MLT ASCP Practice Test Questions Board Practice
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Terms in this set (100)
After experiencing extreme fatigue B;
and polyuria, a patient's basic The correct answer for this question is 1300 mg/dL.
metabolic panel is analyzed in the The laboratorian performed a 1:4 dilution by
laboratory. The result of the glucose adding 0.25 mL (or 250 microliters) of patient
is too high for the instrument to read. sample to 750 microliters of diluent. This creates a
The laboratorian performs a dilution total volume of 1000 microliters. So, the patient
using 0.25 mL of patient sample to sample is 250 microliters of the 1000 microliter
750 microliters of diluent. The result mixed sample, or a ratio of 1:4. Therefore, the result
now reads 325 mg/dL. How should given by the chemistry analyzer must be multiplied
the techologist report this patient's by a dilution factor of 4. 325 mg/dL x 4 = 1300
glucose result? mg/dL.
A. 325 mg/dL
B. 1300 mg/dL
C. 975 mg/dL
D. 1625 mg/dL
,The urease reaction seen in the A;
Christensen's urea agar slant on the Conversion of only the slant to a pink color in a
far right indicates: Christensen's urea agar slant is produced by
bacterial species that have weak urease activity.
A. Weak activity The reaction in the slant to the right is often
B. Strong activity produced by Klebsiella species, as an example.
C. Slant only inoculated Strong urease activity is indicated by conversion of
D. Use of outdated medium the slant and the butt of the tube to a pink color, as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outdated for a prolonged period, either there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.
What is the first step of the PCR D;
reaction? The steps in the PCR process are:
1. Denaturation (Turning double stranded DNA into
A. Hybridization single strands.)
B. Extension 2. Annealing/Hybrization (Attachment of primers to
C. Annealing the single DNA strands.)
D. Denaturation 3. Extension (Creating the complementary strand
to produce new double stranded DNA.)
,The concentration of sodium B;
chloride in an isotonic solution is : Isotonic or normal saline is a 0.85 % solution of
sodium chloride in water.
A. 8.5 %
B. 0.85 %
C. 0.08 %
D. 1 molar
Which of the following laboratory C;
results would be seen in a patient In DIC, or disseminated intravascular coagulation,
with acute Disseminated the prothrombin time is increased due to the
Intravascular Coagulation (DIC)? consumption of the coagulation factors due to the
tiny clots forming throughout the vasculature. This
A. prolonged PT, elevated platelet is also the reason that the fibrinogen levels and
count, decreased FDP platelet levels are decreased. Finally FDP, or fibrin
B. normal PT, decreased fibrinogen, degredation products, are increased due to the
decreased platelet count, decreased formation and subsequent dissolving of many tiny
FDP clots in the vasculature. The FDPs are the pieces of
C. prolonged PT, decreased fibrin that are left after the fibrinolytic processes
fibrinogen, decreased platelet count, take place.
increased FDP
D. normal PT, decreased platelet
count, decreased FDP
A dilution commonly used for a B;
routine sperm count is: A dilution commonly used for a routine sperm
count is a 1:20.
A. 1:2
B. 1:20
C. 1:200
D. 1:400
, The prozone effect ( when B;
performing a screening titer) is most Prozone effect (due to antibody excess) will result
likely to result in: in an initial false negative in spite of the large
amount of antibody in the serum, followed by a
A. False positive positive result as the specimen is diluted.
B. False negative
C. No reaction at all
D. Mixed field reaction
Illustrated in this photograph is an A;
agar quadrant plate containing One of the key characteristics to the identification
casein (A), tyrosine (B), nitrate (C) of Nocardia asteroides is its inability to hydrolyze
and xanthine (D). None of the casein, tyrosine or xanthine, as shown in this
substrates have been hydrolyzed photograph. Nitrates are reduced to nitrites. Both
and nitrate has been reduced. The Nocardia brasiliensis and Actinomadura madurae
most likely identification is: hydrolyze both casein and tyrosine; Streptomyces
griseus hydrolyzes all three of the substrates.
A. Nocardia asteroides
B. Nocardia brasiliensis
C. Streptomyces griseus
D. Actinomadura madurae
On an electronic cell counter, A;
hemoglobin determination may be Since hemoglobin is measured
falsely elevated caused by the spectrophotometrically on hematology analzyers,
presence of: interference from lipemia or icteric specimens can
lead to decreased light detected and measured
A. Lipemic or icteric plasma through the sample and therefore inaccurate
B. Leukocytopenia or Leukocytosis hemoglobin results occur.
C. Rouleaux or agglutinated RBCs
D. Anemia or Polycythemia
2025 Questions and Answers ()
(Verified Answers)
Save
Terms in this set (100)
After experiencing extreme fatigue B;
and polyuria, a patient's basic The correct answer for this question is 1300 mg/dL.
metabolic panel is analyzed in the The laboratorian performed a 1:4 dilution by
laboratory. The result of the glucose adding 0.25 mL (or 250 microliters) of patient
is too high for the instrument to read. sample to 750 microliters of diluent. This creates a
The laboratorian performs a dilution total volume of 1000 microliters. So, the patient
using 0.25 mL of patient sample to sample is 250 microliters of the 1000 microliter
750 microliters of diluent. The result mixed sample, or a ratio of 1:4. Therefore, the result
now reads 325 mg/dL. How should given by the chemistry analyzer must be multiplied
the techologist report this patient's by a dilution factor of 4. 325 mg/dL x 4 = 1300
glucose result? mg/dL.
A. 325 mg/dL
B. 1300 mg/dL
C. 975 mg/dL
D. 1625 mg/dL
,The urease reaction seen in the A;
Christensen's urea agar slant on the Conversion of only the slant to a pink color in a
far right indicates: Christensen's urea agar slant is produced by
bacterial species that have weak urease activity.
A. Weak activity The reaction in the slant to the right is often
B. Strong activity produced by Klebsiella species, as an example.
C. Slant only inoculated Strong urease activity is indicated by conversion of
D. Use of outdated medium 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.
What is the first step of the PCR D;
reaction? The steps in the PCR process are:
1. Denaturation (Turning double stranded DNA into
A. Hybridization single strands.)
B. Extension 2. Annealing/Hybrization (Attachment of primers to
C. Annealing the single DNA strands.)
D. Denaturation 3. Extension (Creating the complementary strand
to produce new double stranded DNA.)
,The concentration of sodium B;
chloride in an isotonic solution is : Isotonic or normal saline is a 0.85 % solution of
sodium chloride in water.
A. 8.5 %
B. 0.85 %
C. 0.08 %
D. 1 molar
Which of the following laboratory C;
results would be seen in a patient In DIC, or disseminated intravascular coagulation,
with acute Disseminated the prothrombin time is increased due to the
Intravascular Coagulation (DIC)? consumption of the coagulation factors due to the
tiny clots forming throughout the vasculature. This
A. prolonged PT, elevated platelet is also the reason that the fibrinogen levels and
count, decreased FDP platelet levels are decreased. Finally FDP, or fibrin
B. normal PT, decreased fibrinogen, degredation products, are increased due to the
decreased platelet count, decreased formation and subsequent dissolving of many tiny
FDP clots in the vasculature. The FDPs are the pieces of
C. prolonged PT, decreased fibrin that are left after the fibrinolytic processes
fibrinogen, decreased platelet count, take place.
increased FDP
D. normal PT, decreased platelet
count, decreased FDP
A dilution commonly used for a B;
routine sperm count is: A dilution commonly used for a routine sperm
count is a 1:20.
A. 1:2
B. 1:20
C. 1:200
D. 1:400
, The prozone effect ( when B;
performing a screening titer) is most Prozone effect (due to antibody excess) will result
likely to result in: in an initial false negative in spite of the large
amount of antibody in the serum, followed by a
A. False positive positive result as the specimen is diluted.
B. False negative
C. No reaction at all
D. Mixed field reaction
Illustrated in this photograph is an A;
agar quadrant plate containing One of the key characteristics to the identification
casein (A), tyrosine (B), nitrate (C) of Nocardia asteroides is its inability to hydrolyze
and xanthine (D). None of the casein, tyrosine or xanthine, as shown in this
substrates have been hydrolyzed photograph. Nitrates are reduced to nitrites. Both
and nitrate has been reduced. The Nocardia brasiliensis and Actinomadura madurae
most likely identification is: hydrolyze both casein and tyrosine; Streptomyces
griseus hydrolyzes all three of the substrates.
A. Nocardia asteroides
B. Nocardia brasiliensis
C. Streptomyces griseus
D. Actinomadura madurae
On an electronic cell counter, A;
hemoglobin determination may be Since hemoglobin is measured
falsely elevated caused by the spectrophotometrically on hematology analzyers,
presence of: interference from lipemia or icteric specimens can
lead to decreased light detected and measured
A. Lipemic or icteric plasma through the sample and therefore inaccurate
B. Leukocytopenia or Leukocytosis hemoglobin results occur.
C. Rouleaux or agglutinated RBCs
D. Anemia or Polycythemia
