MLT ASCP Practice Test Questions board practice with complete solutions 2023/2024
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MLT ASCP
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MLT ASCP
MLT ASCP Practice Test Questions board practice with complete solutions 2023/2024
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 volum...
MLT ASCP Practice Test Questions
board practice with complete solutions
2023/2024
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. - ANSWER-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
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 If the media had outdated for a prolonged period, either there would be no reaction or the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. - ANSWER-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:
1. Denaturation (Turning double stranded DNA into single strands.)
2. Annealing/Hybrization (Attachment of primers to the single DNA strands.)
3. Extension (Creating the complementary strand to produce new double stranded DNA.) - ANSWER-
What is the first step of the PCR reaction?
A. Hybridization
B. Extension
C. Annealing
D. Denaturation
B;
Isotonic or normal saline is a 0.85 % solution of sodium chloride in water. - ANSWER-The concentration
of sodium chloride in an isotonic solution is :
A. 8.5 %
B. 0.85 %
C. 0.08 %
D. 1 molar
C;
In DIC, or disseminated intravascular coagulation, the prothrombin time is increased due to the
consumption of the coagulation factors due to the tiny clots forming throughout the vasculature. This is
also the reason that the fibrinogen levels and platelet levels are decreased. Finally FDP, or fibrin
degredation products, are increased due to the formation and subsequent dissolving of many tiny clots
in the vasculature. The FDPs are the pieces of fibrin that are left after the fibrinolytic processes take
,place. - ANSWER-Which of the following laboratory results would be seen in a patient with acute
Disseminated Intravascular Coagulation (DIC)?
A. prolonged PT, elevated platelet count, decreased FDP
B. normal PT, decreased fibrinogen, decreased platelet count, decreased FDP
C. prolonged PT, decreased fibrinogen, decreased platelet count, increased FDP
D. normal PT, decreased platelet count, decreased FDP
B;
A dilution commonly used for a routine sperm count is a 1:20. - ANSWER-A dilution commonly used for a
routine sperm count is:
A. 1:2
B. 1:20
C. 1:200
D. 1:400
B;
Prozone effect (due to antibody excess) will result in an initial false negative in spite of the large amount
of antibody in the serum, followed by a positive result as the specimen is diluted. - ANSWER-The prozone
effect ( when performing a screening titer) is most likely to result in:
A. False positive
B. False negative
C. No reaction at all
D. Mixed field reaction
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. - ANSWER-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
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. - ANSWER-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. -
ANSWER-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. - ANSWER-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
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