MLT ASCP BOC Exam 2026 – Medical Laboratory Technician Study Guide with Practice Questions and Answers

MLT ASCP BOC EXAM 2026 WITH
100% ACCURATE AND VERIFIED
ANSWERS GRADED A+
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. - answer-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



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 If the media had outdated for a prolonged period, either there would be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D. Use of outdated medium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;

Since hemoglobin is measured spectrophotometrically on hematology analzyers,
interference from lipemia or icteric specimens can lead to decreased light detected and
measured through the sample and therefore inaccurate hemoglobin results occur. -
answer-On an electronic cell counter, hemoglobin determination may be falsely
elevated caused by the presence of:



A. Lipemic or icteric plasma

B. Leukocytopenia or Leukocytosis

C. Rouleaux or agglutinated RBCs

D. Anemia or Polycythemia



False