Cambridge International AS & A Level Biology 9700: Classification, biodiversity and conservation – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover topic 18, Classification, biodiversity and conservation, of the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027, sections 18.1 to 18.3. It is A Level content, examined on Paper 4, and the statistics also suit Paper 5 practice. Calculators are allowed throughout. The Lincoln index, Simpson’s index and correlation formulae are given in the exam, so they are given here too.

Revise first with the study guide and the revision notes. The 9700 Biology hub and the printable checklist cover the whole course.

Formulae: N = (n₁ × n₂) / m₂ ; D = 1 − Σ(n/N)² ; rₛ = 1 − (6 × ΣD²) / (n³ − n) ; r = (Σxy − n x̄ ȳ) / ((n − 1) sₓ sᵧ)

Questions

1.

(a) Define the term niche. [1] (b) State the biological species concept. [1] (c) Give one limitation of the biological species concept. [1]

2. Archaea and Bacteria are both placed in separate domains. State one feature they share and two ways in which they differ. [3]

3. An organism has a body made of branching threads. Its cell walls contain chitin. It secretes enzymes onto dead leaves and absorbs the products, and it reproduces by spores.

(a) State the kingdom to which it belongs. [1] (b) Give two features from the description that show it is not in the kingdom Plantae. [2]

4. Scientists analyse the nucleic acid of two viruses.

(a) Virus P contains RNA with base percentages A 24%, U 31%, G 20%, C 25%. State and explain whether its RNA is single-stranded or double-stranded. [2] (b) Virus Q contains DNA with A 28%, T 28%, G 22%, C 22%. Suggest how virus Q would be classified. [1]

5. To estimate the number of woodlice under logs in a garden, a student caught 36 woodlice, marked each with a dot of non-toxic paint, and released them. Two days later he caught 45 woodlice, 9 of which were marked.

(a) Calculate the estimated population. [2] (b) Some paint marks rubbed off, so only 6 marked woodlice were recorded instead of 9. State and explain the effect on the estimate. [2] (c) State two other assumptions of this method. [2]

6. Two ponds were sampled with random net sweeps. The numbers of five invertebrate species were recorded.

Species	Pond A	Pond B
1	15	40
2	15	4
3	10	3
4	6	2
5	4	1

(a) Simpson’s index for pond A is 0.759. Calculate Simpson’s index for pond B. Show your working. [3] (b) Compare the biodiversity of the two ponds and explain the difference in D. [2] (c) Explain why the sweeps were taken at random positions. [2]

7. Ten quadrats were placed at random in a meadow. Soil moisture and the number of plants of one species were recorded. The table is partly completed.

Quadrat	Moisture / %	Plants	Rank moisture	Rank plants	D²
1	68	31	10	10	0
2	62	27	8	9	1
3	65	24	9	8	1
4	51	22	7	7	0
5	47	15	6	5	1
6	40	17	4	6	4
7	43	12	5	4	1
8	33	9
9	29	11
10	22	5

(a) Complete the ranks and D² values for quadrats 8 to 10 and find ΣD². [2] (b) Calculate Spearman’s rank correlation coefficient. [2] (c) The critical value for 10 pairs at p = 0.05 is 0.648. State a null hypothesis and what you conclude. [3] (d) Suggest why Spearman’s rank was chosen instead of Pearson’s. [1]

8. For ten plots, soil nitrate concentration (x) and dry mass of nettle plants (y) were recorded. A partly completed Pearson’s calculation gives Σxy = 1499.4, x̄ = 23.7, ȳ = 5.77, sₓ = 7.917, sᵧ = 1.869.

(a) Calculate r. [2] (b) State what the value shows. [1] (c) State one condition for using Pearson’s linear correlation. [1] (d) Explain why these results do not prove that nitrate increases nettle growth. [1]

9. A ground-nesting bird lives only on one island. Rats have been introduced, hotels are being built on the coast, and rising temperatures are drying out the wetlands where it feeds.

(a) Explain two reasons for controlling the rats. [2] (b) Outline how a conserved area and a zoo could each help to protect this bird. [3] (c) Explain how climate change could lead to the extinction of this species. [2]

10. Only a few females of an endangered antelope remain. Describe how IVF, embryo transfer and surrogacy, together with a frozen zoo, could help to increase numbers and genetic diversity. [5]

11.

(a) Outline the role of the IUCN in conservation. [2] (b) Outline the role of CITES in conservation. [2] (c) Outline two reasons for maintaining biodiversity. [2]

12. Describe how you would use a belt transect to investigate how plant species change from a sunny path into a wood. [4]

Answers

1. (a) The role of an organism in its ecosystem, including where it lives, what it feeds on and how it interacts with other organisms and its environment. [1] (b) A group of organisms that can interbreed to produce fertile offspring and are reproductively isolated from other groups. [1] (c) It cannot be applied to organisms that reproduce asexually, or to fossils. [1] Examiner insight: “Can breed together” alone does not earn (b); “fertile offspring” is the marking point.

2. Similarity: both are prokaryotes / no nucleus. [1] Difference: Archaea membrane lipids have ether bonds and branched chains, Bacteria have ester bonds and unbranched fatty acids. [1] Difference: Bacteria cell walls contain peptidoglycan, Archaea walls do not; or their rRNA base sequences differ. [1] Examiner insight: Each difference must state both sides; “Archaea have different lipids” is too vague for a mark.

3. (a) Fungi. [1] (b) Cell walls of chitin, not cellulose. [1] Heterotrophic: it digests and absorbs food rather than photosynthesising. [1] Examiner insight: Features must come from the description; “no chlorophyll” is accepted only if linked to not photosynthesising.

4. (a) Single-stranded. [1] A does not equal U and G does not equal C, so the bases are not paired. [1] (b) Double-stranded DNA virus (A = T and G = C suggests complementary pairing). [1] Examiner insight: The explanation mark needs the pairing rule applied to the data, not just “the numbers are different”.

5. (a) N = (36 × 45) / 9 [1] = 180 [1] (b) The estimate rises to (36 × 45) / 6 = 270, an overestimate. [1] Fewer recaptures are recorded as marked, so the marked fraction looks smaller than it really is. [1] (c) Any two: no births, deaths or migration between samples [1]; marked animals mix randomly with the population; marking does not affect survival or chance of recapture. [1] Examiner insight: A bare “180” usually earns both marks, but showing the substitution protects the method mark if you slip.

6. (a) N = 50; (n/N)² values: 0.64, 0.0064, 0.0036, 0.0016, 0.0004 [1]; Σ = 0.652 [1]; D = 1 − 0.652 = 0.348 [1] (b) Pond A has higher biodiversity (0.759 > 0.348). [1] Both have five species, but in pond B one species dominates, while in pond A individuals are spread more evenly. [1] (c) To avoid bias in choosing where to sample. [1] So that the sample is representative of the whole pond. [1] Examiner insight: For (b), same species richness is the key point; “A has more species” is wrong and scores nothing.

7. (a) Ranks: quadrat 8 moisture 3, plants 2, D² = 1; quadrat 9 moisture 2, plants 3, D² = 1; quadrat 10 both 1, D² = 0. [1] ΣD² = 8 + 2 = 10 [1] (b) rₛ = 1 − (6 × 10) / (1000 − 10) [1] = 1 − 60/990 = 0.939 (3 s.f.) [1] (c) Null hypothesis: there is no correlation between soil moisture and number of plants. [1] 0.939 is greater than 0.648. [1] So reject the null hypothesis: there is a significant positive correlation. [1] (d) The plant counts may not be normally distributed. [1] Examiner insight: Allow ECF in (b) and (c) from a wrong ΣD², but the conclusion must compare with the critical value explicitly.

8. (a) r = (1499.4 − 10 × 23.7 × 5.77) / (9 × 7.917 × 1.869) [1] = 131.91 / 133.17 = 0.991 (3 s.f.) [1] (b) A strong positive linear correlation. [1] (c) Any one: continuous data; normally distributed; scatter graph suggests a linear relationship; at least five pairs. [1] (d) Correlation does not show causation; another factor, such as soil moisture, could affect both. [1] Examiner insight: Using (n − 1) = 9, not n = 10, in the denominator is where the method mark is usually lost.

9. (a) Rats eat eggs and chicks of ground-nesting birds. [1] Rats have no natural predators on the island, so their numbers grow unchecked; or they compete for food. [1] (b) Conserved area: protect the wetland and nesting sites, restricting building and disturbance. [1] Zoo: captive breeding to build numbers, with breeding records to avoid inbreeding. [1] Zoo: reintroduce birds once threats are removed; or research and education. [1] (c) Wetlands dry out, so food and habitat are lost. [1] The change is too fast for the species to adapt, and it cannot move off the island. [1] Examiner insight: “Rats kill the birds” gains less than a precise link to eggs and chicks on the ground.

10. Eggs are collected from females, usually after hormone treatment. [1] They are fertilised in a dish with sperm (IVF), which may come from a frozen zoo. [1] Embryos are transferred into the uterus of a female (embryo transfer). [1] A surrogate, possibly of a more common related species, carries the pregnancy, so the rare females can produce many more offspring. [1] Frozen sperm or embryos from other or past populations increase genetic diversity and reduce inbreeding. [1] Examiner insight: Five separate points are needed; describing IVF in detail without mentioning surrogacy or genetic diversity caps the mark.

11. (a) Assesses the conservation status of species. [1] Publishes the Red List to guide conservation priorities. [1] (b) An international agreement between governments. [1] It controls or bans trade in endangered species and their products. [1] (c) Any two: ethical responsibility; ecosystem services such as pollination; source of medicines or crop genes; economic value such as ecotourism. [1] [1] Examiner insight: Confusing IUCN and CITES loses both marks in each part; learn “IUCN assesses, CITES controls trade”.

12. Lay a tape from the path into the wood. [1] Place quadrats at regular intervals along the tape (or continuously). [1] Record the species and their abundance or percentage cover in each quadrat. [1] Measure a factor such as light intensity at each quadrat; repeat with further transects. [1] Examiner insight: A transect is systematic, so “random quadrats” here is a contradiction and costs the method mark.

Where marks are usually lost

• Species definitions without “fertile offspring”.
• Archaea–Bacteria differences given one-sided.
• Swapping n₂ and m₂ in the Lincoln index.
• Forgetting to square n/N, or leaving out “1 −”, in Simpson’s index.
• Explaining D without mentioning evenness when species number is the same.
• No null hypothesis, or no comparison with the critical value, in a correlation test.
• Using n instead of n − 1 in Pearson’s calculation.
• Claiming causation from a correlation.
• Mixing up IUCN and CITES, or zoos and frozen zoos.
• Applying assisted reproduction methods to non-mammals.

Next steps

• Revision notes for topic 18
• Study guide for topic 18
• 9700 Biology course hub
• Printable 9700 checklist
• 9700 A Level diagnostic
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus, for examination in 2025, 2026 and 2027 (version 1), published by Cambridge University Press & Assessment. Topic 18, sections 18.1 to 18.3, and the Mathematical requirements.