Two antibodies, the autocontrol and enzymes
18 min
- Tell an autoantibody from a new alloantibody by the autocontrol, DAT, and eluate
- Read an enzyme-treated panel beside the untreated panel
Try first
| Cell | D | C | E | c | e | K | k | Fya | Fyb | Jka | Jkb | M | N | S | s | IS | 37 | AHG |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 | + | + | 0 | 0 | + | 0 | + | + | 0 | + | + | + | 0 | 0 | + | 0 | 0 | 2+ |
| 2 | + | + | 0 | 0 | + | + | + | 0 | + | + | 0 | + | + | + | 0 | 0 | 0 | 2+ |
| 3 | + | 0 | + | + | 0 | 0 | + | 0 | + | 0 | + | 0 | + | 0 | + | 0 | 0 | 2+ |
| 4 | 0 | + | 0 | + | + | 0 | + | + | + | 0 | + | + | + | + | + | 0 | 0 | 2+ |
| 5 | 0 | 0 | + | + | + | 0 | + | 0 | + | + | + | + | 0 | + | + | 0 | 0 | 2+ |
| 6 | 0 | 0 | 0 | + | + | + | + | 0 | + | + | 0 | 0 | + | 0 | + | 0 | 0 | 2+ |
| 7 | 0 | 0 | 0 | + | + | 0 | + | + | 0 | 0 | + | + | + | + | + | 0 | 0 | 2+ |
| 8 | + | 0 | 0 | + | + | 0 | + | 0 | + | 0 | + | + | 0 | + | 0 | 0 | 0 | 2+ |
| 9 | 0 | 0 | 0 | + | + | 0 | + | 0 | + | + | + | + | + | 0 | + | 0 | 0 | 2+ |
| 10 | + | + | + | + | + | 0 | + | + | + | + | 0 | + | + | 0 | + | 0 | 0 | 2+ |
| 11 | 0 | 0 | 0 | + | + | 0 | + | + | + | + | + | 0 | + | + | + | 0 | 0 | 2+ |
| Autocontrol | 0 | 0 | 2+ |
Get the idea
When one antibody does not explain the panel
Suspect a second antibody when no single antigen accounts for every reactive cell. Reaction strengths or phases that differ from cell to cell, beyond what dosage explains, point the same way.1,2 Work the panel as usual:
- Cross out antigens on the nonreactive cells.
- Test each antigen that is left against the reactive cells.
- If one antigen explains some reactive cells and a second explains the rest, both antibodies are likely present.
Each antibody then needs its own reactive antigen-positive cells and nonreactive antigen-negative cells. The patient should type negative for both antigens.1,2
What the autocontrol adds
The autocontrol tests the patient's plasma with the patient's own red cells. It supports the interpretation. On its own it cannot prove where an antibody came from.1
- Autocontrol nonreactive. An alloantibody is favored.
- Every cell reactive and the autocontrol reactive, no recent transfusion. A warm autoantibody or a drug effect is favored. A DAT follows, and adsorption is used to look for alloantibodies underneath.1,4
- Autocontrol reactive, often weak and mixed field, soon after a transfusion. A new alloantibody may be coating surviving donor cells. An eluate from the patient's cells shows the same specificity as the plasma, and the finding can signal a delayed hemolytic transfusion reaction.1
Enzymes separate antibodies
Ficin and papain cut proteins on the red cell surface. Some antigens lose their reactivity and others react more strongly.1,3
| Effect of ficin or papain | Antigens |
|---|---|
| Enhanced | Rh, Kidd, Lewis, P1, I |
| Destroyed or markedly weakened | Duffy (Fya, Fyb), M, N |
| Variable | S, s |
| Preserved | Kell |
Read the treated panel beside the untreated one. A reaction that disappears after treatment points to an enzyme-sensitive antigen. A reaction that grows stronger points to an enhanced one.1,3 Treated cells have lost their Duffy, M and N antigens. Exclusions for those antibodies therefore come from untreated cells.1,3
References
- Bloch EM, Campbell-Lee S, McKenna DH Jr, Montemayor-Garcia C, Schwartz J, Shaz B, Storry J, eds. Technical Manual. 22nd ed. AABB; 2026.
- Manduzio P. Alloantibody identification: the importance of temperature, strength reaction and enzymes—a practical approach. Hematol Rep. 2024;16(4):815-824. doi:10.3390/hematolrep16040077
- Raman L, Armstrong B, Smart E. Principles of laboratory techniques. ISBT Sci Ser. 2020;15(suppl 1):81-111. doi:10.1111/voxs.12591
- Johnson ST, Puca KE. Evaluating patients with autoimmune hemolytic anemia in the transfusion service and immunohematology reference laboratory: pretransfusion testing challenges and best transfusion-management strategies. Hematology Am Soc Hematol Educ Program. 2022;2022(1):96-104. doi:10.1182/hematology.2022000406
Watch one
A man transfused 2 years ago has a reactive antibody screen. His plasma is tested with an 11-cell panel by tube antiglobulin test. His autocontrol is nonreactive.
The same cells treated with ficin give these antiglobulin reactions: cells 3, 5 and 10 react 3+ to 4+. Every other cell is nonreactive. Which antibodies are present?
| Cell | D | C | E | c | e | K | k | Fya | Fyb | Jka | Jkb | M | N | S | s | IS | 37 | AHG |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 | + | + | 0 | 0 | + | 0 | + | + | 0 | + | + | + | 0 | 0 | + | 0 | 0 | 2+ |
| 2 | + | + | 0 | 0 | + | + | + | 0 | + | + | 0 | + | + | + | 0 | 0 | 0 | 0 |
| 3 | + | 0 | + | + | 0 | 0 | + | 0 | + | 0 | + | 0 | + | 0 | + | 0 | w+ | 3+ |
| 4 | 0 | + | 0 | + | + | 0 | + | + | + | 0 | + | + | + | + | + | 0 | 0 | 1+ |
| 5 | 0 | 0 | + | + | + | 0 | + | 0 | + | + | + | + | 0 | + | + | 0 | 0 | 2+ |
| 6 | 0 | 0 | 0 | + | + | + | + | 0 | + | + | 0 | 0 | + | 0 | + | 0 | 0 | 0 |
| 7 | 0 | 0 | 0 | + | + | 0 | + | + | 0 | 0 | + | + | + | + | + | 0 | 0 | 2+ |
| 8 | + | 0 | 0 | + | + | 0 | + | 0 | + | 0 | + | + | 0 | + | 0 | 0 | 0 | 0 |
| 9 | 0 | 0 | 0 | + | + | 0 | + | 0 | + | + | + | + | + | 0 | + | 0 | 0 | 0 |
| 10 | + | + | + | + | + | 0 | + | + | + | + | 0 | + | + | 0 | + | 0 | w+ | 3+ |
| 11 | 0 | 0 | 0 | + | + | 0 | + | + | + | + | + | 0 | + | + | + | 0 | 0 | 1+ |
| Autocontrol | 0 | 0 | 0 |
- Read the autocontrol first. It is nonreactive at every phase.
A nonreactive autocontrol favors alloantibodies over an antibody that also coats his own cells.
- Cross out on the nonreactive cells 2, 6, 8 and 9. D, C, c, e, K, k, Fyb, Jka, Jkb, M, N, S and s go. E and Fya are left.
A dosage antigen is crossed out only on a nonreactive cell homozygous for it, such as the Jk(a+b−) cells 2 and 6 for Jka.
- Test E against the reactive cells. It explains cells 3, 5 and 10. Cells 1, 4, 7 and 11 react and lack E.
An antigen that leaves reactive cells unexplained means a second antibody is present.
- Test Fya. It explains cells 1, 4, 7 and 11. The Fy(a+b−) cells 1 and 7 react 2+, and the Fy(a+b+) cells 4 and 11 react 1+.
Stronger reactions on homozygous cells are the dosage pattern of a Duffy antibody.
- Read the ficin panel beside the untreated one. The E-positive cells react more strongly. The cells carrying Fya without E turn nonreactive.
Ficin enhances Rh antigens and destroys Duffy antigens, so the treated panel splits the two antibodies.
- Keep anti-Fya in the answer. The ficin panel shows none of it because the antigen is gone.
Treated cells have lost Fya, so their silence cannot exclude anti-Fya.
- Confirm. E has 3 reactive positive cells and 4 nonreactive negative cells. Fya has 5 and 4. His red cells should type E-negative and Fy(a−).
Each antibody needs its own reactive and nonreactive cells, and he cannot make an antibody to an antigen he carries.
Your turn
Use it
- Walter Brandt, 71, a retired letter carrier (MRN 5520931), received 2 units of red cells 8 days ago after hip surgery.
- His antibody screen before that transfusion was nonreactive.
- Today his hemoglobin is 7.9 g/dL, down from 10.1 g/dL two days after the transfusion. A new specimen arrives with an order for 2 more units.
- Today's antibody screen reacts at the antiglobulin phase.
- His autocontrol reacts weakly and looks mixed field, with small agglutinates among many free cells.
| Cell | D | C | E | c | e | K | k | Fya | Fyb | Jka | Jkb | M | N | S | s | IS | 37 | AHG |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 | + | + | 0 | 0 | + | 0 | + | + | 0 | + | + | + | 0 | 0 | + | 0 | 0 | 1+ |
| 2 | + | + | 0 | 0 | + | + | + | 0 | + | + | 0 | + | + | + | 0 | 0 | 0 | 2+ |
| 3 | + | 0 | + | + | 0 | 0 | + | 0 | + | 0 | + | 0 | + | 0 | + | 0 | 0 | 0 |
| 4 | 0 | + | 0 | + | + | 0 | + | + | + | 0 | + | + | + | + | + | 0 | 0 | 0 |
| 5 | 0 | 0 | + | + | + | 0 | + | 0 | + | + | + | + | 0 | + | + | 0 | 0 | 1+ |
| 6 | 0 | 0 | 0 | + | + | + | + | 0 | + | + | 0 | 0 | + | 0 | + | 0 | 0 | 2+ |
| 7 | 0 | 0 | 0 | + | + | + | + | + | 0 | 0 | + | + | + | + | + | 0 | 0 | 0 |
| 8 | + | + | 0 | 0 | + | 0 | + | 0 | + | 0 | + | + | 0 | + | 0 | 0 | 0 | 0 |
| 9 | 0 | 0 | 0 | + | + | 0 | + | 0 | + | + | + | + | + | 0 | + | 0 | 0 | 1+ |
| 10 | + | + | + | + | + | 0 | + | + | + | + | 0 | + | + | 0 | + | 0 | 0 | 2+ |
| 11 | 0 | 0 | 0 | + | + | 0 | + | + | + | + | + | 0 | + | + | + | 0 | 0 | 1+ |
| Autocontrol | 0 | 0 | w+ |
The clue that settled this case is the mixed-field autocontrol 8 days after a transfusion. Only part of his red cells were coated, and those were the donor cells carrying Jka. The eluate confirmed the same anti-Jka that the plasma panel showed.
Results
- Tell an autoantibody from a new alloantibody by the autocontrol, DAT, and eluate
- Read an enzyme-treated panel beside the untreated panel
To review
6 questions from this step will come back in Review.
Next step: Antibody history and antigen-negative unitsReview nowOpen the part
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The rest of this step
A short briefing, a demonstration at the bench, 3 practice problems and a short case.
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