MLT ASCP Practice 2025 latest|
COMPLETE EXAM TEST QUESTIONS AND
VERIFIED ANSWERS
B;
The correct answer for this question is 1300 mg/dL. The laboratorian performed 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
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 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; 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outdated for a prolonged period, either there would be no reaction or the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. -
- 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D. Use of outdated medium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. 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
📌📌-
B;
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
C;
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
📌📌- A dilution
B;
A dilution commonly used for a routine sperm count is a 1:20. -
commonly used for a routine sperm count is:
A. 1:2
B. 1:20
C. 1:200
D. 1:400
B;
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
, C. No reaction at all
D. Mixed field reaction
A;
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; Streptomyces griseus hydrolyzes
all three of the substrates. - - Illustrated in this photograph is an agar
quadrant plate containing casein (A), tyrosine (B), nitrate (C) and xanthine (D).
None of the substrates have been hydrolyzed and nitrate has been reduced. The
most likely identification is:
A. Nocardia asteroides
B. Nocardia brasiliensis
C. Streptomyces griseus
D. Actinomadura madurae
A;
Since hemoglobin is measured spectrophotometrically on hematology analzyers,
interference from lipemia or icteric specimens can lead to decreased light
📌📌
detected and measured through the sample and therefore inaccurate hemoglobin
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
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 can also
lead to a more alkaline urine sample. - - A patient who has a primarily
vegetarian diet will most likely have an acid urine pH.
A;
During primary hypothyroidism, where a defect in the thryoid gland is producing
low levels of T3 and T4, the TSH level is increased. TSH is released in elevated
📌📌
quantities in an attempt to stimulate the thryoid to produce more T3 and T4 as
part of a feedback mechanism. - - Serum TSH levels five-times the upper
limit of normal in the presence of a low T4 and low T3 uptake could mean which
of the following:
A. The thyroid has been established as the cause of hypothyroidism
COMPLETE EXAM TEST QUESTIONS AND
VERIFIED ANSWERS
B;
The correct answer for this question is 1300 mg/dL. The laboratorian performed 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
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 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; 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. -
- 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. 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
📌📌-
B;
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
C;
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
📌📌- A dilution
B;
A dilution commonly used for a routine sperm count is a 1:20. -
commonly used for a routine sperm count is:
A. 1:2
B. 1:20
C. 1:200
D. 1:400
B;
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
, C. No reaction at all
D. Mixed field reaction
A;
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; Streptomyces griseus hydrolyzes
all three of the substrates. - - Illustrated in this photograph is an agar
quadrant plate containing casein (A), tyrosine (B), nitrate (C) and xanthine (D).
None of the substrates have been hydrolyzed and nitrate has been reduced. The
most likely identification is:
A. Nocardia asteroides
B. Nocardia brasiliensis
C. Streptomyces griseus
D. Actinomadura madurae
A;
Since hemoglobin is measured spectrophotometrically on hematology analzyers,
interference from lipemia or icteric specimens can lead to decreased light
📌📌
detected and measured through the sample and therefore inaccurate hemoglobin
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
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 can also
lead to a more alkaline urine sample. - - A patient who has a primarily
vegetarian diet will most likely have an acid urine pH.
A;
During primary hypothyroidism, where a defect in the thryoid gland is producing
low levels of T3 and T4, the TSH level is increased. TSH is released in elevated
📌📌
quantities in an attempt to stimulate the thryoid to produce more T3 and T4 as
part of a feedback mechanism. - - Serum TSH levels five-times the upper
limit of normal in the presence of a low T4 and low T3 uptake could mean which
of the following:
A. The thyroid has been established as the cause of hypothyroidism
