Skip to content

9700 · 18.1

Classification — practice questions

Practice and worked examples for 9700 Classification. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Three species of flightless birds, the Ostrich (Africa), the Rhea (South America), and the Emu (Australia), live on different continents but share similar morphological features (long legs, long neck, large body). Molecular analysis of the protein cytochrome c was performed. The Rhea was found to have 2 amino acid differences compared to the Ostrich, while the Emu had 6 amino acid differences compared to the Ostrich. How does this molecular evidence refine our understanding of their classification?

Show solution outline

This problem demonstrates how molecular data can override conclusions based on morphology.

  1. Initial Hypothesis (Morphology): Based on their similar body plans, one might assume the Ostrich, Rhea, and Emu are all very closely related. This similarity is likely due to convergent evolution as they adapted to similar grassland environments.
  2. Analysis of Molecular Data: Molecular evidence, like amino acid sequences, provides a more direct measure of genetic and evolutionary relatedness. Fewer differences in the sequence imply a more recent common ancestor.
  3. Interpretation:
    • Ostrich vs. Rhea: 2 amino acid differences.
    • Ostrich vs. Emu: 6 amino acid differences.
  4. Conclusion: The smaller number of differences between the Ostrich and the Rhea indicates they are more closely related to each other than either is to the Emu. The molecular data suggests that the Ostrich and Rhea share a more recent common ancestor. The Emu, despite its similar appearance, is more distantly related. This refines the classification by showing that the group of large, flightless birds is not as uniform as morphology suggests.

Worked example 2

A biologist discovers a new species of primate, Species X. To determine its closest relative, a 40 base-pair segment of a mitochondrial gene is sequenced. The number of base differences between Species X and three other known primates are recorded below:

  • Chimpanzee (Pan troglodytes): 2 differences
  • Gorilla (Gorilla gorilla): 5 differences
  • Orangutan (Pongo pygmaeus): 8 differences

Calculate the percentage similarity of Species X to each of the other primates and determine its closest living relative.

Show solution outline

To solve this, we will calculate the percentage similarity for each comparison. The species with the highest percentage similarity is the closest relative.

Formula: Percentage Similarity = ( (Total Bases - Number of Differences) / Total Bases ) * 100%

Step 1: Calculate Similarity with Chimpanzee

  • Total Bases = 40
  • Differences = 2
  • Matching Bases = 40 - 2 = 38
  • Percentage Similarity = (38 / 40) * 100% = 95.0%

Step 2: Calculate Similarity with Gorilla

  • Total Bases = 40
  • Differences = 5
  • Matching Bases = 40 - 5 = 35
  • Percentage Similarity = (35 / 40) * 100% = 87.5%

Step 3: Calculate Similarity with Orangutan

  • Total Bases = 40
  • Differences = 8
  • Matching Bases = 40 - 8 = 32
  • Percentage Similarity = (32 / 40) * 100% = 80.0%

Step 4: Conclusion Comparing the results:

  • Chimpanzee: 95.0%
  • Gorilla: 87.5%
  • Orangutan: 80.0%

Species X has the highest percentage sequence similarity (95.0%) with the Chimpanzee. Therefore, based on this molecular data, the Chimpanzee is the closest living relative of Species X.