Organelles & Their Functions

Question 1

(a) (8 marks) — 1 mark per correct compartment+event, in order:

1. Nucleus — gene transcribed to pre-mRNA; splicing → mature mRNA exits nuclear pore.
2. Ribosome (on RER) — translation begins; signal peptide directs ribosome to RER.
3. Rough ER lumen — polypeptide threaded in; signal peptide cleaved; N-linked glycosylation begins; folding (chaperones); disulfide bonds.
4. RER exit → COPII vesicle — buds off toward Golgi.
5. Golgi (cis face) — receives vesicle.
6. Golgi (medial/trans) — carbohydrate trimming/modification, terminal glycosylation, sorting.
7. Secretory (trans-Golgi) vesicle — protein packaged.
8. Plasma membrane — vesicle fuses (exocytosis) → protein secreted. (Award marks for correct order and correct event; accept ~8 named stages.)

(b) (4 marks)

• RER carries membrane-bound ribosomes and lumenal enzymes (oligosaccharyltransferase) that add the core N-linked glycan co-translationally (2).
• Golgi lacks ribosomes but contains a cisternal gradient of glycosyltransferases/glycosidases that trim and add terminal sugars, plus sorting machinery — so final modification and address-tagging happen here (2).

(c) (4 marks)

• $T = n \cdot t$ (1 mark for linear model).
• $T = 4 \times 45 = 180\text{ s}$ (1).
• $= 180/60 = 3\text{ minutes}$ (2).

(d) (4 marks)

• COPII coats vesicles moving anterograde (ER → Golgi); COPI coats retrograde (Golgi → ER) recycling vesicles (2).
• Directionality arises because coat recruitment, GTPase cycling (Sar1/Arf), and SNARE specificity bias net flux forward; pure random diffusion has no net direction (∆G-driven bias absent), so no sustained net secretion could occur — the system requires energy (GTP hydrolysis) to enforce vectoriality (2).

Question 2

(a) (6 marks)

• Structure: double membrane; smooth outer membrane; folded inner membrane forming cristae; intermembrane space; matrix containing mtDNA, ribosomes, enzymes (3).
• Cristae increase inner-membrane surface area → more ETC complexes and ATP synthase → more ATP (1).
• Ratio = surface-area factor = $5c/c = 5:1$ (2). ATP output scales with membrane area × complex density; density constant so ratio = 5.

(b) (6 marks) Chemical term: $-2.303\,RT\,\Delta pH = -2.303 \times 8.314 \times 310 \times 1.4$ $= -2.303 \times 8.314 \times 310 = -5936.6$; $\times 1.4 = -8311\text{ J/mol} \approx -8.31\text{ kJ/mol}$ (2). Electrical term: $-F\Delta\psi = -96485 \times (-0.15) = +14472.75\text{ J/mol} \approx +14.47\text{ kJ/mol}$ (2).

Wait — sign convention: with $\Delta\psi = -0.15$ V (inside negative), $-F\Delta\psi = +14.47$ kJ. Total: $\Delta G = -8.31 + 14.47 = +6.16\text{ kJ/mol}$.

Because the given $\Delta pH$ and $\Delta\psi$ signs partially oppose, net $\Delta G \approx +6.2\text{ kJ/mol}$ (2). (Full marks for correct substitution and separated terms; accept the arithmetic outcome $\approx +6.2$ kJ/mol with the stated sign convention. Note: physiologically the driving force is written with proton influx favourable; award marks for method + correct magnitudes.)

(c) (6 marks) Analogies (2): both have double membranes; both contain own circular DNA & 70S ribosomes; both make ATP via chemiosmosis / ETC on an internal membrane. (any two) Differences (2): mitochondrion has cristae vs chloroplast has thylakoids/grana + stroma; chloroplast has three membrane systems (adds thylakoid) & pigments (chlorophyll); mitochondrion consumes O₂, chloroplast releases O₂. Endosymbiosis (2): inner membrane = original prokaryote plasma membrane; outer membrane = host phagocytic/vacuolar membrane engulfing it → double membrane predicted for both.

(d) (4 marks) Free energy from 4 protons $= 4 \times |\Delta G_{per\ proton}|$. Using magnitude of proton-motive free energy (taking the favourable-influx convention, energy released per proton ≈ 6.2 kJ… but for adequacy check use standard pmf ≈ 20 kJ/mol × 4).

Using the computed magnitude $6.16$ kJ/mol: $4 \times 6.16 = 24.6\text{ kJ}$ — less than 30.5 kJ needed → insufficient by this dataset (2). (If students use realistic pmf ≈ 20 kJ/mol → $4\times20 = 80$ kJ > 30.5, sufficient.) Award marks for correct comparison and stated conclusion consistent with their (b) value (2).

Question 3

(a) (5 marks)

• 9+2: nine peripheral doublet microtubules surrounding a central pair of singlet microtubules; radial spokes and nexin links (3).
• Motor protein: dynein (ciliary/axonemal dynein) (1).
• Energy source: ATP hydrolysis by dynein drives sliding of doublets → bending (1).

(b) (6 marks) — 1 mark per correct cell:

(c) (7 marks) — valid decision tree reaching all four (4), biological justification of each test (3):

INPUT organelle
IF has_thylakoids(organelle):        # photosynthetic internal membranes
    RETURN "chloroplast"
ELSE:
    IF has_double_membrane_and_cristae(organelle):  # ETC + own DNA
        RETURN "mitochondrion"
    ELSE:
        IF internal_pH_acidic(organelle) AND has_hydrolases(organelle):
            RETURN "lysosome"        # acidic, digestive enzymes
        ELSE:
            RETURN "peroxisome"      # catalase, single membrane, no acid

Justification: thylakoids unique to chloroplast; cristae+double membrane distinguish mitochondrion; acidic pH + acid hydrolases define lysosome; remaining single-membrane organelle with catalase = peroxisome.

[
  {"claim":"Total transit time T = 3 minutes for n=4, t=45s","code":"n=4; t=45; T=n*t; result = (T==180) and (T/60==3)"},
  {"claim":"Chemical term of dG approx -8.31 kJ/mol","code":"R=8.314; T=310; dpH=1.4; chem=-2.303*R*T*dpH; result = abs(chem/1000 - (-8.31)) < 0.05"},
  {"claim":"Electrical term +F*0.15 approx +14.47 kJ/mol","code":"F=96485; dpsi=-0.15; elec=-F*dpsi; result = abs(elec/1000 - 14.47) < 0.05"},
  {"claim":"Net dG approx +6.16 kJ/mol with given signs","code":"R=8.314; T=310; F=96485; chem=-2.303*R*T*1.4; elec=-F*(-0.15); tot=(chem+elec)/1000; result = abs(tot-6.16)<0.1"},
  {"claim":"4 protons at 6.16 kJ each (24.6 kJ) < 30.5 kJ ATP requirement","code":"e=6.16; total=4*e; result = total < 30.5"}
]