MCB 1000L Lab Quiz 1 - Weeks 1-3 Labs -
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Questions and Answers | 2026/2027 Updated
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Practice Question Bank & Answer Rationale
THIS PACK CONTAINS
148 exam-style multiple-choice questions. Each question includes the correct answer and a complete
rationale for focused revision.
COVERAGE
This exam assesses foundational laboratory skills developed during the first three weeks of an introductory
microbiology lab, including aseptic technique, brightfield microscopy, microbial cultivation, and standard
quantification methods. Items require application of laboratory concepts to interpret results, troubleshoot
procedures, and select appropriate techniques.
STUDY GUIDE
Recommended duration: 60 minutes. Passing target: 70%.
HIGHP - Page 1 of 49
,1. During a streak-plate procedure, a student fails to flame the loop between the first and
second quadrants. What is the most likely consequence on the plate after incubation?
[ ] A. No growth will appear because the loop was sterile initially.
[ ] B. Heavy confluent growth will appear in all quadrants, making isolation impossible.
[ ] C. The second quadrant will show reduced growth due to residual bacteria from the first quadrant,
hindering isolation.
[ ] D. Only the first quadrant will have growth; the rest will remain sterile.
CORRECT: C. The second quadrant will show reduced growth due to residual bacteria from the first
quadrant, hindering isolation.
Failure to flame the loop between quadrants carries over high cell numbers from the previous quadrant,
reducing the dilution effect and preventing isolated colonies. Option B is incorrect because some dilution still
occurs; D is incorrect because bacteria are still transferred.
2. A specimen is examined under a compound light microscope with a 10X ocular lens
and a 40X objective lens. What is the total magnification of the image?
[ ] A. 40X
[ ] B. 50X
[ ] C. 400X
[ ] D. 1000X
CORRECT: C. 400X
Total magnification is calculated by multiplying the ocular lens magnification (10X) by the objective lens
magnification (40X), yielding 400X. Options A, B, and D result from incorrect multiplication or using the wrong
lens values.
3. Which statement best describes the purpose of a differential medium such as
MacConkey agar?
[ ] A. It inhibits the growth of Gram-positive bacteria while allowing Gram-negative bacteria to grow.
[ ] B. It allows visual differentiation of lactose-fermenting from non-fermenting bacteria based on colony
color.
[ ] C. It provides enriched nutrients to fastidious organisms that cannot grow on standard media.
[ ] D. It selectively supports the growth of fungi while inhibiting bacterial growth.
CORRECT: B. It allows visual differentiation of lactose-fermenting from non-fermenting bacteria
based on colony color.
Differential media contain indicators that reveal metabolic differences, such as lactose fermentation on
MacConkey agar. Option A describes a selective property, C describes enrichment, and D is unrelated to
MacConkey agar.
HIGHP - Page 2 of 49
,4. In a viable plate count, a 1.0 mL sample of a 10^-5 dilution yields 85 colonies on a
standard plate. What is the original cell concentration (CFU/mL) of the sample?
[ ] A. 8.5 × 10^5 CFU/mL
[ ] B. 8.5 × 10^6 CFU/mL
[ ] C. 8.5 × 10^7 CFU/mL
[ ] D. 8.5 × 10^4 CFU/mL
CORRECT: B. 8.5 × 10^6 CFU/mL
CFU/mL = colonies / (dilution × volume plated) = 85 / (10^-5 × 1.0 mL) = 85 × 10^5 = 8.5 × 10^6 CFU/mL.
The other options reflect errors in exponent arithmetic or volume plated.
5. A student observes a Gram-stained smear under oil immersion and sees pink,
rod-shaped cells arranged in chains. Which conclusion is most accurate?
[ ] A. The organism is a Gram-positive coccus.
[ ] B. The organism is a Gram-negative bacillus.
[ ] C. The organism is a Gram-positive bacillus.
[ ] D. The organism is a Gram-negative coccus.
CORRECT: B. The organism is a Gram-negative bacillus.
Pink color indicates Gram-negative, and rod shape indicates bacillus. Option A is incorrect because
Gram-positive cells appear purple and cocci are spherical; C and D misidentify either color or shape.
6. Which of the following best explains why plates are incubated in an inverted position?
[ ] A. To prevent condensation from dripping onto the agar surface and spreading colonies.
[ ] B. To allow better oxygen exchange for obligate aerobes.
[ ] C. To prevent the agar from drying out during incubation.
[ ] D. To ensure even distribution of nutrients throughout the medium.
CORRECT: A. To prevent condensation from dripping onto the agar surface and spreading colonies.
Inverting plates prevents water condensation from the lid dripping onto the agar, which could cause confluent
growth or contamination. Oxygen exchange, drying, and nutrient distribution are not significantly affected by
inversion.
7. During a serial dilution, a student accidentally plates 0.1 mL instead of 1.0 mL from the
10^-4 dilution. If the original culture had 2.0 × 10^7 CFU/mL, how many colonies would be
expected on the plate?
[ ] A. 200 colonies
[ ] B. 20 colonies
[ ] C. 2000 colonies
HIGHP - Page 3 of 49
, [ ] D. 2 colonies
CORRECT: A. 200 colonies
Expected colonies = original concentration × dilution × volume plated = 2.0 × 10^7 × 10^-4 × 0.1 = 200.
Options B, C, and D result from misplacing the decimal or using the wrong volume.
8. A bacterial culture is grown in a broth medium and then streaked onto a solid agar
plate. After incubation, no colonies appear, but the broth was turbid. What is the most
likely explanation?
[ ] A. The bacteria were nonviable and could not grow on solid medium.
[ ] B. The agar plate was incubated at the wrong temperature.
[ ] C. The bacteria were obligate anaerobes and failed to grow on the aerobic plate.
[ ] D. The streak plate was performed incorrectly, resulting in no transfer of bacteria.
CORRECT: D. The streak plate was performed incorrectly, resulting in no transfer of bacteria.
Turbid broth indicates viable bacteria; failure to grow on the plate most likely results from poor streaking
technique (e.g., loop not cooled, no transfer). Options A, B, and C are less likely because the broth showed
growth.
9. Which of the following is the correct order of steps for a Gram stain?
[ ] A. Crystal violet -> Gram's iodine -> alcohol -> safranin
[ ] B. Safranin -> Gram's iodine -> alcohol -> crystal violet
[ ] C. Crystal violet -> alcohol -> Gram's iodine -> safranin
[ ] D. Gram's iodine -> crystal violet -> alcohol -> safranin
CORRECT: A. Crystal violet -> Gram's iodine -> alcohol -> safranin
The Gram stain sequence is crystal violet (primary stain), Gram's iodine (mordant), alcohol (decolorizer), and
safranin (counterstain). The other options have incorrect order, which would yield erroneous results.
10. A student performs a viable plate count and obtains 45 colonies on a plate from a
10^-6 dilution (1.0 mL plated). The next dilution (10^-7) yields 5 colonies. Which plate
count should be used to calculate the original cell concentration, and why?
[ ] A. The 10^-6 plate, because it has more colonies and is statistically more reliable.
[ ] B. The 10^-7 plate, because it has fewer colonies and is easier to count.
[ ] C. The 10^-6 plate, because it falls within the acceptable range of 30-300 colonies.
[ ] D. The 10^-7 plate, because it falls within the acceptable range of 30-300 colonies.
CORRECT: C. The 10^-6 plate, because it falls within the acceptable range of 30-300 colonies.
Only plates with 30-300 colonies are considered statistically valid for viable counts. The 10^-6 plate with 45
colonies is within range, while the 10^-7 plate with 5 colonies is below the threshold and should not be used.
HIGHP - Page 4 of 49
Requires Respondus LockDown Browser |
Questions and Answers | 2026/2027 Updated
| 100% Correct
Practice Question Bank & Answer Rationale
THIS PACK CONTAINS
148 exam-style multiple-choice questions. Each question includes the correct answer and a complete
rationale for focused revision.
COVERAGE
This exam assesses foundational laboratory skills developed during the first three weeks of an introductory
microbiology lab, including aseptic technique, brightfield microscopy, microbial cultivation, and standard
quantification methods. Items require application of laboratory concepts to interpret results, troubleshoot
procedures, and select appropriate techniques.
STUDY GUIDE
Recommended duration: 60 minutes. Passing target: 70%.
HIGHP - Page 1 of 49
,1. During a streak-plate procedure, a student fails to flame the loop between the first and
second quadrants. What is the most likely consequence on the plate after incubation?
[ ] A. No growth will appear because the loop was sterile initially.
[ ] B. Heavy confluent growth will appear in all quadrants, making isolation impossible.
[ ] C. The second quadrant will show reduced growth due to residual bacteria from the first quadrant,
hindering isolation.
[ ] D. Only the first quadrant will have growth; the rest will remain sterile.
CORRECT: C. The second quadrant will show reduced growth due to residual bacteria from the first
quadrant, hindering isolation.
Failure to flame the loop between quadrants carries over high cell numbers from the previous quadrant,
reducing the dilution effect and preventing isolated colonies. Option B is incorrect because some dilution still
occurs; D is incorrect because bacteria are still transferred.
2. A specimen is examined under a compound light microscope with a 10X ocular lens
and a 40X objective lens. What is the total magnification of the image?
[ ] A. 40X
[ ] B. 50X
[ ] C. 400X
[ ] D. 1000X
CORRECT: C. 400X
Total magnification is calculated by multiplying the ocular lens magnification (10X) by the objective lens
magnification (40X), yielding 400X. Options A, B, and D result from incorrect multiplication or using the wrong
lens values.
3. Which statement best describes the purpose of a differential medium such as
MacConkey agar?
[ ] A. It inhibits the growth of Gram-positive bacteria while allowing Gram-negative bacteria to grow.
[ ] B. It allows visual differentiation of lactose-fermenting from non-fermenting bacteria based on colony
color.
[ ] C. It provides enriched nutrients to fastidious organisms that cannot grow on standard media.
[ ] D. It selectively supports the growth of fungi while inhibiting bacterial growth.
CORRECT: B. It allows visual differentiation of lactose-fermenting from non-fermenting bacteria
based on colony color.
Differential media contain indicators that reveal metabolic differences, such as lactose fermentation on
MacConkey agar. Option A describes a selective property, C describes enrichment, and D is unrelated to
MacConkey agar.
HIGHP - Page 2 of 49
,4. In a viable plate count, a 1.0 mL sample of a 10^-5 dilution yields 85 colonies on a
standard plate. What is the original cell concentration (CFU/mL) of the sample?
[ ] A. 8.5 × 10^5 CFU/mL
[ ] B. 8.5 × 10^6 CFU/mL
[ ] C. 8.5 × 10^7 CFU/mL
[ ] D. 8.5 × 10^4 CFU/mL
CORRECT: B. 8.5 × 10^6 CFU/mL
CFU/mL = colonies / (dilution × volume plated) = 85 / (10^-5 × 1.0 mL) = 85 × 10^5 = 8.5 × 10^6 CFU/mL.
The other options reflect errors in exponent arithmetic or volume plated.
5. A student observes a Gram-stained smear under oil immersion and sees pink,
rod-shaped cells arranged in chains. Which conclusion is most accurate?
[ ] A. The organism is a Gram-positive coccus.
[ ] B. The organism is a Gram-negative bacillus.
[ ] C. The organism is a Gram-positive bacillus.
[ ] D. The organism is a Gram-negative coccus.
CORRECT: B. The organism is a Gram-negative bacillus.
Pink color indicates Gram-negative, and rod shape indicates bacillus. Option A is incorrect because
Gram-positive cells appear purple and cocci are spherical; C and D misidentify either color or shape.
6. Which of the following best explains why plates are incubated in an inverted position?
[ ] A. To prevent condensation from dripping onto the agar surface and spreading colonies.
[ ] B. To allow better oxygen exchange for obligate aerobes.
[ ] C. To prevent the agar from drying out during incubation.
[ ] D. To ensure even distribution of nutrients throughout the medium.
CORRECT: A. To prevent condensation from dripping onto the agar surface and spreading colonies.
Inverting plates prevents water condensation from the lid dripping onto the agar, which could cause confluent
growth or contamination. Oxygen exchange, drying, and nutrient distribution are not significantly affected by
inversion.
7. During a serial dilution, a student accidentally plates 0.1 mL instead of 1.0 mL from the
10^-4 dilution. If the original culture had 2.0 × 10^7 CFU/mL, how many colonies would be
expected on the plate?
[ ] A. 200 colonies
[ ] B. 20 colonies
[ ] C. 2000 colonies
HIGHP - Page 3 of 49
, [ ] D. 2 colonies
CORRECT: A. 200 colonies
Expected colonies = original concentration × dilution × volume plated = 2.0 × 10^7 × 10^-4 × 0.1 = 200.
Options B, C, and D result from misplacing the decimal or using the wrong volume.
8. A bacterial culture is grown in a broth medium and then streaked onto a solid agar
plate. After incubation, no colonies appear, but the broth was turbid. What is the most
likely explanation?
[ ] A. The bacteria were nonviable and could not grow on solid medium.
[ ] B. The agar plate was incubated at the wrong temperature.
[ ] C. The bacteria were obligate anaerobes and failed to grow on the aerobic plate.
[ ] D. The streak plate was performed incorrectly, resulting in no transfer of bacteria.
CORRECT: D. The streak plate was performed incorrectly, resulting in no transfer of bacteria.
Turbid broth indicates viable bacteria; failure to grow on the plate most likely results from poor streaking
technique (e.g., loop not cooled, no transfer). Options A, B, and C are less likely because the broth showed
growth.
9. Which of the following is the correct order of steps for a Gram stain?
[ ] A. Crystal violet -> Gram's iodine -> alcohol -> safranin
[ ] B. Safranin -> Gram's iodine -> alcohol -> crystal violet
[ ] C. Crystal violet -> alcohol -> Gram's iodine -> safranin
[ ] D. Gram's iodine -> crystal violet -> alcohol -> safranin
CORRECT: A. Crystal violet -> Gram's iodine -> alcohol -> safranin
The Gram stain sequence is crystal violet (primary stain), Gram's iodine (mordant), alcohol (decolorizer), and
safranin (counterstain). The other options have incorrect order, which would yield erroneous results.
10. A student performs a viable plate count and obtains 45 colonies on a plate from a
10^-6 dilution (1.0 mL plated). The next dilution (10^-7) yields 5 colonies. Which plate
count should be used to calculate the original cell concentration, and why?
[ ] A. The 10^-6 plate, because it has more colonies and is statistically more reliable.
[ ] B. The 10^-7 plate, because it has fewer colonies and is easier to count.
[ ] C. The 10^-6 plate, because it falls within the acceptable range of 30-300 colonies.
[ ] D. The 10^-7 plate, because it falls within the acceptable range of 30-300 colonies.
CORRECT: C. The 10^-6 plate, because it falls within the acceptable range of 30-300 colonies.
Only plates with 30-300 colonies are considered statistically valid for viable counts. The 10^-6 plate with 45
colonies is within range, while the 10^-7 plate with 5 colonies is below the threshold and should not be used.
HIGHP - Page 4 of 49
