Abstract
Successful male germ cell transplantation requires depletion of the host germ cells to allow efficient colonization of the donor spermatogonial stem cells. Although a sterilizing drug, busulfan, is commonly used for the preparation of recipient models before transplantation, the optimal dose of this drug has not yet been defined in dogs. In this study, 1-year-old mongrel dogs were intravenously injected with three different concentrations of busulfan (10, 15, or 17.5 mg/kg). Four weeks after busulfan treatment, no fully matured spermatozoa were detected in any of the busulfan-treated groups. However, small numbers of PGP9.5-positive spermatogonia were detected in all treatment groups, although no synaptonemal complex protein-3-positive spermatocytes were detected. Of note, acrosin-positive spermatids were not detected in the dogs treated with 15 or 17.5 mg/kg busulfan, but were detected in the other group. Eight weeks after busulfan treatment, the dogs treated with 10 mg/kg busulfan fully recovered, but those in the other groups did not. PGP9.5-positive spermatogonia were detected in the 10 mg/kg group, and at a similar level as in the control group, but these cells were rarely detected in the 15 and 17.5 mg/kg groups. These results suggest that a dose of 15-17.5 mg/kg is optimal for ablative treatment with busulfan to prepare the recipient dogs for male germ cell transplantation. At least eight weeks should be allowed for recovery. The results of this study might facilitate the production of recipient dogs for male germ cell transplantation and can also contribute to studies on chemotherapy.