Pearson Edexcel International A Level Biology: Classification, Biodiversity and Conservation (YBI11) – 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 statements 4.14 to 4.21 of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Biology specification (YBI11), Issue 2. They are Topic 4, International AS (IAS) content. All data are supplied in the questions.

Related: study guide, Edexcel A Level Biology hub and free diagnostics.

Questions

1. State what is meant by the species concept, and give one reason why it cannot be applied to every organism. [2]

2. A research group reports that single-celled organisms from a hot spring have ribosomal RNA sequences unlike those of any known group, and proposes a new classification.

(a) Name the three domains in the current three-domain system. [1] (b) Explain why the new grouping is not accepted immediately, and how it will be evaluated. [3]

3. The same protein of 104 amino acids was sequenced in four species, W, X, Y and Z. The numbers of amino acid differences between pairs are shown.

	W	X	Y
X	4
Y	13	11
Z	9	8	12

(a) Calculate the percentage difference between W and Y. [2] (b) Identify the two most closely related species and justify your answer. [2] (c) Explain why sequence data are used in classification alongside phenotype, and give one limitation of using a single protein. [2]

4. (a) State what is meant by endemism. [1] (b) A bird is endemic to one small island. Explain why it is at greater risk of extinction from an introduced predator than a bird found across a continent. [2] (c) Name the two levels of biodiversity other than species diversity. [1]

5. Genetic diversity at one gene was measured in three populations of one mammal. In the island population, 27 of 90 animals were heterozygous. In a zoo group, 12 of 60 were heterozygous. In a reintroduced group, 6 of 40 were heterozygous. The island value is 0.30.

(a) Calculate the heterozygosity index for the zoo group and for the reintroduced group. [2] (b) Compare the three values and suggest an explanation for the trend. [2] (c) Suggest one management step that could increase the index in the reintroduced group. [1]

6. Two grassland sites were sampled with the same effort. Site A contained 18, 14, 9, 6 and 3 individuals of five plant species. Site B contained four species with 30, 12, 6 and 2 individuals, and its index of diversity was 2.37.

(a) Calculate D for site A using D = N(N − 1) ÷ Σn(n − 1). [3] (b) Compare the diversity of the two sites. [2] (c) Suggest two limitations of using these values for each whole site. [2]

7. Field X holds six species and 100 individuals, of which 94 belong to one species and the other five species have 2, 1, 1, 1 and 1 individuals. Field Y holds four species with 25 individuals each.

(a) State what is meant by species richness. [1] (b) Explain why species richness alone is a poor way to compare these fields, supporting your answer with calculations of D. [3]

8. A small desert mammal shelters in a burrow by day and feeds at night. It has large ears with many blood vessels, and it produces very concentrated urine.

(a) Classify each of the three features as behavioural, anatomical or physiological. [3] (b) Explain how the large ears help the mammal survive. [2] (c) Explain what is meant by the niche of this mammal. [2]

9. A recessive condition affects 1 in 1600 people in a large, randomly mating population.

(a) Calculate the frequency of the recessive allele, q. [2] (b) Calculate the frequency of carriers. [2] (c) Calculate the expected number of carriers in a population of 80 000. [1] (d) State one assumption you made. [1]

10. A population of 300 plants has alleles A (dominant) and a (recessive) at one gene. Genotypes were counted in two generations.

Generation	AA	Aa	aa
1	147	126	27
6	66	138	96

(a) Calculate the frequency of allele a in generation 1 and in generation 6. [3] (b) If the allele frequency had stayed as in generation 1, calculate the expected number of aa plants in generation 6 and compare it with the observed number. [2] (c) Explain how mutation and natural selection could each account for the change. [3]

11. A new lake splits one population of ground beetles into two. Explain how reproductive isolation could lead to the two populations becoming separate species. [4]

12. Evaluate the use of zoos and seed banks in conserving endangered species and their genetic diversity. [8]

Answers

1. Members of a species breed together and produce fertile offspring [1]. It cannot be applied to organisms that reproduce only asexually, or to fossils, because breeding cannot be tested [1]. [2] Examiner insight: Include “fertile”, and give a reason with the limitation, not just the word “difficult”.

2. (a) Archaea, Bacteria and Eukarya [1]. [1] (b) The claim rests on limited data and could contain errors [1]. Other scientists scrutinise the methods and evidence, often through peer review [1]. They repeat the work, and the grouping is accepted only if independent results support it [1]. [3] Examiner insight: A three-mark “explain” needs separate points: why caution is needed, what checking happens, and what acceptance depends on.

3. (a) 13 ÷ 104 × 100 [1] = 12.5% [1]. [2] (b) W and X, with only 4 differences (3.8%) [1], the smallest number, so they share the most recent common ancestor [1]. [2] (c) Sequences show genotype differences not visible in appearance, so they give a more objective measure of relatedness [1]. One protein is a small part of the genome, so it may not reflect whole-genome relationships [1]. [2] Examiner insight: Show the division as well as the answer, and in part (b) quote the figure and say what it means for common ancestry.

4. (a) A species found naturally in one geographical area and nowhere else [1]. [1] (b) The whole population lives in a small area, so one predator can reach all the birds [1]. It has no other population to recover from, and may have no defences against that predator [1]. [2] (c) Genetic diversity and habitat (ecosystem) diversity [1]. [1] Examiner insight: In part (b), compare the island bird with the continental one directly.

5. (a) Zoo: 12 ÷ 60 = 0.20 [1]. Reintroduced: 6 ÷ 40 = 0.15 [1]. [2] (b) The index falls from 0.30 (island) to 0.20 (zoo) to 0.15 (reintroduced) [1]. Each group was founded from a few individuals and may have been inbred, so alleles were lost by genetic drift [1]. [2] (c) Introduce unrelated individuals from the wild population or another group, or plan matings between less related animals [1]. [1] Examiner insight: Quote the values when describing a trend, and make suggestions say what adds new alleles.

6. (a) N = 50, so N(N − 1) = 50 × 49 = 2450 [1]. Σn(n − 1) = 306 + 182 + 72 + 30 + 6 = 596 [1]. D = 2450 ÷ 596 = 4.11 [1]. [3] (b) Site A has the higher D (4.11 against 2.37), so it is more diverse [1]. It has more species and a more even spread, whereas site B is dominated by one species [1]. [2] (c) Each is a sample only, so it may not represent the whole site [1]. D counts individuals but ignores their size or biomass [1]. [2] Examiner insight: Show N(N − 1) and Σn(n − 1) separately, and check that your conclusion points the right way: larger D means more diverse.

7. (a) The number of different species in a habitat [1]. [1] (b) Field X has more species (six against four) [1]. D for X = 9900 ÷ 8744 = 1.13 and D for Y = 9900 ÷ 2400 = 4.13 [1]. Y is far more diverse because its individuals are evenly spread; X is dominated by one species, which richness hides [1]. [3] Examiner insight: Pair each claim with a figure, such as 94 of 100 individuals in one species.

8. (a) Burrowing by day and feeding at night: behavioural [1]. Large ears with many blood vessels: anatomical [1]. Concentrated urine: physiological [1]. [3] (b) Heat is lost from the blood through the large ear surface [1], keeping the body cool and reducing water loss by sweating [1]. [2] (c) The niche is the role of the species in its ecosystem: where it lives and when it is active, such as burrows by day and feeding at night [1]. It also includes its food, interactions with other species and the conditions it tolerates [1]. [2] Examiner insight: State the benefit of an adaptation; “helps it adapt to the desert” only repeats the stem.

9. (a) q² = 1 ÷ 1600 = 0.000625 [1], so q = 0.025 [1]. [2] (b) p = 1 − 0.025 = 0.975 [1]. Carriers = 2pq = 2 × 0.975 × 0.025 = 0.049 (0.04875), about 1 in 20 [1]. [2] (c) 0.04875 × 80 000 = 3900 [1]. [1] (d) Any one of: random mating, no selection, no mutation, no migration, a large population [1]. [1] Examiner insight: 1 in 1600 is q², not q. Writing p + q = 1 and 2pq explicitly makes your method clear.

10. (a) Generation 1: a alleles = 2 × 27 + 126 = 180 of 600, so q = 0.30 [1]. Generation 6: a alleles = 2 × 96 + 138 = 330 [1], so q = 330 ÷ 600 = 0.55 [1]. [3] (b) Expected aa = q² × 300 = 0.09 × 300 = 27 [1]. Observed is 96, over three times the expected number, so the allele frequency has changed [1]. [2] (c) Mutation can produce new copies of allele a, though usually slowly [1]. If aa plants survive and reproduce better, they pass on more copies of a, so its frequency rises by natural selection [1]. Selection is the likelier cause of such a large change in six generations [1]. [3] Examiner insight: Count alleles, not individuals: divide by 600 here, not 300. Compare observed with expected using the numbers.

11. The lake prevents the beetles moving between the two populations, so there is no gene flow [1]. Different mutations arise in each population [1]. The environments differ, so different alleles are selected and allele frequencies diverge [1]. Eventually they can no longer breed together and have fertile offspring: separate species [1]. [4] Examiner insight: Give the mechanism, such as different selection pressures, rather than “they evolve differently”.

12. Zoos: captive breeding protects animals from wild threats [1]; planned matings can keep heterozygosity high [1], but small captive populations still lose diversity through drift and inbreeding [1]. Reintroduction can restore wild populations if the cause of decline is removed and habitat exists [1]. Research and education raise knowledge and awareness [1]. Seed banks store many species and much genetic diversity cheaply in a small space [1]; however, some seeds do not survive drying and freezing, stored seeds must be tested and regrown, and the habitat is not conserved [1]. Conclusion: the methods suit different species and supplement, rather than replace, habitat protection [1]. [8] Examiner insight: An evaluation needs a judgement based on your points, not just a list.

Where marks are usually lost

• Omitting “fertile” from the species definition.
• Describing a trend without quoting values.
• Dividing by individuals, not alleles, for allele frequency.
• Using the recessive phenotype frequency as q rather than q².
• Reading a higher D as lower diversity.
• Naming an adaptation without its benefit.
• Giving only advantages of zoos or seed banks, or saying seed banks protect the habitat.

Next steps

• Read the revision notes and study guide.
• Tick off statements on the printable checklist and visit the course hub.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Biology (YBI11), Specification, Issue 2, February 2021, Pearson Education Limited: Topic 4, statements 4.14 to 4.21.