IB Biology SL Theme B: Form and function questions

Question 1

Paper 2Medium16 marks

Cellulose in plant cell walls and glycoproteins on cell surfaces are examples of biological structures associated with different functions.

In a model of cell-surface recognition, probe A binds only to the A antigen and probe B binds only to the B antigen. The probes represent recognition molecules on another cell. A “+” indicates that binding was detected.

Red-cell sample	Probe A binding	Probe B binding
P	+	−
Q	−	+
R	+	+
S	−	−

For this question, a red-cell sample with A antigen only is group A; with B antigen only is group B; with both antigens is group AB; and with neither antigen is group O.

(a)

Explain how the structure of cellulose makes it suitable for its role as a structural polysaccharide in plant cell walls.

[4]

(b)

Use the results in the table to determine the ABO group of each sample, P, Q, R and S.

[4]

(c)

Explain how glycoproteins contribute to cell-cell recognition. Use ABO antigens and the results in the table in your answer.

[8]

Question 2

Multiple choiceHard1 mark

A plant cell stores starch as represented in the figure. Each hexagon represents one α-glucose monomer; the diagram is schematic and not to scale. Which description correctly identifies P and Q and the structural basis of starch’s storage properties?

Show answer and explanation

Answer: C. In the figure, P joins separate α-glucose monomers into a starch polymer, while Q separates the polymer into monomers. Joining monomers by condensation releases water; hydrolysis uses water to break the polymer apart. Coiling and branching make the polymer compact, while its large molecular size makes it relatively insoluble.

A: reverses the direction of the two processes, so it assigns polymer breakdown to P and polymer building to Q.
B: identifies the processes correctly but reverses their water balance.
D: swaps the structural explanations: coiling and branching account for compactness, while large molecular size accounts for relative insolubility.

Remember: condensation builds glucose polymers and releases water; hydrolysis breaks them down and uses water.

Question 3

Paper 2Easy6 marks

A student exposes equal samples of the same enzyme to different conditions for 10 minutes. The student then measures the activity of each sample at pH 7 and 25 °C.

Conditions during exposure	Relative enzyme activity afterwards (% of untreated control)
pH 7, 25 °C (untreated control)	100
pH 7, 80 °C	8
pH 2, 25 °C	12

(a)

State three features of polypeptide chains that allow a wide variety of polypeptides to exist.

[3]

(b)

Use the results in the table.

(i)

Identify the exposure condition that resulted in the lowest enzyme activity.

[1]

(ii)

State the term for the change in protein structure that can be caused by high temperature or extreme pH, and state one effect of this change on enzyme function.

[2]

Question 4

Multiple choiceMedium1 mark

Two polypeptides each contain 80 amino acids and the same number of each amino-acid type, but the amino acids are in different orders. Which statement correctly links this difference to a possible effect of a large pH change or high temperature?

Show answer and explanation

Answer: B. Polypeptides with the same amino-acid composition can differ when their amino acids are in a different order. A large pH change or high temperature could cause denaturation, changing protein structure.

A: treats composition as sufficient to identify a polypeptide, overlooking amino-acid order.
C: confuses a change in protein structure with a change in amino-acid sequence.
D: assumes structure cannot change unless the amino-acid order changes.

Remember: amino-acid order distinguishes polypeptides, while denaturation changes protein structure.

Question 5

Multiple choiceEasy1 mark

Solutes X and Y each have lower concentrations outside a cell than inside. Membrane protein P transports X into the cell, but not Y; this movement stops when ATP is unavailable. Which description best explains P’s action?

Show answer and explanation

Answer: D. The concentration of X is lower outside the cell, so inward movement is against its concentration gradient. P transports X but not Y, showing selectivity, and its activity depends on ATP. It therefore acts as a selective pump using ATP for active transport.

A: Confuses active transport with facilitated diffusion, which does not use ATP to move particles against a concentration gradient.
B: Reverses the observed direction of X transport.
C: Treats P as non-selective, although it transports X but not Y.
Remember: pumps use ATP to move specific particles against concentration gradients.

Question 6

Paper 2Medium16 marks

A researcher used four model membranes, each consisting of a phospholipid bilayer separating a donor compartment from a receiving compartment. Each donor initially contained one substance at 10.0 mmol dm⁻³; each receiver initially contained none. The donor volume was 20 times the receiver volume. Membrane area and temperature were the same in all four setups. Figure 1 shows the concentration of each substance in the receiving compartment during the experiment.

(a)

At 4 min, identify the substance with the highest concentration in the receiving compartment and state that concentration.

[2]

(b)

Compare the changes in receiving-compartment concentrations of O₂ and CO₂ with those of Na⁺ and glucose over the 8 min. Quote data from Figure 1.

[3]

(c)

Calculate the mean rate of increase in CO₂ concentration in the receiving compartment from 0 to 4 min. Include a unit.

[2]

(d)

Using Figure 1, explain how the structure of the phospholipid bilayer accounts for the pattern of movement of the four substances.

[4]

(e)

A student states: “The bilayer completely blocks Na⁺ and glucose, but allows O₂ and CO₂ to cross.” Assess this statement using Figure 1.

[3]

(f)

The measurements stop at 8 min. Identify one limitation this creates for assessing the movement of Na⁺ and glucose, and suggest a further measurement.

[2]

Question 7

Multiple choiceEasy1 mark

Following ultracentrifugation, a researcher tests separated cell components to relate their structures to their functions. Which component should be classified as an organelle?

Show answer and explanation

Answer: D. Organelles are discrete cell subunits adapted to perform specific functions; the plasma membrane is classified as an organelle.

A: treats the cytoplasm, which is not an organelle, as a discrete subunit.
B: confuses the cytoskeleton with an organelle; it is not classified as one.
C: treats the cell wall as an organelle, although it is not classified as one.
Remember: the plasma membrane is an organelle, but the cell wall, cytoskeleton and cytoplasm are not.

Question 8

Multiple choiceMedium1 mark

A signal forms a gradient across four adjacent regions of an early embryo. Gene G is expressed only where the signal concentration exceeds 6 arbitrary units. Cells expressing gene G differentiate into specialized cell type P; cells not expressing it differentiate into another type. Which outcome is predicted? [1 mark]

Embryo region	Signal concentration / arbitrary units
W	8
X	7
Y	5
Z	3

Show answer and explanation

Answer: C. Gene G is expressed only where the signal concentration exceeds 6 arbitrary units. Regions W and X have concentrations of 8 and 7, so their cells express gene G and differentiate into type P. Regions Y and Z are below the threshold and differentiate into another type.

A: considers only the region with the highest concentration, overlooking that X also exceeds the threshold.
B: reverses the effect of the signal by assigning type P to regions below the threshold.
D: assumes all regions have the same gene expression and fate despite their different signal concentrations.

Remember: a gradient can lead to different gene-expression patterns and cell fates in an early embryo.

Question 9

Paper 2Medium16 marks

Adult human bone-marrow stem cells were divided into equal groups of 100 cells. Each group was exposed for seven days to a different set of signals, M or N, from bone-marrow niche cells. Assume that no cells died, differentiated cells did not divide, and all cells present at day 7 were either stem cells or newly differentiated blood cells.

Signal condition	Stem cells at day 7	Newly differentiated blood cells at day 7
M	240	60
N	60	240

(a)

State two properties of stem cells.

[2]

(b)

Name two adult human locations that contain stem-cell niches, and state one role of a niche.

[3]

(c)

Calculate the percentage of counted cells that were stem cells in condition M and condition N.

[3]

(d)

Compare the outcomes for M and N and infer which set of signals favoured maintenance or proliferation of the stem-cell population and which favoured differentiation.

[4]

(e)

A researcher concludes that these results prove that the stem cells can divide endlessly and differentiate along different pathways. Evaluate this conclusion using the data.

[4]

Question 10

Multiple choiceEasy1 mark

During a spirometry investigation, a student takes a normal breath in and then inhales as much additional air as possible. Which lung volume is represented by the additional volume measured?

Show answer and explanation

Answer: B. Inspiratory reserve volume is the additional volume that can be inhaled after a normal inspiration.

A: confuses the additional inhalation with tidal volume, the volume moved during a normal breath.
C: reverses the direction; expiratory reserve volume is additional air exhaled after a normal expiration.
D: treats the additional volume as vital capacity, which is the volume moved between maximum inspiration and maximum expiration.
Remember: inspiratory reserve is measured beyond the end of a normal inspiration.

Question 11

Paper 2Easy2 marks

A student uses a spirometer to measure the volume of air exhaled during three normal breaths.

Breath	Volume of air exhaled / dm³
1	0.46
2	0.50
3	0.54

(a)

Determine the mean tidal volume.

[1]

(b)

State one change in the diaphragm during inspiration.

[1]

Question 12

Multiple choiceMedium1 mark

In a plant, the rate of water loss from leaf-cell walls increases. Which change in the xylem and role of cohesion are expected?

Show answer and explanation

Answer: B. Water loss from leaf-cell walls draws water out of the xylem through the cell walls, increasing tension (making the pressure potential more negative). Cohesion keeps water molecules connected in a continuous column, allowing this tension to draw water upward.

A: reverses the direction of water movement, so it incorrectly predicts reduced tension.
C: has the correct movement and tension but reverses the role of cohesion.
D: gets the water movement right but reverses the pressure change; tension is negative pressure potential.

Remember: transpiration creates tension, while cohesion maintains the continuous water column.

Question 13

Paper 2Hard16 marks

Six similar, well-watered potted plants were exposed for 2 h to air at either 40% or 90% relative humidity (RH), with six plants in each condition. Light intensity, temperature and air movement were kept the same. Leaf water loss and xylem pressure potential in the upper stem were measured. Figure 1 shows the experimental arrangement and the individual measurements.

(a)

Using the six measurements in Figure 1B, calculate the mean leaf water loss under each RH condition and the percentage decrease from 40% RH to 90% RH.

[3]

(b)

Using Figure 1C, compare the mean upper-stem xylem pressure potential between the two RH conditions, and relate this comparison to leaf water loss.

[2]

(c)

Explain how water loss from cell walls in the leaf leads to movement of water from roots to leaves in the plant.

[5]

(d)

A student concludes: “Figure 1 proves that xylem tension alone is sufficient to maintain upward water transport; cohesion is unnecessary.”

Evaluate this conclusion using Figure 1 and your knowledge of water transport. Include one limitation of the evidence.

[6]

Question 14

Multiple choiceEasy1 mark

A terrestrial region has high temperatures throughout the year and very low rainfall in every month. Which statement(s) identify a biome likely to develop there?

I. A hot desert
II. A tropical forest
III. A tundra

Show answer and explanation

Answer: B. Hot deserts are characterized by high temperatures and very low rainfall, so the stated climate makes a hot desert likely.

A: adds II by treating high temperature alone as sufficient for tropical forest, despite the very low rainfall.

C: includes II, which requires high rainfall, and III, which is characterized by low temperatures.

D: accepts both tropical forest and tundra; the low rainfall excludes the former, and the high temperatures exclude the latter.

Remember: use both temperature and rainfall to distinguish biomes.

Question 15

Paper 2Medium16 marks

A coastal reserve contains upper sand dunes beside a mangrove swamp. Researchers placed ten randomly positioned 1 m² quadrats at each of four sites. The table shows the mean number of living shoots of marram grass (Ammophila arenaria) and seedlings of the mangrove tree Avicennia marina per quadrat.

Site	Habitat	Soil salinity / g kg⁻¹	Soil water content / %	Marram shoots per quadrat	A. marina seedlings per quadrat
A	Upper dune	2	4	13	0
B	Dune–swamp boundary	14	18	6	4
C	Inner mangrove swamp	28	42	0	16
D	Creek edge	36	55	0	8

Field notes describe the following features: marram leaves roll inward in dry air, enclosing the stomata in a humid space, and its roots spread widely through loose, shifting sand. A. marina has upright pneumatophores that project above the waterlogged sediment, and salt glands on its leaves release salt.

To estimate the number of marram shoots in an 800 m² upper-dune patch, researchers counted the shoots in ten additional randomly placed 1 m² quadrats. The counts were 12, 15, 9, 14, 16, 11, 13, 10, 17 and 13.

Researchers also surveyed an estuarine snail in five pools of similar area, counting snails in 10 m² at each pool. The same search time was used at each pool, and other measured conditions were similar.

Water temperature / °C	Snails counted per 10 m²
14	1
18	7
22	15
26	8
30	0

(a)

State one abiotic variable measured in the plant survey and one measured in the snail survey.

[2]

(b)

Describe how the abundance of the two plant species changes across the sites as soil salinity increases.

[2]

(c)

Estimate the number of marram shoots in the 800 m² upper-dune patch.

[3]

(d)

Explain how the features described in the field notes may help marram and A. marina live in their respective habitats.

[4]

(e)

Describe the pattern in the snail counts, including the observed temperature range of pools in which snails were recorded.

[2]

(f)

A student concludes that soil salinity alone determines which of the two plant species occurs at each site. Evaluate this conclusion, using the plant data and field notes.

[3]

Question 16

Paper 2Easy5 marks

A field team records five plant forms in a tropical forest.

Plant	Observation
P	A tree grows until its crown reaches the canopy.
Q	A woody vine rooted in the soil climbs a tree trunk.
R	A plant grows on a branch of a tree.
S	An epiphyte grows on a host tree and later surrounds its trunk.
T	A small herb grows on the shaded forest floor.

(a)

Identify the plant forms represented by Q and R.

[2]

(b)

Identify the plant forms or strategies represented by P, S and T.

[3]

Question 17

Multiple choiceMedium1 mark

The table shows a molar-area index (molar grinding-surface area divided by combined incisor and canine area), measured in the same way for each skull. Diets of the living reference hominids are known from observations. Which inference about the fossil skulls is supported by the data?

Skull	Diet	Molar-area index
Homo sapiens (living reference)	Omnivorous	1.2
Living hominid reference	Herbivorous	2.4
Homo floresiensis (fossil)	Unknown	1.3
Paranthropus robustus (fossil)	Unknown	2.3

Show answer and explanation

Answer: C. The reference skulls associate a lower molar-area index with omnivory and a higher index with herbivory. The index for Homo floresiensis (1.3) is nearer the omnivorous reference (1.2), while that for Paranthropus robustus (2.3) is nearer the herbivorous reference (2.4).

A: reverses the relationship between the index and the reference diets.
B: correctly associates Paranthropus robustus with the herbivorous reference but assigns the lower-index Homo floresiensis to herbivory.
D: correctly associates Homo floresiensis with the omnivorous reference but assigns the higher-index Paranthropus robustus to omnivory.
Remember: comparisons with observed living species can support, but do not prove, dietary inferences from fossil dentition.

IB Biology SL Theme B: Form and function questions

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