MLT ASCP Practice Test Questions board practice Exam 2024 | Questions with 100% Correct Answers | Verified | Updated 2024

The correct answer for this question is 1300 mg/dL. The laboratorian per- formed 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 1. 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producer, both the slant and the butt w Therefore, the reaction is dependent on the strength of urease acti media had outdated for a prolonged period, either there would be n or the appearance of only a faint pink tinge, either in the slant, the b both, again depending on the strength of urease production by the u 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urea agar slant on the far right indicates: A. Weak activity B. Strong activity C. Slant only inoculated D. Use of outdated medium 2. 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 3. 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 4. 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