Biology: Genetics, Populations and Evolution
Genetics, populations and evolution is the A-level Biology topic covering the statistical and mechanistic side of inheritance and evolution: the chi-squared test applied to genetic crosses, gene linkage and recombination frequency, the three types of natural selection (stabilising, directional and disruptive), and the mechanisms of speciation. It extends the cross diagrams and Hardy-Weinberg calculations met in Genetics, Populations, Evolution and Ecosystems with the statistical testing and selection and speciation detail Year 13 papers reward.
Before you start
Make sure you're comfortable with these topics first:
Method
- For a chi-squared test, state the null hypothesis (there is no significant difference between the observed and expected results), then calculate the expected number in each category from the predicted ratio.
- Apply chi-squared = the sum, across all categories, of (observed - expected)^2 / expected.
- Find the degrees of freedom (the number of categories minus 1) and compare the calculated value with the given critical value at the 5% (p = 0.05) significance level: if the calculated value exceeds the critical value, reject the null hypothesis (the observed ratio differs significantly from the predicted one).
- For linkage questions, identify that genes on the same chromosome are described as linked and tend to be inherited together, so offspring ratios depart from those expected from independent assortment.
- Calculate recombination frequency as (number of recombinant offspring / total offspring) x 100, remembering that recombinants arise from crossing over between homologous chromosomes during meiosis I; a value well below 50% indicates linkage.
- For selection questions, identify how the distribution curve changes: stabilising selection favours the mean and narrows the distribution (reduces variance); directional selection favours one extreme and shifts the mean; disruptive selection favours both extremes and can split the distribution into two peaks.
- For speciation questions, identify the type of reproductive isolation described (geographic, leading to allopatric speciation, or behavioural, ecological, temporal or mechanical, leading to sympatric speciation), and explain how stopping gene flow lets allele frequencies in the separated populations diverge until they can no longer interbreed to produce fertile offspring.
Worked example
In a genetics experiment, a dihybrid cross was predicted to produce offspring in the ratio 9 purple round : 3 purple wrinkled : 3 green round : 1 green wrinkled. From 160 offspring, the observed numbers were 81 purple round, 32 purple wrinkled, 37 green round and 10 green wrinkled. Carry out a chi-squared test to determine whether the observed results differ significantly from the predicted 9:3:3:1 ratio. The critical value at the 5% significance level, for the appropriate degrees of freedom, is 7.815.
- Calculate the expected number in each category by applying the 9:3:3:1 ratio to 160 total offspring: expected = 90, 30, 30 and 10 for purple round, purple wrinkled, green round and green wrinkled respectively.
- For each category, calculate (observed - expected)^2 / expected: purple round = (81-90)^2/90 = 0.900; purple wrinkled = (32-30)^2/30 = 0.133; green round = (37-30)^2/30 = 1.633; green wrinkled = (10-10)^2/10 = 0.000.
- Sum these values to find chi-squared: 0.900 + 0.133 + 1.633 + 0.000 = 2.67 (to 2 decimal places).
- Find the degrees of freedom: 4 categories - 1 = 3, matching the given critical value.
- Compare the calculated value with the critical value: 2.67 is less than 7.815.
- Final answer: since the calculated chi-squared value is less than the critical value, the null hypothesis is accepted; there is no significant difference between the observed and expected results, so the data supports a 9:3:3:1 ratio.
Practice questions
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Q1State the null hypothesis used in a chi-squared test comparing observed and expected offspring numbers from a genetic cross.Show answer
Answer: There is no significant difference between the observed and expected results (any difference is due to chance).
Q2A chi-squared test on a cross with 4 phenotype categories gives a calculated value of 9.20. The critical value at p = 0.05 is 7.815. State the conclusion.Show answer
Answer: Since the calculated value (9.20) exceeds the critical value (7.815), the null hypothesis is rejected; there is a significant difference between observed and expected results, so the data does not fit the predicted ratio.
Q3Define the term linkage in genetics.Show answer
Answer: Genes located close together on the same chromosome, which tend to be inherited together because they are not separated by independent assortment.
Q4A test cross between a dihybrid heterozygote and a homozygous recessive individual produced 1000 offspring: 380 parental type A, 365 parental type B, 128 recombinant type C and 127 recombinant type D. Calculate the recombination frequency.Show answer
Answer: 25.5% ((128 + 127) / 1000 x 100).
Q5State the effect of stabilising selection on the variance of a population's phenotype distribution.Show answer
Answer: Variance decreases (the distribution curve becomes narrower and taller around the mean), as individuals with extreme phenotypes are selected against.
Q6Two populations of the same fish species become separated by a waterfall that forms after a landslide, preventing migration between them. State the type of speciation this could lead to.Show answer
Answer: Allopatric speciation.
Q7Give one example of a prezygotic reproductive isolating mechanism that could prevent two populations living in the same area from interbreeding.Show answer
Answer: Any valid example, for instance differences in courtship behaviour, so individuals from the two populations no longer recognise each other as potential mates (behavioural isolation).
Q8In directional selection, describe how the mean of the population's phenotype distribution changes over time.Show answer
Answer: The mean shifts towards one extreme of the distribution (the phenotype being selected for), as individuals with that phenotype have higher survival and reproductive success.
Exam-style questions
Written in the style of a A Level Science exam paper, with a full mark scheme.
A population of insects contains individuals with a wide range of body sizes. A change in the environment means only medium-sized insects can access food efficiently, while very large and very small insects struggle to survive. Name the type of natural selection described, and explain the effect this will have on the distribution of body size in the population over several generations.
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In Drosophila, a cross between a heterozygous fly (with alleles for body colour and wing shape linked on the same chromosome) and a homozygous recessive fly produced 800 offspring: 312 grey body/normal wing, 308 black body/vestigial wing, 92 grey body/vestigial wing and 88 black body/normal wing. (a) Identify which two phenotype classes are the recombinant types. (b) Calculate the recombination frequency between the two genes.
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Two populations of a fish species become geographically separated when a river changes course. Explain how this separation could lead to the evolution of two distinct species over time.
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