Olive Egger with Cream Legbar

Mylied

Crowing
9 Years
Mar 12, 2012
3,507
5,790
492
Middle Georgia
I have a couple olive egger pullets, blue colored, in with my cream legbars (hens and a roo) and I hatched a few eggs from them. They hatched these last couple days. The three chicks came out looking like a barred rock chick. They are the three black ones with yellow spots. They have yellow bellies and wing tips too. That seems so strange to me? Anyone want to give me a quick lesson in genetics?
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Blue olive eggers + cream legbar = barred black or blue.

Blue Olive Eggers =
Blue = (Bl/bl), or a splash and a black gene, which are co-dominant. Splash dilutes black.
Solid = (E/?) E is a dominant pattern gene that should make its carriers all of a single colour.
No barring (b/-)*

Cream Legbar (Roo) =
Black = (bl/bl) Two copies of black, which makes the chicks black.
Duckwing pattern = (e+/e+) which makes the black color of the chicks only show in certain spots. Recessive.
Barred = (B/B) Because he's a purebred legbar rooster, he carries two copies of Barred.

Crossing ensures that the chicks inherit two of each of these genes, one from each parent.
From Mum, they can get (Bl or bl), and (E or ?) and (b or -)
From Dad, they can get (bl or bl), and (e+ or e+) and (B or B) (choices, choices, right?)

So half of the chicks will be black (bl/bl) and half of them will be blue (Bl/bl)
Half of the chicks will be solid (E/e+), and half of them will have the mysterious other gene, which is probably E (?/e+)
All of the chicks will have a single copy of the barring gene from Dad. The cockerels will be (B/b) the pullets will be (B/-), which pretty much comes out to the same thing.

So you get solid(?) barred black or blue chicks

*"-" is, in fact, the absence of a gene. Because the b is stored on the male chromosome, the chicks that inherit it from Mum will be male.

Hope this helps!
 
Blue olive eggers + cream legbar = barred black or blue.

Blue Olive Eggers =
Blue = (Bl/bl), or a splash and a black gene, which are co-dominant. Splash dilutes black.
Solid = (E/?) E is a dominant pattern gene that should make its carriers all of a single colour.
No barring (b/-)*

Cream Legbar (Roo) =
Black = (bl/bl) Two copies of black, which makes the chicks black.
Duckwing pattern = (e+/e+) which makes the black color of the chicks only show in certain spots. Recessive.
Barred = (B/B) Because he's a purebred legbar rooster, he carries two copies of Barred.

Crossing ensures that the chicks inherit two of each of these genes, one from each parent.
From Mum, they can get (Bl or bl), and (E or ?) and (b or -)
From Dad, they can get (bl or bl), and (e+ or e+) and (B or B) (choices, choices, right?)

So half of the chicks will be black (bl/bl) and half of them will be blue (Bl/bl)
Half of the chicks will be solid (E/e+), and half of them will have the mysterious other gene, which is probably E (?/e+)
All of the chicks will have a single copy of the barring gene from Dad. The cockerels will be (B/b) the pullets will be (B/-), which pretty much comes out to the same thing.

So you get solid(?) barred black or blue chicks

*"-" is, in fact, the absence of a gene. Because the b is stored on the male chromosome, the chicks that inherit it from Mum will be male.

Hope this helps!
Thanks so much! So is there a way to tell the male from the female chicks? I'm planning on selling them and I don't want anyone mad if I sell them unsexed but there's a clear difference.
 

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