6.1 ABO Biochemistry and Antigens
Key Takeaways
- FUT1 (H) adds α-1,2-fucose to type 2 chains on red cells to make H, the required substrate for A and B transferases; FUT2 (Se) does the same job on type 1 chains in secretions.
- A transferase adds GalNAc to H; B transferase adds D-galactose to H; group O leaves H unmodified because the common O allele encodes an inactive transferase.
- Bombay hh makes no H, so A and B genes are silent on the red cell; the serum contains anti-H plus anti-A, anti-B, and anti-A,B, and only Oh red cells are compatible.
- Approximate U.S. frequencies are O ~45%, A ~40%, B ~11%, AB ~4%, with higher B in many Asian populations and higher O in many Hispanic and Native American populations.
- Dolichos biflorus is anti-A1 (A1 and A1B positive; A2 and A2B negative). Ulex europaeus is anti-H and ranks leftover H as O > A2 > B > A2B > A1 > A1B.
6.1 ABO Biochemistry and Antigens
Quick Answer: The H gene (FUT1) puts an α-1,2-fucose on a type 2 precursor and creates H antigen, the required substrate for A and B. The A transferase adds N-acetylgalactosamine (GalNAc); the B transferase adds D-galactose. Group O leaves H unmodified. Secretor status is a different enzyme, FUT2 (Se), acting mainly on type 1 chains in saliva and plasma. Bombay (hh) makes no H, so A and B genes are silent on the red cell; the serum contains anti-H plus anti-A, anti-B, and anti-A,B, and only Oh red cells are compatible.
The June 9, 2026 BB outline lists ABO under II.B.1 because almost every later serologic problem — discrepancy, subgroup, HDFN, emergency release — starts with this pathway. If you cannot say which sugar sits on which chain, you will miss Bombay versus group O, A1 versus A2, and why Ulex europaeus ranks phenotypes the way it does.
H antigen comes first
Red-cell ABH synthesis is a two-step story. A precursor oligosaccharide already exists on membrane glycolipids and glycoproteins. The H gene, FUT1, on chromosome 19, encodes an α-1,2-L-fucosyltransferase. That enzyme attaches L-fucose to the terminal galactose of a type 2 chain (Galβ1-4GlcNAc). The product is H antigen. Without H, the A and B transferases have nothing to decorate.
Type 2 chains dominate the red-cell membrane. Type 1 chains (Galβ1-3GlcNAc) dominate secretions and plasma. Do not swap them. FUT1 prefers type 2. FUT2 (Se) prefers type 1. That is why a person can be a red-cell group A and still lack A substance in saliva: the saliva question is Se, not ABO. Type 3 and type 4 chains matter later for the A1 qualitative difference; they are not the everyday precursor on a group O cell.
A person who inherits at least one functional FUT1 allele has red-cell H. The rare person who inherits two inactive alleles is hh and has no red-cell H. That single genotype is Bombay, and it rewrites every ABO rule that follows.
A and B are terminal sugars on H
The ABO locus on chromosome 9q34 encodes glycosyltransferases, not the antigens themselves. The A and B proteins differ by a handful of amino acids that change the donor-sugar preference.
| Gene product | Enzyme | Sugar added to H | Phenotype if H is present |
|---|---|---|---|
| A transferase | α-1,3-N-acetylgalactosaminyltransferase | GalNAc | Group A |
| B transferase | α-1,3-D-galactosyltransferase | D-galactose | Group B |
| Both A and B | Both active transferases | GalNAc and galactose | Group AB |
| Common O (inactive) | No functional transferase | None — H unmodified | Group O |
| Any ABO gene + hh | Transferase may be present | Cannot add — no H substrate | Bombay (Oh) |
Group O is therefore not “no antigen.” Group O red cells are the richest in H. AB red cells convert the most H and keep the least leftover H. A and B transferases compete for the same H substrate. They do not act on a naked type 2 precursor. That single fact explains Bombay: an A or B gene in an hh person cannot paint A or B on the red cell because there is no H to accept the sugar.
Secretor status is FUT2, not ABO
Se (FUT2) encodes a second α-1,2-fucosyltransferase. It fucosylates type 1 chains in saliva, mucus, and plasma, creating type 1 H. If the person also has A or B transferase, those enzymes add GalNAc or galactose in the secretions. About 80% of people of European ancestry are secretors (SeSe or Sese). Nonsecretors (sese) have ABH on red cells if they have FUT1, but not in saliva.
FUT1 and FUT2 sit close together on chromosome 19. Classic Indian Bombay is often hh and sese, so there is no H on red cells and none in secretions. That genotype is why Bombay saliva neutralization tests fail and why the anti-H is so strong. Para-Bombay often keeps a working Se, which is how you separate the two at the bench.
A secretor saliva neutralization test can confirm soluble ABH when you are working an ABO discrepancy or a suspected Bombay. It does not replace red-cell typing, and it does not make a nonsecretor “group O in the mouth.” Nonsecretors still have a perfectly ordinary red-cell ABO type.
Bombay and para-Bombay
Bombay (Oh) is hh. No FUT1 activity means no H, no A, and no B on the red cell, even when A or B genes are inherited. Forward typing looks like group O: anti-A, anti-B, and anti-A,B are nonreactive. Reverse typing does not look like group O. The serum contains anti-H, anti-A, anti-B, and anti-A,B. Group O reagent cells — packed with H — agglutinate strongly, often at immediate spin and often as a hemolysin if complement is present. Ulex europaeus (anti-H lectin) is nonreactive with Bombay cells. The only compatible red cells are Oh. Transfusing group O red cells to a Bombay recipient is an ABO-incompatible, anti-H-mediated hemolytic event, not a universal-donor rescue.
Para-Bombay is a mutated or silenced FUT1 with little or no red-cell H, but Se is often present, so ABH appears in secretions. Residual FUT1 may leave traces of H, A, or B detectable only by adsorption-elution. Anti-H, when present, is usually weaker than classic Bombay anti-H and may not react with all adult O cells. Do not call every H-negative forward type Bombay until you have tested Ulex, O cells in reverse, secretor status, and, when available, molecular FUT1/FUT2.
A practical Bombay workup on a “group O” that reacts with O reverse cells: (1) confirm the extra reverse reaction is anti-H (reacts with O adult cells, weaker or negative with Oh or cord cells that lack H), (2) show Ulex-negative patient cells, (3) test saliva if the patient is old enough and the SOP allows it, (4) do not issue group O red cells.
Frequencies you will be handed as distractors
Approximate United States phenotype frequencies, with the usual teaching set used on BB items:
| Phenotype | Approx. U.S. frequency | What the red cell actually carries |
|---|---|---|
| O | ~45% | Unmodified H |
| A | ~40% | A (mostly A1) plus residual H |
| B | ~11% | B plus residual H |
| AB | ~4% | A and B; least residual H |
Those four numbers are population averages. B and AB rise in many East and South Asian populations (B often approaches the A frequency). O rises in many Hispanic and Native American populations. A is higher in Northern European ancestry. If a stem specifies ancestry, do not force the 45/40/11/4 split onto a population that does not use it.
Subgroups: A1 versus A2, then the weak ones
About 80% of group A is A1; about 20% is A2. The difference is quantitative and qualitative. A1 cells carry more A sites (on the order of 1 × 10^6 versus about 2.5 × 10^5 on A2). A1 transferase also builds type 3 and type 4 A structures that A2 transferase does not make well. Dolichos biflorus lectin is anti-A1: it agglutinates A1 and A1B, not A2 or A2B.
A2 and especially A2B people may form anti-A1. That antibody is usually a room-temperature IgM. It matters in reverse typing (extra reaction with A1 cells) and only becomes a transfusion problem if it reacts at 37 °C. Section 6.2 works the discrepancy pattern; the biochemistry point here is that A2 is not “weak A1.” It is a different transferase product with less A and more leftover H, which is why A2 sits high on the Ulex ladder.
Weaker A subgroups appear on BB items as pattern recognition:
- A3: characteristic mixed-field agglutination with anti-A and anti-A,B. Do not call every mixed-field A3 — recent group O transfusion, stem-cell transplant, and twin chimerism also mix fields.
- Ax: little or no reaction with anti-A; often detected only with anti-A,B; reverse may show anti-A1.
- Am (and Ael): extremely weak A, often found only by adsorption and elution of anti-A.
Parallel B subgroups exist (B3 mixed-field, Bx, Bm) but are less common in U.S. stems. Anti-A,B from group O serum is the reagent that unmasks Ax and Bx when monoclonal anti-A or anti-B looks negative.
Lectins and the H ladder
Plant lectins are still exam tools because they are specific and they do not require complement:
- Ulex europaeus = anti-H. Strength order: O > A2 > B > A2B > A1 > A1B. That order is leftover H, not a random mnemonic. O has only H. A2 converts less H than A1. B convertase leaves more H than A1 convertase. AB converts the most H.
- Dolichos biflorus = anti-A1.
- Griffonia (Bandeiraea) simplicifolia = anti-B.
If Ulex is negative and the reverse has anti-H, you are no longer in ordinary group O — you are in Bombay territory. If Dolichos is negative on a group A patient whose serum agglutinates A1 cells, you are in A2-with-anti-A1 territory, not in a reagent failure.
Exam traps in this section
- Treating group O as “no antigen.” O is unmodified H.
- Giving group O red cells to Bombay because “they are universal.” Bombay needs Oh.
- Using Dolichos to prove group A. Dolichos distinguishes A1 from A2, not A from O.
- Confusing Se with H. FUT1 is red-cell H. FUT2 is secretor type 1 H.
- Memorizing the Ulex order backwards (A1B is weakest, not strongest).
- Calling every mixed-field agglutination A3 without asking about transfusion or transplant.
- Quoting the 45/40/11/4 split as a world constant.
A patient types as group O in the forward type, but the serum agglutinates A1 cells, B cells, and group O screening cells. Ulex europaeus is nonreactive with the patient’s red cells. Which red cells are compatible?
Which enzyme creates H antigen on type 2 chains of the red-cell membrane?
Ulex europaeus lectin (anti-H) is expected to react most strongly with which phenotype?