Category: MICROBIOLOGY Objective Questions

Microbiology Multiple Choice Questions on “Wastewater and Treatment Processes”.

1. Which of the following sewerage systems carry domestic and industrial wastewater?
A. sanitary sewers
B. storm sewers
C. combined sewers
D. storm and combined sewers
Answer: A
Clarification: Sanitary sewers carry domestic and industrial wastewater for its ultimate treatment and disposal.

2. The more oxidizable organic material, the lesser the BOD.
A. True
B. False
Answer: B
Clarification: The magnitude of the BOD is related to the amount of organic material in the wastewater-i.e., the more oxidizable organic material, the higher the BOD.

3. Which of the following is a strict anaerobe?
A. Enterobacter
B. Alcaligenes
C. Pseudomonas
D. Methanosarcina
Answer: D
Clarification: Methane producers are strict anaerobes e.g., Methanobacterium, Methanosarcina, Methanococcus. They produce methane and carbon dioxide as end products.

4. In which of the following treatment involve oxidation of organic constituents of the wastewater?
A. Primary treatment
B. Secondary treatment
C. Advanced treatment
D. Final treatment
Answer: B
Clarification: Secondary or biological treatment is done to adsorb and ultimately oxidize organic constituents of the wastewater, i.e, to reduce the BOD.

5. The upper region of the trickling filter is favorable for the growth of _____________
A. fungi
B. protozoa
C. algae
D. bacteria
Answer: C
Clarification: The upper region of the trickling filter is favorable for the growth of algae, and at times their growth may become so extensive that it impairs the operation of the filter.

6. Activated sludge usually employs an aeration period of ________________
A. 1 hour
B. 24 hours
C. 10-15 hours
D. 4-8 hours
Answer: D
Clarification: The activated sludge process usually employs an aeration period of 4 to 8 hours, after which the mixture is piped to a sedimentation tank.

7. Oxidation ponds are very deep ponds.
A. True
B. False
Answer: B
Clarification: Oxidation ponds also called lagoons are shallow ponds which are 2 to 4 ft in depth designed to allow algal growth on the wastewater effluent.

8. Trickling filter is used in which of the following wastewater treatment processes?
A. Primary treatment
B. Secondary treatment
C. Advanced treatment
D. Final treatment
Answer: B
Clarification: The trickling filter is used in secondary treatment in which a stationary microbial culture is fed by a continuous supply of nutrients. It is used for filtration processes.

9. Belt filter presses are used in which of the following process?
A. Thickening
B. Stabilization
C. Dewatering
D. Disposal
Answer: C
Clarification: Dewatering is done by vacuum filters, belt filter presses and centrifuges and is often enhanced by the addition of polymer or other chemical coagulant aids.

10. Which of the following gases are produced in large amounts during sludge digestion?
A. methane
B. carbon-dioxide
C. hydrogen
D. nitrogen
Answer: A
Clarification: About 60-70% methane is produced during sludge digestion with smaller amounts of carbon dioxide, hydrogen and nitrogen.

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Microbiology Multiple Choice Questions on “Infection Process – Penetration of Epithelial Cell Layers”.

1. Which of the following microorganism causes infection in burns?
A. Pseudomonas aeruginosa
B. Clostridium perfringens
C. Borrelia sp
D. Streptococcus sp
Answer: A
Clarification: Burns often become infected by Pseudomonas aeruginosa or other aerobic or facultatively anaerobic bacteria from the surrounding.

2. V. cholerae does not penetrate into the body.
A. True
B. False
Answer: A
Clarification: V. cholerae, the causative agent of cholera, multiplies on the epithelial layer of the small intestine where it produces a toxin that causes the loss of fluid from the epithelial cells and kills the cells.

3. Which of the following microorganism causes relapsing fever in humans?
A. Pseudomonas aeruginosa
B. Clostridium perfringens
C. Borrelia sp.
D. Streptococcus sp.
Answer: C
Clarification: Borrelia sp. cause relapsing fever in humans when the spirochetes are introduced through the bite of a tick or a body louse.

4. Which of the following microorganism participates in active penetration into the body?
A. Clostridium perfringens
B. Shigella
C. Borrelia sp.
D. Pseudomonas aeruginosa
Answer: B
Clarification: In bacillary dysentery, Shigella bacteria penetrate into and kill the epithelial cells of the colon, then spread to adjacent cells, which are in turn killed.

5. Wounds and burns represent one active mechanism.
A. True
B. False
Answer: B
Clarification: Wounds and burns represent one passive mechanism and they can introduce pathogens directly into the underlying tissues.

6. Bacillary dysentery is caused by which of the following microorganism?
A. Clostridium perfringens
B. Borrelia sp.
C. Shigella
D. Pseudomonas aeruginosa
Answer: C
Clarification: Bacillary dysentery is caused by Shigella bacteria which penetrate into and kill the epithelial cells of the colon, then spread to adjacent cells, which are in turn killed.

7. In which of the following body part does the influenza virus does not penetrate?
A. small intestine
B. nasopharynx
C. trachea
D. bronchi
Answer: A
Clarification: The influenza virus penetrates the epithelial cells lining the nasopharynx, trachea and bronchi.

8. Which of the following is an arthropod?
A. Clostridium perfringens
B. Borrelia sp.
C. Shigella
D. Pseudomonas aeruginosa
Answer: B
Clarification: Arthropods follow passive penetration such as that by Borrelia species which causes relapsing fever in humans.

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Microbiology Multiple Choice Questions on “Morphology and Fine Bacteria Structure – Cell Wall”.

1. Peptidoglycan layer is present in large quantity in?
A. Gram-positive bacteria
B. Gram-negative bacteria
C. Fungi
D. Algae
Answer: A
Clarification: Gram-positive bacteria usually have a much greater amount of peptidoglycan in their cell walls than do Gram-negative bacteria. It may account for 50 percent or more of the dry weight of the wall of some Gram-positive species, but only about 10 percent of the wall of Gram-negative bacteria.

2. Peptidoglycan is made up of __________
A. N-acetylglucosamine
B. N-acetylmuramic acid
C. N-acetylglucosamine, N-acetylmuramic acid
D. N-acetylglucosamine, N-acetylmuramic acid, amino acids
Answer: D
Clarification: Peptidoglycan differs somewhat in composition and structure from one species to another, but it is basically a polymer of N-acetylglucosamine, N-acetylmuramic acid, Amino acids like-L-alanine, D-alanine, D-glutamate, and a diamino acid.

3. Teichoic acid present in Gram-positive bacteria can bind to which ion?
A. Fe ions
B. Phosphorus ions
C. Mg ions
D. Sulphur ions
Answer: C
Clarification: Teichoic acid bind magnesium ions and there is some evidence that they help to protect bacteria from thermal injury by providing an accessible pool of these cations for stabilization of the cytoplasmic membrane.

4. Cord factor is a ___________
A. protein
B. teichoic acid derivative
C. mycolic acid derivative
D. carbohydrate
Answer: C
Clarification: Cord factor (trehalose dimycolate) is a mycolic acid derivative which is toxic and plays an important role in the diseases caused by C.diphtheriae and M.tuberculosis.

5. The outer membrane of the Gram-negative cell wall is anchored to the underlying peptidoglycan by means of which of the following?
A. Braun’s Lipoprotein
B. Phospholipids
C. Proteins
D. Lipopolysaccharide
Answer: a
Clarification: The outer membrane of the Gram-negative cell wall is anchored to the underlying peptidoglycan by means of Braun’s lipoproteins. The membrane is a bilayered structure consisting mainly of phospholipids, proteins and lipopolysaccharides(LPS).

6. Which among the following acts as receptors for bacteriophage attachment in Gram-negative bacteria?
A. Cilia
B. O antigens
C. Lipid A
D. Teichoic acid
Answer: B
Clarification: The polysaccharide O antigens which extend like whiskers from the membrane surface into the surrounding medium. Many of the serological properties of Gram-negative bacteria are attributable to O antigens like they can serve as receptors for bacteriophage attachment.

7. Porins are special proteins act as channels in the outer membrane of Gram-negative bacteria.
A. True
B. False
Answer: A
Clarification: The outer membrane can allow smaller molecules such as nucleosides, oligosaccharides, monosaccharides, peptides and amino acids to pass across by means of channels in special proteins called porins.

8. NAG and NAM of peptidoglycan layer is linked by _________
A. beta-(1,4) glycosidic linkage
B. alpha-(1,4) glycosidic linkage
C. alpha-(1,6) glycosidic linkage
D. beta-(1,6) glycosidic linkage
Answer: A
Clarification: N-acetylglucosamine (NAG) and N-acetylmuramic acid (NAM) of peptidoglycan layer are linked by beta-(1, 4) glycosidic linkage. Each strand contains from 10 to 65 disaccharide units.

9. Gram-negative bacteria are more resistant to antibiotics due to the presence of?
A. Thin peptidoglycan wall
B. Outer lipopolysaccharide layer
C. Porin proteins
D. Teichoic acid
Answer: B
Clarification: Gram-negative bacteria consists of an outer membrane made up of lipopolysaccharides beneath the thin peptidoglycan layer. The outer membrane serves as a barrier to various external chemicals and enzymes that could damage the cell. It also protects the bacteria from antibiotics.

10. Which of the following are present in teichoic acids?
A. ribitol residues
B. glycerol residues
C. glucose residues
D. ribitol or glycerol residues
Answer: D
Clarification: The teichoic acids are water soluble polymers, containing ribitol or glycerol residues joined through phosphodiester linkages. The glycerol or ribitol is joined to a sugar residue such as glucose, galactose or N-acetyl glucosamine.

11. Bayer’s junctions are sites which help in joining which of the following?
A. cytoplasmic membrane and outer membrane
B. outer membrane and capsule
C. cytoplasmic membrane and periplasmic space
D. peptidoglycan layer and cytoplasmic membrane
Answer: A
Clarification: The cytoplasmic membrane and outer membrane are joined at sites termed Bayer’s junctions. In these regions, the outer surface of the cytoplasmic membrane is continuous with the inner surface of the outer membrane creating pores that vary in diameter from 25 to 50 nm.

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Microbiology Multiple Choice Questions on “Microbial Metabolism – Biosynthesis of Deoxyribonucleic Acid”.

1. In DNA molecule, the amount of purines is equal to the amount of pyrimidines.
A. True
B. False
Answer: A
Clarification: In DNA the nitrogenous bases like adenine binds with thymine, and guanine binds with cytosine. As a consequence, the ratio of adenine to thymine or of guanine to cytosine in double stranded DNA is always 1:1. So the amount of purines is equal to the amount of pyrimidines.

2. RNA consists of which combination of bases?
A. Thymine, Guanine, Cytosine
B. Guanine, Cytosine, Uracil
C. Thymine, Adenine, Guanine
D. Adenine, Cytosine, Thymine
Answer: B
Clarification: RNA is composed of nitrogenous bases adenine, guanine, cytosine and uracil. They do not contain thymine as present in DNA.

3. Which of the following are features of semiconservative replication?
A. RNA replicates DNA molecule
B. The DNA molecule produced contains two old templates
C. One new DNA helix is formed
D. DNA duplicates itself and the new chain contains old template strand and new complementary strand
Answer: D
Clarification: In semiconservative replication DNA duplication occurs in which one polynucleotide chain acts as a template to direct the synthesis of a new chain complementary to itself. It results in two daughter helices, each containing one old template strand and one new complementary strand.

4. The chromosome of a typical bacterium is ________________
A. circular double-stranded DNA
B. circular single-stranded DNA
C. double-helix DNA
D. single stranded DNA
Answer: A
Clarification: The chromosome of a typical bacterium is a circular double-stranded DNA molecule; that is, the double helix for a complete genome has no free ends.

5. If the chromosome were extended linearly then it would approximately measure?
A. 13500 micrometre
B. 1000 micrometre
C. 1250 micrometre
D. 500 micrometre
Answer: C
Clarification: If the chromosome were extended linearly then it would measure approximately 1250 micrometers or 1.25 mm which is several hundred times longer than the bacterial cell that contains it.

6. The Lambda Bacteriophage carries out which of the following replication methods?
A. Theta mode
B. Sigma mode
C. Linear mode
D. Does not carry out replication
Answer: B
Clarification: The Lambda Bacteriophage follows sigma mode of replication in which the intermediate structures of DNA have sigma conformation. In this replication begins with the cleavage of a phosphodiester bond in one strand of the circular DNA molecule to produce a nick with 3’-OH and 5’-PO4 ends on that strand.

7. Which of the following replication methods leads to the production of two circular daughter chromosomes?
A. Theta mode
B. Sigma mode
C. Linear mode
D. Rolling circle method
Answer: A
Clarification: In Theta mode of replication a circular parental chromosome is replicated to two circular daughter chromosomes, in each of which one strand of the parental DNA molecule is conserved and a complementary strand is newly synthesized.

8. Eukaryotes replicate their DNA from one origin or growing point per molecule.
A. True
B. False
Answer: B
Clarification: Procaryotes replicate their DNA from one origin or growing point per molecule while eukaryotes replicate from many origins per molecule. Replication may occur in either a unidirectional or bidirectional manner from each origin.

9. DNA gyrase is a ______________ protein.
A. helix-unwinding
B. helix-destabilizing
C. helix-relaxing
D. helix-winding
Answer: C
Clarification: DNA gyrase is a helix-relaxing protein which works along with other enzymes and participates in opening the parental DNA helix ahead of the replication fork.

10. Which of the following enzyme removes the RNA primer with its 5’-nuclease activity?
A. DNA polymerase I
B. DNA polymerase II
C. DNA polymerase III
D. RNA polymerase
Answer: A
Clarification: DNA polymerase I remove the RNA primer with its 5”-nuclease activity; simultaneously it fills in the gap with DNA via its 3’-polymerase activity.

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Microbiology Multiple Choice Questions on “Some Fungi of Special Interest”.

1. Which fungi causes black wart disease of potato?
A. Saprolegnia parasitica
B. Synchytrium endobioticum
C. Rhizopus stolonifer
D. Saccharomyces cerevisiae
Answer: B
Clarification: Synchytrium endobioticum is the most serious parasite of the class Chytridiomycetes that causes black wart disease of potato. The dark warts on the potatoes are galls in which the host cells have been stimulated to divide by the fungus.

2. Zygomycetes have which type of mycelia?
A. nonseptate
B. septate with uninucleate
C. septate with multi muclei
D. white,nonseptate
Answer: D
Clarification: Morphologically, the mycelia of Zygomycetes are usually white or gray and are nonseptate.

3. In the life cycle of Rhizopus stolonifer,when the protoplasts and nuclei of both gametangia coalesce, the structure is known as ________________
A. progametangia
B. coenozygote
C. zygospore
D. zygote
Answer: B
Clarification: The walls between two gametangia dissolve and their protoplasts coalesce. Nuclei of both mating types fuse in pairs, producing many zygote nuclei. The structure that contains them is then called a coenozygote.

4. Schizosaccharomyces versatilis is isolated from ________________
A. grape juice
B. cheese
C. honey
D. beer
Answer: A
Clarification: Schizosaccharomyces versatilis is isolated from grape juice. It grows like yeast but it can also form a true mycelium.

5. Legitimate copulation refers to ___________________
A. fusion between cells of same mating type
B. fusion between cells of different mating type
C. no fusion
D. different sex cells are present within the same body
Answer: B
Clarification: Fusion normally occurs only between cells of difffering mating types, a process termed as legitimate copulation. Such fusions result in diploid cells which form asci containing viable ascospores.

6. Neurospora sp. is important for the study of _____________________
A. human diseases
B. antibiotics
C. genetics and metabolic pathway
D. plant diseases
Answer: C
Clarification: Neurospora sp. which belongs to the class Ascomycetes is of particular interest to biologists because of its wide use in the study of genetics and metabolic pathways. Some species are also responsible for food spoilage and some are used in industrial fermentation.

7. Cryptococcosis is a disease of ________________
A. bacterial infection
B. parasitic infection
C. viral infection
D. mycotic infection
Answer: D
Clarification: Cryptococcus neoformans is an important basidiomycetous pathogen of humans, causing cryptococcosis, a generalized mycotic infection involving the bloodstream as well as lungs, central nervous system and other organs.

8. Aspergillus niger is used in the production of _______________
A. cheese
B. citric acid
C. gluconic acid
D. citric acid and gluconic acid
Answer: D
Clarification: It is an economically important fungi because they are used in a number of industrial fermentations, including the production of citric acid and gluconic acid in abundance.

9. Endocarditis is caused by which of the following fungi?
A. Candida albicans
B. Penicillium notatum
C. Penicillium chrysogenum
D. Agaricus campestris
Answer: A
Clarification: Candida albicans is often isolated from warm-blooded animals, including humans and becomes pathogenic and causes many serious infections. Endocarditis is caused by this organism and it is an infection of the heart.

10. What do you mean by hypertrophy?
A. repeated cell division
B. infection by zoospore
C. increase in cell volume
D. decrease in cell volume
Answer: C
Clarification: In the wart disease of potato, when the host cell is infected by a zoospore, it reacts by undergoing hypertrophy,i.e., an increase in cell volume, and adjacent cells also enlarge to form the characteristic rosette.

11. Which of the following fungi causes severe epidemic of disease among fish in the natural environment?
A. Saprolegnia ferax
B. Saprolegnia parasitica
C. Neurospora crassa
D. Neurospora sitophila
Answer: B
Clarification: Species of Saprolegnia are common in soil and fresh water; hence they are commonly called water molds. S.parasitica causes severe epidemics of disease among fish in the natural environment.

12. Reproductive organs are formed on the rhizoidal hyphae.
A. True
B. False
Answer: B
Clarification: The rhizoidal hyphae enter the substratum and serve to anchor the organism and to absorb nutrients, and the branched hyphae is the one on which reproductive organs are formed.

13. Rhizopus produces clusters of rootlike holdfasts called _____________
A. stolons
B. hyphae
C. runners
D. rhizoids
Answer: D
Clarification: The molds produce clusters of rootlike holdfasts called rhizoids. They also produce stolons or runners capable of taking root where that may give rise to new organisms.

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Microbiology Multiple Choice Questions on “Physical Agents”.

1. Among the following which one is the most effective method of killing microorganisms?
A. High temperature
B. Low temperature
C. High temperature, high moisture
D. Low temperature, high moisture
Answer: C
Clarification: High temperatures combined with high moisture are one of the most effective methods of killing microorganisms. It is important to distinguish between dry heat and moist heat in any procedure for microbial control.

2. How much time is required by spores of Clostridium botulinum to be killed by moist heat at 1200C?
A. 2 hours
B. 4 to 20 minutes
C. 24 hours
D. 6-7 hours
Answer: B
Clarification: Spores of Clostridium botulinum are killed in 4 to 20 minutes by moist heat at 1200C, whereas a 2 hours exposure to dry heat at the same temperature is required.

3. Vegetative cells are much more sensitive to heat than are spores.
A. True
B. False
Answer: A
Clarification: Vegetative cells are much more sensitive to heat than the spores because of the higher level of water activity in the vegetative cells.

4. Which of the following conditions are required for the determination of thermal death time?
A. temperature is varied
B. time is fixed
C. temperature is fixed and time is varied
D. temperature is fixed and time is not selected
Answer: C
Clarification: In thermal death time, the temperature is selected as the fixed point and the time is varied. It refers to the shortest period of time to kill a suspension of bacteria or spores at a prescribed temperature and under specific conditions.

5. Which of the following apparatus is used to provide steam under regulated pressure?
A. autoclave
B. laminar air flow
C. incubator
D. hot oven
Answer: A
Clarification: The laboratory apparatus designed to use steam under regulated pressure is called an autoclave. It is essentially a double-jacketed steam chamber that can maintain steam under pressure and temperature.

6. The general pressure at which the autoclave is operated is ___________
A. 5 Psi
B. 20 Psi
C. 15 Psi
D. 40 Psi
Answer: C
Clarification: Generally the autoclave is operated at a pressure of approximately 15 Psi at 121.50C. It is not the pressure that kills the organisms but the temperature of the steam.

7. Some amount of air is also essential inside the chamber of the autoclave.
A. True
B. False
Answer: B
Clarification: In the operation of an autoclave it is absolutely essential that the air in the chamber be completely replaced by saturated steam. If air is present, it will reduce the temperature obtained within the chamber substantially below.

8. Which of the following methods is used for killing microorganisms of only certain types and not all microorganisms?
A. Pasteurization
B. Incineration
C. Boiling water
D. Fractional Sterilization
Answer: A
Clarification: In pasteurization, milk, cream, and certain alcoholic beverages are subjected to a controlled heat treatment that kills microorganisms of certain types but does not destroy all organisms.

9. Which of the following instruments is used to perform tyndallization?
A. Autoclave
B. Steam Arnold
C. Gas oven
D. Incubator
Answer: B
Clarification: For tyndallization or fractional sterilization the apparatus used is the Steam Arnold however an autoclave can also be used if a free-flowing steam is used.

10. Which of the following actions occur due to low temperature?
A. coagulation of proteins
B. death of microorganisms
C. rate of metabolism is reduced
D. denatures proteins
Answer: C
Clarification: Temperatures below the optimum for growth depresses the rate of metabolism and if the temperature is sufficiently low, growth and metabolism ceases. Low temperature are also useful for the preservation of cultures.

11. Which of the following are resistant to desiccation?
A. gonococci
B. streptococci
C. meningococci
D. bacillus
Answer: B
Clarification: Streptococci are much more resistant to dessication, some survive weeks after being dried but species of Gram-negative cocci are very sensitive to dessication.

12. Passage of water from a low solute concentration into the cell is referred to as ______________
A. plasmoptysis
B. plasmolysis
C. isotonic
D. hypertonic
Answer: A
Clarification: The passage of water from a low solute concentration into the cell is termed plasmoptysis. The pressure built up within the cell as a result of this water intake is termed osmotic pressure.

13. Which of the following radiations have the energy to knock electrons away from molecules and ionize them?
A. Non-ionizing radiations
B. Acoustic radiations
C. Subatomic particles
D. Ionizing radiations
Answer: D
Clarification: Gamma rays and x-rays are called ionizing radiations because they have enough energy to knock electrons away from molecules and ionize them. When such radiations pass through cells, they create free hydrogen radicals, hydroxyl radicals and some peroxides which causes intracellular damage.

14. Wavelengths around ___________ have the highest bactericidal efficiency.
A. 150 Å
B. 3900 Å
C. 2650 Å
d)1500 Å
Answer: C
Clarification: The ultraviolet portion of the spectrum includes all radiations from 150 to 3900 Å. Wavelengths around 2650 Å have the highest bactericidal efficiency.

15. Which of the following inhibits DNA replication?
A. cathode rays
B. UV light
C. x-rays
D. gamma rays
Answer: B
Clarification: Ultraviolet light is absorbed by many cellular materials but most significantly by the nucleic acids where it does the most damage. Pyrimidine dimers are formed due to which DNA replication is inhibited and mutations can result.

16. Which material is used for the Berkefeld filter?
A. diatomaceous earth
B. asbestos pad
C. porcelain
D. sintered glass disks
Answer: A
Clarification: Filters can remove microorganisms from liquids or gases. The Berkefeld filter is made up of diatomaceous earth.

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