Lewis inheritance: Lele HH sese yields which phenotype?

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Multiple Choice

Lewis inheritance: Lele HH sese yields which phenotype?

The key idea is how Lewis antigens are made. The Le gene (FUT3) provides the fucosyltransferase that creates Lewis antigens on the type 1 precursor, producing Le^a. The Se gene (FUT2) adds another fucose to make Le^b, but Le^b only forms if FUT2 is active. So you can have Le^a without Le^b if FUT2 is nonfunctional. With Lele, FUT3 is active, so Lewis antigens can be formed. With sese, FUT2 is inactive, so Le^b cannot be produced. That leaves Le^a present and Le^b absent, giving the Le(a+b-) phenotype on red cells. The Bombay (HH) status affects ABO/H antigens but does not prevent Le^a from appearing here, so the overall phenotype is Le(a+b-).

The key idea is how Lewis antigens are made. The Le gene (FUT3) provides the fucosyltransferase that creates Lewis antigens on the type 1 precursor, producing Le^a. The Se gene (FUT2) adds another fucose to make Le^b, but Le^b only forms if FUT2 is active. So you can have Le^a without Le^b if FUT2 is nonfunctional.

With Lele, FUT3 is active, so Lewis antigens can be formed. With sese, FUT2 is inactive, so Le^b cannot be produced. That leaves Le^a present and Le^b absent, giving the Le(a+b-) phenotype on red cells. The Bombay (HH) status affects ABO/H antigens but does not prevent Le^a from appearing here, so the overall phenotype is Le(a+b-).

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