MLT ASCP Practice Test Questions
board practice
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 microli
...
MLT ASCP Practice Test Questions
board practice
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. - ANS - 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 If the media had outdated for a prolonged period,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.
- ANS - 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D. Use of outdated medium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.) - ANS
- 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. - ANS - 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 intravascul
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