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ASCP PRACTICE QUESTIONS MICROBIOLOGY QUESTIONS AND ANSWERS WITH SOLUTIONS 2024
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ASCP PRACTICE QUESTIONSMICROBIOLOGY QUESTIONS AND
ANSWERS WITH SOLUTIONS 2024
Illustrated in the photograph are colonies that are distinctly pitting the agar surface. This property, in
conjunction with the detection of a bleach-like odor, provides a presumptive identification of Eikenella
corrodens. E. corrodens skin infections are almost always related to human bites, in contrast to
infections with Capnocytophaga canimorsus, which are related to dog bites. In addition, the colonies of
C. canimorsus show and "gliding motility" rather than the pitting seen here. Although the colonies of
Moraxella lacunata (as the species name indicates) may pit the surface of agar, they do not have a bleach
odor and would be extremely rarely associated with wound infections. Although Haemophilus
aphrophilus can cause wound infections, the colonies do not pit the agar nor do they have a bleach odor.
- ANSWER Illustrated in this photograph are colonies growing on the surface of a chocolate agar plate.
The colonies were recovered from a subcutaneous infection of the forearm and had a bleach odor. The
most likely identification is:
Choose the single best answer
a. Moraxella lacunata
b. Haemophilus aphrophilus
c. Eikenella corrodens
d. Capnocytophaga canimorsus
If a blood agar plate and an EMB plate are both inoculated with 0.001 ml of urine, and 90 colonies of
Staph aureus grew on the blood agar plate, what should be reported?
Choose the single best answer
A. 90 colony forming units/mL of Staphylococci aureus
B. 90,000 colony forming units/mL of Staphylococci aureus
C. 9,000 colony forming units/mL of Staphylococci aureus
D. 9 colony forming units/mL of Staphylococci aureus - ANSWER B;
,90,000 colony forming units/mL of Staphylococci aureus should be reported. Calculate the number of
bacteria in the urine sample by multiplying the colony count X the dilution of the calibrated loop
(1/1000). If there are 90 colonies on the EMB, 90 X 1000 dilution = 90,000 colony forming units/ml of
urine.
A;
Salmonella spp. is the correct answer because the isolate is a non-lactose and sucrose fermenter on
TSIA, positive for H2S, lysine decarboxylase positive, urea negative, citrate positive, phenylalanine
deamination negative, indole negative, and ornithine positive. Shigella would be H2S negative, lysine
negative, urea negative, citrate negative, phenylalanine negative, and nonmotile. Proteus vulgaris would
not decarboxylase lysine and ornithine, and is citrate negative, phenylalanine positive, and produce
indole. Citrobacter freundii would usually ferment lactose or sucrose, not decarboxylase lysine, and
would be phenylalanine and indole negative. - ANSWER This member of the Enterobacteriaceae is
shown on triple sugar iron agar, lysine iron agar, urea agar, citrate agar , phenylanine deaminase agar,
and motility indole ornithine agar. It should be identified as:
The correct answer is highlighted below
A. Salmonella spp.
B. Shigella spp.
C. Proteus vulgaris
D. Citrobacter freundii
B;
Of the Mycobacterium species listed in this exercise, M. kansasii and M. marinum are the two
photochromogens, fitting the description of the colonies included in the question. Of these, M. marinum
would not grow at 35°C; therefore, the intended answer in this exercise is M. kansasii. M. tuberculosis is
neither a photochromogen nor scotochromogen, producing only buff-pigmented colonies no matter
what the light exposure. M. gordonae is a scotochromogen, and produces yellow-pigmented colonies,
both in the presence and in the absence of light. - ANSWER Illustrated in this photograph are two
Middlebrook 7H-11 plates incubated at 35°C, on which are growing colonies of an unknown
Mycobacterium species. The plate on the left was incubated in the dark; the plate on the right had been
exposed to light for 24 hours. The most likely identification is:
, The correct answer is highlighted below
A. Mycobacterium gordonae
B. Mycobacterium kansasii
C. Mycobacterium tuberculosis
D. Mycobacterium marinum
A;
Bacteroides fragilis is the anaerobe most commonly recovered from human infections, particularly those
that occur "below the diaphragm". It is commonly associated with parappendiceal abscesses. The
organism grows readily on anaerobic blood agar in an anaerobic atmosphere and produces small, gram
negative bacilli with rounded ends, as seen in the upper photograph. Characteristic of the genus is the
ability to grow in 20% bile and to hydrolyze esculin (shown by the growth in the bile quadrant and black
pigment in the esculin quadrant in the lower photograph), two key characteristics that separate the B.
fragilis group from Prevotella species and Porphyromonas species, which are negative for both.
Fusobacterium nucleatum can also be associated with intra abdominal cavity abscesses; however, the
bacterial cells are slender and tapered and growth is inhibited by 20% bile, esculin is not produced but
the indole reactio - ANSWER An eleven year old boy developed signs and symptoms of acute
appendicitis. At surgery, the appendix was found to be ruptured and encased in a 3 cm in diameter
abscess. Aspiration material was inoculated to an anaerobic blood agar plate. Small, gray, nonhemolytic,
semiopaque colonies grew anaerobically in 36 hours. The upper photograph illustrates the gram stain
features of the organism; the lower photograph is a Presumpto plate indicating the biochemical
reactions (reading clockwise from the 12 o'clock quadrant): growth in 20% bile agar, indole negative,
esculin hydrolysis positive and lecithinase negative. The bacterial species most likely associated with the
abscess is:
The correct answer is highlighted below
A. Bacteroides fragilis
B. Fusobacterium nucleatum
C. Porphyromonas asaccharolytica
D. Prevotella intermedia
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