breeding snafu? Help! - Page 6

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by eichenluft on 02 September 2009 - 16:09

I think I'll stick to the simple GSD color gene genetics (with their variations within each color) and leave the chapters of mouse color genetics and mutations to you -

molly

darylehret

by darylehret on 02 September 2009 - 20:09

It wasn't about that, I don't really care about coat color so much.  Don't know diddly about whites, livers, blues, or pandas, nor do I intend to, unless it can teach me some related knowledge about something I find more important, like in character and drives or something useful that I can apply.   I just don't think you handle it well when someone disagrees with you.  Maybe I don't either, but I didn't make this stuff up, and I'm not too stubborn to think science couldn't be wrong, either.  It's a complex world, and oversimplifying things is human nature, but be prepared to be occasionally surprised.  "Never say never", no matter how much you think you know.

by AnjaBlue on 02 September 2009 - 21:09

Color Genetics for Dummies (like me.) Thank you Molly, very clearly explained!

by eichenluft on 03 September 2009 - 03:09

I'd like to ask the mouse color expert if black is recessive in mice as it is in GSDs, and sable is dominant?  Or is spotted/white dominant?  Is any color in mice dominant?  How are the color genes in mice passed on to the offspring?  In GSDs the parents each carry two, each puppy gets one from each parent.  Variations within each color can depend on which recessive gene the pup inherits (or doesn't inherit) - so how can it be even remotely similar when mice carry many more colors, and variations of each color including spots and other markings? 
even in dogs, some breeds are dominant black, like labs.  Or let's pick a breed that is always black and never any other color - dominant black - Patterdale Terriers?  Or, flat coated retrievers?  Newfoundlands - then they come in white spotted "Landseers" but black is dominant, right?  So, they are not related in color genetics to GSDs, yet they are canine, closer in genetics to GSDs than mice, right?

Heres another one I was thinking about today.  Different breed - Belgian Shepherds.  Originally all "the same breed" with different coat types and colors.  Malinois and Lakenois are always (nearly always?) fawn with black points.  With variations of the fawn color and black points - some a bit darker, some a bit lighter.  But they are <almost> never black, or dark mahogany like Belgian Sheepdogs (long coated solid black always) and Tervurans (long coated deep mahogany sable always) - how come Tervs aren't usually fawn (I've never seen a fawn Terv) - and how come Belgian Sheepdogs are always black and long-coated?  They are all the same base breed right?  And within a litter of each can come a puppy of the other - two Malinois parents (fawn, short-coated) can produce a Tervuran (long coat mahogany sable) or Sheepdog (long coat black) or Lakenois (fawn wire-coat).  And then you can pursue the thought that each "breed" isn't different just because of the coat type and color - even when born in the same litter or with similar bloodlines - the temperament, drives and nerve strengths are also quite different with each, making each an individual breed, even though they are still considered in some countries to be the same breed, with different coat colors and types! 
And where did that Brindle coat on the Dutch Shepherd come from?  Another  Interesting, that.

I'll keep sticking to the simple GSD color genetics - two colors per parent, one gene given from each to each pup, dominant color always rules, recessive influences the variation and offspring of each pup.   See how simple that can be?  One sentence and it's done.  Unless someone can produce a sable dog proven to be born from two non-sable parents?  None yet..... not holding my breath.

molly


darylehret

by darylehret on 03 September 2009 - 04:09

The order of dominance in mice is the same, for the same series, of the same chromosome locus.  As you know, other loci can affect its expression.  For example the masking gene that will produce the dark muzzle phenotype is NOT controlled from the agouti locus.  Dominant black in other canine breeds is NOT controlled from the agouti locus.

At ANY Chromosome location, there are two alleles, and the offspring inherit one copy from each parent.  This is not unique to german shepherds, but to all heterosexually producing animals.  If you can't understand this fundamental principle, I can't help you.

There are many other sources for information of other canine colors, but this thread was pertaining to color patterns which are controlled from the agouti locus.  So for brevety, let's stay relevant.

Each of the agouti series has evolved (by Loss Of Function mutation) from the original wild-type (the Aw allele), and reverse mutations have been documented to occur, as well as intentionally reproduced results in a lab.  What is the difficulty in "getting" that?

Are you so stubborn, to notice I was challenging not only your own statement, but my own as well?  What may be true most of the time, just isn't true all of the time.  So get over it.

"THERE IS NO, ZERO, ZILCH, NADA, NEVER IN A MILLION TRILLION BAZILLION CHANCES IN THE DEPTHS OF HELL that this sable puppy - if it's indeed a sable - came from that sire. NO CHANCE."

Sounds so overzealous and opinionated, doesn't it?


by eichenluft on 03 September 2009 - 06:09

But I'm right.  So how is that opinionated when what I've posted is fact, not opinion - and proven fact.  in FACT, I rarely post anything that is not fact... - prove I'm wrong.  Give me a sable dog that is proven to be produced from two non-sable parents.

molly

darylehret

by darylehret on 03 September 2009 - 12:09

Oh, I proved something alright.  What you may consider as the basis of proof (dna analysis) isn't perfect either.  Epigenetics is a fast growing field of study for factors that affect hereditary functions outside what dna perscribes.  It has been proven that histones permanently activate or deactivate gene expression via a process known as "DNA methylation."  These small chemical groups within the cells latch on to areas along the dna strand and result in changes that can be passed hereditarily for generations, without any alteration to the dna structure.

Just how many gsd's do you think have been dna tested for proof ofparentage in the last hundred years?  A small handful, I'd bet.  How many times have you done it?  Where would one find that kind of information on record?  Give me a b&t produced from to bicolors while you're at it (with "proof").  Demonstrate how factual your "tagged gene" theory can be.

by eichenluft on 03 September 2009 - 13:09

ask a german on this list what their theory of bicolors are - why is it not a color in germany?  It is considered black/tan.  I consider it a black/tan with a melanistic gene "tag" (my own descriptive word).  I don't think a black/tan can be produced from two bicolors - the melanistic gene would cover both genes given to the pups, making 100% bicolors or solid blacks from two bicolor matings.

The  point of this thread is it possible for a sable to be produced from two non-sable parents.  The simple answer is no.  You make things much too complicated.

molly

darylehret

by darylehret on 03 September 2009 - 13:09

Again, YES, it is possible, and has at least been proven in mice if not dogs.  What is significant, is that it is due to a reverse mutation from the same locus that controls coat color pattern in the gsd breed.

by eichenluft on 03 September 2009 - 14:09

Again, NO it is not possible in GSDs.  Never proven to be possible.  Many many times proven to not be possible.  If there is a sable puppy from two non-sable parents, it is certain that the puppy will be found to have a sable oops breeding (or intentional perhaps) parent. 

Unless the owner of the sable pup wants to assume that his pup is the only GSD in the history of the breed who is the "mutation" Daryl speaks of.  I certainly wouldn't.

molly





 


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