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7.3.5Genetics, Populations, Evolution and Ecosystems (A Level only)

Investigating Allele Frequencies — AQA A Level Biology 7402 Revision Notes & Topical Questions

5 min read 9 parts Lesson 13 of 25 1 quiz

📚 AQA A Level Biology 7402 Hub / Topical Revision
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Lesson summary

7.3.5 Investigating Allele Frequencies — AQA A Level Biology 7402 notes: Describe how allele frequencies can be estimated and compared and use the Hardy-Weinberg equations and statistical tests to interpret data about changes in populations.Evolution is a change in allele frequen…

  • phenotype counts
  • DNA analysis
  • chi-squared test
  • Student's t-test
  • Spearman rank
  • Exam-style question 1

Part 1 of 7Why investigate allele frequencies?

Evolution is a change in allele frequencies in a population, so measuring allele frequencies at different times or in different places gives direct evidence of evolutionary change. Examples include monitoring the spread of antibiotic or insecticide resistance alleles, comparing human populations for alleles for lactase persistence or sickle cell trait, and tracking the loss of diversity in endangered species.

Allele frequencies can be estimated from phenotype counts (when the phenotype reveals the genotype) or from DNA analysis (PCR, gel electrophoresis or DNA sequencing), which can identify heterozygotes directly.

Part 2 of 7Calculating frequencies from data

If the genotypes of every individual are known, count alleles: the frequency of an allele = number of copies of the allele / total number of alleles (twice the number of individuals).

If only the recessive phenotype can be distinguished from the others, use Hardy-Weinberg.

Calculating frequencies from data
The frequency of the recessive phenotype = q2, so q = square root of this value, p = 1 - q, and the heterozygote frequency = 2pq.
This assumes random mating, a large population, no selection, no mutation and no migration, so the values are estimates only.

Worked examplein a sample of 400 moths, 36 are pale (recessive) and 364 dark. q2 = 36/400 = 0.09 so q = 0.3, p = 0.7, and 2pq = 0.42, so about 168 of the dark moths are carriers.

Part 3 of 7Comparing samples: sampling and statistics

To compare allele or phenotype frequencies in two areas or times, take large, random samples to reduce bias and the effect of chance. Use random sampling (random number tables to select coordinates) rather than choosing sites by eye. The results are then analysed statistically.

Chi-squared test
The chi-squared test compares observed and expected frequencies for categorical data.
  • The null hypothesis is that there is no significant difference between observed and expected values.
  • Calculate chi-squared = sum of (O - E)2/E, find the degrees of freedom (number of categories - 1) and compare with the critical value at p = 0.05.
    • If the calculated value is greater than the critical value, reject the null hypothesis: the difference is significant and not due to chance alone.
Student's t-test
The Student's t-test compares the means of two samples of continuous data. The Spearman rank test or Pearson's test examines correlation.

Part 4 of 7Investigating selection in practice

A typical investigationcompare the frequency of a phenotype (e.g. banding or shell colour in snails) in different habitats.

Investigating selection in practice
A sample is collected with random sampling in each habitat, counts are recorded for each morph,
  • and a chi-squared test determines whether the frequencies in the two habitats differ more than expected by chance.
If significant, a possible explanation is that predators remove the more conspicuous morph in each habitat (differential predation as a selection pressure).
A correlation does not prove causation, so controls and repeated trials strengthen conclusions.

Ethical considerations include minimising harm to organisms and replacing animals after sampling, and handling the rare species with permission.

Part 5 of 7Evaluating and linking

When evaluating a data set consider

  • sample size (larger is better), randomness, whether the population satisfies Hardy-Weinberg assumptions, possible confounding variables (different habitats, ages), and whether the statistical test is correct for the data type.
  • Link this topic to genetic variation (7.3.1), the Hardy-Weinberg principle (7.2.3), types of selection (7.3.3) and to genetic technologies such as PCR and electrophoresis in 8.4.

Data typeTest
Observed v expected counts in categoriesChi-squared
Difference between two meansStudent's t-test
Association between two variablesCorrelation (Spearman's rank)

Part 6 of 7Using DNA analysis to find allele frequencies

Where the phenotype does not reveal the genotype, DNA techniques can be used.

1
A sample of DNA is amplified at the gene of interest by PCR, using primers that flank the gene.
2
The fragments are then separated by gel electrophoresis,
  • since DNA is negatively charged and moves towards the positive electrode, with smaller fragments travelling further.
3
Different alleles often give fragments of different sizes,
  • so each individual's genotype can be read from the band pattern: one band for a homozygote and two bands for a heterozygote.
4
Counting the alleles from a large, random sample gives the allele frequency directly, with no need for the Hardy-Weinberg assumptions about dominance.
5
Whole genome sequencing can detect single nucleotide polymorphisms across the entire genome and allows frequencies to be compared between populations at many loci simultaneously.

Part 7 of 7Exam-style questions

  1. In a population of 500 beetles, 45 show a recessive phenotype. Use the Hardy-Weinberg equation to estimate the number of heterozygotes. [4]4 marks

  2. A student counts 60 striped and 40 plain snails in a woodland. The expected ratio on a hypothesis is 1:1. Calculate chi-squared and state the conclusion (critical value 3.84 at p = 0.05, 1 degree of freedom). [4]4 marks

  3. Suggest two reasons why the allele frequency calculated by the Hardy-Weinberg equation may differ from the real frequency in a wild population. [2]2 marks

Key terms

TermMeaning
Allele frequencyProportion of copies of a gene in a population that are a given allele.
Null hypothesisStatement that there is no significant difference or association between variables.
Chi-squared testStatistical test comparing observed and expected frequencies of categories.
Degrees of freedomNumber of independent categories, usually categories minus one.
Random samplingSelecting samples so that every part of the population has an equal chance of selection.
Critical valueValue of a test statistic that must be exceeded for a result to be significant.
Student's t-testTest for a significant difference between means of two samples.

Avoid these mistakes

1
Using percentages instead of actual counts in a chi-squared test.
2
Forgetting that q2 is the recessive phenotype frequency, not q.
3
Using the wrong degrees of freedom (categories minus 1).
4
Saying that a significant result proves the cause; it only shows that the difference is unlikely to be due to chance.
5
Not describing random sampling when collecting samples.

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