HT (ASCP) Practice Questions (2026) — Free Sample Questions with Answers

Answer: C

A volume of fixative about equal to the tissue volume: Equal volumes let the formaldehyde be depleted by the tissue, leaving the center poorly fixed and prone to autolysis.

A volume of fixative about three times the tissue volume: A 3:1 ratio is still too little; reactive formaldehyde is consumed and diluted by tissue fluids before fixation is complete.

A volume of fixative about 15 to 20 times the tissue volume: Correct. Standard practice is a fixative-to-tissue ratio of at least 10:1, ideally 15–20:1, so the fixative is not exhausted or diluted.

A volume of fixative about 100 times the tissue volume: Such a large excess adds no benefit, wastes reagent and increases formaldehyde exposure and disposal burden.

Use roughly 15–20 volumes of fixative per volume of tissue (never less than about 10:1) and keep specimens thin enough for formalin to penetrate.

Answer: C

Xylene over-hardened the tissue because exposure was prolonged: Over-exposure to xylene makes tissue hard and brittle, not soft; it also would not turn the xylene milky.

Formalin carried over and reacted with the xylene: Formalin was removed by the earlier aqueous and alcohol steps; the milkiness reflects water, not a formalin reaction.

Residual water blocked clearing and paraffin infiltration: Correct. Water left by incomplete dehydration does not mix with xylene, so the xylene clouds and paraffin cannot infiltrate the tissue, leaving soft, mushy areas.

The paraffin was too cool, so it hardened before entering the tissue: Paraffin temperature was not changed; the clouded xylene points to an earlier dehydration failure.

Processing order is dehydrate (to 100% alcohol) → clear (xylene, miscible with both alcohol and paraffin) → infiltrate. Milky xylene means water is still present.

Answer: A

On edge, so the epidermis, dermis and subcutis appear in each section: Correct. Skin is embedded on edge (perpendicular to the cutting face) so that every layer is shown in cross-section.

Flat with the epidermis facing down, so the surface is cut first: Flat embedding yields tangential sections through one layer and loses the layer relationships the pathologist needs.

Flat with the epidermis facing up, so the deep margin is cut first: This still gives tangential sections; the full thickness of the skin would not be seen.

At a 45° angle, so the surface area of each section is increased: Oblique orientation distorts layer thickness and is not standard for skin.

Orient skin, tubular structures and walls of hollow organs so all layers are cut in true cross-section.

Answer: B

Heat-induced epitope retrieval in citrate buffer: Failed retrieval causes weak or absent specific staining; it does not create brown staining in the negative control.

Quenching of endogenous peroxidase activity: Correct. RBCs and granulocytes contain peroxidase activity that converts DAB to a brown product unless quenched (e.g., 3% hydrogen peroxide) before detection.

Hematoxylin counterstaining after the chromogen: Counterstain only adds blue nuclei; leaving it out cannot produce brown cytoplasmic staining.

Blocking of endogenous biotin with avidin–biotin: Biotin blocking matters for avidin–biotin systems; a polymer detection system does not use biotin, so this is not the cause.

Brown cells in an HRP/DAB negative control point to endogenous peroxidase; quench it with hydrogen peroxide before the primary or detection step.

Answer: D

Mucicarmine on a single section: Mucicarmine demonstrates epithelial mucin, not glycogen.

Alcian blue at pH 2.5 on a single section: Alcian blue stains acid mucopolysaccharides; glycogen is a neutral polysaccharide and is negative.

Congo red with polarized light: Congo red with apple-green birefringence demonstrates amyloid, not glycogen.

PAS with and without diastase digestion: Correct. Glycogen is PAS positive and is removed by diastase (amylase), so staining that disappears after digestion confirms glycogen.

Glycogen = PAS positive, diastase sensitive. Neutral mucins stay PAS positive after diastase.

Answer: C

About 475 mL: This multiplies 500 mL by 0.95, using the stock concentration instead of solving C1V1 = C2V2; the result would be about 90% ethanol.

About 350 mL: 350 mL is 70% of 500 mL, which would be correct only if the stock were 100% ethanol.

About 368 mL: Correct. V1 = (C2 × V2) ÷ C1 = (70 × 500) ÷ 95 = 368.4 mL of 95% ethanol, diluted to 500 mL.

About 679 mL: This inverts the ratio (95 × 500 ÷ 70) and exceeds the final volume, which is impossible.

Dilutions: C1V1 = C2V2. Solve V1 = C2V2/C1, then bring to final volume.