This is copyrighted product of GATEHELPLINE

GATE

HELPLINE

BIOPROCESS

ENGINEERING

MCQs-I

This is copyrighted product of GATEHELPLINE

1. Diauxic patter of the biomass growth is associated with

P) Multiple lag phases

Q) Sequential utilization of multiple substrates

R) Simultaneous utilization of multiple substrates

S) Absence of lag phase

a) P,R c) R,S

b) P,Q d) Q,S

2. Match the following

P) Sedimentation coefficient 1. Aqueous two-phase extraction

Q) Partition coefficient 2. Ultracentrifugation

R) Rejection coefficient 3. Dialysis

S) Activity coefficient 4. Centrifugation

a) P-3, Q-1, R-4, S-2 c) P-2, Q-1, R-4, S-3

b) P- 4, Q-3, R-1, S-2 d) P-4, Q-1, R-2, S-3

3. Match the following

A) Marine type impeller 1. Recirculation of medium

B) Draft tube 2. Aeration of medium

C) Diaphragm valve 3. Animal cell cultivation

D) Sparger 4. Sterile operation

a) A-4, B-2, C-1, D-3 c) A-3, B-1, C-4, D-2

b) A-3, B-4, C-2, D-1 d) A-2, B-1, C-4, D-3

4. In a chemostat, evaluation the dilution rate at the cell wash out condition by applying monod’s

model with the given set of data µmax = 1 h-1

; YX/S = 0.5gg-1

; Ks = 0.2 gL-1

; S0 = 10g L-1

a) 1.00 h-1

c) 0.98 h-1

b) 0.49 h-1

d) 1.02 h-1

By the given data:

µmax = 1 h-1

maximum growth rate

Ks = 0.2 gL-1

monods constant

S0 = 10g L-1

initial substrate concentration

The dilution rate at wash out condition is Dcrit=?

Dcrit= µmax S0/ (Ks+ S0)

Substituting the values in equation

This is copyrighted product of GATEHELPLINE

We get 0.98 hr-1

5. A bacterial culture with an approximate biomass composition of CH1.8O0.5N0.2 is grown

aerobically on a defined medium containing glucose as the sole carbon source and ammonia

being the nitrogen source. In this fermentation, biomass is formed with a yield coefficient of

0.35 gram dry cell weight per gram of glucose and acetate is produced with a yield coefficient of

0.1 gram acetate per gram of glucose. The respiratory coefficient for the above culture will be

a) 0.90 c)1.00

b) 0.95 d)1.05

6. A bacterial culture having a specific oxygen uptake rate of 5 mmol O2 (g-DCW)-1

hr -1

is being

grown aerobically in a fed- batch bioreactor. The maximum value of the volumetric oxygen

transfer coefficient is 0.18s-1

for the stirred tank bioreactor and the critical dissolved oxygen

concentration is 20% of the saturation concentration (8 mg/ml). the maximum density to which

the cells can be grown in the fed-batch process without the growth being limited by oxygen

transfer is approximately

a) 14 g/l c) 32 g/l

b) 26 g/l d) 65 g/l

A double reciprocal plot was created from the specific growth rate and limiting substrate

concentration data obtained from a chemostat experiment. A linear regression gave of 1.25 hr and

100 mg-hr-1

for the intercept and slope, respectively.

7. The respective values of the Monod kinetic constants µm (hr-1

) and Ks (mg/l) are as follows:

a) 0.08,8 c) 0.8,80

b) 0.8,0.8 d) 8,8

Solution:

Intercept = 1.25 hr

Slope = 100 mg hr l-1

Now we have to find the maximum growth rate and Monod’s constant

From the reciprocal plot of Monod’s equation

1/µ= Ks/ µm*1/S+1/ µm

Y=mX+C

C = intercept = 1/ µm => 1.25 = 1/ µm => µm = 0.8

m = slope = Ks/ µm=> 100= Ks/ µm=> 100*0.8=Ks => Ks = 80

8. The same culture (with the µm and Ks values as computed above) is cultivated in a 10-litre

chemostat being operated with a 50 ml/min sterile feed containing 50g/l of substrate. Assuming

an overall yield coefficient of 0.3 g-DCW / g- substrate, the respective values of the outlet

biomass and substrate concentrations are

a) 15 g/L, 48 mg/L c) 15g/L, 0.48 g/L

b) 48 g/L,15 g/L d) 4.8 g/L, 4.8 g/L

This is copyrighted product of GATEHELPLINE

Solution:

Given data is some for above problem & hence

µm = 0.8 hr-1

Ks = 80 mg/l

Total volume (V) = 10 lit

Flow rate (F)= 50 ml/ min

S0 = 50g/ lit

Yield Y = 0.3 g-DCW/ g-substrate

To find: biomass X and limiting substrate concentration

Monod equation

S = Ks D/ µm -D

D= F/V

F= 50 ml/ min = 3000ml/ hr

D= F/V = 3000/10000 = 0.3

S= 80*0.3/0.8-0.3 = 48 mg/l

X = y [So - S]

X = 0.3 *[50-0.048] = 0.3 [49.952] = 14.985 = 15 g/l

9. The respiratory coefficient for the reaction

aCHmOn + b O2 + cNH3 d CHαOβNγ + eH2O+ fCO2 is defined as

a) f/a c) b/f

b) e/b d) f/b

10. match the following

A. Pressure 1. photometer

B. Foam 2. rotameter

C. Turbidity 3. diaphragm gauge

D. Flow rate 4. rubber sheathed electrode

a) A-3 B-4 C-1 D-2 c) A-1 B-3 C-2 D-4

b) A-4 B-1 C-2 D-3 d) A-1 B-2 C-3 D-4