8/29/25, 9:36 AM MLT ASCP Practice EXAM, QUESTIONS WITH ACCURATE ANSWERS | MULTIPLE CHOICES |2025\2026!! Flashcards | Quizlet
MLT ASCP Practice EXAM, QUESTIONS WITH
ACCURATE ANSWERS | MULTIPLE CHOICES
|2025\2026!!
Save
Terms in this set (100)
After experiencing B;
extreme fatigue and The correct answer for this question is 1300 mg/dL.
polyuria, a patient's basic The laboratorian performed a 1:4 dilution by adding
metabolic panel is 0.25 mL (or 250 microliters) of patient sample to 750
analyzed in the laboratory. microliters of diluent. This creates a total volume of
The result of the glucose 1000 microliters. So, the patient sample is 250
is too high for the microliters of the 1000 microliter mixed sample, or a
instrument to read. The ratio of 1:4. Therefore, the result given by the
laboratorian performs a chemistry analyzer must be multiplied by a dilution
dilution using 0.25 mL of factor of 4. 325 mg/dL x 4 = 1300 mg/dL.
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
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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
The urease reaction seen conversion of the slant and the
in the Christensen's urea butt of the tube to a pink color,
agar slant on the far right as seen in the tube to the left.
indicates: The slant only reaction in the
right tube may be seen early on
A. Weak activity if only the slant had been
B. Strong activity inoculated; however, with a
C. Slant only inoculated strong urease producer, both
D. Use of outdated the slant and the butt would
medium 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outdated for a prolonged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.
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D;
What is the first step of the
The steps in the PCR process are:
PCR reaction?
1. Denaturation (Turning double stranded DNA into
single strands.)
A. Hybridization
2. Annealing/Hybrization (Attachment of primers to
B. Extension
the single DNA strands.)
C. Annealing
3. Extension (Creating the complementary strand to
D. Denaturation
produce new double stranded DNA.)
The concentration of B;
sodium chloride in an Isotonic or normal saline is a 0.85 % solution of
isotonic solution is : sodium chloride in water.
A. 8.5 %
B. 0.85 %
C. 0.08 %
D. 1 molar
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Which of the following C;
laboratory results would In DIC, or disseminated intravascular coagulation, the
be seen in a patient with prothrombin time is increased due to the consumption
acute Disseminated of the coagulation factors due to the tiny clots
Intravascular Coagulation forming throughout the vasculature. This is also the
(DIC)? reason that the fibrinogen levels and platelet levels
are decreased. Finally FDP, or fibrin degredation
A. prolonged PT, elevated products, are increased due to the formation and
platelet count, decreased subsequent dissolving of many tiny clots in the
FDP vasculature. The FDPs are the pieces of fibrin that are
B. normal PT, decreased left after the fibrinolytic processes take place.
fibrinogen, decreased
platelet count, decreased
FDP
C. prolonged PT,
decreased fibrinogen,
decreased platelet count,
increased FDP
D. normal PT, decreased
platelet count, decreased
FDP
A dilution commonly used B;
for a routine sperm count A dilution commonly used for a routine sperm count is
is: a 1:20.
A. 1:2
B. 1:20
C. 1:200
D. 1:400
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MLT ASCP Practice EXAM, QUESTIONS WITH
ACCURATE ANSWERS | MULTIPLE CHOICES
|2025\2026!!
Save
Terms in this set (100)
After experiencing B;
extreme fatigue and The correct answer for this question is 1300 mg/dL.
polyuria, a patient's basic The laboratorian performed a 1:4 dilution by adding
metabolic panel is 0.25 mL (or 250 microliters) of patient sample to 750
analyzed in the laboratory. microliters of diluent. This creates a total volume of
The result of the glucose 1000 microliters. So, the patient sample is 250
is too high for the microliters of the 1000 microliter mixed sample, or a
instrument to read. The ratio of 1:4. Therefore, the result given by the
laboratorian performs a chemistry analyzer must be multiplied by a dilution
dilution using 0.25 mL of factor of 4. 325 mg/dL x 4 = 1300 mg/dL.
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
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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
The urease reaction seen conversion of the slant and the
in the Christensen's urea butt of the tube to a pink color,
agar slant on the far right as seen in the tube to the left.
indicates: The slant only reaction in the
right tube may be seen early on
A. Weak activity if only the slant had been
B. Strong activity inoculated; however, with a
C. Slant only inoculated strong urease producer, both
D. Use of outdated the slant and the butt would
medium 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.
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D;
What is the first step of the
The steps in the PCR process are:
PCR reaction?
1. Denaturation (Turning double stranded DNA into
single strands.)
A. Hybridization
2. Annealing/Hybrization (Attachment of primers to
B. Extension
the single DNA strands.)
C. Annealing
3. Extension (Creating the complementary strand to
D. Denaturation
produce new double stranded DNA.)
The concentration of B;
sodium chloride in an Isotonic or normal saline is a 0.85 % solution of
isotonic solution is : sodium chloride in water.
A. 8.5 %
B. 0.85 %
C. 0.08 %
D. 1 molar
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Which of the following C;
laboratory results would In DIC, or disseminated intravascular coagulation, the
be seen in a patient with prothrombin time is increased due to the consumption
acute Disseminated of the coagulation factors due to the tiny clots
Intravascular Coagulation forming throughout the vasculature. This is also the
(DIC)? reason that the fibrinogen levels and platelet levels
are decreased. Finally FDP, or fibrin degredation
A. prolonged PT, elevated products, are increased due to the formation and
platelet count, decreased subsequent dissolving of many tiny clots in the
FDP vasculature. The FDPs are the pieces of fibrin that are
B. normal PT, decreased left after the fibrinolytic processes take place.
fibrinogen, decreased
platelet count, decreased
FDP
C. prolonged PT,
decreased fibrinogen,
decreased platelet count,
increased FDP
D. normal PT, decreased
platelet count, decreased
FDP
A dilution commonly used B;
for a routine sperm count A dilution commonly used for a routine sperm count is
is: a 1:20.
A. 1:2
B. 1:20
C. 1:200
D. 1:400
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