Genetics, Populations, Evolution and Ecosystems
Genetics, Populations, Evolution and Ecosystems is the A-level Biology topic covering inheritance (monohybrid, dihybrid and epistatic crosses), chi-squared testing, the Hardy-Weinberg principle, genetic drift, natural selection, speciation, and ecosystem sampling. It draws together genetics with statistics and whole-population thinking. Expect chi-squared and Hardy-Weinberg calculations.
Before you start
Make sure you're comfortable with these topics first:
Method
- For any genetic cross, define the phenotypes, assign allele letters, write the parental genotypes, then use a Punnett square or genetic diagram to find the offspring ratio.
- For a chi-squared test, calculate expected values from the predicted ratio, find (O-E)^2/E for every category, sum them, then compare the total to the critical value at the given degrees of freedom to accept or reject the null hypothesis.
- For Hardy-Weinberg questions, use p + q = 1 and p^2 + 2pq + q^2 = 1, starting from whichever genotype/phenotype frequency the question gives you (usually the homozygous recessive, q^2).
- Distinguish genetic drift (a random change in allele frequency, e.g. the founder effect or a population bottleneck) from natural selection (a non-random, directional change driven by differential survival/reproduction).
- For natural selection questions, name the correct type (stabilising, directional or disruptive) from the description of how the trait distribution changes, and explain it in terms of survival and reproduction of individuals with the advantageous allele.
- For ecosystem sampling questions, describe how the sampling method removes bias (e.g. using random number coordinates to place quadrats) before calculating any diversity index.
Worked example
A plant breeder crossed two pea plants that were both heterozygous for seed shape (Rr) and seed colour (Yy). The two genes assort independently. From 240 offspring, a chi-squared test was used to test whether the results fitted the expected 9:3:3:1 ratio. The calculated chi-squared value was 6.10. The critical value of chi-squared at p = 0.05 for 3 degrees of freedom is 7.82. State and explain the conclusion that should be drawn.
- Compare the calculated value to the critical value: 6.10 is less than 7.82.
- Because the calculated value does not exceed the critical value, the null hypothesis is not rejected.
- This means there is no significant difference between the observed and expected (9:3:3:1) numbers of offspring.
- Final answer: the null hypothesis is accepted; any difference between the observed and expected results is due to chance, so the data are consistent with the two genes assorting independently.
Practice questions
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Q1State the genotype of an organism showing a recessive phenotype for a gene with alleles B (dominant) and b (recessive).Show answer
Answer: bb (homozygous recessive).
Q2What is meant by the term epistasis in genetics?Show answer
Answer: The interaction of genes at different loci, in which one gene (the epistatic gene) masks or suppresses the expression of another gene (the hypostatic gene).
Q3State the Hardy-Weinberg equation that links genotype frequencies.Show answer
Answer: p^2 + 2pq + q^2 = 1
Q4In a population, 9% of individuals show a recessive phenotype. Calculate the frequency of the recessive allele.Show answer
Answer: 0.3 (q^2 = 0.09, so q = sqrt(0.09) = 0.3)
Q5Identify the type of natural selection in which individuals with an extreme trait value are favoured over those with the average trait value, shifting a population's mean over time.Show answer
Answer: Directional selection.
Q6A dihybrid cross (RrYy x RrYy) produced 320 offspring. Calculate the expected number of offspring in the round-yellow phenotype class, which makes up 9/16 of the expected 9:3:3:1 ratio.Show answer
Answer: 180 (9/16 x 320 = 180)
Exam-style questions
Written in the style of a A Level Science exam paper, with a full mark scheme.
Explain what is meant by the term genetic drift.
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A population of 40000 fur seals was reduced to just 25 individuals after a period of intense hunting, before recovering to over 150000 today. Explain why the genetic diversity of the modern population remains low despite its large current size, using the term population bottleneck.
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In a species of squash, fruit colour is controlled by two genes. A dominant allele (A) at gene 1 suppresses colour production, giving a white fruit regardless of genotype at gene 2; the recessive allele (a) allows colour to be produced. When colour is not suppressed, a dominant allele (B) at gene 2 gives yellow fruit and the recessive allele (b) gives green fruit. Two squash plants were crossed: one heterozygous at both genes (AaBb) and the other homozygous recessive at both genes (aabb). Using a Punnett square or equivalent working, determine the phenotypic ratio of the offspring from this cross, stating how many of every 4 offspring would be white, yellow and green, and explain why this ratio differs from the 12:3:1 ratio produced by crossing two AaBb heterozygotes together.
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Free printable worksheet
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