Breeding Policy

Color and Pattern

Here at Ausdauer, we breed dogs in the most realistic way possible, giving all dogs bred a set of genetics based on their appearance and pedigrees. Two tri-color Australian shepherds will NOT produce a merle dog. They will probably produce additional tri-colors. Two black German Shepherds will also ONLY produce black puppies, because the gene is recessive, while black labs may produce other colors.

As all litters at Ausdauer are individually drawn and colored, emphasize on color/pattern differences that are genetically impossible to produce by two dogs will NOT occur. Instead, each dog will appear unique through pose, fur-type, markings, and expression.

Ausdauer also breeds its dogs for specific coat colors, markings, and attributes. Dogs with the genetic potential to produce desirable traits will be bred together to achieve these breeding goals.

We do not breed any dogs that would have the potential for producing puppies with health problems. This includes:

• Flashy Irish x Flashy Irish (which can produce predominantly white dogs, causing deafness if white is present over the ears, especially in boxers)
  
• Harlequin x Harlequin
  
• Merle x Merle

Foundations

All of Ausdauer's foundation pairs are created for the purpose of breeding. They are not to complete a collection of NCC registrable dog breeds. Therefore all foundation dogs are bred at some point it time, once required (and preferred) titles and certificates are achieved. Furthermore, Ausdauer strives to provide the NCC with "unpopular" and rare breeds. For this purpose, Ausdauer does not create foundation dogs who have more than 50 individuals registered with the NCC.*
*A few exceptions do occur, commonly regarding Ausdauers specializations.

The Process

Rather than choosing litter number, gender, and color/pattern, Ausdauer leaves all attributes up to chance, as it would be in real life. This is to create a realistic breeding process, and increase the chance of varied offspring. All attributes are chosen using an 8-sided die.

Number
Litter size is normally between 3 and 6 puppies. Our 8-sided die is cast, and the results correspond to the following:
1 & 3 = 3 Puppies
2 & 4 = 4 Puppies
5 & 7 = 5 Puppies
6 & 8 = 6 Puppies

Gender
Even = XX, or female
Odd = XY, or male

Color & Pattern
This depends on the genes of the parents. I will set up an example breeding;

Dam: ee/Kk^br/bb/a^sa/dd/Gg/ii/sⁱs^p/tt/Hh/c^chc^ch/Mm
Cream female with fancy irish harlequin, carrying brown, brindle, merle, and saddle.

Sire: Ee/k^brk^y/bb/Aa^w/Dd/Gg/II/s^ps^w/tt/hh/Cc^u/mm
Piebald brindle male, carrying recessive red, wolf-sable, brown, dilution, graying, and urajiro.

So, first, we roll to see the number of offspring.
1d8 = 1
So 3 puppies are born.

Puppy One:
First, we roll to find out it's gender.
1d8 = 2, even, so XX/female

Then we begin to determine its color and pattern. First, we look at the recessive red gene. Because the dam is red (ee) this puppy has to have at least one 'e' gene. All we have to roll for is what gene comes from the father.
Odds = E, the first gene listed
Even = e, the second gene listed
1d8 = 7, so "E" is the gene passed down from the father.
This means Puppy One is Ee, not-red.

Next, we have the "K"-gene, or dominant black. Because both parents are heterozygous at this loci, we most roll for each parent. Using the formula above (Odds = the first allele, evens = the second allele), we roll:
From Dam: 1d8 = 2 = k^br
From Sire: 1d8 = 2 = k^y
So Puppy One is k^brk^y, brindle carrying clear agouti.

We continue to do this for each of the different loci, until we have an entire genotype, then proceed to do the same for each puppy in the litter.

Here are the results from this breeding:

Puppy One:
Ee/k^brk^y/bb/a^sA/dd/gG/Ii/sⁱs^w/tt/Hh/c^chC/mm
Piebald brindle female

Puppy Two:
Ee/k^brk^br/bb/a^sa^w/Dd/gg/Ii/sⁱs^p/tt/Hh/c^chC/mm
Flashy irish brindle female

Puppy Three:
Ee/K^br/bb/a^sA/dd/Gg/Ii/s^ps^p/tt/hh/c^chC/Mm
Diluted blue merle piebald male

And that's how Ausdauer creates its litters! =D