Punnit Squares showing color inheritance theories:
_________________________________________________

If theory #1 - that the colors are alleles that occur at the same gene
location is correct, this is the only way that one could expect all three 
colors in 1 litter:

(the letters I will use to indicate color genes are arbitrary here!)

B = black
s = salt & pepper
b = bicolor (black&silver)

Parent #1 = Bb (Heterozygous black, carrying black and silver)
Parent #2 = sb (Heterozygous s&p, carrying black and silver)

     B     b     
_____________
   |    |                    On average, one could expect:
s  | Bs |  sb
___|____|____                    Bs - 1 black carrying s&p
b  | Bb |  bb                    Bb - 1 black carrying b&s
   |    |                        sb - 1 s&p carrying b&s
                                 bb - b b&s
_________________________________________________________________________

However, there are experienced breeders reporting all three colors coming
from black to black breedings, and black to black/silver breedings.
* This is impossible under theory #1 *
_________________________________________________________________________
Theory #2:

This is only possible if the b&s gene is independant (not an allele) and 
is situated at a second gene location. (locus)

In fact, the b&s gene may be one that simply modifies the s&p color 
pattern, turning the hairs solid black. (which may explain how 
banded hairs still occur in the coats of b&s dogs from time to time)

This is a much more complicated chart, and there are actually two
variations in genotype in black to black breeding could produce all three
colors. This is just one. 

Black and salt& pepper are alleles. Thus, the dog is always
Bs, BB or ss. There is no b at this alelle.

The gene creating black & silver is at a different location. It does not
produce solid black hair unless it is homozygous. Even it the homozygous 
state, it is still recessive to black.

*for the purposes of this chart, I will assign it the character '+' to
indicate the black and silver gene and assign the character '-' to
indicate it's absence. 
 
Parent #1 is heterozygous black - "Bs" and carries black & silver "+-"

Bs+-

Parent #2 has identical genetic makeup.

When creating this chart, you must pair each allele with both alleles of 
the other pair, thus the chart is more complicated. (there are more 
combinations possible that must be considered)
 
       B+    B-    s+    s-  
_______________________________     | Remember: B is dominant - all 
   |                                | pups with B or BB will be black.  
B+ |  BB++  BB+-  Bs++  Bs+-        | 
___|___________________________     | Those pups with ss, in combination   
   |                                | with -- will be pure for salt&pepper
B- |  BB+-  BB--  Bs+-  Bs--        |
______________________________      | Those pups with ss, in combination
   |                                | with +- will be salt&pepper and
s+ |  Bs++  Bs+-  ss++  ss+-        | carry black and silver.
___|___________________________     |
   |                                | Pups with ss, in combination with
s- |  Bs+-  Bs--  ss+-  ss--        | with ++will be black and silver. The
   |                                | homozygous pair is dominant over the
                                    | salt&pepper gene.

So, if this theory is correct, blacks (Bs++ to Bs+-) and (Bs+- to Bs+-)
would be capable of producing all three colors if bred together. 

Note that in Bs+- to Bs+-, only one black and silver would be
expected in 16 puppies, with three s&p and 12 black.

In Bs+- to Bs++, there would be three black and silvers, one s&p
and 12 black.

___________________________________________________________________

 Black & Silver to Black  - 3 color chart
___________________________________________________________________


Black and silver parent is ss++
Black parent is Bs+-
 
       B+    B-    s+    s-  
_______________________________     | Remember: B is dominant - all 
   |                                | pups with B or BB will be black.  
s+ |  Bs++  Bs+-  ss++  ss+-        | 
___|___________________________     | Those pups with ss, in combination   
   |                                | with -- will be pure for salt&pepper
s+ |  Bs++  Bs+-  ss++  ss+-        |
___|__________________________      | Those pups with ss, in combination
   |                                | with +- will be salt&pepper and
s+ |  Bs++  Bs+-  ss++  ss+-        | carry black and silver.
___|___________________________     |
   |                                | Pups with ss, in combination with
s+ |  Bs++  Bs+-  ss++  ss+-        | with ++will be black and silver. The
   |                                | homozygous pair is dominant over the
                                    | salt&pepper gene.


(Actually. only one line was required as the black and silver parent can 
only contribute S+ to each breeding, but for comparison, I've made it 
identical to the other)

Thus in such a pairing, you could expect four black & silver, four salt
& pepper and eight blacks. (statistical averages!)

*************************************************************************
In the end, however, it is the genetic makeup of the black parent that is 
the important factor. 

All black and silvers would have the same genetic makeup. 
(ss++)

Salt and peppers could have only two combinations. 
(ss+-) S&P carrying b&s
(ss--) pure for salt&pepper

Only blacks would be capable of carrying all three colors in their genetic
makeup. 
(BB--) Pure for black. All puppies would be black regardless of mate
(BB+-) Homozygous black, all pups black, but some carrying b&s)
(BB++) Homozygous black, all pups black, all carrying b&s)
(Bs--) Black, can produce only black and s&p pups
(Bs+-) Black, carrying s&p and b&s - can produce all 3 colors
(Bs++) Black, carrying s&p and b&s - can produce all 3 colors

Just theory! But interesting, eh?

-- 
Catherine McMillan (c) 2000



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