OCR A Level Biology A: Water, carbohydrates and lipids (H420)

This guide covers learning outcomes 2.1.2(a) to 2.1.2(j) of the OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025. It teaches water, monomers and polymers, the elements in biological molecules, glucose and ribose, disaccharides, starch, glycogen and cellulose, triglycerides, phospholipids and cholesterol. Amino acids, proteins, inorganic ions and the biochemical tests (2.1.2(k) onwards) are separate units.

Use it with the revision notes and the practice set. The wider module is in Foundations in Biology, with more questions in Foundations practice. Find the course on the OCR A Level Biology hub and tick off outcomes on the printable checklist. Not sure where you stand? Try the free diagnostics.

Outcomes covered

Outcome	You must be able to
2.1.2(a)	Explain hydrogen bonding in water and link its properties to roles as solvent, transport medium, coolant and habitat, with prokaryote and eukaryote examples
2.1.2(b)	Use the ideas of monomer and polymer; explain condensation and hydrolysis
2.1.2(c)	State the elements in carbohydrates, lipids, proteins and nucleic acids
2.1.2(d)	Describe glucose (a hexose, with alpha and beta forms) and ribose (a pentose)
2.1.2(e)	Explain how glycosidic bonds form and break; sucrose, lactose, maltose
2.1.2(f), (g)	Describe amylose, amylopectin, glycogen and cellulose, and relate structure to function
2.1.2(h), (i)	Describe triglycerides and phospholipids; saturated and unsaturated fatty acids; ester bonds
2.1.2(j)	Relate triglyceride, phospholipid and cholesterol properties to function

Practise sketching each structure, because the specification says this section should include diagrams of molecular structure and bonding.

2.1.2(a) Water

A water molecule is H2O: an oxygen atom covalently bonded to two hydrogen atoms. The shared electrons lie nearer the oxygen, so oxygen carries a small negative charge (δ−) and each hydrogen a small positive charge (δ+). The molecule is polar.

The δ+ hydrogen of one molecule is attracted to the δ− oxygen of a neighbour. This attraction is a hydrogen bond. Each bond is weak, but a large number of them gives water its properties.

Property	Cause	Biological role
Good solvent for ions and polar molecules	Polar water molecules surround charged or polar particles and keep them apart	Solvent: metabolic reactions happen in solution in cytoplasm
Cohesion and adhesion	Hydrogen bonds between water molecules (cohesion) and to other polar surfaces (adhesion)	Transport medium: continuous columns of water move in xylem; blood plasma carries dissolved substances
High specific heat capacity	Much energy is needed to break hydrogen bonds and raise the temperature	Stable temperature inside cells and in aquatic habitats
High latent heat of vaporisation	Energy is needed to break hydrogen bonds so molecules escape as vapour	Coolant: sweating, transpiration
Ice is less dense than liquid water	In ice the molecules sit in a fixed, open lattice held by hydrogen bonds	Ice floats and insulates the water below, so habitats stay liquid
Surface tension	Cohesion at the surface	Surface of a pond is a habitat for organisms such as pond skaters

Examples across prokaryotes and eukaryotes

• Solvent. The cytoplasm of a bacterium is an aqueous solution in which enzymes, ions and substrates dissolve and move. The same is true of the cytosol of a eukaryotic cell.
• Transport medium. Plasma carries glucose, ions and urea in a mammal. Water in xylem carries mineral ions upwards in a plant.
• Coolant. A mammal loses heat as sweat evaporates. A plant loses heat as water evaporates from leaf surfaces during transpiration.
• Habitat. Lakes and oceans support bacteria, algae and animals. Water’s high specific heat capacity means these habitats change temperature slowly.

Worked example: cooling by sweat

Assume that each gram of sweat that evaporates removes 2.4 kJ from the body, and that a 60 kg person has an average specific heat capacity of 3.5 J g⁻¹ °C⁻¹. By how much would body temperature have risen if 60 g of sweat had not evaporated?

1. Heat removed by evaporation = 60 g × 2.4 kJ g⁻¹ = 144 kJ.
2. Heat needed to raise the body by 1 °C = 60 000 g × 3.5 J = 210 000 J = 210 kJ.
3. Temperature rise avoided = 144 ÷ 210 = 0.69 °C (2 s.f.).

Quote supplied data in your working, and link the answer to the property: the high latent heat of vaporisation.

2.1.2(b) Monomers, polymers, condensation and hydrolysis

A monomer is a small molecule that can bond to others of the same kind. A polymer is a long chain of many monomers.

Monomer	Polymer
Monosaccharide (e.g. glucose)	Polysaccharide (e.g. starch)
Amino acid	Polypeptide
Nucleotide	Polynucleotide

A condensation reaction links two molecules and gives off a molecule of water. A hydrolysis reaction uses one molecule of water to break a bond. They are opposite reactions, and in cells enzymes catalyse both. Polymers are built by condensation and broken down by hydrolysis, so the same pair of reactions links carbohydrates, proteins and nucleic acids. Triglycerides are also made by condensation, but they are not chains of repeating monomers.

2.1.2(c) Elements in biological molecules

Molecule	Elements
Carbohydrates	C, H, O
Lipids	C, H, O
Proteins	C, H, O, N, S
Nucleic acids	C, H, O, N, P

2.1.2(d) Glucose and ribose

Glucose has the formula C6H12O6. It is a hexose (six carbons) and a monosaccharide. In solution it forms a ring of five carbons plus one oxygen; carbon 6 is a CH2OH group attached outside it. Carbon 1, next to the ring oxygen, carries an –OH group.

• α-glucose: the –OH on carbon 1 is below the plane of the ring.
• β-glucose: the –OH on carbon 1 is above the plane of the ring.

This one difference decides which polymers glucose forms. Glucose is soluble because it is polar and carries several –OH groups.

Ribose has the formula C5H10O5. It is a pentose (five carbons). Its ring contains four carbons and one oxygen, with the fifth carbon outside the ring. The difference between a hexose and a pentose is the number of carbon atoms. Ribose is the sugar in RNA nucleotides; nucleotides are covered in 2.1.3.

2.1.2(e) Glycosidic bonds and disaccharides

A glycosidic bond is the covalent bond between two monosaccharides. It forms by condensation between –OH groups, releasing water, and breaks by hydrolysis.

Disaccharide	Made from	Formed in or by
Maltose	Two α-glucose units	Product of starch breakdown
Sucrose	One α-glucose and one fructose	Transport sugar in plants
Lactose	One glucose and one galactose	Milk sugar in mammals

All three have the formula C12H22O11. Two C6H12O6 molecules contain C12H24O12; removing H2O gives C12H22O11. The relative molecular mass is 2 × 180 − 18 = 342.

Worked example: polysaccharides and water

An amylose chain contains 400 glucose residues. How many glycosidic bonds and water molecules are involved in making it, and what is its relative molecular mass? (Glucose = 180; water = 18.)

1. Joining 400 residues in a line needs 400 − 1 = 399 bonds.
2. Each condensation releases one water, so 399 water molecules are released.
3. Mr = 400 × 180 − 399 × 18 = 72 000 − 7 182 = 64 818.

2.1.2(f), (g) Starch, glycogen and cellulose

All three are polysaccharides of glucose. Their different bonding and shape give them different jobs.

	Amylose	Amylopectin	Glycogen	Cellulose
Monomer	α-glucose	α-glucose	α-glucose	β-glucose
Linkages	α-1,4	α-1,4 with α-1,6 branch points	α-1,4 with α-1,6 branch points	β-1,4
Shape	Unbranched, coiled into a helix	Branched	Highly branched	Straight, unbranched
Where	Plant starch grains	Plant starch grains	Animal and fungal cells; some bacteria	Plant cell walls

Starch is a mixture of amylose and amylopectin. In amylose the α-1,4 links make the chain curl into a compact helix. In amylopectin α-1,6 links create branches about every 24 to 30 residues.

Glycogen is similar to amylopectin, but branches occur about every 8 to 12 residues, so it is more highly branched and more compact.

Cellulose has β-1,4 links. Because of the –OH position on carbon 1, each β-glucose is rotated 180° relative to its neighbour, so the chain is straight. Hydrogen bonds form between –OH groups on parallel chains, building microfibrils, which build fibres.

Structure and function

• Glucose is soluble, so it dissolves in blood plasma or cell sap and is transported. Its bonds release energy in respiration.
• Starch is a store in plants. It is insoluble, so it does not alter the water potential of cells. The helix of amylose is compact, and amylopectin has many free ends, so enzymes can hydrolyse many bonds at once and release glucose quickly.
• Glycogen is the store in animals and fungi. Its many branches give a very large number of free ends for rapid hydrolysis, which suits the high energy demand of muscle. Some bacteria also store glycogen.
• Cellulose is the main component of plant cell walls. Many hydrogen bonds between chains make microfibrils strong, so the wall resists the pressure of the cell contents and the cell does not burst.

2.1.2(h), (i) Fatty acids and triglycerides

A fatty acid is a hydrocarbon chain with a carboxyl group (–COOH) at one end.

• Saturated: no C=C double bonds. The chain is straight, so molecules pack closely.
• Unsaturated: one (monounsaturated) or more (polyunsaturated) C=C double bonds. A double bond puts a kink in the chain, so molecules pack less closely.

A triglyceride is one glycerol molecule joined to three fatty acids. Each joint is an ester bond, formed by condensation where a glycerol –OH meets a fatty acid –COOH. Three water molecules are released. Hydrolysis of a triglyceride (using lipase) needs three water molecules and gives glycerol and three fatty acids.

Worked example: Mr of a triglyceride

Stearic acid has the formula C18H36O2. Calculate the Mr of the triglyceride made from glycerol (C3H8O3) and three stearic acid molecules. (C = 12, H = 1, O = 16.)

1. Glycerol Mr = 36 + 8 + 48 = 92.
2. Stearic acid Mr = 216 + 36 + 32 = 284.
3. Triglyceride = 92 + 3 × 284 − 3 × 18 = 92 + 852 − 54 = 890 (formula C57H110O6).

A phospholipid has glycerol joined to two fatty acids by ester bonds, and a phosphate group in place of the third fatty acid. The phosphate head is charged and hydrophilic. The two fatty acid tails are hydrophobic.

2.1.2(j) Function of triglycerides, phospholipids and cholesterol

Triglycerides

• Energy store. Fatty acid chains have many C–H bonds, so lipids store more energy per gram than carbohydrates. Use about 38 kJ g⁻¹ for lipid and 17 kJ g⁻¹ for carbohydrate when data are supplied. For 10 g, that is 380 kJ against 170 kJ, a ratio of 2.2.
• Insulation and protection. A fat layer under the skin reduces heat loss, and fat around organs cushions them.
• Hydrophobic. Triglycerides are insoluble, so they do not change the water potential of cells.

Phospholipids

In water, phospholipids arrange themselves so that the hydrophilic heads point outwards into the water and the hydrophobic tails point inwards towards each other, forming a bilayer. This is the basis of the cell-surface membrane in prokaryotes and eukaryotes, and of the membranes around organelles in eukaryotes. Ions and polar molecules cannot easily cross the hydrophobic core, though small non-polar molecules can. Kinked unsaturated tails help keep it fluid in the cold.

Cholesterol

Cholesterol is a small, flat, four-ring steroid with one polar –OH group. In animal cell membranes it sits between phospholipids with its –OH near the heads and its ring structure among the tails. It makes the membrane less permeable to ions and small polar molecules, and it stabilises the membrane: it limits movement of the tails at high temperature and stops them packing tightly at low temperature. Most prokaryotes make no cholesterol, so this role applies to eukaryotic membranes.

Common errors

• Writing that hydrogen bonds hold the atoms inside a water molecule. Those are covalent bonds.
• Saying water is “a good coolant” with no property.
• Writing “n water molecules” for a chain of n monomers. A chain needs n − 1 condensations.
• Calling a triglyceride a polymer made of identical units.

Official syllabus

OCR, A Level Biology A (H420) specification, Version 4.1 (April 2026), first assessment 2025: section 2.1.2, learning outcomes (a) to (j).