This set of Bioseparation Technology Multiple Choice Questions & Answers (MCQs) focuses on “Cell Disruption using Ultrasonic Vibrations”.
1. At what frequency the ultrasonic vibrations can disrupt the cells in cell suspension?
a) >18 kHz
b) <18 kHz
c) 18 kHz
d) 18 Hz
View Answer
Explanation: The ultrasonic vibrations >18 kHz can disrupt the cells in cell suspension. The ultrasonic waves utilizes the sinusoidal movement of probe in the liquid and since it characterized by frequency of 18 kHz to 1 MHz it undergoes small displacements of about < 50 μm with moderate velocities.
2. The cells are subjected to ultrasonic vibrations by introducing an ultrasonic vibration ___
a) collecting into vessel
b) emitting tips
c) into cell suspension
d) emitting and collecting
View Answer
Explanation: The cells of the cell suspension which need to be disrupted are subjected to ultrasonic vibrations by introducing an ultrasonic vibration emitting tips in to the suspension and emitting tips used are of different sizes based on the volume of the sample being used.
3. The duration of ultrasound provided for cell disruption is ___
a) sample size
b) cell type
c) sample size, cell type, cell concentration
d) cell concentration
View Answer
Explanation: The duration of ultrasound used for cell disruption is based on the size of cells being disrupted; type of cell for example, thermophilic cells use cold environment; concentration of cell for example, if there is more amount of cells the ultrasonic waves should be subjected more than less amount of cell.
4. What is the impact of high frequencies on the cells?
a) Ultrafiltration
b) Movement of cell
c) Aggregation of cell
d) Cavitations in cells
View Answer
Explanation: The high frequencies of the ultrasonic vibrations leads to cavitations in the cell which allows the formation of bubbles in the liquid medium and the bubbles reaches the resonance size and it collapses due to shock waves released in the form of mechanical energy. The shock waves leads to disruption of cells in the cell suspension.
5. What is the time taken by bacterial cells to be disrupted?
a) 30 – 60sec
b) 2 – 10 min
c) 10 – 20min
d) 10 – 20sec
View Answer
Explanation: 30 – 60 seconds is sufficient for bacterial cells to be disrupted by the ultrasonic vibrations. These cells are small and can be easily ruptured by the shock waves produced by the ultrasonicator therefore; it takes less time for these cells to be disrupted.
6. What is the time taken by yeast cells to be disrupted?
a) 30 – 60sec
b) 2 – 10 min
c) 10 – 20min
d) 10 – 20sec
View Answer
Explanation: 2 – 10 minutes are required for yeast cells to be disrupted. These cells are bigger than bacterial cells and it need more time to be ruptured since it cannot be ruptured easily.
7. When is ultrasonic vibration used in conjunction with chemical cell disruption method?
a) Fungal cells
b) Plant cells
c) Fungal and Bacterial cells
d) bacterial cells
View Answer
Explanation: The ultrasonic vibrations are used in conjunction with chemical method of disruption during fungal and bacterial cell disruption. It removes the barriers around the cells by weakening the exposing them to chemicals like enzymes or detergents.
8. What is the effect of ultrasonic vibrations on the kinetics of the reaction?
a) Rate of reaction increases
b) Rate of reaction decreases
c) Rate of reaction decreases with increasing the sonic amplitude
d) Rate of reaction increases with increasing the sonic amplitude
View Answer
Explanation: The reaction rate increases with increasing sonic amplitude but the varying frequency will have negligible effect on the rate of reaction. The Arrhenius equation shows that the ultrasonic vibrations increased the frequency factor of the reaction rate and activation energy remains unchanged.
9. Estimate the concentration of product when the yeast cell disruption is carried out for 240 sec. The yeasts cells disrupted to release the intracellular product and the time taken for 3.49mg/ml cells was 60 sec and 4.56mg/ml for 120sec.
a) 4.96mg/ml
b) 5.98mg/ml
c) 1.5mg/ml
d) 2.4mg/ml
View Answer
Explanation: \(\frac{C}{C_{max}}\) = 1 – exp(-\(\frac{t}{\theta}\)) is used to see the cell disruption in the yeast cells. 3.49 = Cmax (1 – exp(-\(\frac{60}{\theta}\)), 4.56 = Cmax (1 – exp(-\(\frac{120}{\theta}\)) ∴ Cmax = 5 mg/ml. At θ = 50 sec, t = 240sec. So, C = 5(1 – exp(-\(\frac{240}{50}\))) = 4.96 \(\frac{mg}{ml}\).
10. Which instrument is used for generating ultrasonic vibrations?
a) Centrifuge
b) Ultrasonic disruptor
c) UV spectrometer
d) Laminar Air flow
View Answer
Explanation: Ultrasonic disruptor is used for generating ultrasonic vibrations. Centrifuge is used for centrifugation. UV spectrometer is used for observing the growth of the particle. Laminar air flow is used for avoiding contamination while working with microbial cells.
Sanfoundry Global Education & Learning Series – Bioseparation Technology.
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