Question 5 of 9: Ammonia and Oxygen Demand for Recombinant Protein Production
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2016 — 04-BS-13, Biology. Three-hour, closed-book exam (one double-sided aid sheet permitted, approved calculator allowed). Format: Part I offers 5 questions (any 3 constitute a complete answer, 20 marks each; Q2 itself offers two alternative sub-problems) and Part II offers 3 questions (any 2 constitute a complete answer, 20 marks each) — a full paper is 5 questions. All 8 numbered questions (with both alternatives of Q2) are solved below for completeness. Q2–Q7 are calculation/derivation questions; Q1, Q6(a)(b)(d), and Q8 are essay questions.
Reference texts: Shuler & Kargi, Bioprocess Engineering: Basic Concepts (2nd ed., Prentice Hall) — elemental/electron balances, yield coefficients, fermenter mass and energy balances, growth kinetics; Madigan et al., Brock Biology of Microorganisms (15th ed., Pearson) — bacterial/viral structure, rapid methods, MPN; Toledo, Fundamentals of Food Process Engineering (3rd ed., Springer) — plant tissue structure and mechanical properties.
Question 4: Ammonia and Oxygen Demand for Recombinant Protein Production (20 marks)
Find. (a) ammonia demand (g/g glucose); (b) O2 demand (g/g glucose); (c) both requirements at $f=0$ (wild type) and the % difference.
Approach. Because both biomass and protein formulas are already normalized to 1 carbon atom (C1 formulas), their coefficients $c,f$ are directly obtainable from the mass yields (cmol = mol here). The N-balance then gives $b$ directly; the degree-of-reduction (electron) balance gives $a$ without needing the full C/H/O atom balances. Part (c) repeats both balances with $f=0$, holding $c$ (and hence $Y_{xs}$) fixed as stated.
Convert mass yields to molar coefficients (per mole glucose).
$$c=\frac{Y_{xs}\,\text{MW}_{glu}}{\text{MW}_X}=\frac{0.48(180.2)}{25}=3.460,\qquad f=\frac{Y_{ps}\,\text{MW}_{glu}}{\text{MW}_P}=\frac{0.096(180.2)}{22.03}=0.785\ \text{mol/mol glucose}.$$
Part (a): nitrogen balance. NH3 is the only N source, split between biomass and protein:
$$b=0.24c+0.25f=0.24(3.460)+0.25(0.785)=\boxed{1.027\ \text{mol NH}_3\text{/mol glucose}}.$$
Converting to mass per gram of glucose (MW NH3 = 17):
$$\frac{1.027(17)}{180.2}=\boxed{0.0969\ \text{g NH}_3\text{/g glucose}}.$$
Part (b): degree-of-reduction (electron) balance. Biomass degree of reduction $\gamma_B=4.07$ (given); protein's is computed the same way, $\gamma_P=4(1)+1.55-2(0.31)-3(0.25)=4.18$. The substrate's total available electrons ($\gamma_S\times 6$ C-atoms per mole glucose) split between biomass, protein, and O2:
$$\gamma_S(6)=\gamma_Bc+\gamma_Pf+4a \;\Rightarrow\; a=\frac{4.0(6)-4.07(3.460)-4.18(0.785)}{4}=\boxed{1.659\ \text{mol O}_2\text{/mol glucose}}.$$
In mass terms (MW O2 = 32): $a(32)/180.2=\boxed{0.2946\ \text{g O}_2\text{/g glucose}}$.
Part (c): wild-type strain, $f=0$, $Y_{xs}$ unchanged. $c$ is unchanged at 3.460 (same biomass yield). Nitrogen balance: $b_{wt}=0.24c=0.24(3.460)=0.830$ mol/mol glucose $\Rightarrow \dfrac{0.830(17)}{180.2}=\boxed{0.0783\ \text{g NH}_3\text{/g glucose}}$. Electron balance: $4a_{wt}=\gamma_S(6)-\gamma_Bc=24-4.07(3.460)=9.918\Rightarrow a_{wt}=2.480$ mol/mol glucose $\Rightarrow \dfrac{2.480(32)}{180.2}=\boxed{0.4403\ \text{g O}_2\text{/g glucose}}$.
Compare. Relative to the wild type, the recombinant strain needs
$$\frac{0.0969-0.0783}{0.0783}=\boxed{+23.6\%\ \text{more ammonia}}\quad\text{(extra N for the protein)},$$
$$\frac{0.2946-0.4403}{0.4403}=\boxed{-33.1\%\ \text{less oxygen}}$$
because, with the biomass yield held fixed, the protein is additional product: its carbon comes out of what the wild type would respire to CO2 (zero available electrons), so the $\gamma_Pf=4.18(0.785)=3.28$ mol of available electrons it captures are no longer discharged to O2 — any product with $\gamma_P>0$ lowers the O2 demand in this comparison (carbon check: CO2 falls from 2.540 to 1.755 mol/mol glucose).
