UPDATED 2024 MLT ASCP PRACTICE
TEST QUESTIONS BOARD PRACTICE
WITH ACCURATE ANSWERS
,UPDATED 2024 MLT ASCP PRACTICE
TEST QUESTIONS BOARD PRACTICE
WITH ACCURATE ANSWERS
Plasma concentrations of creatinine are used to assess renal function. Creatinine clearance is based on
the serum creatinine level and is used to measure glomerular filtration rate, or GFR. - CORRECT
ANSWERSAn increased serum level of which of the following analytes is MOST commonly associated
with decreased glomerular filtration?
A. Creatinine
B. Uric acid
C. Urea
D. Ammonia
E. Glucose
D;
The last choice in this question is false. Rocky mountain spotted fever is NOT geographically restricted to
the Rocky mountain region. In fact, according to the CDC, this disease is widely distributed east of the
Rocky Mountains and also occurs in limited areas on the Pacific Coast. - CORRECT ANSWERSWhich one
of the following statements is false?
A. Rickettsial infections are usually best diagnosed by specific serology, rather than the Weil-Felix test
B. Ehrlichiosis is transmitted by the lonestar tick
C. Rickettsia rickettsii causes Rocky mountain spotted fever
D. Rocky mountain spotted fever is almost always geographically restricted to the Rocky mountain
region
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
B;
TEST QUESTIONS BOARD PRACTICE
WITH ACCURATE ANSWERS
,UPDATED 2024 MLT ASCP PRACTICE
TEST QUESTIONS BOARD PRACTICE
WITH ACCURATE ANSWERS
Plasma concentrations of creatinine are used to assess renal function. Creatinine clearance is based on
the serum creatinine level and is used to measure glomerular filtration rate, or GFR. - CORRECT
ANSWERSAn increased serum level of which of the following analytes is MOST commonly associated
with decreased glomerular filtration?
A. Creatinine
B. Uric acid
C. Urea
D. Ammonia
E. Glucose
D;
The last choice in this question is false. Rocky mountain spotted fever is NOT geographically restricted to
the Rocky mountain region. In fact, according to the CDC, this disease is widely distributed east of the
Rocky Mountains and also occurs in limited areas on the Pacific Coast. - CORRECT ANSWERSWhich one
of the following statements is false?
A. Rickettsial infections are usually best diagnosed by specific serology, rather than the Weil-Felix test
B. Ehrlichiosis is transmitted by the lonestar tick
C. Rickettsia rickettsii causes Rocky mountain spotted fever
D. Rocky mountain spotted fever is almost always geographically restricted to the Rocky mountain
region
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
B;
