2/8/25, 7:26 PM MLT ASCP Practice 2025 latest| COMPLETE EXAM TEST QUESTIONS AND VERIFIED ANSWERS (multiple choices) WITH R…
MLT ASCP Practice 2025 latest| COMPLETE
EXAM TEST QUESTIONS AND VERIFIED
ANSWERS (multiple choices) WITH
RATIONALES| get it right!!
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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 microliters of diluent. This creates a total volume of
laboratory. The result of 1000 microliters. So, the patient sample is 250
the glucose is too high microliters of the 1000 microliter mixed sample, or a
for the instrument to ratio of 1:4. Therefore, the result given by the
read. The laboratorian chemistry analyzer must be multiplied by a dilution
performs a dilution using factor of 4. 325 mg/dL x 4 = 1300 mg/dL.
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
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,2/8/25, 7:26 PM MLT ASCP Practice 2025 latest| COMPLETE EXAM TEST QUESTIONS AND VERIFIED ANSWERS (multiple choices) WITH R…
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
A. Weak activity on 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D. Use of outdated the slant and the butt would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had outdated for a prolonged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.
D;
What is the first step of
The steps in the PCR process are:
the 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.)
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,2/8/25, 7:26 PM MLT ASCP Practice 2025 latest| COMPLETE EXAM TEST QUESTIONS AND VERIFIED ANSWERS (multiple choices) WITH R…
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
Which of the following C;
laboratory results would In DIC, or disseminated intravascular coagulation,
be seen in a patient with the prothrombin time is increased due to the
acute Disseminated consumption of the coagulation factors due to the
Intravascular Coagulation tiny clots forming throughout the vasculature. This is
(DIC)? also the reason that the fibrinogen levels and
platelet levels are decreased. Finally FDP, or fibrin
A. prolonged PT, degredation products, are increased due to the
elevated platelet count, formation and subsequent dissolving of many tiny
decreased FDP clots in the vasculature. The FDPs are the pieces of
B. normal PT, decreased fibrin that are left after the fibrinolytic processes
fibrinogen, decreased take place.
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 B;
used for a routine sperm A dilution commonly used for a routine sperm count
count is: is a 1:20.
A. 1:2
B. 1:20
C. 1:200
D. 1:400
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, 2/8/25, 7:26 PM MLT ASCP Practice 2025 latest| COMPLETE EXAM TEST QUESTIONS AND VERIFIED ANSWERS (multiple choices) WITH R…
The prozone effect ( B;
when performing a Prozone effect (due to antibody excess) will result in
screening titer) is most an initial false negative in spite of the large amount
likely to result in: of antibody in the serum, followed by a positive
result as the specimen is diluted.
A. False positive
B. False negative
C. No reaction at all
D. Mixed field reaction
Illustrated in this A;
photograph is an agar One of the key characteristics
quadrant plate to the identification of
containing casein (A), Nocardia asteroides is its
tyrosine (B), nitrate (C) inability to hydrolyze casein,
and xanthine (D). None of tyrosine or xanthine, as shown
the substrates have been in this photograph. Nitrates are
hydrolyzed and nitrate reduced to nitrites. Both
has been reduced. The Nocardia brasiliensis and
most likely identification Actinomadura madurae
is: hydrolyze both casein and
tyrosine; Streptomyces griseus
A. Nocardia asteroides hydrolyzes all three of the
B. Nocardia brasiliensis substrates.
C. Streptomyces griseus
D. Actinomadura
madurae
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MLT ASCP Practice 2025 latest| COMPLETE
EXAM TEST QUESTIONS AND VERIFIED
ANSWERS (multiple choices) WITH
RATIONALES| get it right!!
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 microliters of diluent. This creates a total volume of
laboratory. The result of 1000 microliters. So, the patient sample is 250
the glucose is too high microliters of the 1000 microliter mixed sample, or a
for the instrument to ratio of 1:4. Therefore, the result given by the
read. The laboratorian chemistry analyzer must be multiplied by a dilution
performs a dilution using factor of 4. 325 mg/dL x 4 = 1300 mg/dL.
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
https://quizlet.com/1003233925/mlt-ascp-practice-2025-latest-complete-exam-test-questions-and-verified-answers-multiple-choices-with-rational… 1/47
,2/8/25, 7:26 PM MLT ASCP Practice 2025 latest| COMPLETE EXAM TEST QUESTIONS AND VERIFIED ANSWERS (multiple choices) WITH R…
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
A. Weak activity on 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.
D;
What is the first step of
The steps in the PCR process are:
the 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.)
https://quizlet.com/1003233925/mlt-ascp-practice-2025-latest-complete-exam-test-questions-and-verified-answers-multiple-choices-with-rational… 2/47
,2/8/25, 7:26 PM MLT ASCP Practice 2025 latest| COMPLETE EXAM TEST QUESTIONS AND VERIFIED ANSWERS (multiple choices) WITH R…
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
Which of the following C;
laboratory results would In DIC, or disseminated intravascular coagulation,
be seen in a patient with the prothrombin time is increased due to the
acute Disseminated consumption of the coagulation factors due to the
Intravascular Coagulation tiny clots forming throughout the vasculature. This is
(DIC)? also the reason that the fibrinogen levels and
platelet levels are decreased. Finally FDP, or fibrin
A. prolonged PT, degredation products, are increased due to the
elevated platelet count, formation and subsequent dissolving of many tiny
decreased FDP clots in the vasculature. The FDPs are the pieces of
B. normal PT, decreased fibrin that are left after the fibrinolytic processes
fibrinogen, decreased take place.
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 B;
used for a routine sperm A dilution commonly used for a routine sperm count
count is: is a 1:20.
A. 1:2
B. 1:20
C. 1:200
D. 1:400
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, 2/8/25, 7:26 PM MLT ASCP Practice 2025 latest| COMPLETE EXAM TEST QUESTIONS AND VERIFIED ANSWERS (multiple choices) WITH R…
The prozone effect ( B;
when performing a Prozone effect (due to antibody excess) will result in
screening titer) is most an initial false negative in spite of the large amount
likely to result in: of antibody in the serum, followed by a positive
result as the specimen is diluted.
A. False positive
B. False negative
C. No reaction at all
D. Mixed field reaction
Illustrated in this A;
photograph is an agar One of the key characteristics
quadrant plate to the identification of
containing casein (A), Nocardia asteroides is its
tyrosine (B), nitrate (C) inability to hydrolyze casein,
and xanthine (D). None of tyrosine or xanthine, as shown
the substrates have been in this photograph. Nitrates are
hydrolyzed and nitrate reduced to nitrites. Both
has been reduced. The Nocardia brasiliensis and
most likely identification Actinomadura madurae
is: hydrolyze both casein and
tyrosine; Streptomyces griseus
A. Nocardia asteroides hydrolyzes all three of the
B. Nocardia brasiliensis substrates.
C. Streptomyces griseus
D. Actinomadura
madurae
https://quizlet.com/1003233925/mlt-ascp-practice-2025-latest-complete-exam-test-questions-and-verified-answers-multiple-choices-with-rational… 4/47
