Question 1 · Hard
A 68-year-old man has adenocarcinoma of the pancreatic body, with several liver metastases on CT and biopsy. His bilirubin is normal. Results: CA 19-9 below 2 U/mL (reference < 37 U/mL), unchanged when the sample is retested at a 1:10 dilution; CEA 22 ng/mL [22 µg/L] (reference < 5); assay controls within limits. He takes no supplements. A blood bank sample sent before a planned procedure types as group O, D-positive, Le(a-b-), with a negative antibody screen. The CA 19-9 result is best explained by:
- AA high-dose hook effect
- BA Lewis-negative phenotype
- CBiotin interference
- DHeterophile antibody interference
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Answer: B · A Lewis-negative phenotype
CA 19-9 is sialyl Lewis a, a carbohydrate that can be made only when the Lewis fucosyltransferase (FUT3) adds fucose to a type 1 chain. About 5 to 10% of people, more in some populations, lack a working FUT3 gene and type Le(a-b-), and they cannot make CA 19-9 however large their tumor. In such patients a normal CA 19-9 is meaningless, and CEA or imaging is used to follow the cancer. A hook effect is excluded because the 1:10 dilution did not raise the result, and biotin is unlikely with no supplements. Heterophile antibodies more often raise sandwich assay results falsely and would not explain the Lewis type.
- A: A hook effect would give a higher result after dilution, and the 1:10 retest was unchanged.
- C: He takes no supplements, so biotin interference is unlikely.
- D: Heterophile antibodies usually raise sandwich results falsely and do not explain the Le(a-b-) type.