Biotechnology Applications

Question 1

(a) (6 marks)

• E. coli cannot process eukaryotic proteins with signal peptides/introns; the A (21 aa) and B (30 aa) chains are cloned separately (1).
• Each chain fused to $\beta$-galactosidase to protect from proteolysis and aid expression/solubility (1).
• Fusion proteins expressed on separate plasmids/genes; the fusion partner cleaved chemically (CNBr at methionine) to release chains (1).
• Chains purified separately (1).
• A and B chains combined in vitro and oxidised to form correct disulfide bonds (1).
• Two interchain disulfides (A7–B7, A20–B19) plus one intrachain (A6–A11) give the mature, active hormone (1).

(b) (5 marks) Solution of $X = X_0 e^{\mu t}$ (1): $X(8) = 0.8\,e^{0.35\times 8} = 0.8\,e^{2.8} = 0.8 \times 16.4446 = 13.16\ \text{g L}^{-1}$ (2) Doubling time $t_d = \dfrac{\ln 2}{\mu} = \dfrac{0.6931}{0.35} = 1.98\ \text{h}$ (2).

(c) (6 marks) $\frac{dP}{dt}=q_p X = q_p X_0 e^{\mu t}$ (1) Integrate $0\to t$: $P(t) = \frac{q_p X_0}{\mu}\left(e^{\mu t}-1\right)$ (2, setup + integration). Evaluate at $t=8$: $P = \frac{0.045 \times 0.8}{0.35}\left(e^{2.8}-1\right) = \frac{0.036}{0.35}(16.4446-1)$ $= 0.102857 \times 15.4446 = 1.588\ \text{g L}^{-1}$ (3). (≈ 1.59 g of insulin per litre.)

(d) (4 marks) $0.35 = 0.40\frac{S}{0.5+S} \Rightarrow 0.35(0.5+S)=0.40S$ (1) $0.175 + 0.35S = 0.40S \Rightarrow 0.175 = 0.05S \Rightarrow S = 3.5\ \text{g L}^{-1}$ (2). Comment: fed-batch drip-feeds glucose so $S$ stays above $K_s$ (non-limiting for growth) yet low enough to avoid overflow metabolism (acetate/Crabtree) and substrate inhibition (1).

Question 2

(a) (6 marks)

• Rice endosperm makes geranylgeranyl diphosphate (GGPP) but the carotenoid pathway is switched off/absent there (1).
• Transgene 1: phytoene synthase (psy) — from daffodil (later maize in GR2) — converts GGPP → phytoene (2).
• Transgene 2: carotene desaturase (crtI) from Erwinia uredovora — carries out the desaturation steps converting phytoene → lycopene (2).
• Endogenous cyclases convert lycopene → $\beta$-carotene (provitamin A), giving golden colour (1).

(b) (4 marks) Required $\beta$-carotene $= 300 \times 12 = 3600\ \mu g$/day (1). Rice mass $= \dfrac{3600}{35} = 102.9\ \text{g/day} \approx 103\ \text{g}$ (3).

(c) (4 marks) Effective $\beta$-carotene per gram after losses: $35 \times 0.80 \times 0.70 = 19.6\ \mu g/g$ (2). Rice mass $= \dfrac{3600}{19.6} = 183.7\ \text{g/day} \approx 184\ \text{g}$ (2).

(d) (4 marks)

• $184\ \text{g} < 200\ \text{g}$, so a realistic diet CAN meet full RAE even with losses (2).
• Limitation (any one, 2): variable absorption depends on dietary fat; $\beta$-carotene degrades in storage; genetic/nutritional status affects conversion; single nutrient does not address broader malnutrition (protein, Fe, Zn).

Question 3

(a) (6 marks)

• Immunise mouse; harvest antigen-specific B-cells from spleen (1).
• Fuse B-cells with immortal myeloma cells (PEG) → hybridomas combining antibody production + immortality (2).
• Myeloma line is HGPRT⁻; HAT medium (hypoxanthine, aminopterin, thymidine) blocks de novo nucleotide synthesis so only fused cells (with B-cell HGPRT via salvage pathway) survive (2).
• B-cells alone are mortal/senescent and die out; unfused myelomas die in HAT — selection leaves only hybridomas, then single clones screened (1).

(b) (6 marks)

• Reproductive cloning: SCNT → embryo implanted → whole organism (genetic copy, e.g. Dolly) (1).
• Therapeutic cloning: SCNT → blastocyst → harvest embryonic stem cells for tissue repair (embryo destroyed) (1).
• iPSCs: adult somatic cells reprogrammed to pluripotency by factor expression — no embryo, no egg needed (1).
• Ethical advantage of iPSC: avoids destruction of embryos / no egg donation (1).
• Yamanaka factors: Oct4, Sox2, Klf4, c-Myc (2 — all four for full marks).

(c) (6 marks) Pseudocode (2):

function cells(t):          # t in hours
    N0 = 5e4
    Td = 18
    return N0 * 2 ** (t / Td)

Compute for $t = 72$ h (3 days) (2): $N = 5\times10^4 \times 2^{72/18} = 5\times10^4 \times 2^4 = 5\times10^4 \times 16$ $= 8.00\times10^5\ \text{cells}$ (2).

[
  {"claim":"Biomass after 8h ~13.16 g/L","code":"X=0.8*exp(0.35*8); result = abs(float(X)-13.156)<0.05"},
  {"claim":"Doubling time ~1.98 h","code":"td=ln(2)/Rational(35,100); result = abs(float(td)-1.980)<0.01"},
  {"claim":"Insulin produced ~1.588 g/L","code":"P=(0.045*0.8/0.35)*(exp(2.8)-1); result = abs(float(P)-1.588)<0.02"},
  {"claim":"Monod substrate S=3.5 g/L","code":"S=symbols('S'); sol=solve(Eq(0.35,0.40*S/(0.5+S)),S)[0]; result = abs(float(sol)-3.5)<1e-6"},
  {"claim":"Rice needed with losses ~183.7 g","code":"m=3600/(35*0.8*0.7); result = abs(float(m)-183.67)<0.5"},
  {"claim":"Hybridoma count after 72h = 8e5","code":"N=5e4*2**(72/18); result = abs(float(N)-800000)<1"}
]