The division of Department of Health and Human Services (DHHS) responsible for protecting public health by ensuring the safety and efficacy of foods, drugs, biological products, medical devices, and cosmetics is:
A. Health Insurance Portability and Accountability Act (HIPPA)
B. the Joint Commission
C. Food and Drug Administration (FDA)
D. Clinical and Laboratory Standards Institute (CLSI)
Provide the equivalent measurement for one pint.
A. 4.2 liters
B. 0.25 liters
C. 0.1 liter
D. 0.474 liters
This antibody is found in the serum of Le(a-b-) secretors.
Which one of these Lewis blood group system phenotypes usually produces anti-Lea?
A. Le(a+b+)
B. Le(a+b-)
C. Le(a--b+)
D. Le(a--b-)
Phase of reactivity is primarily at immediate spin (4+) and reactions get weaker at AHG (w+). There is no specific pattern of reactivity and the auto control is negative which rules out an autoantibody. This is a strong cold antibody which is still
slightly present after incubation and washing.
Activation and binding of the antibody takes place at room temperature or colder. Eliminating this phase will prevent the antibody from binding. Cold antibodies usually are more of a nuisance to blood bankers and are not clinically significant.
When performing an antibody screen, both the screen cells are 4+ at immediate spin and W+ at AHG. The antibody panel shows 4+ reactions at immediate spin and W+ reactions at AHG and there is no specific match to the reaction pattern.
The auto control is negative.
What would be a logical next step?
A. Have patient redrawn
B. Repeat testing using warmed patient sample and reagents and just do AHG reading
C. Run an enzyme panel
Francisella is slow growing on primary isolation culture media and produces poorly-staining coccobacilli on Gram stain. Because Francisella tularensis is included on the list of bioterrorism agents, suspicious isolates should be referred immediately to a public health laboratory in lieu of attempting an in-laboratory identification.
One of the BIGGEST problems with isolating Francisella is that these organisms:
A. produce wide zones of hemolysis
B. grow so slowly
C. overgrow other organisms
D. produce greenish colonies like E. coli
Microbiology
Matching: The detection of a distinct odor is often helpful in the presumptive identification of bacterial culture isolates. Match each of the odors listed with its corresponding bacterial species name.
1.
Streptococcus anginosus (milleri)
2.
Pseudomonas aeruginosa
3.
Eikenella corrodens
4.
Alcaligenes faecalis
A. Pared apples
B. Grapes
C. Butterscotch
D. Bleach
Clostridium difficile is a gram-positive, anaerobic, spore-forming bacillus that produces a strong toxin. C difficile is a common cause of antibiotic-associated diarrhea and pseudomembranous colitis.
Microbiology
A recto-sigmoidoscopy revealed pseudomembranes in a patient with severe diarrhea following prolonged treatment with ampicillin. Which of the following organisms is MOST likely to be isolated?
A. Bacteroides fragilis
B. Clostridium botulinum
C. Clostridium difficile
D. Fusobacterium nucleatum
E. Klebsiella oxytoca
Serological diagnosis of active or recent infection generally requires the demonstration of IgM antibody, or the demonstration of a fourfold rise in the titer of specific IgG antibody. Which of the following would be considered most significant as it relates to serological testing:
A. Presence of an antibody titer is generally diagnostic
B. Rise of antibody titers is diagnostic
C. Concentration of antibody is diagnostic
D. Cross reactivity is not significant
Match the type of media with the phrase that best describes that media:
1.
Differential
2.
Selective
3.
Enrichment
A. Media that contains agents that inhibit all but one specific organism.
B. Contains certain factors that allow colonies of specific organisms to appear different than other colonies.
C. Encourages the growth of specific types of organisms.
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
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