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” is this statement true or false?

a) True

b) False

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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?

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3. Calculate the stoichiometric coefficients of the following biological reaction:

Hexadecane: C_{16}H_{34} + a O_{2} + b NH_{3} = c (C_{4}.4H_{7}.3N0.86O1.2) + d H2O + e CO_{2}

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

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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: C_{6}H_{12}O_{6} + aO_{2} + bNH_{3} = c (C_{4}.4H_{7}.3N0.86O1.2) + d H_{2}O + e CO_{2}

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

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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 CO

_{2}= 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. H_{10}O_{5}) and ammonium hydroxide (NH_{4}OH) with a respiratory quotient of 1.4. Estimate the stoichiometric coefficient of the equation:

aC_{5}H_{10}O_{5} + bO_{2} + cNH_{4}OH → CH_{1.66}N_{0.13}O_{0.40} + dCO_{2} + eH_{2}O

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

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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:

3C_{6}H_{12}O_{6} + 😯_{2} + 2NH_{3} -> 2C_{5}H_{7}O_{2}N + 8CO_{2} +14H_{2}O

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

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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

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Explanation: Yx/s= 0.59xMW biomass/1xMWglucose

= 0.59×23.74/1×180 = 0.078 g. g⁻¹.

8. Refer Q7, 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

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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

View Answer

Explanation: Glucose is C

_{6}H

_{12}O

_{6}. 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

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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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