Biology HL

1. 1.

   Cell X has ribosomes, a plasma membrane and circular DNA not enclosed by a nuclear envelope. Cell Y has a nucleus and mitochondria. Identify the cell types and give two structures they share with a function for each, explaining why lack of a nucleus does not mean X lacks genetic information.

   [4 marks] · no calculator

   Worked explanation

   1. Classify using the location of DNA and presence of membrane-bounded organelles rather than cell size alone.
   2. Choose any two shared structures and their functions, for example ribosomes for protein synthesis and a plasma membrane for selective exchange; cytoplasm as the medium for metabolic reactions is also valid.
   3. Nucleus and DNA are not synonyms: prokaryotic DNA occupies a non-membrane-bounded region.

   Common mistake: A prokaryote has genetic material; absence of a nuclear envelope is not absence of DNA.

   Marking points

   • X is prokaryotic and Y is eukaryotic.
   • One valid shared structure named with a function, such as ribosomes for protein synthesis, plasma membrane for a selective boundary, or cytoplasm as the medium for metabolic reactions.
   • A second distinct valid shared structure with a function; accept any two appropriate examples, not only ribosomes and membrane.
   • X contains DNA even though it is not enclosed in a nucleus.

   Examiner tip: A prokaryote has genetic material; absence of a nuclear envelope is not absence of DNA.

2. 2.

   A micrograph shows a cell 36 mm long at a magnification of x1200. Calculate its actual length in micrometres. Explain the difference between increasing magnification and improving resolution, and why cell length alone cannot establish whether the cell is prokaryotic.

   [4 marks]

   Worked explanation

   1. Divide image length by magnification to obtain actual length in the same units.
   2. Multiply 0.030 mm by 1000 micrometres per mm to obtain 30 micrometres.
   3. Use structural evidence, not a size cutoff; a larger blurred image has not necessarily gained resolution.

   Common mistake: Multiplying by magnification reverses the relationship and produces an implausibly huge cell.

   Marking points

   • Actual length = 36/1200 = 0.030 mm.
   • 0.030 mm = 30 micrometres.
   • Magnification enlarges an image; resolution is the ability to distinguish nearby points as separate, so enlargement alone need not reveal more detail.
   • Cell-size ranges overlap; nuclear/organellar or genetic organization is needed for secure classification.

   Examiner tip: Multiplying by magnification reverses the relationship and produces an implausibly huge cell.

3. 3.

   Mitochondria contain DNA and ribosomes and divide within cells. Explain how these observations support endosymbiosis, give one further relevant structural feature, and explain why no single feature alone proves the historical explanation.

   [4 marks] · no calculator

   Worked explanation

   1. Treat each observation as evidence consistent with an ancestral cell rather than claiming modern mitochondria live independently.
   2. A surrounding membrane system and an internal lineage of division fit engulfment followed by integration.
   3. Historical explanations are evaluated by converging evidence, not logically proved by one resemblance.

   Common mistake: Modern mitochondria depend on nuclear genes; ancestral independence does not imply present-day independence.

   Marking points

   • Internal DNA and ribosomes are consistent with descent from a once independently functioning prokaryote.
   • Division of existing mitochondria is consistent with inheritance of an engulfed cellular lineage.
   • A double membrane is a further feature consistent with engulfment/endosymbiosis.
   • Individual features can admit alternative explanations; multiple independent lines, especially molecular comparisons, strengthen the inference.

   Examiner tip: Modern mitochondria depend on nuclear genes; ancestral independence does not imply present-day independence.

4. 4.

   A cell measures 48 mm across in a printed micrograph. A 12 mm scale bar represents 3.0 micrometres. Calculate the cell's actual width and the magnification of the printed image. Explain why the scale bar remains useful after uniform resizing of the entire image.

   [4 marks]

   Worked explanation

   1. Use the scale bar as a ratio before converting units: the cell is four times its length.
   2. For magnification, convert the printed 12 mm bar to 12000 micrometres; magnification is dimensionless.
   3. A copied numeric magnification becomes invalid when image size changes, but an embedded bar scales together with the specimen.

   Common mistake: Do not divide millimetres by micrometres without conversion, or assume a printed magnification stays valid after resizing.

   Marking points

   • The cell spans 48/12 = 4 scale-bar lengths.
   • Actual width = 4 x 3.0 = 12 micrometres.
   • Magnification = 12000 micrometres / 3.0 micrometres = 4000.
   • Uniform resizing changes cell image and bar lengths by the same factor, preserving their ratio, although numerical magnification changes.

   Examiner tip: Do not divide millimetres by micrometres without conversion, or assume a printed magnification stays valid after resizing.

5. 5.

   A hypothetical cell contains membrane-bound nuclei and mitochondria. Compare its DNA organisation with that of a typical bacterium, and explain how mitochondrial DNA provides evidence consistent with endosymbiosis without proving it from this observation alone.

   [4 marks] · no calculator

   Worked explanation

   1. Compare the main genomic compartment rather than assuming all DNA in a eukaryotic cell is inside its nucleus.
   2. Separate the typical bacterial pattern from exceptions instead of making a universal claim about all prokaryotes.
   3. An inherited organellar genome fits endosymbiosis, but a historical explanation is strengthened by converging evidence rather than one resemblance.

   Common mistake: Bacteria do contain DNA; absence of a membrane-bound nucleus is not absence of genetic material.

   Marking points

   • The eukaryotic nuclear DNA is organised in linear chromosomes enclosed by a nuclear envelope.
   • A typical bacterium has its main circular DNA molecule in a nucleoid without a nuclear envelope; additional plasmids may occur.
   • Mitochondria possess their own DNA, commonly circular, consistent with descent from an endosymbiotic prokaryote.
   • Multiple independent lines of evidence, such as ribosomes, division and sequence relationships, are needed; DNA presence alone is not decisive.

   Examiner tip: Bacteria do contain DNA; absence of a membrane-bound nucleus is not absence of genetic material.

Original, unofficial practice questions from SubjectScout — not official past-paper questions or complete syllabus coverage. Added English teaching explanations are AI-authored; independent teacher review is pending.