250+ TOP MCQs on Stoichiometry of Growth and Product Formation and Answers

Bioprocess Engineering Multiple Choice Questions on “Stoichiometry of Growth and Product Formation”.

1. ”Bacteria have slightly higher nitrogen content than fungi”.
A. True
B. False

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

Answer: C

3. Calculate the stoichiometric coefficients of the following biological reaction:
Hexadecane: C16H34 + a O2 + b NH3 = c (C4.4H7.3N0.86O1.2) + 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

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 (C4.4H7.3N0.86O1.2) + 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

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 (CH1.66N0.13O0.40) 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 → CH1.66N0.13O0.40 + 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

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)

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

Answer: C

7. The growth of S.cerevisiae on glucose under anaerobic conditions can be described by the following reaction:
C6H12O6 + aNH3 → 0.59 CH1.64N0.16O0.52 (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

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 CH1.64N0.16O0.52 (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

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 (CH1.64N0.16O0.52)
∴ a = (0.59)(0.16) = 0.094.

9. What is the degree of reduction of glucose?
A. 4
B. 3
C. 12
D. 24

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

Answer: D