What is the mechanism for supF selection in E. coli carrying the P3 plasmid? And what is meant by the ''amber'' notation in the genotype - are these cells resistant to tetracycline and ampicillin?
E. coli harboring the plasmid P3 enable the selection and maintenance of plasmids that encode the tRNA suppressor F gene (supF). P3, a low-copy 60 kb episomal plasmid, encodes the kanamycin resistance gene as well as amber mutants of the tetracyline and ampicillin resistance genes. The amber mutations are point mutations in the resistance markers which inactivate the expression of resistance in the strain unless the tRNA supressor F (supF) is present. Therefore, strains that harbor P3 alone are resistant to kanamycin, but sensitive to both tetracycline and ampicillin. But when the E. coli carrying the P3 plasmid are transformed with supF-containing plasmids (e.g. pcDNA1 and pCDM8), they are rendered resistant to both tetracycline and ampicillin (as well as kanamycin) by suppression of the amber mutations.
The rate of spontaneous reversion of the amber point mutations on the P3 episome is fairly high, so it is important to select supF clones with both tetracycline and ampicillin resistance to reduce background growth. To avoid enriching for revertants, it is recommended that relatively low concentrations of tetracycline (7.5 to 10 ug/ml) and ampicillin (25 to 40 ug/ml) be used.