Pearson Edexcel International A Level Biology: Classification, Biodiversity and Conservation (YBI11) – Revision Notes

Quick recall notes for statements 4.14 to 4.21 of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Biology specification (YBI11), Issue 2. These statements are in Topic 4, part of the International AS (IAS) content. The study guide explains everything in detail; afterwards try the practice questions. Course pages: Edexcel A Level Biology hub, printable checklist and free diagnostics.

Definitions to know word for word

Term	Definition
Classification	Organising the variety of life into groups using relationships shown by phenotype and genotype similarities and differences (4.14)
Species (working definition)	A group whose members breed together and produce fertile offspring
Three domains	Archaea, Bacteria, Eukarya
Biodiversity	The variety of life in an area: species, genetic and habitat diversity
Species richness	The number of different species in a habitat
Endemism	A species found naturally only in one geographical area
Niche	How a species fits into its ecosystem: habitat, food, relationships with other species and active period
Allele frequency	How common an allele is among all the alleles of that gene in a population

4.14 Classification in brief

• Evidence: phenotype (structure, behaviour) and genotype (DNA, mRNA and protein sequences).
• Species concept limits: no test on fossils, not applicable to asexual organisms, some hybrids.
• New groupings are proposed, then critically evaluated: published, peer reviewed, repeated by others, and accepted only if independent evidence supports them.
• Ribosomal RNA sequence comparison led to three domains: Archaea, Bacteria, Eukarya.
• Fewer sequence differences means a more recent common ancestor.

Method: percentage difference

1. Count the differences between two sequences.
2. Divide by the number of bases (or amino acids) compared.
3. Multiply by 100.
4. The smaller the percentage, the closer the relationship.

Reminder: 8 differences in 250 bases is 8 ÷ 250 × 100 = 3.2%.

4.15-4.16 Variety of life, biodiversity, endemism

• The variety of life is extensive but is being threatened by human activity (4.15).
• Threats to quote: habitat loss, overexploitation, pollution, invasive species, climate change.
• Three levels of biodiversity: species, genetic, habitat.
• Endemic species are confined to one area, so one disturbance can cause extinction.

4.17 and 4.18 Formulae

Measure	Formula	Notes
Species richness	count of species	Ignores numbers of individuals
Heterozygosity index	number of heterozygotes ÷ number of individuals in the population	Between 0 and 1; higher means more genetic diversity
Index of diversity	D = N(N − 1) ÷ Σn(n − 1)	N = all individuals; n = individuals of one species; higher D means greater diversity

Method: calculating D

1. List each species with its count n.
2. Work out n(n − 1) for each.
3. Add them: Σn(n − 1).
4. Add all counts for N, then work out N(N − 1).
5. Divide N(N − 1) by Σn(n − 1).

Worked reminder: counts 6, 3 and 1 give N = 10 and N(N − 1) = 90. Σn(n − 1) = 30 + 6 + 0 = 36. D = 90 ÷ 36 = 2.5.

Worked reminder: heterozygosity index

A captive group of 25 animals contains 4 heterozygotes at the gene tested. Index = 4 ÷ 25 = 0.16. A wild group with an index of 0.40 has more genetic diversity. A low captive value suggests few founders, inbreeding or drift, and a risk that the group cannot adapt.

Facts about D

• One species only: D = 1 (the minimum).
• A habitat dominated by one species has a low D even if richness is high.
• Compare habitats only with comparable, random samples of similar effort.
• D does not take body size or biomass into account.

4.19 Niche and adaptations

Type	Test question	Example
Behavioural	Is it something the organism does?	Nocturnal feeding; burrowing; migration; hibernation
Anatomical	Is it a structure you can see?	Rolled leaves with sunken stomata; thick fur; thick blubber
Physiological	Is it an internal process?	Concentrated urine; antifreeze glycoproteins in blood; enzymes working at low temperatures

• Always add the benefit: “reduces water loss by trapping humid air round the stomata”.
• Species with the same niche compete, and one tends to be excluded.

4.20 Hardy-Weinberg

Item	Expression
Alleles	p + q = 1
Genotypes	p² + 2pq + q² = 1
Homozygous dominant	p²
Heterozygous	2pq
Homozygous recessive	q²

Conditions for the equation to hold: no selection, no mutation, no migration, random mating, large population.

Method: from the recessive phenotype

1. Convert the recessive phenotype frequency to a decimal. This is q².
2. Take the square root for q.
3. p = 1 − q.
4. Work out 2pq for carriers and p² for homozygous dominant.
5. Multiply by the population size for numbers.

Worked reminder: 4 in 10 000 have a recessive condition. q² = 0.0004, q = 0.02, p = 0.98, carriers = 2 × 0.98 × 0.02 = 0.0392, about 1 in 26.

Method: has the allele frequency changed?

1. Count alleles: A = 2 × (AA) + (Aa); a = 2 × (aa) + (Aa).
2. Divide each by twice the number of individuals.
3. Compare generations. A difference means a change is occurring.

Worked reminder: a population of 200 has 98 AA, 84 Aa and 18 aa. A alleles = 2 × 98 + 84 = 280 of 400, so p = 0.70 and q = 0.30. If a later generation of 200 has q = 0.40, the allele frequency has changed.

Causes and speciation

• Mutation introduces new alleles, slowly.
• Natural selection raises the frequency of alleles that improve survival and reproduction.
• Reproductive isolation stops gene flow. The populations accumulate different mutations and face different selection pressures. Eventually they cannot breed together and have fertile offspring: new species.

4.21 Zoos and seed banks

Method	Best point	Main weakness
Captive breeding	Safe from wild threats; planned matings protect heterozygosity	Small populations; loss of genetic diversity; loss of natural behaviour
Reintroduction	Rebuilds wild populations	Needs habitat and threat removed; survival may be poor
Research	Knowledge of biology and breeding	May not apply to wild animals
Education	Awareness and funding	Hard to measure effect
Seed banks	Many species and alleles cheaply stored	Some seeds do not tolerate drying and freezing; regular testing needed; habitat not conserved

Evaluation checklist

For a 4.21 question, work through these prompts in order.

1. Name the method and say what it achieves (for example, keeping heterozygosity high by planned breeding).
2. Give a strength linked to genetic diversity or population size.
3. Give a limitation, such as small founder numbers, cost, or the need for habitat.
4. Judge: which method suits which species, or how the methods work together.

Useful examples to quote

• Endemic: lemurs in Madagascar; kakapo in New Zealand.
• Reintroduction: Arabian oryx returned to the wild in Oman after captive breeding.
• Seed bank: the Millennium Seed Bank at Kew stores dried, frozen seeds.
• Isolation: populations on separate islands, or on either side of a new barrier, adapting separately.
• Niche conflict: two similar species in one habitat tend to differ in food, feeding place or time of activity.

Must-know distinctions

• Species richness (a count) versus index of diversity (count and evenness).
• Heterozygosity index (individuals) versus allele frequency (alleles).
• q (allele) versus q² (genotype).
• Behavioural versus physiological adaptation: acts versus internal process.
• Captive breeding (animals in a zoo) versus reintroduction (return to the wild).

Quick self-test

1. Name the three domains.
2. Why can the species concept not be applied to all organisms?
3. Why must a new taxonomic grouping be critically evaluated by the scientific community?
4. Define endemism.
5. A population of 40 has 14 heterozygotes at a locus. Calculate the heterozygosity index.
6. A sample has species counts 5, 3 and 2. Calculate D.
7. State the five conditions needed for the Hardy-Weinberg equation to hold.
8. If q = 0.2, what is the frequency of heterozygotes?
9. Classify as behavioural, anatomical or physiological: a fish producing antifreeze proteins.
10. Give one limitation of a seed bank.

Answers

1. Archaea, Bacteria, Eukarya.
2. It cannot be tested on fossils or applied to asexual organisms, and some different species hybridise.
3. Independent checking and repetition show the evidence supports the conclusion before the classification changes.
4. A species found naturally in only one geographical area.
5. 14 ÷ 40 = 0.35.
6. N = 10, N(N − 1) = 90; Σn(n − 1) = 20 + 6 + 2 = 28; D = 90 ÷ 28 = 3.2 (to 2 significant figures).
7. No selection, no mutation, no migration, random mating, large population.
8. p = 0.8, so 2pq = 2 × 0.8 × 0.2 = 0.32.
9. Physiological.
10. Some seeds cannot be dried and frozen without dying, or stored seeds lose viability and need regrowing; the habitat is not conserved.

Where marks are usually lost

• Leaving “fertile” out of the species definition.
• Giving an adaptation without saying how it helps survival.
• Describing a niche as just “where it lives”.
• Reading a larger D as less diverse.
• Using N² or N(N − 1) inside the sum.
• Giving q as the recessive phenotype frequency instead of its square root.
• Forgetting the carrier calculation needs 2pq, not pq.
• Evaluating zoos with only advantages or only disadvantages and no conclusion.
• Saying seed banks conserve habitats.

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.