AQA AS Level Biology Paper 2, June 2025: Question 4

📐 Step-by-Step Calculation

1. Count the stomata in Figure 7:
   Carefully counting every pore/guard cell pair gives exactly 12 stomata.
2. Calculate radius (r) from diameter:
   Radius r = diameter ÷ 2 = 0.36 mm ÷ 2 = 0.18 mm
3. Calculate area of the circular field of view:
   Area = πr² = 3.14 × (0.18 mm)²
   Area = 3.14 × 0.0324 mm² = 0.101736 mm² (accepts 0.1017 – 0.102 mm²)
4. Calculate stomatal density:
   Density = Number of stomata ÷ Area
   Density = 12 ÷ 0.101736 ≈ 117.95 stomata mm⁻²
5. Apply required significant figures (2 s.f.):
   117.95 rounded to 2 significant figures = 120

✅ Correct Answer & Mark Scheme

120 stomata mm⁻² (3 marks)

🧠 Exam Technique

• Cross off as you count: Mark each stoma with a pen or pencil directly on Figure 7 so you do not double-count or miss any around the periphery.
• Keep full precision in calculator: Store the area value in your calculator memory and divide 12 by the unrounded value before rounding only at the very final step to 2 s.f.

❌ Common Errors

• Using diameter instead of radius: Forgetting to divide 0.36 mm by 2 before squaring leads to an area 4× too large.
• Rounding too early: Rounding 0.101736 to 0.1 mm² introduces substantial rounding error into the final quotient.
• Ignoring significant figure instructions: Leaving the answer as 117.95 or 118 loses the third mark.

✅ Mark Scheme Requirement

Count the number of stomata in many parts of the leaf AND calculate a mean/average.

💡 Key Knowledge

Reliability in biological sampling requires:

• Representative sampling: Stomata are not distributed perfectly uniformly across a leaf epidermis.
• Replication & mean: Taking multiple samples allows calculation of a mean, reducing the effect of anomalous fields of view.

❌ Examiner Pitfall

• Incomplete statement: Writing only "look at more areas" scores 0. You must state both repeating/sampling many parts and calculating a mean.
• Wrong level of replication: Suggesting "use more leaves" or "test different plants" changes the sample material rather than finding a reliable value for this leaf.

✅ Any ONE of the following paired statements (1 mark):

• Feature: Hairs
  How: 'Traps' water vapour / creates a layer of humid air and decreases / reduces the water potential gradient (between inside and outside of leaf).
• Feature: Stomata in pits / sunken stomata / stomatal grooves
  How: 'Traps' water vapour / reduces air movement / maintains high humidity and decreases / reduces the water potential gradient.
• Feature: Thick waxy cuticle
  How: Increases diffusion distance (for water vapour).
• Feature: Thick waxy cuticle
  How: Prevents / reduces evaporation / transpiration (since wax is waterproof/impermeable).

💡 Key Knowledge: The Transpiration Gradient

Water moves down a water potential (or concentration) gradient from high water potential inside air spaces (ψ ≈ 0) to lower water potential in ambient air.

• Trapping moist air next to stomata reduces this gradient, drastically slowing the rate of diffusion.
• The cuticle is composed of hydrophobic cutin/wax, which creates an impermeable barrier to water molecules.

❌ Common Errors

• "Traps water": Examiners reject "traps water" or "traps moisture" alone. It must be specified as water vapour or humid/moist air.
• Stating "stops all water loss": Adaptations reduce water loss, they never stop it completely, as stomata must open for CO₂ uptake.
• Missing the water potential link: Merely stating "hairs trap air" without mentioning that this reduces the water potential (or diffusion) gradient fails to gain full credit in questions worth explanation marks.

✅ Any TWO of the following points (2 marks max):

1. As wind speed increases, the rate of transpiration increases, but at a slower / decreasing rate (plateaus / levels off).
2. Rate of transpiration / evaporation is higher in the non-xerophyte than in the xerophyte (at all wind speeds).
   OR Higher wind speed has a greater effect on the non-xerophyte than the xerophyte.
3. Some transpiration / evaporation occurs in still air / at zero wind speed (in both plants).

🧠 Exam Technique for Graphical Conclusions

• Look at the intercept (x = 0): Neither curve starts at the origin (0,0); both have a positive y-intercept, meaning transpiration occurs even with no wind.
• Look at the comparative heights: The non-xerophyte curve is consistently above the xerophyte curve across the entire range.
• Describe the gradient change: Both curves show a steep initial rise that gradually flattens out (rate of increase slows).

❌ Common Errors

• Writing "directly proportional": The relationship is non-linear; describing it as "directly proportional" is a severe biological error.
• Vague comparisons: Stating "wind affects transpiration" without specifying direction (increases) or comparison between the two plant types.