Silkied D’Uccles Project

[JacinLarkwell]

How will all of these chicks be mottled if both parents only have 1 mottled gene?

 

[WVduckchick]

Neat!

 

[Egg Snatcher]

Thanks for the tag! Following!
Cute little chickies!

Glad to have you!

Following! Those chicks look wonderful and I love the idea. I can't wait to see what they look like as they grow.

glad to have you!

How will all of these chicks be mottled if both parents only have 1 mottled gene?

The parents of these chicks (F1) both carry the mottleing gene because their mother, a Millie Fleur, carries the mottled gene (Millie Fleurs show the mottleing bc they are 2 gened mottled). She gave each chick a mottling gene and a vulture hock gene when I breed her to the silkie, the silkie passed on the silkie feather gene to all the chicks. Because these 3 genes are recessive they do not show with out a second copy. Breeding the F1s together will make two copies of the mottled gene, silkie feather gene and vulture hock gene, so the chicks i have righ know will show the mottled gene, silkie feather gene and vulture hock gene.
Long story short. It takes to copies for a recessive gene to show. Both the F1s carry the mottled gene so these chicks (F2) will be mottled.

 

[pipdzipdnreadytogo]

Just a small correction, but if the parents of these chicks are your f1 Silkie / d'Uccle mixes and thus only carry one copy of the mottling gene from the d'Uccle side of the mix, then only about one in four of these chicks should express mottling, not all of them. The parents in this case should both be Mo+/mo, which means that the pattern of inheritance from breeding them together is that their chicks can inherit Mo+ from both parents, Mo+ from the father and mo from the mother, mo from the father and Mo+ from the mother, or mo from both parents. That means that about 1 in 4 will not be mottled and not carry the gene (Mo+/Mo+), 2 in 4 will not be mottled but carry the gene (Mo+/mo), and 1 in 4 will be mottled with two copies of the gene (mo/mo).

Same goes for the silkied feathering gene and the vulture hock gene. If both parents are H+/h for silkied feathering, then the offspring will be about 25% not silkied and not carrying the gene, 50% not silkied but carrying the gene, and 25% silkied. And if both parents are V+/v for vulture hocks, then you get the same ratio of 25% without the gene, 50% heterozygous for it, 25% homozygous with the gene.

The only way to be sure your chicks will express a recessive trait is if both parents also express it. The parents just carrying the gene for it means that some offspring may express it and others won't.

Unless I'm misunderstanding what you're saying, in which case disregard!

 

[my spoiled chickens]

OMG! TWO OF MY FAVE BREEDS TOGETHER!!!! Silkies and d'uccles are right up there with OEGB's and Sebrights

 

[JacinLarkwell]

Glad to have you!

glad to have you!

The parents of these chicks (F1) both carry the mottleing gene because their mother, a Millie Fleur, carries the mottled gene (Millie Fleurs show the mottleing bc they are 2 gened mottled). She gave each chick a mottling gene and a vulture hock gene when I breed her to the silkie, the silkie passed on the silkie feather gene to all the chicks. Because these 3 genes are recessive they do not show with out a second copy. Breeding the F1s together will make two copies of the mottled gene, silkie feather gene and vulture hock gene, so the chicks i have righ know will show the mottled gene, silkie feather gene and vulture hock gene.
Long story short. It takes to copies for a recessive gene to show. Both the F1s carry the mottled gene so these chicks (F2) will be mottled.

Yeah, but Mm x Mm will only make 25% mottled. 50% will carry but not show and 25% won't carry it at all

 

[Egg Snatcher]

Yeah, but Mm x Mm will only make 25% mottled. 50% will carry but not show and 25% won't carry it at all

Yes, I just didn’t tell the percent of them. I will fix this in my first post. Thanks

 

[Egg Snatcher]

Just a small correction, but if the parents of these chicks are your f1 Silkie / d'Uccle mixes and thus only carry one copy of the mottling gene from the d'Uccle side of the mix, then only about one in four of these chicks should express mottling, not all of them. The parents in this case should both be Mo+/mo, which means that the pattern of inheritance from breeding them together is that their chicks can inherit Mo+ from both parents, Mo+ from the father and mo from the mother, mo from the father and Mo+ from the mother, or mo from both parents. That means that about 1 in 4 will not be mottled and not carry the gene (Mo+/Mo+), 2 in 4 will not be mottled but carry the gene (Mo+/mo), and 1 in 4 will be mottled with two copies of the gene (mo/mo).

Same goes for the silkied feathering gene and the vulture hock gene. If both parents are H+/h for silkied feathering, then the offspring will be about 25% not silkied and not carrying the gene, 50% not silkied but carrying the gene, and 25% silkied. And if both parents are V+/v for vulture hocks, then you get the same ratio of 25% without the gene, 50% heterozygous for it, 25% homozygous with the gene.

The only way to be sure your chicks will express a recessive trait is if both parents also express it. The parents just carrying the gene for it means that some offspring may express it and others won't.

Unless I'm misunderstanding what you're saying, in which case disregard!

Yes you are correct. Thanks for adding that!

 

[Egg Snatcher]

Ok here is the plan with the F2 chicks….
Keep mottled and silkied feathered pullets. Breed the pullets to a Millie Fleur next year and continue selecting for D’Uccles traits until I get D’Uccles that are silkied.
Next year when breeding the pullets I hope to get rid of black skin a crest. I know it can be hard to get rid of the 5 toe gene but I’m gonna try my best.

 

[NagemTX]

Following!