ASCP Hematology MLT Exam 1 & 2 Questions and
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ANSWER STARTS THEN QUESTION
ANSWER- A;
One of the key characteristics to the identification of Nocardia asteroides is its inability to
hydrolyze casein, tyrosine or xanthine, as shown in this photograph. Nitrates are reduced to
nitrites. Both Nocardia brasiliensis and Actinomadura madurae hydrolyze both casein and
tyrosine; Streptomyces griseus hydrolyzes all three of the substrates.
Illustrated in this photograph is an agar quadrant plate containing casein (A), tyrosine (B),
nitrate (C) and xanthine (D). None of the substrates have been hydrolyzed and nitrate has
been reduced. The most likely identification is:
A. Nocardia asteroides
B. Nocardia brasiliensis
C. Streptomyces griseus
D. Actinomadura madurae
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A;
Since hemoglobin is measured spectrophotometrically on hematology analzyers, interference
from lipemia or icteric specimens can lead to decreased light detected and measured through
the sample and therefore inaccurate hemoglobin results occur.
On an electronic cell counter, hemoglobin determination may be falsely elevated caused by
the presence of:
A. Lipemic or icteric plasma
B. Leukocytopenia or Leukocytosis
C. Rouleaux or agglutinated RBCs
D. Anemia or Polycythemia
False
A patient who has a primarily vegetarian diet will most likely have an alkaline urine pH. A
low-carbohydrate diet as well as the ingestion of citrus fruits can also lead to a more alkaline
urine sample.
A patient who has a primarily vegetarian diet will most likely have an acid urine pH.
A;
During primary hypothyroidism, where a defect in the thryoid gland is producing low levels of
T3 and T4, the TSH level is increased. TSH is released in elevated quantities in an attempt to
stimulate the thryoid to produce more T3 and T4 as part of a feedback mechanism.
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Serum TSH levels five-times the upper limit of normal in the presence of a low T4 and low T3
uptake could mean which of the following:
A. The thyroid has been established as the cause of hypothyroidism
B. The thyroid is ruled-out as the cause of hypothyroidism
C. The pituitary has been established as the cause of hypothyroidism
D. The diagnosis is consistent with secondary hyperthyroidism
B;
The correct answer for this question is 1300 mg/dL. The laboratorian performed a 1:4 dilution
by adding 0.25 mL (or 250 microliters) of patient sample to 750 microliters of diluent. This
creates a total volume of 1000 microliters. So, the patient sample is 250 microliters of the
1000 microliter mixed sample, or a ratio of 1:4. Therefore, the result given by the chemistry
analyzer must be multiplied by a dilution factor of 4. 325 mg/dL x 4 = 1300 mg/dL.
After experiencing extreme fatigue and polyuria, a patient's basic metabolic panel is analyzed in
the laboratory. The result of the glucose is too high for the instrument to read. The laboratorian
performs a dilution using 0.25 mL of patient sample to 750 microliters of diluent. The result
now reads 325 mg/dL. How should the techologist report this patient's glucose result?
A. 325 mg/dL
B. 1300 mg/dL
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C. 975 mg/dL
D. 1625 mg/dL
A;
Conversion of only the slant to a pink color in a Christensen's urea agar slant is produced by
bacterial species that have weak urease activity. The reaction in the slant to the right is often
produced by Klebsiella species, as an example. Strong urease activity is indicated by
conversion of the slant and the butt of the tube to a pink color, as seen in the tube to the left.
The slant only reaction in the right tube may be seen early on if only the slant had been
inoculated; however, with a strong urease producer, both the slant and the butt would turn.
Therefore, the reaction is dependent on the strength of urease activity. If the media had
outdated for a prolonged period, either there would be no reaction or the appearance of only
a faint pink tinge, either in the slant, the butt or both, again depending on the strength of
urease production by the unknown organism.
The urease reaction seen in the Christensen's urea agar slant on the far right indicates:
A. Weak activity
B. Strong activity
C. Slant only inoculated
D. Use of outdated medium
D;
The steps in the PCR process are: