Cambridge International AS & A Level Biology 9700: Carbohydrates and lipids – Revision Notes

New to this content? Start with the Carbohydrates and lipids study guide, which explains everything here in full with worked examples.

These notes cover section 2.2 Carbohydrates and lipids (learning outcomes 1 to 11) of Topic 2, Biological molecules, in the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027. It is AS Level content, examined on Paper 1 and Paper 2; Paper 4 can draw on it because A Level questions require knowledge of the AS content.

Other pages for this topic: Biological molecules revision notes (all of Topic 2, including the biochemical tests) · Carbohydrates and lipids practice questions · Cambridge A Level Biology hub · 9700 checklist · Free 9700 AS diagnostic

Definitions (2.2.2 and 2.2.3)

Term	Definition to learn
Monomer	Small molecule that joins with others like it to form a polymer
Polymer	Large molecule made of many repeating monomer units
Macromolecule	Very large biological molecule (polysaccharide, protein, nucleic acid)
Monosaccharide	Single sugar unit, e.g. glucose, fructose
Disaccharide	Two monosaccharides joined by a glycosidic bond, e.g. maltose, sucrose
Polysaccharide	Many monosaccharides joined by glycosidic bonds, e.g. starch, glycogen, cellulose

Role of covalent bonds: they join monomers together to make polymers. Glycosidic bonds (sugars) and ester bonds (lipids) are covalent. They are strong and need hydrolysis to break.

Triglycerides are large but not polymers (no repeating monomer chain).

Glucose rings (2.2.1)

• Ring = 5 C + 1 O. Carbon 6 is the CH₂OH outside the ring, on carbon 5.
• Ring OH pattern (Haworth view): C2 below, C3 above, C4 below.
• Carbon 1 decides the form: OH below = α; OH above = β.
• In β-glucose, the C1 OH is on the same side of the ring as the CH₂OH.

Reducing and non-reducing sugars (2.2.4)

Sugar	Type	Reducing?
Glucose	Monosaccharide	Yes
Fructose	Monosaccharide	Yes
Maltose	Disaccharide (α-glucose + α-glucose)	Yes
Sucrose	Disaccharide (α-glucose + fructose)	No

Condensation and hydrolysis (2.2.5 and 2.2.6)

Method in steps: forming a glycosidic bond (condensation)

1. OH on carbon 1 of one sugar lines up with an OH on the other sugar.
2. H from one OH and the whole OH from the other leave as H₂O.
3. The sugars are left joined by a C–O–C glycosidic bond.

Method in steps: breaking it (hydrolysis)

1. Water is added across the glycosidic bond.
2. The bond breaks; each sugar gets back an OH group.
3. In cells, enzymes catalyse this; in the lab, boiling with dilute HCl does it.

Product	Monomers	Bond
Maltose	α-glucose + α-glucose	α-1,4 glycosidic
Sucrose	α-glucose + fructose	1,2 glycosidic
Amylose	α-glucose	α-1,4
Amylopectin, glycogen	α-glucose	α-1,4 chains, α-1,6 branch points
Cellulose	β-glucose	β-1,4

Counting rule: an unbranched chain of n monomers has n − 1 bonds, so n − 1 waters released on making it and n − 1 used to fully hydrolyse it.

Worked reminder: an amylose chain of 300 glucose units forms by 299 condensation reactions.

Non-reducing sugar test in steps

1. Benedict’s test on the sample: stays blue (no reducing sugar).
2. Fresh sample + dilute HCl, heat in a boiling water bath: hydrolyses the glycosidic bond.
3. Cool and neutralise with sodium hydrogencarbonate.
4. Benedict’s again and heat: orange-red / brick-red precipitate = non-reducing sugar was present.

Polysaccharides (2.2.7 and 2.2.8)

	Amylose	Amylopectin	Glycogen	Cellulose
Monomer	α-glucose	α-glucose	α-glucose	β-glucose
Branched?	No	Yes	Yes, more than amylopectin	No
Shape	Helix	Branched	Compact, highly branched	Straight chains
Role	Plant store	Plant store	Animal and fungal store	Plant cell wall

Storage polysaccharides: structure → function

• Compact (coils, branches) → lots of glucose in small space.
• Insoluble → no effect on water potential; stays in the cell.
• Branches → many chain ends → enzymes release glucose quickly (glycogen fastest).
• Glycosidic bonds hydrolysed easily by enzymes when energy is needed.

Cellulose: arrangement → cell wall function

• β-glucose: every other unit rotated 180° → straight chain.
• Parallel chains linked by many hydrogen bonds → microfibrils → fibres.
• Fibres in layers in a matrix → high tensile strength: wall resists bursting when water enters by osmosis (cell becomes turgid).
• Gaps between fibres → wall is freely permeable.

Triglycerides (2.2.9 and 2.2.10)

• 1 glycerol + 3 fatty acids, joined by 3 ester bonds, releasing 3 H₂O (condensation).
• Non-polar and hydrophobic. Insoluble in water.
• Saturated fatty acid: no C=C, straight chain, formula CₙH₂ₙO₂.
• Unsaturated fatty acid: one or more C=C; each double bond = kink, and 2 fewer H atoms.

Worked reminder: tripalmitin = glycerol (Mr 92) + 3 palmitic acid (Mr 256) − 3 water: 92 + 768 − 54 = 806.

Structure	Function
Many C–H bonds, little O	More energy per gram than carbohydrate
Insoluble	Large stores with no osmotic effect
Low density	Buoyancy
Poor heat conductor	Insulation (fat under skin, blubber)
Oxidation produces water	Metabolic water
Cushioning layers	Protection of organs

Phospholipids (2.2.11)

• Glycerol + 2 fatty acids (ester bonds) + phosphate group.
• Phosphate head: charged, polar, hydrophilic.
• Fatty acid tails: non-polar, hydrophobic.
• Bilayer formation is Topic 4: Cell membranes and transport revision notes.

Drawing and calculation checklists

Drawing a glucose ring (check every box)

• Hexagon with the O in the top right corner.
• Carbons 1 to 5 in the ring, numbered clockwise from the right-hand corner.
• CH₂OH (carbon 6) above carbon 5.
• An H and an OH on each of carbons 1 to 4, on opposite sides of the ring.
• C1 OH placed correctly for the form asked (α below, β above).

Drawing a glycosidic bond

• Show both rings, with only one O bridging carbon 1 of the first sugar and the correct carbon of the second.
• Show H₂O leaving (condensation) or joining (hydrolysis) with an arrow.
• Label the bond “glycosidic” and give its number, e.g. α-1,4.

Mr method in steps

1. Add the Mr values of all the monomers.
2. Count the bonds formed (n − 1 for an unbranched chain; 3 ester bonds for a triglyceride).
3. Subtract 18 for each bond.

Worked reminder: maltose = 180 + 180 − 18 = 342.

Must-know distinctions

Don’t confuse	Difference
α-glucose vs β-glucose	C1 OH below vs above the ring
Condensation vs hydrolysis	Releases water to make a bond vs uses water to break one
Glycosidic vs ester bond	Sugar to sugar vs fatty acid to glycerol
Amylose vs amylopectin	Unbranched helix, α-1,4 only vs branched, α-1,4 and α-1,6
Amylopectin vs glycogen	Plants, fewer branches vs animals, more branches
Starch vs cellulose	α-glucose, energy store vs β-glucose, structural
Bonds within vs between cellulose chains	Glycosidic within; hydrogen bonds between
Triglyceride vs phospholipid	3 fatty acids, wholly hydrophobic vs 2 fatty acids + phosphate, hydrophilic head
Saturated vs unsaturated	No C=C vs at least one C=C

Quick self-test

1. How many carbon atoms are in the ring of a glucose molecule?
2. Which carbon atom differs between α-glucose and β-glucose, and how?
3. Name the two monosaccharides that make up sucrose.
4. Which of these are reducing sugars: maltose, sucrose, fructose?
5. How many water molecules are released when a chain of 50 glucose units is made?
6. Name the glycosidic bonds at the branch points of glycogen.
7. Give two reasons why glycogen is a better energy store than glucose.
8. What holds neighbouring cellulose molecules together in a microfibril?
9. Why must the acid be neutralised in the non-reducing sugar test?
10. A fatty acid has the formula C₁₈H₃₂O₂. How many C=C double bonds does it have?
11. How many ester bonds are there in one phospholipid molecule, linking fatty acids to glycerol?
12. Explain why a triglyceride is described as hydrophobic.

Answers

1. Five (the sixth ring atom is oxygen).
2. Carbon 1: OH below the ring in α, above in β.
3. Glucose and fructose.
4. Maltose and fructose. Sucrose is non-reducing.
5. 49 (n − 1).
6. α-1,6 glycosidic bonds.
7. Any two: insoluble, so no effect on water potential; compact; does not leave the cell; branched with many ends for rapid hydrolysis.
8. Hydrogen bonds.
9. Benedict’s solution only works in alkaline conditions.
10. Two. Saturated C₁₈ would be C₁₈H₃₆O₂; each C=C removes 2 H, and 36 − 32 = 4 H, so 2 double bonds.
11. Two.
12. It is non-polar overall: its long hydrocarbon tails cannot form hydrogen bonds with water, so it does not mix with or dissolve in water.

Where marks are usually lost

• Putting six carbons in the glucose ring, or forgetting the CH₂OH on carbon 5.
• Drawing α-glucose with the C1 OH above the ring.
• Saying hydrolysis “releases water” or condensation “uses water”.
• Writing “glucose” without α or β when the question asks about a specific polysaccharide.
• Saying cellulose molecules are joined to each other by glycosidic bonds.
• Writing that starch is “a branched molecule”. Only amylopectin branches; amylose does not.
• Missing the neutralisation step in the non-reducing sugar test.
• Giving “fats are an energy store” without the reason (more C–H bonds per gram, more energy released per gram).
• Saying triglycerides are polar or that phospholipid tails are hydrophilic.
• Calling a triglyceride a polymer.

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027, Cambridge University Press & Assessment – Topic 2, Biological molecules (section 2.2 Carbohydrates and lipids).