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ABO, H and secretor status

16 min

Skills you'll build4 skills
  • List the possible genotypes behind a blood-group phenotype in a person or family
  • Predict red-cell A and B antigens from H substance and the inherited transferase
  • Determine secretor status from ABH substances in saliva and match it to Lewis type
  • Recognize the Bombay phenotype and choose Bombay red cells for its anti-H

At the bench: Resolve a blood group discrepancy before selecting compatible components

Try first

Try first

A group A mother and a group B father have a group O child. Does that fit ABO inheritance?

Keep going to see the reasoning.

Keep going to see the reasoning.

Right. Keep going to follow the reasoning.

Keep going to see the reasoning.

Get the idea

Use the explanation to check your first decision.

One phenotype, more than one genotype

Routine serology shows the phenotype. In the common ABO model, A and B are codominant and O is silent. Group A can be AA or AO, and group B can be BB or BO. Groups O and AB each have one common genotype.1,3 Each parent passes one of two alleles with equal chance. An AO × BO pairing can therefore give children of group A, B, AB or O, each with a chance of 1 in 4. Family studies or molecular testing establish a genotype.1,3

A and B are built on H

ABO and H antigens are sugars at the ends of chains on red-cell glycoproteins and glycolipids. Each is added by its own enzyme:2

Gene productSugar addedProduct
FUT1 (H transferase)L-fucoseH on red cells
FUT2 (secretor transferase)L-fucoseH in secretions
A transferaseN-acetylgalactosamineA, built on H
B transferaseD-galactoseB, built on H

The A and B transferases use H as their substrate. Group O cells keep the most unconverted H, and A1B cells the least.2,3

Secretor status and saliva testing

A person with at least one functional FUT2 allele makes soluble H in saliva and other secretions, and the inherited ABO transferases add A or B to it. In hemagglutination inhibition, saliva is incubated with a known antibody, and indicator red cells are added. Soluble antigen neutralizes the antibody, so the indicator cells stay unagglutinated.2,3 Le(a−b−) red cells occur in secretors and nonsecretors, so Lewis typing cannot establish secretor status.3

The Bombay phenotype

In the classical Bombay phenotype (Oh), FUT1 and FUT2 are both inactive. The red cells lack H, so an inherited A or B transferase has nothing to modify, and forward grouping looks like group O. The plasma contains anti-A, anti-B and anti-H. The anti-H has a broad thermal range and reacts strongly with ordinary group O cells, which are rich in H. The patient needs compatible Bombay red cells.1-3

Follow the bench demonstration

Sources for this step
References
  1. Dean L. Blood Groups and Red Cell Antigens. National Center for Biotechnology Information; 2005. Accessed September 27, 2026.
  2. Stanley P, Wuhrer M, Lauc G, Stowell SR, Cummings RD. Structures common to different glycans. In: Varki A, Cummings RD, Esko JD, et al, eds. Essentials of Glycobiology. 4th ed. Cold Spring Harbor Laboratory Press; 2022. Accessed September 27, 2026.
  3. Bloch EM, Campbell-Lee S, McKenna DH Jr, Montemayor-Garcia C, Schwartz J, Shaz B, Storry J, eds. Technical Manual. 22nd ed. AABB; 2026.

Watch one

Read each decision and its explanation.

A genetic counselor asks the blood bank which ABO groups a couple's next child could have. The mother, Priya Parham, is group A. The father, Desmond Dillon, is group B. Their first child is group O. The family relationships are confirmed.

What is the chance that the next child is group AB?

  1. List the options: the mother is AA or AO, and the father is BB or BO.

    Serology shows phenotypes, and groups A and B each allow two common genotypes.

  2. The group O first child establishes both genotypes: the mother is AO and the father is BO.

    A group O child carries two O alleles, one from each parent.

  3. Write the four combinations: A with B gives AB, A with O gives AO, O with B gives BO, and O with O gives OO.

    Each parent passes either allele with equal chance.

  4. Each combination has a chance of 1/2 × 1/2 = 1/4, or 25%.

    Two independent draws of 1/2 each multiply.

  5. Report the four possible groups, A, B, AB and O, each at 25%.

    Each pregnancy has the same chances, so the report gives probabilities.

The parents are AO and BO. The chance of a group AB child is 1/2 × 1/2 = 1/4, or 25%.

Practice this skill

Your turn

Problem 1 of 3 · Hints available

Which enzyme activity forms H antigen on red-cell type 2 chains, the substrate that A and B transferases modify?

Incorrect. FUT2 creates H on type 1 chains in secretory tissues, the source of soluble ABH in saliva and other secretions. Red-cell ABH depends on FUT1.

Correct. FUT1 adds L-fucose to the red-cell type 2 precursor to form H, and an inherited A or B transferase then adds its sugar to H. In classical Bombay, with FUT1 and FUT2 both inactive, the red cells lack H, so A and B cannot form.

Incorrect. FUT3 adds fucose to type 1 chains and forms the Lewis antigens, Lea from the precursor and Leb from type 1 H.

Hint
  1. Red-cell H sits on type 2 chains.
  2. Secretory tissue and red cells use different enzymes to make H.

Review Antigen construction

Problem 2 of 3 · Hints available

A patient's red cells type as group O. The plasma strongly agglutinates group O reagent cells and group O donor cells at 37 °C, and the autocontrol is nonreactive. Which explanation fits?

Incorrect. An autoantibody such as anti-I would also react with the patient's own cells, and this autocontrol is nonreactive. Anti-I also reacts best in the cold.

Correct. Bombay red cells lack H and type as group O, but the plasma contains anti-A, anti-B, and anti-H. The anti-H has a broad thermal range and reacts strongly with H-rich group O cells, so red-cell support requires compatible Bombay units.

Incorrect. Anti-A1 reacts with A1 reagent cells. Group O cells carry no A antigen, so anti-A1 leaves them unagglutinated.

Hint
  1. The autocontrol is nonreactive, so the antibody does not react with the patient's own cells.
  2. Ask what these reagent cells and donor cells have in common that her own cells lack.

Review Bombay and para-Bombay phenotypes

Problem 3 of 3 · On your own

A group B adult's red cells type Le(a−b−). Saliva is tested by hemagglutination inhibition. Saline controls in place of saliva agglutinate each indicator cell 3+.

Saliva incubated withIndicator cellsReaction
Anti-AA1 cells3+
Anti-BB cells0
Anti-H lectinO cells0

What is the secretor status?

Le(a−b−) cells occur in secretors and nonsecretors alike. The saliva inhibition test determines secretor status.

Called secretor status from a Le(a−b−) type

Le(a−b−) red cells occur in both secretors and nonsecretors, because the phenotype results from an inactive FUT3. Secretor status is determined by testing saliva for ABH substances, for example by hemagglutination inhibition. Red-cell A, B, and H typing depends on FUT1 and the ABO transferases and is unaffected by secretor status.

A group B person has no A transferase, so even a secretor has no A substance in saliva. The anti-A result is the expected one.

Red-cell H depends on FUT1. Secretor status depends on FUT2 and is determined by testing saliva.

The saliva stopped anti-B and anti-H from agglutinating their indicator cells, so it contains soluble B and H. Soluble ABH in saliva makes this person a secretor.

Review Secretor status

Use it

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

  • Rosalind Blank, 41, who restores antique clocks, has a type and screen before planned surgery.
  • She has never been transfused or pregnant.
  • Her father is group AB, and her mother is group A.
  • Her specimen was tested by forward and reverse grouping, antibody screen, autocontrol and anti-H lectin.
Decision 1 of 3
TestReaction
Anti-A, patient cells0
Anti-B, patient cells0
A1 cells, patient plasma4+
B cells, patient plasma4+
Screening cells I and II, group O4+ at immediate spin, 37 °C and AHG
Autocontrol0
Anti-H lectin, patient cells0

What do her results show?

An autoantibody would react with her own cells, and the autocontrol is nonreactive.

Group O cells always carry H. Her cells are nonreactive with anti-H lectin, so they are not ordinary group O.

Her cells lack A, B and H. The plasma reacts strongly with H-rich group O cells at every phase, the pattern expected with anti-H.

Review Bombay and para-Bombay phenotypes

Decision 2 of 3

Her father is group AB, so she inherited an A or B allele from him. Why do her red cells carry neither A nor B?

A and B transferases add their sugar to H. Without FUT1 activity her red cells make no H, so no A or B can form.

A weak subgroup still carries H and reacts with anti-H lectin. Her cells have no H for a transferase to build on.

Assumed an A transferase can create A antigen without H

A and B transferases add their sugar to H, so red cells without H carry no A or B antigen even when an A or B allele is inherited. If that step is missed, a patient with the Bombay phenotype is called ordinary group O and the anti-H in the plasma is overlooked.

Ordinary group O cells react strongly with anti-H lectin. Hers are nonreactive, so the lack of H explains the result.

Review Antigen construction

Decision 3 of 3

She may need red cells during surgery. Which units does the transfusion service look for?

Group O cells carry the most H of any group. Her anti-H reacts strongly with them at 37 °C.

Assumed group O red cells are compatible with anti-H

Bombay red cells type as group O, but the plasma contains anti-H, which reacts strongly with the abundant H on ordinary group O cells over a broad thermal range. Group O red cells can therefore cause a hemolytic transfusion reaction, and red-cell support requires compatible Bombay units.

Bombay red cells lack A, B and H, so her anti-A, anti-B and anti-H do not react with them. Compatibility testing confirms each unit.

AB cells carry A, B and H. Her plasma reacts with all three.

Review Bombay and para-Bombay phenotypes

Her cells typed as group O but were nonreactive with anti-H lectin. Group O cells carry the most H of any group, so cells with no H lack the substrate for A and B. The strong reactions with the group O screening cells were from anti-H.

For reference

Sources checked