MLT ASCP Practice Test Board Exam
Questions & Answers (Grade A+)
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 -
correct answer ✅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.
The urease reaction seen in the Christensen's urea agar slant on the
far right indicates:
,MLT ASCP Practice Test Board Exam
Questions & Answers (Grade A+)
A. Weak activity
B. Strong activity
C. Slant only inoculated
D. Use of outdated medium -
correct answer ✅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 If the media had outdated for a prolonged period,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.
What is the first step of the PCR reaction?
A. Hybridization
,MLT ASCP Practice Test Board Exam
Questions & Answers (Grade A+)
B. Extension
C. Annealing
D. Denaturation -
correct answer ✅D;
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.)
The concentration of sodium chloride in an isotonic solution is :
A. 8.5 %
B. 0.85 %
C. 0.08 %
D. 1 molar -
correct answer ✅B;
Isotonic or normal saline is a 0.85 % solution of sodium chloride in
water.
, MLT ASCP Practice Test Board Exam
Questions & Answers (Grade A+)
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 -
correct answer ✅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.
A dilution commonly used for a routine sperm count is:
A. 1:2
Questions & Answers (Grade A+)
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 -
correct answer ✅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.
The urease reaction seen in the Christensen's urea agar slant on the
far right indicates:
,MLT ASCP Practice Test Board Exam
Questions & Answers (Grade A+)
A. Weak activity
B. Strong activity
C. Slant only inoculated
D. Use of outdated medium -
correct answer ✅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.
What is the first step of the PCR reaction?
A. Hybridization
,MLT ASCP Practice Test Board Exam
Questions & Answers (Grade A+)
B. Extension
C. Annealing
D. Denaturation -
correct answer ✅D;
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.)
The concentration of sodium chloride in an isotonic solution is :
A. 8.5 %
B. 0.85 %
C. 0.08 %
D. 1 molar -
correct answer ✅B;
Isotonic or normal saline is a 0.85 % solution of sodium chloride in
water.
, MLT ASCP Practice Test Board Exam
Questions & Answers (Grade A+)
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 -
correct answer ✅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.
A dilution commonly used for a routine sperm count is:
A. 1:2
