Interview MCQ Set 1
1. Which of the following is an indirect method for measuring bacterial growth?
a) Cell count
b) Cell mass
c) Cell activity
d) Both Cell mass and Cell activity
Answer: c [Reason:] Cell activity is an indirect method for measuring bacterial growth by relating the degree of biochemical activity to the size of the population.
2. Which of the following instrument is used for bacterial count?
a) Petroff-Hausser counting chamber
Answer: a [Reason:] Bacteria can be counted easily and accurately with the Petroff-Hausser counting chamber. This is a special slide accurately ruled into squares that are 1/400 mm2 in area. A suspension of unstained bacteria can be counted in the chamber, using a phase contrast microscope.
3. Which of the following method is used for viable count of a culture?
a) Direct microscopic count
b) Plate-count method
c) Membrane-filter count
d) Plate-count method and membrane-filter count
Answer: d [Reason:] The main disadvantage of direct counting of cell numbers is that there is no way to determine whether the cells being counted are viable.To determine the viable count of a culture, we must use a technique that allows viable cells to multiply, such as the plate-count method or membrane-filter method.
4. The number of bacteria per ml depends on the dilution of the sample.
Answer: a [Reason:] In the plate count method, the formula used is as follows:
Number of bacteria per ml= Number of colonies counted on plate X dilution of sample.
Thus number of bacteria directly depends on the dilution of sample.
5. Which of the following is the relationship between optical density and cell mass?
a) exponentially proportional
b) linearly proportional
c) inversely proportional
d) not related
Answer: b [Reason:] The photoelectric colorimeter used for measuring bacterial population, measures optical density (a function of light intensity) which is almost linearly proportional to cell mass.
6. How many cells present per millilitre in a bacterial culture can make the culture turbid?
c)1 lakh cells
Answer: d [Reason:] Bacteria in a suspension absorb and scatter the light passing through them, so that a culture of more than 107 to 108 cells per millilitre appears turbid to the naked eye. Then a spectrophotometer or colorimeter can be used for turbidimetric measurements.
7. A dead cell does not contribute to turbidity in the culture medium.
Answer: b [Reason:] Both dead as well as living cells contribute to turbidity. However, turbidity cannot be measured for cultures grown in deeply colored media or cultures that contain suspended material other than bacteria.
8. Which of the following is a direct measurement of growth?
a) Determination of nitrogen content
b) Turbidimetric methods
c) Determination of Dry weight of cells
d) Measurement of a specific chemical change produced on a constituent of the medium
Answer: c [Reason:] To measure the dry weight of cells is the most direct approach for quantitative measurement. All others are indirect methods and is applicable only in special circumstances.
9. Which of the following method is used for enumeration of bacteria in vaccines and cultures?
a) Microscopic Count
b) Membrane filter
c) Plate count
d) Dry weight determination
Answer: a [Reason:] Membrane count is used for enumeration of bacteria in vaccines and cultures. Even electronic enumeration is used in this application.
10. Colony-forming units per ml is the unit of _____________
a) Microscopic count
b) Electronic enumeration
c) Plate count
d) Turbidimetric measurement
Answer: c [Reason:] Since the plate count method is used in the enumeration of bacteria in milk, water,foods,soil,etc, the unit in which growth is measured is colony-forming units per ml as the bacteria forms colonies in the petri dish.
Interview MCQ Set 2
1. In which of the following situations will the logarithmic plot of survivors be constant?
a) physiological conditions are different
b) cells of microbial population vary in size
c) age and physiological conditions are uniform
d) temperature conditions are different
Answer: c [Reason:] Logarithmic plot reveals that the death rate is constant when all conditions are strictly uniform, including the age and the physiological condition of all the microorganisms in the population.
2. A suspension of E.coli exposed to a heat treatment may yield a greater number of survivors if a plating medium of _____________ is used.
b) trypticase soy agar
c) deoxycholate agar
d) potato dextrose
Answer: b [Reason:] A suspension of E.coli exposed to a heat treatment may yield a greater number of survivors if a plating medium of trypticase soy agar is used rather than a medium containing bile salts such as deoxycholate agar.
3. We require less time to kill the population of many cells.
Answer: b [Reason:] It takes time to kill the population and if we have many cells, we must treat them for a longer time to be reasonably sure that all of the bacteria are dead.
4. The effectiveness of heat in killing microorganisms is much greater in _____________
d) acid and alkali
Answer: a [Reason:] The environment has a profound influence on the rate as well as the efficacy of microbial destruction. The effectiveness of heat is much greater in acid than in alkali material.
5. A high concentration of which of the following compounds in the material generally increases the thermal resistance of microorganisms?
Answer: b [Reason:] The consistency of the material will markedly influence the penetration of the agent and high concentration of carbohydrates generally increases the thermal resistance of organisms.
6. Which of the following show the maximum resistance to physical and chemical agents?
b) Mold spores
c) Bacterial spores
Answer: c [Reason:] In spore forming species, the bacterial spores are extremely resistant in comparison to growing vegetative cells to physical and chemical agents.
7. How much time is required by the young cells to be killed by a lethal agent?
a) 24 hours
b) 3-4 hours
c) 30 mins
d) 5 mins
Answer: c [Reason:] The young cells are all killed within 25 minutes but a considerable part of the more resistant, older cells still survives. Young, actively metabolizing cells are apt to be more easily destroyed than old, dormant cells.
8. Which of the following actions are not affected by antimicrobial agents?
a) cell wall synthesis
b) nucleic acid synthesis
c) protein synthesis
d) capsule formation
Answer: d [Reason:] Antimicrobial agents inhibit or kill microorganisms by any of these actions like damaging cell wall synthesis, inhibiting protein or nucleic acid synthesis and inhibition of enzyme action.
Interview MCQ Set 3
1. Which fungi causes black wart disease of potato?
a) Saprolegnia parasitica
b) Synchytrium endobioticum
c) Rhizopus stolonifer
d) Saccharomyces cerevisiae
Answer: b [Reason:] 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?
b) septate with uninucleate
c) septate with multi muclei
Answer: d [Reason:] 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 ________________
Answer: b [Reason:] 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
Answer: a [Reason:] Schizosaccharomyces versatilis is isolated from grape juice.It grows like a 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 [Reason:] 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
c) genetics and metabolic pathway
d) plant diseases
Answer: c [Reason:] 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 [Reason:] 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 _______________
b) citric acid
c) gluconic acid
d) citric acid and gluconic acid
Answer: d [Reason:] 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 [Reason:] Candida albicans is often isolated from warm-blooded animals, including humans and becomes pathogenic and causes many serious infections.Endocaritis is caused by this organism and it is an infection of 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 [Reason:] In the wart disease of potato, when the host cell is infected by a zoospore, it reacts by undergoing hypertrophy,i.e., 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 [Reason:] 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.
Answer: b [Reason:]The rhizoidal hyphae enters 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 _____________
Answer: d [Reason:] The molds produce clusters of rootlike holdfasts called rhizoids.They also produce stolons or runners capable of taking root where thay may give rise to new organisms.
Interview MCQ Set 4
1. Bacterial chromosome has the capacity to code for ____________ different proteins.
Answer: d [Reason:] It has been estimated that bacterial chromosomes have the capacity to code for approximately 3500 different proteins.
2. Which is the functional unit of inheritance?
Answer: c [Reason:] A gene is a functional unit of inheritance; it specifies the formation of a particular polypeptide as well as various types of RNA. The chromosome is divided into genes.
3. Populations of cells descending from a single cell are known as ____________
c) wild-type cells
d) parent cell
Answer: b [Reason:] Populations of cells descending from a single cell are known as clones but a cell or an organism which shows the effects of a mutation is called a mutant.
4. The replacement of a purine by a pyrimidine or vice versa is known as ____________
c) base-pair substitution
d) missense mutation
Answer: a [Reason:] Transversion is a type of point mutation is the replacement of a purine by a pyrimi-dine or vice versa.
5. Sickle cell anemia is which type of mutation?
a) nonsense mutation
b) missense mutation
c) deletion mutation
d) insertion mutation
Answer: b [Reason:] A good example of a missense mutation in humans is the disease sickle cell ane-mia. A single base substitution in the codon for the sixth amino acid of normal hemoglobin A changes the sixth amino acid from glutamic acid to valine, thus forming the characteristic hemo-globin S of sickle cell anemia.
6. In nonsense mutation, the new mRNA codon specifies the same amino acid as that from the original codon.
Answer: b [Reason:] In neutral mutation the altered gene triplet produces a mRNA codon which specifies the same amino acid because the codon resulting from mutation is a synonym for the original codon.
7. Which of the following mutations result in frameshift mutation?
a) missense mutation
Answer: d [Reason:] Insertion and deletion mutations result in a shift of the reading frame caused by addition or loss of one or more nucleotides.
8. Which of the following mutagen causes a break in the phosphodiester backbone of the nucleic acid?
a) UV light
c) Nitrous acid
d) base analogs
Answer: b [Reason:] When x-rays interact with DNA, the result is usually a break in the phosphodiester backbone of the nucleic acid.
9. Base analogs have the same hydrogen-bonding properties as the normal bases.
Answer: b [Reason:] Although similar in structure, base analogs do not have the same hydrogen-
bonding properties as the normal bases.They can therefore introduce errors in replication which results in mutation.
10. Acridine orange is which type of mutagen?
a) base analog
b) chemical compounds
c) intercalating agents
Answer: c [Reason:] Acridine orange is an intercalating agent that can intercalate between the base pairs in the central stack of the DNA helix and distort the structures and cause subsequent replication errors.
11. Photoreactivation can cause repair of DNA damaged by which of the following mutagens?
a) UV rays
c) Nitrous acid
d) Base analogs
Answer: a [Reason:] Photoreactivation mechanism is used for repairing damage caused by UV radia-tion.A special enzyme designated PRE, induced by visible light, splits or unlinks the dimers formed because of exposure to UV light and restores the DNA to its original state.
12. Which of the following enzyme is used to join the DNA fragments together?
Answer: d [Reason:] Ligase enzyme is used to join the fragments together which are formed by poly-merases to fill in the gaps in the DNA strand.
13. What is the mutation rate for E.coli per bacterium per cell division?
a) 5.8 x 10-6
b) 5.8 x 10-9
c) 5.8 x 10-8
d) 5.8 x 10-7
Answer: c [Reason:] Mutation rate for E.coli maybe given as 5.8 x 10-8 mutations per bacterium per cell division.
Interview MCQ Set 5
1. The cocci belonging to the family Dienococcaceae occur mainly in _______________
Answer: a [Reason:] The cocci belonging to the family Deinococcaceae occur mainly in tetrads or cubical packets.
2. The organisms belonging to the genus Deinococcus forms colonies of which color?
Answer: c [Reason:] The family Deinococcaceae contains a single genus Deinococcus which forms red colonies.
3. The family Micrococcaceae do not exhibit any unusual resistance to gamma and ultraviolet radiation.
Answer: a [Reason:] The family Micrococcaceae do not exhibit any unusual resistance to gamma and ultraviolet radiation. But the organisms belonging to family Deinococcaceae have an unusually high resistance to gamma and ultraviolet radiations.
4. Among the following which is positive for coagulase test?
a) Staphylococcus aureus
b) Staphylococcus epidermidis
c) Staphylococcus saprophyticus
Answer: a [Reason:] Staphylococcus aureus is positive for coagulase test which is a test for the ability of the bacteria to cause blood plasma to clot.
5. Streptococcus are catalase-positive.
Answer: b [Reason:] Streptococcus are catalase-negative.The organisms are homofermentative i.e., the predominant end product of sugar fermentation is lactic acid.
6. Which of the following species of streptococci belongs to Lancefield group N?
Answer: d [Reason:] The streptococci are divided into categories known as the Lancefield groups based on differences in their cell-wall polysaccharides. S.cremoris belong to Lancefield group N and are harmless contaminants of milk and dairy products.
7. S.pneumoniae produces which of the following type of colonies?
d) alpha-, beta-haemolytic
Answer: a [Reason:] S.pneumoniae are alpha-haemolytic i.e. colonies are surrounded by a cloudy, colorless or greenish zone of partially lysed erythrocytes.
8. Which of the following is the causative agent of dental caries?
Answer: c [Reason:] S.mutans inhabits the human oral cavity and is the major causative agent of dental carries.
9. Which of the following genus of species occurs in human feces?
Answer: b [Reason:] Coprococcus occurs in human feces and uses carbohydrates as a major source of carbon and energy.