MLT ASCP Practice Test Questions Board Exam.
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 The result of the glucose
volume of 1000 microliters. So, the patient is too high for the
sample is 250 microliters of the 1000 instrument to read. The
microliter mixed sample, or a ratio of 1:4. laboratorian per- forms a
Therefore, the result given by the 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. 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
2. A; tube may be
Conversion of only the slant to a pink color in seen early on
a Christensen's urea agar slant is produced if only the
by bac- terial species that have weak urease slant had
activity. The reaction in the slant to the right been
is often produced by Klebsiella species, as an inoculated;
example. Strong urease activity is indicated however,
, MLT ASCP Practice Test Questions Board Exam.
with a strong urease producer, both the microliters of diluent. The
slant and the butt would turn. Therefore, re- sult 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
The urease reaction seen
in the Christensen's urea
agar slant on the far right
indi- cates:
A. Weak activity
B. Strong activity
C. Slant only inoculated
D. Use of outdated
medium
, MLT ASCP Practice Test Questions Board Exam.
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outdated for a pro- longed period, either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.
What is the first step of the PCR
3. D; reaction?
The steps in the PCR process are:
1. Denaturation (Turning double stranded DNA A. Hybridization
into single strands.)
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; forming
Isotonic or normal saline is a 0.85 % solution throughout
of sodium chloride in water. the
vasculature.
This is also
the reason
that the
fibrinogen
levels and
5. C;
platelet levels
In DIC, or disseminated intravascular
are
coagulation, the prothrombin time is
decreased.
increased due to the con- sumption of the
Finally FDP, or
coagulation factors due to the tiny clots
fib-
, MLT ASCP Practice Test Questions Board Exam.
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 The result of the glucose
volume of 1000 microliters. So, the patient is too high for the
sample is 250 microliters of the 1000 instrument to read. The
microliter mixed sample, or a ratio of 1:4. laboratorian per- forms a
Therefore, the result given by the 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. 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
2. A; tube may be
Conversion of only the slant to a pink color in seen early on
a Christensen's urea agar slant is produced if only the
by bac- terial species that have weak urease slant had
activity. The reaction in the slant to the right been
is often produced by Klebsiella species, as an inoculated;
example. Strong urease activity is indicated however,
, MLT ASCP Practice Test Questions Board Exam.
with a strong urease producer, both the microliters of diluent. The
slant and the butt would turn. Therefore, re- sult 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
The urease reaction seen
in the Christensen's urea
agar slant on the far right
indi- cates:
A. Weak activity
B. Strong activity
C. Slant only inoculated
D. Use of outdated
medium
, MLT ASCP Practice Test Questions Board Exam.
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.
What is the first step of the PCR
3. D; reaction?
The steps in the PCR process are:
1. Denaturation (Turning double stranded DNA A. Hybridization
into single strands.)
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; forming
Isotonic or normal saline is a 0.85 % solution throughout
of sodium chloride in water. the
vasculature.
This is also
the reason
that the
fibrinogen
levels and
5. C;
platelet levels
In DIC, or disseminated intravascular
are
coagulation, the prothrombin time is
decreased.
increased due to the con- sumption of the
Finally FDP, or
coagulation factors due to the tiny clots
fib-
, MLT ASCP Practice Test Questions Board Exam.
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-
