Stoichiometry of Growth & Product Formation Questions and Answers

This set of Bioprocess Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Stoichiometry of Growth and Product Formation”.

1. ”Bacteria have slightly higher nitrogen content than fungi”.
a) True
b) False
View Answer

Answer: a
Explanation: Bacteria tend to have slightly higher nitrogen contents (11-14%) than fungi (6.3-9.0%).

2. Estimate the degree of reduction of Methane, Glucose and Ethanol?
a) γ(CH₄) = 6, γ(C₆H₁₂O₆) = 4, γ(C₂H₅OH) = 6
b) γ(CH₄) = 6, γ(C₆H₁₂O₆) = 4, γ(C₂H₅OH) = 8
c) γ(CH₄) = 8, γ(C₆H₁₂O₆) = 4, γ(C₂H₅OH) = 6
d) γ(CH₄) = 4, γ(C₆H₁₂O₆) = 6, γ(C₂H₅OH) = 8
View Answer

Answer: c
Explanation:

The degree of reduction of Methane, Glucose & Ethanol is 6

3. Calculate the stoichiometric coefficients of the following biological reaction:
Hexadecane: C₁₆H₃₄ + a O₂ + b NH₃ = c (C_4.4H_7.3N_0.86O_1.2) + d H₂O + e CO₂
a) a = 12.427, b = 2.085, c = 2.42, d = 12.43, e = 5.33
b) a = 12.345, b = 3.456, c = 2.42, d = 12.46, e = 5.44
c) a = 12.594, b = 2.345, c = 3.42, d = 12.49, e = 5.66
d) a = 12.345, b = 3.560, c = 3.42, d = 12.46, e = 5.44
View Answer

Answer: a
Explanation: Amount of carbon in 1 mole of substrate = 16 (12) = 192 g
Amount of carbon converted to biomass = 192 (2/3) = 128 g
Then, 128 = c (4.4)(12); c = 2.42
Amount of carbon converted to CO₂ = 192 – 128 = 64 g
64 = e (12)
e = 5.33
The nitrogen balance yields
14b = c (0.86)(14)
b = (2.42)(0.86) = 2.085
The hydrogen balance is
34 (1) + 3b = 7.3c + 2d
d = 12.43
The oxygen balance yields
2a(16) = 1.2c(16) + 2e(16) + d(16)
a = 12.427.

4. Calculate the stoichiometric coefficients of the following biological reaction:
Glucose: C₆H₁₂O₆ + aO₂ + bNH₃ = c (C_4.4H_7.3N_0.86O_1.2) + d H₂O + e CO₂
a) a = 1.573, b = 0.685, c = 0.470, d = 2.564, e = 2
b) a = 2.789, b = 1.896, c = 0.438, d = 1.395, e = 1
c) a = 1.473, b = 0.782, c = 0.909, d = 3.854, e = 2
d) a = 2.390, b = 1.295, c = 0.943, d = 2. 564, e = 1
View Answer

Answer: c
Explanation: Amount of carbon in 1 mole of substrate = 72 g
Amount of carbon converted to biomass = 72(2/3) = 48 g
Then, 48 = 4.4c (12); c= 0.909
Amount of carbon to CO₂ = 72-48 = 24 g
24 = 12e
e = 2
The nitrogen balance yields
14b = 0.86c (14)
b = 0.782
The hydrogen balance is
12 + 3b = 7.3c + 2d
d = 3.854
The oxygen balance yields
6(16) + 2(16)a = 1.2(16)c + 2(16)e +16d
a = 1.473.

5. A yeast (CH_1.66N_0.13O_0.40) is growing aerobically on arabinose (C₅H₁₀O₅) and ammonium hydroxide (NH₄OH) with a respiratory quotient of 1.4. Estimate the stoichiometric coefficient of the equation:
aC₅H₁₀O₅ + bO₂ + cNH₄OH → CH_1.66N_0.13O_0.40 + dCO₂ + eH₂O
a) a = 0.2823, b = 0.2938, c = 0.13, d = 0.4113, e = 0.9065
b) a = 0.2893, b = 0.3638, c = 0.13, d = 0.4321, e = 0.9478
c) a = 0.2843, b = 0.2590, c = 0.13, d = 0.4321, e = 0.9576
d) a = 0.2823, b = 0.2130, c = 0.13, d = 0.4113, e = 0.8743
View Answer

Answer: a
Explanation: Equations for coefficients:
C atom balance: 5a = 1 + d —————— (1)
H atom balance: 10a + 5c = 1.66 + 2e ——– (2)
O atom balance: 5a + 2b + c = 0.40 + 2d + e —- (3)
N atom balance: c = 0.13 ———————— (4)
Respiratory quotient: RQ = 1.4 = d/b ———– (5)

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Therefore, eq.2× eq.(3) – eq.(2), we get,
b = 0.2938
From eq.(5), we get,
d = 0.4113
From eq.(1), we get,
a = 0.2823
From eq.(3), we get,
e = 0.9065.

6. Estimating Yield from Stoichiometry:
3C₆H₁₂O₆ + 8 O₂ + 2NH₃ -> 2C₅H₇O₂N + 8CO₂ +14H₂O
Given:3(180) 8(32) 2(17) 2(113)
Calculate the yield of glucose.
a) 0.53 g cells / g glucose used
b) 0.27 g cells / g glucose used
c) 0.42 g cells / g glucose used
d) 0.51 g cells / g glucose used
View Answer

Answer: c
Explanation:

7. The growth of S.cerevisiae on glucose under anaerobic conditions can be described by the following reaction:
C₆H₁₂O₆ + aNH₃ → 0.59 CH_1.64N_0.16O_0.52 (biomass) + 0.43 C₃H₈O₃ + 1.54CO₂ + 1.3C₂H₅OH +0.036H₂O
Determine the biomass (MW= 23.74) yield coefficient Y x/s
a) 0.078 g. g⁻¹
b) 0.070 g. g⁻¹
c) 0.068 g. g⁻¹
d) 0.060 g. g⁻¹
View Answer

Answer: a
Explanation: Yx/s = 0.59xMW biomass/1xMWglucose
= 0.59×23.74/1×180 = 0.078 g. g⁻¹.

8. The growth of S.cerevisiae on glucose under anaerobic conditions can be described by the following reaction:
C₆H₁₂O₆ + aNH₃ → 0.59 CH_1.64N_0.16O_0.52 (biomass) + 0.43 C₃H₈O₃ + 1.54CO₂ + 1.3C₂H₅OH + 0.036H₂O and determine the product yield coefficient Y etoh/s and determine coefficient a.
a) 0.57 g. g⁻¹, 0.078
b) 0.33 g. g⁻¹, 0.094
c) 0.45 g. g⁻¹, 0.086
d) 0.44 g. g⁻¹, 0.056
View Answer

Answer: b
Explanation: Yetoh/s = 1.3 x MW EtOH / 1x MW glucose
= 0.59 x 46/ 1×180 = 0.33 g.g⁻¹
N in NH₃ = N in biomass (CH_1.64N_0.16O_0.52)
∴ a = (0.59)(0.16) = 0.094.

9. What is the degree of reduction of glucose?
a) 4
b) 3
c) 12
d) 24
View Answer

Answer: d
Explanation: Glucose is C₆H₁₂O₆. The degree of reduction = 6 x (+4) + 12 x (+1) + 6 x (-2) = 24.

10. What is the COD (chemical oxygen demand) of ethanol, expressed as g COD/g ethanol?
a) 1.30 g COD/g ethanol
b) 32 g COD/g ethanol
c) 2.24 g COD/g ethanol
d) 2.09 g COD/g ethanol
View Answer

Answer: d
Explanation: Ethanol (C₂H₅OH) has a degree of reduction of 12 (mol electrons/mol ethanol). Oxygen has a degree of reduction of -4 (mol electrons/mol oxygen)and a molecular weight of 32 g/mol, which corresponds to 8 g oxygen (=COD)/mol electrons. Multiplying 12 by 8, this results in 96 g COD/mol ethanol, or, 96/46 = 2.09 g COD/g ethanol.

Sanfoundry Global Education & Learning Series – Bioprocess Engineering.

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