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Influence of pigments (breast coloring) on the general plumage phenotype

 

 

Two types of melanin are present in the feathers of green colored birds, black eumelanin and brown-red pheomelanin. Both influence the final feather coloring or its shade. The loss of one of these pigments, partial or complete, will express on the entire plumage. Pheomelanin acts as a darkening factor.

 

                                                     Reduction of pheomelanin (is not sex linked)

 

2 factor – (white-breasted birds) – the absence of pheomelanin will show e.g. in the coloring of the head, where a significant loss of pigmentation occurs in the middle section of a feather. Complete (double-factor) loss of pheomelanin is required to breed lutino mutations.

 

1 factor – loss of pheomelanin (pink-breasted birds) – a partial fading of feather color occurs. The lack of a single factor of pheomelanin means that one factor still remains, so a certain amount of pheomelanin is preserved, which enables the existence of blue-breasted birds, but its levels are so low, that its own coloring is insufficient to influence the structural coloring and the purple breast coloring does not occur.

 

Presence or absence of pheomelanin, either single or double-factor always concerns the plumage coloring of the entire body, not only the breast-coloring. In white-breasted birds pheomelanin is also absent in the head feather coloring. In purple-breasted birds it is preserved (as per head coloring chapter).

                                                      Reduction of eumelanin (is sex linked)

 

 

1 factor – a partial fading of feather color occurs. In purple-breasted birds (where darkening factor pheomelanin is present) the phenotype appears as: 1f yellow purple-breasted males – green pastel (these birds have the collar and tailbone color ca. 30% lighter than green birds). In white-breasted birds the phenotype appears as: 1f yellow males – gold (these birds have the collar and tailbone color ca. 50% lighter than green birds)

 

2 factor – a more significant fading of feather color occurs. In purple-breasted birds (where darkening factor pheomelanin is present) the phenotype appears as: 2f yellow males – gold (these birds have the collar and tailbone color ca. 50% lighter than green birds). In white-breasted birds the phenotype appears as: 2f yellow males – practically yellow (these birds have white/off-white collar and tailbone color)

 

Presence or absence of eumelanin, either single or double-factor always concerns the plumage coloring of the entire body not only breast coloring. On the example of breast feathers from green and blue purple-breasted as well as white-breasted birds we see that bottom segments (1) of the feathers contain little amounts of eumelanin, which however does not influence the phenotype of breast coloring.

In yellow and silver birds the presence of eumelanin is not visible in this area (1), neither in purple-breasted nor white-breasted birds. At the same time the intensity of breast coloring is the same in yellow, green, silver and blue birds. This is clear proof that eumelanin and pheomelanin are not linked to each other in any way and are present in the entire plumage.

  Females are phenotypically alike 2 factor males.

 

 

Influence of breast coloring on plumage phenotype

 

 

As pheomelanin acts as a darkening factor, in birds with the same genetic dispositions (back color) a changed phenotype will show. So either its presence darkens or its absence lightens the appearance of the feathers.

 

 

Difference in feather coloring and structure in males and females

 

 

Similarly as in the structure of head feathers there are differences between the sexes in the feathers of the breast. Every breeder surely noticed that females have a less distinct and somewhat matte purple breast compared to males. The basis for this surely lies in the species’ phylogeny where it is at least partially necessary that the females are less prominently colored.

 

If we compare the feathers of males and females we find that the barbs of the feather in its upper segment (3) are significantly longer in males as in females. It is in these parts where in the medulla due to interference the optic structural color is created. In the females the structural color is less prominent, as the barbs are shorter and the feather coloring is less distinct as in males. The feathers of females, as they show less blue structural iridescence are more influenced by the color of pheomelanin. Similar observation can be made for the barb length on pictures of white-breasted birds. As pheomelanin is absent in these birds, the medulla is not activated and no structural coloring is present. The plumage is neither influenced by structural coloring nor by the color of the pigment and it appears white. Some little amount of pheomelanin remains however and a light blue shade may be observed in the tips of the barbs (2) of white-breasted birds. When observing in breeds this shade is only visible under certain lighting conditions.

 

Breast coloring

 

As it is implied by the above statements breast coloring comprises pigment color (brown-red pheomelanin) and optic structural color (blue), which is produced by means of interference in the medulla of the feather’s barb.

 

 

Inheritance of breast color

 

 

The alleles (different forms of a gene) for breast color are localized on autosomal chromosomes. That means that breast color is not sex linked and is inherited freely, in the same way for both males and females.

 

♂ males as well as ♀ females may be single and double-factor for the specified trait:

 

Purple PP: dominant.

 

Lilac ww: recessive to purple, but dominant over white.

 

White pp: recessive to both purple and lilac.

 

Lilac breast color may range from almost white to almost purple. Lilac-breasted birds are better left out from a breed, or the breeder might eliminate white-breasted birds by their usage.

 

 

Breeding examples

 

 

1^st breeding – white-breasted × purple-breasted

 

 

The genotype of a bird with white breast color must be pp. Let’s assume that the other parent is purple-breasted homozygous (non-splitting), so its genotype will be PP. We must keep in mind that gametes (ovum or sperm) only carry one allele from the pair. All of them that stem from the white-breasted bird will carry allele p. All of them that stem from the purple-breasted bird will carry allele P. In this case the result of fertilization will always be the same. If the gametes of the first parent (naturally colored) – P fuse with those of the second one (white-breasted) – p, the resulting zygotes will have a genotype of Pp. Since P is dominant over p the phenotype of the first generation of offspring will be 100% purple-breasted, splitting to white-breasted.

 

2^nd breeding – white-breasted × purple-breasted, splitting to white-breasted (as in breeding no. 1)

 

The parents in this case will have the following genotypes: white-breasted parent will be pp. All of its gametes will carry a single p allele. The genotype of the parent, which is purple-breasted, splitting to white-breasted is Pp. In this case 50% of the gametes will carry allele P and the other 50% will carry allele p. Two results are possible from this breeding:

1.)  The gamete carrying p may fuse with another one carrying p, the result will be a zygote with a pp genotype. This bird will be a white-breasted specimen.

2.)  The gamete carrying P will fuse with one carrying p. This offspring will have a Pp genotype and will be purple-breasted, splitting to white breast color. In other words, the result of this breeding will be 50% white-breasted young and 50% purple-breasted, splitting to white-breasted.

 

 

 

3^rd breeding – purple-breasted, splitting to white-breasted × purple-breasted, splitting to white-breasted

 

 

 

Both parents will have genotype Pp, so 50% of the gametes from both parents will carry allele P and 50% allele p. We get four different options for merging gametes. The possible results are given in the table.

 

 

 

In other words, genotypes of the offspring are: 1 PP : 2 Pp : 1 pp (or 25% : 50% : 25%).

The phenotypical ratio is 3:1, meaning 75% purple-breasted (comprising 50% splitting to white-breasted and 25% homozygous purple-breasted) and 25% white breasted (pp).

 

 

 

4^th breeding – purple-breasted × lilac-breasted

 

 

The genotype of a lilac-breasted bird is ww. Let’s assume, that the purple-breasted bird is homozygous (non-splitting), so its genotype will be PP. We must keep in mind that gametes (egg or sperm) only carry one allele from the pair. All of them that stem from the lilac-breasted bird will carry allele w. All of them that stem from the purple-breasted bird will carry allele P. In this case the result of fertilization will always be the same. If the gametes of the first parent (naturally colored) – P fuse with those of the second one (lilac-breasted) – w, the resulting zygotes will have a genotype of Pw. Since P is dominant over w the phenotype of the first generation of offspring will be 100% purple-breasted, splitting to lilac-breasted.

 

 

5^th breeding – purple-breasted, splitting to white-breasted × purple-breasted, splitting to lilac-breasted

 

 

 

The parents will have a genotype of Pp and Pw. The possible results of offspring are given in the table.

 

 

 

Phenotypical ratio is 3 : 1,

meaning 75% purple-breasted (comprising 25% purple-breasted, splitting to white-breasted, 25% purple-breasted, splitting to lilac-breasted and 25% homozygous purple-breasted) and 25% lilac-breasted, splitting to white-breasted.