MLT ASCP Practice Test Questions board practice
1. B -Answer
The correct answer for this question is 1300 mg/dL. The laboratorian per-
formed 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.: 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
2. A -Answer
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 -
Answer however, with a strong urease producer, both the slant and the ould turn.
butt w
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media had outdated for a prolonged period, either there would be no reaction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 utt
b or
both, again depending on the strength of urease production by the unknown
organism.: 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,D. Use of outdated medium
3. D -Answer
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.): What is the first step of the PCR reaction?
A. Hybridization
B. Extension
C. Annealing
D. Denaturation
4. B -Answer
Isotonic or normal saline is a 0.85 % solution of sodium chloride in water.: The
concentration of sodium chloride in an isotonic solution is :
A. 8.5 %
B. 0.85 %
C. 0.08 %
D. 1 molar
5. C -Answer
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.: 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
6. B -Answer
A dilution commonly used for a routine sperm count is a 1:20.: A dilution
commonly used for a routine sperm count is:
A. 1:2
B. 1:20
C. 1:200
D. 1:400
, 7. B -Answer
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.: 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
8. A -Answer
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 -Answer
hotograph
Streptomyces griseus hydrolyzes all three of the substrates.: Illustrated in is
this p and xanthine
an agar quadrant plate containing casein (A), tyrosine (B), nitrate (C)en reduced.
(D). None of the substrates have been hydrolyzed and nitrate has be
The most likely identification is:
A. Nocardia asteroides
B. Nocardia brasiliensis
C. Streptomyces griseus
D. Actinomadura madurae
9. A -Answer
Since hemoglobin is measured spectrophotometrically on hematology analzy-
ers, interference from lipemia or icteric specimens can lead to decreased light
detected and measured through the sample and therefore inaccurate hemo-
globin results occur.: 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
10. 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
1. B -Answer
The correct answer for this question is 1300 mg/dL. The laboratorian per-
formed 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.: 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
2. A -Answer
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 -
Answer however, with a strong urease producer, both the slant and the ould turn.
butt w
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 utt
b or
both, again depending on the strength of urease production by the unknown
organism.: 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
3. D -Answer
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.): What is the first step of the PCR reaction?
A. Hybridization
B. Extension
C. Annealing
D. Denaturation
4. B -Answer
Isotonic or normal saline is a 0.85 % solution of sodium chloride in water.: The
concentration of sodium chloride in an isotonic solution is :
A. 8.5 %
B. 0.85 %
C. 0.08 %
D. 1 molar
5. C -Answer
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.: 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
6. B -Answer
A dilution commonly used for a routine sperm count is a 1:20.: A dilution
commonly used for a routine sperm count is:
A. 1:2
B. 1:20
C. 1:200
D. 1:400
, 7. B -Answer
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.: 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
8. A -Answer
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 -Answer
hotograph
Streptomyces griseus hydrolyzes all three of the substrates.: Illustrated in is
this p and xanthine
an agar quadrant plate containing casein (A), tyrosine (B), nitrate (C)en reduced.
(D). None of the substrates have been hydrolyzed and nitrate has be
The most likely identification is:
A. Nocardia asteroides
B. Nocardia brasiliensis
C. Streptomyces griseus
D. Actinomadura madurae
9. A -Answer
Since hemoglobin is measured spectrophotometrically on hematology analzy-
ers, interference from lipemia or icteric specimens can lead to decreased light
detected and measured through the sample and therefore inaccurate hemo-
globin results occur.: 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
10. 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
