ASCP MLT Exam Prep: Free Practice Questions and Study Plan

Answer: D. Potassium. Red cells hold far more potassium than plasma does, so when they rupture, potassium leaks into the serum and the result reads high. That's why a hemolyzed sample usually means recollecting for potassium. Sodium and chloride are lower inside red cells than in plasma, so hemolysis doesn't raise them. Hemolysis doesn't release glucose; if anything, delayed separation lowers glucose because cells keep consuming it. Source: Simon LV, Rout P, "Hyperkalemia" (StatPearls/NCBI).

Answer: A. Infants begin producing their own anti-A and anti-B over the first months of life, so a newborn's serum either lacks ABO antibodies or contains maternal IgG that crossed the placenta. Either way, reverse grouping wouldn't reflect the baby's own type, so ABO typing in infants under 4 months relies on forward grouping. Newborn red cells do carry A and B antigens, though they're weaker than adult antigens (B is wrong). Anticoagulant isn't the issue (C). Reverse grouping detects mainly IgM ABO antibodies at room temperature, not only IgG (D). Source: Canadian Blood Services, Clinical Guide to Transfusion, ch. 8 "Pre-transfusion testing"; Dean L, "The ABO blood group" (NCBI Bookshelf).

Answer: B. Microcytic, hypochromic. MCV = (Hct × 10) ÷ RBC = (30 × 10) ÷ 4.0 = 75 fL, below ASCP's 76–100 fL range, so the cells are microcytic. MCHC = (Hgb × 100) ÷ Hct = (9.0 × 100) ÷ 30 = 30 g/dL, below 32–36 g/dL, so they're hypochromic. Iron deficiency is the classic cause of this pattern. Choosing D means you calculated MCV but skipped MCHC; A and C misread the MCV. Source: ASCP MLT Content Guideline (RBC indices; reference ranges); Sarma PR, "Red Cell Indices," Clinical Methods (NCBI Bookshelf).

Answer: D. Staphylococcus aureus. Clusters plus a positive catalase point to staphylococci (or micrococci); a positive coagulase separates S. aureus from the coagulase-negative staphylococci. S. epidermidis is coagulase negative. Streptococci grow in chains or pairs and are catalase negative. Micrococcus is catalase positive but coagulase negative. Source: Taylor TA, Tobin EH, "Staphylococcus aureus Infection" (StatPearls/NCBI).

Answer: B. 2.2 mg/dL. Indirect = total − direct = 3.0 − 0.8 = 2.2 mg/dL. Labs don't measure unconjugated bilirubin directly; they calculate it this way, and it's one of the calculations ASCP expects you to know. D adds instead of subtracting. Source: ASCP MLT Content Guideline (required calculations); Guerra-Ruiz AR et al., "Measurement and clinical usefulness of bilirubin in liver disease," Adv Lab Med 2021.

Answer: D. Free hemoglobin or myoglobin. The strip's blood pad detects the peroxidase-like activity of the heme group, whether it comes from intact red cells, free hemoglobin, or myoglobin. With no cells on the microscopic, free heme pigment is the likely cause, either from intravascular hemolysis, muscle injury, or red cells that lysed in dilute or alkaline urine. Ascorbic acid causes false negatives on the blood pad, not positives. Bilirubin doesn't react on the blood pad. High specific gravity crenates cells but doesn't destroy them. Source: Simerville JA, Maxted WC, Pahira JJ, "Urinalysis: A Comprehensive Review," Am Fam Physician 2005.

Answer: B. Acute myeloid leukemia. Auer rods are needle-shaped clumps of azurophilic (primary) granules, and they show up in myeloid blasts. Finding them points to a myeloid, not lymphoid, lineage. Lymphoblasts (A) and mature CLL lymphocytes (C) don't form them, and the reactive lymphocytes of mononucleosis (D) don't either. Source: Vakiti A, Reynolds SB, Mewawalla P, "Acute Myeloid Leukemia" (StatPearls/NCBI).

Answer: A. Continuous gentle agitation. At 20–24 °C, platelets must be gently and continuously agitated, which keeps them suspended and supports gas exchange through the bag. Leaving them still (B) isn't permitted at that temperature. Irradiation (C) is done only for specific patients, such as those at risk of transfusion-associated graft-versus-host disease, not as a storage rule. Freezing (D) is the plasma rule, not a platelet one. Federal regulation also permits platelet storage at 1–6 °C; for that temperature range, agitation is optional. Source: 21 CFR 640.25, General requirements (eCFR); 21 CFR 640.24 (eCFR).

Answer: C. Immunity from vaccination. The vaccine contains only surface antigen, so a vaccinated person makes anti-HBs but never anti-HBc. Past natural infection (D) leaves both anti-HBs and anti-HBc positive. Acute and chronic infection (A, B) both show HBsAg positive. Source: CDC, "Interpretation of Hepatitis B Serologic Test Results".

Answer: B. Metabolic acidosis, partially compensated. Work it in order. pH 7.30 is below 7.35, so the patient is acidemic. Bicarbonate is low (14, below 23–29), which explains the acidosis, so the primary problem is metabolic. pCO₂ is also low (30, below 35–44): the lungs are blowing off CO₂ to compensate. Because pH hasn't returned to normal, compensation is partial. A respiratory acidosis (A) would need a high pCO₂. Source: ASCP MLT Content Guideline (blood gas reference ranges); Castro D, Zubair M, "Arterial Blood Gas" (StatPearls/NCBI).

Answer: C. Pseudomonas aeruginosa. Oxidase positive, non-lactose fermenting, fruity or grape-like odor, and pyocyanin's blue-green color together are the classic P. aeruginosa picture. E. coli and Klebsiella ferment lactose and are oxidase negative. Proteus doesn't ferment lactose either, but it's oxidase negative and known for swarming, not pigment. Source: Lecturio Medical Concept Library, "Pseudomonas".

Answer: B. 12.0 × 10³/µL. Corrected WBC = observed WBC × 100 ÷ (100 + nRBCs) = 15.0 × 100 ÷ 125 = 12.0. Instruments can count nRBCs as white cells, so the raw count runs high. ASCP expects this correction when more than 10 nRBCs are present. A subtracts 25% instead of using the formula; D multiplies the wrong way. Source: ASCP MLT Content Guideline (required calculations); NSCC Hematology Laboratory Manual, "WBC Correction for the Presence of NRBCs".

Answer: B. A single control outside ±3 SD violates the 1₃ₛ rule, a rejection rule that usually signals random error (sometimes a large systematic error). Hold patient results, find and fix the cause, then rerun. The ±2 SD rule (1₂ₛ) is the warning rule (A). Rerunning the control until it passes (C) hides the problem. Source: Westgard QC, "Westgard Rules" and Multirules.

Answer: A. 140 mg/dL. LDL = total cholesterol − HDL − (triglycerides ÷ 5) = 220 − 50 − 30 = 140 mg/dL. The TG ÷ 5 term estimates VLDL cholesterol in mg/dL. C forgets the VLDL term; D forgets HDL. The equation isn't valid when triglycerides are above 400 mg/dL. Source: Friedewald WT, Levy RI, Fredrickson DS, Clin Chem 1972;18:499–502; MDCalc, "LDL Calculated"; ASCP MLT Content Guideline.

Answer: B. 35 days. CPDA-1 red cells last 35 days at 1–6 °C. ACD, CPD, or CP2D without additive gives 21 days (A). Adding an additive solution extends red cells to 42 days (C). Five days (D) is the dating for platelets stored at 20–24 °C. Source: 21 CFR 610.53, Dating periods for blood components (eCFR).

Answer: A. Factor VII. The PT tests the extrinsic and common pathways; the aPTT tests the intrinsic and common pathways. Factor VII is the only factor measured by the PT but not the aPTT, so it alone produces an isolated prolonged PT. Deficiencies of VIII, IX, or XII prolong the aPTT instead. Source: The Blood Project, "Isolated Elevation in Prothrombin Time (PT)".

Answer: D. For bacterial culture, CSF should be sent to the laboratory immediately and processed promptly. If it cannot be processed right away, do not refrigerate the specimen; hold it at room temperature. Current WHO surveillance guidance uses 35–37 °C incubation for CSF that has been inoculated into Trans-Isolate transport medium when processing or transport is delayed, not as a blanket storage rule for every raw CSF tube. Source: WHO, Surveillance Guide for Vaccine-Preventable Diseases, Module 11: Invasive Bacterial Vaccine-Preventable Diseases (2023), Table 11.2.

Answer: B. 1.5. Globulin = total protein − albumin = 7.0 − 4.2 = 2.8 g/dL. A/G = 4.2 ÷ 2.8 = 1.5. D is the globulin value itself, a stop-one-step-short error. A inverts the ratio (2.8 ÷ 4.2). C divides total protein by albumin. Source: ASCP MLT Content Guideline (required calculations); Busher JT, "Serum Albumin and Globulin," Clinical Methods (NCBI Bookshelf).

Answer: A. pH increases. Bacteria multiply in unrefrigerated urine and break urea down into ammonia, raising the pH. The same bacteria consume glucose (it falls, not rises) and convert nitrate to nitrite (nitrite rises, not falls). Bilirubin breaks down with light exposure and time, so it falls. Source: Queremel Milani DA, Jialal I, "Urinalysis" (StatPearls/NCBI).

Answer: C. 70 × 10³/µL. Absolute retic = RBC count × retic fraction = 3.5 × 10⁶/µL × 0.020 = 70,000/µL = 70 × 10³/µL. That sits inside ASCP's 20–115 × 10³/µL range. A is a decimal slip (multiplying by 0.002). D divides by the retic fraction instead of multiplying (3.5 ÷ 0.02). Source: ASCP MLT Content Guideline (absolute counts; reticulocyte range).

Answer: D. The most dangerous acute hemolytic reactions are ABO-incompatible transfusions, and most of those trace back to identification or labeling errors. A clerical mix-up may mean another patient is about to receive the wrong unit too. So the investigation starts with a clerical check, followed by a visual check of the post-transfusion sample for hemolysis and a DAT. Repeat typing, antibody testing, and cultures come later as indicated. Source: Suddock JT, Crookston KP, "Transfusion Reactions" (StatPearls/NCBI); Children's Minnesota, "Transfusion Reaction Evaluation".

Answer: C. 100 mL/min. First convert volume to a flow rate: 1,440 mL ÷ 1,440 minutes (24 h) = 1.0 mL/min. Then CrCl = (urine creatinine × urine flow) ÷ plasma creatinine = (120 × 1.0) ÷ 1.2 = 100 mL/min. D uses volume per day without converting to minutes. Source: ASCP MLT Content Guideline (creatinine clearance; timed urine); Shahbaz H, Rout P, Gupta M, "Creatinine Clearance" (StatPearls/NCBI).

Answer: C. 50,000 CFU/mL. A 1-µL loop delivers 0.001 mL, so multiply colonies by 1,000: 50 × 1,000 = 50,000 CFU/mL. With a 10-µL loop (0.01 mL), you'd multiply by 100 instead. Source: St. Agnes Hospital Microbiology, "Urine Cultures" (MICR 6380).

Answer: D. Falsely prolonged clotting times. Citrate tubes are designed for a 9:1 blood-to-anticoagulant ratio. Underfilling leaves too much citrate for the plasma, which binds the calcium added during testing and drags clotting times out. Reject and recollect. Source: Fritsma GA, "Underfilled or overfilled coagulation specimen tubes" (Fritsma Factor).

Answer: D. Prozone (antibody excess). When antibody vastly outnumbers antigen, each antigen site gets saturated by separate antibody molecules, so the lattice that produces visible agglutination or precipitation can't form. Diluting the serum brings the two back into the zone of equivalence, and the reaction appears. Postzone is the opposite problem, too much antigen. Source: Sidana R et al., "Prozone phenomenon in secondary syphilis," Indian J Sex Transm Dis AIDS 2011.

Answer: B. 15%. %Transferrin saturation = (serum iron ÷ TIBC) × 100 = (60 ÷ 400) × 100 = 15%. A low saturation with a high TIBC is the pattern seen in iron deficiency. A comes from dividing TIBC by iron (400 ÷ 60); D subtracts the right answer from 100. Source: ASCP MLT Content Guideline (required calculations); StatPearls, "Total Iron Binding Capacity".

Answer: A. Anti-D. HDFN needs an IgG antibody that can cross the placenta and target an antigen well developed on fetal red cells. Anti-D is IgG and causes severe HDFN. Lewis, I, and P1 antibodies are usually IgM, which can't cross the placenta, and Lewis antigens aren't well developed at birth anyway. Source: Dean L, "Hemolytic disease of the newborn," Blood Groups and Red Cell Antigens (NCBI Bookshelf).

Answer: C. 8.0 × 10³/µL. Cells/µL = cells counted × dilution ÷ volume counted. Volume = 4 squares × 1 mm² × 0.1 mm = 0.4 µL (1 mm³ = 1 µL). So 160 × 20 ÷ 0.4 = 8,000/µL = 8.0 × 10³/µL. Forgetting the dilution gives A (160 ÷ 0.4 = 400). Forgetting the 0.1 mm depth gives B (160 × 20 ÷ 4 = 800). D counts the volume as if only two squares were read. Source: ASCP MLT Content Guideline (manual hemocytometer counts); NSCC Hematology Laboratory Manual, "Manual White Blood Cell Count Procedure".

Answer: A. Mueller-Hinton agar; 0.5 McFarland. Standard disk diffusion uses Mueller-Hinton agar and an inoculum adjusted to a 0.5 McFarland turbidity standard (about 1–2 × 10⁸ CFU/mL, usually quoted as 1.5 × 10⁸). A heavier inoculum shrinks zones and can make a susceptible organism look resistant. Blood agar and MacConkey aren't standardized for disk diffusion. Source: Hudzicki J, "Kirby-Bauer Disk Diffusion Susceptibility Test Protocol," ASM.

Answer: B. 75 mg/dL. Beer's law says absorbance is proportional to concentration, so concentration of unknown = (absorbance of unknown ÷ absorbance of standard) × concentration of standard = (0.30 ÷ 0.40) × 100 = 75 mg/dL. D inverts the ratio. This only holds within the method's linear range. Source: ASCP MLT Content Guideline (Beer's law); Newcastle University Biomedical Sciences Wiki, "Beer-Lambert Law".

Answer: A. Increased red cell destruction. This is the classic hemolysis pattern. Destroyed red cells release LD; their hemoglobin is broken down into unconjugated bilirubin; haptoglobin is used up binding free hemoglobin; and the marrow responds by pushing out reticulocytes. Marrow suppression (B) would lower the reticulocyte count, not raise it. Iron deficiency (C) doesn't produce the destruction markers. Dehydration (D) concentrates blood; it doesn't cause anemia with this chemistry. The pattern tells you hemolysis is happening, not why: immune, inherited, and mechanical causes all need further testing. Source: ARUP Consult, "Hemolytic Anemias".

Answer: A. Donor red cells with recipient plasma. The crossmatch asks one question: does anything in this patient's plasma react with this specific donor unit's red cells? B reverses the pairing. C is the antibody screen, which tests the patient's plasma against commercial group O reagent cells before any unit is chosen. D is forward ABO typing. Source: Canadian Blood Services, Clinical Guide to Transfusion, ch. 8 "Pre-transfusion testing" ("Antibody screening," "Crossmatch").

Answer: C. Acid-fastness is a staining property shared by M. tuberculosis complex and other mycobacteria, so a positive smear says acid-fast organisms are there, not which ones. That's why every specimen still goes to culture, the gold standard, whatever the smear shows. A and D claim a species-level answer the smear can't give. B assigns a disease state that a smear can't determine. Source: CDC, Tuberculosis: "Clinical and Laboratory Diagnosis" ("AFB smear classification").

Answer: D. 400 mg/dL. Dilution factor = total volume ÷ specimen volume = (0.10 + 0.40) ÷ 0.10 = 5. Original concentration = 80 × 5 = 400 mg/dL. C is the most common trap: it uses the diluent volume alone (0.40 ÷ 0.10 = 4) instead of the total volume. A divides by the factor instead of multiplying. B multiplies by the diluent's share of the total (0.8). The "no correction applied" detail matters: if the analyzer had already corrected, multiplying again would double-count. Source: Derived calculation from the values given; dilution correction falls under laboratory mathematics in the ASCP MLT content guideline.

Answer: D. CDC's sequence is: antigen/antibody combination assay first; if reactive, a supplemental HIV-1/HIV-2 antibody differentiation assay; and if the screen is reactive but the differentiation assay is negative, HIV RNA testing to check for acute infection. A CD4 count (A) measures immune status, not diagnosis. B is a monitoring test, not the next diagnostic step. C stops before the algorithm is complete. Source: CDC STI Treatment Guidelines, "HIV Infection: Diagnostic Considerations".

Answer: C. Nitroprusside chemistry detects acetoacetate but does not measure β-hydroxybutyrate, the predominant ketone in diabetic ketoacidosis. So a nitroprusside strip can read low or negative even when β-hydroxybutyrate is increased. That makes A and B wrong. D changes both the specimen and the kind of result: this is a semi-quantitative urine test. Source: Sacks DB et al., "Guidelines and Recommendations for Laboratory Analysis in the Diagnosis and Management of Diabetes Mellitus," Clinical Chemistry 2023 ("Ketone Testing").

Answer: D. Thin films keep red cells in a single layer, so you can see parasite and red cell morphology to identify the species and count infected cells. Thick films (A) examine more blood and are more sensitive for detection, but they're harder to read. Microscopy doesn't measure drug susceptibility (C). One negative set can't rule out malaria (B): CDC advises repeating smears every 12–24 hours for a total of three sets. Source: CDC, Malaria: "Evaluation and Diagnosis" (clinical guidance).

Answer: C. Under OSHA's bloodborne pathogens standard, contaminated needles aren't recapped, bent, or removed except in narrow situations, and even then only with a mechanical device or one-handed technique. Two-handed recapping (A) is the classic needlestick setup. Breaking needles (B) is prohibited outright. Sharps go into a closable, puncture-resistant, leakproof, labeled sharps container as soon as feasible, never the regular trash (D). Source: OSHA, 29 CFR 1910.1030, paragraphs (d)(2)(vii) and (d)(4)(iii)(A).

Answer: A. Group B. Forward typing finds the B antigen on the cells; reverse typing finds anti-A in the plasma. The two halves agree, which is what makes the type valid. Group A (B) is the mirror image. Group AB (C) would react with both reagents and have no ABO antibodies. Group O (D) would react with neither reagent and have both antibodies. If forward and reverse results ever disagree, that's a discrepancy to resolve before reporting. Source: Canadian Blood Services, Clinical Guide to Transfusion, ch. 8 "Pre-transfusion testing" ("ABO typing," Table 1).

Answer: C. A1c reflects average glucose over roughly the past 3 months because it's tied to the life span of red cells. Anything that changes that life span or replaces the red cells, such as recent blood loss or a transfusion, can throw the result off. A1c doesn't require fasting and can be drawn any time of day, so A, B, and D don't affect it. Source: NIDDK, "A1C Test".