Category: MICROBIOLOGY Objective Questions

Microbiology Multiple Choice Questions on “Soil Microbiology – Carbon Cycle & Sulphur Cycle”.

1. Which among the following are the most important agents for carbon dioxide fixation?
A. bacteria
B. fungi
C. algae
D. protozoa
Answer: C
Clarification: Green plants and algae are the most important agents of carbon dioxide fixation.

2. Fresh air contains approximately ______ percent carbon dioxide by volume.
A. 0.01
B. 2
C. 5
D. 0.03
Answer: D
Clarification: Fresh air contains approximately 0.03 percent carbon dioxide by volume. The end product, carbon dioxide, from the degradation of organic carbon compounds are released into the air and soil.

3. Cellulose is degraded to cellobiose by the enzyme __________________
A. cellulase
B. beta-glucosidase
C. hexokinase
D. cellulose dehydrogenase
Answer: A
Clarification: The most abundant organic material in plants is cellulose. The initial enzymatic at-tack is by cellulase which splits this long-chain polymer of glucose to cellobiose which contains two glucose units.

4. The amount of organic material in the forest soil remains approximately the same from year to year.
a)True
B. False
Answer: A
Clarification: Under most natural systems of vegetation, e.g., forests, the amount of organic material in the soil remains approximately the same from year to year. This results from a balance established between the annual litterfall and death of the plants and the capacity of microorganisms to degrade these tissues.

5. In carbon cycle flow of energy is _________________
A. bidirectional
B. linear
C. cyclic
D. irreversible
Answer: B
Clarification: In carbon cycle flow of energy is linear in unidirectional way through carbon dioxide in the atmosphere to organic forms of carbon in soil.

6. Cellobiose is metabolized directly by many microorganisms.
A. True
B. False
Answer: B
Clarification: Cellobiose is split into glucose by the enzyme beta-glucosidase; glucose is metabolized directly by microorganisms.

7. Cysteine breaks down in presence of cysteine desulfurase to give _________________
A. oxaloacetic acid
B. sulphuric acid
C. pyruvic acid
D. glyoxalate
Answer: C
Clarification: Degradation of proteins liberates amino acids, some of which contain sulphur released by the enzymatic activity of many heterotrophic bacteria. Cysteine breaks down in presence of cysteine desulfurase to give pyruvic acid, hydrogen sulphide and ammonia.

8. Which of the following processes is performed by Thiobacillus thiooxidans?
A. converting sulphur to sulphates
B. converting sulphur to sulphides
C. converting sulphur to sulphites
D. converting organic sulphur to inorganic sulphur
Answer: A
Clarification: Thiobacillus thiooxidans, an autotroph, is capable of oxidizing elemental sulphur to sulphates.

9. Sulphates are reduced to hydrogen sulphide by _____________________
A. Desulfotomaculum sp.
B. Thiobacillus thiooxidans
C. Photosynthetic sulfur bacteria
D. Rhodospirillum
Answer: A
Clarification: Sulphates are reduced to hydrogen sulphide by soil microorganisms like Desulfotomaculum species. Like calcium sulphate gives us hydrogen sulphide and calcium hydroxide.

10. The Winogradsky column experiment is done in the dark.
A. True
B. False
Answer: B
Clarification: A laboratory technique which facilitates isolation of various sulfur-metabolilizing bacteria is the Winogradsky column. The column contains mud, CaSO₄, plant tissue, and water. It is exposed to daylight and incubated at room temperature.

11. The reduction of sulphates and sulphites to hydrogen sulphide is done by which group of bacteria?
A. aerobic sulfate-reducing bacteria
B. photosynthetic sulphur bacteria
C. anaerobic sulphate-reducing bacteria
D. heterotrophic bacteria
Answer: C
Clarification: Organic acids serve as the electron donors for the reduction of sulfates to sulfites to hydrogen sulfide by anaerobic sulfate-reducing bacteria.

12. Which among the following develop in the upper portion of the Winogradsky column?
A. Sulfate-reducing bacteria
B. Green-sulfur bacteria
C. Purple-sulfur bacteria
D. Thiobacilli
Answer: D
Clarification: The aerobic sulfur-metabolizing bacteria, Thiobacillus sp.., develop in the upper portion of the column and oxidize reduced sulfur compounds.

13. Purple and green sulfur bacteria use ___________________ as the electron donor to reduce carbon dioxide.
A. S^2-
B. SO₄^2-
C. H²S
D. Organic acids
Answer: C
Clarification: Photosynthetic microorganisms such as the purple and green sulfur bacteria use hydrogen sulphide as the electron donor to reduce carbon dioxide.

14. Which among the following is a non-sulfur purple bacteria?
A. Rhodomicrobium
B. Thiobacillus
C. Chromatium
D. Chlorobium
Answer: A
Clarification: The non-sulfur bacteria purple bacteria like Rhodomicrobium, Rhodospirillum. Rhodopseudomonas are facultative phototrophs and can utilize sulfide at low levels.

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Microbiology Multiple Choice Questions on “Bacterial Conjugation”.

1. The first demonstration of recombination in bacteria was achieved by _______________
A. Lederberg and Tatum
B. Luria and Delbruck
C. Joshua and Lederberg
D. Luria and Tatum
Answer: A
Clarification: The first demonstration of recombination in bacteria was achieved by Lederberg and Tatum in 1946 in a brilliant and remarkable experiment that opened the door to a whole new world of microbiology.

2. Mating between species is necessary for recombination to take place.
A. True
B. False
Answer: A
Clarification: In a regular cycle of sexual reproduction, there is an opportunity for different mutants of a species to mate with each other and produce new individuals with new combinations of mutations, i.e., to produce recombinants.

3. Which type of E.coli strain was chosen to prove the experiment of conjugation?
A. prototrophs
B. auxotrophs
C. polyauxotrophs
D. autotrophs
Answer: C
Clarification: When Lederberg and Tatum did their experiments, they used polyauxotrophs (mutants with more than one nutritional requirement) so that back mutation to the wild type does not occur.

4. In conjugation only small fragments of the bacterial chromosome are transferred.
A. True
B. False
Answer: B
Clarification: In conjugation it is possible for large segments of the chromosome, and in special cases the entire chromosome to be transferred.

5. What is the shape of DNA in the male cells of E.coli?
A. linear
B. supercoiled
C. circular
D. relaxed
Answer: C
Clarification: Male cells contain a small circular piece of DNA, which is in the cytoplasm and not part of a chromosome.

6. Which of the following cells of E.coli are referred to as F^—
A. Male cells
B. Female cells
C. Both male and female cells
D. Neither male nor female cells
Answer: B
Clarification: Female cells lack the sex factor or F factor (fertility factor) and are labeled as F^—. They are recipient cells.

7. What is the frequency of formation of recombinants in a F⁺ X F^— cross?
A. 100
B. 40
C. 10
D. 1
Answer: D
Clarification: The formation of recombinants in a F⁺ X F^— cross occurs at a low frequency— about one recombinant per 10⁴ to 10⁵ cells.

8. The F factor DNA is sufficient to specify how many genes?
A. 2
B. 10
C. 40
D. 100
Answer: C
Clarification: The F factor DNA is only sufficient to specify about 40 genes that control sex-factor replication and synthesis of sex pili.

9. Which of the following is true for an Hfr X F— cross?
A. frequency of recombination high, transfer of F factor low
B. frequency of recombination high, transfer of F factor high
C. frequency of recombination low, transfer of F factor high
D. frequency of recombination low, transfer of F factor low
Answer: A
Clarification: In an Hfr X F^— cross, the frequency of recombination is high and the transfer of F factor is low as here the F factor of the Hfr is rarely transferred during recombination.

10. Which of the following can be used as a measure to construct a linkage map of the Hfr chromosome?
A. frequency of recombination
B. time of entry
C. locus of mutation
D. transfer of F factor
Answer: B
Clarification: Each gene enters the F^— cell at a characteristic time, and a linkage map of the Hfr chromosome can be constructed using time of entry as a measure.

11. The Hfr chromosome is transferred to the F^— cell in a _____________ fashion.
A. circular
B. coiled
C. dimer
D. linear
Answer: D
Clarification: The Hfr chromosome is transferred to the F^— cell in a linear fashion even though it is a circular chromosome.

12. How much time is required to inject a copy of the whole Hfr E.coli genome?
A. 24 hrs
B. 30 mins
C. 100 mins
D. 48 hrs
Answer: C
Clarification: It takes about 100 min to inject a copy of the whole Hfr E.coli genome (i.e., the chromosome and the integrated F factor.

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Microbiology Questions on “Therapeutic Drugs for Treatment of Fungal and Protozoan Diseases”.

1. Nystatin is effective in curing ____________
A. Deep mycoses
B. Dermatophyte infections
C. Systemic mycoses
D. Candida infections

Answer: D
Clarification: Nystatin is effective in topical Candida infections but not in deep mycoses or even dermatophyte infections.

2. Polyene antibiotics act on the plasma membrane of the fungus.
A. True
B. False

Answer: A
Clarification: Polyene antibiotics like the nystatin and amphotericin B act on the plasma membrane of the fungus, combining with membrane sterols, and cause leakage of intracellular potassium and other metabolites.

3. 5-fluorocytosine is which type of antibiotic?
A. polyene
B. pyrimidine antimetabolites
C. imidazole
D. antibacterial

Answer: B
Clarification: 5-fluorocytosine is a type of pyrimidine antimetabolite which is especially effective in the treatment of yeast infections.

4. Which of the following antibiotics causes inhibition of ergosterol biosynthesis?
A. amphotericin B
B. 5-fluorocytosine
C. ketoconazole
D. miconazole

Answer: C
Clarification: The mechanism of action for ketoconazole is inhibition of ergosterol biosynthesis, an important step in the synthesis of fungal cell walls and membranes.

5. Miconazole is active against which of the following fungal infections?
A. coccidioidomycosis
B. cryptococcosis
C. moniliasis
D. sporotrichosis

Answer: A
Clarification: Miconazole is very active against coccidioidomycosis and paracoccidioidomycosis. It reacts with cytoplasmic membranes and causes them to leak.

6. The antibiotic griseofulvin is active against deep mycoses.
A. True
B. False

Answer: B
Clarification: The antibiotic griseofulvin is active against the dermatophyte infections (superficial skin infections) but not against the deep mycoses (systemic diseases caused by fungi).

7. Which drug is used for the treatment of leishmaniasis?
A. chloroquine phosphate
B. metronidazole
C. suramin
D. sodium stibogluconate

Answer: D
Clarification: Leishmaniasis can be treated with the drug sodium stibogluconate. This drug is used for the treatment of protozoan diseases.

8. Which of the following animallike flagellates has a flagellum originating from the midportion of the cell?
A. amastigote
B. promastigote
C. epimastigote
D. trypomastigote

Answer: C
Clarification: The epimastigote has a flagellum originating from the midportion of the cell. It is a morphological stage of Leishmania and Trypanosoma.

Microbiology Multiple Choice Questions on “Microorganisms Microscopic Examination – Light Microscope”.

1. Which part of the compound microscope helps in gathering and focusing light rays on the specimen to be viewed?
A. Eyepiece lens
B. Objective lens
C. Condenser lens
D. Magnifying lens
Answer: C
Clarification: Compound microscope contains three separate lens systems. The condenser lens is placed between the light source and the specimen and it gathers and focuses the light rays in the plane of the microscopic field to view the specimen.

2. What is the minimum distance for the eye to focus any object?
A. 11 cm
B. 25 cm
C. 32 cm
D. 4 2 cm
Answer: B
Clarification: The eye cannot focus on objects brought closer to it less than 25 cm; this is, accordingly the distance of maximal effective magnification. An object must also subtend an angle at the eye of 1 degree or greater.

3. Resolving power of a microscope is a function of ____________
A. Wavelength of light used
B. Numerical aperture of lens system
C. Refractive index
D. Wavelength of light used and numerical aperture of lens system
Answer: D
Clarification: The ability of a microscope to distinguish two adjacent points as distinct and separate is known as resolving power. Resolving power is a function of wavelength of light used and the numerical aperture (NA. of the lens system. NA refers to the refractive index of the medium multiplied with the sine value of the half-aperture angle.

4. The greatest resolution in light microscopy can be obtained with ___________
A. Longest wavelength of visible light used
B. An objective with minimum numerical aperture
C. Shortest wavelength of visible light used
D. Shortest wavelength of visible light used and an objective with the maximum numerical aperture
Answer: D
Clarification: The relationship between numerical aperture (NA. and resolution is:-
Resolution (D. = wavelength / 2(NA)
Thus maximum resolution is obtained with the shortest wavelength of visible light and an objective with the maximum NA.

5. Oil immersion objective lens has an NA value of____________
A. 0.65
B. 0.85
C. 1.33
D. 1.00
Answer: C
Clarification: NA = refractive index * sine (half-aperture angle).
The maximum NA for a dry objective is less than 1.0 as the refractive index of air is 1. The values of NA for oil immersion lens is slightly greater than 1.0 in the range of (1.2 to 1.4) as the refractive index of oil is 1.56.

6. In fluorescence microscopy, which of the following performs the function of removing all light except the blue light?
A. Exciter filter
B. Barrier filter
C. Dichroic mirror
D. Mercury arc lamp
Answer: A
Clarification: In fluorescence microscopy, the function of the exciter filter is to remove all but the blue light; the barrier filter blocks out blue light and allows any other light emitted by the fluorescing specimen to pass through and reach the eye.

7. Total Magnification is obtained by __________
A. Magnifying power of the objective lens
B. Magnifying power of eyepiece
C. Magnifying power of condenser lens
D. Magnifying power of both the objective lens and eyepiece
Answer: D
Clarification: The total magnification is determined by multiplying the magnifying power of the objective by that of the eyepiece. Generally, an eyepiece having a magnification of 10X is used although eyepieces of higher or lower magnifications are available.

8. In light microscopy, which of the following is used as fixatives prior to staining technique?
A. Osmic acid
B. Glutaraldehyde
C. Heat
D. Osmic acid, glutaraldehyde, heat
Answer: C
Clarification: Most staining techniques kill cells and so preliminary to staining, the cells are sometimes fixed. Commonly used chemical fixatives include osmic acid and mainly glutaraldehyde. But for light microscopy heat is the most commonly used fixative.

9. In Phase contrast microscopy, the rate at which light enters through objects is __________
A. Constant
B. Inversely proportional to their refractive indices
C. Directly proportional to their refractive indices
D. Exponentially related to their refractive indices
Answer: B
Clarification: Phase contrast microscopy is based on the fact that the rate at which light travels through objects is inversely related to their refractive indices. Since the frequency of light waves is independent of the medium through which they travel, the phase of a light ray passing through an object of the higher refractive index than the surrounding medium will be relatively retarded.

10. Which part of the light microscope controls the intensity of light entering the viewing area?
A. Coarse adjustment screw
B. Fine adjustment screw
C. Diaphragm
D. Condenser lens
Answer: C
Clarification: On the condenser is mounted a shutter like an apparatus called the diaphragm which opens and closes to permit more or less light into the viewing area. Condenser lens just helps in condensing the light rays. Coarse and fine adjustment screws are used for focusing under different power lens.

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Microbiology Multiple Choice Questions on “Energy Production – Some Principles of Bioenergetics”.

1. The respiratory chain of bacteria is associated with the _____________
A. mitochondrial membrane
B. cytoplasmic membrane
C. cell wall
D. cytoplasm
Answer: B
Clarification: The respiratory chain of bacteria is associated with the cytoplasmic membrane and that of eukaryotes is present in the mitochondrial membrane.

2. The prosthetic group of a cytochrome contains how many iron atoms?
A. one
B. two
C. three
D. four
Answer: A
Clarification: Cytochromes are a major class of oxidative enzymes whose prosthetic group is a derivative of heme and contains a single iron atom, which is responsible for the oxidative or reductive properties of the enzyme.

3. Among the following which can directly react with oxygen?
A. cytochrome c
B. cytochrome c₁
C. cytochrome a
D. cytochrome a₃
Answer: D
Clarification: Cytochrome a and a₃ together are called cytochrome oxidase, both of them contain copper. But only cytochrome a₃ can react directly with oxygen.

4. How many ATP molecules are formed for per molecule of FADH₂ reoxidized?
A. 3
B. 1
C. 2
D. 4
Answer: C
Clarification: Three ATP are formed per molecule of NADH₂ reoxidized but only two ATP molecules are formed per molecule of FADH₂ reoxidized, because of the high energy of NADH₂.

5. In aerobic respiration the terminal electron acceptor is oxygen, nitrate, sulfate, etc.
A. True
B. False
Answer: B
Clarification: In aerobic respiration the terminal electron acceptor is oxygen; in anaerobic respiration the final electron acceptor is nitrate, sulphate and carbonate.

6. In photosynthesis by green plants, algae, cyanobacteria which of the following acts as terminal electron acceptor?
A. Water
B. Oxygen
C. NADP+
D. FAD+
Answer: C
Clarification: In photosynthesis by green plants, algae and cyanobacteria, water serves as a primary electron donor and NADP+ as a terminal electron acceptor.

7. A reducing agent will accept electrons and an oxidizing agent will donate electrons.
A. True
B. False
Answer: B
Clarification: An oxidizing agent will absorb electrons and will therefore become reduced whereas a reducing agent donates electrons, becoming oxidized in the process.

8. K_eq is greater than 1.0 depending on which of the following conditions?
A. standard free energy change is negative
B. standard free energy change is positive
C. chemical reaction proceeds in reverse direction
D. products are not formed
Answer: A
Clarification: When standard free energy change is negative value then K_eq is greater than 1.0 and the formation of products is favoured.

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Microbiology test focuses on “World of Bacteria II – Endospore – Forming Gram – Positive Bacteria”.

1. Which of the following are mesophilic saprophytes?
A. B.polymyxa
B. B.anthracis
C. B.subtilis
D. B.thuringiensis
Answer: C
Clarification: Bacillus subtilis are mesophilic saprophytes and are widely distributed in nature. This means they can grow best within a temperature range of approximately 25 to 40 degrees Celsius.

2. Which of the following causes “milky disease” of Japanese beetle grubs?
A. B.thuringiensis
B. B.popilliae
C. B.sphaericus
D. B.anthracis
Answer: B
Clarification: B.popilliae is a pathogenic species of Bacillus species that causes “milky disease” of Japanese beetle grubs.

3. Which of the following species is associated with spoilage of canned goods?
A. B.stearothermophiles
B. B.cereus
C. B.subtilis
D. B.sphaericus
Answer: A
Clarification: B.stearothermophilus is a thermophilic species having a maximum of 65 to 75 degrees Celsius. The endospores are highly resistant to heat and, therefore, this species is one of those associated with spoilage of canned goods.

4. Which of the following genus of species play an active role in the decomposition of urea?
A. Bacillus
B. Sporosarcina
C. Clostridium
D. Desulfotomaculum
Answer: B
Clarification: Sporosarcinae are widely distributed in fertile soil, where they play an active role in the decomposition of urea.

5. C.perfringens is the major causative agent of __________________
A. botulism
B. tetanus
C. gas gangrene
D. anthrax
Answer: C
Clarification: Clostridium perfringens is the major causative agent of the wound infection known as gas gangrene.

6. Which of the following Clostridium species has the ability to fix Nitrogen?
A. C.difficile
B. C.pasteurianum
C. C.tetani
D. C.thermosaccharolyticum
Answer: B
Clarification: C.pasteurianum is a mesophilic soil clostridium that is particularly noted for its ability to fix Nitrogen.

7. The genus Desulfotomaculum obtain energy by anaerobic respiration.
A. True
B. False
Answer: A
Clarification: The members of the genus Desulfotomaculum obtain energy by anaerobic respiration, with sulphate serving as the terminal electron acceptor and organic substrates such as lactic or pyruvic acid serving as the electron donors.

8. Pseudomembranous colitis is a disease of _________________
A. stomach
B. wounds
C. bowel
D. limbs
Answer: C
Clarification: C.difficile causes pseudomembranous colitis, a severe disease of the bowel.

9. Bacillus species cannot fix Nitrogen.
A. True
B. False
Answer: B
Clarification: Bacillus polymyxa has the ability to fix Nitrogen under anaerobic conditions.

To practice all areas of Microbiology for tests, .