This set of Bioseparation Processes Multiple Choice Questions & Answers (MCQs) focuses on “Centrifuge Principle and Types”.
1. What is the density of microbial cells?
a) 1.05 to 1.15 g/cm3
b) 1.04 to 1.10 g/cm3
c) 1.10 to 1.60 g/cm3
d) 1.30 g/cm3
View Answer
Explanation: The density of microbial cells is 1.05 to 1.15 g/cm3. The density of mammalian cells is 1.04 to 1.10 g/cm3. The density of organelles is 1.10 to 1.60 g/cm3. The density of proteins is 1.30 g/cm3.
2. A centrifuge separates is a device which separates particles on the basis of _________
a) size
b) shape
c) Density
d) size, shape, density
View Answer
Explanation: A centrifuge separates the particles on the basis of size, shape and density of the particles under the artificial induced gravitational fields. The particles are either in the form of suspension or the macromolecules in a solution.
3. Centrifugation can be used when the dispersed material is denser than the medium in which they are dispersed.
a) False
b) True
View Answer
Explanation: Centrifugation can only be used when the particles from suspension or solutions are dispersed in denser medium than the liquid in which they are dispersed. The biological substances are separated on the basis of the densities in the aqueous form i.e. in association with a large number of water molecules.
4. What is sedimentation coefficient?
a) s = –\(\frac{ϑ}{\omega^2 R}\)
b) s = \(\frac{ϑ}{R}\)
c) s = \(\frac{ϑ}{\omega^2 R}\)
d) s = \(\frac{ϑ}{\omega^2}\)
View Answer
Explanation: The sedimentation coefficient is constant resulting as a property of both the particles and the medium in which the body force is applied and the velocity through a viscous medium is proportional to the accelerating field i.e. s = \(\frac{ϑ}{\omega^2 R}\).
5. What is the unit of sedimentation coefficient?
a) Svedberg unit
b) Atmosphere
c) Pascal
d) M2
View Answer
Explanation: The unit of sedimentation coefficient is expressed in terms of Svedberg unit which is denoted by S, where 1 svedberg unit (S) = 10-13 s. Svedberg unit was discovered by Theodor Svedberg.
6. Estimate the time taken to complete the clarification of suspension of ribosome in a centrifuge at 10,000rpm having diameter of 1cm and the initial distance from the centre of rotation is 4cm. The measured sedimentation coefficient was s20.w at 70S ribosome.
a) 5.6 h
b) 6.5h
c) 8.1h
d) 1.8h
View Answer
Explanation: s = \(\frac{dR}{dt} \frac{1}{ω^2 R}\), at t = 0 and R = R0 ω2 st = ln\(\frac{R}{R_0}\) and time taken for the centrifugation is t = \(\frac{ln(\frac{R}{R_0})}{ω^2 S} = \frac{ln(\frac{5}{4})1h/3600s}{(10,000\, rev/min × \frac{2\pi\, rad}{rev} × \frac{1\,min}{60sec})^{2}(70 × 10^{-13}s)}\) = 8.1 h. Since time is inversely proportional to the square of rotation speed, the time can be reduced to 2h by doubling the speed.
7. What will be the speed of centrifuge when the bacterial cell debris has Gt = 54 × 106 s with diameter of 10cm and which type of centrifuge can be used for full sedimentation in appropriate amount of time?
a) 11590 rpm
b) 12596 rpm
c) 13495 rpm
d) 16989 rpm
View Answer
Explanation: ω = \((\frac{Gtg}{Rt})^{\frac{1}{2}}\) = (\(\frac{54 × 10^6 × 9.81}{0.05 × 2(3600)})^{\frac{1}{2}}\) = 1213 \(\frac{rad}{s} × \frac{1\,rev}{2\pi\, rad}\) × 60s/min = 11590 rpm. The speed can be achieved in a production tubular bowl centrifuge.
8. hich type of centrifuge works well with for particles of relatively low sedimentation coefficient?
a) Disk centrifuge
b) Tubular bowl centrifuge
c) Multi chamber centrifuge
d) Basket centrifuge
View Answer
Explanation: In tubular bowl centrifuge, the solids deposit on the walls of the bowl and the feed continues until the bowl is almost full at the time when operation has stopped and solids are removed therefore, it is suitable for the particles of relatively low sedimentation coefficient such as protein precipitates.
9. How can you estimate the sedimentation velocity at 1g?
a) vg = \(\frac{ln(\frac{R}{R_0})}{ω^2 t}\)
b) vg = \(\frac{g\, ln(\frac{R}{R_0})}{t}\)
c) vg = \(\frac{g\, ln(\frac{R}{R_0})}{ω^2 t}\)
d) vg = –\(\frac{g\, ln(\frac{R}{R_0})}{ω^2 t}\)
View Answer
Explanation: The sedimentation velocity at 1g is estimated by using vg = \(\frac{g\, ln(\frac{R}{R_0})}{ω^2 t}\), it is used for the determination of vg in the laboratory scale of centrifuge. The minimum time for the clarification of the sample in a laboratory at a particular speed can be determined by the experimental work.
10. Which centrifuge can be used for the separation of proteins and nucleic acid?
a) Normal centrifuge
b) Laboratory centrifuge
c) Ultra centrifuge
d) Preparative centrifuge
View Answer
Explanation: Macromolecules like proteins, nucleic acids, and carbohydrates cannot be separated using normal centrifuges because of the heavy molecular weight of the molecules, these can be separated using ultracentrifuges which can generate very strong artificial gravitational field.
11. Which analysis is used commonly in the industries for the characterization of centrifuge?
a) Beta analysis
b) Sigma analysis
c) Gamma analysis
d) Theta analysis
View Answer
Explanation: Sigma analysis is used for the analysis in the industries and it uses the constant operation which characterizes the centrifuge into the feed flows of volumetric flow rate and is determined by Flow rate = sedimentation coefficient * constant operation i.e. Q = vg × Σ.
12. Calculate the velocity of a centrifuge in a centrifugal field at a steady state when radius of particle is 2.5cm and density of the particle is 1.2 g/cm3 and the initial density is 1.0 g/cm3, the viscosity is 0.01poise.
a) 14500 cm/s
b) 13611 cm/s
c) 12430 cm/s
d) 11459 cm/s
View Answer
Explanation: The velocity of the centrifuge can be calculated using the equation,v = \(\frac{2a^2 (\rho – \rho_0)g}{9\mu}\) ∴ v = \(\frac{2 × 2.5^2(1.2 – 1.0)9.8}{9 × 0.01}\) = 13611 \(\frac{cm}{s}\).
13. Calculate the Reynolds number for the spherical particles when the radius of the particle is 2.5cm, velocity is 13611 cm/s, viscosity is 0.001 poise, density is 1.2 g/cm3.
a) 81.66 × 105
b) 81.66 × 104
c) 81.66 × 106
d) 81.66 × 107
View Answer
Explanation: Reynolds number Re = \(\frac{2av\rho}{\mu} = \frac{2 × 2.5 × 13611 × 1.2}{0.001}\) = 81.66 × 106.
14. Calculate the geometry and the speed of the centrifuge when the flow rate is 120 cm/s and the sedimentation coefficient is 13611 cm/s.
a) 0.0088
b) 0.0099
c) 0.0100
d) 0.0150
View Answer
Explanation: The geometry and the speed of centrifuge can be represented by sigma analysis. Q = vg × Σ ∴ Σ = \(\frac{Q}{v_g}\) = \(\frac{120}{13611}\) = 0.0088.
Sanfoundry Global Education & Learning Series – Bioseparation Processes.
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