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Edexcel A-level Biology A (Salters-Nuffield) 9BN0 · Topic 4: Biodiversity and Natural Resources
Mini-Lesson · A-level

Biodiversity & Natural Resources

SNAB Topic 4 is about the variety of life and how we measure, classify, use and conserve it. Expect real index calculations, the Hardy–Weinberg equation, plant cell ultrastructure, and the biology behind plant fibres and modern drug testing.

measuring biodiversity selection & classification plants as a resource three strands you must be able to link together

Work through each screen, answer the questions as you go — several are A-level calculations — and collect ⭐ stars. Press Start when you are ready.

Biodiversity · 4.1–4.2

Measuring biodiversity

Biodiversity can be considered at three levels: the diversity of habitats, the number of species, and the genetic diversity within a species. An endemic species is one found naturally in one geographical area and nowhere else — endemics are especially vulnerable to extinction.

H = number of heterozygotes ÷ number of individuals in the populationheterozygosity index — a measure of genetic diversity WITHIN a species
D = N(N − 1) ÷ Σ n(n − 1)index of diversity — N is the total number of organisms of ALL species; n is the number of organisms of EACH species

Species richness is just a count of species. An index of diversity is better because it also takes abundance into account: a wood with 100 oaks and 1 birch is far less diverse than one with 50 of each, even though both have a richness of 2.

Calculate

Your turn — heterozygosity index

1In a population of 200 individuals, 48 are heterozygous at a particular locus. Calculate the heterozygosity index (H). Give your answer to 2 decimal places.
H
Hint: H = 48 ÷ 200.
Calculate

Your turn — index of diversity

2A quadrat contains 12 of species A, 8 of species B and 5 of species C. Calculate the index of diversity D = N(N−1) ÷ Σ n(n−1). Give your answer to 2 decimal places.
D
Hint: N = 25 so N(N−1) = 600. Σ n(n−1) = (12×11) + (8×7) + (5×4) = 132 + 56 + 20 = 208.
Niche & adaptation · 4.3–4.4

Niche, adaptation and natural selection

A niche is the role and position a species has in its environment: what it eats, when it is active, what eats it, what it tolerates. If two species occupy the same niche, one will out-compete the other (competitive exclusion) — so coexisting species always differ somewhere.

Adaptations come in three flavours:

  • Anatomical — e.g. the marram grass rolled leaf, sunken stomata and thick cuticle that reduce transpiration.
  • Physiological — e.g. the kangaroo rat producing extremely concentrated urine.
  • Behavioural — e.g. nocturnality to avoid daytime heat.

Natural selection, in order: (1) there is genetic variation caused by mutation; (2) a selection pressure acts; (3) individuals with the advantageous allele are more likely to survive and reproduce; (4) they pass the allele on, so the allele frequency in the population increases. Evolution is that change in allele frequency.

Never write that an organism "adapts to" or "wants to" survive. Individuals do not adapt; populations evolve, because some individuals already happened to carry a useful allele.

Hardy–Weinberg · 4.5

The Hardy–Weinberg equation

Hardy–Weinberg lets you calculate allele and genotype frequencies and, crucially, lets you test whether a population is evolving.

p + q = 1
p² + 2pq + q² = 1p = frequency of the dominant allele · q = frequency of the recessive allele
p² = homozygous dominant · 2pq = heterozygous · q² = homozygous recessive

You almost always start from , because the homozygous recessive is the only genotype you can identify from the phenotype alone.

The assumptions matter: a large population, random mating, no migration in or out, no mutation, and no selection. If the observed frequencies differ from the predicted ones, one of those conditions has been broken — the population is evolving. Reproductive isolation lets allele frequencies in two populations diverge, which is the first step towards speciation.

Calculate

Your turn — Hardy–Weinberg

3A recessive genetic condition affects 1 in 2500 people. Use p² + 2pq + q² = 1 to calculate the percentage of the population who are carriers (heterozygous). Give your answer to 2 decimal places.
%
Hint: q² = 1/2500 = 0.0004, so q = 0.02 and p = 0.98. Carriers = 2pq = 2 × 0.98 × 0.02.
Quick check

Testing the assumptions

?A population’s observed genotype frequencies differ significantly from those predicted by Hardy–Weinberg. What is the best conclusion?
Classification · 4.6

Classification and the three domains

Taxonomic hierarchy: Domain, Kingdom, Phylum, Class, Order, Family, Genus, Species. Each group is nested inside the one above, and each species has a binomial name (genus + species), e.g. Homo sapiens.

Woese’s comparison of ribosomal RNA sequences showed that the "bacteria" were really two profoundly different groups, so a level above kingdom was added — the three domains: Bacteria, Archaea and Eukarya.

How science works: the change was not accepted because Woese said so. It was accepted after the data were published, peer reviewed, presented at conferences, criticised, and independently reproduced using other molecular evidence (DNA sequences, protein sequences, immunology). That is the process SNAB wants you to describe.

Plant cells · 4.7–4.10

Plant cell ultrastructure, starch and cellulose

On top of the standard eukaryotic organelles, a plant cell has: a cellulose cell wall with a middle lamella (calcium pectate) gluing adjacent walls together, pits and plasmodesmata (cytoplasmic channels between cells), chloroplasts (grana of thylakoids in a stroma), amyloplasts (starch stores) and a large vacuole with a tonoplast.

  • Starch — a polymer of α-glucose. Amylose has 1,4-glycosidic bonds only and coils into a compact helix; amylopectin is branched (1,6 bonds), so it can be hydrolysed quickly. Insoluble, so it does not affect water potential — a perfect store.
  • Cellulose — a polymer of β-glucose. Alternate residues are flipped 180°, giving straight, unbranched chains. Many hydrogen bonds cross-link them into microfibrils, which are laid down in different directions to give enormous tensile strength.

Core practical 6/7/8 territory: the tensile strength of plant fibres; the effect of mineral deficiency (nitrate for amino acids and nucleotides, calcium for the middle lamella, magnesium for chlorophyll) on plant growth; and the antimicrobial properties of plant extracts.

Sort it

Starch or cellulose?

Tap a statement, then tap where it belongs.

🥔 Starch only

🌾 Cellulose only

🔁 Both

Match it

Match the measure to its meaning

Tap a definition on the left, then the correct term.

Definition
Term
Drug testing · 4.13

From folklore to clinical trials

Historically, drugs were tested by trial and error — William Withering used a digitalis soup from foxglove, adjusting the dose until it worked without killing the patient. Modern testing is far more rigorous:

  • Pre-clinical — the drug is tested on cells, tissues and then animals to check for basic toxicity and to establish a dose.
  • Phase I — a small number of healthy volunteers: is it safe?
  • Phase II — a larger group of patients: does it work, and at what dose?
  • Phase III — a very large, randomised, double-blind, placebo-controlled trial comparing it with the current best treatment.

Double-blind means neither the patient nor the doctor knows who has the drug — that removes both the placebo effect and unconscious bias in reporting outcomes.

Calculate

Your turn — bacterial growth

4A single bacterium is placed on nutrient agar and divides every 20 minutes under optimal conditions. Assuming no limiting factors, calculate how many bacteria there are after 3 hours.
bacteria
Hint: 3 hours = 180 min ÷ 20 = 9 divisions. Number = 2⁹.
Conservation · 4.14–4.16

Growing microbes and conserving species

Bacterial growth (4.14) requires nutrients (a carbon source, nitrogen, minerals), a suitable temperature and pH, and — for aerobes — oxygen. A culture shows a lag phase (enzymes being made), an exponential (log) phase, a stationary phase (nutrients running out, waste accumulating) and a death phase. Aseptic technique keeps the culture pure. Core practical 9 uses this to test the antimicrobial activity of plant extracts, measuring the zone of inhibition.

Conservation (4.16):

  • Zoos — captive breeding, studbooks to maximise genetic diversity, reintroduction, research and education. Criticisms: the small gene pool, behavioural problems in captivity, poor reintroduction success, and cost.
  • Seed banks — cheap, store enormous genetic diversity in a tiny space, and seeds stay viable for decades when dried and frozen. Limitations: some species (e.g. many tropical trees) have recalcitrant seeds that cannot be dried; viability must be tested by periodic germination; a seed bank preserves the species but not the habitat.
Quick check

Judging a conservation strategy

?Which is the strongest argument that a seed bank is a valuable conservation tool?
Quick check

Why is cellulose so strong?

?Which feature explains the tensile strength of cellulose?
Recap

The big ideas to know

Biodiversity: measured as species richness (how many species), genetic diversity (heterozygosity index H) and an index of diversity D that also accounts for abundance.

D = N(N − 1) ÷ Σ n(n − 1), where N = total organisms and n = organisms of each species. A larger D means greater diversity.

Niche: the role and position of a species in its habitat. Two species cannot occupy the same niche indefinitely.

Natural selection: variation → selection pressure → differential survival and reproduction → change in allele frequency.

Hardy–Weinberg: p + q = 1 and p² + 2pq + q² = 1. Assumes a large population, random mating, no migration, no mutation and no selection.

Classification: domain, kingdom, phylum, class, order, family, genus, species. Three domains: Bacteria, Archaea, Eukarya (from rRNA evidence).

Plant cells: cellulose cell wall, chloroplasts, amyloplasts, vacuole, plasmodesmata, middle lamella, pits.

Starch vs cellulose: α-glucose (1,4 links, coiled, compact store) vs β-glucose (alternate residues flipped 180°, straight chains, H-bonded into microfibrils).

You have covered the whole of SNAB Topic 4. Press Finish to see your score.

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