MLT ASCP Practice Test Exam 2026
Questions and Answers 100% Pass
Guaranteed
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 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?
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,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 If the media had outdated for a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 answer-The
urease reaction seen in the Christensen's urea agar slant on the far right indicates:
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,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D. Use of outdated medium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 answer-What is the first step of the PCR reaction?
A. Hybridization
B. Extension
C. Annealing
D. Denaturation
B;
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, Isotonic or normal saline is a 0.85 % solution of sodium chloride in water. -
Correct 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. - Correct answer-Which of the following
laboratory results would be seen in a patient with acute Disseminated Intravascular
Coagulation (DIC)?
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Questions and Answers 100% Pass
Guaranteed
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 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?
©COPYRIGHT 2025, ALL RIGHTS RESERVE 1
,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 answer-The
urease reaction seen in the Christensen's urea agar slant on the far right indicates:
©COPYRIGHT 2025, ALL RIGHTS RESERVE 2
,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 answer-What is the first step of the PCR reaction?
A. Hybridization
B. Extension
C. Annealing
D. Denaturation
B;
©COPYRIGHT 2025, ALL RIGHTS RESERVE 3
, Isotonic or normal saline is a 0.85 % solution of sodium chloride in water. -
Correct 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. - Correct answer-Which of the following
laboratory results would be seen in a patient with acute Disseminated Intravascular
Coagulation (DIC)?
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