relate the molecular structure of triglycerides to their functions in living organisms

Carbohydrates and Lipids – Cambridge IGCSE/A‑Level (Syllabus 2)

Learning Objectives

• Explain how the molecular structure of carbohydrates and lipids determines their biological functions.
• Identify the main macromolecules in a sample using the qualitative tests required by syllabus 2.1.
• Describe the synthesis, breakdown and utilisation of triglycerides (syllabus 2.2 & 2.3).
• Apply knowledge to typical AO1–AO3 exam questions.

1. Key Terminology (Syllabus 2.2)

Term	Cambridge Definition
Monomer	The smallest repeat unit that can join with others to form a polymer (e.g., glucose, glycerol, fatty acid).
Polymer	A macromolecule formed by the linkage of many monomers (e.g., starch, triglyceride).
Condensation (dehydration) reaction	Joining of two monomers with loss of a water molecule, producing a covalent bond (glycosidic or ester).
Hydrolysis	The reverse of condensation – water breaks a covalent bond, releasing the monomers.
Ester bond (–CO–O–)	Covalent link formed between the –OH of glycerol and the –COOH of a fatty acid.
Saturation	Presence (saturated) or absence (unsaturated) of C=C double bonds in a fatty‑acid chain.
cis / trans configuration	Geometric arrangement of substituents around a C=C double bond; cis = same side, trans = opposite sides.

2. Carbohydrates

2.1 Glucose – the basic monomer

• Empirical formula: C₆H₁₂O₆
• Predominantly exists as cyclic hemiacetals in aqueous solution.

2.2 Formation & hydrolysis of glycosidic bonds

• Condensation: –OH on C‑1 of one glucose reacts with –OH on C‑4 (or C‑6) of another, releasing H₂O and giving an α‑ or β‑glycosidic bond depending on the orientation of the reacting –OH.
• Hydrolysis: Addition of H₂O breaks the bond, regenerating the monosaccharides.

2.3 Major polysaccharides – structure ↔ function

Polysaccharide	Linkage type	Macro‑structure	Biological role
Starch (amylose + amylopectin)	α‑(1→4) in amylose; α‑(1→4) with α‑(1→6) branches in amylopectin	Coiled helices (amylose) and branched network (amylopectin)	Short‑term energy reserve in plants; readily hydrolysed by amylase.
Glycogen	α‑(1→4) with frequent α‑(1→6) branches (≈ every 8–12 residues)	Highly branched, compact granules	Rapidly mobilisable glucose store in animals; high surface‑area for enzyme access.
Cellulose	β‑(1→4)	Linear, rigid chains that hydrogen‑bond to neighbours → microfibrils	Structural component of plant cell walls; resistant to enzymatic hydrolysis.

2.4 Structure‑function summary for carbohydrates

• α‑linkages → flexible, soluble, easily hydrolysed (energy storage).
• β‑linkages → straight, hydrogen‑bonded, insoluble (structural support).
• Branching increases surface area → faster enzymatic breakdown (glycogen vs. starch).

3. Lipids

3.1 What are lipids?

• Broad class of hydrophobic or amphipathic organic molecules.
• Relevant syllabus classes: triglycerides (neutral lipids) and phospholipids (amphiphilic lipids).

3.2 Triglyceride (triacylglycerol) – detailed structure

Formed by esterification of one glycerol molecule with three fatty‑acid molecules.

Component	Structural feature	Resulting property
Glycerol	Three‑carbon backbone; each carbon bears a –OH group	Provides three attachment points; gives a small polar “head”.
Fatty‑acid chain	Linear hydrocarbon chain (C14–C22) terminating in –COOH	Non‑polar tail; length & unsaturation control melting point, fluidity and energy content.
Ester bond (–CO–O–)	Condensation product of –OH (glycerol) and –COOH (fatty acid)	Stable covalent link; masks the polarity of the carboxyl group.

3.3 Fatty‑acid classification

Type	Structural characteristic	Effect on physical properties
Saturated	No C=C double bonds; all C–C single bonds	Straight chains → tight packing → higher melting point (solid at 25 °C).
Monounsaturated (cis)	One C=C double bond in cis configuration	Single kink → looser packing → lower melting point (liquid at 25 °C).
Polyunsaturated (cis)	Two or more C=C double bonds, all cis	Multiple kinks → very low melting point; essential fatty acids (ω‑3, ω‑6).
Trans‑unsaturated	One or more C=C double bonds in trans configuration	Chain behaves almost like saturated → higher melting point; rare in natural fats.

3.4 How structure determines function

Function	Structural basis	Biological significance
Long‑term energy storage	Long hydrocarbon tails rich in C–H bonds; esterification hides polar –COOH.	≈ 9 kcal g⁻¹ – more than twice the energy of carbohydrates.
Thermal insulation	Hydrophobic tails form a layer that traps air.	Reduces heat loss in mammals and birds.
Mechanical protection & cushioning	Dense, non‑polar droplets surrounding organs.	Protects kidneys, eyes, fetal membranes, etc.
Pre‑cursor for membrane phospholipids	Glycerol backbone can be re‑esterified with two fatty acids + a phosphate‑containing head group.	Provides the hydrophobic core of the lipid bilayer.
Transport of lipophilic substances	Incorporation into lipoprotein particles (chylomicrons, VLDL, LDL, HDL).	Solubilises triglycerides and cholesterol in the aqueous bloodstream.

3.5 Phospholipids – structure and membrane role

• Glycerol backbone esterified with two fatty‑acid chains (hydrophobic tails) and a phosphate group linked to a polar head group (e.g., choline, serine, ethanolamine).
• Amphipathic: hydrophilic head interacts with water; hydrophobic tails face each other, forming a bilayer.
• Key functions: cell‑membrane structure, vesicle formation, signal transduction, and acting as substrates for second‑messenger pathways.

4. Metabolism of Triglycerides (Syllabus 2.3)

4.1 Hydrolysis (lipolysis)

1. Hormone‑sensitive lipase (activated by epinephrine, glucagon) cleaves ester bonds.
2. Result: one glycerol molecule + three free fatty acids (FFAs).

4.2 Fate of glycerol

1. Phosphorylated by glycerol kinase → glycerol‑3‑phosphate.
2. Oxidised by glycerol‑3‑phosphate dehydrogenase → dihydroxyacetone phosphate (DHAP).
3. DHAP enters glycolysis or gluconeogenesis.

4.3 Activation of fatty acids

Fatty acid + ATP + CoA → fatty‑acyl‑CoA + AMP + PP_i (requires acyl‑CoA synthetase, occurs in the cytosol).

4.4 β‑Oxidation (mitochondrial matrix)

Step	Reaction	Products per cycle
1. Oxidation	Acyl‑CoA + FAD → trans‑Δ²‑enoyl‑CoA + FADH₂	1 FADH₂
2. Hydration	Enoyl‑CoA + H₂O → L‑β‑hydroxyacyl‑CoA	–
3. Oxidation	L‑β‑hydroxyacyl‑CoA + NAD⁺ → β‑ketoacyl‑CoA + NADH + H⁺	1 NADH
4. Thiolysis	β‑ketoacyl‑CoA + CoA‑SH → acetyl‑CoA + shortened acyl‑CoA	1 acetyl‑CoA

Each round removes a 2‑C fragment; the final round yields two acetyl‑CoA.

4.5 Energy yield (per 16‑C palmitate)

• Activation: 2 ATP (actually 2 ATP equivalents).
• β‑Oxidation: 7 cycles → 7 NADH, 7 FADH₂, 8 acetyl‑CoA.
• Oxidative phosphorylation: 7 NADH × 2.5 ATP + 7 FADH₂ × 1.5 ATP = 28 + 10.5 = 38.5 ATP.
• Citric‑acid cycle (8 acetyl‑CoA): 8 × (3 NADH + 1 FADH₂ + 1 GTP) → 24 NADH + 8 FADH₂ + 8 GTP → 60 ATP.
• Total ≈ 106 ATP per palmitate (≈ 2.6 ATP per carbon).

4.6 Transport in the bloodstream

• Triglycerides are packaged into chylomicrons (intestinal) or VLDL (liver).
• Lipoprotein lipase on capillary endothelium hydrolyses TGs, releasing FFAs for uptake by muscle or adipose tissue.
• Remnant particles become LDL (cholesterol delivery) or HDL (reverse‑cholesterol transport).

5. Practical Qualitative Tests (Syllabus 2.1)

Emulsion test

Test	Target molecule(s)	Observed result	Reaction principle
Benedict’s solution	Reducing sugars (e.g., glucose, maltose)	Blue → green → orange → brick‑red precipitate (amount‑dependent)	Cu²⁺ reduced to Cu⁺ (Cu₂O) under alkaline conditions.
Iodine solution	Starch (amylose helices)	Deep blue‑black colour	I₂ fits into the helical cavity forming a charge‑transfer complex.
Lipids (triglycerides, phospholipids)	Milky/cloudy suspension when shaken with water	Lipid droplets are insoluble in water but disperse as tiny spheres.
Biuret test	Proteins (included for completeness)	Purple colour	Cu²⁺ complexes with peptide bonds under alkaline conditions.

Sample AO2 practical question

A student adds Benedict’s solution to an unknown liquid, heats it, and observes a brick‑red precipitate. Which biomolecule is present and why?

Answer: A reducing sugar is present. The aldehyde group of the monosaccharide reduces Cu²⁺ to Cu⁺, giving a red Cu₂O precipitate.

6. Summary of Structure ↔ Function

• Carbohydrates: α‑glycosidic bonds → flexible, soluble, quick energy (starch, glycogen). β‑glycosidic bonds → straight, hydrogen‑bonded, structural (cellulose).
• Triglycerides: Long, reduced hydrocarbon tails → high energy density; saturation & chain length dictate melting point and physical state.
• Phospholipids: Amphipathic nature → spontaneous bilayer formation, essential for cell membranes.
• Fatty‑acid composition (saturated vs unsaturated, cis vs trans) directly influences fluidity of membranes and the physical state of stored fat.
• Transport of lipids in blood relies on lipoprotein particles; defects lead to clinical conditions (e.g., atherosclerosis).

7. Exam‑style Questions (AO1–AO3)

1. Structure & melting point – Explain how the degree of unsaturation in fatty‑acid chains influences the melting point of a triglyceride. Give a real‑world example (e.g., butter vs. olive oil).
2. Metabolic pathway – Describe step‑by‑step hydrolysis of a stored triglyceride and show how the resulting glycerol and fatty acids are used to generate ATP during prolonged exercise.
3. Energy comparison – Compare the ATP yield from 1 g of glucose with that from 1 g of a typical dietary triglyceride. Relate the difference to the molecular structures.
4. Carbohydrate drawing – Draw the Haworth projections of α‑D‑glucose and β‑D‑glucose. Indicate which form is present in starch and which in cellulose.
5. Interpretation of test results – Benedict’s – blue (no change); iodine – deep blue‑black; emulsion – milky suspension. Identify the biomolecule(s) present and justify your answer.
6. Transport question – Explain why triglycerides must be packaged into lipoproteins for transport in the bloodstream and describe the role of lipoprotein lipase.

8. Revision Tips

• Use colour‑coded tables: green = energy storage, blue = structural, red = membrane‑related.
• Draw one complete diagram of a triglyceride and label each part (glycerol, ester bond, fatty‑acid tail, saturated/unsaturated).
• Memorise the three key carbohydrate linkages (α‑1,4; α‑1,6; β‑1,4) and associate each with its function.
• Practice converting the β‑oxidation table into a concise flow‑chart – this is a common AO2/3 requirement.
• Link every test to the functional group it detects (e.g., aldehyde in reducing sugars, helical cavity in amylose, non‑polarity of lipids).