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How to foresee the probability to have brindle, fawn, pied etc. puppies in a litter.
 
Some definitions:
Phenotype:what we can see, what express itself
Genotype:what we don't always see, what is here, but don't express itself.
Homozygote:when dog caring only one gene(phenotype and genotype the same)
Heterozygote:when dog caring two different genes(phenotype and genotype are different)
 
To cut short, we will call colors by letters. We write in capital letter dominante colors and in small letters-the recessive colors.
So, "B"-for black/brindle and "F" for fawn (or pied etc. color). Each dog carry two genes for the color (in this type of subject) and in mating each dog (Sire and Dam) gives one of his two genes to it's puppies.
 
Fawn dog is "ff", because fawn it's recessive, so a fawn dog-gomozygote. If he was geterozygote(two different genes), he will be "Bf" and as a brindle is dominante, he will be a brindle.
Brindle dog can be:either "BB" brindle (homozygote)-he is carring only the brindle and can give only this color to his puppies or "Bf" Brindle (heterozygote) he is carringthe fawn and can give either brindle or fawn gene to his puppies.
 
So, we have 3 solutions about the parents:
"ff"-fawn
"Bf"-brindle, carring fawn
"BB"-brindle, not carring fawn or true brindle
 
 Next illustration cell will show chance to have this or that color.
 
Mating #1. Both parents are fawn.
Female/Male
 f             
f              
f
f f
f f
f
f f
f f
Each parent in this example has only one fawn to transmit. So 100% puppie will be fawn.
 
Mating #2. Both parents are brindle. It is more difficalt case, because there is three possible cases!
Case #1-Both parents are homozygotes
Case #2-Both parents are heterozygotes
Case #3-One parent homozygote, another-geterozygote.
 
Case #1. Both parents are homozygotes (BB), not carring the fawn.
.
Female/Male
 B        
B          
B
BB
BB
B
BB
BB
Each parent has only brindle gene to transmit to his puppies. So, all puppies will be homozygotes as their parents are. This is pure brindle.
 
Case #2. Both parents are heterozygotes(Bf) and carring fawn.
 
Female/Male
 B        
        
B
BB
Bf
f
Bf
ff
 
Take a look at four central squares in this cell. There is possibility of color repartition. One square out of four-"BB", so 1/4 of litter will be brindle homozygote(not caring the fawn). We have one square out of four "ff"-it's means 1/4 of litter will be fawn and two squares out of four-"Bf"-half of the litter will be brindle heterozygote (carring fawn).
 
Case #3.One parent homozygote (BB), another-geterozygote(Bf).
 
Female/Male
 B        
        
B
BB
BB
f
Bf
Bf
 
Whole litter, 100%, will be brindle! But one half will be homozygote(not carring fawn gene "BB"), and other half will be heterozygote(carring fawn gene "Bf"). Not easy to chooes a puppy if you want him to transmit fawn gene, when he will be mated.
 
Third mating. Cross fawn "ff" and brindle geterozygote "Bf"
 
Female/Male
 B        
        
f
Bf
Bf
f
Bf
Bf
 
100% puppies will be brindle, but all carry the fawn. You can easy to choose your puppy, who will carry fawn gene and will produce fawn puppies at his mature age!
 
 Forth and last mating. Crossing between fawn "ff" and brindle heterozygote "Bf"
 
Female/Male
 f        
        
B
Bf
Bf
f
ff
ff
 
50% puppies will be fawn and 50% puppies will be brindle. All the brindle one will be heterozygote (carring fawn)
 
Of course, it is only possibilities. If you take the last mating there is chances to have only brindle or only fawns by the struck of luck. If you have 8 puppies out of this mating, nothing says, that it will be 4 brindle and 4 fawn puppies. It could be 5 brindle and 3 fawns!
Two fawns will always give you the fawn puppies and two brindle homozygotes-only brindle homozygotes...
 
Pied gene. How does it works???
The pied gene is recessive. The puppy must heritate it frm his both parents to express it. As the gene is recessive, we will write it in a small letter "c". A pied fawn dog would called "fc fc", an a homozygote pied/brindle will be called "Bc fc"
A dog who has two "c" in his gene code will be pied, a dog who has only one "c" on his gene code will be a pied carrier.
Color cod recapitulation:
BB-brindle homozygote (does not carring pied or fawn)
Bf-brindle heterozygote (fawn carrier)
ff-Fawn
BcBc-homozygote pied/brindle (not carried fawn)
BcFc-heterozygote pied/brindle (fawn carrier)
fcfc-pied/fawn
Bcf-brindle heterozygote (fawn and pied carrier)
BcB-brindle homozygote (pied carrier only)
fcf-fawn carried pied
Next illustration cell will show chance to have this or that color.
 
Two pied/fawn mating
Female/Male
 fc        
fc         
fc
fcfc
fcfc
fc
fcfc
fcfc
All puppies will be pied/fawn
 
Two homozygote pied/brindles (not carring fawn) mating
Female/Male
 Bc        
 Bc        
Bc
BcBc
BcBc
Bc
BcBc
BcBc
All puppies will be pied/brindle and won't carry tha fawn.
 
Mating between pied/brindle and pied/fawn
Case#1. Cross between homozygote pied/brindle and pied/fawn.
Female/Male
 Bc        
 Bc        
fc
Bcfc
Bcfc
fc
Bcfc
Bcfc
All puppies will be pied/brindle carring fawn and pied/fawn.
 
Case#2. Cross between heterozygote pied/brindle and pied/fawn
Female/Male
 Bc        
 fc        
fc
Bcfc
fcfc
fc
Bcfc
fcfc
50% of puppies will be heterozygote pied/brindle(fawn carrier) and  50% of puppies will be pied/fawn.
 
 
 
 
 
                                                                                                                                                                                                                                                                                                                                                                                                                                                                                 

 

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