MLT ASCP Practice Test Questions board practice with
100% Verified Answers
After experiencing extreme fatigue and polyuria, reaction is dependent on the strength of urease
a patient's basic metabolic panel is analyzed in activity. If the media had outdated for a prolonged
the laboratory. The result of the glucose is too period, either there would be no reaction or the
high for the instrument to read. The laboratorian appearance of only a faint pink tinge, either in the
performs a dilution using 0.25 mL of patient slant, the butt or both, again depending on the
sample to 750 microliters of diluent. The result strength of urease production by the unknown
now reads 325 mg/dL. How should the organism.
techologist report this patient's glucose result?
A. 325 mg/dL
B. 1300 mg/dL What is the first step of the PCR reaction?
C. 975 mg/dL
D. 1625 mg/dL - ANSWER -B; A. Hybridization
The correct answer for this question is 1300 B. Extension
mg/dL. The laboratorian performed a 1:4 dilution C. Annealing
by adding 0.25 mL (or 250 microliters) of patient D. Denaturation - ANSWER -D;
sample to 750 microliters of diluent. This creates The steps in the PCR process are:
a total volume of 1000 microliters. So, the patient 1. Denaturation (Turning double stranded DNA
sample is 250 microliters of the 1000 microliter into single strands.)
mixed sample, or a ratio of 1:4. Therefore, the 2. Annealing/Hybrization (Attachment of primers
result given by the chemistry analyzer must be to the single DNA strands.)
multiplied by a dilution factor of 4. 325 mg/dL x 4 3. Extension (Creating the complementary strand
= 1300 mg/dL. to produce new double stranded DNA.)
The urease reaction seen in the Christensen's The concentration of sodium chloride in an
urea agar slant on the far right indicates: isotonic solution is :
A. Weak activity A. 8.5 %
B. Strong activity B. 0.85 %
C. Slant only inoculated C. 0.08 %
D. Use of outdated medium - ANSWER -A; D. 1 molar - ANSWER -B;
Conversion of only the slant to a pink color in a Isotonic or normal saline is a 0.85 % solution of
Christensen's urea agar slant is produced by sodium chloride in water.
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 Which of the following laboratory results would be
of the slant and the butt of the tube to a pink seen in a patient with acute Disseminated
color, as seen in the tube to the left. The slant Intravascular Coagulation (DIC)?
only reaction in the right tube may be seen early
on if only the slant had been inoculated; A. prolonged PT, elevated platelet count,
however, with a strong urease producer, both the decreased FDP
slant and the butt would turn. Therefore, the B. normal PT, decreased fibrinogen, decreased
,MLT ASCP Practice Test Questions board practice with
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platelet count, decreased FDP
C. prolonged PT, decreased fibrinogen,
decreased platelet count, increased FDP Illustrated in this photograph is an agar quadrant
D. normal PT, decreased platelet count, plate containing casein (A), tyrosine (B), nitrate
decreased FDP - ANSWER -C; (C) and xanthine (D). None of the substrates
In DIC, or disseminated intravascular have been hydrolyzed and nitrate has been
coagulation, the prothrombin time is increased reduced. The most likely identification is:
due to the consumption of the coagulation
factors due to the tiny clots forming throughout A. Nocardia asteroides
the vasculature. This is also the reason that the B. Nocardia brasiliensis
fibrinogen levels and platelet levels are C. Streptomyces griseus
decreased. Finally FDP, or fibrin degredation D. Actinomadura madurae - ANSWER -A;
products, are increased due to the formation and One of the key characteristics to the identification
subsequent dissolving of many tiny clots in the of Nocardia asteroides is its inability to hydrolyze
vasculature. The FDPs are the pieces of fibrin casein, tyrosine or xanthine, as shown in this
that are left after the fibrinolytic processes take photograph. Nitrates are reduced to nitrites. Both
place. Nocardia brasiliensis and Actinomadura madurae
hydrolyze both casein and tyrosine;
Streptomyces griseus hydrolyzes all three of the
substrates.
A dilution commonly used for a routine sperm
count is:
A. 1:2 On an electronic cell counter, hemoglobin
B. 1:20 determination may be falsely elevated caused by
C. 1:200 the presence of:
D. 1:400 - ANSWER -B;
A dilution commonly used for a routine sperm A. Lipemic or icteric plasma
count is a 1:20. B. Leukocytopenia or Leukocytosis
C. Rouleaux or agglutinated RBCs
D. Anemia or Polycythemia - ANSWER -A;
Since hemoglobin is measured
The prozone effect ( when performing a spectrophotometrically on hematology analzyers,
screening titer) is most likely to result in: interference from lipemia or icteric specimens
can lead to decreased light detected and
A. False positive measured through the sample and therefore
B. False negative inaccurate hemoglobin results occur.
C. No reaction at all
D. Mixed field reaction - ANSWER -B;
Prozone effect (due to antibody excess) will
result in an initial false negative in spite of the A patient who has a primarily vegetarian diet will
large amount of antibody in the serum, followed most likely have an acid urine pH. -
by a positive result as the specimen is diluted. ANSWER -False
A patient who has a primarily vegetarian diet will
, MLT ASCP Practice Test Questions board practice with
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most likely have an alkaline urine pH. A low- associated with sinusitis.
carbohydrate diet as well as the ingestion of
citrus fruits can also lead to a more alkaline urine
sample.
Which of the following blood additives is most
useful for serum collection:
Serum TSH levels five-times the upper limit of A. Polymer barrier
normal in the presence of a low T4 and low T3 B. Oxalate
uptake could mean which of the following: C. EDTA
D. Citrate - ANSWER -A;
A. The thyroid has been established as the Oxalate, EDTA, and citrate are anticoagulants
cause of hypothyroidism that inhibit clot formation.
B. The thyroid is ruled-out as the cause of
hypothyroidism
C. The pituitary has been established as the
cause of hypothyroidism The laboratorian completed the mixing study
D. The diagnosis is consistent with secondary ordered for John Doe. The results are as follows:
hyperthyroidism - ANSWER -A;
During primary hypothyroidism, where a defect in Initial aPTT result: 167 seconds
the thryoid gland is producing low levels of T3 Initial 1:1 Mix with Normal Pooled Plasma: 158
and T4, the TSH level is increased. TSH is seconds
released in elevated quantities in an attempt to Incubated 1:1 Mix with Normal Pooled Plasma:
stimulate the thryoid to produce more T3 and T4 150 seconds
as part of a feedback mechanism.
Which of the choices below would most likely
explain the results for this patient?
Which of the following species or organisms is A. Factor VIII deficiency
the most likely to be the cause of mycotic B. Immediate-acting coagulation inhibitor
keratitis (fungal eye infection)? C. Time/temperature-dependent coagulation
inhibitor
A. Fusarium species D. Factor VII deficiency - ANSWER -B;
B. Trichophyton rubrum. This patient is most likely suffering from an
C. Scedosporium apiospermum immediate-acting coagulation inhibitor; most
D. Aspergillus niger - ANSWER -A; commonly, lupus anticoagulant. Notice that the
Fusarium species is the most likely associated addition of normal pooled plasma does not
with mycotic keratitis. correct upon initial or incubated mix, which
means that the inhibitor is not time or
Trichophyton rubrum is a dermatophyte that temperature-dependent.
commonly causes an itching, scaling skin
infection of the feet, known as tinea pedis. Factor VIII is not the correct answer as a factor
Scedosporium apiospermum is commonly deficiency would have corrected upon the
associated with sinusitis. Aspergillus niger addition of normal pooled plasma. Factor VII is
typically causes otitis externa and can also be not the correct answer, as the aPTT assay does
100% Verified Answers
After experiencing extreme fatigue and polyuria, reaction is dependent on the strength of urease
a patient's basic metabolic panel is analyzed in activity. If the media had outdated for a prolonged
the laboratory. The result of the glucose is too period, either there would be no reaction or the
high for the instrument to read. The laboratorian appearance of only a faint pink tinge, either in the
performs a dilution using 0.25 mL of patient slant, the butt or both, again depending on the
sample to 750 microliters of diluent. The result strength of urease production by the unknown
now reads 325 mg/dL. How should the organism.
techologist report this patient's glucose result?
A. 325 mg/dL
B. 1300 mg/dL What is the first step of the PCR reaction?
C. 975 mg/dL
D. 1625 mg/dL - ANSWER -B; A. Hybridization
The correct answer for this question is 1300 B. Extension
mg/dL. The laboratorian performed a 1:4 dilution C. Annealing
by adding 0.25 mL (or 250 microliters) of patient D. Denaturation - ANSWER -D;
sample to 750 microliters of diluent. This creates The steps in the PCR process are:
a total volume of 1000 microliters. So, the patient 1. Denaturation (Turning double stranded DNA
sample is 250 microliters of the 1000 microliter into single strands.)
mixed sample, or a ratio of 1:4. Therefore, the 2. Annealing/Hybrization (Attachment of primers
result given by the chemistry analyzer must be to the single DNA strands.)
multiplied by a dilution factor of 4. 325 mg/dL x 4 3. Extension (Creating the complementary strand
= 1300 mg/dL. to produce new double stranded DNA.)
The urease reaction seen in the Christensen's The concentration of sodium chloride in an
urea agar slant on the far right indicates: isotonic solution is :
A. Weak activity A. 8.5 %
B. Strong activity B. 0.85 %
C. Slant only inoculated C. 0.08 %
D. Use of outdated medium - ANSWER -A; D. 1 molar - ANSWER -B;
Conversion of only the slant to a pink color in a Isotonic or normal saline is a 0.85 % solution of
Christensen's urea agar slant is produced by sodium chloride in water.
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 Which of the following laboratory results would be
of the slant and the butt of the tube to a pink seen in a patient with acute Disseminated
color, as seen in the tube to the left. The slant Intravascular Coagulation (DIC)?
only reaction in the right tube may be seen early
on if only the slant had been inoculated; A. prolonged PT, elevated platelet count,
however, with a strong urease producer, both the decreased FDP
slant and the butt would turn. Therefore, the B. normal PT, decreased fibrinogen, decreased
,MLT ASCP Practice Test Questions board practice with
100% Verified Answers
platelet count, decreased FDP
C. prolonged PT, decreased fibrinogen,
decreased platelet count, increased FDP Illustrated in this photograph is an agar quadrant
D. normal PT, decreased platelet count, plate containing casein (A), tyrosine (B), nitrate
decreased FDP - ANSWER -C; (C) and xanthine (D). None of the substrates
In DIC, or disseminated intravascular have been hydrolyzed and nitrate has been
coagulation, the prothrombin time is increased reduced. The most likely identification is:
due to the consumption of the coagulation
factors due to the tiny clots forming throughout A. Nocardia asteroides
the vasculature. This is also the reason that the B. Nocardia brasiliensis
fibrinogen levels and platelet levels are C. Streptomyces griseus
decreased. Finally FDP, or fibrin degredation D. Actinomadura madurae - ANSWER -A;
products, are increased due to the formation and One of the key characteristics to the identification
subsequent dissolving of many tiny clots in the of Nocardia asteroides is its inability to hydrolyze
vasculature. The FDPs are the pieces of fibrin casein, tyrosine or xanthine, as shown in this
that are left after the fibrinolytic processes take photograph. Nitrates are reduced to nitrites. Both
place. Nocardia brasiliensis and Actinomadura madurae
hydrolyze both casein and tyrosine;
Streptomyces griseus hydrolyzes all three of the
substrates.
A dilution commonly used for a routine sperm
count is:
A. 1:2 On an electronic cell counter, hemoglobin
B. 1:20 determination may be falsely elevated caused by
C. 1:200 the presence of:
D. 1:400 - ANSWER -B;
A dilution commonly used for a routine sperm A. Lipemic or icteric plasma
count is a 1:20. B. Leukocytopenia or Leukocytosis
C. Rouleaux or agglutinated RBCs
D. Anemia or Polycythemia - ANSWER -A;
Since hemoglobin is measured
The prozone effect ( when performing a spectrophotometrically on hematology analzyers,
screening titer) is most likely to result in: interference from lipemia or icteric specimens
can lead to decreased light detected and
A. False positive measured through the sample and therefore
B. False negative inaccurate hemoglobin results occur.
C. No reaction at all
D. Mixed field reaction - ANSWER -B;
Prozone effect (due to antibody excess) will
result in an initial false negative in spite of the A patient who has a primarily vegetarian diet will
large amount of antibody in the serum, followed most likely have an acid urine pH. -
by a positive result as the specimen is diluted. ANSWER -False
A patient who has a primarily vegetarian diet will
, MLT ASCP Practice Test Questions board practice with
100% Verified Answers
most likely have an alkaline urine pH. A low- associated with sinusitis.
carbohydrate diet as well as the ingestion of
citrus fruits can also lead to a more alkaline urine
sample.
Which of the following blood additives is most
useful for serum collection:
Serum TSH levels five-times the upper limit of A. Polymer barrier
normal in the presence of a low T4 and low T3 B. Oxalate
uptake could mean which of the following: C. EDTA
D. Citrate - ANSWER -A;
A. The thyroid has been established as the Oxalate, EDTA, and citrate are anticoagulants
cause of hypothyroidism that inhibit clot formation.
B. The thyroid is ruled-out as the cause of
hypothyroidism
C. The pituitary has been established as the
cause of hypothyroidism The laboratorian completed the mixing study
D. The diagnosis is consistent with secondary ordered for John Doe. The results are as follows:
hyperthyroidism - ANSWER -A;
During primary hypothyroidism, where a defect in Initial aPTT result: 167 seconds
the thryoid gland is producing low levels of T3 Initial 1:1 Mix with Normal Pooled Plasma: 158
and T4, the TSH level is increased. TSH is seconds
released in elevated quantities in an attempt to Incubated 1:1 Mix with Normal Pooled Plasma:
stimulate the thryoid to produce more T3 and T4 150 seconds
as part of a feedback mechanism.
Which of the choices below would most likely
explain the results for this patient?
Which of the following species or organisms is A. Factor VIII deficiency
the most likely to be the cause of mycotic B. Immediate-acting coagulation inhibitor
keratitis (fungal eye infection)? C. Time/temperature-dependent coagulation
inhibitor
A. Fusarium species D. Factor VII deficiency - ANSWER -B;
B. Trichophyton rubrum. This patient is most likely suffering from an
C. Scedosporium apiospermum immediate-acting coagulation inhibitor; most
D. Aspergillus niger - ANSWER -A; commonly, lupus anticoagulant. Notice that the
Fusarium species is the most likely associated addition of normal pooled plasma does not
with mycotic keratitis. correct upon initial or incubated mix, which
means that the inhibitor is not time or
Trichophyton rubrum is a dermatophyte that temperature-dependent.
commonly causes an itching, scaling skin
infection of the feet, known as tinea pedis. Factor VIII is not the correct answer as a factor
Scedosporium apiospermum is commonly deficiency would have corrected upon the
associated with sinusitis. Aspergillus niger addition of normal pooled plasma. Factor VII is
typically causes otitis externa and can also be not the correct answer, as the aPTT assay does
