Cocker Spaniel Color Genetics 101
With this page, I hope to share some knowledge of color genetics in Cocker Spaniels. I will not be using technical terminology and will try to keep it simple and usable. I have found that it is more valuable to be able to apply information than it is to try to impress people with it.
Please look in the following books for more information
about color genetics in general or in specific to the Cocker
Spaniel.
The American Cocker Spaniel by Dr Alvin Grossman
The New Art of Breeding Better Dogs by Kyle Onstott
When I am recording colors in my records, I often use the genotype (genetic codes that determine what color a dog will be and can produce) instead of the phenotype (color that you see when you look at the dog.) I find that they are more useful when trying to choose possible breedings. The abbreviated codes that I use are as follows:
BASE COLOR
bl - black
br - brown
bf - buff (from silver to red)
bnr - brown nosed red (from silver to red)
PATTERNS
t - tan points (tan pattern like on a Doberman)
p - parti (base color and white, this white pattern is the piebald gene)
s - sable (2 tone color pattern like on a Collie or Beagle)
First, a disclaimer.... I have not yet bred or owned a sable. I have had few buffs and partis. I think I can explain the basics, but if you think there may be an error, please contact an expert in those areas for clarification.
Base Colors
These colors are easiest to understand on
solid colored Cockers so I will start with those.
Base colors are like the background color.
Considering only solids (blacks, buffs, browns
and brown nosed reds) No tan points, no partis,
and no sables.
To help keep things clear, I will use the
abbreviated code for all references to genotype
and will write out the color for all phenotypes.
For example a "black" dog will carry two base
colors that it can pass on to its offspring such
as (bl - br). Such a dog would be able to produce
black and/or brown puppies.
Dominant VS Recessive
With a breed like Cockers that has so many base colors, color dominance is better discribed as a diamond shape.
bl
br.........bf
bnr
Bl is the most dominant. If a dog has one bl gene it will be black. If it has 2 bl genes it will only produce black puppies. This looks like (bl-bl) and is said to be "black dominant." A dog may look black, have one bl gene and still carry any of the other genes as recessive. For example (bl-br), a black dog that carries br as a recessive and can produce black and brown puppies.
Bnr is the most recessive. It can be carried
with any other color gene but will only appear
as a brown nosed red if the dog's genotype is
bnr-bnr. Such a dog could produce any of the
other more dominant colors to include...
black (bl-bnr)
brown (br-bnr)
buff (bf-bnr)
Brown nosed red (bnr-bnr)
Both bf and br can be considered recessive to bl and both are dominant to bnr. Neither is dominant over the other.
If you breed a brown (br-br) to a buff (bf-bf) you will get 100% black (br-bf) puppies. Each of those puppies will carry (br-bf) genes and will be able to produce brown and buff puppies. Some breeders who would argue this point say that their experience shows that this is not so. They bred a buff to a brown and got buff (or brown). This is explained when you consider the bnr gene. This could happen in the following....
brown (br-bnr) X buff (bf-bf) = 50% black (br-bf), 50% buff (bf-bnr)
brown (br-br) X buff (bf-bnr) = 50% black (br-bf), 50% brown (br-bnr)
brown (br-bnr) X buff (bf-bnr) = 25% black (br-bf), 25% brown (br-bnr),
25% buff (bf-bnr), 25% brown nosed red (bnr-bnr)
This shows very well how important it is to know your breeding stock's genotypes. With out that knowledge it is nearly impossible to predict the colors of future puppies.
How about a quiz....
I did a breeding with Hope, a brown (br-bnr) bitch to a black and white (bf-br) dog. What should the resulting puppies' base colors be?
brown (br-bnr) X black/w (bf-br) = ANSWER
What about a breeding between a buff and a brown?
buff (bf-bnr) X brown (br-bnr) = ANSWER
Just a note on determining the color genotypes of dogs.
black (bl-bnr) X black (br-bf) =
25% black (bl-br), 25% black (bl-bf), 25% buff (bf-bnr), 25% brown (br-bnr)
See that 50% of the puppies are black but will not produce the same color of puppies. The only way to tell whether a black is (bl-bl), (bl-bf), (bl-br), (bl-bnr) or (br-bf) is through pedigree analysis and test breeding.
Lesson 2 Particolors
The piebald gene (parti) is a simple recessive. In other words to get a parti puppy, said puppy must get one p (parti gene) from each of its parents. Every parti has a genotype of (p-p) meaning it has 2 parti genes. Every parti must pass on one p gene to every puppy it produces.
p = parti gene
s = solid (is really more like the absence of parti pattern)
For example...
parti (pp) X solid (sp) =50% parti (pp), 50% solid (sp)
parti (pp) X solid (ss) = 100% solid (sp)
parti (pp) X parti (pp) = 100% parti (pp)
solid (sp) X solid (sp) = 25% solid (ss), 50% solid (sp), 25% parti (pp)
notes
1. to be parti a puppy MUST have 2 parti genes
2. any solids carrying parti (sp) will be able to produce parti
3. solids which are (ss) will never produce parti because they are unable to pass on a parti gene.
Quiz time.....
The above mentioned breeding between Hope, the brown bitch and black and white dog is again the feature of the question. The dog being a particolor is (pp). The brown bitch is an unknown. Both of Hope's parents are solid. Hope's paternal granddam is a brown and white parti. What are the possible outcomes of the future litter?
(br-bnr)(s?) X (br-bf)(pp) =?
Did you try to find all the possible colors that Hope will produce? Keep that in mind for the next lesson.
Lesson 3 Tan Points
The tan points pattern is inherited in the same way as the particolor gene. It is a simple recessive and will require two tan genes to produce the tan pattern on a puppy.
t = tan piont pattern gene
(tt) = dog that has 2 tan point genes and is tan pointed
(t-) = dog that has only one tan point gene, is not tan pointed but will be able to produce tan pointed puppies.
(--) = dog that does not carry any tan point genes and will never produce a tan pointed puppy.
You might ask how you can tell if a dog that is not tan pointed, carries the gene. Earlier I mentioned pedigree research and knowledge of resulting puppies. These are some facts that can be used to determine if a dog carries the tan point gene.
1. A tan pointed dog WILL carry 2 tan genes (tt)
2. All dogs that have at least 1 tan pointed parent will be (tt) or (t-) and will be able to produce tan pointed puppies.
3. When two tan pointed dogs are bred the resulting litter will all be tan pointed puppies.
Hope (br-bnr) (s?) (t-)
Hope is a solid brown. How can I know that she carries exactly one tan point gene? Look at Hope's pedigree to see if you can tell.
Hope's sire, Crash is a brown and tan (br-bnr) (sp) (tt). All of his get will get one tan gene from him. This was also confirmed in the fact that Hope produced a brown and tan puppy (Gus) in her first litter. Gus is pictured at the bottom of Hope's page.
Quiz #3
Hope (t-) X Paxil (??) = ANSWER
I hope you enjoyed these short color lessons. If you have any questions you can e-mail me at mccrack@highvision.net
Now for a cheat sheet. Use this to determine color possibilities.
BASE COLORS
black X black
(bl-bl) X (bl-bl) = 100% black (bl-bl)
(bl-bl) X (bl-br) = 100% black (50%bl-bl, 50%bl-br)
(bl-bl) X (bl-bf) = 100% black (50%bl-bl, 50% bl-bf)
(bl-bl) X (bl-bnr) = 100% black (50%bl-bl, 50%bl-bnr)
(bl-br) X (bl-br) = 75%black (25%bl-bl, 50%bl-br), 25% brown (br-br)
(bl-br) X (bl-bf) = 100% black (25%bl-bl, 25%bl-bf, 25%bl-br, 25%br-bf)
(bl-br) X (bl-bnr) = 75%black (25%bl-bl, 25%bl-br, 25%bl-bnr), 25% brown (br-bnr)
(bl-bf) X (bl-bnr) = 75%black (25%bl-bl, 25%bl-bf, 25%bl-bnr), 25% buff (bf-bnr)
(bl-bf) X (bl-bf) = 75%black (25%bl-bl, 50%bl-bf), 25% buff (bf-bf)
(bl-bnr) X (bl-bnr) = 75%black (25%bl-bl, 50%bl-bnr), 25% brown nosed red (bnr-bnr)
(br-bf) X (br-bf) = 25%brown (br-br), 25% buff (bf-bf), 50% black (br-bf)
black X brown
(bl-bl) X (br-br) = 100% black (bl-br)
(bl-bl) X (br-bnr) = 100% black (50%bl-br, 50%bl-bnr)
(bl-br) X (br-br) = 50%black (bl-br), 50% brown (br-br)
(bl-br) X (br-bnr) = 50%black (25%bl-br, 25%bl-bnr) 50%brown (25%br-br, 25%br-bnr))
(bl-bf) X (br-br) = 100%black (50%bl-br, 50%bf-br)
(bl-bf) X (br-bnr) = 75%black (25%bl-br, 25%bf-br, 25%bl-bnr), 25% buff (bf-bnr)
(bl-bnr) X (br-br) = 50%black (bl-br), 50% brown (br-bnr)
(bl-bnr) X (br-bnr) = 50%black (25%bl-br, 50%bl-bnr), 25% brown(br-bnr), 25% brown nosed red (bnr-bnr)
(br-bf) X (br-br) = 50% black (br-bf), 25% brown (br-br), 25% buff (bf-bf)
(br-bf) X (br-bnr) = 50%brown (25%br-br, 25%br-bnr), 25% black (br-bf), 25% buff (bf-bnr)
black X buff
(bl-bl) X (bf-bf) = 100% black (bl-bf)
(bl-bl) X (bf-bnr) = 100% black (50%bl-bf, 50%bl-bnr)
(bl-br) X (bf-bf) = 100%black (50%bl-bf, 50%br-bf)
(bl-br) X (bf-bnr) = 75%black (25%bl-bf, 25%bf-bnr, 25%br-bf) 25%brown(br-bnr))
(bl-bf) X (bf-bf) = 50%black (bl-bf), 50% buff (bf-bf)
(bl-bf) X (bf-bnr) = 50%black (25%bl-bf, 25%bl-bnr, 50% buff (25%bf-bf, 25%bf-bnr)
(bl-bnr) X (bf-bf) = 50%black (bl-bf), 50% buff (bf-bnr)
(bl-bnr) X (bf-bnr) = 50%black (25%bl-bf, 50%bl-bnr), 25% buff(bf-bnr), 25% brown nosed red (bnr-bnr)
(br-bf) X (bf-bf) = 50% black (br-bf), 50% buff (bf-bf)
(br-bf) X (bf-bnr) = 50%buff (25%bf-bf, 25%bf-bnr), 25% black (br-bf), 25% brown (br-bnr)
black X brown nosed red
(bl-bl) X (bnr-bnr) = 100% black (bl-bnr)
(bl-br) X (bnr-bnr) = 50%black (bl-bnr), 50%brown(br-bnr)
(bl-bf) X (bnr-bnr) = 50%black (bl-bnr), 50% buff (bf-bnr)
(bl-bnr) X (bnr-bnr) = 50%black (bl-bnr), 50% brown nosed red (bnr-bnr)
(br-bf) X (bnr-bnr) = 50%buff (bf-bnr), 50% brown (br-bnr)
brown X brown
(br-br) X (br-br) = 100% brown (br-br)
(br-br) X (br-bnr) = 100%brown (50%br-br, 50%br-bnr)
(br-bnr) X (br-bnr) = 75% brown (25%br-br, 50%br-bnr), 25% brown nosed red (bnr-bnr)
brown X buff
(br-br) X (bf-bf) = 100% black (br-bf)
(br-bnr) X (bf-bf) = 50%black (br-bf), 50% buff (bf-bnr)
(br-br) X (bf-bnr) = 50% black, 50% brown (br-bnr)
(br-bnr) X (bf-bnr) = 25% black (br-bf), 25% brown (br-bnr), 25% buff (bf-bnr), 25% brown nosed red (bnr-bnr)
brown X brown nosed red
(br-br) X (bnr-bnr) = 100% brown (br-bnr)
(br-bnr) X (bnr-bnr) = 50 % brown (br-bnr), 50% brown nosed red (bnr-bnr)
buff X buff
(bf-bf) X (bf-bf) = 100% buff (bf-bf)
(bf-bf) X (bf-bnr) = 100%buff (50%bf-bf, 50%bf-bnr)
(bf-bnr) X (bf-bnr) = 75% buff (25%bf-bf, 50%bf-bnr), 25% brown nosed red (bnr-bnr)
buff X brown nosed red
(bf-bf) X (bnr-bnr) = 100% buff (bf-bnr)
(bf-bnr) X (bnr-bnr) = 50 % buff (bf-bnr), 50% brown nosed red (bnr-bnr)
brown nosed red X brown nosed red
(bnr-bnr) X (bnr-bnr) = 100% brown nosed red (bnr-bnr)
PATTERNS
solid X solid
(s-s) X (s-s) = 100% solid (s-s)
(s-s) X (s-p) = 100% solid (50% s-s, 50% s-p)
(s-p) X (s-p) = 75% solid (50% s-s, 25%s-p), 25% parti (p-p)
solid X parti
(s-s) X (p-p) = 100% solid (s-p)
(s-p) X (p-p) = 50% solid (s-p), 50% parti (p-p)
parti X parti
(p-p) X (p-p) = 100% parti (p-p)
tan X tan
(tt) X (tt) = 100% tan (tt)
tan X no tan
(tt) X (t-) = 50% tan (tt), 50% no tan (t-)
(tt) X (--) = 100% no tan (t-)
no tan X no tan
(t-) X (t-) = 25% tan (tt), 75% no tan ( 25% (--), 50% (t-)
(t-) X (--) = 100% no tan ( 50% (--), 50% (t-)
(--) X (--) = 100% no tan (--)
