What are the main types of movement of substances in biological systems?

In biological systems, the main types of movement of substances include diffusion, osmosis, and active transport. Diffusion is the passive movement of molecules from an area of higher concentration to an area of lower concentration. Osmosis is a specific type of diffusion involving the movement of water molecules across a semi-permeable membrane. Active transport, on the other hand, requires energy to move substances against their concentration gradient, from a region of lower concentration to a region of higher concentration.

How does diffusion differ from active transport in cells?

Diffusion is a passive process that does not require energy, allowing molecules to move from an area of higher concentration to an area of lower concentration. In contrast, active transport requires energy, usually in the form of ATP, to move molecules against their concentration gradient, from a region of lower concentration to a region of higher concentration. This process is essential for maintaining concentration gradients across cell membranes, which are crucial for various cellular functions.

Why is osmosis important for maintaining cell health?

Osmosis is crucial for maintaining cell health because it regulates the water balance within cells and their surrounding environment. By controlling the movement of water across cell membranes, osmosis helps maintain cell turgor, which is essential for structural integrity and function. It also plays a vital role in nutrient absorption and waste removal, ensuring that cells remain in an optimal state for biological processes.

What role does the cell membrane play in the movement of substances?

The cell membrane plays a critical role in the movement of substances by acting as a selective barrier that regulates what enters and exits the cell. It is composed of a phospholipid bilayer with embedded proteins that facilitate the transport of molecules. These proteins include channels and carriers that assist in passive transport, as well as pumps that are involved in active transport. The cell membrane's selective permeability is essential for maintaining homeostasis and allowing cells to respond to changes in their environment.

How does the movement of substances relate to the Edexcel International A Levels Biology curriculum?

The movement of substances is a fundamental concept in the Edexcel International A Levels Biology curriculum, particularly under the topic of 'Molecules, Diet, Transport and Health.' Understanding how substances move across cell membranes is essential for comprehending various biological processes, including nutrient uptake, waste removal, and cellular respiration. This knowledge is crucial for students to grasp more complex topics such as metabolism, homeostasis, and the functioning of different organ systems.

Why is "Defining osmosis using 'water concentration' instead of water potential" a common mistake in Movement of Substances?

Why it happens: Students learn a simplified GCSE definition ('water moves from high to low water concentration') and carry it into IAL. How to avoid it: At IAL osmosis MUST be defined with **water potential**: net movement of water from **higher (less negative)** to **lower (more negative) water potential** across a partially permeable membrane. 'Water concentration' is not creditworthy. Source: Edexcel IAL Biology Unit 1 examiner reports

Why is "Getting the direction wrong because of the negative sign of water potential" a common mistake in Movement of Substances?

Why it happens: Water potentials are negative, so −150 kPa feels 'smaller' than −400 kPa, when in fact −150 kPa is HIGHER. How to avoid it: Remember pure water = 0 kPa (the maximum). Less negative = higher water potential. Water always moves to the **more negative** value. Compare numbers on a number line: −150 is higher (further right) than −400. Source: Edexcel IAL Biology Unit 1 examiner reports

Why is "Claiming facilitated diffusion uses ATP" a common mistake in Movement of Substances?

Why it happens: Because it involves proteins, students assume it must require energy like active transport. How to avoid it: Facilitated diffusion is **passive** — it moves substances **down** the gradient with **no ATP**. Only **active transport** moves substances **against** the gradient and uses ATP. The presence of a carrier protein alone does not mean ATP is used. Source: Edexcel IAL Biology Unit 1 examiner reports

Why is "Saying a substance is moved 'down the gradient' in active transport" a common mistake in Movement of Substances?

Why it happens: The phrase 'down the gradient' is over-learned from diffusion. How to avoid it: Active transport moves substances **against** the concentration gradient, from **lower to higher** concentration. This is the defining feature that distinguishes it from diffusion — state it explicitly. Source: Edexcel IAL Biology Unit 1 examiner reports

Why is "Treating diffusion distance as directly proportional to rate in Fick's law" a common mistake in Movement of Substances?

Why it happens: Students apply all three factors as multipliers without noticing that distance is on the bottom of the equation. How to avoid it: In Fick's law, distance is the **denominator**: rate ∝ (SA × conc. diff.) ÷ distance. **Halving the distance doubles the rate**; doubling the distance halves it. Surface area and concentration difference are directly proportional; distance is inversely proportional. Source: Edexcel IAL Biology Unit 1 examiner reports

Why is "Stating that the glucose co-transporter directly uses ATP" a common mistake in Movement of Substances?

Why it happens: Glucose moves against its gradient, so students assume the carrier must hydrolyse ATP itself. How to avoid it: In co-transport the ATP is used by the **Na⁺/K⁺ pump** to create the sodium gradient; the co-transporter then uses that **Na⁺ gradient** (not ATP directly) to drag glucose in. This is **secondary (indirect) active transport**. Source: Edexcel IAL Biology Unit 1 examiner reports

Edexcel International A Levels Biology (XBI11-YBI11)

Movement of Substances

0% Complete

3 tasks remaining

Short Notes - Movement of Substances

Page 1/0

Detailed Study Notes

Detailed notes on Molecules, Diet, Transport and Health for Edexcel International A Levels Biology, covering key concepts, explanations, examples, and exam-focused revision points.

Movement of Substances — Pearson Edexcel International A Level Biology (XBI11/YBI11) Study Notes

How substances cross the partially permeable cell-surface membrane: simple diffusion and Fick's law, facilitated diffusion through channel and carrier proteins, osmosis defined by water potential, active transport, bulk transport (endocytosis and exocytosis) and co-transport.

At a glance

Simple diffusion — net movement of particles down a concentration gradient; no proteins, no ATP. Only small, non-polar molecules (O₂, CO₂) cross the bilayer this way.
Fick's law — rate of diffusion ∝ (surface area × concentration difference) ÷ diffusion distance.
Facilitated diffusion — passive, down a gradient, but uses channel or carrier proteins for polar molecules and ions; no ATP.
Osmosis — net movement of water down a water potential gradient across a partially permeable membrane (NEVER say 'water concentration').
Active transport — movement against a gradient using carrier proteins and ATP from respiration.
Bulk transport — endocytosis (in) and exocytosis (out) move large quantities/large molecules via vesicles; both need ATP.
Co-transport — a carrier moves two substances together; e.g. Na⁺ and glucose into intestinal epithelial cells.
Water potential (ψ) of pure water = 0 kPa; adding solute makes ψ more negative.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XBI11-YBI11 syllabus (2018-onwards).

2.3 — Describe simple diffusion as the net passive movement of molecules down a concentration gradient, and use Fick's law to relate rate to surface area, concentration difference and diffusion distance.
2.3 — Describe facilitated diffusion through channel and carrier proteins as a passive process for polar molecules and ions.
2.4 — Define osmosis in terms of water potential and explain water movement across a partially permeable membrane.
2.4 — Explain the effect of solute concentration on the water potential of a solution and predict net water movement between cells/solutions.
2.5 — Describe active transport as movement against a concentration gradient requiring carrier proteins and ATP from respiration.
2.5 — Describe endocytosis and exocytosis as bulk transport processes, and explain co-transport with reference to glucose absorption.

Simple diffusion and Fick's law

Passive movement down a gradient — and the equation that controls its rate.

Simple diffusion is the net movement of particles (molecules or ions) from a region of higher concentration to a region of lower concentration — that is, down a concentration gradient — as a result of their random kinetic motion. It is passive: the energy comes from the kinetic energy of the particles themselves, not from ATP.

Across the cell-surface membrane, only certain substances can diffuse directly through the phospholipid bilayer:

Small, non-polar (lipid-soluble) molecules — oxygen ( $O_2$ ), carbon dioxide ( $CO_2$ ) and other lipophilic molecules dissolve in the hydrophobic tail region and pass straight through.
Large molecules and charged/polar particles (glucose, amino acids, ions) cannot cross the bilayer freely — they need transport proteins (facilitated diffusion or active transport).

Fick's law summarises what controls the rate of diffusion:

$\text{rate of diffusion} \;\propto\; \dfrac{\text{surface area} \times \text{concentration difference}}{\text{diffusion distance}}$

So the rate increases when:

the surface area of the membrane is larger (e.g. microvilli, alveoli, root hair cells),
the concentration difference (gradient) across the membrane is steeper (maintained by blood flow or ventilation), or
the diffusion distance is shorter (e.g. a single squamous epithelial cell wall in the alveolus).

Fick's law: a larger surface area and steeper concentration difference speed diffusion, while a longer diffusion distance slows it.

Fick's law is the basis for almost every "explain why this exchange surface is efficient" question — flat lungs, folded gut, thin gills all map onto one of the three terms.

Simple diffusion = net passive movement down a concentration gradient; no ATP.
Only small, non-polar/lipid-soluble molecules (O₂, CO₂) cross the bilayer directly.
rate ∝ (surface area × concentration difference) ÷ diffusion distance.

See the full worked example for movement of substances →

Facilitated diffusion — channel and carrier proteins

Passive transport for polar molecules and ions through membrane proteins.

Polar molecules (e.g. glucose, amino acids) and charged ions cannot pass through the hydrophobic core of the bilayer. They cross by facilitated diffusion: still down a concentration gradient and still passive (no ATP), but assisted by intrinsic transport proteins.

There are two kinds of protein:

Channel proteins — form a water-filled hydrophilic pore through the membrane. Many are gated (open or close in response to a signal) and are highly specific to one type of ion (e.g. a Na⁺ channel, a K⁺ channel).
Carrier proteins — bind the specific solute on one side; the binding causes the protein to change shape (conformation), releasing the solute on the other side. In facilitated diffusion this shape change is driven by the concentration gradient, not by ATP.

Because facilitated diffusion depends on a finite number of proteins, its rate plateaus at high solute concentration once all the carriers/channels are saturated — unlike simple diffusion, whose rate keeps rising linearly with the gradient. Examiners love a graph comparing the two.

A channel protein provides an open hydrophilic pore for specific ions; a carrier protein binds its solute and changes shape to ferry it across. Both work down the gradient with no ATP.

For polar molecules and ions that cannot cross the bilayer.
Down a gradient, passive, NO ATP.
Channel proteins = pores (often gated); carrier proteins change shape.
Rate saturates when all proteins are occupied.

See the full worked example for movement of substances →

Osmosis and water potential

Water moves down a water potential gradient — never 'water concentration'.

Osmosis is the net movement of water molecules from a region of higher (less negative) water potential to a region of lower (more negative) water potential, through a partially permeable membrane.

Water potential ( $\psi$ ) is the tendency of water molecules to move out of a system; it is measured in kilopascals (kPa).

Pure water has the highest water potential, defined as $\psi = 0$ kPa.
Adding solute lowers the water potential, making it negative. The more solute, the more negative $\psi$ becomes (the more it draws water in).

So a dilute solution has a higher (less negative) water potential than a concentrated solution. In osmosis, water moves from the dilute side to the concentrated side — i.e. down the water potential gradient — until the water potentials equalise.

Solute potential and pressure potential. In plant cells, $\psi = \psi_s + \psi_p$ , where $\psi_s$ is the (negative) solute potential and $\psi_p$ is the (usually positive) pressure potential generated by the cell wall pushing back on the turgid protoplast.

Effect on cells:

Surrounding solution	Animal cell	Plant cell
Higher ψ (hypotonic, dilute)	water in → swells, may lyse	water in → turgid (wall resists)
Equal ψ (isotonic)	no net change	flaccid / incipient plasmolysis
Lower ψ (hypertonic, concentrated)	water out → crenated/shrinks	water out → plasmolysed (membrane pulls from wall)

Water moves down the water potential gradient — from ψ = −200 kPa (less negative) to ψ = −500 kPa (more negative) — across the partially permeable membrane.

Osmosis = net movement of water down a WATER POTENTIAL gradient across a partially permeable membrane.
Pure water ψ = 0 kPa; adding solute makes ψ more negative.
Water moves from higher (less negative) ψ to lower (more negative) ψ.
Plant cell: ψ = ψs + ψp.

See the full worked example for movement of substances →

Active transport

Against the gradient — carrier proteins powered by ATP from respiration.

Active transport is the movement of a substance against its concentration gradient (from a region of lower to a region of higher concentration), using a carrier protein and energy supplied by ATP from respiration.

Mechanism:

The specific solute binds to a carrier protein on one side of the membrane.
ATP is hydrolysed to ADP + Pi at the protein; the phosphate may bind to the carrier (phosphorylation).
This causes the carrier to change shape, carrying the solute across and releasing it on the other side — even though the concentration is already higher there.
The protein returns to its original shape, ready to repeat.

Evidence that it needs respiration: active transport stops when a cell is treated with a respiratory inhibitor (e.g. cyanide) or deprived of oxygen, because no ATP is made. Cells specialised for active transport (e.g. kidney proximal tubule cells, root hair cells) contain many mitochondria.

Examples: reabsorption of glucose and amino acids in the kidney; uptake of mineral ions by root hair cells; the sodium–potassium (Na⁺/K⁺) pump that maintains the resting potential in neurones (3 Na⁺ out, 2 K⁺ in per ATP).

All four mechanisms compared across the bilayer. Simple and facilitated diffusion and osmosis are passive (down the gradient / down the water potential gradient, no ATP); only active transport pumps a solute against its gradient using a carrier protein and ATP (hydrolysed to ADP + Pi).

Moves substances AGAINST the gradient.
Uses a carrier protein + ATP from respiration.
Stops if respiration is inhibited (no ATP); cells have many mitochondria.
Examples: kidney reabsorption, root hair ion uptake, Na⁺/K⁺ pump.

See the full worked example for movement of substances →

Bulk transport and co-transport

Endocytosis, exocytosis and coupled co-transport of glucose.

Very large molecules (proteins, polysaccharides) and bulk quantities of material are too big for transport proteins. They move by bulk transport using vesicles, which requires ATP.

Endocytosis — the cell-surface membrane invaginates around the material, pinching off to form a vesicle inside the cell. Phagocytosis engulfs solids (e.g. a phagocyte engulfing a bacterium); pinocytosis takes in liquids.
Exocytosis — a vesicle (often from the Golgi apparatus) fuses with the cell-surface membrane, releasing its contents to the outside. Used to secrete enzymes, hormones and neurotransmitters.

Co-transport couples the movement of two substances through a single carrier, using the gradient of one to drive the other. The classic example is glucose absorption in the small intestine:

The Na⁺/K⁺ pump on the basal membrane actively pumps Na⁺ out of the epithelial cell into the blood — lowering Na⁺ inside the cell.
This creates a Na⁺ gradient into the cell from the gut lumen.
A co-transporter (symporter) on the membrane facing the lumen carries Na⁺ and glucose together into the cell — glucose is dragged in against its own concentration gradient, powered indirectly by the Na⁺ gradient.
Glucose then leaves into the blood by facilitated diffusion.

This is secondary (indirect) active transport: the ATP is spent on the Na⁺ pump, not on the glucose carrier itself.

Endocytosis: the membrane invaginates around material and pinches off a vesicle into the cell. Exocytosis: a vesicle fuses with the membrane and releases its contents outside. Both are bulk transport and need ATP.

Bulk transport uses vesicles + ATP for very large molecules/quantities.
Endocytosis = membrane invaginates inward (phagocytosis/pinocytosis).
Exocytosis = vesicle fuses with membrane, secretes contents out.
Co-transport: Na⁺ gradient (set by Na⁺/K⁺ pump) drags glucose in against its gradient.

See the full worked example for movement of substances →

Quick recap

Simple diffusion: net passive movement down a concentration gradient; rate ∝ (SA × conc. diff.) ÷ distance (Fick's law).
Facilitated diffusion: passive, down a gradient, through channel/carrier proteins; no ATP.
Osmosis: net water movement down a WATER POTENTIAL gradient across a partially permeable membrane.
Pure water ψ = 0 kPa; solute makes ψ more negative; water moves to the more negative ψ.
Active transport: against the gradient, carrier protein + ATP from respiration.
Bulk transport (endocytosis/exocytosis) moves large molecules via vesicles and needs ATP.
Co-transport: a Na⁺ gradient set up by the Na⁺/K⁺ pump drives glucose uptake against its gradient.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Simple diffusion = net passive movement of particles down a concentration gradient (no ATP).
Fick's law: rate ∝ (surface area × concentration difference) ÷ diffusion distance.
Facilitated diffusion = passive, down a gradient, via channel/carrier proteins, NO ATP.
Osmosis = net movement of water from higher (less negative) to lower (more negative) water potential across a partially permeable membrane.
Pure water ψ = 0 kPa; adding solute makes ψ more negative.
Active transport = against the gradient, carrier protein + ATP from respiration.
Co-transport: Na⁺/K⁺ pump lowers cell Na⁺ → Na⁺ gradient drags glucose in via a symporter.

How it’s examined

Movement of substances is examined in Unit 1 (Lifestyle, Transport, Genes and Health) and underpins exchange and transport across the whole course. Typical questions: define osmosis using water potential (2-3 marks, AO1 — water-potential wording is essential); distinguish the transport mechanisms by gradient direction, proteins and ATP use, often as a comparison table (4-6 marks); apply Fick's law to explain why an exchange surface is efficient or to perform a simple proportional calculation (AO2); explain active transport evidence from respiratory-inhibitor data (AO2/AO3); and describe co-transport of glucose in the intestine (4-5 marks). 'Explain' questions must give the mechanism, not just name the process.

Sources: Pearson Edexcel International Advanced Level Biology (YBI11/XBI11) specification (Issue 2, 2018 onwards); Edexcel IAL Biology Unit 1 examiner reports; Edexcel IAL Biology Unit 1 sample assessment materials and mark schemes. Last reviewed 2026-06-24.

Take this whole topic with you

Step-by-step worked examples — Movement of Substances

Step-by-step solutions to past-paper-style questions on movement of substances, written exactly the way a tutor would explain them at the board.

Question type:

1Identifying the transport mechanism (2 marks)
Getting startedIdentify & classify• diffusion, active transport, Unit 1
Question
A cell takes up potassium ions from a solution where the potassium ion concentration is lower than inside the cell. State the name of this process and one reason why it cannot be diffusion. (2 marks)
Step-by-step solution
1. Step 1
  The ions move from a lower concentration outside to a higher concentration inside — i.e. against the concentration gradient.
2. Step 2
  Diffusion only moves particles down a gradient (high to low). Moving up a gradient requires energy, so the process is active transport (carrier protein + ATP).
Answer
Active transport — because the ions move against the concentration gradient, which diffusion cannot do.
Examiner tip
Mark scheme: (1) active transport; (2) movement is against the concentration gradient / from low to high concentration. Saying only 'it uses energy' without referencing the gradient is not enough for the second mark.
2Predicting the direction of osmosis (2 marks)
Getting startedWord problem• osmosis, water potential, Unit 1
Question
Cell A has a water potential of −400 kPa. It is placed in a solution with a water potential of −150 kPa. State the direction of net water movement and explain your answer. (2 marks)
Step-by-step solution
1. Step 1
  Compare the two water potentials: the solution (−150 kPa) is higher (less negative) than the cell (−400 kPa).
2. Step 2
  Water moves by osmosis down the water potential gradient — from higher (less negative) to lower (more negative) water potential — so water moves from the solution INTO cell A.
Answer
Water moves into cell A, because the cell has a lower (more negative) water potential than the surrounding solution.
Examiner tip
Mark scheme: (1) water moves into the cell; (2) cell has lower / more negative water potential than the solution / water moves down a water potential gradient. Any reference to 'water concentration' is not credited.
3Channel versus carrier proteins (3 marks)
Building confidenceIdentify & classify• facilitated diffusion, membrane proteins, Unit 1
Question
Glucose and chloride ions both enter a cell by facilitated diffusion. Explain how facilitated diffusion of these substances differs from simple diffusion of oxygen, and describe the roles of channel and carrier proteins. (3 marks)
Step-by-step solution
1. Step 1
  Oxygen is small and non-polar, so it dissolves in and crosses the phospholipid bilayer directly by simple diffusion. Glucose and chloride ions are polar/charged, so they cannot cross the hydrophobic bilayer and need transport proteins.
2. Step 2
  Channel proteins form a hydrophilic pore through the membrane that allows specific ions (e.g. chloride) to pass through.
3. Step 3
  Carrier proteins bind the specific solute (e.g. glucose) and change shape to move it across. Both are passive — down the gradient, no ATP — the difference from simple diffusion is only the use of proteins.
Answer
Polar/charged solutes need proteins: channels form pores for ions; carriers change shape to move solutes. Both are passive, like simple diffusion, but require proteins.
Examiner tip
Mark scheme: (1) glucose/ions are polar/charged so cannot cross the bilayer (whereas O₂ is small/non-polar and can); (2) channel protein = pore for ions; (3) carrier protein changes shape; both passive/no ATP.
4Why a respiratory inhibitor stops uptake (3 marks)
Building confidenceWord problem• active transport, ATP, Unit 1
Question
Root hair cells take up nitrate ions from the soil even when the soil nitrate concentration is very low. When the roots are treated with a respiratory inhibitor, nitrate uptake stops almost completely. Explain these observations. (3 marks)
Step-by-step solution
1. Step 1
  Soil nitrate concentration is lower than inside the cell, so the ions are taken up against the concentration gradient — this is active transport.
2. Step 2
  Active transport uses a carrier protein and energy from ATP, which is produced by respiration.
3. Step 3
  The respiratory inhibitor stops respiration, so no ATP is made; without ATP the carrier proteins cannot change shape, so active transport — and therefore nitrate uptake — stops.
Answer
Uptake is by active transport against the gradient, powered by ATP from respiration; the inhibitor stops respiration → no ATP → no active transport.
Examiner tip
Mark scheme: (1) uptake is against the gradient / by active transport; (2) needs ATP from respiration / carrier protein; (3) inhibitor stops respiration so no ATP, hence no active transport. A common error is to say the inhibitor 'kills' the cell — it stops respiration, not the cell.
5Applying Fick's law — a proportional calculation (2 marks)
StretchDirect calculation• Fick's law, calculation, Unit 1
Question
A gas exchange surface has a relative diffusion rate of 1.0 unit. A second surface has twice the surface area, the same concentration difference, and half the diffusion distance. Use Fick's law to calculate the relative diffusion rate of the second surface. Show your working. (2 marks)
Step-by-step solution
1. Step 1
  Fick's law: rate ∝ (surface area × concentration difference) ÷ diffusion distance. Express the new rate as the old rate multiplied by each factor's change.
  $\text{rate} \propto \dfrac{\text{SA} \times \Delta C}{\text{distance}}$
2. Step 2
  Surface area ×2, concentration difference ×1 (unchanged), distance ÷2 means dividing by ½ which multiplies by 2. New rate = 1.0 × 2 × 1 × 2 = 4.0 units.
  $\text{new rate} = 1.0 \times \dfrac{2 \times 1}{0.5} = 4.0$
Answer
Relative diffusion rate = 4.0 units (4× the original).
Examiner tip
Mark scheme: (1) correct method — applies all three factors of Fick's law; (2) correct answer 4.0 (units). Halving the distance DOUBLES the rate — many candidates wrongly halve it. Show working to secure the method mark even if the arithmetic slips.
6Co-transport of glucose in the intestine (4 marks)
StretchMulti-step problem• co-transport, active transport, Unit 1
Question
Explain how glucose is absorbed from the lumen of the small intestine into an epithelial cell when the glucose concentration in the lumen is lower than in the cell. (4 marks)
Step-by-step solution
1. Step 1
  The sodium–potassium (Na⁺/K⁺) pump in the basal membrane uses ATP to actively pump Na⁺ out of the epithelial cell into the blood.
2. Step 2
  This lowers the Na⁺ concentration inside the cell, creating a Na⁺ concentration gradient from the gut lumen into the cell.
3. Step 3
  A co-transporter (symporter) protein in the membrane facing the lumen carries Na⁺ and glucose together into the cell. Na⁺ moves down its gradient and drags glucose in with it, against glucose's own gradient.
4. Step 4
  Glucose then leaves the cell into the blood by facilitated diffusion. The ATP is spent on the Na⁺/K⁺ pump, so glucose uptake is secondary (indirect) active transport.
Answer
Na⁺/K⁺ pump (ATP) lowers cell Na⁺ → Na⁺ gradient → co-transporter brings Na⁺ and glucose in together → glucose moved against its gradient → exits to blood by facilitated diffusion.
Examiner tip
Mark scheme: (1) Na⁺/K⁺ pump moves Na⁺ out using ATP; (2) creates Na⁺ gradient into cell; (3) co-transporter carries Na⁺ AND glucose together; (4) glucose moves against its gradient / leaves by facilitated diffusion. The key high-band idea is that glucose is moved indirectly using the Na⁺ gradient, not by its own ATP.

Model Answers — Movement of Substances

High-scoring sample answers for movement of substances on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Edexcel IAL Biology Unit 1 style2 marks
Q (2 marks). Define simple diffusion.
Model answer
Simple diffusion is the net movement of molecules or ions from a region where they are at a higher concentration to a region where they are at a lower concentration — that is, down a concentration gradient.

It is a passive process: the energy comes from the random kinetic motion of the particles themselves and no ATP is required.
Why this scores
Why this scores 2/2. (1) net movement down a concentration gradient / from high to low concentration; (2) passive / no energy (ATP) required. The word 'net' is important — particles move both ways, but there is a net movement from high to low.
Question 2
Edexcel IAL Biology Unit 1 style3 marks
Q (3 marks). Define osmosis. Your answer should refer to water potential.
Model answer
Osmosis is the net movement of water molecules from a region of higher (less negative) water potential to a region of lower (more negative) water potential, through a partially permeable membrane.

It is a passive process; no metabolic energy (ATP) is needed. Water continues to move until the water potentials on both sides are equal.
Why this scores
Why this scores 3/3. (1) net movement of water; (2) from higher (less negative) to lower (more negative) WATER POTENTIAL; (3) across a partially permeable membrane. Defining osmosis in terms of 'water concentration' or a 'water concentration gradient' is the most common error and scores 0 for the water-potential mark — examiners require water-potential language at IAL.
Question 3
Edexcel IAL Biology Unit 1 style6 marks
Q (6 marks). Compare simple diffusion, facilitated diffusion and active transport with respect to (i) the direction of movement relative to the concentration gradient, (ii) whether membrane proteins are involved, and (iii) whether ATP is required.
Model answer
Simple diffusion moves particles down the concentration gradient (high to low). It does not require membrane transport proteins — small non-polar molecules such as oxygen and carbon dioxide pass directly through the phospholipid bilayer — and it does not require ATP, so it is a passive process.

Facilitated diffusion also moves particles down the concentration gradient and is likewise passive (no ATP). However, it does require membrane proteins: channel proteins form hydrophilic pores for specific ions, and carrier proteins bind a specific solute and change shape to move it across. It is used for polar molecules and ions that cannot cross the bilayer directly.

Active transport moves particles against the concentration gradient (from low to high concentration). It requires a carrier protein and the input of ATP from respiration, which causes the carrier to change shape and move the solute to the side where it is already more concentrated.

In summary, simple and facilitated diffusion are both passive and follow the gradient, differing only in whether proteins are used; active transport is the only one of the three that moves solutes against the gradient and uses ATP.
Why this scores
Why this scores 6/6. Marks available: (1) simple diffusion is down the gradient; (2) simple diffusion uses no proteins / no ATP; (3) facilitated diffusion is down the gradient and passive; (4) facilitated diffusion uses channel/carrier proteins; (5) active transport is against the gradient; (6) active transport uses a carrier protein and ATP. A clear comparison table or three structured paragraphs both gain full credit; vague statements such as 'they all move substances across membranes' do not.
Question 4
Edexcel IAL Biology Unit 1 style5 marks
Q (5 marks). Use Fick's law to explain how the structure of an alveolus makes it an efficient surface for gas exchange.
Model answer
Fick's law states that the rate of diffusion is proportional to (surface area × concentration difference) ÷ diffusion distance, so an efficient exchange surface maximises surface area and the concentration difference while minimising the diffusion distance.

The alveoli are numerous and folded, giving the lungs a very large total surface area for diffusion of oxygen and carbon dioxide; the larger the surface area, the higher the rate of diffusion.

The wall of the alveolus is a single layer of flattened (squamous) epithelial cells, and the capillary wall is also one cell thick. This gives a very short diffusion distance between the air and the blood, which increases the rate of diffusion.

A steep concentration gradient is maintained: ventilation continually replaces the air in the alveoli with fresh, oxygen-rich air, and the blood flow in the capillaries continually carries oxygenated blood away and brings deoxygenated blood in. This keeps the concentration difference of oxygen (and carbon dioxide) across the surface as large as possible.

Together, the large surface area, short diffusion distance and maintained concentration gradient maximise every term in Fick's law, making the alveolus a highly efficient exchange surface.
Why this scores
Why this scores 5/5. (1) states Fick's law correctly; (2) large surface area (many folded alveoli); (3) short diffusion distance (single squamous epithelium); (4) steep concentration gradient maintained by ventilation and blood flow; (5) explicit link of each feature to increasing the RATE of diffusion. Naming the features without linking them to Fick's terms loses the explanation marks.
Question 5
Edexcel IAL Biology Unit 1 style6 marks
Q (6 marks). Cells lining the small intestine absorb glucose from the gut even when the glucose concentration in the gut is lower than inside the cell. Explain how this is achieved, and explain why the process would stop if the cell's mitochondria were poisoned.
Model answer
Glucose is absorbed against its concentration gradient by co-transport with sodium ions, a form of secondary active transport.

First, the sodium–potassium (Na⁺/K⁺) pump in the basal membrane of the epithelial cell uses ATP to actively transport sodium ions out of the cell into the blood. This lowers the sodium ion concentration inside the cell and so establishes a steep sodium ion concentration gradient from the gut lumen into the cell.

A co-transporter protein in the membrane facing the lumen then carries sodium ions and glucose molecules together into the cell. The sodium ions move down their concentration gradient and, as they do so, the glucose is carried in with them — even though glucose is moving against its own concentration gradient. The glucose then passes from the cell into the blood by facilitated diffusion.

If the mitochondria were poisoned, aerobic respiration would stop, so little or no ATP would be produced. Without ATP, the Na⁺/K⁺ pump cannot pump sodium ions out of the cell, so the sodium ion gradient is not maintained and collapses. With no sodium gradient to drive the co-transporter, glucose can no longer be carried in against its gradient, so absorption stops.

This shows that although the glucose carrier itself does not directly hydrolyse ATP, glucose uptake still depends entirely on ATP because it relies on the sodium gradient created by the ATP-powered pump.
Why this scores
Why this scores 5-6/6. (1) Na⁺/K⁺ pump moves Na⁺ out using ATP; (2) creates a Na⁺ gradient into the cell; (3) co-transporter moves Na⁺ and glucose together; (4) glucose moved against its own gradient / leaves by facilitated diffusion; (5) poisoning mitochondria stops respiration → no ATP; (6) no ATP → pump stops → Na⁺ gradient collapses → no co-transport. The high-band insight is recognising that glucose uptake is INDIRECTLY ATP-dependent via the Na⁺ gradient.
Question 6
Edexcel IAL Biology Unit 1 style7 marks
Q (7 marks). A plant cell with a solute potential of −800 kPa and a pressure potential of +300 kPa is placed in a solution with a water potential of −400 kPa. Calculate the water potential of the cell, state the direction of net water movement, and explain what will happen to the cell, using water potential terms throughout.
Model answer
The water potential of the cell is found using ψ = ψ_s + ψ_p, where ψ_s is the solute potential and ψ_p is the pressure potential.

ψ = (−800) + (+300) = −500 kPa.

The cell water potential (−500 kPa) is lower (more negative) than the surrounding solution (−400 kPa). Water moves by osmosis down the water potential gradient, from higher (less negative) to lower (more negative) water potential, so water moves into the cell across the partially permeable cell-surface membrane.

As water enters, the volume of the protoplast increases and it pushes harder against the cellulose cell wall, so the pressure potential (ψ_p) rises. The entering water also dilutes the cell contents slightly, but the rising pressure potential makes the overall cell water potential less negative, moving it closer to the water potential of the external solution.

Water continues to enter until the cell water potential equals the solution water potential (−400 kPa), at which point there is no further net movement. The wall prevents the cell from bursting, so the cell becomes turgid rather than lysing. (A plant cell cannot reach a water potential less negative than the surrounding solution because the rigid wall halts net water entry once the water potentials are equal.)
Why this scores
Why this scores 6-7/7. (1) correct equation ψ = ψ_s + ψ_p; (2) correct value −500 kPa; (3) cell is more negative than the solution; (4) water moves into the cell down the water potential gradient; (5) across a partially permeable membrane; (6) pressure potential rises / cell water potential becomes less negative; (7) cell becomes turgid (wall prevents bursting) and net movement stops when water potentials are equal. Throughout, water-potential language is required — any use of 'water concentration' loses credit.
Question 7
Edexcel IAL Biology Unit 1 style3 marks
Q (3 marks). State three features that distinguish active transport from simple diffusion.
Model answer
Active transport moves a substance against its concentration gradient (from a lower to a higher concentration), whereas simple diffusion moves substances down the gradient.

Active transport requires a carrier protein, whereas simple diffusion moves substances directly through the phospholipid bilayer with no protein.

Active transport requires ATP from respiration, whereas simple diffusion is passive and uses no metabolic energy.
Why this scores
Why this scores 3/3. Any three of: (1) against vs down the gradient; (2) carrier protein required vs no protein; (3) ATP/energy required vs passive. Each comparison must make the contrast explicit — stating only what active transport does, without the simple-diffusion comparison, can be self-penalising on a 'distinguish' question.
Question 8
Edexcel IAL Biology Unit 1 style5 marks
Q (5 marks). Both facilitated diffusion and active transport can use carrier proteins to move solutes across the cell-surface membrane. Explain how the two processes differ.
Model answer
In facilitated diffusion a carrier protein binds a specific solute and changes shape to move it across the membrane, but the solute moves down its concentration gradient, from a higher to a lower concentration. The energy for the conformational change comes from the binding of the solute and the gradient itself, so the process is passive and requires no ATP.

In active transport a carrier protein also binds a specific solute and changes shape, but the solute is moved against its concentration gradient, from a lower to a higher concentration. This is not spontaneous, so it requires energy: ATP from respiration is hydrolysed to ADP + Pi at the carrier (often phosphorylating it), and this drives the shape change that pumps the solute to the side where it is already more concentrated.

The key distinction is therefore the direction of movement relative to the gradient and the need for ATP: facilitated diffusion is passive and follows the gradient, whereas active transport is an energy-requiring process that works against it. A consequence is that active transport stops if respiration is inhibited (e.g. by cyanide or lack of oxygen), whereas facilitated diffusion continues as long as a gradient exists.
Why this scores
Why this scores 5/5. (1) facilitated diffusion moves solute down the gradient; (2) facilitated diffusion is passive / no ATP; (3) active transport moves solute against the gradient; (4) active transport requires ATP (from respiration) hydrolysed to ADP + Pi; (5) named consequence — active transport stops without respiration whereas facilitated diffusion does not. The A*-distinguishing point is naming the ATP→ADP+Pi hydrolysis at the carrier as the energy source for the conformational change.
Question 9
Edexcel IAL Biology Unit 1 style5 marks
Q (5 marks). Two gas-exchange surfaces, P and Q, were compared. Surface Q has 3 times the surface area of P, the same concentration difference, and a diffusion distance that is 0.5 times that of P. (a) Using Fick's law, calculate how many times faster the rate of diffusion is across Q than across P. Show your working. (b) Suggest one structural feature that gives a gas-exchange surface a short diffusion distance.
Model answer
(a) Fick's law: rate of diffusion ∝ (surface area × concentration difference) ÷ diffusion distance.

Taking the rate across P as the baseline, the relative rate across Q is the product of the change in each factor:

relative rate = (surface area factor × concentration-difference factor) ÷ distance factor relative rate = (3 × 1) ÷ 0.5 relative rate = 3 ÷ 0.5 = 6

So the rate of diffusion across Q is 6 times faster than across P. (No unit is required because this is a ratio of rates.)

(b) A wall made of a single layer of flattened (squamous) epithelial cells gives a very short diffusion distance between the air and the blood. (Alternatively: capillary walls that are one cell thick, lying close to the exchange surface.)
Why this scores
Why this scores 5/5. (a) (1) correct method applying all factors; (2) surface area ×3 and concentration difference ×1 handled correctly; (3) dividing by 0.5 doubles the rate; (4) correct answer 6 (×). (b) (5) named feature giving a short distance (squamous epithelium / single cell layer). Most marks are lost by treating distance as a multiplier — halving the distance must DOUBLE the rate. Showing working secures the method mark even if the final figure is wrong.
Question 10
Edexcel IAL Biology Unit 1 style6 marks
Q (6 marks). Identical 5 g cylinders of potato tissue were placed in sucrose solutions of different concentration for 24 hours, then reweighed. The percentage change in mass was: distilled water +18%; 0.2 mol dm⁻³ +4%; 0.4 mol dm⁻³ −3%; 0.6 mol dm⁻³ −11%. Describe the trend shown by the data and explain the results in terms of water potential. State what the sucrose concentration at which there would be no change in mass would tell you about the cells.
Model answer
Trend. As the sucrose concentration increases from distilled water to 0.6 mol dm⁻³, the percentage change in mass falls from +18% to −11%; the tissue gains mass in the more dilute solutions (distilled water and 0.2 mol dm⁻³) and loses mass in the more concentrated solutions (0.4 and 0.6 mol dm⁻³).

Explanation (water potential). In distilled water and the 0.2 mol dm⁻³ solution, the surrounding solution has a higher (less negative) water potential than the cells. Water therefore moves into the cells by osmosis, down the water potential gradient, across the partially permeable cell-surface membranes, so the cells gain water, become more turgid and the tissue gains mass.

In the 0.4 and 0.6 mol dm⁻³ solutions the surrounding solution has a lower (more negative) water potential than the cells. Water therefore moves out of the cells down the water potential gradient, the cells lose water (and become flaccid or plasmolysed), and the tissue loses mass. The more concentrated the external solution, the more negative its water potential, so the steeper the gradient and the greater the loss of mass — which is why the −11% loss occurs at 0.6 mol dm⁻³.

No-change point. The concentration at which there is no change in mass (between 0.2 and 0.4 mol dm⁻³, where the line crosses 0%) is where there is no net movement of water. At this point the water potential of the external solution equals the water potential of the cells, so this concentration can be used to estimate the water potential of the potato tissue.
Why this scores
Why this scores 6/6 (Level 3). (1) describes the trend with direction and quoted values (+18% → −11%); (2) gain in dilute, loss in concentrated; (3) in dilute solutions external ψ is higher/less negative so water enters by osmosis; (4) in concentrated solutions external ψ is lower/more negative so water leaves; (5) movement is down the water potential gradient across a partially permeable membrane; (6) the no-change point is where external ψ = cell ψ, used to find the tissue's water potential. Reaches Level 3 because it links specific data values to a fully water-potential-based mechanism. Any 'water concentration' phrasing forfeits the water-potential marks.

Key Definitions and Keywords — Movement of Substances

Definitions to memorise and the exact keywords mark schemes credit for movement of substances answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Simple diffusion
Examiner keyword
The net passive movement of molecules or ions from a region of higher concentration to a region of lower concentration (down a concentration gradient), resulting from their random kinetic motion. No ATP and no transport proteins are required.
Concentration gradient
Examiner keyword
The difference in concentration of a substance between two regions. Particles diffuse down a concentration gradient (from high to low); the steeper the gradient, the faster the rate of diffusion.
Fick's law
Examiner keyword
The rate of diffusion is proportional to (surface area × concentration difference) ÷ diffusion distance. Rate increases with larger surface area or steeper gradient, and decreases with longer diffusion distance.
Facilitated diffusion
Examiner keyword
The net passive movement of polar molecules and ions down a concentration gradient, through specific channel or carrier proteins in the membrane. Passive (no ATP) but, unlike simple diffusion, requires transport proteins.
Channel protein
Examiner keyword
An intrinsic membrane protein forming a water-filled hydrophilic pore that allows specific ions to pass through by facilitated diffusion. Many are gated (open or close in response to a signal).
Carrier protein
Examiner keyword
An intrinsic membrane protein that binds a specific solute and changes shape to transport it across the membrane. Used in both facilitated diffusion (passive) and active transport (ATP-driven).
Osmosis
Examiner keyword
The net movement of water molecules from a region of higher (less negative) water potential to a region of lower (more negative) water potential, through a partially permeable membrane. A passive process requiring no ATP.
Water potential (ψ)
Examiner keyword
The tendency of water molecules to move out of a system, measured in kilopascals (kPa). Pure water has a water potential of 0 kPa; adding solute lowers it, making it more negative.
Partially permeable membrane
Examiner keyword
A membrane that allows water and some small molecules to pass through but restricts the movement of larger solute molecules. It is across such a membrane that osmosis occurs.
Active transport
Examiner keyword
The movement of a substance against its concentration gradient (from lower to higher concentration) using a carrier protein and energy from ATP produced by respiration.
Endocytosis
Examiner keyword
A bulk transport process in which the cell-surface membrane invaginates around material, pinching off to form a vesicle inside the cell. Includes phagocytosis (solids) and pinocytosis (liquids); requires ATP.
Exocytosis
Examiner keyword
A bulk transport process in which a vesicle fuses with the cell-surface membrane and releases its contents to the outside of the cell, used to secrete substances such as enzymes and hormones. Requires ATP.
Co-transport (co-transporter)
Examiner keyword
A process in which a single carrier protein moves two substances across a membrane at the same time, using the concentration gradient of one substance (e.g. Na⁺) to move the other (e.g. glucose) against its gradient. A form of secondary active transport.
Passive transport
Examiner keyword
Movement of substances across a membrane that does not require ATP, because it occurs down a concentration or water potential gradient. Includes simple diffusion, facilitated diffusion and osmosis.

Common Mistakes and Misconceptions — Movement of Substances

The traps other students keep falling into on movement of substances questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Defining osmosis using 'water concentration' instead of water potential
Edexcel IAL Biology Unit 1 examiner reports
Why it happens
Students learn a simplified GCSE definition ('water moves from high to low water concentration') and carry it into IAL.
How to avoid it
At IAL osmosis MUST be defined with water potential: net movement of water from higher (less negative) to lower (more negative) water potential across a partially permeable membrane. 'Water concentration' is not creditworthy.
✕Getting the direction wrong because of the negative sign of water potential
Edexcel IAL Biology Unit 1 examiner reports
Why it happens
Water potentials are negative, so −150 kPa feels 'smaller' than −400 kPa, when in fact −150 kPa is HIGHER.
How to avoid it
Remember pure water = 0 kPa (the maximum). Less negative = higher water potential. Water always moves to the more negative value. Compare numbers on a number line: −150 is higher (further right) than −400.
✕Claiming facilitated diffusion uses ATP
Edexcel IAL Biology Unit 1 examiner reports
Why it happens
Because it involves proteins, students assume it must require energy like active transport.
How to avoid it
Facilitated diffusion is passive — it moves substances down the gradient with no ATP. Only active transport moves substances against the gradient and uses ATP. The presence of a carrier protein alone does not mean ATP is used.
✕Saying a substance is moved 'down the gradient' in active transport
Edexcel IAL Biology Unit 1 examiner reports
Why it happens
The phrase 'down the gradient' is over-learned from diffusion.
How to avoid it
Active transport moves substances against the concentration gradient, from lower to higher concentration. This is the defining feature that distinguishes it from diffusion — state it explicitly.
✕Treating diffusion distance as directly proportional to rate in Fick's law
Edexcel IAL Biology Unit 1 examiner reports
Why it happens
Students apply all three factors as multipliers without noticing that distance is on the bottom of the equation.
How to avoid it
In Fick's law, distance is the denominator: rate ∝ (SA × conc. diff.) ÷ distance. Halving the distance doubles the rate; doubling the distance halves it. Surface area and concentration difference are directly proportional; distance is inversely proportional.
✕Stating that the glucose co-transporter directly uses ATP
Edexcel IAL Biology Unit 1 examiner reports
Why it happens
Glucose moves against its gradient, so students assume the carrier must hydrolyse ATP itself.
How to avoid it
In co-transport the ATP is used by the Na⁺/K⁺ pump to create the sodium gradient; the co-transporter then uses that Na⁺ gradient (not ATP directly) to drag glucose in. This is secondary (indirect) active transport.

Movement of Substances — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Across the cell-surface membrane, only certain substances can diffuse directly through the phospholipid bilayer:

Small, non-polar (lipid-soluble) molecules — oxygen ( $O_2$ ), carbon dioxide ( $CO_2$ ) and other lipophilic molecules dissolve in the hydrophobic tail region and pass straight through.
Large molecules and charged/polar particles (glucose, amino acids, ions) cannot cross the bilayer freely — they need transport proteins (facilitated diffusion or active transport).

Fick's law summarises what controls the rate of diffusion:

$\text{rate of diffusion} \;\propto\; \dfrac{\text{surface area} \times \text{concentration difference}}{\text{diffusion distance}}$

So the rate increases when:

the surface area of the membrane is larger (e.g. microvilli, alveoli, root hair cells),
the concentration difference (gradient) across the membrane is steeper (maintained by blood flow or ventilation), or
the diffusion distance is shorter (e.g. a single squamous epithelial cell wall in the alveolus).

Fick's law: a larger surface area and steeper concentration difference speed diffusion, while a longer diffusion distance slows it.

Fick's law is the basis for almost every "explain why this exchange surface is efficient" question — flat lungs, folded gut, thin gills all map onto one of the three terms.

Surrounding solution

Animal cell

Plant cell

Higher ψ (hypotonic, dilute)

water in → swells, may lyse

water in → turgid (wall resists)

Equal ψ (isotonic)

no net change

flaccid / incipient plasmolysis

Lower ψ (hypertonic, concentrated)

water out → crenated/shrinks

water out → plasmolysed (membrane pulls from wall)

Movement of Substances

Short Notes - Movement of Substances

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Identifying the transport mechanism (2 marks)

2Predicting the direction of osmosis (2 marks)

3Channel versus carrier proteins (3 marks)

4Why a respiratory inhibitor stops uptake (3 marks)

5Applying Fick's law — a proportional calculation (2 marks)

6Co-transport of glucose in the intestine (4 marks)

Simple diffusion

Concentration gradient

Fick's law

Facilitated diffusion

Channel protein

Carrier protein

Osmosis

Water potential (ψ)

Partially permeable membrane

Active transport

Endocytosis

Exocytosis

Co-transport (co-transporter)

Passive transport

✕Defining osmosis using 'water concentration' instead of water potential

✕Getting the direction wrong because of the negative sign of water potential

✕Claiming facilitated diffusion uses ATP

✕Saying a substance is moved 'down the gradient' in active transport

✕Treating diffusion distance as directly proportional to rate in Fick's law

✕Stating that the glucose co-transporter directly uses ATP

Movement of Substances — frequently asked questions

Movement of Substances

Short Notes - Movement of Substances

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Identifying the transport mechanism (2 marks)

2Predicting the direction of osmosis (2 marks)

3Channel versus carrier proteins (3 marks)

4Why a respiratory inhibitor stops uptake (3 marks)

5Applying Fick's law — a proportional calculation (2 marks)

6Co-transport of glucose in the intestine (4 marks)

Simple diffusion

Concentration gradient

Fick's law

Facilitated diffusion

Channel protein

Carrier protein