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HLA typing and crossmatching

17 min

  • Calculate sibling HLA-haplotype sharing probabilities
  • Distinguish recipient attack on a graft from graft-versus-host injury
  • Report HLA typing only to the allele resolution the typing method reached
  • Tell single-antigen bead HLA antibody results from a donor-cell crossmatch

Read the full reference

Try first

Try first

A patient needs a hematopoietic stem cell donor, and her one full brother is tested. Ignoring recombination, what is the chance he is HLA-identical to her?

The next section explains it.

The next section explains it.

Right. The next section explains why.

The next section explains it.

The next section explains it.

Get the idea

HLA is inherited as haplotypes

The HLA genes sit close together on chromosome 6 and pass from parent to child as a haplotype. Each child receives one haplotype from each parent. Ignoring recombination, two full siblings have a 25% chance of sharing both parental HLA haplotypes, 50% of sharing one, and 25% of sharing neither. Each sibling is a separate draw, so a patient with several siblings has a better chance that at least one is identical.1,2

Which way the attack runs

In rejection, the recipient's T cells and antibodies attack donor tissue. In graft-versus-host disease after an allogeneic hematopoietic transplant, mature donor T cells attack the recipient's skin, gut and liver.1,3

Typing reports the resolution it reached

DNA typing methods differ in how much of the sequence they read.4

MethodWhat it readsWhat can stay unresolved
PCR-SSPWhich primer pairs amplifyAlleles outside the primer panel
PCR-SSOWhich probes bind the amplified DNAAllele combinations with the same probe pattern
Sanger sequencingThe sequence of the amplified regionPhase and unsequenced regions
Next-generation sequencingLong regions, often phasedCoverage, software and the reference database

When a probe pattern or sequence fits more than one allele combination, the report lists the ambiguity. Further testing resolves it when donor matching needs that level of detail.2,4

Single-antigen beads and the crossmatch

Single-antigen beads carry purified HLA molecules and show which HLA specificities the recipient's antibody recognizes. Bead fluorescence is semiquantitative and depends on the method, so no single value works as a universal cutoff. A flow crossmatch mixes recipient serum with the donor's own T and B lymphocytes and detects bound IgG with fluorescent anti-human IgG. Resting T cells carry HLA class I only. B cells carry class I and class II, so an antibody to a class II antigen such as DQ binds donor B cells alone.1,4,5

References
  1. Abbas AK, Lichtman AH, Pillai S, Henrickson S. Cellular and Molecular Immunology. 11th ed. Elsevier; 2025. Elsevier.
  2. Marsh SGE, Osoegawa K, Bodmer WF, et al. Nomenclature for factors of the HLA system, 2026. HLA. 2026;107(3):e70595. doi:10.1111/tan.70595
  3. Jagasia MH, Greinix HT, Arora M, et al. National Institutes of Health Consensus Development Project on criteria for clinical trials in chronic graft-versus-host disease: I. The 2014 Diagnosis and Staging Working Group report. Biol Blood Marrow Transplant. 2015;21(3):389-401.e1. doi:10.1016/j.bbmt.2014.12.001
  4. Spierings E, Madrigal JA, Fleischhauer K. Histocompatibility. In: Sureda A, Corbacioglu S, Greco R, Kröger N, Carreras E, eds. The EBMT Handbook: Hematopoietic Cell Transplantation and Cellular Therapies. 8th ed. Springer; 2024. doi:10.1007/978-3-031-44080-9_9
  5. Kongtim P, Vittayawacharin P, Zou J, et al. ASTCT consensus recommendations on testing and treatment of patients with donor-specific anti-HLA antibodies. Transplant Cell Ther. 2024;30(12):1139-1154. doi:10.1016/j.jtct.2024.09.005

Watch one

Farid Nazari, 38, needs an allogeneic hematopoietic transplant. He has two full sisters, and both agree to HLA typing. His family asks the chance that at least one sister is HLA-identical to him.

  1. Find the chance for one sister: 1/2 × 1/2 = 1/4 = 0.25.

    Each child receives one of two haplotypes from each parent.

  2. Find the chance one sister is not identical: 1 − 0.25 = 0.75.

    It is easier to count the ways every sister fails to match.

  3. Find the chance neither sister is identical: 0.75 × 0.75 = 0.5625.

    Each sister's inheritance is a separate draw.

  4. Subtract from 1: 1 − 0.5625 = 0.4375.

    At least one match is everything except no match.

  5. Report 43.8% as a chance and send both sisters' specimens for high-resolution typing.

    A probability describes the family before typing. Only typing shows whether a given sister matches.

P(at least one identical) = 1 − 0.75 × 0.75 = 0.4375, or 43.8%.

Your turn

Problem 1 of 3

Which direction of immune attack defines graft-versus-host disease after an allogeneic hematopoietic transplant?

Incorrect. Recipient immunity attacking donor tissue is the host-versus-graft direction of rejection.

Incorrect. Recipient cells attacking recipient self-antigens is autoimmunity. Graft-versus-host disease runs from donor cells toward the recipient.

Correct. Mature donor T lymphocytes attack recipient tissue, commonly skin, gastrointestinal tract, and liver. The same donor response can attack residual malignant cells as a graft-versus-leukemia effect.

Hint
  1. Name whose cells arrive in the transplant.
  2. The name of the disease says which side attacks which.

Review Allorecognition and graft injury

Problem 2 of 3

What does a flow-cytometric donor lymphocyte crossmatch directly assess?

Incorrect. Molecular HLA typing determines sequence. The crossmatch asks whether recipient antibody binds donor cells.

Correct. Fluorescent anti-human IgG detects recipient antibody bound to donor T or B lymphocytes, and the flow method is generally more sensitive than complement-dependent cytotoxicity.

Incorrect. Single-antigen beads assign specificity against purified HLA molecules. The crossmatch tests binding to the donor's own lymphocytes.

Hint
  1. Ask what the recipient's serum is mixed with in this test.
  2. Beads carry purified molecules. A crossmatch uses something from the donor.

Review HLA antibody assessment and crossmatching

Problem 3 of 3

A patient has three full siblings. Ignoring recombination, what is the chance that at least one of them is HLA-identical to the patient, to one decimal place?

Show the answer

57.8 %

Chance that one sibling is not identical = 1 − 0.25 = 0.75. Chance that none of the three is identical = 0.75 × 0.75 × 0.75 = 0.421875. Chance of at least one = 1 − 0.421875 = 0.578, or 57.8%. Adding 25% three times gives 75%, which counts the families with two or three identical siblings more than once.

Review The histocompatibility barrier

Use it

  • Wendell Ashgrove, 44, a mail carrier who has walked the same route for 20 years, is a kidney transplant candidate.
  • His cousin offers to be a living donor.
  • The donor was typed by PCR-SSO. At HLA-DQB1 the probe pattern fits two allele combinations.
  • He received several transfusions after a car crash years ago.
  • Single-antigen beads show antibody to HLA-DQ7, which the donor carries, with a bead signal of 3,200.
  • A flow crossmatch uses donor lymphocytes.
TestResultPreviousReference intervalFlag
Flow crossmatch, T cellsNegativeNegative
Flow crossmatch, B cellsPositiveNegative

Specimen: H 1, L 6, I 1. Recipient serum collected 13:10; donor lymphocytes from EDTA whole blood collected 12:45

Decision 1 of 3

How is the donor's HLA-DQB1 typing reported?

Frequency does not decide what this donor carries. Reporting one combination claims a resolution the probes did not reach.

Assumed every DNA typing result resolves all alleles

Primer or probe coverage, sequencing phase, and unsequenced regions leave some DNA typing results ambiguous. Reporting an ambiguous result as a single allele overstates the resolution that donor matching relies on.

The report states the resolution the method reached. Sequencing can resolve the ambiguity when the program needs it.

The typing worked. A repeat with the same probes gives the same ambiguous pattern.

Review HLA typing

Decision 2 of 3

Why is the B-cell crossmatch positive and the T-cell crossmatch negative?

Nothing points to a failed tube. The split fits the antibody's target.

Class I antigens sit on both T and B cells. A class I antibody would usually make the T-cell crossmatch positive too.

HLA-DQ is class II. B cells express it and resting T cells do not, so a class II antibody gives a B-cell-only positive crossmatch.

Review HLA antibody assessment and crossmatching

Decision 3 of 3

A colleague says a bead signal of 3,200 is below 5,000, so the anti-DQ7 does not matter. What does the laboratory report?

The report gives the specificity, the signal and the crossmatch together. The transplant program weighs them under its own protocol.

Bead signal is semiquantitative and depends on the method. The donor's own B cells bound the antibody.

Used one bead intensity as a universal rejection threshold

Single-antigen bead results show which HLA specificities are present, and the physical crossmatch tests binding to the donor's own cells. Bead signal is semiquantitative and method-dependent, so one fluorescence value read as a rejection cutoff can overstate or understate risk.

The bead results name the specificity that explains the crossmatch. The program needs both.

Review HLA antibody assessment and crossmatching

The clue that settled this case is the B-cell-only positive crossmatch beside a bead result for HLA-DQ7. The antibody is donor-specific and binds the donor's own cells, whatever its bead signal. The donor's DQB1 typing is reported at the resolution the probes reached.

Keep

Sources checked