This set of Separation Processes Multiple Choice Questions & Answers (MCQs) focuses on “Mechanical Separations in Biotechnology”.
1. Which of the following changes does not induce precipitation?
a) pH
b) Temperature
c) Addition of salts
d) Gravity
View Answer
Explanation: Gravity does not induce precipitation.
2. What is a special case of precipitation?
a) Distillation
b) Fractional distillation
c) Sedimentation
d) Crystallization
View Answer
Explanation: Crystallization is a special case of precipitation where the produc is crystallized and produced slowly under very controlled conditions.
3. Why is the pH change to induce precipitation not useful for commercialization?
a) Proteins do not precipitate easily
b) The minimum pH is unknown
c) There are large differences in the isoelectric points
d) Carrying a pH meter is not helpful
View Answer
Explanation: The pH change to induce precipitation not useful for commercialization because there are large differences in the isoelectric points.
4. How does the salting out mechanism not occur?
a) Salt removes water by associating with water molecules
b) Few amount of water is left for proteins
c) Shielding the electrostatic protein-protein changes that account for protein-protein repulsion
d) Creating a barrier between the proteins and water
View Answer
Explanation: The salting out mechanism occurs partly because salt removes water by associating with water molecules and partly by shielding the electrostatic protein-protein changes that account for protein-protein repulsion.
5. When are organic solvents not commonly used for precipitation?
a) DNA precipitation
b) RNA precipitation
c) Plasma-protein precipitation
d) Protein separation
View Answer
Explanation: The organic solvents ethanol and acetone are used for, DNA, RNA and plasma-protein precipitation.
6. How is the physical process of sedimentation not enhanced?
a) Coagulation
b) Flocculation
c) Agglomeration
d) Decantation
View Answer
Explanation: Decantation does not enhance sedimentation whereas coagulation and flocculation do.
7. How is flocculation defined as?
a) Further agglomeration of small, slowly settling floc formed during coagulation to form a larger floc
b) Cause (a fluid) to change to a solid or semi-solid state
c) Gradually pour (wine, port, or another liquid) from one container into another, typically in order to separate out sediment
d) Deagglomeration to break into smaller particles
View Answer
Explanation: Flocculation is defined as further agglomeration of small, slowly settling floc formed during coagulation to form a larger floc.
8. Which particles have the largest diameter?
a) Colloidal
b) Flocculated
c) Coagulated
d) Dispersed
View Answer
Explanation: The flocculated particles have the largest diameter from 1-10 millimeters.
9. If the passes through the device is n=1
The time constant T=1345s
Expected release percent= 95%
Maximum concentration C= 10
Calculate the time required
a) 2344s
b) 4345s
c) 2345s
d) 4028s
View Answer
Explanation: Since c/C= (1-exp(-t/T)n, hence t=4028s.
10. If the passes through the device is n=1
The time constant T=1345s
Expected release percent= 0.85%
Maximum concentration C= 10
Calculate the time required
a) 2344s
b) 4345s
c) 2500s
d) 4028s
View Answer
Explanation: Since c/C= (1-exp(-t/T)n, hence t=2500s.
11. If the passes through the device is n=1
The time constant T=1345s
Expected release percent= 0.89%
Maximum concentration C= 10
Calculate the time required
a) 2000s
b) 2500s
c) 3000s
d) 3400s
View Answer
Explanation: Since c/C= (1-exp(-t/T)n, hence t=3000s.
12. If the passes through the device is n=1
The time constant T=1345s
Expected release percent= 77%
Maximum concentration C= 10
Calculate the time required
a) 2000s
b) 2500s
c) 3000s
d) 3400s
View Answer
Explanation: Since c/C= (1-exp(-t/T)n, hence t=2000s.
13. If the passes through the device is n=1
The time constant T=1345s
Expected release percent= 92%
Maximum concentration C= 10
Calculate the time required
a) 3100s
b) 3200s
c) 3300s
d) 3400s
View Answer
Explanation: Since c/C= (1-exp(-t/T)n, hence t=3400s.
14. If the passes through the device is n=1
The time constant T=1345s
Expected release percent= 87.5%
Maximum concentration C= 10
Calculate the time required
a) 2344s
b) 4566s
c) 2800s
d) 4454s
View Answer
Explanation: Since c/C= (1-exp(-t/T)n, hence t=2800s.
15. If the passes through the device is n=1
The time constant T=1345s
Expected release percent= 83%
Maximum concentration C= 10
Calculate the time required
a) 2344s
b) 4345s
c) 2345s
d) 4028s
View Answer
Explanation: Since c/C= (1-exp(-t/T)n, hence t=2345s.
Sanfoundry Global Education & Learning Series – Separation Processes.
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