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Eduqas A-level Biology (A400QS) · Component 2: Continuity of Life
Mini-Lesson · A-level

Component 2 — Continuity of Life

Eduqas Component 2 covers the continuity of life: classification and evolutionary history, biodiversity, mitosis and meiosis, sexual reproduction in humans and plants, inheritance and the chi-squared test, variation and evolution with the Hardy–Weinberg principle, and the applications of reproduction and genetics.

classification reproduction & inheritance evolution 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.

Topic 1 · Evolutionary history

Classification and biodiversity

Classification is hierarchical: domain, kingdom, phylum, class, order, family, genus, species, with a binomial name for each species. Modern classification is phylogenetic — it aims to reflect evolutionary relationships, and is increasingly based on molecular evidence (DNA and rRNA sequences, protein comparison, immunology) rather than appearance, which can mislead through convergent evolution.

The three-domain system (Bacteria, Archaea, Eukarya) replaced the five-kingdom system when ribosomal RNA sequencing revealed that the archaea are as different from bacteria as either is from us.

Biodiversity can be assessed at three levels: within a habitat (species richness and diversity indices), within a species (the proportion of polymorphic loci — the variety of alleles in the gene pool), and at the molecular level (DNA fingerprinting and sequencing).

Simpson’s Diversity Index: D = 1 − Σ (n ÷ N)²n = number of individuals of each species · N = total number of individuals
D lies between 0 and 1: the closer to 1, the greater the diversity
Calculate

Your turn — Simpson’s Diversity Index

1A sample of 50 organisms contains 25 of species A, 15 of species B and 10 of species C. Calculate Simpson’s Diversity Index, D = 1 − Σ(n/N)². Give your answer to 2 decimal places.
D
Hint: (25/50)² = 0.25; (15/50)² = 0.09; (10/50)² = 0.04. Σ = 0.38. D = 1 − 0.38.
Topic 2 · Copying genetic information

DNA replication and mitosis

Semi-conservative replication: DNA helicase unwinds the helix and breaks the hydrogen bonds; free nucleotides pair with the exposed bases; DNA polymerase forms the phosphodiester bonds, always working 5′→3′. Each daughter molecule keeps one parental strand.

The cell cycle: interphase (G1 → S, where DNA replicates → G2), then mitosis, then cytokinesis.

  • Prophase — chromosomes condense and become visible as two sister chromatids; the nuclear envelope breaks down; the spindle forms.
  • Metaphase — chromosomes align on the equator.
  • Anaphase — centromeres divide and the chromatids are pulled to opposite poles.
  • Telophase — nuclear envelopes re-form; chromosomes decondense.

Cancer is a failure of the controls on this cycle: mutations in proto-oncogenes (making them oncogenes) or tumour suppressor genes allow uncontrolled mitosis. Carcinogens and mutagens increase the mutation rate and so the risk.

Topics 3–4 · Sexual reproduction

Meiosis and sexual reproduction in humans and plants

Meiosis halves the chromosome number and generates variation. Two divisions produce four haploid, genetically different cells.

  • Crossing over (prophase I) — homologous chromosomes pair; chiasmata form and sections of non-sister chromatids are exchanged, creating new allele combinations.
  • Independent assortment (metaphase I) — each homologous pair aligns independently: 2ⁿ combinations for n pairs.
  • Random fertilisation — any sperm may fuse with any egg.

In humans: spermatogenesis produces four small motile sperm from each primary spermatocyte and runs continuously from puberty; oogenesis produces one large ovum plus polar bodies, and is arrested part-way until ovulation. Fertilisation: the acrosome reaction digests a path through the zona pellucida; the membranes fuse; the cortical reaction then hardens the zona pellucida to prevent polyspermy; the haploid nuclei fuse to give a diploid zygote.

In flowering plants: the anther makes pollen; the ovule contains the embryo sac. Pollination is followed by the growth of the pollen tube down the style, and then by double fertilisation — one male nucleus fuses with the egg cell to form the diploid zygote, and the other fuses with two polar nuclei to form the triploid endosperm, the food store for the embryo.

Sort it

Mitosis or meiosis?

Tap a statement, then tap the process it describes.

🔁 Mitosis

🎲 Meiosis

↔️ Both

Topic 5 · Inheritance

Monohybrid, dihybrid, linkage and sex linkage

  • Monohybrid — one gene. Heterozygous × heterozygous gives a 3:1 phenotypic ratio.
  • Codominance — both alleles are expressed in the heterozygote (e.g. the AB blood group; roan cattle). Heterozygous × heterozygous gives 1:2:1.
  • Dihybrid — two unlinked genes. Double heterozygotes give 9:3:3:1.
  • Linkage — genes on the same chromosome are inherited together, so the parental combinations are far more common than expected and the recombinants are rare (they only arise by crossing over). A dihybrid cross that gives a strongly non-9:3:3:1 ratio is the classic sign of linkage.
  • Sex linkage — genes on the X chromosome. Males are XY and so hemizygous: a single recessive allele is expressed. Haemophilia and Duchenne muscular dystrophy are therefore far more common in males, and a carrier mother passes the allele to half of her sons.

Mutation (spec 5(f)): a gene mutation can be a single base substitution — in sickle cell anaemia, one base change substitutes valine for glutamic acid in the β-globin chain, so the haemoglobin polymerises and distorts the red cell. A chromosome mutation such as the non-disjunction that gives three copies of chromosome 21 causes Down’s syndrome.

Topic 5 · the chi-squared test

Testing your genetic ratios

χ² = Σ (O − E)² ÷ Edegrees of freedom = number of classes − 1
  1. Null hypothesis: there is no significant difference between the observed and expected results — any difference is due to chance.
  2. Work out the expected numbers from the predicted ratio.
  3. Calculate χ².
  4. Compare with the critical value at p = 0.05 for the correct degrees of freedom.
  5. Below the critical value → accept the null hypothesis. At or above it → reject it: the difference is significant, so something else (such as linkage) is operating.

p = 0.05 means that a difference this large would arise by chance alone in fewer than 1 in 20 experiments. It is the conventional threshold in biology — not a natural law.

Calculate

Your turn — chi-squared

2A monohybrid cross predicts a 3:1 ratio. From 200 offspring, 140 show the dominant phenotype and 60 the recessive. Calculate χ² = Σ (O − E)² ÷ E. Give your answer to 2 decimal places.
χ²
Hint: Expected = 150 and 50. (140−150)²/150 = 0.667; (60−50)²/50 = 2.000.
Quick check

Interpreting your χ²

?Your χ² is 2.67. With 1 degree of freedom, the critical value at p = 0.05 is 3.84. What do you conclude?
Topic 6 · Variation and evolution

Natural selection, Hardy–Weinberg and speciation

Variation is continuous (polygenic, strongly influenced by the environment — height, mass) or discontinuous (controlled by one or a few genes — blood group). Its ultimate source is mutation; sexual reproduction shuffles it.

Natural selection: variation → selection pressure → the better-adapted individuals survive and reproduce → the frequency of the advantageous allele increases. Stabilising selection favours the intermediate; directional selection favours one extreme; disruptive selection favours both extremes.

p + q = 1
p² + 2pq + q² = 1p² = homozygous dominant · 2pq = heterozygous · q² = homozygous recessive

The Hardy–Weinberg principle applies only if the population is large, mating is random, and there is no mutation, no migration and no selection. If observed frequencies differ from predicted, one of those conditions is broken — the population is evolving.

Speciation: allopatric — a geographical barrier prevents gene flow; the two populations diverge under different selection pressures until they are reproductively isolated. Sympatric — reproductive isolation arises without a physical barrier (polyploidy, a change in flowering time, a behavioural change). Genetic drift — chance changes in allele frequency — has a far larger effect in small populations.

Calculate

Your turn — Hardy–Weinberg

3In a population, 16 % of individuals are homozygous recessive. Use p² + 2pq + q² = 1 to calculate the percentage that is homozygous dominant.
%
Hint: q² = 0.16, so q = 0.4 and p = 0.6. Homozygous dominant = p² = 0.6².
Match it

Where does the variation come from?

Tap a source of variation, then when it happens.

Source of variation
When it happens
Topic 7 · Applications

DNA profiling, PCR and genetic engineering

PCR amplifies DNA: denature (95 °C)anneal primers (50–65 °C)extend with Taq polymerase (72 °C). Each cycle doubles the amount of DNA, so n cycles gives 2ⁿ copies.

Gel electrophoresis: DNA is negatively charged, so all fragments move towards the anode; the gel sieves them, so shorter fragments travel further. Comparing the pattern of bands from short tandem repeats gives a DNA profile, used in forensics, paternity testing and in establishing evolutionary relationships.

Genetic engineering: the gene is cut out with a restriction enzyme (leaving sticky ends), joined into a plasmid vector cut with the same enzyme using DNA ligase, and taken up by a host cell (transformation). Marker genes identify the transformed cells. This is how human insulin is made.

Applications and ethics: gene therapy, genetically modified crops (pest resistance, higher yield, Golden Rice), and stem cell therapy. The debate is real and you must be able to argue both sides: benefits to health and food security against the escape of transgenes, the loss of biodiversity, the welfare of GM animals, and questions about the destruction of embryos and about who controls and profits from the technology.

Calculate

Your turn — PCR

4A single DNA molecule is amplified through 20 cycles of PCR. Assuming 100 % efficiency, calculate the number of DNA molecules produced.
molecules
Hint: Each cycle doubles it: 2²⁰. Note 2¹⁰ = 1024, so 2²⁰ = 1024².
Quick check

Reading a DNA profile

?On a gel, a child has a band that is present in neither the mother’s nor the alleged father’s profile. What does this suggest?
Recap

The big ideas to know

Classification: domain, kingdom, phylum, class, order, family, genus, species. Three domains (Bacteria, Archaea, Eukarya) from rRNA evidence. Modern classification is phylogenetic and molecular.

Simpson’s Diversity Index: D = 1 − Σ(n/N)². A value closer to 1 means greater diversity.

Mitosis: two genetically identical diploid cells — growth, repair, asexual reproduction.

Meiosis: four genetically different haploid cells. Variation from crossing over (prophase I), independent assortment (metaphase I) and random fertilisation.

Inheritance: monohybrid (3:1), dihybrid (9:3:3:1), codominance, linkage, sex linkage (haemophilia, Duchenne muscular dystrophy).

χ² = Σ(O − E)²/E. If χ² is less than the critical value at p = 0.05, accept the null hypothesis — the difference is due to chance.

Mutation: gene mutation (sickle cell anaemia — a single base substitution) and chromosome mutation (Down’s syndrome — non-disjunction). Epigenetics controls gene expression without changing the base sequence.

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

You have covered the whole of Eduqas Component 2. Press Finish to see your score.

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