IB Biology SL Theme B: Form and function questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

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 sampleProbe A bindingProbe 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 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 3

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.

0123456780123456789Time / minConcentration in receiving compartment / mmol dm-3● O2○ CO2▲ Na^+□ Glucose
(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 4

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 regionSignal concentration / arbitrary units
W8
X7
Y5
Z3
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 5

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 conditionStem cells at day 7Newly differentiated blood cells at day 7
M24060
N60240
(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 6

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 7

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.

SiteHabitatSoil salinity / g kg⁻¹Soil water content / %Marram shoots per quadratA. marina seedlings per quadrat
AUpper dune24130
BDune–swamp boundary141864
CInner mangrove swamp2842016
DCreek edge365508

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 / °CSnails counted per 10 m²
141
187
2215
268
300
(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 8

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?

SkullDietMolar-area index
Homo sapiens (living reference)Omnivorous1.2
Living hominid referenceHerbivorous2.4
Homo floresiensis (fossil)Unknown1.3
Paranthropus robustus (fossil)Unknown2.3
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.

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