MLT ASCP CERTIFICATION PRACTICE TEST QUESTIONS & ANSWERS 2026-2027 | H…
P R O F E S S I O N A L P R A C T I C E M AT E R I A L S
MLT ASCP Certification
Practice Test Questions &
Answers 2026-2027 | High-
Yield Board Review
(Rationales)
Verified Answers Exam Ready With Rationales 100 QUESTIONS
DOCUMENT OVERVIEW
This document provides 100 high-yield MLT ASCP certification practice questions,
complete with correct answers and detailed rationales, along with diagrams. It offers a
comprehensive review of medical laboratory technology principles and is suitable for board
certification preparation and in-depth subject matter study.
TOPICS
Clinical Chemistry and Urinalysis Q1–Q39
Microbiology and Immunology Q40–Q66
Molecular Diagnostics and Genetics Q67–Q69
Hematology and Coagulation Q70–Q99
Quality Control and Laboratory Operations Q100–Q100
Page 1
, E XA M Q U EST I O N S
Q1 QUESTION 1 OF 100
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?
A. 325 mg/dL
B. 1300 mg/dL
C. 975 mg/dL
D. 1625 mg/dL
Q2 QUESTION 2 OF 100
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
Page 2
, RATIONALE
The polymerase chain reaction (PCR) is a method used to amplify a specific segment of DNA. The
process begins with denaturation, where the double-stranded DNA template is heated to separate
it into two single strands.
Q3 QUESTION 3 OF 100
B;
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
RATIONALE
The concentration of sodium chloride in an isotonic solution is 0.85%, which is clinically significant
for fluid and electrolyte balance.
Q4 QUESTION 4 OF 100
B;
A dilution commonly used for a routine sperm count is a 1:20.
CORRECT ANSWER
A dilution commonly used for a routine sperm count is:
A. 1:2
B. 1:20
C. 1:200
D. 1:400
Page 3
, RATIONALE
A dilution commonly used for a routine sperm count is:
A. 1:2
B. 1:20
C. 1:200
D. 1:400:20 is standard for routine sperm counts because it effectively dilutes the sample to a
concentration that is easily quantifiable using a hemocytometer. This dilution factor ensures that
sperm are not too numerous to count accurately while still maintaining a sufficient number for a
statistically reliable assessment of sperm concentration.
Q5 QUESTION 5 OF 100
A;
During primary hypothyroidism, where a defect in the thryoid gland is producing low levels
of T3 and T4, the TSH level is increased. TSH is released in elevated quantities in an attempt
to stimulate the thryoid to produce more T3 and T4 as part of a feedback mechanism.
CORRECT ANSWER
Serum TSH levels five-times the upper limit of normal in the presence of a low T4 and low T3
uptake could mean which of the following:
A. The thyroid has been established as the cause of hypothyroidism
B. The thyroid is ruled-out as the cause of hypothyroidism
C. The pituitary has been established as the cause of hypothyroidism
D. The diagnosis is consistent with secondary hyperthyroidism
Q6 QUESTION 6 OF 100
B
2. D
3. A
4. C
Red to Brown Urine: porphobilinogen, hematuria, myoglobinuria, etc.
Green: Food colorings; Increased carotene in the diet;
Pseudomonas aeruginosa infection
Yellow: bilirubin, bile pigments
White: phosphates, other crytals
Page 4
P R O F E S S I O N A L P R A C T I C E M AT E R I A L S
MLT ASCP Certification
Practice Test Questions &
Answers 2026-2027 | High-
Yield Board Review
(Rationales)
Verified Answers Exam Ready With Rationales 100 QUESTIONS
DOCUMENT OVERVIEW
This document provides 100 high-yield MLT ASCP certification practice questions,
complete with correct answers and detailed rationales, along with diagrams. It offers a
comprehensive review of medical laboratory technology principles and is suitable for board
certification preparation and in-depth subject matter study.
TOPICS
Clinical Chemistry and Urinalysis Q1–Q39
Microbiology and Immunology Q40–Q66
Molecular Diagnostics and Genetics Q67–Q69
Hematology and Coagulation Q70–Q99
Quality Control and Laboratory Operations Q100–Q100
Page 1
, E XA M Q U EST I O N S
Q1 QUESTION 1 OF 100
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?
A. 325 mg/dL
B. 1300 mg/dL
C. 975 mg/dL
D. 1625 mg/dL
Q2 QUESTION 2 OF 100
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
Page 2
, RATIONALE
The polymerase chain reaction (PCR) is a method used to amplify a specific segment of DNA. The
process begins with denaturation, where the double-stranded DNA template is heated to separate
it into two single strands.
Q3 QUESTION 3 OF 100
B;
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
RATIONALE
The concentration of sodium chloride in an isotonic solution is 0.85%, which is clinically significant
for fluid and electrolyte balance.
Q4 QUESTION 4 OF 100
B;
A dilution commonly used for a routine sperm count is a 1:20.
CORRECT ANSWER
A dilution commonly used for a routine sperm count is:
A. 1:2
B. 1:20
C. 1:200
D. 1:400
Page 3
, RATIONALE
A dilution commonly used for a routine sperm count is:
A. 1:2
B. 1:20
C. 1:200
D. 1:400:20 is standard for routine sperm counts because it effectively dilutes the sample to a
concentration that is easily quantifiable using a hemocytometer. This dilution factor ensures that
sperm are not too numerous to count accurately while still maintaining a sufficient number for a
statistically reliable assessment of sperm concentration.
Q5 QUESTION 5 OF 100
A;
During primary hypothyroidism, where a defect in the thryoid gland is producing low levels
of T3 and T4, the TSH level is increased. TSH is released in elevated quantities in an attempt
to stimulate the thryoid to produce more T3 and T4 as part of a feedback mechanism.
CORRECT ANSWER
Serum TSH levels five-times the upper limit of normal in the presence of a low T4 and low T3
uptake could mean which of the following:
A. The thyroid has been established as the cause of hypothyroidism
B. The thyroid is ruled-out as the cause of hypothyroidism
C. The pituitary has been established as the cause of hypothyroidism
D. The diagnosis is consistent with secondary hyperthyroidism
Q6 QUESTION 6 OF 100
B
2. D
3. A
4. C
Red to Brown Urine: porphobilinogen, hematuria, myoglobinuria, etc.
Green: Food colorings; Increased carotene in the diet;
Pseudomonas aeruginosa infection
Yellow: bilirubin, bile pigments
White: phosphates, other crytals
Page 4
