MLT ASCP Practice UPDATED Exam Questions and CORRECT Answers

MLT ASCP Practice UPDATED Exam
Questions and CORRECT Answers
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


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 If the media had outdated for a prolonged period,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:


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D. Use of outdated medium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;
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)?


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. - CORRECT 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. -
CORRECT 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. - CORRECT
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. - CORRECT 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
A patient who has a primarily vegetarian diet will most likely have an alkaline urine pH. A low-
carbohydrate diet as well as the ingestion of citrus fruits can also lead to a more alkaline urine
sample. - CORRECT ANSWER - A patient who has a primarily vegetarian diet will most
likely have an acid urine pH.


A;