Cost and colors...

[AquaEyes]

Quote:

 

[arherp]

Ok. That wasn't the kind of mix I was referring to. I actually raise Muscovies in those colors. What I am talking about is if you breed color "X" to white recessive, all babies come out looking like a sunbleached version of X. When bred together one quarter of the offspring will be rec. white, the other three quarters will be the crappy X again. I show waterfowl, and some of these things are vary small variations, they are all taken into account when put on a show bird.

The examples I think you were showing was bird X is bred to bird Y and produces bird XY. I am just referring to maintaining the same shades or qualities of a color, even with certain mixed pairings.

 

[AquaEyes]

Quote:
OK, well, I'll offer a possible explanation....not sure if this is why in this specific case, but it makes sense biologically.

Perhaps the gene in which a mutation causes "recessive white" is expressed in both copies. What I mean is that each copy of the gene on the two chromosomes makes its protein product, contributing to the depth of pigment in the bird. In a "wild-type" color, that is more than enough protein product to give the full depth of color, and in fact, only one copy of the gene is necessary to make enough pigment for full coloring. When a bird has a dilution gene, the total protein product is reduced, and the pigment is lighter. If one of those genes is knocked out (in other words, the bird is het for recessive white), now you have less protein product being made, combined with some of what IS made being bound up by the dilution gene. There wasn't a difference noticed in the wild-type because one copy of the gene was still more than enough to make the amount of protein required for "wild-type" coloration, and there was no dilution gene binding any of it up. But when another gene "binds up" some of that product, now you notice a difference in birds with two "protein makers" (i.e. copies of the normal gene) versus those with only one (i.e. het for white).

I can explain it mathematically.....

For full color, you need 100 units of pigment. Each copy of the gene makes up to 120 units by itself. Birds with one copy can thus still meet the 100 units of pigment requirement, and the excess just gets reabsorbed (or the mechanism shuts down before reaching maximum).

Then you add a dilution gene, which produces a protein that eats up 30% of however much pigment is floating around. Birds with just one functioning copy of the normal pigment gene (i.e. het for white) will have less net pigment produced than birds with two functioning copies of the normal pigment gene (i.e. not carrying the recessive white gene). The result is that diluted birds het for white will be a touch lighter than diluted birds NOT het for white.

It's a model of how it could work. If that's for sure how it works IN THIS CASE I do not know. But the mechanism I described is real in other cases.

ETA -- or a simpler, slightly different explanation. There have been studies that show that there actually IS a difference in appearance of creatures that are heterozygous for a recessive dilution or white mutation as compared to creatures that don't have such a gene. I'll give another example -- black cats that are heterozygous for siamese-albinism might look (to our eyes) just as "black" as those cats that aren't het for siamese-albinism. But there is a difference -- it's just that we can't distinguish it with our eyes. I remember reading about studies that measured the actual quantity of pigment in hairs from homozygous black cats and compared it to black cats het for siam-alb. There was less pigment quantity on a chemical basis measured in the hets, but the difference wasn't enough to be distinct to our eyes. Perhaps the addition of another dilution gene brings the amount of pigment down enough that we are able to distinguish the difference.

 

[Arbor]

I have several offspring from an opal male x blue hen split white. Oddly enough, most offspring show the odd white flight feather. I've also had the same when this hen was bred to a bronze male. All offspring, however, do not look much different from a standard blue. That said, I do have two young males from the opal x blue split white that look different from eachother. The one looks like a regular blue and does not show any white flights. The other, with the white flights has a lot less barring on the feathers. I also have a young male that is bronze split IB blackshoulder. His barring is excactly the same as the regular blue and no traces of darker flights is visible.

 

[deerman]

Quote:
The chicks from the blue hen split white, not all will be split white. So you should get some without white flights(not split white)


Now the chicks from the bronze X blue hen and Opal X blue hen.......will all look alike, blue. Only good records will you be able to tell which ones are split opal or bronze.

Yes those split blackshoulder will sometime show a small area on the wing that is not barred.

note small area on this, purple peacock ,he is split to solid wing(blackshoulder). BS in other colors really don't have BLACK shoulder, brown in cameo ,peach, purple, and silver/grey in opal ,charcoal and etc.

 

[arherp]

Quote:
OK, well, I'll offer a possible explanation....not sure if this is why in this specific case, but it makes sense biologically.

Perhaps the gene in which a mutation causes "recessive white" is expressed in both copies. What I mean is that each copy of the gene on the two chromosomes makes its protein product, contributing to the depth of pigment in the bird. In a "wild-type" color, that is more than enough protein product to give the full depth of color, and in fact, only one copy of the gene is necessary to make enough pigment for full coloring. When a bird has a dilution gene, the total protein product is reduced, and the pigment is lighter. If one of those genes is knocked out (in other words, the bird is het for recessive white), now you have less protein product being made, combined with some of what IS made being bound up by the dilution gene. There wasn't a difference noticed in the wild-type because one copy of the gene was still more than enough to make the amount of protein required for "wild-type" coloration, and there was no dilution gene binding any of it up. But when another gene "binds up" some of that product, now you notice a difference in birds with two "protein makers" (i.e. copies of the normal gene) versus those with only one (i.e. het for white).

In a het to het breeding though, only 66% of the offspring produced will be het. The non-het birds still exhibit the same lightness I am talking about, and it usually takes around four years to breed out.

I can explain it mathematically.....

For full color, you need 100 units of pigment. Each copy of the gene makes up to 120 units by itself. Birds with one copy can thus still meet the 100 units of pigment requirement, and the excess just gets reabsorbed (or the mechanism shuts down before reaching maximum).

Then you add a dilution gene, which produces a protein that eats up 30% of however much pigment is floating around. Birds with just one functioning copy of the normal pigment gene (i.e. het for white) will have less net pigment produced than birds with two functioning copies of the normal pigment gene (i.e. not carrying the recessive white gene). The result is that diluted birds het for white will be a touch lighter than diluted birds NOT het for white.

It's a model of how it could work. If that's for sure how it works IN THIS CASE I do not know. But the mechanism I described is real in other cases.

ETA -- or a simpler, slightly different explanation. There have been studies that show that there actually IS a difference in appearance of creatures that are heterozygous for a recessive dilution or white mutation as compared to creatures that don't have such a gene. I'll give another example -- black cats that are heterozygous for siamese-albinism might look (to our eyes) just as "black" as those cats that aren't het for siamese-albinism. But there is a difference -- it's just that we can't distinguish it with our eyes. I remember reading about studies that measured the actual quantity of pigment in hairs from homozygous black cats and compared it to black cats het for siam-alb. There was less pigment quantity on a chemical basis measured in the hets, but the difference wasn't enough to be distinct to our eyes. Perhaps the addition of another dilution gene brings the amount of pigment down enough that we are able to distinguish the difference.

 

[arherp]

Quote:
The chicks from the blue hen split white, not all will be split white. So you should get some without white flights(not split white)


Now the chicks from the bronze X blue hen and Opal X blue hen.......will all look alike, blue. Only good records will you be able to tell which ones are split opal or bronze.

Yes those split blackshoulder will sometime show a small area on the wing that is not barred.

note small area on this, purple peacock ,he is split to solid wing(blackshoulder). BS in other colors really don't have BLACK shoulder, brown in cameo ,peach, purple, and silver/grey in opal ,charcoal and etc.

http://i205.photobucket.com/albums/bb128/cdeerman/birds002.jpg

Thanks for that bit, it has helped me a lot.

 

[AquaEyes]

Quote:
Hmmm...I'm not sure I understand what's going on. My understanding is that Aa X Aa = 50% Aa (het X het = 50% het). Do you know for sure the non-het birds are non-het? Has this been confirmed through test breeding? Is it possible that additional diluting genes are at play?

 

[arherp]

Sorry, I call them 66% hets... it refers to the three normal colored birds out of a four clutch, het to het breeding. two will be het, one will not be.
50% of the offspring will be het.

 

[AquaEyes]

Quote:
Ah, of course...I understand...they are the non-whites with a 2/3 chance of being split to white.