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MLT ASCP ACTUAL EXAM PAPER 2026 QUESTIONS WITH SOLUTIONS GRADED A+

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MLT ASCP ACTUAL EXAM PAPER 2026 QUESTIONS WITH SOLUTIONS GRADED A+

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January 16, 2026
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2025/2026
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MLT ASCP ACTUAL EXAM PAPER 2026
QUESTIONS WITH SOLUTIONS GRADED A+

⫸ 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If the media had outdated for a prolonged period, either there would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.. Answer: 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D. Use of outdated medium


⫸ 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.). Answer: 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.. Answer: 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.. Answer: 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


⫸ B;
A dilution commonly used for a routine sperm count is a 1:20..
Answer: A dilution 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 as the specimen is diluted.. Answer: The
prozone effect ( when performing a screening titer) is most likely to
result in:


A. False positive
B. False negative
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.. Answer: 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;
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