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.

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Question 1

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 2

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 3

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 4

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]

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