Bioprocess Engineering Questions and Answers – Ideal Reactor Operation

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This set of Bioprocess Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Ideal Reactor Operation”.

1. Aerobic reactions are not batch operations.
a) True
b) False
View Answer

Answer: a
Explanation: Aerobic reactions are not batch operations in the strictest sense; the low solubility of oxygen in aqueous media means that it must be supplied continuously while carbon dioxide and other off-gases are removed.

2. In a perfectly mixed reactor _________
a) The output composition is different from input composition
b) The output composition is identical from input composition
c) Both output and input composition are constant
d) Both output and input composition are not constant
View Answer

Answer: b
Explanation: In a perfectly mixed reactor, the output composition is identical to composition of the material inside the reactor, which is a function of residence time and rate of reaction.

3. Convert vmax = 2.5 mmol m-3 s-1 into mM h-1.
a) 2500
b) 900
c) 25
d) 9
View Answer

Answer: d
Explanation:
FConversion of vmax 2.5 mmol m-3 s-1 into mM h-1
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4. Convert vmax = 7 mmol m-3 s-1 into mM h-1.
a) 25
b) 20
c) 25.20
d) 20.25
View Answer

Answer: c
Explanation:
Conversion of vmax 7 mmol m-3 s-1 into mM h-1

5. In batch reaction time with enzyme deactivation, calculate the first-order deactivation rate constant.
(Given – so = 12 mM; vmax = 9 mM h-1; Km = 8.9 mM; sf = 1.2 mM; th = 4.4 h).
a) 0.150 h-1
b) 0.158 h-1
c) 0.155 h-1
d) 0.154 h-1
View Answer

Answer: b
Explanation:
Find the first-order deactivation rate constant with enzyme deactivation

6. By taking the parameters of Q5, and the batch reaction time tb.
a) 5.0 h
b) 10.0 h
c) 15.0 h
d) 20.0 h
View Answer

Answer: a
Find the batch reaction time tb for the parameters

7. Zymomonas mobilis is used to convert glucose to ethanol in a batch fermenter under anaerobic conditions. The yield of biomass from substrate is 0.06 g g-1; YPX is 7.7 g g-1. The maintenance coefficient is 2.2 g g-1 h-1; the specific rate of product formation due to maintenance is 1.1 h-1. The maximum specific growth rate of Z. mobilis is approximately 0.3 h-1.5 g bacteria are inoculated into 50 litres of medium containing 12 g l-1 glucose. Determine batch culture times required to produce 10 g biomass.
a) 3.1 h
b) 3.3 h
c) 3.5 h
d) 3.7 h
View Answer

Answer: d
Explanation: Yxs = 0.06 gg-l; Ypx= 7. 7 gg-l; μmax = 0. 3 h-l; ms=2.2 gg-l h-l; mp= 1.1 h-l; x0=5g/501= 0.1 gl-l;s0= 12 gl-l.

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If 10 g biomass are produced by reaction, the final amount of biomass present is (10 + 5) g = 15 g. Therefore xf = 15 g / 50 l = 0.3 g l-1.

The batch culture times required to produce 10 g biomass is 3.7 h

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8. By taking the pararmeters of Q7 and Determine batch culture times required to achieve 90% substrate conversion.
a) 5.1 h
b) 5.3 h
c) 5.5 h
d) 5.7 h
View Answer

Answer: d
The batch culture times required to achieve 90% substrate conversion

9. By taking the parameters question 8 and question 7, determine batch culture times required to produce 100 g ethanol.
a) 3.2 h
b) 3.4 h
c) 3.6 h
d) 3.8 h
View Answer

Answer: b
The batch culture times required to produce 100 g ethanol

10. The Zymomonas mobilis cells are used for chemostat culture in a 60 m3 fermenter. The feed contains 12 g l-1 glucose; Ks for the organism is 0.2 g l-1. What flow rate is required for a steady-state substrate concentration of 1.5 g l-1?
a) 15.6 m3 h-1
b) 15.8 m3 h-1
c) 15.4 m3 h-1
d) 15.2 m3 h-1
View Answer

Answer: a
Explanation: Yxs = 0.06 g g-1; Ypx = 7.7 g g-1max = 0.3 h-1; Ks = 0.2 g-1; ms = 2.2 g g-1 h-l; si = 12 g 1-1; V=60 m3. qp = 3.4 h-1, YPS = Ypx Yxs = 0.46 g g-1.
s = 1.5 g l-1
Flow rate required for steady-state substrate concentration of 1.5 g l-1

11. By taking Q10 and at the flow rate of Q10, what is the cell density?
a) 0.42 g l-1
b) 0.44 g l-1
c) 0.46 g l-1
d) 0.48 g l-1
View Answer

Answer: a
Explanation: When synthesis of product is coupled with energy metabolism as for ethanol, x is evaluated. Therefore:
The synthesis of product coupled with energy metabolism as for ethanol

12. By taking Q10 and Q11 into account and at the flow rate of Q10, what concentration of ethanol is produced?
a) 5.1 g l-1
b) 5.3 g l-1
c) 5.5 g l-1
d) 5.7 g l-1
View Answer

Answer: c
Explanation: Assuming ethanol is not present in the feed, PI = 0. Steady-state product concentration is given by:
The concentration of ethanol produced for PI = 0

13. Which of the following type is of the perfusion culture?
a) Batch
b) Conc. Batch
c) Continuous
d) Conc. Fed-Batch
View Answer

Answer: d
Explanation: Concentrated Fed-Batch- In a process that can be considered a form of perfusion, the culture is perfused to generate ultrahigh cell concentration, greater than 108 cells/mL; and the product is also retained in the vessel.

14. Immobilised lactase is used to hydrolyse lactose in dairy waste to glucose and galactose. Enzyme is immobilised in resin particles and packed into a 0.5 m3 column. The total effectiveness factor for the system is close to unity; Km for the immobilised enzyme is 1.32 kg m-3; Vmax is 45 kg m-3 h-1. The lactose concentration in the feed stream is 9.5 kg m-3; a substrate conversion of 98% is required. The column is operated with plug flow for a total of 310 d per year. At what flow rate should the reactor be operated?
a) 1.56 m3 h-1
b) 1.58 m3 h-1
c) 1.50 m3 h-1
d) 1.54 m3 h-1
View Answer

Answer: a
Explanation: For 98% substrate conversion, sf = (0.02 sI) = 0.19 kg m-3
Flow rate for the reactor to be operated for 98% substrate conversion

15. By taking the parameters of Q14 into consideration, estimate how many tonnes of glucose are produced per year?
a) 56.3 tonnes yr-1
b) 56.6 tonnes yr-1
c) 56.7 tonnes yr-1
d) 56.5 tonnes yr-1
View Answer

Answer: c
Explanation: The rate of lactose conversion is equal to the difference between inlet and outlet mass flow rates of lactose:
F(si – sf) = 1.56 m3 h-1 (9.5 – 0.19) kg m-3 = 14.5 kg h-1
Converting this to an annual rate based on 310 d per year and a molecular weight for lactose of 342:
Find the tonnes of glucose produced per year for lactose of 342

The enzyme reaction is:
lactose + H20 → glucose + galactose.
Therefore, from reaction stoichiometry, 315 kg mol glucose are produced per year. The molecular weight of glucose is 180; therefore:

Rate of lactose conversion is equal to inlet & outlet mass flow rates of lactose

Sanfoundry Global Education & Learning Series – Bioprocess Engineering.

To practice all areas of Bioprocess Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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