Biology SL

1. 1.

   Marram grass lives on exposed sand dunes. Explain how a rolled leaf, hairs inside the rolled surface and sunken stomata can reduce water loss in this habitat.

   [4 marks] · no calculator

   Worked explanation

   1. Link the exposed, freely draining habitat to the risk of losing water faster than roots replace it.
   2. Rolling, hairs and recessed pores shelter air next to the stomata, reducing removal of water vapour.
   3. The more humid boundary layer reduces the vapour gradient; these structures reduce rather than abolish transpiration.

   Common mistake: Saying hairs absorb all the water needed misses their role in sheltering the humid air near stomata.

   Marking points

   • Exposed dunes experience wind and limited available water, favouring water conservation.
   • Rolling shelters the stomatal surface from moving air.
   • Hairs and sunken stomata retain a humid boundary layer near the pores.
   • Higher local humidity reduces the water-vapour concentration gradient and therefore transpiration.

   Examiner tip: Saying hairs absorb all the water needed misses their role in sheltering the humid air near stomata.

2. 2.

   A mangrove grows in waterlogged sediment with little oxygen. Some roots project above the sediment and have gas-permeable regions exposed at low tide. Explain how these features benefit living root cells and why submerged roots may still be oxygen-limited.

   [4 marks] · no calculator

   Worked explanation

   1. Abundant water does not imply abundant oxygen available to root cells.
   2. The projecting roots provide an air-entry route and internal aeration supports ATP production by aerobic respiration.
   3. At high tide the air-entry route may be covered, so the adaptation does not guarantee unlimited oxygen supply.

   Common mistake: Do not say roots photosynthesise to obtain oxygen; this task concerns gas entry and transport to respiring cells.

   Marking points

   • Waterlogging restricts oxygen supply through the sediment.
   • Exposed gas-permeable root regions allow oxygen to enter from air.
   • Internal air spaces can conduct oxygen towards living submerged root tissues for aerobic respiration.
   • When submerged, access to air is restricted and oxygen diffusion through water is relatively slow.

   Examiner tip: Do not say roots photosynthesise to obtain oxygen; this task concerns gas entry and transport to respiring cells.

3. 3.

   Hypothetical equal-age seedlings are grown for 14 days at soil salinities of 0, 5, 10 and 20 g dm^-3. Mean dry-mass gains are 0.8, 1.2, 0.6 and 0.0 g respectively. Other conditions are matched. Identify the best tested salinity, describe the supported tolerance pattern and explain why these results cannot establish the species' exact natural distribution.

   [4 marks] · no calculator

   Worked explanation

   1. Compare the four means rather than assuming a salt-adapted plant always grows faster at higher salinity.
   2. The optimum among tested treatments is 5 g dm^-3; an untested intermediate could perform better.
   3. A 14-day growth endpoint measures one life stage and response, not survival and reproduction throughout the natural habitat.

   Common mistake: A zero dry-mass gain is not evidence that every seedling died, and four treatments cannot locate an exact threshold.

   Marking points

   • 5 g dm^-3 gives the greatest tested mean gain, 1.2 g.
   • Growth is lower at both 0 and 10 g dm^-3 and no mean dry-mass gain occurs at 20 g dm^-3.
   • Zero growth does not by itself demonstrate death or establish the precise upper tolerance boundary.
   • Natural distribution also depends on other abiotic factors, biotic interactions and/or dispersal; one short seedling trial is insufficient.

   Examiner tip: A zero dry-mass gain is not evidence that every seedling died, and four treatments cannot locate an exact threshold.

4. 4.

   Consider a hypothetical reef-building coral with photosynthetic symbionts. For this exercise, sustained growth requires temperature 23-29 degrees C, salinity 32-37 g kg^-1 and at least 15% of surface light. Site A has 26 degrees C, 35 g kg^-1 and 25% light; B has 27 degrees C, 34 g kg^-1 and 5% light; C has 31 degrees C, 35 g kg^-1 and 30% light. Identify which site passes all supplied requirements, explain the constraint at each other site, and explain why passing these three conditions does not guarantee a reef will form. These ranges are invented for this organism, not universal coral limits.

   [4 marks] · no calculator

   Syllabus placement evidence: Biology guide, first assessment 2025, B4.1.5 Conditions required for coral reef formation; B4.1.3 Abiotic variables affecting species distribution. Applies B4.1.3 and B4.1.5 on abiotic tolerance and reef conditions through supplied organism-specific thresholds. Selected SL written practice, not field identification, universal coral thresholds, evidence of reef formation or complete statement-level coverage.

   Worked explanation

   1. Test each site's three values against the supplied intervals rather than ranking sites by one favourable measurement. B fails the light threshold and C fails the temperature ceiling.
   2. Coral growth depends on interacting conditions and persistence through time. The table screens possible habitat suitability for one hypothetical organism; it neither predicts community establishment nor establishes universal thresholds.

   Common mistake: Meeting a temperature range alone does not establish a suitable habitat, and the invented thresholds must not be reported as universal coral requirements.

   Marking points

   • Only A meets all three stated requirements.
   • B has insufficient light, potentially limiting symbiont photosynthesis despite suitable temperature and salinity.
   • C exceeds the stated temperature tolerance; abundant light cannot compensate for failure of that requirement.
   • Other conditions such as suitable substrate, pH, water clarity over time, recruitment and disturbance remain untested, so necessary conditions are not sufficient evidence of reef formation.

   Examiner tip: Meeting a temperature range alone does not establish a suitable habitat, and the invented thresholds must not be reported as universal coral requirements.

5. 5.

   In a hypothetical dune transect, ten equal quadrats at each of soil pH 5, 6 and 7 contain a target grass in 2, 8 and 5 quadrats respectively. Sampling effort is equal and identification is assumed correct. Calculate percentage occupancy at each pH and identify the highest observed association. Explain why absence from a quadrat is not proof of physiological intolerance, and propose a design change that would help distinguish a pH effect from position along the dune. No fieldwork has been performed here.

   [4 marks]

   Syllabus placement evidence: Biology guide, first assessment 2025, B4.1.3 Abiotic variables affecting species distribution; B4.1.4 Range of tolerance of a limiting factor, application of skills: transect data. Analyses supplied transect occupancy and abiotic association at SL under B4.1.3-B4.1.4. Written data interpretation does not fulfil the requirement to collect actual transect data in a natural or semi-natural habitat, nor certify complete statement-level coverage.

   Worked explanation

   1. Occupancy counts quadrats containing the species, not individual abundance: divide each count by ten and multiply by 100.
   2. A field association combines physiological tolerance with access, interactions and detection. Replication across locations with recorded covariates helps test whether the pH pattern persists; it does not automatically establish a causal pH response.

   Common mistake: Eighty percent occupied quadrats is not eighty percent of all individual plants, nor proof of an exact pH optimum.

   Marking points

   • Occupancies are 20%, 80% and 50% for pH 5, 6 and 7 respectively.
   • Highest observed occupancy is at pH 6, not proof that the exact physiological optimum is 6.
   • Absence may reflect limited dispersal, competition or missed plants rather than inability to tolerate that pH.
   • Repeat independently placed transects or sample matched pH values across positions while recording moisture/exposure, reducing confounding between pH and dune location.

   Examiner tip: Eighty percent occupied quadrats is not eighty percent of all individual plants, nor proof of an exact pH optimum.

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.