MLT ASCP Practice Test Questions – Board Exam Practice

MLT ASCP Practice Test Questions – Board
MLT
Exam
ASCP
Practice.pdf
Practice Test Questions – Board
MLT
Exam
ASCP
Practice.pdf
Practice Test Questions – Board Exam Practice.pdf




MLT ASCP Practice Test
Questions – Board Exam
Practice




MLT ASCP Practice Test Questions – Board
MLT
Exam
ASCP
Practice.pdf
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, MLT ASCP Practice Test Questions board practice.pdf
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MLT ASCP Practice Test Questions board practice
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1. After experiencing extreme fa- B;
tigue and polyuria, a patient's The correct answer for this question is
basic metabolic panel is ana- 1300 mg/dL. The laboratorian performed
lyzed in the laboratory. The re- a 1:4 dilution by adding 0.25 mL (or 250
sult of the glucose is too high microliters) of patient sample to 750 mi-
for the instrument to read. The croliters of diluent. This creates a total
laboratorian performs a dilution volume of 1000 microliters. So, the pa-
using 0.25 mL of patient sam- tient sample is 250 microliters of the 1000
ple to 750 microliters of diluent. microliter mixed sample, or a ratio of 1:4.
The result now reads 325 mg/dL. Therefore, the result given by the chem-
How should the techologist re- istry analyzer must be multiplied by a
port this patient's glucose re- dilution factor of 4. 325 mg/dL x 4 = 1300
sult? mg/dL.

A. 325 mg/dL
B. 1300 mg/dL
C. 975 mg/dL
D. 1625 mg/dL

2. The urease reaction seen in the A;
Christensen's urea agar slant on Conversion of only the slant to a pink
the far right indicates: color in a Christensen's urea agar slant
is produced by bacterial species that
A. Weak activity have weak urease activity. The reaction
B. Strong activity in the slant to the right is often produced
C. Slant only inoculated by Klebsiella species, as an example.
D. Use of outdated medium Strong urease activity is indicated by con-
version 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 out-
dated for a prolonged period, either there
would be no reaction or the appearance
of only a faint pink tinge, either in the

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slant, the butt or both, again depending
on the strength of urease production by
the unknown organism.




3. What is the first step of the PCR D;
reaction? The steps in the PCR process are:
1. Denaturation (Turning double stranded
A. Hybridization DNA into single strands.)
B. Extension 2. Annealing/Hybrization (Attachment of
C. Annealing primers to the single DNA strands.)
D. Denaturation 3. Extension (Creating the complemen-
tary strand to produce new double
stranded DNA.)

4. The concentration of sodium B;
chloride in an isotonic solution Isotonic or normal saline is a 0.85 % so-
is : lution of sodium chloride in water.

A. 8.5 %
B. 0.85 %
C. 0.08 %
D. 1 molar

5. Which of the following laborato- C;
ry results would be seen in a pa-In DIC, or disseminated intravascular co-
tient with acute Disseminated In-agulation, the prothrombin time is in-
travascular Coagulation (DIC)? creased due to the consumption of the
coagulation factors due to the tiny clots
A. prolonged PT, elevated forming throughout the vasculature. This
platelet count, decreased FDP is also the reason that the fibrinogen
B. normal PT, decreased fibrino- levels and platelet levels are decreased.
gen, decreased platelet count, Finally FDP, or fibrin degredation prod-
decreased FDP ucts, are increased due to the formation

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C. prolonged PT, decreased and subsequent dissolving of many tiny
fibrinogen, decreased platelet clots in the vasculature. The FDPs are
count, increased FDP the pieces of fibrin that are left after the
D. normal PT, decreased platelet fibrinolytic processes take place.
count, decreased FDP

6. A dilution commonly used for a B;
routine sperm count is: A dilution commonly used for a routine
sperm count is a 1:20.
A. 1:2
B. 1:20
C. 1:200
D. 1:400

7. The prozone effect ( when per- B;
forming a screening titer) is Prozone effect (due to antibody excess)
most likely to result in: will result in an initial false negative in
spite of the large amount of antibody in
A. False positive the serum, followed by a positive result
B. False negative as the specimen is diluted.
C. No reaction at all
D. Mixed field reaction

8. Illustrated in this photograph is A;
an agar quadrant plate contain- One of the key characteristics to the iden-
ing casein (A), tyrosine (B), ni- tification of Nocardia asteroides is its in-
trate (C) and xanthine (D). None ability to hydrolyze casein, tyrosine or
of the substrates have been hy- xanthine, as shown in this photograph.
drolyzed and nitrate has been re- Nitrates are reduced to nitrites. Both
duced. The most likely identifi- Nocardia brasiliensis and Actinomadura
cation is: madurae hydrolyze both casein and tyro-
sine; Streptomyces griseus hydrolyzes all
A. Nocardia asteroides three of the substrates.
B. Nocardia brasiliensis
C. Streptomyces griseus
D. Actinomadura madurae





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