UPDATED 2024 MLT ASCP PRACTICE TEST QUESTIONS BOARD PRACTICE WITH ACCURATE ANSWERS

UPDATED 2024 MLT ASCP PRACTICE
TEST QUESTIONS BOARD PRACTICE
WITH ACCURATE ANSWERS

,UPDATED 2024 MLT ASCP PRACTICE
TEST QUESTIONS BOARD PRACTICE
WITH ACCURATE ANSWERS
B;

The presence of hyaline, septate hyphae, and a young conidiophore with a foot cell and a swollen vesicle
in a fungus culture are excellent clues indicative of Aspergillus. Aspergillus has the ability to infect
primarily immunocompromised hosts, and causes pneumonia and/or disseminates to other organs.
Clinical specimens should be inoculated onto primary isolation media, such as Sabouraud's dextrose
agar. Aspergillus spp. are fast growing and can be white, yellow, yellow-brown, brown to black or green
in color. - CORRECT ANSWERSThe presence of hyaline, septate hyphae, and a young conidiophore with a
foot cell and a swollen vesicle in a fungus culture are excellent clues indicative of:



A. Acremonium

B. Aspergillus

C. Paecilomyces

D. Penicillium

E. Scopulariopsis



A;

Bacitracin susceptibility is a presumptive test for the identification of Lancefield group A Streptococcus. -
CORRECT ANSWERSWhich of the following is a presumptive test for the identification of Lancefield
group A Streptococcus:



A. Bacitracin susceptibility

B. Positive coagulase

C. Hippurate hydrolysis

D. Gram stain



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. - CORRECT ANSWERSAfter 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. - CORRECT ANSWERSThe 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.) - CORRECT
ANSWERSWhat 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. - CORRECT ANSWERSThe
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. - CORRECT ANSWERSWhich 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