The gene combination for a chicken with black and white feathers typically involves the presence of multiple alleles, particularly those related to the auto-somal dominant gene for feather color. In many cases, the black feather color can be attributed to the dominant allele "B," while white feathers may result from the presence of the recessive allele "b." A chicken with a genotype such as "BB" or "Bb" could express black feathers, while a genotype of "bb" would typically lead to white feathers. However, the specific interactions may vary depending on other genetic factors and modifiers.
An Erminette chicken is a chicken with a gene for black feathers and a gene for white feathers. Since the genes are co-dominant, the Erminette chicken has black and white feathers, rather than one or the other or grey.
The offspring will likely be a mix of black and white feathers, as erminette coloring is a pattern that combines both black and white. The specific distribution of colors will depend on the genetics of the parent chickens.
The genetics of chicken color involves more than one gene. There are the genes for the down and genes for the barbed feathers. There are enhancers and diluters of color, barring and mottling and all numbers and different types of genes involved in the final coloration of a chicken. It would be very difficult to get a predominately homozygous recessive white chicken and not have any leakage. The White Leghorn is a dominant white chicken, and it even has color leakage problems. There's just too many variables as chicken coloration is more than just a black hen versus a red hen with an occasional white one thrown in the mix.
Allele is the term that describes the form of a gene that produces a specific trait such as long feathers or short feathers. Traits are passed in discrete units from parents to there offspring.
genetics
It is not just one gene that causes the color of eyes. Eye color genetics is complicated and it takes many genes to create an eye color. Regardless of the parents' and children's eye color any combination can occur even for the same eye color.
An erminette chicken would carry both a black gene and a white gene. The erminette chicken represents the heterozygous phenotype.
The probability is 0 (but the daughter will be a carrier of the color blind gene). This is because the gene dictating whether someone is color blind or not is linked to the X chromosome (and not the Y). The color blind gene is a recessive gene whilst the normal color vision gene is a dominant gene. Hence if a girl (XX) has one normal vision gene (from one parent) and one color blind gene (from the other parent), her normal vision gene will be dominant to the recessive color blind gene and hence she will have normal vision (but she will be a carrier of the color blind gene). If both her parents contribute the recessive color blind gene to her, then she will be color blind. For a woman (XX) to be color blind, she needs to be have both genes to be recessive (ie where there is no dominant normal color vision gene to dominate). For a man (XY), as long as the X gene contributed by his mother is a color blind gene, he will be color blind because he has no other X chromosome where a dominant normal color gene could reside. Hence, to answer the question, a man with normal color vision (XY, with a dominant normal color vision X gene since the gene can't be the recessive color blind gene otherwise he will be colorblind) and a colorblind woman (XX, both recessive color blind genes), will each contribute an X each the child. The man will contribute his only X chromosome which carries the normal color vision X gene and the woman can only contribute a recessive color blind gene. The man's normal color vision X gene will be dominant, and hence the daughter will definitely have normal vision (despite being a carrier).
The gene will code for a different protein than it should.
It is a coat color gene in animals that is responsible for the deep orange color as in tigers.
The base color is black/red factors, controlled by the Extension gene. A horse with one or two copies of the Extension gene (Ee or EE) will have black factors. A horse with no copies of the Extension gene (ee) will be chestnut.The Agouti gene acts on the Extension gene by limiting where the black factors are displayed, into points such as the legs, mane, tail, muzzle, and ears. A horse with at least one copy of the Extension gene that doesn't not have a copy of the Agouti gene will be black, provided that no other gene modify the color. One or two copies of the Agouti gene in combination with the Extension gene will result in a bay horse. A chestnut horse (no copies of the Extension gene) can have copies of the Agouti gene, but it will not affect the horse's color, as the horse has no black factors for the Agouti gene to limit.