MLT ASCP Practice Test Questions for Board Exam Prep.
1. B; After experiencing extreme
The correct answer for this question is 1300 mg/dL. fatigue and polyuria, a pa-
The laboratorian performed a 1:4 dilution by adding tient's basic metabolic panel
0.25 mL (or 250 microliters) of patient sample to 750 is analyzed in the laboratory.
microliters of diluent. This creates a total volume The result of the glucose is
of 1000 microliters. So, the patient sample is 250 too high for the instrument
microliters of the 1000 microliter mixed sample, or to read. The laboratorian per-
a ratio of 1:4. Therefore, the result given by the forms a dilution using 0.25
chemistry analyzer must be multiplied by a dilution mL of patient sample to 750
factor of 4. 325 mg/dL x 4 = 1300 mg/dL.
, MLT ASCP Practice Test Questions for Board Exam Prep.
microliters of diluent. The re-
sult now reads 325 mg/dL.
How should the techologist
report this patient's glucose
2. A; result?
Conversion of only the slant to a pink color in a
Christensen's urea agar slant is produced by bac- A. 325 mg/dL
terial species that have weak urease activity. The B. 1300 mg/dL
reaction in the slant to the right is often produced C. 975 mg/dL
by Klebsiella species, as an example. Strong urease D. 1625 mg/dL
activity is indicated by conversion of the slant and
the butt of the tube to a pink color, as seen in the The urease reaction seen in
tube to the left. The slant only reaction in the right the Christensen's urea agar
tube may be seen early on if only the slant had been slant on the far right indi-
inoculated; however, with a strong urease producer, cates:
both the slant and the butt would turn. Therefore,
A. Weak activity
B. Strong activity
C. Slant only inoculated
D. Use of outdated medium
, MLT ASCP Practice Test Questions for Board Exam Prep.
the reaction is dependent on the strength of ure-
ase activity. If the media had outdated for a pro-
longed 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.
3. D; What is the first step of the
The steps in the PCR process are: PCR reaction?
1. Denaturation (Turning double stranded DNA into
single strands.) A. Hybridization
2. Annealing/Hybrization (Attachment of primers to B. Extension
the single DNA strands.) C. Annealing
3. Extension (Creating the complementary strand to D. Denaturation
produce new double stranded DNA.)
4. B; platelet levels
Isotonic or normal saline is a 0.85 % solution of are decreased.
sodium chloride in water. Finally FDP, or
fib-
5. C;
In DIC, or disseminated intravascular coagulation,
the prothrombin time is increased due to the con-
sumption of the coagulation factors due to the
tiny clots forming throughout the vasculature. This
is also the reason that the fibrinogen levels and
, MLT ASCP Practice Test Questions for Board Exam Prep.
The concentration of sodium chloride in an isotonic solu- tion is :
A. 8.5 %
B. 0.85 %
C. 0.08 %
D. 1 molar
Which of the following labo- ratory results would be seen in a patient with acute
Dis- seminated Intravascular Co- agulation (DIC)?
A. prolonged PT, elevat-
1. B; After experiencing extreme
The correct answer for this question is 1300 mg/dL. fatigue and polyuria, a pa-
The laboratorian performed a 1:4 dilution by adding tient's basic metabolic panel
0.25 mL (or 250 microliters) of patient sample to 750 is analyzed in the laboratory.
microliters of diluent. This creates a total volume The result of the glucose is
of 1000 microliters. So, the patient sample is 250 too high for the instrument
microliters of the 1000 microliter mixed sample, or to read. The laboratorian per-
a ratio of 1:4. Therefore, the result given by the forms a dilution using 0.25
chemistry analyzer must be multiplied by a dilution mL of patient sample to 750
factor of 4. 325 mg/dL x 4 = 1300 mg/dL.
, MLT ASCP Practice Test Questions for Board Exam Prep.
microliters of diluent. The re-
sult now reads 325 mg/dL.
How should the techologist
report this patient's glucose
2. A; result?
Conversion of only the slant to a pink color in a
Christensen's urea agar slant is produced by bac- A. 325 mg/dL
terial species that have weak urease activity. The B. 1300 mg/dL
reaction in the slant to the right is often produced C. 975 mg/dL
by Klebsiella species, as an example. Strong urease D. 1625 mg/dL
activity is indicated by conversion of the slant and
the butt of the tube to a pink color, as seen in the The urease reaction seen in
tube to the left. The slant only reaction in the right the Christensen's urea agar
tube may be seen early on if only the slant had been slant on the far right indi-
inoculated; however, with a strong urease producer, cates:
both the slant and the butt would turn. Therefore,
A. Weak activity
B. Strong activity
C. Slant only inoculated
D. Use of outdated medium
, MLT ASCP Practice Test Questions for Board Exam Prep.
the reaction is dependent on the strength of ure-
ase activity. If the media had outdated for a pro-
longed 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.
3. D; What is the first step of the
The steps in the PCR process are: PCR reaction?
1. Denaturation (Turning double stranded DNA into
single strands.) A. Hybridization
2. Annealing/Hybrization (Attachment of primers to B. Extension
the single DNA strands.) C. Annealing
3. Extension (Creating the complementary strand to D. Denaturation
produce new double stranded DNA.)
4. B; platelet levels
Isotonic or normal saline is a 0.85 % solution of are decreased.
sodium chloride in water. Finally FDP, or
fib-
5. C;
In DIC, or disseminated intravascular coagulation,
the prothrombin time is increased due to the con-
sumption of the coagulation factors due to the
tiny clots forming throughout the vasculature. This
is also the reason that the fibrinogen levels and
, MLT ASCP Practice Test Questions for Board Exam Prep.
The concentration of sodium chloride in an isotonic solu- tion is :
A. 8.5 %
B. 0.85 %
C. 0.08 %
D. 1 molar
Which of the following labo- ratory results would be seen in a patient with acute
Dis- seminated Intravascular Co- agulation (DIC)?
A. prolonged PT, elevat-
