cDNA, genomic sequence cloning and sequence analysis of ribosomal protein L18A gene( RPL18A ) from the Giant Panda (Ailuropoda melanoleuca)

RPL18A is a component of the 60S large ribosomal subunit encoded by RPL18A gene, which plays an important role in ribosome. To explore the structure characteristic of ribosomal protein L18 (RPL18A) gene of the Giant Panda (Ailuropoda melanoleuca) and the differences with other species reported,we have designed primers which based on the already-known information of RPL18A gene. Then, the cDNA and the genomic sequence of RPL18A were cloned successfully from the Giant Panda using RT-PCR technology respectively, followed by sequencing and analyzing. The result indicated that the cDNA fragment of the RPL18A cloned from Giant Panda is 571 bp in size, containing an open reading frame of 531bp, and deduced protein was composed of 176 amino acids with an estimated molecular weight of 20.76 kD and PI of 11.42. The length of the genomic sequence is 1999bp, which was found to possess four exons and three introns. Alignment analysis indicated that the nucleotide sequence and the deduced amino acid sequence are highly conserved to other four mammals studied, including Homo sapiens, Mus musculus, Rattus norvegicus and Bos taurus. The homologies for nucleotide sequences of Giant Panda RPL18A with that of these species are 91.53%, 92.66%, 89.27% and 89.83%, respectively, while the homologies for amino acid sequences are 98.86%, 98.86%, 98.30% and 98.30%, respectively. Topology prediction showed there are 4 different patterns of functional sites in the RPL18A protein of the Giant Panda. The cDNA and the genomic sequence of RPL18A were cloned successfully from the Giant Panda in this research, which provided scientific material to enrich and improve the mammals RPL18A gene database.