MLT ASCP Practice Test Questions board practice

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. - - - ✔✔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 If the media had outdated for a prolonged period, either there would be no reaction or the appearance of only a faint pink tinge, either in the slant, the butt or both, again depending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urea agar slant on the far right indicates: A. Weak activity B. Strong activity C. Slant only inoculated D. Use of outdated medium D; The steps in the PCR process are: 1.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 concentration. - - - ✔✔The laboratorian completed the mixing study ordered for John Doe. The results are as follows: Initial aPTT result: 167 seconds Initial 1:1 Mix with Normal Pooled Plasma: 158 seconds Incubated 1:1 Mix with Normal Pooled Plasma: 150 seconds Which of the choices below would most likely explain the results for this patient? A. Factor VIII deficiency B. Immediate-acting coagulation inhibitor C. Time/temperature-dependent coagulation inhibitor D. Factor VII deficiency A; HbsAg is positive in acute and chronic Hepatitis B infections, since the antigen is found on the actual surface of the virus. HbeAg is present in the blood when the hepatitis B viruses are replicating, indicating an active infection. Anti-Hbc IgM is present due to the immune response to the presence of the hepatitis core antigen and indicates an acute infection. Anti-HBs is generally interpreted as indicating recovery and immunity from hepatitis B virus infection, according to the CDC. - - - ✔✔Given the following results, what is the immune status of the patient? HbsAg: positive HbeAg: positive Anti-HBc IgM: positive Anti-HBs: negative A. acute infection B. chronic infection C. immunization D. susceptible A; Albumin is a "negative" acute phase protein since it is found in decreased levels during acute phase response. Alpha-1-antitrypsin, fibrinogen, and ceruloplasmin are all "positive" acute phase proteins that are found in increased levels during acute phase response. - - - ✔✔Which one of the following usually shows a decrease during an acute phase response? A. Albumin B. Alpha-1 Antitrypsin C. Fibrinogen D. Ceruloplasmin False Because lower titers could be due to both passive and immune anti-D, in the absence of results that suggest immune anti-D, routine antibody titration is not a good use of time compared to assuming that anti-D is passive. Best practice guidelines do NOT recommend routine titration for women known to be injected with RhIg and exhibiting a 2+ or less reaction with D+ red cells consistent with passive anti-D from RhIg. - - - ✔✔True/False A pregnant female who received RhIg at 28 weeks gestation has a positive antibody screen at delivery. If the antibody has been confirmed as anti-D alone and reacts 1+ in the indirect antiglobulin test with D+ red cells, performing a titration to investigate if the anti-D is immune is good practice.