The whole nucleotide sequence and chromosomal localization of the gene for human metabotropic glutamate receptor subtype 6.

Metabotropic glutamate receptor subtype 6 (mGluR6) is restrictedly expressed in the retinal ON bipolar cells and ablation of mouse mGluR6 by gene targeting results in a loss of ON responses to light stimulus and impairs the detection of visual contrasts. We have isolated genomic clones containing the human mGluR6 gene and determined the whole nucleotide sequence of the mGluR6 gene. The transcription initiation site of the human mGluR6 gene has been identified using primer extension analysis in combination with reverse transcriptase-mediated polymerase chain reaction analysis of human retinal RNA, while the termination of the mGluR6 mRNA has been assigned by the analysis of rapid amplification of 3'-cDNA ends. The human mGluR6 gene consists of 16,742 base pairs with 10 exons separated by nine introns. The human mGluR6 is composed of 877 amino acid residues with a signal peptide of 24 amino acid residues and the mature protein shows a 94.6% homology with the rat counterpart. A CpG-rich island is present at exon 1 and its preceding putative promoter region and this unusual sequence, like several tissue-specific genes, may be important for a specific expression of the mGluR6 gene in the retinal bipolar cells. The human mGluR6 gene has been mapped to chromosome 5q35 by the analyses of blot hybridization of a DNA panel of human/mouse/hamster somatic cell hybrids and fluorescence in situ hybridization of human chromosomes. This study should provide the genetic basis for not only better understanding the molecular mechanism underlying a tissue-specific expression of the mGluR6 gene but also exploring a potential defect in human mGluR6 in a certain inherited eye disease.