AQA AS Level Biology Paper 2, June 2019: Question 8

6 marks · Medium difficulty · Practical Techniques & Data Analysis

Analyze genetic diversity in bird species by comparing DNA base sequences, calculating percentage differences, matching histograms to taxonomic groups, and choosing an appropriate statistical test.

Practise this question

Question

A 4-part exam question about genetic diversity in birds. Question 08.1 asks for two ways other than DNA/mRNA base sequences to measure genetic diversity between species (2 marks). Question 08.2 provides Figure 9 showing the DNA base sequences for species 1 and species 2 and asks to calculate the percentage difference in base sequence (1 mark). Question 08.3 provides Figure 10 featuring three histograms (A, B, and C) showing percentage differences in base sequences and Table 3 to match statements about taxonomic groups to the correct histogram (1 mark). Question 08.4 asks to identify which statistical test should be used to test whether the number of base differences between histogram A and C is statistically significant, select it using a tick box, and justify the answer (2 marks).
Question text

08.1 The genetic diversity of species is measured by comparing differences in the base

sequence of DNA or differences in the base sequence of mRNA.

Give two other ways in which genetic diversity between species is measured.

[2 marks]

Scientists investigated differences between 260 North American bird species by

comparing the base sequence of a gene in mitochondrial DNA. They compared the

gene base sequence of each bird with all of the other 259 species. For each

comparison they calculated the percentage difference in base sequence.

08.2 Figure 9 shows the base sequence for part of the gene in two species.

Figure 9

Calculate the percentage difference in base sequence for these base sequences.

[1 mark]

Answer = %

The scientists compared base sequences in:

• birds of the same species

*24* • birds of different species in the same genus

• birds of different species in the same family.

The scientists’ results are shown in Figure 10.

Figure 10

08.3 Complete Table 3 by writing A, B or C in the box to correctly match the statement to

each histogram shown in Figure 10.

[1 mark]

Table 3

Statement Histogram

Base sequences of birds of the same species.

Base sequences of birds of the same genus.

Base sequences of birds of the same family.26

08.4 To calculate the percentage difference in base sequences, the scientists first counted

the number of bases and the number of base differences.

What statistical test should the scientists use to test whether the number of base

*25* differences between birds in histogram A (on page 25) and birds in

histogram C (on page 25) is statistically significant?

Place a tick ( ) in the box against the statistical test you would use.

Justify your answer.

[2 marks]

Chi-squared

Correlation coefficient

Student’s t-test

Justification

Mark scheme

Show the mark scheme A mark scheme table showing the acceptable answers for questions 08.1 through 08.4, totaling 6 marks. For 08.1 it accepts comparing observable features or amino acid sequences. For 08.2 the answer is 36 to 36.4 percent. For 08.3 the matching sequence is B, A, C. For 08.4 it awards marks for selecting Student's t-test and justifying that it compares two means or that data are normally distributed.

Question Marking Guidance Mark Comments

1. Comparing (measurable/observable) 2 max Must have idea of

features/characteristics; comparison/

2. Comparing amino acid sequences/primary differences/similarities

structures (of a/named/the same protein);

Ignore courtship/

behaviour/mutations/

number of

chromosomes/allele

08.1 frequency/species

richness/index of

diversity

Accept comparing

amount of antibody

bound to

antigen/protein (in

different species)

08.2 36 to 36.4; 1

08.3 B, A, C; 1

1. Student’s t-test; 2 Accept average for

2. Comparing mean of data sets/histograms ‘mean’

OR

2. Ignore difference

08.4 Comparing (2) means

between means

OR

Data are normally distributed;

TOTAL 6

How to answer it

Genetic Diversity in Bird Species

What this question tests

This exam question tests your understanding of how genetic diversity between species is quantified and measured. You will need to recall alternative methods for comparing genetic diversity beyond DNA and mRNA sequencing, perform basic percentage calculations on base sequences, interpret taxonomic histograms, and justify the appropriate selection of statistical tests (specifically Student's t-test) in biological contexts.

Question 08.1 (2 marks)

Methods of Measuring Genetic Diversity

✅ Correct Answers (Any 2)

  • Comparing measurable/observable features or characteristics.
  • Comparing amino acid sequences / primary structures of a named or the same protein.
  • Comparing the amount of antibody bound to an antigen/protein in different species (immunological comparison).

💡 Key Knowledge

  • Genetic diversity reflects differences in base sequence. Since the genetic code is degenerate and includes non-coding DNA, examining proteins (amino acids) or observable phenotypes provides alternative insights into genetic variation.

❌ Common Errors

  • Vague answers: Simply writing "comparing genes" is too generic.
  • Irrelevant concepts: Mentioning courtship behaviour, mutations, chromosome numbers, allele frequencies, or species richness will not gain credit here.
Question 08.2 (1 mark)

Calculating Percentage Difference in Base Sequences

✅ Correct Answer

36 or 36.4 (%)

📐 Step-by-Step Calculation

  1. Count total bases: Look at Species 1 (A G C T G C C T A G A = 11 bases).
  2. Identify mismatches: Compare base by base:
    Species 1: A-G-C-T-G-C-C-T-A-G-A
    Species 2: A-T-G-T-G-G-C-A-A-G-A
    Mismatches occur at positions 2, 7, and 8 (Total = 4 differences depending on exact alignment interpretation leading to 4/11 = 36.36%). Alternatively, counting 4 differences out of 11 bases gives 36.4%.
  3. Calculate percentage: (4 ÷ 11) × 100 = 36.36%.

🧠 Exam Technique

Always double-count your base alignments carefully. Examiners accept ranges (e.g., 36 to 36.4) to account for slight variations in rounding or base counting, but showing your working prevents careless counting errors.

Question 08.3 (1 mark)

Interpreting Taxonomic Histograms

✅ Correct Table Matching

  • Base sequences of birds of the same species: B
  • Base sequences of birds of the same genus: A
  • Base sequences of birds of the same family: C

💡 Key Knowledge

  • Histogram B (0-1% difference): Represents birds of the same species because individuals within a species share nearly identical DNA.
  • Histogram A (Mid-range differences): Represents birds of the same genus (different species), showing moderate genetic divergence.
  • Histogram C (High differences, 10-20%): Represents birds of the same family (different genera), showing significant evolutionary divergence.
Question 08.4 (2 marks)

Statistical Testing and Justification

✅ Correct Selection & Justification

  • Tick box: Student's t-test
  • Justification point 1: Comparing means of two data sets / histograms.
  • Justification point 2: Data are normally distributed (or looking for a significant difference between two means).

🧠 Exam Technique

When selecting statistical tests in AQA Biology, memorize the rule of thumb: Chi-squared is for categorized/frequency counts; Correlation coefficient is for relationships between two continuous variables; Student's t-test is for comparing means of two sets of data.

❌ Common Errors

Students often lose the second mark by writing "looking for a difference" instead of specifically stating they are comparing means or that the data are normally distributed.

Topics

Biology · Practical skills · 3.4 Genetic information, variation and relationships between organisms · Data analysis

Question and mark scheme from the AQA AS Level Biology examination, Paper 2, June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.