So am I right in assuming that if you bred a lavender hen to a lavender roo, you would get all lavender offspring? Or would it produce something else as when you breed two blues together and get a mix of colors from the offspring?
Before one can have a lavender bird, the bird must first be black. The two E locus alleles (genes) that are normally used to make a bird black are extended black (E) and birchen (E^R). The birchen allele needs some help to make a self black bird so they will have genes like melanotic or other black extenders to make them black. The autosomal recessive gene called lavender (lav)dilutes the black to a blue color. Birds that are self blue must be able to be produced by crossing a blue bird to a blue bird which is true for the lavender phenotype. The blue gene ( Bl) is an autosomal incompletely dominant gene that also dilutes black to blue. It takes two recessive genes (lav/lav) in a bird to produce the lavender phenotype but only one blue gene to make a blue bird. When a bird inherits two blue genes (Bl/Bl), both blue genes work together to dilute the black to a splash color. You can not cross two blue birds and get 100 % blue offspring therefore blue are not called self blue. Blue X blue = a 1:2:1 phenotypic ratio or 1 black (bl+/bl+) to 2 blue (Bl/bl+) to 1 splash ( Bl/Bl)
A single blue gene sometimes will cause a bird to look black. For some reason the blue gene does not dilute the black. You can get different shades of blue in blue birds also. Lots to be learned about the penetrance problem ( different shades of blue) associated with the blue gene.
The genotype of a blue bird would be like E/E, Ml/Ml, S/_ or S/S, Bl/bl+.
A lavender bird would be like E/E, Ml/Ml, S/_ or S/S, lav/lav.
Non laced blue birds would have the same genotype as a black bird with the exception of the blue gene (Bl). Blue birds that are laced with black would also have the pattern gene (Pg) and may or may not have gene called columbian (Co).