Free ASCP MB Practice Test: 60 Questions With Answers

Answer: B. 22%

In double-stranded DNA, A pairs with T and G pairs with C, so A = T and G = C (Chargaff's rule). If A is 28%, T is also 28%, which accounts for 56%. The remaining 44% is split equally between G and C, so G = 22%.

• A (28%) is the adenine value, not guanine.
• C (44%) is G + C combined.
• D (56%) is A + T combined.

Source: Core base-pairing principle; see Buckingham, Molecular Diagnostics (ASCP MB reading list).

Answer: C. The 2′-hydroxyl group on ribose

Ribose carries a hydroxyl at the 2′ carbon; deoxyribose does not. Under alkaline conditions that 2′-OH can attack the neighboring phosphodiester bond and cleave the backbone. This is one reason RNA specimens need faster handling and more careful storage than DNA.

• A. Uracil vs. thymine changes base pairing and repair, not backbone lability.
• B. Single-strandedness exposes RNA to ribonucleases but is not the chemical basis of alkaline hydrolysis.
• D. DNA uses the same 3′→5′ phosphodiester linkage, so this can't explain the difference.

Source: Core nucleic acid chemistry; see Alberts, Molecular Biology of the Cell, and Buckingham (ASCP MB reading list).

Answer: D. Adenine and guanine

Purines have a fused double-ring structure: adenine and guanine. Pyrimidines have a single ring: cytosine, thymine (DNA), and uracil (RNA).

• A and C are pyrimidine pairs.
• B is a complementary base pair (one purine, one pyrimidine), not two purines.

Source: Core nucleic acid chemistry; see Buckingham (ASCP MB reading list).

Answer: A. DNA ligase

DNA ligase forms the final phosphodiester bond between adjacent fragments, joining the 3′-OH of one to the 5′-phosphate of the next.

• B. Primase lays down the short RNA primers; it doesn't join fragments.
• C. Helicase unwinds the double helix ahead of the fork.
• D. Topoisomerase relieves supercoiling tension; it doesn't seal Okazaki fragment nicks.

Source: Core replication biology; see Alberts (ASCP MB reading list).

Answer: B. Disruption of the splice donor site

Most introns begin with GT (GU in pre-mRNA) and end with AG. The 5′ GT is part of the splice donor consensus. Altering it usually prevents normal recognition of the exon–intron boundary, which can cause the exon to be skipped or the intron to be retained.

• A. Introns are removed, but the splice signals themselves sit at the intron ends; changing them changes splicing.
• C. The variant doesn't alter an exonic codon directly; any downstream stop would be a consequence of abnormal splicing, not a direct effect.
• D. Promoters are upstream regulatory regions, not intron splice sites.

Source: Core RNA processing; see Alberts and Buckingham (ASCP MB reading list).

Answer: D. c.100_101del

A frameshift happens when the number of inserted or deleted coding nucleotides is not a multiple of three. c.100_101del removes two nucleotides, shifting the reading frame for every codon after it, usually producing a premature stop.

• A. A single-nucleotide substitution keeps the frame (it may be silent, missense, or nonsense).
• B. Deleting three nucleotides removes one codon but keeps the frame (in-frame deletion).
• C. Inserting three nucleotides adds one codon but keeps the frame (in-frame insertion).

Source: HGVS Nomenclature, sequence variant descriptions

Answer: C. Nonsense

TAG on the coding strand corresponds to UAG in mRNA, a stop codon. A substitution that converts an amino acid codon to a stop codon is a nonsense variant.

• A. Silent variants change the codon but not the amino acid.
• B. Missense variants change the amino acid to a different amino acid, not to a stop.
• D. A single-base substitution doesn't shift the reading frame.

Source: Standard genetic code; see Buckingham (ASCP MB reading list).

Answer: A. RNA-dependent DNA polymerase activity

Reverse transcriptase reads an RNA template and synthesizes DNA, which is exactly what RT-PCR needs before amplification can begin.

• B. A DNA-dependent RNA polymerase does the opposite: it makes RNA from DNA (transcription).
• C. 5′→3′ exonuclease activity degrades nucleotides ahead of a polymerase; it is used in hydrolysis-probe detection, not cDNA synthesis.
• D. Ligase joins DNA ends; it doesn't synthesize DNA from RNA.

Source: Core enzymology; see Buckingham (ASCP MB reading list).

Answer: B. 3′→5′ exonuclease (proofreading) activity

Proofreading polymerases remove a misincorporated nucleotide from the growing 3′ end before continuing, which lowers the error rate. Standard Taq polymerase lacks 3′→5′ exonuclease activity, so high-fidelity enzymes are preferred when sequence accuracy of the product matters.

• A. 5′→3′ exonuclease activity degrades downstream strands or probes; it doesn't correct misincorporations.
• C. Every PCR polymerase must be thermostable; that alone says nothing about fidelity.
• D. RNase H degrades RNA in RNA:DNA hybrids and is relevant to reverse transcription, not DNA replication fidelity.

Source: Core enzymology; see Buckingham (ASCP MB reading list).

Answer: C. 5′-GAATTC-3′

Write the complement and read it 5′→3′. GAATTC pairs with CTTAAG; read 5′→3′ that strand is GAATTC, identical to the original. (This is the EcoRI site.)

• A. GAAGTC reverse-complements to GACTTC.
• B. GATTCC reverse-complements to GGAATC.
• D. GCATTC reverse-complements to GAATGC.

Source: Core restriction enzyme principle; see Buckingham (ASCP MB reading list).

Answer: D. Transmitted by the mother; affected fathers do not transmit it

Mitochondria, and their DNA, are inherited through the egg, so mtDNA variants pass from mothers to their children. Severity can vary among siblings because cells may carry a mix of normal and variant mtDNA (heteroplasmy).

• A describes X-linked dominant transmission from an affected father.
• B describes autosomal dominant inheritance.
• C describes autosomal recessive inheritance.

Source: Core genetics; see Buckingham and Thompson & Thompson (ASCP MB reading list).

Answer: A. *mecA

mecA encodes an altered penicillin-binding protein (PBP2a) with low affinity for beta-lactams, the classic mechanism of methicillin resistance in S. aureus. (Some assays also detect the related mecC gene.)

• B. vanA confers vancomycin resistance, classically in enterococci.
• C. blaKPC encodes a carbapenemase, typically in Enterobacterales.
• D. tcdB is the Clostridioides difficile toxin B gene.

Source: See Persing, Molecular Microbiology, and Buckingham (ASCP MB reading list).

Answer: B. 58 °C

Wallace rule: Tm ≈ 2 °C × (A + T) + 4 °C × (G + C) = 2(11) + 4(9) = 22 + 36 = 58 °C. The Wallace rule is an empirical quick estimate for short oligonucleotides; primer-design software generally uses more detailed nearest-neighbor calculations.

• A, C, and D come from miscounting or swapping the 2 and 4 multipliers (e.g., 4 × 11 + 2 × 9 = 62).

Source: Stephenson, Calculations for Molecular Biology and Biotechnology (ASCP MB reading list).

Answer: C. 250 µg/mL

For dsDNA, an A260 of 1.0 ≈ 50 µg/mL. Concentration = 0.250 × 50 µg/mL × 20 (dilution factor) = 250 µg/mL.

• A. 12.5 µg/mL is the diluted concentration; the dilution factor was left out.
• B. 200 µg/mL uses the RNA factor (40 µg/mL).
• D. 165 µg/mL uses the single-stranded DNA factor (33 µg/mL).

Source: Stephenson, Calculations for Molecular Biology and Biotechnology; Buckingham (ASCP MB reading list).

Answer: A. Protein or phenol contamination

Relatively pure DNA typically gives an A260/A280 near 1.8. Protein contamination can lower the ratio because aromatic amino acids absorb near 280 nm; residual phenol can also depress the ratio. A value of 1.5 therefore points to contamination rather than pure DNA, but the ratio alone does not identify a single contaminant with certainty.

• B. RNA contamination tends to move the ratio toward the value expected for pure RNA (~2.0).
• C. 1.5 is below the usual value for relatively pure DNA.
• D. Concentration affects absolute absorbance, not the purity ratio itself when readings remain in the instrument's linear range.

Source: Buckingham, Molecular Diagnostics (ASCP MB reading list).

Answer: D. Application of a low-ionic-strength buffer or water

Nucleic acid binds silica in the presence of high concentrations of chaotropic salt, stays bound through ethanol washes that remove salts and contaminants, and is eluted when a low-salt buffer or water rehydrates it off the membrane.

• A. Chaotropic salt promotes binding, not release.
• B. Proteinase K digests proteins during lysis, before binding.
• C. The ethanol wash removes contaminants while the nucleic acid stays bound.

Source: Buckingham, Molecular Diagnostics (ASCP MB reading list).

Answer: B. Comparing protein spectral profiles with a reference database

MALDI-TOF ionizes organism proteins (largely abundant ribosomal proteins) and separates them by mass-to-charge ratio based on time of flight. The resulting spectral fingerprint is matched to a library.

• A. 16S rRNA sequencing is a nucleic acid method.
• C. Target amplification is PCR or another NAAT, not MALDI-TOF.
• D. Biochemical panels are conventional phenotypic identification.

Source: Persing, Molecular Microbiology (ASCP MB reading list).

Answer: A. Capillary electrophoresis with a denaturing polymer

Capillary electrophoresis of fluorescently labeled fragments achieves single-base resolution in the size range used for STR and fragment analysis, with precise sizing against an internal size standard.

• B and D. Standard agarose gels can't reliably resolve 1-bp differences in this size range.
• C. Pulsed-field electrophoresis is designed for very large DNA molecules, such as in bacterial strain typing.

Source: Buckingham, Molecular Diagnostics (ASCP MB reading list).

Answer: C. Raise the temperature and lower the salt concentration

Higher temperature and lower salt destabilize imperfectly matched duplexes more than perfectly matched ones, so only the specific probe–target hybrids survive.

• A. That combination lowers stringency and would add more nonspecific bands.
• B. More probe doesn't improve specificity and can raise background.
• D. Shorter washes remove less nonspecifically bound probe.

Source: Buckingham, Molecular Diagnostics (ASCP MB reading list).

Answer: D. Cleaved by 5′→3′ exonuclease activity during extension

In a hydrolysis assay, the polymerase degrades the bound probe as it extends the primer. Separating reporter from quencher produces signal that accumulates each cycle.

• A describes a molecular beacon.
• B describes FRET (dual hybridization) probes.
• C describes a Scorpions primer-probe.

Source: ASCP MB content guideline lists TaqMan, FRET, beacon, and Scorpions probe structures; principles in Buckingham (ASCP MB reading list).

Answer: B. Primer-dimer formation

SYBR Green fluoresces with any double-stranded DNA, including short primer-dimers. Primer-dimers are short and melt at a noticeably lower temperature than the specific product, which is why melt-curve analysis is used with intercalating dyes.

• A. Target contamination would produce a product with the same 85 °C melting peak.
• C. A SYBR Green assay doesn't use a hydrolysis probe.
• D. Inhibition suppresses amplification; it doesn't create a new product in the NTC.

Source: Buckingham, Molecular Diagnostics (ASCP MB reading list).

Answer: A. About 10 times more target

At 100% efficiency the target doubles each cycle. A 3.3-cycle difference is 2^3.3 ≈ 10-fold. The lower Ct belongs to the sample with more starting template, because it crosses threshold sooner.

• B reverses the relationship between Ct and quantity.
• C confuses the number of cycles with the fold change.
• D would correspond to a 1-cycle difference.

Source: Buckingham, Molecular Diagnostics; Stephenson (ASCP MB reading list).

Answer: C. About 90%

Efficiency = 10^(−1/slope) − 1 = 10^(1/3.6) − 1 ≈ 1.90 − 1 ≈ 0.90, or about 90%. A slope of −3.32 corresponds to 100% efficiency; a steeper (more negative) slope means lower efficiency.

• A would need a slope near −3.9.
• B corresponds to −3.32.
• D corresponds to a slope near −3.1.

Source: Stephenson, Calculations for Molecular Biology and Biotechnology (ASCP MB reading list).

Answer: D. Increased risk of amplicon contamination

Nested PCR uses a second primer pair inside the first amplicon, which raises sensitivity and specificity. But opening tubes and moving amplified product into a new reaction creates a serious carryover risk.

• A. Nested PCR can follow reverse transcription.
• B. Its main advantage is higher sensitivity.
• C. Detection can be by gel, dye, or probe; a probe isn't inherent to the method.

Source: Buckingham, Molecular Diagnostics (ASCP MB reading list).

Answer: B. Competition or primer interactions

Primer pairs in a multiplex compete for dNTPs, polymerase, and magnesium, and can interact with one another. Efficient targets can outcompete weaker ones. The usual fix is rebalancing primer concentrations or redesigning primers during optimization.

• A. The target amplified well in singleplex with the same specimen.
• C. A denaturation failure would affect all targets, including in singleplex.
• D. Multiplex assays routinely amplify many targets.

Source: Buckingham, Molecular Diagnostics (ASCP MB reading list).

Answer: A. Absolute quantification without a standard curve

Digital PCR splits the reaction into thousands of partitions, amplifies to endpoint, and counts positive vs. negative partitions. Poisson statistics correct for partitions that received more than one copy.

• B. Partitioning is the defining feature of digital PCR.
• C. Ct-based quantification against standards describes real-time qPCR.
• D. Digital PCR measures predefined targets; discovering unknown variants requires sequencing.

Source: ASCP MB content guideline lists digital PCR under PCR variations; principles in Buckingham (ASCP MB reading list).

Answer: C. Transcription-mediated amplification (TMA)

TMA uses reverse transcriptase to make a DNA copy with an RNA polymerase promoter, then the RNA polymerase transcribes many RNA copies, all at one temperature.

• A. LAMP is isothermal but uses a strand-displacing DNA polymerase and produces DNA.
• B. PCR requires thermal cycling and makes DNA.
• D. Hybrid Capture is signal amplification, not target amplification.

Source: ASCP MB content guideline (TMA, NASBA, LAMP, Hybrid Capture); Buckingham (ASCP MB reading list).

Answer: B. It lacks a 3′-hydroxyl group

Polymerases add each new nucleotide to the 3′-OH of the growing strand. A ddNTP is incorporated normally but has no 3′-OH, so nothing can be added after it.

• A. ddNTPs carry the triphosphate needed for incorporation.
• C. ddNTPs pair normally; that's why their positions reveal the sequence.
• D. The polymerase remains active; only that strand stops.

Source: Buckingham, Molecular Diagnostics (ASCP MB reading list).

Answer: A. Heterozygous A/G single-nucleotide variant

A heterozygous SNV produces two overlapping peaks at a single position, confirmed in both directions, with clean sequence on either side.

• B. A homozygous variant shows a single, full-height peak of the new base.
• C. Clean, reproducible double peaks in both directions argue against a random artifact.
• D. A heterozygous insertion or deletion causes overlapping, out-of-phase sequence from that point onward, not a single mixed position.

Source: Buckingham, Molecular Diagnostics (ASCP MB reading list).

Answer: C. 1 in 1,000

Q = −10 × log10(P error). For Q30, P = 10^−3, or 1 in 1,000 (99.9% base-call accuracy).

• A misreads the score as a direct ratio.
• B is Q20.
• D is Q40.

Source: ASCP MB content guideline lists quality score under bioinformatics; Kulkarni & Roy, Clinical Genomics (ASCP MB reading list).

Answer: D. VCF

The Variant Call Format lists variant positions, reference and alternate alleles, genotype, and quality metrics.

• A. FASTQ holds raw reads with per-base quality scores.
• B. BAM holds reads aligned to a reference (binary form of SAM).
• C. BED lists genomic intervals, such as the regions targeted by a panel.

Source: Kulkarni & Roy, Clinical Genomics (ASCP MB reading list).

Answer: B. It was unmethylated

Bisulfite deaminates unmethylated cytosine to uracil, which PCR copies as thymine. 5-methylcytosine resists conversion and still reads as C. Comparing C vs. T at CpG sites reveals methylation status.

• A. A methylated cytosine would remain C.
• C. Bisulfite analysis is designed to read methylation, and a converted CpG is the expected signature of an unmethylated cytosine. A true C>T sequence variant is possible in principle, but it would have to be shown in untreated DNA; it isn't the best interpretation here.
• D. Conversion of unmethylated C to T is the expected result of successful treatment.

Source: ASCP MB content guideline (bisulfite conversion, epigenetic modification detection); Buckingham (ASCP MB reading list).

Answer: A. Separated areas with one-way workflow

Prevention starts with design: reagent preparation and specimen processing happen in clean areas, amplification and analysis in separate post-amplification areas, and people, equipment, and materials don't move backward from post- to pre-amplification spaces.

• B. NTCs detect contamination; they don't prevent it.
• C. Each area should have dedicated equipment; filter tips alone don't stop equipment-borne carryover.
• D. Waiting for a contamination event is reactive, not preventive.

Source: ASCP MB content guideline (contamination prevention, monitoring, elimination); Buckingham (ASCP MB reading list).

Answer: C. Uracil-containing carryover amplicons

Because earlier products contain uracil, UNG can excise those uracils and make the old amplicons unamplifiable. Native template contains thymine, not uracil, so it's spared. UNG is then inactivated by heat.

• A. Patient DNA contains T, not U, and isn't targeted.
• B. UNG acts on uracil in DNA, not on RNA as such.
• D. Standard primers don't contain uracil.

Source: Buckingham, Molecular Diagnostics (ASCP MB reading list).

Answer: D. Treat as invalid and retest

The internal control shows whether amplification could have occurred in that specific tube. If it fails, a negative target result can't be trusted. Common causes are inhibitors or extraction failure; the lab's procedure typically calls for re-extraction, dilution, or a new specimen.

• A. Reporting negative risks a false negative.
• B. Nothing indicates the target is present.
• C. The NTC already performed as expected; the problem is in the patient sample.

Source: ASCP MB content guideline (assay controls, troubleshooting); Buckingham (ASCP MB reading list).

Answer: B. Heparin (green top)

Heparin can inhibit PCR amplification, including by interfering with DNA polymerase activity. EDTA and ACD have been used as alternatives without the same classic PCR-inhibitory effect. Specimen requirements still depend on the performing laboratory and assay.

• A. EDTA is commonly used for molecular blood testing.
• C. ACD is also used for some molecular tests.
• D. The anticoagulants are not equivalent for PCR workflows.

Source: Holodniy et al., Journal of Clinical Microbiology, “Inhibition of human immunodeficiency virus gene amplification by heparin” (heparin inhibited PCR; EDTA and ACD lacked the inhibitory activity in this study).

Answer: A. Formalin-induced deamination artifacts

Formalin fixation fragments and chemically damages DNA; cytosine deamination appears as low-level C>T (G>A) changes. These artifacts are a known reason FFPE variant calls need quality thresholds and, sometimes, confirmation.

• B. True subclonal variants would generally be present in matched fresh tissue as well.
• C. No bisulfite step is involved in routine FFPE processing.
• D. Alignment problems don't typically produce a systematic C>T signature.

Source: Buckingham, Molecular Diagnostics; Kulkarni & Roy, Clinical Genomics (ASCP MB reading list).

Answer: C. −70 °C or colder in single-use aliquots

RNA is chemically labile and vulnerable to ubiquitous RNases. Ultra-low storage and aliquoting to avoid freeze–thaw cycles best preserve it unless the lab's validated procedure specifies otherwise.

• A and B. Refrigerator and room temperature storage allow degradation over two weeks.
• D. Repeated freeze–thaw cycles degrade RNA.

Source: ASCP MB content guideline (specimen storage); Buckingham (ASCP MB reading list).

Answer: D. Test in-house like patient specimens; never refer it

CLIA requires PT samples to be tested in the same manner as patient specimens and prohibits sending PT samples to another laboratory for any analysis the lab is certified to perform. The motive doesn't matter; PT referral can lead to serious sanctions.

• A and B. Both are prohibited PT referral.
• C. PT samples must be tested the same number of times as patient samples are routinely tested.

Source: 42 CFR 493.801(b) and (b)(4), as applied in HHS Departmental Appeals Board Decision No. 1870

Answer: B. Semiannually in year one, then annually

CLIA requires competency evaluation at least twice during the first year an individual tests patient specimens and at least annually after that, covering the six required procedures (such as direct observation of testing and assessment of problem-solving skills).

• A. Once in the first year is insufficient.
• C. Monthly assessment may be a lab's choice but isn't the regulatory minimum.
• D. Competency assessment is routine, not just a response to PT failure.

Source: CMS CLIA brochure: What Do I Need to Do to Assess Personnel Competency?; 42 CFR 493.1451(b)(8)–(9)

Answer: A. Research use only (RUO)

That labeling statement is the regulatory hallmark of RUO products under FDA's IVD labeling rule.

• B. ASRs are building-block reagents (such as specific primers or probes) sold for use in laboratory-developed tests and carry their own labeling requirements.
• C. Cleared or approved IVDs carry full diagnostic labeling with an intended use.
• D. An LDP is a test designed and validated within a laboratory, not a product label category.

Source: 21 CFR 809.10(c)(2)(i), described in FDA, Bioresearch Monitoring Inspections: In Vitro Diagnostics Devices

Answer: C. Freshly prepared dilute bleach, then a rinse

Hypochlorite oxidatively damages nucleic acids so they can no longer be amplified. A follow-up rinse (water or ethanol) removes corrosive residue.

• A. Ethanol disinfects but doesn't reliably destroy DNA.
• B. Water only spreads contamination.
• D. Quaternary ammonium products are disinfectants, not reliable nucleic acid destroyers.

Source: ASCP MB content guideline (contamination elimination; safety); Buckingham (ASCP MB reading list).

Answer: B. 2.0 log10 decrease

log10(50,000) ≈ 4.7 and log10(500) ≈ 2.7. The difference is 2.0 log10, a 100-fold decrease. Viral load changes are compared on the log scale because quantitative assays have meaningful variability at low differences.

• A would be a 10-fold drop (to 5,000 copies/mL).
• C is the log of the final value, not the change.
• D is the log of the starting value.

Source: Persing, Molecular Microbiology; Stephenson (ASCP MB reading list).

Answer: A. A toxin gene from toxigenic C. difficile

Molecular tests detect nucleic-acid targets associated with a toxin-producing C. difficile organism. A positive NAAT shows that a toxigenic organism is present; it does not by itself prove that toxin is being produced or that the patient's symptoms are caused by C. difficile.

• B. Glutamate dehydrogenase is detected by an antigen assay, not NAAT.
• C. mecA is associated with methicillin resistance in staphylococci.
• D. vanA is associated with vancomycin resistance in enterococci and other organisms.

Source: CDC, Clinical Testing and Diagnosis for C. diff Infection (molecular assays detect the presence of a toxin-producing C. difficile organism).

Answer: D. A carbapenemase gene

KPC (Klebsiella pneumoniae carbapenemase) hydrolyzes carbapenems and many other beta-lactams. Its detection is important for treatment and infection-control decisions.

• A. Methicillin resistance in staphylococci is mecA/mecC.
• B. Vancomycin resistance is typically vanA/vanB.
• C. Fluoroquinolone resistance commonly involves gyrA/parC mutations, not blaKPC.

Source: Persing, Molecular Microbiology (ASCP MB reading list).

Answer: C. BCR::*ABL1

CML is defined by t(9;22)(q34;q11), the Philadelphia chromosome, which creates BCR::ABL1. The p210 transcript is typical of CML. Quantitative monitoring tracks response to tyrosine kinase inhibitors.

• A. PML::RARA, t(15;17), defines acute promyelocytic leukemia.
• B. IGH::BCL2, t(14;18), is characteristic of follicular lymphoma.
• D. RUNX1::RUNX1T1, t(8;21), occurs in a subset of AML.

Source: ASCP MB content guideline (leukemias/lymphomas, e.g., CML, translocations); Coleman & Tsongalis, Diagnostic Molecular Pathology (ASCP MB reading list).

Answer: B. Follicular lymphoma

t(14;18) juxtaposes BCL2 with the immunoglobulin heavy-chain locus, causing overexpression of the anti-apoptotic BCL2 protein.

• A. Burkitt lymphoma is characterized by MYC rearrangement, most often t(8;14).
• C. Mantle cell lymphoma is characterized by t(11;14), CCND1::IGH.
• D. APL is characterized by t(15;17), PML::RARA.

Source: Coleman & Tsongalis, Diagnostic Molecular Pathology (ASCP MB reading list).

Answer: A. A clonal B-cell population

Each B cell has a unique rearrangement of a characteristic length. A reactive (polyclonal) population gives a bell-shaped spread of peaks; a clonal expansion produces one or two dominant peaks. Results are interpreted with morphology and other findings, because clonality alone does not equal malignancy.

• B. A polyclonal population produces a Gaussian-like distribution without a dominant peak.
• C. IGH assays assess B cells; T-cell clonality uses T-cell receptor gene targets.
• D. A clear dominant peak with background is an interpretable result.

Source: ASCP MB content guideline (clonal rearrangements); Coleman & Tsongalis, Diagnostic Molecular Pathology (ASCP MB reading list).

Answer: D. Lack of benefit from anti-EGFR antibodies

Activating RAS mutations act downstream of EGFR, so blocking EGFR doesn't shut down the signaling pathway. RAS testing is used to select patients for anti-EGFR therapy.

• A. Lynch syndrome is a germline mismatch repair disorder; somatic KRAS mutation does not indicate it.
• B. HER2 status, not KRAS, guides anti-HER2 therapy.
• C. Tumor KRAS mutations are somatic and aren't inherited.

Source: Coleman & Tsongalis, Diagnostic Molecular Pathology; DeVita Primer of the Molecular Biology of Cancer (ASCP MB reading list).

Answer: C. MLH1 promoter hypermethylation analysis

Many sporadic MSI-high colorectal cancers result from acquired MLH1 promoter hypermethylation, which silences MLH1 (and secondarily PMS2). Finding hypermethylation (often alongside BRAF V600E) favors a sporadic tumor; absence keeps germline testing for Lynch syndrome on the table.

• A. KRAS status guides therapy, not Lynch triage.
• B. BRCA1 is associated with hereditary breast and ovarian cancer.
• D. HER2 FISH isn't part of mismatch repair evaluation.

Source: ASCP MB content guideline (hereditary cancer syndromes, e.g., colon); Coleman & Tsongalis, Diagnostic Molecular Pathology (ASCP MB reading list).

Answer: B. Hemoglobin H disease

With only one functional alpha gene, excess beta chains form tetramers (HbH), causing a moderate hemolytic anemia.

• A. A silent carrier has one gene deleted.
• C. Alpha-thalassemia trait (two genes deleted) usually causes mild microcytosis, and it isn't a three-gene deletion.
• D. Hb Bart's hydrops fetalis results from deletion of all four genes.

Source: ASCP MB content guideline (hemoglobinopathies, e.g., thalassemias); Thompson & Thompson Genetics and Genomics in Medicine (ASCP MB reading list).

Answer: A. F5 c.1601G>A (p.Arg534Gln), legacy R506Q

The variant removes an activated protein C cleavage site, so factor Va is inactivated more slowly, raising venous thrombosis risk. Labs often report both the current HGVS name and the legacy name because older literature uses R506Q.

• B describes the prothrombin variant (legacy 20210G>A), a separate thrombophilia.
• C describes a factor V deficiency, which causes bleeding rather than thrombosis.
• D. Factor V Leiden is a single-nucleotide substitution, not a repeat expansion.

Source: Mayo Clinic Laboratories, Factor V Leiden (R506Q) Mutation, Blood (F5 c.1601G>A, p.Arg534Gln; legacy R506Q)

Answer: D. 50%

Each parent passes on either the variant or the typical allele, each with a 1-in-2 chance. Across one pregnancy the outcomes are: 1 in 4 affected (two variant copies), 1 in 2 unaffected carriers (one of each), and 1 in 4 with two typical copies.

• A. 25% is the chance a child is affected, and also the chance of two typical copies.
• B. 2/3 (about 67%) is the carrier chance for a sibling already known to be unaffected. That is a different question, because knowing the child is unaffected removes the affected outcome from the denominator.
• C. 75% is the chance a child inherits at least one variant copy (affected or carrier).

Source: NHLBI, Cystic Fibrosis: Causes (inheritance when both parents carry one variant gene); ASCP MB content guideline (single gene disorders, e.g., cystic fibrosis)

Answer: D. Premutation

GeneReviews categories: normal about 5–44 repeats, intermediate about 45–54, premutation about 55–200, and full mutation more than 200 (usually with abnormal methylation). A premutation carrier is at risk for FXPOI and FXTAS, and her premutation can expand to a full mutation when transmitted to her children.

• A, B, and C don't match 120 repeats.

Source: GeneReviews, FMR1 Disorders, "Allele Size" (last revision May 16, 2024)

Answer: C. Reduced-penetrance allele

GeneReviews categories: normal 26 or fewer repeats, intermediate 27–35, reduced penetrance 36–39, and full penetrance 40 or more. A person with 38 repeats is at risk for HD but may not develop symptoms.

• A and B are smaller ranges.
• D begins at 40 repeats.

Source: GeneReviews, Huntington Disease, "CAG repeat sizes" (last update February 12, 2026)

Answer: A. Prader-Willi syndrome

Prader-Willi syndrome results from absent expression of paternally derived imprinted genes in 15q11.2-q13, whether by paternal deletion, maternal uniparental disomy, or an imprinting defect. Methylation analysis detects all three because each leaves only the maternal pattern.

• B. Angelman syndrome shows the opposite: only the paternal pattern (loss of maternal contribution).
• C. A normal result shows both maternal and paternal patterns.
• D. Fragile X involves FMR1 on the X chromosome.

Source: GeneReviews, Prader-Willi Syndrome, Diagnosis/testing and Establishing the Diagnosis (maternal-only imprinting at 15q11.2-q13); ASCP MB content guideline (epigenetic disorders).

Answer: B. 15

The child's 12 can come from the mother, so the other allele, 15, must come from the biological father. The alleged father carries 15, so he isn't excluded at this locus. (Conclusions in real cases depend on many loci and statistical calculations.)

• A. 12 is explained by the mother.
• C. 14 isn't present in the child.
• D. 13 isn't present in the child.

Source: Derived Mendelian/STR interpretation under the stated one-locus parentage scenario; ASCP MB content guideline establishes genetic identity/parentage as in scope. Real parentage conclusions require multilocus statistical analysis.

Answer: D. 75%

At a locus with no shared alleles, % donor = donor area ÷ (donor area + recipient area) = 6,000 ÷ 8,000 = 75%. Labs average results across several informative loci and use formulas adapted to shared-allele patterns.

• A. 25% is the recipient fraction.
• B. 33% comes from dividing recipient area by donor area (2,000 ÷ 6,000), which isn't a fraction of the total.
• C. 80% doesn't follow from these values.

Source: Derived calculation from the stated no-shared-alleles STR peak-area assumptions; ASCP MB content guideline establishes engraftment as in scope. Clinical chimerism laboratories use validated multilocus methods and assay-specific calculations.

Answer: C. Poor metabolizer; avoid clopidogrel if possible

*2 is a no-function allele, so *2/*2 predicts a poor metabolizer. Clopidogrel is a prodrug activated partly by CYP2C19, so poor metabolizers form less active drug. CPIC's 2022 guideline recommends avoiding clopidogrel if possible and using prasugrel or ticagrelor at standard dose when there's no contraindication for cardiovascular indications. The laboratory reports genotype and phenotype; prescribing decisions belong to the clinical team.

• A. *2/*2 is not ultrarapid (that's associated with *17/*17).
• B. Two no-function alleles aren't a normal genotype.
• D. *2/*2 is poor, not intermediate, and dose doubling isn't the CPIC recommendation.

Source: NCBI Medical Genetics Summaries, Clopidogrel Therapy and CYP2C19 Genotype, Table 2 (CPIC 2022 update)

Answer: B. Increased risk of severe cutaneous reactions, including SJS/TEN

FDA's pharmacogenetic association table links HLA-B15:02 positivity with a higher risk of severe skin reactions from carbamazepine and advises avoiding carbamazepine unless the potential benefits outweigh the risks. HLA-B15:02 is also associated with increased risk of Stevens-Johnson syndrome/toxic epidermal necrolysis (SJS/TEN).

• A. HLA type does not primarily predict carbamazepine absorption.
• C. Carbamazepine has hematologic risks, but HLA-B*15:02 is used for severe cutaneous-reaction risk.
• D. HLA variation can affect drug-hypersensitivity risk as well as transplant matching.

Source: FDA, Table of Pharmacogenetic Associations, carbamazepine / HLA-B*15:02 row (higher risk of severe skin reactions; avoid unless benefits outweigh risks).