Level 4 — ApplicationPlant Biology

Plant Biology

60 minutes50 marksprintable — key stays hidden on paper

Level 4 (Application: novel problems, no hints) Time limit: 60 minutes Total marks: 50

Answer all questions. Show reasoning. Use ...... for any calculations.


Question 1 — Water transport under stress (12 marks)

A botanist studies a tall tree on a hot, dry, windy afternoon. Sap in the xylem is under tension.

(a) The xylem sap column has an effective tension (negative pressure) of 1.5 MPa-1.5\text{ MPa} at the top of a 30 m30\text{ m} tree. Gravity alone contributes a pressure drop of approximately 0.01 MPa0.01\text{ MPa} per metre of height. Calculate the additional pressure drop (beyond gravity) that must be supplied by transpirational pull to sustain this tension, given that water enters the roots at +0.1 MPa+0.1\text{ MPa}. (4)

(b) Explain, using the cohesion–tension theory, why a single air bubble (embolism) introduced into one xylem vessel does not necessarily stop water transport in the whole stem. (4)

(c) Predict and justify what happens to the rate of transpiration if the wind suddenly stops but temperature stays high. (4)


Question 2 — Stomata and hormones (10 marks)

A researcher exposes two batches of leaves to identical light but sprays batch B with abscisic acid (ABA).

(a) Predict which batch shows greater transpiration and explain the mechanism of ABA action on guard cells at the level of ions and water. (5)

(b) The guard cells of the sprayed leaves are found to have lost turgor. Draw the logical link between loss of guard-cell turgor and stomatal aperture, referring to the unequal thickening of guard-cell walls. (5)


Question 3 — Translocation puzzle (10 marks)

In a labelling experiment, radioactive sucrose is applied to a mature source leaf. Isotope later appears in growing root tips and developing fruit but not in other mature leaves.

(a) Using the pressure-flow model, explain why sucrose moves toward root tips and fruit but not toward mature leaves. (6)

(b) A ring of bark (containing phloem) is removed from around the trunk below the source leaves. Predict where sugar accumulates and what happens to the roots over time. Justify. (4)


Question 4 — Tropisms and growth (10 marks)

A seedling is laid horizontally in the dark.

(a) State which tropism(s) operate and predict the direction of bending of the root and of the shoot. (3)

(b) Explain the differential response of root and shoot to auxin that produces these opposite bends. (4)

(c) Design a simple control experiment to show that the shoot bending is due to auxin redistribution and not simply to gravity acting mechanically on the tissue. (3)


Question 5 — Reproduction and flowering (8 marks)

(a) A "short-day" plant flowers only when nights exceed a critical length. A grower interrupts the middle of a long night with a brief flash of red light and finds flowering is prevented. Explain this result in terms of the photoperiodic mechanism. (4)

(b) After successful double fertilization in a flowering plant, name the ploidy and fate of (i) the zygote and (ii) the primary endosperm nucleus, and state what the ovule and ovary each become. (4)

Answer keyMark scheme & solutions

Question 1 (12)

(a) (4 marks) Gravity drop over 30 m: 0.01×30=0.30 MPa0.01 \times 30 = 0.30\text{ MPa}. (1) Total pressure change from root (+0.1+0.1 MPa) to top (1.5-1.5 MPa): ΔP=0.1(1.5)=1.6 MPa\Delta P = 0.1 - (-1.5) = 1.6\text{ MPa} (1) Additional (transpirational) drop beyond gravity: 1.60.30=1.3 MPa1.6 - 0.30 = 1.3\text{ MPa} (1) So transpirational pull must supply 1.3 MPa\approx 1.3\text{ MPa} of the tension. (1)

(b) (4 marks)

  • Xylem consists of many parallel vessels/tracheids, not a single tube (1).
  • Water can move laterally through pits/bordered pits into adjacent functional vessels, bypassing the embolised one (1).
  • Cohesion (H-bonding) keeps the water columns in other vessels intact so tension is transmitted (1).
  • Redundancy of the network means overall flow continues, though slightly reduced (1).

(c) (4 marks)

  • Wind normally removes the humid boundary layer at the leaf surface (1).
  • When wind stops, water vapour accumulates near stomata, raising local humidity (1).
  • This reduces the water-vapour concentration gradient between leaf air spaces and outside air (1).
  • Therefore transpiration rate decreases (even though temperature is still high) (1).

Question 2 (10)

(a) (5 marks)

  • Batch A (no ABA) shows greater transpiration (1).
  • ABA triggers efflux of K+K^+ (and anions/malate) from guard cells (1).
  • Solute loss raises guard-cell water potential (makes it less negative) (1).
  • Water leaves the guard cells by osmosis (1).
  • Guard cells lose turgor → stomata close → transpiration falls in batch B (1).

(b) (5 marks)

  • Guard-cell walls are unevenly thickened: inner wall (facing pore) thick/rigid, outer wall thin/elastic (1).
  • When turgid, the thin outer wall stretches more, bowing the cells outward and opening the pore (1).
  • When turgor is lost, cells become flaccid and straighten (1).
  • The paired cells collapse together, closing the pore (1).
  • Hence loss of turgor → closed stoma → reduced water loss (1).

Question 3 (10)

(a) (6 marks)

  • Source (mature leaf) loads sucrose into phloem sieve tubes actively (1).
  • High solute lowers water potential at source → water enters from xylem → high hydrostatic pressure (1).
  • Sinks (root tips, fruit) unload sucrose; solute drops, water leaves, pressure falls (1).
  • A pressure gradient from source to sink drives bulk (mass) flow of phloem sap (1).
  • Mature leaves are sources, not sinks — they export, not import, so no net inflow to them (1).
  • Flow is thus directional toward regions of active growth/storage (roots, fruit) that are net sinks (1).

(b) (4 marks)

  • Sugar accumulates in the bark/phloem above the ring (near the source) (1) because downward transport is blocked (1).
  • Roots below the ring are starved of sugar (1) and eventually die/stop growing; tree may die over time (ringing/girdling effect) (1).

Question 4 (10)

(a) (3 marks)

  • Gravitropism (geotropism) operates (1) (phototropism excluded—it is dark).
  • Root bends downward → positively gravitropic (1).
  • Shoot bends upward → negatively gravitropic (1).

(b) (4 marks)

  • Auxin accumulates on the lower side of the horizontal organ (1).
  • In the shoot, high auxin promotes cell elongation → lower side grows faster → shoot bends up (1).
  • In the root, the same high auxin concentration on the lower side inhibits elongation (roots are sensitive to lower auxin levels) (1).
  • So the upper side of the root elongates more → root bends down (1).

(c) (3 marks)

  • Control: place a mica/impermeable barrier vertically down the middle of the shoot tip (or remove tip), preventing lateral auxin movement (1).
  • Keep everything else (gravity, orientation) identical (1).
  • If bending is abolished/altered, the response depends on auxin redistribution, not mere mechanical pull of gravity (1).

Question 5 (8)

(a) (4 marks)

  • Photoperiod is measured by phytochrome, which detects continuous night length (1).
  • Short-day plants need an uninterrupted long night to convert enough PfrP_{fr} back to PrP_r (flowering form) (1).
  • A red-light flash converts PrPfrP_r \to P_{fr}, effectively breaking the night into two short nights (1).
  • Sufficient PfrP_{fr} remains to inhibit flowering, so flowering is prevented (1).

(b) (4 marks)

  • (i) Zygote: diploid (2n2n) → develops into the embryo (1).
  • (ii) Primary endosperm nucleus: triploid (3n3n) → develops into endosperm (nutritive tissue) (1).
  • Ovule → becomes the seed (1).
  • Ovary → becomes the fruit (1).

[
  {"claim":"Gravity pressure drop over 30 m is 0.30 MPa","code":"result = (0.01*30 == 0.30)"},
  {"claim":"Total pressure change from root +0.1 to top -1.5 is 1.6 MPa","code":"result = (0.1-(-1.5) == 1.6)"},
  {"claim":"Additional transpirational drop beyond gravity is 1.3 MPa","code":"total=0.1-(-1.5); grav=0.01*30; result = (total-grav == 1.3)"},
  {"claim":"Zygote ploidy 2n and endosperm nucleus 3n differ by one genome set n","code":"n=symbols('n'); zygote=2*n; endosperm=3*n; result = (simplify(endosperm-zygote)==n)"}
]