MLT GlobeDHA exam preparation (Dubai) – page 34
700 practice MCQs for the DHA medical laboratory exam. Level: Basic to intermediate.
For an assay, allowable total error (TEa) is 10%, bias is 2% and CV is 2%. The sigma metric is:
Sigma = (TEa − |bias|) ÷ CV = (10 − 2) ÷ 2 = 4.0. Forgetting to subtract bias gives 5.0. Higher sigma values allow simpler QC rules.
A partially acid-fast aerobic actinomycete from a lung biopsy grows in lysozyme broth, while a Streptomyces control strain does not. This result supports identification as:
Nocardia is lysozyme resistant and keeps growing in lysozyme broth, while Streptomyces is inhibited. This, with partial acid-fastness, helps separate Nocardia from other aerobic actinomycetes.
Pus from a liver abscess grows tiny beta-hemolytic streptococcal colonies with a caramel (butterscotch) odour. The organism most likely belongs to the:
The S. anginosus group forms minute colonies with a caramel odour and is known for causing deep abscesses. Other viridans groups rarely produce this picture.
The in vitro hair perforation test is used to separate:
T. mentagrophytes complex forms wedge-shaped perforations in hair in vitro, whereas T. rubrum does not. It is a classic test for these two look-alike dermatophytes.
In the shell vial culture method for CMV, the virus is detected within 24–48 hours by:
Centrifugation enhances infection of cells on a coverslip, and fluorescent antibodies detect early antigen before cytopathic effect appears. Conventional CMV culture may take up to 3 weeks.
A pregnant woman has positive rubella IgM and IgG. The IgG avidity is high. This most likely means:
IgG avidity increases with time after infection. High avidity rules out recent primary infection; IgM may persist or be falsely positive.
Amniotic fluid lamellar body count is used to assess fetal lung maturity. How is it usually measured?
Lamellar bodies, which store surfactant, are similar in size to platelets and are counted on the platelet channel. Thin-layer chromatography is used for the L/S ratio, and 450 nm for bilirubin.
Red cells are not agglutinated by anti-A; A is shown only by adsorption–elution. The plasma often has an anti-A1-like antibody, and secretor saliva contains H only. The subgroup is:
Ael is shown only by adsorption–elution, has no A in saliva and often has anti-A in plasma. Am has A substance in the saliva of secretors.
A patient with severe pneumococcal infection has red cells agglutinated by almost all adult ABO-compatible plasmas but not by cord plasma. The cells react with peanut (Arachis hypogaea) lectin. This is:
Bacterial neuraminidase exposes the T antigen; almost all adults have anti-T but newborns do not. Arachis lectin reacts with T; Tn is persistent and reacts with Salvia lectins.
A group A patient receives a stem cell transplant from a group O donor. Until conversion is complete, which components are appropriate?
This is a minor ABO mismatch (donor plasma has anti-A). Red cells must be compatible with the donor's anti-A (group O), and plasma must be compatible with the recipient's A cells (A or AB).
A group O patient receives a stem cell transplant from a group A donor. Which red cells should be given early after transplant?
In a major ABO mismatch, the recipient still makes anti-A, so group O red cells are given until anti-A disappears. Plasma should be group A or AB to protect the new donor-derived A cells.
A pregnant woman is found to have partial D category DVI. How should she be managed for RhIG?
People with partial D, such as DVI, lack parts of the D antigen and can make anti-D. Pregnant women with partial D should receive RhIG and D-negative red cells.
Measured sodium is 130 mmol/L with glucose 600 mg/dL (33.3 mmol/L). Correcting by 1.6 mmol/L sodium for each 100 mg/dL glucose above 100 mg/dL, the corrected sodium is about:
Glucose is 500 mg/dL above 100, so 5 × 1.6 = 8; 130 + 8 = 138 mmol/L. Hyperglycemia pulls water out of cells and dilutes sodium; this is a real (translocational) change, not an analytical artifact.
A 65-year-old with a large mesenchymal tumor has fasting hypoglycemia with low insulin, low C-peptide, low β-hydroxybutyrate and a clear glucose rise after glucagon. The most likely cause is:
IGF-2 acts on insulin receptors, suppressing insulin, C-peptide and ketogenesis while preserving liver glycogen. Insulinoma raises insulin and C-peptide; injected insulin raises insulin.
Serum 1,5-anhydroglucitol falls during periods of hyperglycemia because:
1,5-anhydroglucitol is normally almost fully reabsorbed by the kidney. When glycosuria occurs, glucose competes for reabsorption, so it is lost in urine and serum levels drop, reflecting recent glucose peaks.
A freezing point osmometer reads a sample freezing point of −0.558 °C. Given 1 Osm/kg water lowers the freezing point by 1.86 °C, the osmolality is:
Osmolality = 0.558 ÷ 1.86 = 0.300 Osm/kg = 300 mOsm/kg. Reading the freezing point value directly as 558 ignores the 1.86 °C constant.
A child has tendon xanthomas and early atherosclerosis with only moderately raised cholesterol. Plasma sitosterol and other plant sterols are markedly high. The defect is in:
Sitosterolemia results from loss of ABCG5/ABCG8, which normally pump plant sterols from gut cells and liver into the gut lumen and bile. Plant sterols accumulate and cause xanthomas.
A day after cardiac arrest and resuscitation, a patient has ALT 3000 U/L, AST 4000 U/L and LD 6000 U/L. The values fall by half within 2–3 days. The most likely diagnosis is:
Hypoperfusion causes rapid centrilobular necrosis with very high aminotransferases, a very high LD (ALT/LD ratio often below 1.5) and a quick fall. Viral hepatitis rises and falls more slowly with lower LD.
Bull's moving average algorithm monitors hematology analyzer performance by tracking which parameters in batches of patient samples?
Red cell indices are very stable across a patient population, so a shift in their moving average suggests an analyzer problem. WBC and platelet counts vary too much between patients for this use.
A patient with PNH on a C5 inhibitor (eculizumab) still has mild anemia, a raised reticulocyte count and a new positive DAT with anti-C3 only. The best explanation is:
Blocking C5 prevents membrane attack complex lysis, but C3 still deposits on CD55-deficient cells, and these are cleared by macrophages. This extravascular hemolysis gives a C3-only positive DAT.