Deck 9: Final Identification
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Deck 9: Final Identification
1
A microbiologist sets up a set of decarboxylase tubes for identification of a glucose fermenting, gram-negative rod. After 24 hours, the tubes reveal
Base control: gray
Arginine: purple
Lysine: yellow
Ornithine: purple
These tubes should be interpreted as
A) all positive.
B) invalid.
C) positive for arginine and ornithine.
D) positive for lysine.
Base control: gray
Arginine: purple
Lysine: yellow
Ornithine: purple
These tubes should be interpreted as
A) all positive.
B) invalid.
C) positive for arginine and ornithine.
D) positive for lysine.
invalid.
2
A yellow ONPG test indicates production of
A) beta-galactosidase.
B) beta-galactoside permease.
C) lactase.
D) both a and b.
A) beta-galactosidase.
B) beta-galactoside permease.
C) lactase.
D) both a and b.
beta-galactosidase.
3
The cytochrome oxidase test is used primarily for the identification of
A) gram-positive organisms.
B) gram-negative organisms.
C) Veillonella species.
D) Micrococcus species.
A) gram-positive organisms.
B) gram-negative organisms.
C) Veillonella species.
D) Micrococcus species.
gram-negative organisms.
4
The metabolism of carbohydrates by Pseudomonas aeruginosa is determined using
A) the ONPG test.
B) a phenol red carbohydrate broth with a Durham tube.
C) duplicate semisolid carbohydrate tubes with and without oil.
D) TSI agar.
A) the ONPG test.
B) a phenol red carbohydrate broth with a Durham tube.
C) duplicate semisolid carbohydrate tubes with and without oil.
D) TSI agar.
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5
The detection of acetyl-methyl carbinol or acetoin is the basis of the _________ test.
A) Voges-Proskauer
B) methyl red
C) IMVIC
D) citrate
A) Voges-Proskauer
B) methyl red
C) IMVIC
D) citrate
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6
A positive citrate test is due to the organism's ability to
A) break down tryptophan.
B) use citrate as the sole source of carbon.
C) extract nitrogen from ammonium salt.
D) both b and c.
E) all of the above.
A) break down tryptophan.
B) use citrate as the sole source of carbon.
C) extract nitrogen from ammonium salt.
D) both b and c.
E) all of the above.
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7
Which biochemical test requires the use of 10% ferric chloride to detect a positive reaction?
A) Tryptophan deaminase
B) Phenylalanine deaminase
C) Deamination of lysine
D) Both a and b
E) Both b and c
A) Tryptophan deaminase
B) Phenylalanine deaminase
C) Deamination of lysine
D) Both a and b
E) Both b and c
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8
Which of the following reagents are required for testing of nitrate reduction?
1) Alpha-naphthylamine
2) Sulfanilic acid
3) Zinc dust
4) Alpha-naphthol
5) 40% KOH
A) 1 and 2
B) 1, 2, and 3
C) 4 and 5
D) 3, 4 and 5
1) Alpha-naphthylamine
2) Sulfanilic acid
3) Zinc dust
4) Alpha-naphthol
5) 40% KOH
A) 1 and 2
B) 1, 2, and 3
C) 4 and 5
D) 3, 4 and 5
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9
Phenol red carbohydrate broth should be incubated at/in ___________________
1) -25°C.
2) 35°C.
3) ambient air.
4) CO2.
A) 1 and 3
B) 1 and 4
C) 2 and 3
D) 2 and 4
1) -25°C.
2) 35°C.
3) ambient air.
4) CO2.
A) 1 and 3
B) 1 and 4
C) 2 and 3
D) 2 and 4
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10
Prior to adding reagent(s) and interpretation, the conventional methyl red test requires incubation
A) at 42°C.
B) for 24 hours.
C) at 25°C.
D) for 48 hours.
A) at 42°C.
B) for 24 hours.
C) at 25°C.
D) for 48 hours.
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11
The proper amount and order of addition of reagent(s) for the test for the production of acetoin is
A) 0.6 mL of 5.0% alpha naphthol and 0.2 mL of 40% KOH.
B) 0.6 mL of 40% KOH and 0.2 mL of 5.0% alpha-naphthol.
C) 0.2 mL of 5.0% alpha naphthol and 0.6 mL of 40% KOH.
D) 0.2 mL of 5.0% 40% KOH and 0.6 mL of 5.0% alpha naphthol.
A) 0.6 mL of 5.0% alpha naphthol and 0.2 mL of 40% KOH.
B) 0.6 mL of 40% KOH and 0.2 mL of 5.0% alpha-naphthol.
C) 0.2 mL of 5.0% alpha naphthol and 0.6 mL of 40% KOH.
D) 0.2 mL of 5.0% 40% KOH and 0.6 mL of 5.0% alpha naphthol.
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12
After the addition of reagents for the nitrate test, the tube remains colorless. To determine if the organism reduced nitrates to nitrogen gas, you must now add
A) 10% ferric chloride.
B) zinc dust.
C) TTC.
D) nothing-the test is complete.
A) 10% ferric chloride.
B) zinc dust.
C) TTC.
D) nothing-the test is complete.
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13
A member of the Enterobacteriaceae is non-motile. The other biochemical reactions do not correlate with the identification of Klebsiella or Shigella species. What is the next step in the workup of this organism?
A) Repeat the biochemical tests
B) Perform serotyping
C) Repeat the motility test at 25°C
D) Perform an oxidase test
A) Repeat the biochemical tests
B) Perform serotyping
C) Repeat the motility test at 25°C
D) Perform an oxidase test
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14
A day shift microbiologist encounters an oxidase negative, non-lactose fermenter in a urine culture. To determine if it is a member of the Enterobacteriaceae or a nonfermenter, he sets up a TSI agar slant. After 10 hours of incubation at 35°C in ambient air, the evening shift microbiologist reads the tube as a yellow slant and yellow deep or butt. This result
A) indicates a member of the Enterobacteriaceae.
B) indicates a nonfermenter.
C) could be due to a gram-positive organism. Perform a Gram stain.
D) is invalid.
A) indicates a member of the Enterobacteriaceae.
B) indicates a nonfermenter.
C) could be due to a gram-positive organism. Perform a Gram stain.
D) is invalid.
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15
An oxidase-negative, gram-negative rod turns the urea agar slant pink within 4 hours. This indicates
A) an invalid reaction.
B) a possible Proteus species.
C) that the organism will probably also be VP positive.
D) that it will be susceptible to vancomycin.
A) an invalid reaction.
B) a possible Proteus species.
C) that the organism will probably also be VP positive.
D) that it will be susceptible to vancomycin.
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16
A non-motile, VP-positive, gram-negative rod is a possible
A) Shigella species.
B) Enterobacter species.
C) Serratia species.
D) Klebsiella species.
A) Shigella species.
B) Enterobacter species.
C) Serratia species.
D) Klebsiella species.
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17
One potential disadvantage of an automated identification system is
A) the need for extensive training.
B) the cost.
C) a lack of specificity.
D) a lack of rapid turnaround.
A) the need for extensive training.
B) the cost.
C) a lack of specificity.
D) a lack of rapid turnaround.
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18
In nucleic acid synthesis, which enzyme is responsible for replication of DNA?
A) Restriction endonucleases
B) DNA ligase
C) Isomerase
D) DNA polymerase
A) Restriction endonucleases
B) DNA ligase
C) Isomerase
D) DNA polymerase
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19
Which procedure is best suited for determining the HIV viral load of an individual?
A) Direct fluorescent antibody assay
B) Monoclonal antibodies
C) Real-time polymerase chain reaction
D) DNA fingerprinting
A) Direct fluorescent antibody assay
B) Monoclonal antibodies
C) Real-time polymerase chain reaction
D) DNA fingerprinting
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20
The end result of a positive enzyme immunoassay is read as
A) agglutination.
B) fluorescence.
C) a lack of color.
D) a colored compound.
A) agglutination.
B) fluorescence.
C) a lack of color.
D) a colored compound.
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