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Erianthus arundinaceus is an important,closely related genus of Saccharum officinarura L.It is therefore important to understand how the chromosomes are transmitted when it hybridizes with sugarcane.The hybrids and backcross progenies of S.officinarum and E.arundinaceus and their parents were used for Karyotype analysis and to study the law of chromosome transmission.The results showed that the somatic chromosome number of both of the E.arundinaceus Hainan92-105 and Hainan92-77 were 2n =60=60sm,belonging to type 1 A,and the BC1 YC01-21 was 2n=104=100m+4sm,belonging to type 2C.The other six tested clones belonged to type 2B.The both F1s YC96-66 and YC96-40 that originated from Badila(2n=80=70m+10sm)with E.Arundinaceus were 2n=70=68m+2sm,which suggests an n+n transmission.The cross between YC96-66(female parent)and CP84-1198(male parent,2n=120=114m+6sm)also followed the same genetic law and the somatic chromosome number of their progeny,YC01-3(2n = 105 = 95m+10sm).The cross derived from YC96-40(female)and CP84-1198(male),YC01-21 had 2n= 104= 100m+4sm chromosomes,following the same genetic law of n+n.However,YC01-36 had 2n = 132 = 130m+2sin chromosomes,which suggests a 2n+n chromosome transmission.It can be inferred that the inheritance of chromosomes was very complex in the BC1.The difference in chromosome number between clones was as high as 28.This could be explained by the 2n+n transmission of chromosomes.In addition,as there was not be a regular number of haploids,this phenomenon is termed as disequilibrium hybridization.