[ SHOWGSD-L ] Re: ~~Black bred to a Black question

  • From: "Carolyn" <marhaven@xxxxxxxxxxxxxx>
  • To: <rbuffdogs@xxxxxxxxx>, <Erynbrook@xxxxxxx>, <showgsd-l@xxxxxxxxxxxxx>
  • Date: Mon, 21 Jan 2008 23:01:46 -0800

I guess I have no idea what color would "appear"  to be black and is not in 
fact black??
I've never seen a bi color mistaken to be a black?
A bi color is very defined in it's markings and that is actually what makes 
them a bi color.
There is no 'shading' or blending or sprinkling of color.....it is literally a 
two color dog.....the 
black, and the defined markings?    At least that has always been my 
understanding.

Carolyn    marhaven@xxxxxxxxxxxxxx
www.Marhaven.com
  ----- Original Message ----- 
  From: Kathy Tank To: Erynbrook@xxxxxxx ; showgsd-l@xxxxxxxxxxxxx Cc: 
VSabiaB@xxxxxxx 
  Sent: Monday, January 21, 2008 6:31 PMSubject: [ SHOWGSD-L ] Re: ~~Black bred 
to a Black question


  We may be forgetting however that what "appears" black, may not infact be a
  genetic black but a very very dark bi color.  So that opens up the color 
question.
  Kathy

  On 21/01/2008, Erynbrook@xxxxxxx <Erynbrook@xxxxxxx> wrote:
  >
  > Black bred to black can only produce black. Think it through: Since> black  
is
  > the most recessive color in GSDs, to BE a black a dog must have two  black
  > genes. If he had a black gene & a b/t gene, he would BE a  b/t,  if he had
  > a> black gene and a sable gene he would BE a sable,  because black is
  > recessive to> all other recognized colors.
  >
  > So now we know that both black dogs in a black-to-black breeding have only
  > black genes to contribute to the litter. What other color could the pups
  > be but> black? There are simply no other color genes in play!
  >

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