Complete immunological tolerance to sheep cells can be induced in mice when cyclophosphamide is injected together with sheep cells or up to 72 hr before or 48 hr after the antigen. As is true for radiation-induced immune suppression, the drug is most effective when given in the 24 hr prior to antigen. Complete cyclophosphamide-induced immunological suppression requires large doses of sheep cells (6.2 x 10(9) cells), presumably to enable antigen to reach sequestered receptor sites. The cyclophosphamide tolerance system has been analyzed with the Jerne technique to determine plaque-forming cells and with isotopic methods to measure rates of nucleic acid synthesis. This drug suppression has been found to consist of two components. The first is nonspecific injury to the lymphoid system caused by the cytotoxic drug and is related to the proportion of spleen cells killed. The second is antigen-specific immunological tolerance and appears to correlate with profound depression of deoxyribonucleic acid synthesis in the surviving cells. This tolerance is thought to be most consistent with a mechanism in which antigenic stimulation in the presence of cyclophosphamide-inhibited DNA synthesis and mitosis leads to the elimination or death of the specific immunological clone. Tolerance induction with cyclophosphamide is associated with loss of the 19S hemolysin plaques which are seen in nonstimulated mouse spleen, implicating these cells in immune responsiveness. The ability to induce tolerance is lost on the 3rd postantigen day at the end of a 24-hr period in which 19S cells have increased 8-fold and 7S cells 200-fold. The data suggest that loss of sensitivity is due to the emergence on day 3 of drug-resistant plaque-forming cells, particularly those of the 19S variety. In the succeeding days after antigen injection there is a progressive increase in the resistance of plaque-forming cells to cyclophosphamide administration.
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