IB Biology HL 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 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)XOOHOHOHOHCH2OH(b)Glycogen fragmentYOOHOHOHCH2OHα-glucoseα-glucoseα-glucoseα-glucoseα-glucoseα-glucoseα-glucoseα-glucoseα-glucose
(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]

Question 2

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 3

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 4

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 5

Paper 2Easy5 marks

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

AXmatrixYBlysosomephagocytic vacuolefusiondigestive enzymesengulfed material
(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 6

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 7

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.

012345600.511.522.533.544.555.566.5Side length of cube / cmSurface-area-to-volume ratio / cm⁻¹
(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 8

Multiple choiceEasy1 mark

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

012345678900.511.522.533.544.555.56Time / sLung volume / dm³

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 9

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 10

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.

SpecimenWall thickness / mm (micrographs 1, 2, 3)Lumen diameter / mm (micrographs 1, 2, 3)
A0.48, 0.50, 0.523.0, 3.1, 3.2
B0.38, 0.40, 0.426.0, 6.2, 6.4

The relative wall thickness index is calculated as:

mean wall thicknessmean lumen diameter×100%\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 11

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

HabitatSoil water content (%)Pore-water salinity (g kg⁻¹)Soil oxygen (%)
Foredune6218
Mangrove swamp68253

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:

Quadrat123456
Trees counted81191079

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

SpeciesDune-like substrate (g)Swamp-like substrate (g)
Marram grass2.40.5
Grey mangrove0.72.2
(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 12

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?

LightCO₂ suppliedNet biomass change / mg in 24 h
YesYes+12
YesNo−2
NoYes−2
NoNo−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 13

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.

CultureA at start / cells mL⁻¹A after 10 days / cells mL⁻¹P at start / cells mL⁻¹P after 10 days / cells mL⁻¹
A alone, light100800——
A alone, dark100100——
P alone, light——100600
P alone, dark——100100
A and P together, light100200100500
A and P together, dark100100100100

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:

CultureA aloneP aloneA and P together
Light reaching base / %767224

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]

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