IB Biology HL Theme A: Unity and diversity 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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Multiple choice Written

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Question 1

Paper 2Medium16 marks

A student investigated the effect of water availability on seed germination. Five dishes were used for each treatment, with 20 seeds in each dish. Each dish contained filter paper and the same type of water. The dishes were kept at the same temperature and light conditions. After 72 hours, a seed was counted as germinated if its radicle had emerged. At 20 mL, the seeds were fully submerged; at the other volumes, the filter paper was moist but the seeds were not submerged.

Water added / mL	Number of seeds germinated out of 100
0	0
2	38
5	82
10	86
20	42

(a)

Calculate the difference in germination percentage points between the 10 mL and 20 mL treatments.

[2]

(b)

Describe the relationship between water added and germination.

[3]

(c)

Explain why germination was absent without added water but high in the 5 mL and 10 mL treatments.

[3]

(d)

Explain how water acts as a transport medium from roots to leaves in a seedling, including one adaptation of xylem for this role.

[3]

(e)

The syllabus statement is that water is the medium for life. Using the results and your biological knowledge, explain how water can be essential as a medium without increasing water volume always increasing germination. Refer to the environment in which the first cells originated and to a role of water in present-day cells.

[5]

Question 2

Multiple choiceMedium1 mark

A study finds the same genetic code and shared genes across bacteria, archaea and eukaryotes. The oldest known fossil evidence of cells is dated to 3.5 billion years ago, while a molecular-clock estimate places LUCA at 3.8 billion years ago. Which interpretation is supported?

Show answer and explanation

Answer: B. A universal genetic code and genes shared across the three domains support common ancestry and LUCA. A molecular clock estimates LUCA’s age, while fossils provide evidence that cells existed by the fossil date, not the exact beginning of life. Earlier lineages could have become extinct through competition with LUCA and its descendants. A: Treats the oldest known fossil as the exact origin of life; fossil evidence does not establish that life began then. C: Reverses the dating approaches: molecular clocks estimate LUCA’s age, while fossils provide evidence of ancient cells. D: Equates LUCA with the first life and treats its estimated age as excluding earlier forms. Remember: LUCA is the latest common ancestor of extant life, not necessarily the first life.

Question 3

Multiple choiceMedium1 mark

A filamentous eukaryote has a cell wall containing chitin. Microscopy of one tube shows many nuclei in a continuous length of cytoplasm, with no cross-walls. Which interpretation accounts for both observations?

Show answer and explanation

Answer: B. Chitin is characteristic of fungal cell walls. The absence of cross-walls and presence of many nuclei in continuous cytoplasm identify an aseptate fungal hypha.

A: associates a cell wall with plants but overlooks that plant walls contain cellulose, not chitin.
C: matches the multiple nuclei but overlooks that skeletal muscle cells have no cell wall.
D: identifies a fungus from the chitin wall but contradicts the observation that there are no cross-walls.
Remember: aseptate fungal hyphae contain multiple nuclei in continuous cytoplasm.

Question 4

Paper 2Medium16 marks

A student makes a temporary mount of leaf epidermal cells and examines it using a light microscope. At the objective used, a length of 100 μm on a stage micrometer spans 40 eyepiece graticule divisions. One cell has a width of 24 eyepiece graticule divisions. A printed image of the same cell has a width of 6.0 mm. Assume that the printed image has not been resized.

(a)

Calculate the actual length represented by one eyepiece graticule division.

[2]

(b)

Calculate the actual width of the cell and the magnification of the printed image. State the length on the printed image of a scale bar representing 10 μm.

[5]

(c)

A fluorescent stain is used to make cell boundaries visible. To locate protein R, a fluorescently labelled antibody against R is used. Describe how immunofluorescence reveals the location of R and distinguish it from the use of the fluorescent stain.

[3]

(d)

In a further test, 100 cells were scored according to the location of the strongest detectable fluorescence signal.

Preparation	Plasma membrane (%)	Cytoplasm (%)	No detectable signal (%)
Fluorescently labelled antibody against R	70	25	5
Fluorescently labelled non-specific antibody	6	7	87

Assess the claim that protein R is located exclusively at the plasma membrane, using the data.

[3]

(e)

Another team wants to investigate the internal surfaces of cell membranes and the detailed structure of a protein complex while preserving it close to its native state. Outline how electron microscopy and its developments can be used to address these aims.

[3]

Question 5

Multiple choiceMedium1 mark

Bacteriophage lambda is undergoing its lytic cycle in bacterial cells. The table shows mean measurements at successive times after infection, all normalized per initially infected bacterium.

Time after infection / min	Phage DNA copies per initially infected bacterium	Assembled phages inside bacteria	Free infectious phages
0	1	0	0
10	28	0	0
20	28	20	0
30	28	0	18

Which interval is most consistent with release of assembled phages from the bacteria?

Show answer and explanation

Answer: B. From 20 to 30 min, assembled phages inside bacteria fall from 20 to 0 while free infectious phages rise from 0 to 18. This indicates release; the phage DNA copy number remains unchanged during this interval.

A: selects the earlier increase in phage DNA, which indicates replication rather than release.
C: identifies the interval when assembled particles appear inside bacteria, indicating assembly before release.
D: assigns replication to an interval when the phage DNA copy number is unchanged.

Remember: distinguish replication, assembly and release by tracking what changes inside and outside the host cell.

Question 6

Paper 2Medium4 marks

A culture of Escherichia coli containing bacteriophage lambda was monitored. At 40 min, the culture conditions were changed. Figure 1 shows the measured concentrations of intact bacterial cells and free lambda particles.

(a)

Using Figure 1, identify the viral cycle most consistent with each interval, 0–40 min and 40–80 min. Support each choice with one feature of the graph.

[2]

(b)

Explain how lambda DNA is maintained during the first interval, and state two ways in which the bacterial host supports phage production during the second interval.

[2]

Question 7

Multiple choiceMedium1 mark

The key is used for three local pond species. A field team identifies an observed animal and compares a barcode from pond-water environmental DNA (eDNA) with reference barcodes. The animal has no shell and six legs; the sample barcode is CTAG. Which pair gives the animal’s key identification and the species detected in the water sample, respectively?

Dichotomous key

Feature	Identification or route
Shell present	Pond snail
Shell absent	Go to 2
Six legs	Water beetle
More than six legs	Freshwater shrimp

Reference barcodes

Species	Barcode
Pond snail	GACT
Water beetle	TGCA
Freshwater shrimp	CTAG

Show answer and explanation

Answer: A. The key routes an animal with no shell to the next distinction, where six legs identifies a water beetle. The eDNA barcode CTAG matches the freshwater shrimp reference barcode.

B: identifies the observed animal correctly but assigns the sample barcode to the pond snail, whose barcode is GACT.
C: matches the sample barcode correctly but selects the shell-present route for the observed animal.
D: swaps the key identification and the species identified from the eDNA barcode.
Remember: a dichotomous key uses observable features; an eDNA barcode is compared with reference barcodes.

Question 8

Multiple choiceMedium1 mark

A conservation team compares four candidate species. The tree shows their relationships and relative branch lengths; the status labels show whether each species is globally endangered. Which candidate best fits the EDGE programme’s conservation-prioritization rationale?

Show answer and explanation

Answer: B. EDGE prioritization considers both evolutionary distinctness and global endangerment. The tree shows that Species L has a longer branch not shared with close relatives than Species M: 5 units compared with 1 unit. Both are globally endangered, whereas Species K’s longer independent branch is labelled not globally endangered. Species L therefore fits both considerations.

A: selects the longest independent lineage but overlooks that Species K is not globally endangered.
C: uses the global-endangerment label but overlooks that Species M shares most of its recent lineage with Species N.
D: could result from misreading Species N’s status; it is not globally endangered and has only a short unshared branch.

Remember: EDGE combines evolutionary distinctness with global endangerment.

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