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B1.1 Carbohydrates and lipids

Carbohydrates and lipids connect carbon chemistry, condensation and hydrolysis reactions, molecular structure, solubility, membrane formation, and energy storage functions in organisms.

Syllabus
First assessment 2025
Topic
B1.1
Level
HL

Carbon's Four Bonds Create Biological Variety

Carbon is central to biological molecules because one carbon atom can form four covalent bonds, allowing stable chains, branches and rings.

Its four outer-shell electrons can be shared with carbon, hydrogen, oxygen, nitrogen or sulfur. Changing the carbon skeleton or attached groups changes shape and chemical behaviour, so structure can produce different biological functions.

Useful consequences:

  • carbon atoms link into chains, branches and rings
  • single and double bonds change shape and reactivity
  • large molecules can contain many different carbon arrangements

Glycogen and cellulose both contain glucose units, but their bonding arrangement gives glycogen a compact storage form and cellulose strong fibres.

Four bonds explain carbon's versatility; they do not make every carbon compound chemically identical. Always connect the particular structure to the claimed function.

Carbon atom properties

Assessment in practice

1–4 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline / Identify.

Command terms

Outline / Identify

What earns marks

Build the answer around this relationship: Carbon can form four covalent bonds with carbon and other non-metal elements.

Representative question

Question 1

[Maximum number: 4]

Outline the chemical properties of carbon that allow it to form diverse compounds.

Condensation Builds Macromolecules

A condensation reaction joins smaller molecules into a larger molecule while removing water.

A new covalent bond forms between reacting groups; the removed H and OH combine as water. Repeating this process links monomers into a polymer, so the bond-forming mechanism explains molecular growth.

For a condensation step, identify:

  • the two starting subunits
  • the new covalent bond
  • the water molecule released

Two monosaccharides join to form a disaccharide. One contributes H and the other OH; water leaves and the remaining atoms are connected by a new glycosidic bond.

Condensation is not simply ‘mixing monomers’. A covalent bond must form and water must be removed; breaking the bond by adding water is hydrolysis.

Macromolecules by condensation

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline / State / Identify.

Command terms

Outline / State / Identify

What earns marks

Build the answer around this relationship: Condensation reactions build larger molecules from smaller subunits.

Watch for

Reversing condensation and hydrolysis in reaction equations.

Representative question

Question 1

[Maximum number: 5]

Outline the production of a dipeptide by a condensation reaction, showing the structure of a generalized dipeptide.

Hydrolysis Splits Biological Polymers

Hydrolysis breaks a covalent bond in a larger biological molecule by using water.

The water molecule separates into H and OH, which attach to the two products. This reverses the bond-forming logic of condensation and lets digestive enzymes release absorbable smaller molecules.

Trace the reaction as:

  • water enters the reaction
  • a polymer bond is cleaved
  • H and OH cap the two products

Hydrolysing a disaccharide produces two monosaccharides: one receives H and the other OH. The products are smaller than the starting molecule and can be transported for metabolism.

Hydrolysis is not the same as physically dissolving a food. It changes a covalent bond; water alone may be present, but enzymes usually control the biological rate.

Digestion by hydrolysis

Assessment in practice

1–3 marks
How it is assessed

This objective is assessed through structured response, commonly using State / Outline / Describe.

Command terms

State / Outline / Describe / Explain / Identify

What earns marks

Build the answer around this relationship: Hydrolysis breaks covalent bonds by adding water.

Watch for

Confusing hydrolysis with condensation when identifying reaction type.

Representative question

Question 1

[Maximum number: 6]

Describe the importance of hydrolysis in digestion.

Monosaccharide Structure Shapes Its Role

A monosaccharide is a single sugar unit whose arrangement of atoms determines how it behaves and what larger molecules it can form.

Its functional groups and carbon skeleton provide sites for covalent bonding and affect solubility. Different arrangements therefore change how sugars are joined, transported or used in cells.

When comparing monosaccharides, check:

  • carbon skeleton and ring/chain form
  • positions of functional groups
  • which bonds the arrangement can form

Glucose and fructose have the same molecular formula but different arrangements. That structural difference affects how each participates in condensation reactions and metabolic pathways.

Having the same formula does not mean having the same structure or function. Do not infer biological role from formula alone.

Form and function of monosaccharides

Assessment in practice

1–3 marks
How it is assessed

This objective is assessed through structured response, commonly using State / Draw / Identify.

Command terms

State / Draw / Identify / Outline

What earns marks

Build the answer around this relationship: Glucose, fructose, galactose and ribose are monosaccharides.

Watch for

Classifying disaccharides or polysaccharides as monosaccharides.

Representative question

Question 1

[Maximum number: 5]

Outline how the properties of glucose are linked to their uses in organisms.

Polysaccharides Store Glucose Compactly

Starch and glycogen store glucose as large, compact polysaccharides so many glucose units can be held with relatively little osmotic effect.

Joining glucose units into a polymer reduces the number of dissolved particles compared with storing every glucose molecule separately. Branching also creates several ends where enzymes can release glucose quickly.

A useful storage design combines:

  • many glucose units in one polymer
  • compact packing to reduce osmotic impact
  • branching for faster mobilisation

When an animal needs glucose between meals, enzymes can remove units from many glycogen branches at once, making release faster than working from one unbranched end.

Storage polysaccharides are not just ‘energy’. Their compactness and branching explain the storage advantage; cellulose has a different structure and role.

Polysaccharides as energy storage

Assessment in practice

1–3 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline / Describe / Distinguish.

Command terms

Outline / Describe / Distinguish / Compare

What earns marks

Build the answer around this relationship: Starch stores glucose energy in plants as amylose and amylopectin.

Watch for

Confusing glycogen with glucagon or glucose.

Representative question

Question 1

[Maximum number: 4]

Outline how and where energy is stored in plants.

Cellulose Fibres Gain Strength from Parallel Chains

Cellulose supports plant cell walls because its straight glucose chains align and hydrogen-bond to form strong fibres.

The β-linked arrangement keeps each chain extended. Many weak hydrogen bonds between neighbouring chains add together, producing tensile strength without covalently joining every chain.

Link structure to strength:

  • β-linked glucose forms straight chains
  • chains align side by side
  • many hydrogen bonds reinforce the microfibril

A cellulose microfibril can resist pulling because force is shared across many aligned chains; the same glucose monomers in a branched storage polymer do not form that fibre architecture.

Hydrogen bonds are individually weak, but their combined effect is strong. Do not say cellulose is strong because its chains are covalently bonded to one another.

Cellulose structure and function

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Label / Describe / Outline.

Command terms

Label / Describe / Outline / Identify

What earns marks

Build the answer around this relationship: Cellulose is made from beta-glucose monomers joined by 1,4 glycosidic bonds.

Watch for

Describing cellulose as branched or made from alpha-glucose.

Representative question

Question 1

[Maximum number: 3]

Describe how cellulose is formed from monosaccharides.

Non-Polar Lipids Avoid Water

A non-polar lipid has no substantial charge separation, so it interacts weakly with polar water and tends to separate from it.

Water forms favourable interactions with polar or charged groups. A non-polar hydrocarbon region cannot make equivalent interactions, so lipid molecules cluster away from water; this hydrophobic behaviour shapes membranes and storage droplets.

Check the structure for:

  • mostly hydrocarbon bonds
  • few or no charged/polar groups
  • poor interaction with water

Oil forms a separate layer on water because its non-polar molecules do not mix favourably with the polar water network.

Hydrophobic does not mean ‘repelled by every substance’ or ‘contains no oxygen’. Judge the overall polarity and the interactions available to water.

Hydrophobic properties of lipids

Assessment in practice

1–4 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Explain.

Command terms

Explain

What earns marks

Build the answer around this relationship: Lipids are generally hydrophobic because large parts of their molecules are non-polar.

Watch for

Calling oils insoluble without linking this to non-polar hydrocarbon chains and lack of hydrogen bonding.

Representative question

Question 1

[Maximum number: 1]

Which substance must be transported in the blood by lipoprotein complexes?

A

Cholesterol

B

Oxygen

C

Sodium chloride

D

Amino acids

Ester Bonds Assemble Triglycerides and Phospholipids

Triglycerides and phospholipids form when glycerol reacts with fatty acids in condensation reactions that create ester bonds.

Each ester bond joins a hydroxyl group of glycerol to a carboxyl group of a fatty acid and releases water. The number and type of attached groups then determine whether the molecule is mainly for storage or membrane structure.

For a synthesis diagram, count:

  • glycerol backbone
  • fatty-acid tails attached
  • ester bonds and water molecules released

Attaching three fatty acids to glycerol forms a triglyceride and releases three water molecules; replacing one tail with a phosphate-containing group gives a phospholipid.

A phospholipid is not simply a triglyceride with fewer tails. Its phosphate-containing head changes polarity and therefore its behaviour in water.

Formation of triglycerides and phospholipids

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using State / Identify / Describe.

Command terms

State / Identify / Describe

What earns marks

Build the answer around this relationship: Triglycerides contain glycerol joined to three fatty acids.

Watch for

Naming hydrolysis instead of condensation when fatty acids join glycerol.

Representative question

Question 1

[Maximum number: 1]

Identify the molecule that was used to form part Y of the triglyceride.

Fatty-Acid Double Bonds Change Chain Shape

Fatty acids are long hydrocarbon chains with a carboxyl group; double bonds in the chain change its shape and affect packing.

A cis double bond introduces a bend, preventing neighbouring chains from packing as tightly. Less tight packing generally lowers the temperature at which a lipid becomes fluid.

Compare fatty acids by checking:

  • chain length
  • number and position of double bonds
  • whether a bend interrupts packing

Two chains of equal length can differ in melting behaviour: the chain with a cis double bond bends and packs less efficiently, so it tends to remain fluid at a lower temperature.

Unsaturation is not the only factor: chain length and the type/position of double bonds also affect packing. Do not equate ‘more double bonds’ with a universal numerical change.

Fatty acids

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline / State / Draw.

Command terms

Outline / State / Draw / Distinguish / Compare / Identify

What earns marks

Build the answer around this relationship: Saturated fatty acids have no carbon-carbon double bonds in the hydrocarbon chain.

Watch for

Calling saturated fatty acids unsaturated because they contain a carboxyl group.

Representative question

Question 1

[Maximum number: 4]

Distinguish between the structures of the different types of fatty acids in food.

Triglycerides Store Energy Efficiently

Triglycerides are effective long-term energy stores because their reduced hydrocarbon bonds release much energy and they can be stored with little associated water.

Their non-polar nature means they do not attract a large hydration shell, so storage is compact. When hydrolysed, fatty acids can be oxidised to transfer chemical energy to ATP-producing pathways.

The storage advantage comes from:

  • energy-rich reduced carbon
  • hydrophobic, compact storage
  • separation from the aqueous cytoplasm

A fat droplet can store substantial chemical energy in a small volume without dissolving throughout the cytoplasm, unlike a large pool of free glucose.

Triglycerides are storage molecules, not membrane bilayers. Their three tails and lack of a strongly polar head explain why they form droplets rather than a surface sheet.

Triglycerides functions

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Explain / Outline / State.

Command terms

Explain / Outline / State

What earns marks

Build the answer around this relationship: Triglycerides store more energy per gram than carbohydrates.

Watch for

Saying lipids are more efficient because they are easier to transport than carbohydrates.

Representative question

Question 1

[Maximum number: 4]

Outline the use of lipids to store energy in humans.

Phospholipids Self-Assemble into Bilayers

Phospholipids are amphipathic: a polar phosphate-containing head interacts with water while non-polar tails avoid it, so they self-assemble into bilayers.

In water, heads face the aqueous environments and tails pack away from water in the interior. This arrangement creates a flexible hydrophobic barrier that separates compartments while allowing selected molecules to cross.

A bilayer requires:

  • hydrophilic heads facing water
  • hydrophobic tails facing inward
  • a continuous, dynamic sheet rather than a solid wall

In a cell membrane, water contacts both outer and inner head surfaces, while the tail core makes it difficult for many ions and polar molecules to pass without transport proteins.

The bilayer is not formed with tails facing water. Reversing the orientation would expose the least water-compatible part and would not create a stable membrane barrier.

Phospholipid bilayers

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline / Annotate.

Command terms

Outline / Annotate

What earns marks

Build the answer around this relationship: Phospholipid heads are hydrophilic and phosphate-containing.

Watch for

Reversing the hydrophilic phosphate head and hydrophobic fatty-acid tails.

Representative question

Question 1

[Maximum number: 2]

Annotate the diagram to illustrate the amphipathic nature of phospholipids.

Steroids Have a Non-Polar Hydrocarbon Core

Steroids are lipids built from four fused carbon rings; their largely hydrocarbon core makes them mostly non-polar, although attached groups can change the overall polarity.

The fused-ring skeleton is dominated by C–C and C–H bonds, which interact weakly with water. A hydroxyl, carbonyl or other polar group can add a local interaction, so classification depends on the whole molecule.

To judge polarity, inspect:

  • the fused-ring hydrocarbon skeleton
  • the number and position of polar groups
  • whether the whole molecule can interact favourably with water

A steroid with one hydroxyl group still has a mostly non-polar ring system, so it may associate with membranes while the hydroxyl provides a limited polar interaction.

‘Steroid’ names a structural family, not an absolute statement that every member is completely non-polar. Do not ignore attached functional groups.

Non-polar steroids

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Identify.

Command terms

Identify

What earns marks

Build the answer around this relationship: Steroids are lipids with four fused carbon rings.

Representative question

Question 1

[Maximum number: 1]

Testosterone is a hormone that is important for male reproductive development.

To which group of compounds does testosterone belong?

A

Nucleotides

B

Carbohydrates

C

Lipids

D

Amino acids

Structure To Function

B1.1 becomes easy when every answer follows structure -> property -> function. Carbon skeletons and functional groups create molecular diversity. Condensation builds larger molecules and hydrolysis breaks them. Alpha-glucose stores energy as starch and glycogen; beta-glucose forms strong cellulose. Surface carbohydrates enable recognition. Lipids are hydrophobic, triglycerides store energy, phospholipids self-assemble into bilayers, and steroids cross membranes because they are mostly non-polar.

  • Carbon bonding and functional groups explain molecular diversity.
  • Condensation releases water; hydrolysis uses water.
  • Carbohydrates can store energy, build cell walls, and mark cell surfaces.
  • Lipids are hydrophobic and not true polymers.
  • Triglycerides store energy; phospholipids form membranes; steroids signal across membranes.
ConceptIB Biology HL