Bacteriophage phi 6 has a genome of three segments of double-stranded RNA, designated L, M, and S. A 1.2-kbp kanamycin resistance gene was inserted into segment M but was shown to be genetically unstable because of a high recombination rate between segment M and the 3' ends of segments S and L. The high rate of recombination is due to complementary homopolymer tracts bounding the kan gene. Removal of one arm of this potential hairpin stabilizes the insertion. The insertion of a 241- or 427-bp lacZ' gene into segment M leads to a stable Lac+ phage. The insertion of the same genes bounded by complementary homopolymer arms leads to recombinational instability. A stable derivative of this phage was shown to have lost one of the homopolymer arms. Several other conditions foster recombination. The truncation of a genomic segment at the 3' end prevents replication, but such a damaged molecule can be rescued by recombination. Similarly, insertion of the entire 3-kb lacZ gene prevents normal formation of virus, but the viral genes can be rescued by recombination. It appears that conditions leading to the retardation or absence of replication of a particular genomic segment facilitate recombinational rescue.