AbstractDextran B1355 induces separate IgM responses in mice against its α(l→3) and a(l→6)‐linked glucosyl linkages. The maximum injectable dose (10 mg) produces α(l→3) immunity and α(l→6) tolerance in the BALB/c strain. The same amount will tolerize the CBA strain for both epitopes, whereas 1 mg will produce α(l→3) tolerance and α(l→6) immunity. The converse dissociation in these strains between tolerance and immunity to the two intrinsic epitopes of dextran B1355 seems, in the face of the corresponding minimum immunogenic doses, incompatible with B cell tolerance mechanisms invoking either an excess of “one nonspecific signal” or lack of a “second signal”.A basis was sought for the exceptional resistance of BALB/c mice to α(l→3) tolerization, for dextran B1355 behaved like a typical polysaccharide antigen in the CBA strain. The induction time for α(l→3) tolerance in CBA was even shorter (<6 h) than that for α(l→6), and the unresponsive state was as stable on cell transfer as α(l→6) tolerance in either strain. Macrophage involvement in this strain difference had been previously sought but not found (Immunology 1976.30: 221.). Although BALB/c mice are “high responders” for the α( l→3) epitope, the B cell clone did not show any superior capacity to regenerate. Indeed, the slow recovery of responsiveness to both epitopes in bone marrow‐reconstituted, irradiated mice was notably prolonged in the case of BALB/c.PFC inhibition by oligosaccharides of the nigerose series revealed that the α(l→3) response corresponded to at least a tetrasaccharide determinant in CBA but to only a trisaccharide in BALB/c. As B cells specific for isomalto‐tetraose have a higher tolerance dose threshold than those for isomaltohexa‐ose (Immunochemistry1977.14: 253), it seems probable that genetic resistance of BALB/c to α(l→3) tolerance depends on recognition of a relatively small epitope which could involve instabili
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