Chapter 6: Microbial Culture — Online MCQ Test
BIOTECHNOLOGY · Grade 12 · CBSE(NCERT)
Practice Chapter 6: Microbial Culture with a free chapter-wise online MCQ test.
This chapter covers: Microbial Culture Nutrient Media Growth Kinetics Pure Culture Sterilization Fermentation Microorganisms Culture Media Aseptic Techniques Industrial Microbiology.
AI-generated questions from basic to board-exam level, with instant results and explanations.
Chapter 6: Microbial Culture — Important Questions & Answers
What is a microbial culture medium?
- A. A nutrient preparation used to grow microorganisms under controlled conditions
- B. A chemical used only to kill microorganisms
- C. A microscope slide used for observing bacteria
- D. A dye used for staining microbial cells
Answer: A. A nutrient preparation used to grow microorganisms under controlled conditions
A culture medium supplies nutrients and suitable conditions for microbial growth. It may be liquid, semi-solid or solid.
A culture medium supplies nutrients and suitable conditions for microbial growth. It may be liquid, semi-solid or solid.
Which substance is most commonly used as a solidifying agent in microbiological media?
- A. Gelatin
- B. Agar
- C. Cellulose
- D. Starch
Answer: B. Agar
Agar is commonly used because most microorganisms do not degrade it and it remains solid at incubation temperatures. It melts near 100°C and solidifies around 40-45°C.
Agar is commonly used because most microorganisms do not degrade it and it remains solid at incubation temperatures. It melts near 100°C and solidifies around 40-45°C.
The standard condition commonly used for sterilization in an autoclave is:
- A. 100°C for 5 minutes
- B. 121°C at 15 psi for about 15-20 minutes
- C. 60°C for 30 minutes
- D. 37°C for 24 hours
Answer: B. 121°C at 15 psi for about 15-20 minutes
Autoclaving uses steam under pressure, generally 121°C at 15 psi for 15-20 minutes. This kills vegetative cells as well as spores.
Autoclaving uses steam under pressure, generally 121°C at 15 psi for 15-20 minutes. This kills vegetative cells as well as spores.
A pure culture contains:
- A. Only one type of microorganism
- B. Microorganisms growing without any nutrient
- C. Only dead microorganisms
- D. A mixture of bacteria, fungi and viruses
Answer: A. Only one type of microorganism
A pure culture is derived from a single cell or colony and contains only one microbial species or strain. It is essential for studying specific microbial properties.
A pure culture is derived from a single cell or colony and contains only one microbial species or strain. It is essential for studying specific microbial properties.
Which method is most suitable for sterilizing heat-sensitive solutions such as vitamin or antibiotic solutions?
- A. Autoclaving
- B. Hot air oven
- C. Membrane filtration
- D. Direct flaming
Answer: C. Membrane filtration
Heat-sensitive solutions may be damaged by high temperature, so membrane filtration is used. The filter physically removes microorganisms from the liquid.
Heat-sensitive solutions may be damaged by high temperature, so membrane filtration is used. The filter physically removes microorganisms from the liquid.
A student wants to isolate well-separated bacterial colonies from a mixed sample on an agar plate. Which technique is most appropriate?
- A. Streak plate method
- B. Gram staining
- C. Autoclaving
- D. Centrifugation
Answer: A. Streak plate method
The streak plate method dilutes cells over the agar surface to obtain isolated colonies. Each isolated colony can originate from a single cell or clump.
The streak plate method dilutes cells over the agar surface to obtain isolated colonies. Each isolated colony can originate from a single cell or clump.
A plate made from 0.1 mL of a 10^-5 dilution shows 80 colonies. What is the viable count in the original sample?
- A. 8.0 x 10^5 CFU/mL
- B. 8.0 x 10^6 CFU/mL
- C. 8.0 x 10^7 CFU/mL
- D. 8.0 x 10^8 CFU/mL
Answer: C. 8.0 x 10^7 CFU/mL
CFU/mL = colonies divided by volume plated multiplied by dilution factor. Thus, 80 divided by 0.1 multiplied by 10^5 = 8.0 x 10^7 CFU/mL.
CFU/mL = colonies divided by volume plated multiplied by dilution factor. Thus, 80 divided by 0.1 multiplied by 10^5 = 8.0 x 10^7 CFU/mL.
A broth culture shows high turbidity, but plating gives very few colonies. Which explanation is most likely?
- A. All cells are actively dividing
- B. The medium contains no nutrients
- C. Many cells are dead or non-viable
- D. The culture is necessarily pure
Answer: C. Many cells are dead or non-viable
Turbidity measures total suspended cells and particles, including dead cells. Plate counts measure only viable cells capable of forming colonies.
Turbidity measures total suspended cells and particles, including dead cells. Plate counts measure only viable cells capable of forming colonies.
In a chemostat, if the dilution rate is increased beyond the maximum specific growth rate of the microorganism, what is the likely outcome?
- A. Cells grow faster than their physiological limit
- B. Washout of cells from the vessel occurs
- C. The culture automatically becomes a pure culture
- D. Nutrient concentration becomes zero permanently
Answer: B. Washout of cells from the vessel occurs
If cells are removed faster than they can divide, the population cannot be maintained. This leads to washout in a continuous culture system.
If cells are removed faster than they can divide, the population cannot be maintained. This leads to washout in a continuous culture system.
Two cultures have the same optical density. Culture X is in exponential phase, while Culture Y has just been heat-treated. Which conclusion is most accurate?
- A. Both cultures must have equal viable counts
- B. Culture Y may show high optical density but low viable count
- C. Culture X must contain only dead cells
- D. Optical density measures only colony-forming cells
Answer: B. Culture Y may show high optical density but low viable count
Optical density estimates turbidity due to suspended cells and does not distinguish live from dead cells. Heat-treated cells may remain turbid but fail to form colonies.
Optical density estimates turbidity due to suspended cells and does not distinguish live from dead cells. Heat-treated cells may remain turbid but fail to form colonies.