to the formation of cell membranes.
When phospholipids are placed in an aqueous environment, they spontaneously arrange into a bilayer:
The bilayer acts as a barrier to most water-soluble (polar) molecules and ions, while allowing small non-polar molecules (e.g. O₂, CO₂) to pass through freely. This selective permeability is essential for maintaining different environments inside and outside the cell.
The degree of saturation of the fatty acid tails affects membrane fluidity:
| Fatty Acid Type | Effect on Membrane |
|---|---|
| Saturated fatty acids | Straight tails pack closely — membrane is less fluid (more rigid) |
| Unsaturated fatty acids | Kinked tails prevent close packing — membrane is more fluid |
Organisms adapt their membrane lipid composition in response to temperature. For example, plants growing in colder climates have a higher proportion of unsaturated fatty acids in their membranes to maintain fluidity at low temperatures.
Cholesterol (a steroid lipid) is also found in animal cell membranes, where it inserts between phospholipid molecules to regulate fluidity — reducing fluidity at high temperatures and preventing the membrane from becoming too rigid at low temperatures.
Exam Tip: When discussing membrane structure, always link the amphipathic nature of phospholipids to the formation of the bilayer. Explain that the hydrophilic heads interact with water on both the extracellular and intracellular sides, while the hydrophobic tails are shielded from water in the interior.
The emulsion test is used to detect the presence of lipids in a food sample.
Method:
Results:
| Observation | Conclusion |
|---|---|
| A cloudy white emulsion forms | Lipid is present |
| Solution remains clear | Lipid is absent |
The emulsion forms because the lipid dissolves in ethanol but is insoluble in water. When the ethanol solution is added to water, tiny droplets of lipid come out of solution and scatter light, creating the cloudy appearance.
Exam Tip: Do not confuse the emulsion test with the biuret test (for proteins) or Benedict's test (for reducing sugars). Each test is specific to one type of biological molecule. Make sure you can describe the method, the positive result and the biochemical principle for each test.
Exam Tip: When answering questions about lipids, always distinguish clearly between triglycerides (energy storage) and phospholipids (membrane structure). Many students confuse the two.
This lesson sits in Edexcel 9BI0 Topic 1 — Biological Molecules, on the structure and properties of lipids. Content statements paraphrase to: describe the structure of a triglyceride as a condensation product of glycerol and three fatty acids linked by ester bonds; distinguish saturated fatty acids (no C=C double bonds) from unsaturated fatty acids (one or more C=C double bonds, with cis and trans geometric isomers); describe the structure of a phospholipid with its hydrophilic phosphate-containing head and two hydrophobic fatty-acid tails; explain how amphipathicity drives bilayer formation; describe the emulsion test for lipids (refer to the official Pearson Edexcel 9BI0 specification for exact wording). The material is examined directly on Paper 1 and reactivated synoptically on Paper 2 (cell-surface membrane structure, Topic 2), Paper 2 (lipid metabolism and respiratory substrates, Topic 5) and Paper 3 (transport of lipids by lipoproteins and steroid-hormone signalling, Topic 7). Synoptic reach extends from triglyceride hydrolysis through β-oxidation into the Krebs cycle and from cholesterol structure into both membrane-fluidity buffering and steroid-hormone biosynthesis.
Question (8 marks): A triglyceride is hydrolysed completely in the lumen of the small intestine.
(a) Write a word equation for the hydrolysis of one triglyceride molecule, identifying all products and the number of water molecules consumed. (3)
(b) Explain quantitatively why triglycerides yield more ATP per gram on complete oxidation than carbohydrates of equivalent mass, with reference to the chemistry of the fatty-acid tail. (5)
Solution with mark scheme:
(a) Step 1 — write the equation. Triglyceride + 3H₂O → glycerol + 3 fatty acids.
M1 (AO1) — naming all products correctly. Many candidates lose marks here by writing "glycerol + fatty acids" without the stoichiometric 3 in front of fatty acids, or by writing "fats" rather than "fatty acids." A common pitfall is to omit the water reactant entirely, treating hydrolysis as if it were spontaneous bond-breaking.
Step 2 — quantify the water requirement. Three ester bonds are cleaved, so three molecules of water are consumed per triglyceride molecule.
A1 (AO1) — explicit "three water molecules" with the link to three ester bonds. The mirror-image stoichiometry of the condensation step (which released three waters) is the discriminator.
Step 3 — name the enzyme. Lipase (pancreatic lipase) catalyses the hydrolysis, with bile salts acting as emulsifiers to increase the surface area of lipid droplets accessible to the enzyme.
A1 (AO2) — naming lipase plus a synoptic mention of bile-salt emulsification.
(b) Step 1 — state the empirical figures. Lipid yields approximately 37 kJ g⁻¹ (or ~9 kcal g⁻¹) on complete oxidation; carbohydrate yields approximately 17 kJ g⁻¹ (or ~4 kcal g⁻¹) — a roughly two-fold difference per gram.
M1 (AO1) — citing the figures with units.
Step 2 — explain via H content. Fatty-acid tails are predominantly –CH₂– chains with very few oxygens; carbohydrate is roughly (CH₂O)ₙ, already partially oxidised. A typical 16-carbon palmitic-acid tail (C₁₅H₃₁COOH) has 31 hydrogen atoms on 16 carbons, whereas glucose (C₆H₁₂O₆) has only 12 hydrogens on 6 carbons and 6 oxygens already attached. The lipid is therefore more reduced — there is more chemical potential to extract by oxidative removal of hydrogens.
A1 (AO2) — connecting H:C ratio to "more reduced," not just to "more H."
Step 3 — link to the electron-transport chain. Each pair of hydrogens removed during β-oxidation feeds NADH or FADH₂ into the electron-transport chain (Topic 5), generating ~2.5 ATP and ~1.5 ATP respectively via oxidative phosphorylation. More hydrogens per gram therefore means more reduced coenzymes per gram and more ATP per gram.
A1 (AO2) — explicit reduced-coenzyme-to-ATP linkage.
Step 4 — evaluate the consequence. This is why mammals store long-term energy as adipose triglyceride: anhydrous, more energy-dense per gram, no osmotic cost. Migratory birds, hibernating mammals and seeds all converge on lipid storage for the same chemical reason.
A1 (AO3) — evaluative synthesis. The AO3 discriminator and top-band mark.
Total: 8 marks (a: M1 A1 A1; b: M1 A1 A1 A1). A clean A* response will tie the per-gram yield to reduction state (not just H count), and to anhydrous storage as the evolutionary corollary.
Question (6 marks): Phospholipids spontaneously form a bilayer when dispersed in water. Explain how the molecular structure of a phospholipid drives this self-assembly, and why the bilayer is the energetically favoured arrangement.
Mark scheme decomposition by AO:
| Mark | AO | Awarded for |
|---|---|---|
| 1 | AO1 | Identifying a phospholipid as glycerol + 2 fatty acids + a phosphate-containing head group linked by ester bonds. |
| 2 | AO1 | Stating that the molecule is amphipathic: hydrophilic phosphate head, hydrophobic fatty-acid tails. |
| 3 | AO2 | Linking amphipathicity to the hydrophobic effect — water molecules cannot hydrogen-bond with the non-polar tails, so they reorganise to minimise contact. |
| 4 | AO2 | Describing the bilayer geometry: hydrophilic heads facing the aqueous phase on both faces, tails sequestered in the interior. |
| 5 | AO2 | Linking bilayer self-assembly to entropy gain in surrounding water (release of "ordered" hydration shells around hydrophobic tails) — the thermodynamic driver. |
| 6 | AO3 | Evaluative synthesis: bilayers form without enzymes or templating because the structure is the global free-energy minimum — a self-assembling architecture from which all cellular compartmentalisation derives. |
Total: 6 marks split AO1 = 2, AO2 = 3, AO3 = 1. The AO3 mark rewards the candidate who frames bilayer formation as an entropic (water-ordering) phenomenon rather than as a simple "tails hate water" story.
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