Plant Biology

Question 1

(a) (5)

• Xylem consists of dead, hollow tracheids/vessel elements with lignified walls forming continuous open tubes — no cytoplasm to obstruct flow (1); lignin gives mechanical strength to resist the negative pressure (1). Phloem is living and used for sugar transport, not bulk water ascent (1).
• Cohesion: H-bonding between water molecules holds the column together as a continuous thread (1). Adhesion of water to hydrophilic xylem walls + tension generated by evaporation at leaf air spaces pulls the whole column up (1).

(b) (3) $\Delta P_{grav}=\rho g h = 1000\times9.81\times100 = 9.81\times10^{5}\ \text{Pa} = 0.981\ \text{MPa}$

• Correct formula (1), substitution (1), answer ≈ 0.98 MPa (1).

(c) (4)

• Total water potential difference driving flow: $\Delta\Psi = \Psi_{root}-\Psi_{leaf} = (-0.3)-(-1.5) = +1.2\ \text{MPa}$ (2)
• This 1.2 MPa gradient (leaf more negative) drives water upward. Since 1.2 MPa > 0.98 MPa required to counter gravity (part b), there remains ~0.22 MPa to overcome frictional resistance — cohesion–tension can account for ascent to 100 m (2).

(d) (6) $Q=\frac{\pi r^4 \Delta P}{8\eta L}$

• $r^4=(20\times10^{-6})^4 = (2\times10^{-5})^4 = 1.6\times10^{-19}\ \text{m}^4$ (1)
• Numerator: $\pi\times1.6\times10^{-19}\times1.0\times10^{6} = \pi\times1.6\times10^{-13}=5.027\times10^{-13}$ (1)
• Denominator: $8\times1.0\times10^{-3}\times100 = 0.8$ (1) $Q = \frac{5.027\times10^{-13}}{0.8}=6.28\times10^{-13}\ \text{m}^3\text{s}^{-1}$ (2)
• Comment: one vessel carries a tiny flow (~$6\times10^{-13}\,\text{m}^3\text{s}^{-1}$); plants bundle thousands of parallel vessels so total flow adds — and because $Q\propto r^4$, small radius increases resistance sharply, favouring many narrow safe vessels over few wide ones (cavitation trade-off) (1).

(e) (4)

• Under high tension the water thread can break, forming a gas bubble = cavitation/embolism, breaking continuity and blocking that vessel (2).
• Adaptation: bordered pits between vessels confine the embolism and let water bypass through adjacent conduits; narrow vessels / pit membranes resist air-seeding (2).

Question 2

(a) (6)

• Blue light activates H⁺-ATPase → protons pumped out of guard cells → membrane hyperpolarises → K⁺ influx through voltage-gated channels, plus malate/Cl⁻ accumulation (2).
• Lowered solute potential draws water in by osmosis → guard cells become turgid; unevenly thickened inner walls make cells bow apart → pore opens (2).
• ABA (drought signal) triggers efflux of K⁺/anions → water leaves → guard cells lose turgor → stomata close, conserving water (2).

(b) (4) — critical night length = 9 h; SDP flowers when continuous dark > 9 h.

• (i) 10 h dark > 9 h → flowers (1)
• (ii) 8 h dark < 9 h → no flowering (1)
• (iii) 9 h dark but interrupted by red flash splits it into <9 h continuous segments → no flowering (2)

(c) (4)

• Phytochrome exists as $P_r$ (absorbs red) and $P_{fr}$ (absorbs far-red, biologically active) (1).
• Red flash converts $P_r\to P_{fr}$; the presence of active $P_{fr}$ during the night is "read" as daylight, resetting the dark measurement (2).
• Red/far-red reversibility: a subsequent far-red flash converts $P_{fr}\to P_r$, cancelling the effect → flowering restored, proving phytochrome control (1).

(d) (6) Reset logic: dark timer accumulates hours; each red flash resets the continuous dark counter to zero, so the longest uninterrupted dark run must exceed critical length.

def will_flower(light_h, dark_h, flashes, plant_type, critical=9):
    # flashes: list of times (hours into the dark period) that reset dark timer
    # Determine the longest continuous dark stretch
    if not flashes:
        longest_dark = dark_h
    else:
        pts = [0] + sorted(flashes) + [dark_h]
        longest_dark = max(pts[i+1] - pts[i] for i in range(len(pts)-1))
 
    if plant_type == "short_day":
        return longest_dark >= critical      # needs long uninterrupted night
    elif plant_type == "long_day":
        return longest_dark < critical       # needs short night
    else:
        raise ValueError("unknown plant_type")

• Correct longest-continuous-dark computation (3); correct SDP vs LDP condition (2); flash-reset handled (1).

Check: (i) SDP, dark 10, no flash → 10≥9 → True ✓; (iii) 9 h dark, flash at 4.5 → longest = 4.5 < 9 → False ✓.

Question 3

(a) (5)

• At the source (mature leaf), sugars are actively loaded into sieve tubes → lowers $\Psi_s$ → water enters from xylem osmotically → high hydrostatic (turgor) pressure (2).
• At the sink (root/fruit), sugars are unloaded → water leaves → low pressure (1).
• The resulting pressure gradient drives bulk (mass) flow of phloem sap from source to sink through sieve tubes (1). Flow direction depends only on source–sink relationship, hence bidirectional in the plant overall (1).

(b) (5)

• Shaded side auxin = 57 units, lit = 43 units (1).
• $E_{shaded}=k\cdot 57 = 0.02\times57 = 1.14\ \text{mm h}^{-1}$; over 2 h → $2.28\ \text{mm}$ (1.5)
• $E_{lit}=0.02\times43 = 0.86\ \text{mm h}^{-1}$; over 2 h → $1.72\ \text{mm}$ (1.5)
• Difference $\Delta L = 2.28-1.72 = 0.56\ \text{mm}$ (1)

(c) (4) $\theta \approx \frac{\Delta L}{w} = \frac{0.56}{2}=0.28\ \text{rad}$ (2) $\theta = 0.28\times\frac{180}{\pi}=16.0^\circ$ (1)

• Assumption: the lit side acts as the (near-inextensible) inner arc of radius ≈ $w$, bending is uniform, small-angle/arc approximation holds (1).

(d) (4)

• In shoots, auxin accumulating on the shaded side promotes cell elongation → shaded side grows faster → bends toward light (positive phototropism) (2).
• In roots, auxin accumulating on the lower side (gravitropism) inhibits elongation (roots are far more auxin-sensitive; high auxin is inhibitory) → upper side grows faster → root bends downward (positive gravitropism) (2). Same signal, opposite dose–response.

[
  {"claim":"Hydrostatic support pressure over 100 m is ~0.981 MPa","code":"rho=1000; g=9.81; h=100; dP=rho*g*h; result = abs(dP-9.81e5) < 1"},
  {"claim":"Available water potential difference 1.2 MPa exceeds gravity term 0.98 MPa","code":"dpsi=(-0.3)-(-1.5); grav=1000*9.81*100/1e6; result = (abs(dpsi-1.2)<1e-9) and (dpsi>grav)"},
  {"claim":"Poiseuille flow Q approx 6.28e-13 m^3/s","code":"import sympy as sp; r=20e-6; dP=1.0e6; eta=1.0e-3; L=100; Q=sp.pi*r**4*dP/(8*eta*L); result = abs(float(Q)-6.283e-13) < 1e-15"},
  {"claim":"Auxin elongation difference 0.56 mm and bend angle 0.28 rad","code":"k=0.02; t=2; Es=k*57*t; El=k*43*t; dL=Es-El; theta=dL/2; result = (abs(dL-0.56)<1e-9) and (abs(theta-0.28)<1e-9)"}
]