MLT ASCP PRACTICE TEST QUESTIONS (BOARD PRACTICE)
ACTUAL EXAM QUESTIONS AND ANSWERS WITH CERTIFIED
RATIONALES (A+ GUIDE SOLUTION) NEWEST 2025/2026
Terms in this set (100)
After experiencing extreme B;
fatigue and polyuria, a The correct answer for this question is 1300 mg/dL.
patient's basic metabolic The laboratorian performed a 1:4 dilution by adding 0.25
panel is analyzed in the mL (or 250 microliters) of patient sample to 750 microliters
of diluent. This creates a total volume of 1000 microliters.
laboratory. The result of the
So, the patient sample is 250 microliters of the 1000
glucose is too high for the microliter mixed sample, or a ratio of 1:4. Therefore, the
instrument to read. The result given by the chemistry analyzer must be multiplied by
laboratorian performs a a dilution factor of 4. 325 mg/dL x 4 = 1300 mg/dL.
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/dLC.
975 mg/dL
D. 1625 mg/dL
,The urease reaction seen in A;
the Christensen's urea agar Conversion of only the slant to a pink
slant on the far right color in a Christensen's urea agar slant
is produced by bacterial species that
indicates:
have weak urease activity. The reaction
in the slant to the right is often produced by Klebsiella
A. Weak activity species, as an example. Strong urease activity is indicated by
B. Strong activity 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
C. Slant only
in the right tube may be seen early on if only the slant had
inoculatedD. Use of been inoculated; however, with a strong urease producer,
outdated both the slant and the butt would turn. Therefore, the
medium 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the media had outdated for a prolonged period, either there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.
,What is the first step of the D;
PCR reaction? The steps in the PCR process are:
1. Denaturation (Turning double stranded DNA intosingle
A. Hybridization strands.)
B. Extension 2. Annealing/Hybrization (Attachment of primers tothe single
C. Annealing DNA strands.)
D. Denaturation 3. Extension (Creating the complementary strand toproduce
new double stranded DNA.)
The concentration of sodium B;
chloride in an isotonic Isotonic or normal saline is a 0.85 % solution of sodium
chloride in water.
solution is :
A. 8.5 %
B. 0.85 %
C. 0.08 %
D. 1 molar
, Which of the following C;
laboratory results would be In DIC, or disseminated intravascular coagulation, the
seen in a patient with acute prothrombin time is increased due to the consumption of
the coagulation factors due to the tiny clots forming
Disseminated Intravascular
throughout the vasculature. This is also the reason that the
Coagulation (DIC)? fibrinogen levels and platelet levels are decreased. Finally
FDP, or fibrin degredation products, are increased due to the
A. prolonged PT, formation and subsequent dissolving of many tiny clots in
the vasculature. The FDPs are the pieces of fibrin that are
elevatedplatelet count,
left after the fibrinolytic processes take place.
decreased
FDP
B. normal PT,
decreasedfibrinogen,
decreased platelet count,
decreased
FDP
C. prolonged
PT,decreased fibrinogen,
decreased platelet count,
increased FDP
D. normal PT,
decreasedplatelet count,
decreased
FDP
B;
A dilution commonly used
A dilution commonly used for a routine sperm count is a 1:20.
for a routine sperm count is:
A. 1:2
B. 1:20
C. 1:200
D. 1:400
ACTUAL EXAM QUESTIONS AND ANSWERS WITH CERTIFIED
RATIONALES (A+ GUIDE SOLUTION) NEWEST 2025/2026
Terms in this set (100)
After experiencing extreme B;
fatigue and polyuria, a The correct answer for this question is 1300 mg/dL.
patient's basic metabolic The laboratorian performed a 1:4 dilution by adding 0.25
panel is analyzed in the mL (or 250 microliters) of patient sample to 750 microliters
of diluent. This creates a total volume of 1000 microliters.
laboratory. The result of the
So, the patient sample is 250 microliters of the 1000
glucose is too high for the microliter mixed sample, or a ratio of 1:4. Therefore, the
instrument to read. The result given by the chemistry analyzer must be multiplied by
laboratorian performs a a dilution factor of 4. 325 mg/dL x 4 = 1300 mg/dL.
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/dLC.
975 mg/dL
D. 1625 mg/dL
,The urease reaction seen in A;
the Christensen's urea agar Conversion of only the slant to a pink
slant on the far right color in a Christensen's urea agar slant
is produced by bacterial species that
indicates:
have weak urease activity. The reaction
in the slant to the right is often produced by Klebsiella
A. Weak activity species, as an example. Strong urease activity is indicated by
B. Strong activity 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
C. Slant only
in the right tube may be seen early on if only the slant had
inoculatedD. Use of been inoculated; however, with a strong urease producer,
outdated both the slant and the butt would turn. Therefore, the
medium 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.
,What is the first step of the D;
PCR reaction? The steps in the PCR process are:
1. Denaturation (Turning double stranded DNA intosingle
A. Hybridization strands.)
B. Extension 2. Annealing/Hybrization (Attachment of primers tothe single
C. Annealing DNA strands.)
D. Denaturation 3. Extension (Creating the complementary strand toproduce
new double stranded DNA.)
The concentration of sodium B;
chloride in an isotonic Isotonic or normal saline is a 0.85 % solution of sodium
chloride in water.
solution is :
A. 8.5 %
B. 0.85 %
C. 0.08 %
D. 1 molar
, Which of the following C;
laboratory results would be In DIC, or disseminated intravascular coagulation, the
seen in a patient with acute prothrombin time is increased due to the consumption of
the coagulation factors due to the tiny clots forming
Disseminated Intravascular
throughout the vasculature. This is also the reason that the
Coagulation (DIC)? fibrinogen levels and platelet levels are decreased. Finally
FDP, or fibrin degredation products, are increased due to the
A. prolonged PT, formation and subsequent dissolving of many tiny clots in
the vasculature. The FDPs are the pieces of fibrin that are
elevatedplatelet count,
left after the fibrinolytic processes take place.
decreased
FDP
B. normal PT,
decreasedfibrinogen,
decreased platelet count,
decreased
FDP
C. prolonged
PT,decreased fibrinogen,
decreased platelet count,
increased FDP
D. normal PT,
decreasedplatelet count,
decreased
FDP
B;
A dilution commonly used
A dilution commonly used for a routine sperm count is a 1:20.
for a routine sperm count is:
A. 1:2
B. 1:20
C. 1:200
D. 1:400
