IB Biology 2025 · Theme B · Form and function

B2.1 Membranes and membrane transport. Practice questions with markscheme.

68 original IB-style questions on B2.1, written from the 2025 guide: 31 multiple-choice, 25 short-answer, 7 data-based, 2 drawing, 2 extended-response parts, 1 labelling. Below are two ready-made practice papers built from them, one at standard level (22 marks) and one at higher level (20 marks); switch between them above the paper. Hand one out as a class quiz or homework, or sit it yourself and mark against the scheme. Print it, project it, or build a fresh one on the same topic.

What the guide asks for

10 statements at SL and HL, 7 additional higher level.

B2.1.1SL / HL Lipid bilayers as the basis of cell membranes
B2.1.2SL / HL Lipid bilayers as barriers
B2.1.3SL / HL Simple diffusion across membranes
B2.1.4SL / HL Integral and peripheral proteins in membranes
B2.1.5SL / HL Movement of water molecules across membranes by osmosis and the role of aquaporins
B2.1.6SL / HL Channel proteins for facilitated diffusion
B2.1.7SL / HL Pump proteins for active transport
B2.1.8SL / HL Selectivity in membrane permeability
B2.1.9SL / HL Structure and function of glycoproteins and glycolipids
B2.1.10SL / HL Fluid mosaic model of membrane structure
B2.1.11HL Relationships between fatty acid composition of lipid bilayers and their fluidity
B2.1.12HL Cholesterol and membrane fluidity in animal cells
B2.1.13HL Membrane fluidity and the fusion and formation of vesicles
B2.1.14HL Gated ion channels in neurons
B2.1.15HL Sodium–potassium pumps as an example of exchange transporters
B2.1.16HL Sodium-dependent glucose cotransporters as an example of indirect active transport
B2.1.17HL Adhesion of cells to form tissues

In the bank for B2.1

31 multiple-choice
25 short-answer
7 data-based
2 drawing
2 extended-response parts
1 labelling
29 higher level only

Every question is original and tagged to a guide statement. See the whole bank →

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The practice paper

22 marks at SL · 20 at HL · each has its own code

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Biology · topic quiz

Standard level · topic practice, not an exam format

35 minutes22 marks

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Covers B2.1 Membranes and membrane transport · SL

Name:

A plant cell is placed in a solution with a higher solute concentration than its cytoplasm. What is the expected result?

[1]

Solutes leave the cell until concentrations are equal
Water leaves and the membrane pulls away from the wall
The cell bursts as water rushes in
The cell swells and becomes fully turgid

In an early investigation of membrane structure, researchers extracted the lipids from the plasma membranes of a counted sample of red blood cells from a mammal, so that all of the lipid in the extract came from plasma membranes. The lipids were dissolved in a solvent and dropped onto the surface of water in a shallow tray. The solvent evaporated, leaving the lipids as a layer one molecule thick on the water surface. A movable barrier then pushed the molecules together until they were tightly packed, and the area of the layer was measured. The table shows the results.

Measurement	Value
Number of red blood cells in the sample	3.2 × 10⁹
Mean surface area of one red blood cell / μm²	95
Area of the tightly packed lipid layer / m²	0.61

(a)Calculate the ratio of the area of the lipid layer to the total surface area of the red blood cells in the sample (1 μm² = 10⁻¹² m²).[2]

(b)Explain how these results support the model that the lipids of a plasma membrane form a bilayer.[3]

(c)The investigation was repeated later with improved methods. From the same number of cells, 1.5 times as much lipid was extracted, and the mean surface area of a red blood cell was measured as 143 μm². Deduce whether the later results support the same conclusion.[2]

(d)The first investigation underestimated both the amount of lipid in the membranes and the surface area of the cells, yet it reached the same ratio as the later one. Discuss what this episode shows about how scientific conclusions are reached and tested.[3]

Two substances, X and Y, both enter a type of cell. The initial rate of uptake of each was measured at a range of external concentrations, with all other conditions held constant. The results are shown.

External concentration / mmol dm⁻³	Uptake rate of X / μmol min⁻¹	Uptake rate of Y / μmol min⁻¹
2	8	4
4	16	7
8	32	12
16	64	13
24	96	13

(a)Analyse the data to deduce which substance crosses the membrane by facilitated diffusion. Justify your answer.[2]

(b)Explain the pattern in the uptake rate of the substance that crosses by facilitated diffusion.[2]

(c)Predict, with reasons, the effect of adding a metabolic inhibitor that stops ATP production on the uptake of X and of Y.[2]

(d)State one variable that must be kept constant for this comparison to be valid.[1]

Describe the fluid mosaic model of membrane structure.

[4]

Original practice questions © Biology by Bradford · CC BY-NC-SA 4.0 · Not affiliated with or endorsed by the International Baccalaureate Organization.
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Markscheme BbB-EAAACAAAABQAOq89

One mark per point; / separates alternative wording within a point, OR separates alternative answers, words in brackets are not required, underlined words are essential. OWTTE = or words to that effect.

1. B2.1.5 [1]

B: the surroundings are hypertonic, so water leaves by osmosis and the protoplast shrinks; the wall prevents overall collapse;

2. B2.1.1

(a) [2]
- total surface area of the cells = 3.2 × 10⁹ × 95 × 10⁻¹² = 0.30 m² (0.304 m²);
- ratio = 0.61 ÷ 0.304 = 2.0 (: 1);
Accept 2.01. Award [1] for the total surface area alone. ECF from a wrong total surface area.
(b) [3 max]
- the lipid on the water formed a layer one molecule thick / a monolayer;
- its area was (about) twice the surface area of the cells, so there is enough lipid to cover the surface of each cell twice / to form two layers;
- in the membrane the hydrophobic (hydrocarbon) tails of the two layers face inwards, towards each other, away from water;
- the hydrophilic (phosphate) heads face the aqueous solutions on both sides of the membrane (the cytoplasm and the fluid outside the cell), OWTTE;
- (phospholipids are amphipathic, so) this arrangement forms spontaneously in water, OWTTE;
(c) [2]
- new area of lipid = 0.61 × 1.5 = 0.92 m² (0.915 m²) and new total surface area of the cells = 3.2 × 10⁹ × 143 × 10⁻¹² = 0.46 m² (0.458 m²);
- ratio still (about) 2.0 (: 1), so the later results support the same conclusion / a bilayer;
Both new values are needed for the first mark. ECF from (a).
(d) [3 max]
- the two errors were of about the same size (both about 1.5 times) and cancelled out, so the original ratio was right partly by chance, OWTTE;
- a correct conclusion can come from flawed measurements, so agreement with later findings does not show that a method was reliable;
- repeating investigations with improved techniques / methods tests earlier conclusions and can reveal hidden (systematic) errors, OWTTE;
- scientists should evaluate the methods and uncertainties behind a result, not only whether the result fits a model, OWTTE;
- a model is held with more confidence when it is supported by several independent lines of evidence, and stays open to revision, OWTTE;
Accept other valid points about the nature of science that are linked to this episode.

3. B2.1.6-B2.1.8

(a) [2]
- substance Y;
- its uptake rate rises then levels off/plateaus (saturates) at ~13 μmol min⁻¹, whereas X keeps rising (roughly proportionally), OWTTE;
(b) [2]
- a limited number of channel/carrier proteins move the substance across;
- at high concentrations all the transport proteins are occupied/saturated, so the rate reaches a maximum, OWTTE;
(c) [2 max]
- uptake of X would not be reduced because simple diffusion (between the phospholipids) does not use ATP;
- uptake of Y would not be reduced because facilitated diffusion is passive / the channel or carrier proteins do not use ATP;
- (neither would be affected) unlike active transport, which stops when ATP is unavailable, OWTTE;
(d) [1]
- temperature / cell type / cell (membrane) surface area / time / pH;
Mark the first answer only.

4. B2.1.10 [4 max]

membrane is a phospholipid bilayer with hydrophilic heads outward and hydrophobic tails inward;
phospholipids (and some proteins) move laterally, making the membrane fluid;
proteins are scattered through the bilayer like a mosaic;
integral proteins span/embed in the bilayer, whereas peripheral proteins sit on a surface;
glycoproteins/glycolipids occur on the outer surface / cholesterol among the phospholipids (in animal cells);

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Biology · topic quiz

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30 minutes20 marks

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Covers B2.1 Membranes and membrane transport · HL

Name:

Which row correctly describes facilitated diffusion?

[1]

Row	Protein required	Direction relative to concentration gradient	ATP used
A.	yes	down the gradient	no
B.	yes	against the gradient	yes
C.	no	down the gradient	no
D.	no	against the gradient	yes

Row A
Row B
Row C
Row D

Glucose is absorbed into cells lining the small intestine together with sodium ions by a cotransporter protein. Why is this described as indirect (secondary) active transport?

[1]

Glucose moves down its concentration gradient without a protein
The cotransporter hydrolyses ATP directly
Glucose moves against its gradient using the sodium gradient
Sodium ions move against their gradient through it

In a tissue, neighbouring cells are joined by two types of cell–cell junction, J1 and J2, which use different forms of cell-adhesion molecule (CAM). The CAMs in J1 bind to the CAMs of the neighbouring cell only when calcium ions are present; the CAMs in J2 bind whether or not calcium ions are present. A piece of the tissue was placed in a solution from which all calcium ions had been removed. What is the most likely result?

[1]

Neither junction is affected, as CAMs are proteins, not ions
J1 junctions come apart, while J2 junctions keep the cells joined
Both junctions come apart, since calcium ions form the junctions
J2 junctions come apart, while J1 junctions keep the cells joined

Outline how a cell secretes a protein by exocytosis.

[3]

Cells that carry out large amounts of active transport contain many mitochondria, the organelles where most of a cell's ATP is produced. Suggest why.

[2]

Explain how a plasma membrane can be selectively permeable, allowing the cell to control which substances enter and leave.

[3]

Researchers investigated five proteins, V, W, X, Y and Z, in the plasma membrane of liver cells from a rat (Rattus norvegicus), using three tests. Test 1: plasma membranes were isolated and washed in a concentrated salt solution, which weakens the attractions between charged groups but leaves the phospholipid bilayer intact. Proteins released into the salt solution were identified. Test 2: intact cells were treated with marker M, a chemical that attaches to any part of a protein exposed at the surface it reaches but cannot cross membranes. Proteins carrying M were then identified. Test 3: plasma membranes were broken up into small sealed vesicles that had turned inside out, so the surface that had faced the cytoplasm now faced outwards. Marker M was added to the liquid around the vesicles. The table shows the results.

Protein	Released by the salt wash (test 1)	Labelled by M on intact cells (test 2)	Labelled by M on inside-out vesicles (test 3)
V	yes	no	yes
W	no	yes	yes
X	no	no	yes
Y	yes	yes	no
Z	no	yes	no

(a)Identify the two peripheral proteins.[1]

(b)Deduce, with reasons, the position of protein W and of protein X in the membrane.[3]

(c)Predict, with a reason, how the result for protein Z would have changed if many of the vesicles in test 3 had not been sealed, and the effect this would have on the conclusion about its position.[2]

(d)Membrane proteins carry out many different functions. Suggest, with reasons, which of the proteins V to Z could be a channel protein, and which could be a glycoprotein.[3]

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Markscheme BbB-EAEACAAAABQAEsJr

1. B2.1.6 [1]

A: channel/carrier protein, passive, down the gradient;

2. B2.1.16 [1]

C: the cotransporter itself uses no ATP; the Na⁺ gradient (built by the ATP-driven Na⁺/K⁺ pump) powers glucose uptake against its gradient;

3. B2.1.17 [1]

B: each type of junction uses its own form of CAM; without calcium ions the CAMs of J1 can no longer bind to those of the neighbouring cell, so J1 junctions separate, while the CAMs of J2 still bind and hold the cells together; calcium ions are needed for J1 CAMs to bind (not A) but do not form the junctions themselves (not C), and D reverses the two junctions;

4. B2.1.13 [3 max]

protein is packaged/carried in a (secretory) vesicle (from the Golgi apparatus);
vesicle moves to / fuses with the plasma membrane;
fusion is possible because of membrane fluidity, OWTTE;
contents are released outside the cell (without the protein crossing the bilayer);

Do not accept diffusion of the protein through the bilayer.

5. B2.1.7 [2 max]

active transport requires energy from ATP (to pump particles against their concentration gradients);
ATP is used to change the conformation of the pump proteins;
more mitochondria supply more ATP, so more pumps can operate / more pumping can occur, OWTTE;

6. B2.1.8 [3 max]

the hydrophobic core/centre of the phospholipid bilayer blocks ions and (larger) polar molecules, while small non-polar molecules (O₂/CO₂) diffuse through freely (simple diffusion is not selective);
channel proteins and pumps/carriers are each specific to one solute (a binding site/pore complementary in shape/charge);
(so) a substance only crosses at a significant rate if the membrane contains a matching transport protein;
the cell controls permeability through the types and numbers of channels/pumps in the membrane / by opening and closing gated channels, OWTTE;

7. B2.1.4

(a) [1]
- V and Y;
Both needed for the mark.
(b) [3 max]
- neither W nor X is released by the salt wash, so both are integral / embedded in the bilayer (not attached to a surface);
- W is labelled both on intact cells and on inside-out vesicles, so it is exposed at both surfaces / spans the whole membrane;
- X is labelled only on inside-out vesicles, so it is exposed only at the cytoplasmic surface / is embedded in the inner (cytoplasmic) layer only;
- the salt does not release them because part of each protein lies among the hydrophobic hydrocarbon tails in the core of the bilayer, which the salt leaves intact, OWTTE;
(c) [2 max]
- Z would (also) be labelled in test 3;
- because M could enter unsealed vesicles and reach the surface that had faced the outside of the cell (now on the inside of the vesicles);
- so Z would wrongly appear to be exposed at both surfaces / to span the membrane (like W), OWTTE;
Accept a prediction that Y would also be labelled in test 3.
(d) [3 max]
- W could be the channel protein, as it alone is exposed at both surfaces / spans the membrane;
- a channel must form a (hydrophilic) pore through the hydrophobic core, from one side of the membrane to the other;
- W or Z could be the glycoprotein, as they are integral proteins exposed at the outer surface of the cell;
- the carbohydrate chains of glycoproteins are on the extracellular side of the membrane, so V and X, which are exposed only at the cytoplasmic surface, could not be glycoproteins, OWTTE;
Content pivot. Do not award a letter given without a reason. Accept Y for the glycoprotein, as it is exposed only at the outer surface. Credit a reference to the carbohydrate chains acting in cell recognition or adhesion outside the cell within the fourth point.