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MLT ASCP Practice Test Questions board practice Questions with Correct Answers
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MLT ASCP Practice Test Questions boardpractice Questions with 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. - -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 If the media had outdated for a prolonged period, either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. - -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D. Use of outdated medium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.) - -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. - -
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. - -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. - -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. - -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. - -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. - -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. - -A patient who has a primarily
vegetarian diet will most likely have an acid urine pH.
-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. - -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
-A;
Fusarium species is the most likely associated with mycotic keratitis.
Trichophyton rubrum is a dermatophyte that commonly causes an itching,
scaling skin infection of the feet, known as tinea pedis. Scedosporium
apiospermum is commonly associated with sinusitis. Aspergillus niger
typically causes otitis externa and can also be associated with sinusitis. - -
Which of the following species or organisms is the most likely to be the cause
of mycotic keratitis (fungal eye infection)?
A. Fusarium species
B. Trichophyton rubrum.
C. Scedosporium apiospermum
D. Aspergillus niger
-A;
Oxalate, EDTA, and citrate are anticoagulants that inhibit clot formation. - -
Which of the following blood additives is most useful for serum collection:
A. Polymer barrier
B. Oxalate
C. EDTA
D. Citrate
-B;
This patient is most likely suffering from an immediate-acting coagulation
inhibitor; most commonly, lupus anticoagulant. Notice that the addition of
normal pooled plasma does not correct upon initial or incubated mix, which
means that the inhibitor is not time or temperature-dependent.
Factor VIII is not the correct answer as a factor deficiency would have
corrected upon the addition of normal pooled plasma. Factor VII is not the
correct answer, as the aPTT assay does not account for factor VII activity or
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