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MLT ASCP Practice Test Questions Board PracticeUpdated 2023-2024 100% Verified Answers
Answer :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. - Quiz :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
Answer :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the media had outdated for a prolonged period, either there would be no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. - Quiz :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D. Use of outdated medium
, Answer :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.) - Quiz :What is the first step of the PCR reaction?
A. Hybridization
B. Extension
C. Annealing
D. Denaturation
Answer :B;
Isotonic or normal saline is a 0.85 % solution of sodium chloride in water. -
Quiz :The concentration of sodium chloride in an isotonic solution is :
A. 8.5 %
B. 0.85 %
C. 0.08 %
D. 1 molar
Answer :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. - Quiz :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
Answer :B;
,A dilution commonly used for a routine sperm count is a 1:20. - Quiz :A
dilution commonly used for a routine sperm count is:
A. 1:2
B. 1:20
C. 1:200
D. 1:400
Answer :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. - Quiz :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
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. - Quiz :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
Answer :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. - Quiz :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
Answer :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. - Quiz :A patient who has a primarily
vegetarian diet will most likely have an acid urine pH.
Answer :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. - Quiz :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
Answer :A;
Fusarium species is the most likely associated with mycotic keratitis.
Trichophyton rubrum is a dermatophyte that commonly causes an itching,
scaling skin infection of the feet, known as tinea pedis. Scedosporium
apiospermum is commonly associated with sinusitis. Aspergillus niger typically
causes otitis externa and can also be associated with sinusitis. - Quiz :Which of
the following species or organisms is the most likely to be the cause of mycotic
keratitis (fungal eye infection)?
A. Fusarium species
B. Trichophyton rubrum.
C. Scedosporium apiospermum
D. Aspergillus niger
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