IB Biology HL Theme B: Form and function questions

Question 1

Paper 2Easy6 marks

In a mammal, glucose is transported in the blood and stored in liver cells as glycogen. Figure 1 shows two monosaccharides in ring form and a fragment of glycogen.

(a)

State which of X and Y is a pentose and which is a hexose.

[2]

(b)

State one feature visible in Figure 1(b) that contributes to the compact nature of glycogen.

[1]

(c)

Outline how the solubility and chemical stability of glucose relate to its transport in the blood, and state the benefit of glucose oxidation in cells.

[3]

A molecule is formed by condensation reactions linking one glycerol molecule to three fatty acid molecules. It dissolves in a non-polar solvent but is only sparingly soluble in water. Which statements are correct?

I. Its solubility pattern is consistent with it being a lipid.
II. The molecule is a triglyceride.
III. The molecule is a phospholipid.

Show answer and explanation

Answer: A. Lipids dissolve in non-polar solvents and are only sparingly soluble in aqueous solvents, so the observed pattern is consistent with a lipid. A glycerol molecule linked by condensation to three fatty acids forms a triglyceride; a phospholipid has two fatty acids and a phosphate group.

B: recognizes the solubility pattern but confuses the three-fatty-acid product with a phospholipid.
C: identifies the triglyceride but mistakes it for a phospholipid and omits the matching solubility evidence.
D: accepts the correct statements but also labels the product a phospholipid, despite its specified composition.

Remember: three fatty acids form a triglyceride; two fatty acids and a phosphate group form a phospholipid.

Question 3

Multiple choiceEasy1 mark

A student is considering possible polypeptides assembled from the 20 amino acids represented in the genetic code. Which statements describe the possible chains?

I. Two chains can contain the same amino-acid types in different orders.
II. Each chain must contain all 20 amino-acid types.
III. Chains can range in length from a few amino acids to thousands.

Show answer and explanation

Answer: C. The 20 amino-acid types are possible building blocks, but their order and the length of a polypeptide chain can vary. Thus I and III are correct; a chain need not contain every type.

A: accepts variation in order but incorrectly assumes chain length cannot vary.
B: accepts variation in length but assumes amino-acid order is fixed.
D: mistakes the 20 amino acids available for a set that every chain must contain.

Remember: a polypeptide can use a subset of the 20 amino-acid types in different orders and chain lengths.

Question 4

Paper 2Easy2 marks

A researcher compares two proteins that contain the same types of amino acids but in a different order.

(a)

Explain why the proteins may have different three-dimensional conformations.

[1]

(b)

State the role of hydrogen bonds in stabilizing alpha helices and beta-pleated sheets.

[1]

Question 5

Multiple choiceMedium1 mark

A researcher measures cell adhesion and cell recognition before and after selectively removing carbohydrate chains from glycoproteins and glycolipids on the extracellular surface of the plasma membrane. The results are shown.

Cell condition	Cells adhering to other cells (%)	Cells correctly recognized (%)
Untreated	82	91
Carbohydrate chains removed	46	52

What conclusion is supported by the results?

Show answer and explanation

Answer: A. Removing the extracellular carbohydrate chains lowered adhesion from 82% to 46% and correct recognition from 91% to 52%. The paired decreases support a role for these structures in both processes.
B: accounts for the decrease in recognition but overlooks the decrease in adhesion.
C: accounts for the decrease in adhesion but overlooks the decrease in recognition.
D: reverses the relationship: removal decreased both responses, rather than increasing them.
Remember: membrane carbohydrate chains can contribute to cell adhesion and cell recognition.

Question 6

Paper 2Medium16 marks

A researcher investigated the movement of different solutes across a model phospholipid bilayer. The same apparatus was used for separate trials, one solute per trial. Figure 1 shows the initial conditions for each trial. After 30.0 minutes, the amount of solute detected in chamber B was measured.

Solute tested	Description	Amount detected in chamber B after 30.0 min / nmol
Steroid	Non-polar molecule	15.2
Glucose	Polar molecule	1.6
Sodium ions	Ions	0.8
Polymer	Large, hydrophilic molecule	0.2

(a)

State the independent variable, the dependent variable and one variable controlled in the investigation.

[3]

(b)

Calculate the percentage of the initial amount of steroid detected in chamber B after 30.0 minutes.

[2]

(c)

Compare the amounts detected in chamber B for the four solutes.

[3]

(d)

Explain how the structure of the phospholipid bilayer accounts for the results for the steroid and the other three solutes.

[4]

(e)

A student concludes that the bilayer is an effective barrier to every solute tested. Evaluate this conclusion using the data.

[4]

Question 7

Multiple choiceEasy1 mark

A phagocytic cell encloses an engulfed particle and digestive enzymes inside a membrane-bound vacuole. Which effect of this arrangement helps protect the cell’s cytoplasm?

Show answer and explanation

Answer: D. The vacuole confines digestive enzymes with the engulfed particle, separating them from cytoplasmic molecules and reducing unwanted digestion.

A: confuses the vacuole’s role with the increased inner-membrane surface area of mitochondria. B: reverses compartmentalization by spreading the enzymes through the cytoplasm. C: describes a mitochondrial adaptation, not the protective effect of a phagocytic vacuole.

Remember: compartments can keep biochemical processes apart or concentrate their components.

Question 8

Paper 2Easy5 marks

Figure 1 shows a schematic mitochondrion and the fusion of a lysosome with a phagocytic vacuole.

(a)

Identify the mitochondrial adaptations indicated by X and Y.

[2]

(b)

In which mitochondrial compartment are enzymes and substrates of the Krebs cycle concentrated?

[1]

(c)

Explain one advantage of the compartmentalization shown in panel B for the cell.

[2]

Question 9

Multiple choiceEasy1 mark

A researcher is studying a stem-cell niche in an adult human hair follicle. Which statement describes a function of this niche?

Show answer and explanation

Answer: A. A stem-cell niche can maintain stem cells or promote their proliferation and differentiation; hair follicles are an example of a niche in adult humans.
B: assigns the stem cells’ capacity for continued division and differentiation to the niche.
C: recognizes the niche as a location but wrongly denies its role in maintaining or influencing stem cells.
D: confuses the niche with stem cells, which can differentiate along different pathways.
Remember: distinguish a stem cell’s properties from the function of its niche.

Question 10

Paper 2Easy5 marks

Cubes of different side lengths are used as simplified models of cells. Figure 1 shows the surface-area-to-volume ratio of the cubes.

(a)

Read the surface-area-to-volume ratio for a cube with a side length of 2 cm. Give the unit.

[1]

(b)

Determine the side length of a cube with a surface-area-to-volume ratio of 1.5 cm⁻¹. Give the unit.

[1]

(c)

Explain how the pattern in Figure 1 suggests a constraint on the size of a cell.

[3]

Question 11

Multiple choiceEasy1 mark

A student records a spirometer trace. Which statements correctly identify lung volumes from the trace?

I. Tidal volume is 0.5 dm³. II. Inspiratory reserve volume is 2.0 dm³. III. Vital capacity is 4.0 dm³.

Show answer and explanation

Answer: D. Tidal volume is the change between ordinary expiration and inspiration; inspiratory reserve volume is measured from ordinary inspiration to maximum inspiration; vital capacity is the difference between maximum inspiration and maximum expiration.

Ordinary breaths range from 2.5 to 3.0 dm³, so tidal volume is 0.5 dm³.
Inspiratory reserve volume is 5.0 − 3.0 = 2.0 dm³.
Vital capacity is 5.0 − 1.0 = 4.0 dm³.

A: Confusing inspiratory reserve volume with vital capacity makes the 2.0 dm³ reserve seem to be the capacity. B: Measuring inspiratory reserve from ordinary expiration gives 5.0 − 2.5 = 2.5 dm³. C: Reading the ordinary inspiration peak as tidal volume gives 3.0 dm³ instead of subtracting the expiration level.

Remember: Reserve volumes are measured from the appropriate level of ordinary breathing to a forced maximum.

Question 12

Paper 2Easy6 marks

A student prepares a transverse section of a dicotyledonous leaf to examine the arrangement of tissues involved in gas exchange.

(a)

Draw a plan diagram of the section. Label the waxy cuticle, upper and lower epidermis, palisade mesophyll, spongy mesophyll, air spaces and a vein. Do not draw individual cells.

[4]

(b)

State one role in gas exchange of the air spaces and one role of the guard cells of a stoma.

[2]

Question 13

Multiple choiceMedium1 mark

Wall thickness and lumen diameter were measured from four vessel micrographs. Which specimen is most consistent with an artery?

Show answer and explanation

Answer: D. Arteries have a thicker wall relative to lumen diameter than veins. The wall-thickness-to-lumen-diameter ratios are P: 150/1000 = 0.15, Q: 30/300 = 0.10, R: 40/800 = 0.05 and S: 80/400 = 0.20. S has the greatest relative wall thickness.

A: chooses P by comparing absolute wall thickness alone; its wall is 150 µm, but its ratio is 0.15.
B: chooses Q by using the smallest lumen diameter alone rather than comparing wall thickness relative to diameter.
C: chooses R by reversing the criterion and selecting the smallest wall-to-lumen ratio.
Remember: identify an artery by its relatively thick wall, not by absolute wall thickness alone.

Question 14

Paper 2Hard16 marks

Histology technicians are testing whether vessel type can be identified from transverse micrographs. Specimen A has been independently identified as an artery and specimen B as a vein. Wall thickness and lumen diameter were measured in three micrographs of each specimen.

Specimen	Wall thickness / mm (micrographs 1, 2, 3)	Lumen diameter / mm (micrographs 1, 2, 3)
A	0.48, 0.50, 0.52	3.0, 3.1, 3.2
B	0.38, 0.40, 0.42	6.0, 6.2, 6.4

The relative wall thickness index is calculated as:

\frac{\text{mean wall thickness}}{\text{mean lumen diameter}}\times 100\%

(a)

Calculate the relative wall thickness index for specimens A and B. Give each answer to one decimal place and show your working.

[4]

(b)

Describe the structural difference between specimen A and specimen B indicated by your results.

[3]

(c)

A further unknown specimen, C, is examined in three transverse micrographs. Its wall thickness measurements are 0.58, 0.60 and 0.62 mm; its lumen diameters are 7.8, 8.0 and 8.2 mm.

Calculate the relative wall thickness index for specimen C, giving your answer to one decimal place.

[2]

(d)

Identify specimen C as an artery or a vein. Explain why comparing only its absolute wall thickness with that of specimen A could lead to a different conclusion.

[3]

(e)

A colleague proposes using a universal cutoff of 10.0%: any vessel with an index above 10.0% is an artery and any vessel with an index below 10.0% is a vein. Evaluate this proposal using the results for A, B and C, and suggest one way to test the cutoff before using it to identify further micrographs.

[4]

Question 15

Multiple choiceMedium1 mark

An arthropod and a vertebrate each move an appendage when a skeletal muscle contracts. Which comparison correctly describes how their skeletons contribute?

Show answer and explanation

Answer: C. Arthropods have exoskeletons, while vertebrates have endoskeletons. Arthropod muscles attach to the inner surface of the exoskeleton, whereas vertebrate muscles attach to bones. When muscles contract, the skeleton provides anchorage and acts as a lever to move the appendage.
A: reverses which group has an exoskeleton and which has an endoskeleton.
B: places the arthropod muscle attachment on the outer, rather than inner, surface of the exoskeleton.
D: reverses the roles: the skeleton acts as the lever, while muscles provide the force.
Remember: skeletons provide muscle anchorage and act as levers during movement.

Question 16

Paper 2Medium16 marks

A field team records three consecutive dive cycles of a bottlenose dolphin. Each cycle consists of one dive followed by a surface interval. No breaths are observed during dives; at least one breath is observed during each surface interval.

Dive cycle	Time underwater / min	Surface interval for breathing / min
1	4.0	0.5
2	5.0	0.5
3	3.0	0.5

(a)

Describe the pattern in the timing of the dives and surface intervals, using data from the table.

[2]

(b)

Calculate the percentage of the total observation time that the dolphin spends underwater.

[3]

(c)

Explain how the dolphin’s body shape, limbs and tail are adapted for swimming. Include the direction of tail movement.

[6]

(d)

Explain how airway adaptations allow the dolphin to breathe in the pattern recorded.

[5]

Question 17

Multiple choiceMedium1 mark

Two regions on separate continents have the climate patterns shown. What is likely about their biomes and communities?

Region	Temperature through the year	Rainfall through the year
P	Consistently high	Consistently high
Q	Consistently high	Consistently high

Show answer and explanation

Answer: A. Biome distribution is associated with patterns of temperature and rainfall. Consistently high values for both are characteristic of tropical forests; similar abiotic conditions can favor similar communities through convergent evolution.
B: selects hot desert from the high temperature but overlooks the consistently high rainfall.
C: classifies rainfall as biotic; it is an abiotic factor that helps determine biome distribution.
D: matches the high rainfall but overlooks the consistently high temperature.
Remember: terrestrial biome distribution reflects both temperature and precipitation.

Question 18

Paper 2Hard16 marks

At Kestrel Bay, marram grass (Ammophila arenaria) grows on exposed foredunes above the high-tide line, where the sand is dry and shifting and winds are strong. Grey mangrove trees (Avicennia marina) grow in the lower intertidal estuary, in a mangrove swamp with waterlogged sediment. Marram leaves roll inward during dry, windy periods; the stomata are on the inner surface of the rolled leaves. Grey mangroves have aerial roots that project above the sediment and have openings exposed to the air.

Table 1. Conditions measured in the two habitats

Habitat	Soil water content (%)	Pore-water salinity (g kg⁻¹)	Soil oxygen (%)
Foredune	6	2	18
Mangrove swamp	68	25	3

To estimate the number of grey mangrove trees in a 0.80 ha area of swamp, six non-overlapping 100 m² quadrats were selected at random. The numbers of trees counted were:

Quadrat	1	2	3	4	5	6
Trees counted	8	11	9	10	7	9

Seedlings of both species were also grown for eight weeks under two substrate conditions. The “dune-like” and “swamp-like” treatments reproduced the corresponding moisture, salinity and soil oxygen conditions in Table 1. Light, temperature and nutrient availability were kept constant. Each value is the mean dry-mass increase per seedling for 10 seedlings.

Table 2. Mean dry-mass increase after eight weeks

Species	Dune-like substrate (g)	Swamp-like substrate (g)
Marram grass	2.4	0.5
Grey mangrove	0.7	2.2

(a)

Describe the habitat of each species, including its physical location and ecosystem type.

[2]

(b)

Calculate the estimated number of grey mangrove trees in the 0.80 ha area.

[3]

(c)

Explain how the stated leaf feature of marram grass and root feature of grey mangrove help each species survive in its habitat.

[6]

(d)

An ecologist concludes that each species is adapted to the abiotic conditions of its own habitat and that matching the three soil conditions in Table 1 is sufficient to ensure successful long-term establishment after transplantation. Evaluate this conclusion using the data and the study context.

[5]

Question 19

Multiple choiceHard1 mark

A prokaryote was cultured with identical mineral nutrients and no organic compounds; CO₂ availability was controlled as shown. The table gives the net change in biomass after 24 h. Which conclusion is supported?

Light	CO₂ supplied	Net biomass change / mg in 24 h
Yes	Yes	+12
Yes	No	−2
No	Yes	−2
No	No	−2

Show answer and explanation

Answer: B. Photosynthetic nutrition uses light energy to make organic matter from inorganic carbon. Biomass increases only when both light and CO₂ are supplied (+12 mg in 24 h); each other condition gives a decrease. With no organic compounds supplied, this supports photosynthetic nutrition.

A: treats CO₂ as sufficient without light; the dark, CO₂-supplied culture loses biomass.
C: treats light as sufficient without CO₂; the illuminated culture without CO₂ loses biomass.
D: assumes either factor alone is sufficient; both single-factor conditions show biomass loss.
Remember: photosynthetic nutrition uses light energy and an inorganic carbon source.

Question 20

Paper 2Hard16 marks

A unicellular alga, A, and a photosynthetic prokaryote, P, were cultured separately or together. Each species present in a culture had an initial density of 100 cells mL⁻¹. Cultures were kept at the same temperature and pH, with inorganic carbon and mineral nutrients maintained at non-limiting concentrations. Light treatments received the same illumination for 12 hours each day; dark treatments received no light. The table shows mean cell densities after 10 days. Each mean is based on five replicate cultures.

Culture	A at start / cells mL⁻¹	A after 10 days / cells mL⁻¹	P at start / cells mL⁻¹	P after 10 days / cells mL⁻¹
A alone, light	100	800	—	—
A alone, dark	100	100	—	—
P alone, light	—	—	100	600
P alone, dark	—	—	100	100
A and P together, light	100	200	100	500
A and P together, dark	100	100	100	100

A dash indicates that the species was not present.

(a)

State the mode of nutrition shared by A and P, and explain how this mode supports population growth.

[3]

(b)

Calculate the percentage decrease in A’s mean density after 10 days when grown with P in light, compared with A grown alone in light.

[2]

(c)

Compare the changes in density of the two species when grown alone in light and in darkness. Use the data.

[3]

(d)

In the light treatments, the mean percentage of incident light reaching the base of the culture after 10 days was:

Culture	A alone	P alone	A and P together
Light reaching base / %	76	72	24

Evaluate the claim that P reduces the growth of A by competing with it for light. Use the population and light data, link the light data to photosynthetic nutrition, and identify and explain up to two limitations before reaching a conclusion.

[8]

IB Biology HL Theme B: Form and function questions

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