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Choosing and reading molecular tests

15 min

Skills you'll build2 skills
  • Choose FISH, karyotype, or a molecular test for the change being sought
  • Calculate an expected variant allele fraction from the abnormal-cell fraction

Try first

Try first

A KMT2A rearrangement is suspected in a marrow specimen. Conventional G-banded chromosome analysis shows a normal karyotype. Does that result exclude the rearrangement?

Keep going to see the reasoning.

Right. Keep going to follow the reasoning.

Keep going to see the reasoning.

Keep going to see the reasoning.

Get the idea

Use the explanation to check your first decision.

Choosing a method

The suspected abnormality and the required analytic sensitivity determine the assay.1

MethodMain strengthMain limit
KaryotypeBroad survey that can reveal an unexpected cloneNeeds dividing cells, at megabase-level resolution
FISHRapid, cell-by-cell analysis of selected lociDetects only abnormalities the probe covers
RT-PCRIdentifies an expressed fusion transcriptDetects only the sequence the primers bracket
Targeted next-generation sequencingMany targets in parallel, small clones detected at depthPanel content and specimen quality limit what is detectable

A karyotype surveys all chromosomes at megabase-level resolution, and a targeted method examines selected loci. A normal karyotype therefore does not exclude an abnormality a targeted probe can detect, and a negative FISH or PCR result covers only the loci or sequence the assay examined.1 A break-apart FISH probe shows that a target gene is rearranged, but it does not identify the partner gene. A dual-fusion probe or a sequence-level method is needed for that.1 DNA-level findings and expressed RNA are complementary. FISH can show a genomic rearrangement without showing whether it is expressed, and a validated reverse-transcription assay shows expression.3 Each assay also needs the right controls. A no-template control checks for reagent contamination, and a positive control confirms that the assay detects its target.1

Interpreting a variant allele fraction

Variant allele fraction (VAF) = variant copies ÷ all copies at that locus.2 VAF counts copies of the locus, and the percentage of abnormal cells is a different number. Each diploid cell carries two copies of the locus, so 30 heterozygous cells among 100 cells contribute 30 variant copies out of 200 total copies, an expected VAF of 15%.2 Treating the cell fraction itself as the VAF doubles the expected value for a simple heterozygous clone. Two variant copies in each of those cells would contribute 60 of 200, or 30%, with no change in the cell fraction. Copy-number changes, subclones, and specimen admixture can change this relationship, and sequencing alone cannot establish which model applies.2

Follow the bench demonstration

Sources for this step
References
  1. Keohane EM, Preston MM, Mirza KM, Walenga JM, eds. Rodak's Hematology: Clinical Principles and Applications. 7th ed. Elsevier; 2025. Accessed September 26, 2026. https://www.us.elsevierhealth.com/rodaks-hematology-9780323936507.html
  2. Shao L, Akkari Y, Cooley LD, et al. Chromosomal microarray analysis, including constitutional and neoplastic disease applications, 2021 revision: a technical standard of the American College of Medical Genetics and Genomics (ACMG). Genet Med. 2021;23:1818-1829. doi:10.1038/s41436-021-01214-w
  3. Apperley JF, Milojkovic D, Cross NCP, et al. 2025 European LeukemiaNet recommendations for the management of chronic myeloid leukemia. Leukemia. 2025;39(8):1797-1813. doi:10.1038/s41375-025-02664-w

Watch one

Read each decision and its explanation.

A marrow specimen from a patient with suspected CML has a normal karyotype. The clinical picture and blood film still support CML. What test do you request next, and why?

  1. Assess the karyotype result. Normal banding does not exclude BCR::ABL1, because cryptic and variant rearrangements occur.

    A karyotype's resolution is generally measured in megabases, so it can miss a cryptic or variant rearrangement.

  2. Request FISH or RT-PCR targeted to BCR::ABL1, either of which can detect the fusion the karyotype missed.

    FISH or RT-PCR can each confirm the fusion without needing a visible translocation on the karyotype.

  3. Choose RT-PCR when transcript identification is also needed, because FISH shows the rearrangement but not which transcript is expressed.

    The transcript identified now is the one the monitoring assay must target later.

  4. State the result together with its coverage, so a negative finding is not interpreted as excluding an untested abnormality.

    A negative result from either method applies only to the abnormality it was designed to detect.

Request FISH or RT-PCR targeted to BCR::ABL1. RT-PCR also identifies the transcript for later monitoring, and either result is reported together with what it does and does not cover.

Practice this skill

Your turn

Problem 1 of 3 · Hints available

FISH with a break-apart probe for a suspected gene rearrangement shows separated signals in the abnormal population. Does this identify the rearrangement's partner gene?

A break-apart probe shows that the target locus is rearranged. It does not identify which gene it is joined to.

A break-apart probe supports a rearrangement involving that one gene. Identifying the specific partner requires a dual-fusion probe or a sequence-level method.

A break-apart probe is designed to detect a rearrangement at its target locus. It cannot identify the partner gene.

Hint
  1. A break-apart probe places two colors on opposite sides of one gene.
  2. Compare what a break-apart probe shows with what a dual-fusion probe shows.

Review Choosing a method

Problem 2 of 3 · Hints available

In a simple diploid mixture, 30 of 100 cells each carry one copy of a variant at a locus, and the other cells carry none. What is the expected variant allele fraction?

Incorrect. That is the percentage of abnormal cells. Each cell contributes two copies of the locus, so the 30 variant copies are divided by 200.

Incorrect. This doubles the cell fraction. The allele fraction divides the variant copies by all copies at the locus: 30 ÷ 200.

Correct. Each of the 100 diploid cells contributes two locus copies, giving 200 total. Thirty variant copies ÷ 200 locus copies × 100 = 15%.

Hint
  1. Each diploid cell carries two copies of the locus.
  2. Divide the variant copies by all copies at the locus. The number of cells is a different denominator.
  3. Work out the variant copies and the total copies separately before dividing.

Review Copy number, allele fraction, and transcript identity

Problem 3 of 3 · On your own

Sequencing finds 20 of 80 cells heterozygous for a variant at one locus, with the remaining cells carrying no variant. What is the expected variant allele fraction?

Show the answer

12.5 %

Twenty heterozygous cells contribute 20 variant copies out of 160 total copies (80 cells × 2 copies each). 20 ÷ 160 = 12.5%.

Review Copy number, allele fraction, and transcript identity

Use it

Read the case evidence, then make each decision. The step is complete without it.

  • A marrow aspirate is submitted for a patient with cytopenias and suspected myelodysplastic neoplasm.
  • The karyotype is normal.
  • A targeted next-generation sequencing panel detects a TET2 variant at 20% VAF and an SF3B1 variant at 35% VAF.
  • A copy-number microarray shows no copy-number change at either locus.
Decision 1 of 3

Assuming a single heterozygous variant copy per abnormal cell and no copy-number change, about what fraction of cells carry the SF3B1 variant?

Thirty-five percent is the VAF, the fraction of locus copies. It differs from the fraction of cells, because each abnormal cell contributes one variant copy out of two at the locus.

Reported the abnormal-cell fraction as the VAF

Each diploid cell carries two copies of the locus, so 30 heterozygous cells among 100 contribute 30 variant copies out of 200, a VAF of 15%. Treating the cell fraction as the VAF doubles the expected value for a simple heterozygous clone.

A VAF of 35% with one variant copy per abnormal cell and two copies per cell means the cell fraction is about double the VAF, close to 70%.

With a single heterozygous copy per cell and no copy-number change, as stated here, the cell fraction is about twice the VAF.

Review Copy number, allele fraction, and transcript identity

Decision 2 of 3

The normal copy-number microarray result is reported with the sequencing findings. What does it show?

A copy-number microarray detects gains and losses of DNA. It does not detect the sequence-level variants the panel already found, so a normal result here does not contradict them.

A copy-number microarray and targeted sequencing detect different kinds of change. The normal array result shows dosage is preserved at these loci, and it does not affect the sequence variants already detected.

The array and the sequencing panel examine different kinds of change. A normal array does not call the sequencing result into question.

Review Choosing a method

Decision 3 of 3

A colleague asks whether this workup has ruled out every possible genetic abnormality in this specimen. What do you say?

The panel detects only its included targets, and the array detects only copy-number change. Neither method covers every possible abnormality outside what it was designed to examine.

Took a negative FISH as excluding untargeted changes

A FISH probe set detects only the loci it targets, and a karyotype misses changes below its megabase resolution, such as cryptic fusions or point mutations. A negative result covers only the regions the method examined.

The sequencing panel's coverage and the array's resolution both have limits. A normal or negative result from either is limited to what that method examines.

Repeating the same panel would examine the same targets again. The limitation is the panel's coverage, and running it again would not change that.

Review Choosing a method

Each result was interpreted within the coverage of its method. Sequencing gives the VAF for each variant, and a VAF of 35% with a single heterozygous copy per cell means about 70% of cells carry the SF3B1 variant. The normal copy-number array shows that dosage is preserved. It does not call the sequencing findings into question, and neither method examines abnormalities outside its own targets.

For reference

Sources checked