B3.1 Gas exchange. Practice questions with markscheme.

• D: paired guard cells flank each stoma; turgor changes open (turgid) or close (flaccid) the pore, balancing CO₂ uptake against transpiration;

• D: in quiet exhalation the diaphragm and external intercostals relax, so the diaphragm domes upwards and the ribs fall, reducing the volume of the thorax and raising the pressure in the lungs above atmospheric, so air flows out; at the end of each inhalation and exhalation air flow stops because the pressures are equal;

• A: vital capacity = tidal volume + inspiratory reserve + expiratory reserve = 0.5 + 2.7 + 1.1 = 4.3 dm³; during exercise, expiratory reserve = 4.3 − 2.1 − 1.6 = 0.6 dm³ (the larger tidal volume eats into both reserves);

• (a) [1]
  • 16 mm;

  Accept 15.5 to 17 mm.

• (b) [3]
  • as the larva grows its volume increases faster than its surface area, so its surface area-to-volume ratio falls;
  • the oxygen needed depends on the volume/mass of respiring tissue, whereas uptake through the skin depends on the area of skin;
  • the distance from the skin to the innermost tissues increases, and diffusion is too slow over longer distances to supply them;
  • (so) the skin can no longer meet the demand and the gills, a large, thin exchange surface, take over, OWTTE;

  Do not accept 'larger larvae need more oxygen' without reference to surface area, volume or distance.

• (c) [1]
  • the skin becomes thicker / gains more cell layers / develops scales, increasing the diffusion distance;
  • fewer/less dense blood capillaries just beneath the skin, so the concentration gradient is not maintained as well, OWTTE;

  Mark the first answer only.

• (a) [2]
  • area of one airway = π × (0.030 cm)² = 2.83 × 10⁻³ cm² / π × (0.30 mm)² = 0.283 mm²;
  • total = 65 536 × 2.83 × 10⁻³ cm² = 185 cm²;

  Accept 180 to 190 cm², or 1.85 × 10⁴ mm² with the unit stated. Award [1] for 741 cm² (diameter used as the radius). ECF for a correct method from one wrong value.

• (b) [2]
  • the number of airways doubles at every generation, so air is delivered to millions (over 8 million) of alveolar sacs;
  • (so) a very large total alveolar surface area is packed into the small volume of the chest;
  • every alveolus lies at the end of an airway, so all of the exchange surface is ventilated, OWTTE;
• (c) [2]
  • the same volume of air per second is spread over a total cross-sectional area over 4 000 times greater (11 100 cm² compared with 2.5 cm²), so it moves far more slowly;
  • over the very short distances inside the alveolar sacs, gases move quickly enough by diffusion, so fast air flow is not needed there, OWTTE;

• (a) [1]
  • (2.52 − 2.33) ÷ 2.52 × 100 = 7.5 %;

  Accept 7.54 %.

• (b) [2]
  • the lower surface;
  • coating the lower surface cut the loss in mass far more (to 7.5 %) than coating the upper surface did (34.7 %, close to the 35.8 % of the uncoated leaf);
  • water vapour leaves the leaf mainly by diffusion through the stomata, so the surface that loses more water has more stomata, OWTTE;

  The reason marks require 'lower surface'.

• (c) [1]
  • some water evaporates through the cuticle, which is not completely waterproof;
  • the coating or the wax seal on the leaf stalk was not perfect, so some water vapour escaped through gaps / unsealed stomata;
  • some mass is lost as carbon dioxide released by cell respiration, OWTTE;

  Mark the first answer only.

• (d) [2]
  • the rate of photosynthesis falls / is lower than in an uncoated leaf;
  • carbon dioxide enters through the stomata, the same pores through which water vapour is lost, and most of them are blocked by the jelly;
  • (so) less carbon dioxide reaches the air spaces and mesophyll cells, OWTTE;

  The prediction mark requires a reason.

• A: both volumes are 64 mm³; P: surface area = 2(8×4 + 8×2 + 4×2) = 112 mm², ratio 1.75 mm⁻¹; Q: surface area = 2(8×8 + 8×1 + 8×1) = 160 mm², ratio 2.5 mm⁻¹; the centre of Q is 0.5 mm from its upper and lower surfaces, whereas the centre of P is 1 mm from the nearest surface, so flattening raises the ratio and shortens the diffusion distance;

• C: ventilation with water and a continuous flow of blood keep bringing oxygen-rich water to the surface and carrying oxygenated blood away, so the gradient is never lost; A would abolish the gradient rather than maintain it; B fails because oxygen crosses by diffusion, not active transport; D fails because gills do not store oxygen;

• B: 12 stomata are visible; area of the field of view = π × (0.20 mm)² = 0.126 mm²; stomatal density = 12 ÷ 0.126 = 95 stomata mm⁻²;

• D: CO₂ (lowering pH) reduces haemoglobin's O₂ affinity, promoting release exactly where respiration is fastest;

• (a) [1]
  • (97 − 85) ÷ 97 × 100 = 12 %;

  Accept 12.4 %.

• (b) [3 max]
  • each haemoglobin molecule has four haem groups, each able to bind one oxygen molecule;
  • binding is cooperative: oxygen binding to one haem group changes the shape/conformation of the molecule, raising the affinity of the other haem groups for oxygen;
  • (so) at 7 kPa and above most of the haem groups are still occupied, because cooperative binding keeps the affinity high: the flat upper part (plateau) of the S-shaped curve, OWTTE;
  • large falls in saturation happen only below about 6 kPa, on the steep part of the curve, where losing some oxygen lowers the affinity of the remaining haem groups, OWTTE;

  Do not accept answers that only describe the shape of the curve without reference to haem groups or cooperative binding.

• (c) [2 max]
  • deeper, faster breathing increases ventilation, so the alveolar air is replaced with fresh air more often;
  • (so) the alveolar pO₂ rises closer to the 12 kPa of the air breathed in (for example from 7 to 8 kPa);
  • (so) the concentration gradient into the blood is steeper and the blood leaves the lungs at a higher pO₂, raising saturation (from 85 % to 90 % for 7 to 8 kPa), OWTTE;
• (d) [2]
  • at sea level 97 − 57 = 40 % of the oxygen capacity is released, whereas at 4 500 m 85 − 57 = 28 % is released;
  • (so) the blood releases 30 % less oxygen / 0.7 times as much at 4 500 m, a much larger fall than the 12 % fall in saturation, OWTTE;

  ECF from misread table values. Accept "about a third less" or "about two thirds as much".

• an upper and a lower epidermis, each a single protective cell layer covered by a (waxy) cuticle (thicker on the upper surface);
• palisade mesophyll of tightly packed column-shaped cells (with many chloroplasts) directly beneath the upper epidermis;
• spongy mesophyll of loosely packed (rounded) cells with (large) air spaces between them, below the palisade layer;
• veins/vascular bundles within the mesophyll (in the middle of the leaf), containing xylem (towards the upper side) and phloem (towards the lower side);
• guard cells forming stomata, mainly in the lower epidermis, OWTTE;

This asks for distribution only; explanations of gas exchange are credited under B3.1.7 questions instead.

• stomata (mostly on the lower epidermis) allow CO₂ in and O₂ out;
• guard cells open/close stomata, controlling exchange (and water loss);
• spongy mesophyll has (interconnected) air spaces, giving a large internal surface area;
• mesophyll cell walls are moist, so gases dissolve before diffusing in;
• thin/flat leaf shape keeps diffusion distances short;
• waxy cuticle prevents (uncontrolled) water loss elsewhere, OWTTE;