AMT MT Practice Questions (2026) — Free Sample Questions with Answers

15 free AMT MT questions from our 150-question bank, grouped by content domain. Each answer explains why the right option is right — and why every wrong option is wrong.

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General Laboratory

Question 1

A serum specimen from an asymptomatic outpatient shows potassium 7.8 mmol/L, calcium 4.1 mg/dL, magnesium 0.9 mg/dL and an ALP below the reference interval. The serum is not hemolyzed. Which is the MOST likely cause?

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Answer: C

In vitro hemolysis during collection: Hemolysis raises K, but it is ruled out by the clear serum and it would not lower Ca, Mg and ALP.

Prolonged tourniquet time with fist pumping: Fist pumping raises K but does not cause a marked fall in calcium, magnesium and ALP.

K2EDTA contamination of the specimen: Correct. Potassium EDTA adds K and chelates Ca and Mg (and the Mg and Zn that ALP needs), giving high K with low Ca, Mg and ALP; drawing the lavender tube before the serum tube is a classic cause.

Delayed separation of a refrigerated whole-blood tube: Cold storage leaks K from cells but does not explain the low calcium, magnesium and ALP.

High K + low Ca, Mg and ALP in a non-hemolyzed serum = EDTA contamination; recollect in the correct order of draw.

Question 2

A patient has orders for blood cultures, PT/INR, a comprehensive metabolic panel in a gold gel tube, a CBC, and a glucose in a gray-top tube, all from one venipuncture. Which tube is drawn THIRD?

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Answer: B

Light-blue sodium citrate tube: The citrate tube is second, right after the blood cultures.

Gold gel separator tube: Correct. CLSI PRE02 order: blood cultures, then light-blue citrate, then serum tubes (gold), then green heparin, lavender EDTA and gray fluoride-oxalate.

Lavender EDTA tube: EDTA follows the serum and heparin tubes so its additive cannot carry over into them.

Gray sodium fluoride/potassium oxalate tube: The glycolytic-inhibitor tube is drawn last.

Order of draw (CLSI PRE02): sterile/blood culture → citrate → serum → heparin → EDTA → fluoride/oxalate.

Question 3

The Levey-Jennings chart plots two levels of glucose control as an SD index (SDI: 0 = the mean, +2 = 2 SD above the mean). Both levels are assayed once in every analytic run. Which interpretation of run 10 is correct?

Glucose QC, SD index by run-3-2-1012312345678910RunSDI
Level 1 Level 2
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Answer: A

Reject the run: 2-2s across levels, a systematic error: Correct. Two control observations in the same run that both exceed +2 SD on the same side violate the 2-2s rule (across materials), which points to systematic error such as a calibration or reagent problem.

Accept the run: each level is only a 1-2s warning: Treats each control on its own; the 1-2s warning triggers the other rules, and two levels beyond +2 SD together break 2-2s.

Reject the run: R-4s across levels, a random error: R-4s needs one control above +2 SD and another below −2 SD; here both are on the positive side.

Reject the run: 1-3s on level 1, a random error: Neither value exceeds 3 SD (level 1 is +2.4 SD), so 1-3s is not violated.

Apply 1-2s as a warning, then check 1-3s, 2-2s and R-4s; two results beyond the same 2 SD limit (within a run or in consecutive runs) signal systematic error.

Chemistry

Question 4

A grossly lipemic specimen (triglycerides 2,500 mg/dL) has a sodium of 126 mmol/L on the chemistry analyzer, which uses an indirect ISE. A whole-blood sodium by direct ISE on the blood gas analyzer is 139 mmol/L, and the measured osmolality is normal. What is the MOST likely explanation?

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Answer: B

Translocational hyponatremia from an infused osmole such as mannitol: An infused osmole such as mannitol draws water out of cells and lowers sodium by both indirect and direct ISE, and it raises the measured osmolality (an osmolal gap).

Electrolyte exclusion by lipid in the diluted indirect ISE sample: Correct. Lipid occupies plasma volume; indirect ISE dilutes a fixed volume that contains less water, so sodium is falsely low (pseudohyponatremia), while direct ISE measures activity in plasma water.

Dilutional hyponatremia from inappropriate ADH secretion: True hyponatremia would be low by direct ISE and would lower the measured osmolality.

Sodium loss through the kidneys during diuretic therapy: Renal sodium loss is a true hyponatremia and would not differ between the two methods.

Pseudohyponatremia (lipemia, very high protein) affects indirect ISE and flame methods, not direct ISE; the measured osmolality stays normal.

Question 5

An asymptomatic adult has a fasting plasma glucose of 128 mg/dL and an A1C of 6.7% measured on the same blood sample. According to the ADA Standards of Care, which conclusion is correct?

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Answer: B

Diabetes is suspected, but a repeat FPG on another day is required first: A repeat on another day is one way to confirm, but two different abnormal tests on the same sample already confirm the diagnosis.

Diabetes is diagnosed, because both tests on one sample meet the criteria: Correct. Without unequivocal hyperglycemia, diagnosis needs two abnormal results; these may be two different tests from the same sample (FPG ≥ 126 mg/dL and A1C ≥ 6.5%).

Prediabetes is present, because the FPG is below 140 mg/dL: An FPG of 126 mg/dL or more meets the diabetes criterion; 140 mg/dL is an outdated cutoff.

No diagnosis is possible until a 2-hour 75-g OGTT is performed: The OGTT is one option, not a requirement when FPG and A1C are both diagnostic.

ADA: FPG ≥ 126 mg/dL, 2-h OGTT ≥ 200, A1C ≥ 6.5%, or random ≥ 200 with symptoms; without symptoms, confirm with a second abnormal result (same or separate sample).

Question 6

An adult with jaundice and pale stools has the results shown. Which condition do they MOST likely indicate?

Results
TestResultReference interval
Total bilirubin8.0 mg/dL0.3–1.2 mg/dL
Direct (conjugated) bilirubin6.5 mg/dL0.0–0.3 mg/dL
ALP610 U/L40–130 U/L
ALT95 U/L7–55 U/L
Urine bilirubinPositiveNegative
Urine urobilinogenDecreased0.2–1.0 mg/dL
Show answer & explanation

Answer: A

Posthepatic biliary obstruction: Correct. Conjugated hyperbilirubinemia with bilirubinuria, a marked ALP rise, and decreased urobilinogen with pale stools shows that bilirubin is not reaching the intestine.

Prehepatic (hemolytic) jaundice: Hemolysis raises unconjugated bilirubin, with no bilirubinuria and increased urobilinogen.

Gilbert syndrome: Gilbert syndrome causes a mild unconjugated rise with normal ALP and urine findings.

Acute viral hepatitis: Hepatitis also causes conjugated bilirubinemia, but with very high ALT, a smaller ALP rise and normal or increased urobilinogen.

Obstruction: conjugated bilirubin and ALP up, urine bilirubin positive, urobilinogen down, pale stools. Hemolysis: unconjugated up, urobilinogen up, urine bilirubin negative.

Hematology

Question 7

An adult has a hemoglobin of 8.0 g/dL, a platelet count of 18 × 10³/µL, many fragmented red cells (helmet cells and small triangular fragments) on the smear and an increased LD. The PT, APTT and fibrinogen are normal. Which condition is MOST likely?

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Answer: A

Thrombotic thrombocytopenic purpura: Correct. Microangiopathic hemolysis (schistocytes, high LD) with severe thrombocytopenia but normal PT, APTT and fibrinogen points to a thrombotic microangiopathy such as TTP.

Disseminated intravascular coagulation: DIC also causes schistocytes and thrombocytopenia, but consumes factors, prolonging PT/APTT and lowering fibrinogen.

Warm autoimmune hemolytic anemia: Warm AIHA shows spherocytes and a positive DAT, not fragmented cells with severe thrombocytopenia.

Hereditary spherocytosis: HS shows spherocytes with a normal platelet count.

Schistocytes + low platelets: normal coagulation tests suggest TTP/HUS; prolonged PT/APTT with low fibrinogen suggests DIC.

Question 8

A healthy adult has the CBC results shown and normal serum ferritin. Which condition do the results MOST likely indicate?

CBC
TestResultReference interval
RBC6.1 × 10⁶/µL4.2–5.4 × 10⁶/µL
Hemoglobin11.0 g/dL12.0–16.0 g/dL
Hematocrit34%37–47%
MCV56 fL80–100 fL
MCHC32.4 g/dL32–36 g/dL
RDW13.5%11.5–14.5%
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Answer: A

Beta-thalassemia minor: Correct. Marked microcytosis with an increased RBC count, a normal RDW and normal ferritin (Mentzer index MCV ÷ RBC = 9, below 13) points to thalassemia trait; HbA2 quantitation would confirm it.

Iron deficiency anemia: Iron deficiency usually gives a low RBC count, a high RDW and a low ferritin.

Anemia of chronic inflammation: Chronic inflammation causes a normocytic or mildly microcytic anemia with a low or normal RBC count, not an MCV of 56 fL.

Sideroblastic anemia: Sideroblastic anemia gives a dimorphic population with a high RDW and increased iron stores.

Microcytosis with a high RBC count and normal RDW suggests thalassemia trait; microcytosis with a low RBC count and high RDW suggests iron deficiency.

Coagulation and Hemostasis

Question 9

A patient with new extensive bruising has a normal PT and the APTT results shown (reference 25–35 seconds). Which finding is MOST likely?

APTT mixing study (1:1 with normal pooled plasma)
SpecimenAPTT (s)
Patient plasma62
Immediate 1:1 mix34
1:1 mix incubated 2 h at 37 °C55
Normal plasma incubated 2 h33
Show answer & explanation

Answer: B

Factor VIII deficiency: A deficiency corrects both immediately and after incubation, because the normal plasma supplies the factor.

A factor VIII inhibitor: Correct. Correction on immediate mixing that is lost after incubation shows a time- and temperature-dependent inhibitor, typical of an acquired factor VIII autoantibody that presents with bleeding.

A lupus anticoagulant: Lupus anticoagulant usually fails to correct even on the immediate mix and is associated with thrombosis, not bruising.

Heparin contamination: Heparin prolongs the immediate mix as well, and the thrombin time would be prolonged.

Mixing study: corrects = deficiency; no immediate correction = immediate-acting inhibitor (lupus anticoagulant, heparin); corrects then prolongs on incubation = time-dependent inhibitor (factor VIII).

Immunohematology

Question 10

A Kleihauer-Betke stain on a postpartum sample from a D-negative mother shows 1.5% fetal red cells. Each 300-µg vial of RhIG covers 30 mL of fetal whole blood. How many vials should be given?

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Answer: C

  1. Fetal bleed = % fetal cells × 50 = 1.5 × 50 = 75 mL fetal whole blood.
  2. Vials = 75 mL ÷ 30 mL per vial = 2.5 vials.
  3. The decimal is 0.5 or more, so round up to 3 vials.
  4. Add one vial: 3 + 1 = 4 vials of 300 µg.

2 vials: Rounds 2.5 down and does not add the extra vial. A decimal of 0.5 or more is rounded up, then one vial is added.

3 vials: Rounds 2.5 correctly to 3 but forgets the extra vial that covers the imprecision of the KB count.

4 vials: Correct. 1.5 × 50 = 75 mL of fetal whole blood. 75 ÷ 30 = 2.5 vials, rounded up to 3, plus 1 = 4 vials.

5 vials: Divides by 15 mL (the fetal red-cell volume one vial covers) while using the whole-blood bleed: 75 ÷ 15 = 5.

KB dose: % fetal cells × 50 = mL fetal whole blood. Divide by 30, round (below 0.5 down, 0.5 or more up) and add one vial.

Question 11

A patient’s antibody screen is positive. The IAT panel results below were obtained, and the auto control is negative. Rule-outs are made on nonreactive cells that are homozygous for the antigen (any nonreactive K+ cell for anti-K). Which antibody is MOST likely present?

Antibody identification panel (+ = antigen present, 0 = absent)
CellDCEceKFyaFybJkaJkbIAT
1++00+++0++0
2+0++00++0+3+
30+0++00++00
400+++0+0++2+
5000++0++0+0
6+0++++0+++2+
7+00++0+0+00
8++0++00+0+0
Auto control: 0
Show answer & explanation

Answer: C

Anti-D: Cells 1, 7 and 8 are D-positive and nonreactive, so anti-D is ruled out.

Anti-Jkb: Cell 2 is Jkb-positive and reactive, but cells 5 and 8 are Jk(a−b+) and nonreactive, so anti-Jkb is ruled out.

Anti-E: Correct. Only the E-positive cells (2, 4 and 6) react, and all five E-negative cells are nonreactive, which meets the rule of three. Every other antigen is ruled out on a nonreactive homozygous cell. Cell 2 (R2R2, EE) reacts more strongly than the Ee cells, which is Rh dosage.

Anti-E and anti-c: Anti-c is ruled out: cell 5 (rr, cc) and cells 3, 7 and 8 are c-positive and nonreactive. A second antibody is not needed to explain the pattern.

Cross out antigens on nonreactive homozygous cells, then match the remaining specificity to the reactive cells and confirm with the rule of three. Stronger reactions with homozygous cells support the identification.

Blood Banking and Transfusion Services

Question 12

A 58-year-old man develops severe respiratory distress 90 minutes after the start of a plasma transfusion. Findings are shown below. Which reaction is MOST likely?

Findings at the time of the reaction
FindingResult
Temperature38.4 °C (before transfusion 37.0 °C)
Blood pressure92/58 mm Hg (before transfusion 124/78)
Jugular venous distensionAbsent
Chest radiographNew bilateral infiltrates
BNPWithin reference interval
Fluid balanceNot positive
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Answer: C

Transfusion-associated circulatory overload: TACO shows hypertension, jugular venous distension, positive fluid balance and a raised BNP. These are all absent here.

Anaphylactic reaction: Anaphylaxis starts within minutes with wheezing, angioedema and hypotension. It does not produce bilateral lung infiltrates with fever.

Transfusion-related acute lung injury: Correct. Hypoxemic distress within 6 hours of a plasma-rich product, with bilateral infiltrates, fever, hypotension and no sign of volume overload (normal BNP, no JVD), fits TRALI. It is often caused by donor anti-HLA or anti-HNA antibodies.

Febrile nonhemolytic transfusion reaction: The fever fits, but FNHTR does not cause respiratory failure or lung infiltrates.

TRALI: noncardiogenic pulmonary edema within 6 hours, often with hypotension and fever. TACO: cardiogenic, with hypertension and raised BNP.

Microbiology

Question 13

A throat culture grows beta-hemolytic colonies of gram-positive cocci in chains that are catalase-negative. The isolate is PYR-positive and does not grow or blacken bile esculin agar. Which organism is MOST likely?

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Answer: B

Streptococcus agalactiae: Group B streptococci are PYR-negative and are identified by a positive CAMP test and hippurate hydrolysis.

Streptococcus pyogenes: Correct. Among beta-hemolytic, catalase-negative cocci in chains, PYR positivity points to group A streptococci. A negative bile esculin result excludes enterococci, which are also PYR-positive.

Enterococcus faecalis: Enterococci are PYR-positive too, but they grow on bile esculin agar, blacken it, and tolerate 6.5% NaCl.

Streptococcus pneumoniae: S. pneumoniae is alpha-hemolytic, PYR-negative, and identified by optochin susceptibility and bile solubility.

PYR positive: group A streptococci and enterococci. Bile esculin separates them, because only enterococci are positive.

Urinalysis and Body Fluids

Question 14

A 30-year-old woman has fever, flank pain and costovertebral-angle tenderness. Her urine sediment shows many WBCs and bacteria, plus cylinders with a protein matrix packed with granulocytes that have lobed nuclei. Which condition is MOST consistent with these findings?

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Answer: A

Acute pyelonephritis: Correct. Casts packed with WBCs form in the renal tubules, so they place the infection in the kidney. With bacteria, fever and flank pain, this is acute pyelonephritis.

Acute cystitis: Cystitis causes WBCs and bacteria, but casts form only in the tubules. WBC casts point to the kidney, not the bladder.

Acute glomerulonephritis: Glomerulonephritis gives RBC casts and dysmorphic RBCs, not casts full of neutrophils with bacteria.

Acute interstitial nephritis: Interstitial nephritis can give WBC casts, but usually with eosinophils after a drug and without bacteria or fever with flank pain.

WBC casts mean tubular (renal) inflammation. With bacteria, they indicate pyelonephritis. RBC casts point to glomerular disease.

Question 15

A urine collected after an imaging procedure earlier in the day gives the results below. The refractometer passes its distilled-water check. What is the MOST likely explanation for the SG discrepancy?

Urinalysis results
TestResult
Color / clarityPale yellow / clear
Reagent strip glucoseNegative
Reagent strip proteinNegative
Reagent strip pH6.0
Reagent strip SG1.015
Refractometer SG1.048
Show answer & explanation

Answer: A

Radiographic contrast media in the urine: Correct. Iodinated contrast media raise the refractive index, and so the refractometer SG, often above 1.040. The reagent strip measures ionic solutes only and is not affected.

Glucose spilling into the urine: Glucose raises the refractometer SG more than the strip SG, but the glucose pad is negative and could not explain a gap this large.

Highly alkaline urine on the reagent strip: Alkaline urine lowers the strip SG (0.005 is added when the pH is 6.5 or higher). The pH here is 6.0.

Refractometer calibration error: The distilled-water check reads correctly, which rules out a calibration error.

A refractometer SG above 1.040 with a normal strip SG points to radiographic contrast media (or another high-molecular-weight solute), which the ionic strip method does not detect.

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