German Black

German Black

What happens when a male black german shepherd mates with a female white german shepherd.?

I have this question because i wan't stud service done on my white german shepherd and i wondewr if she mates with a black german shepherd. Will the puppies by black and white?

"~Chalbri~-etc":
Your comments are true, but at no stage do you attempt to actually answer the QUESTION.

"Mom of Three":
The dominant M^ that produces merle is not part of the GSD gene pool.
Too many GSDs have a chest blaze, but the only DNA-proven GSD line with Collie-like markings on face-&-neck are Frankie von Phenom and about half her descendants (only half because it is the effect of a dominant that is lethal when homozygous) - and that so-called "Panda" marking is banned in GSDs because of the amount of white it contains. All other splash-marked GSDs are assumed to be cross-breeds, although I hear tales of the odd line that carries the s^p allele for piebald spotting. Either way, those with more white than an inconspicuous chest spot and maybe some white toes (but the nails of those toes must be black) have too much white to be acceptable as GSDs

Okay, to "Kevin C"s question:
First thing is that, despite "Victor K"s comment (his awareness of GSD genetics is more freshman than honours at present, but he has a few years ahead in which to get it right.), there will be NO gray pups - (1) eumelanin (the dark pigment in canine coats) is not like the pigments used in paints, and (2) the dominant G^ for gray coats is not present in the GSD gene pool.

Next thing is to appreciate how the self-black GSD and the self-white pooches are created, so that you understand why there will be no black-&-white pups, either..

In GSDs, all actual self-colours (whether black or liver or blue; white is not a colour, it is a LACK of colour) are set by the recessive pairing a^ a^ in the Agouti Pattern series. The default colour from a^ a^ is self-black, but a pair of the recessive modifier b^ in the Black-brown series will convert that to liver; a pair of the recessive modifier d^ in the Dilute series will convert the black to the blue-gray referred to as blue. If the pup inherits both the b^ b^ and the d^ d^ their combined effect is a creamy-blue known as Isabella. (A pair of the b^ and/or d^ will also affect the dark area of dogs whose Agouti genes produce sable or saddle or bi-colour.)

In all self-white breeds so far DNA tested the recessive pairing e^ e^ is present. But its effect is not to produce self-white, just to prevent eumelanin (dark pigment) from being formed by the hair follicles - something else has to block the phaeomelanin (tan pigment). However, as the factors for self-white behave as though they were a simple recessive, we can be sure that the second gene involved must be on the same chromosome as the e^, and located VERY close to it. My guess is that it will turn out to be the Int^ postulated by Iljin about 70 years ago, his explanation for why the phaeomelanin (tan pigment) gets degraded to pale yellow or chalky white; but if its DNA has already been located the news has yet to reach me.

So the situation, were you unwise enough to continue with your foolish plan, is that you would be mating a self-black (thereby proven to be a^ a^ plus B^ ?^ plus D ?^ plus E^m ?^ plus ?^ ?^) to a self-white (thereby proven to be ?^ ?^ plus ?^ ?^ plus ?^ ?^ plus e^ e^ plus Int^ ?^).
(PS: That E^m could be an E, but that is very unlikely in our breed.)

The ?^s are where we have no evidence from your information as to which allele of that series is present to make up the pair of alleles that every individual inherits in every gene location except those on just the X chromosome. To work it out just remember that the series I am mentioning are A for Agouti, B for Black-brown, D for Dilute, E for Extension, and Int for Iljin's Intensity series. There are other colour series (you'll find all the known ones in the Files section of http://pets.groups.yahoo.com/group/The_GSD_Source/ but will need to become an instant-member, as Yahoo doesn't let me make that section "Public access") but they have no significant effect in your case.

You would find a Punnett Square too complicated - it is a marvellous tool for showing the possibilities from a single autosomal character where the genotype for that character is known for both parents, usable for the interaction of 2 simple autosomal characters, but after that you start to need a multidimensional "page" to draw on!

In this case, we have no idea what the self-white possesses in the Agouti series, but a high percentage of self-whites are genetically sables, so it is highly probable that she is A^w a^t (wolf-sable + saddle), very possible that she is a^t a^t. It is possible but unlikely that she is A^w a^ or a^t a^ - unlikely because people who understand genetics do NOT want the genes from a self-white degrading the depth & shine of the self-black's all-over darkness, and those who understand genetics don't want the all-over spread of eumelanin (dark pigment) making it harder for the e^ e^ to have its epistatic effect of blocking the eumalin from being formed (most whites have some "badger markings).

What it boils down to is that if the self-black is E^m e^ half the pups ( ± whatever random chance "decides" during the moments of fertilisation) will be self-white, about half won't. If the self-black is the E^m E^m that is preferred, all the pups won't be self-whites.

Either way, of the pups that are not self-white, about half are likely to be sables (assuming that the dam carries just one A^w) with the others probably being saddles. But they could ALL be saddles, or about half could be self-blacks. But if one parent is a self-black it is impossible to get all 3 of sables, saddles and self-blacks in the same litter, just any 2 of those.

And regardless of their displayed colours, ALL the pups will be carriers of an a^ for self-colour from dad and an e^ for no-black from mum, plus are almost certain to have an Int^ or an int^ from mum.
The effect of that last gene will be that all the non-whites will have poor tans - from chalky"silver" to a washed-out creamy-fawn. If the sire is carrying an e^, then about half the coloured pups will be "dirty yella dawgs" with no black - maybe when they're born, but definitely by the time they are about 2½ years old.

To complicate it:
Many of the people who aim to produce pet-level self-whites are very happy to use carriers of liver or blue - either reduces the density of the black so can be regarded as making the epistasis (in this case, blocking of eumalin) easier.
Although the DogSport fanciers of blacks would avoid carriers of liver & blue, again, pet-level BYBs might include them. So it is possible that the non-whites might be blue or liver instead of black where black should be in whatever their Agouti pattern is.

The possibilities are predictable (always allowing for random chance when the pairing at a locus is not homozygous), but you haven't given us enough evidence as to what alleles each parent is likely to possess, apart from the series that produced the black in one case, the white in the other case, for 'us' to work out the probabilities.

But all genuine GSD breeders know that the black-to-white mating is a bad idea on colour grounds.

In addition to the inadvisability of muddling those two sets of coat-colour together, I say your plan is unwise because I am certain that your bit.ch is not of breeding quality.
If she were, you would be asking her breeder, not asking here where a high proportion of the so-called answers come from know-it-all kids pretending to be adult & expert.
If the stud you had in mind was of breeding quality, you would be asking HIS owner or his breeder.

But in case I'm wrong about your bit.ch and that stud:
The ultimate proof of a GSD being of breeding quality is possessing a Breed Survey report (BS.Cl. in English, KKl. in German). No self-white is eligible for that, as all whites and blues and livers are disqualified in FCI#166: http://www.fci.be/uploaded_files/166A1991_en.doc which is the ONLY GSD Standard approved by the WUSV (World Union of GSD Clubs, which has at least one seat-&-vote for every GSD nation and meets in conjunction with each SiegerSchau to sort out GSD issues).

If you live anywhere except Australia, Britain, Canada, New Zealand, the USofA there is a possibility that - IF your bit.ch has a 6 generation pedigree where EVERY ancestor is a KC-registered self-white - you might manage to have her re-registered as a Berger Blanc Suisse (see http://www.fci.be/uploaded_files/347a2002_en.doc ), but in those 5 countries no breed can be recognised that is based on what is a fault in its parent breed. In Australia and Germany your white GSD couldn't even get registered as a GSD - they take the idea of a "disqualification" seriously, there.

But whatever the breed, the MINIMUM requirements for proving a medium-sized pooch breed-worthy are:
(1) recorded on the Breed Register without any limitations
(2) possesses official certificates for both hips & elbows
(3) possesses show certificates (placings or Excellent gradings) to prove that it looks typical of its breed.
Highly desirable are:
(4) a character certificate - BH or CGC
(5) a training qualification - for a GSD the ideal is HGH or SchH1, but all training qualifications prove that at least the pooch is highly willing to take notice of what its owner wants.

Sadly, almost every KC cares only whether both parents are in the same Breed Register - no checking on character or conformation or health or size or trainability...

Unless your bit.ch measures up, and so does the stud you choose to compensate for her flaws, forget about breeding from her.

And I haven't even touched on the financial risks every breeder faces. Nor the time wasted checking up on liars who claim - sometimes even believe - that they are God's gift to dog breeders but have no fences, no income, no intention of attending weekly training classes, just