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Old 09-28-2008, 04:49 PM   #1 (permalink)
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Default ldominant/recessive coloring

Suppose there is a breeding between two dogs. One carries the dominant black and tan and the recessive solid black genes. The other carries the blk/tan or blk/red double recessive dilution gene.

There must be two recessive solid black genes to produce a solid black or two recessive dilution genes to produce a blue.

Am I right in saying that there can only be blk/tan pups? OR if one of the dogs passes a solid black and the other passes a dilution, could there be some solid blues? Or could there be blue/tans?
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Old 09-28-2008, 06:04 PM   #2 (permalink)
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Default Re: ldominant/recessive coloring

I believe both dogs must have the dilution gene for there to be blue puppies? So you could produce carriers of the blue gene but not actual blue puppies. Someone correct me if I'm wrong...

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http://www.total-german-shepherd.com...at-Colors.html
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Old 09-29-2008, 10:18 AM   #3 (permalink)
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Default Re: ldominant/recessive coloring

The only way you'd get black pups out of this litter is if both parents were black recessive. Since one parent is black tan/black and the other is homozygous blue/liver tan, you could get black tan pups or blue/liver tan pups. The pups would either be black and tan carrying the dilution or blue/liver and tan carrying the black. Assuming I did my Punnett square correctly.
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Old 09-29-2008, 10:25 AM   #4 (permalink)
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Default Re: ldominant/recessive coloring

Quote:
Originally Posted By: PaschaSuppose there is a breeding between two dogs. One carries the dominant black and tan and the recessive solid black genes. The other carries the blk/tan or blk/red double recessive dilution gene.

There must be two recessive solid black genes to produce a solid black or two recessive dilution genes to produce a blue.

Am I right in saying that there can only be blk/tan pups? OR if one of the dogs passes a solid black and the other passes a dilution, could there be some solid blues? Or could there be blue/tans?
So parent 1 is heterozygous Black & Tan/Black.
Parent 2 is homozygous Black & Tan. I'm unclear when you say "double dilution gene" if you mean it carries 2 copies of one dilution gene, either 2 blue or 2 liver, (in which case the dog would BE a dilute and every pup produced will be a carrier) or if it carries 1 copy of each of the 2 dilution genes, 1 blue and 1 liver.

Either way, ALL pups will be Black & Tan. There cannot be solid blacks without both parents carrying at least one gene for solid black. Statistically, half the pups would be homozygous Black & Tan and the other half would be heterozygous Black & Tan/Black.

As far as the dilutes, it depends on what you mean whe you say "double dilution gene".
If the parent carries 2 copies of one dilution gene, and therefore IS Blue or Liver & Tan, not Black & Tan, ALL pups will inherit one dilution gene. They will not be blue or liver, but all will be carriers of the gene.
If the parent carries 1 copy of each dilution gene (1 for blue, 1 for liver), most of the pups will probably inherit one or both of the dilution genes and be carriers, some will not, but none of them will be dilutes themselves.

Either way, none of the pups produced will be dilutes themselves, since only one parent carries dilute genes, but many will be carriers of dilute genes.
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Old 09-30-2008, 09:41 PM   #5 (permalink)
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Default Re: ldominant/recessive coloring

The only dominant coat gene(s) in the breed are for sable, all other patterns including self colour are recessive genes so be careful about your terminology. Genes for dilution in blue or liver are also recessive but separate from the coat colour/patterns. White is recessive.

Two black and tan dogs may only produce black offspring if they each carry a gene for black, but not all offspring in a litter may be black. If one carries a gene for black, some of the offspring may carry the gene, but none will express that trait. Two black dogs will always produce black, they can offer no other colour genes. This basic principle applies to the dilutes, but dilutes can be seen in all patterns/colours including sable and solid black and occur on a different locus than the colour/pattern genes. Read about colour genes here:

http://www.jmadesign.com/Frankenhaus/colorgen01.shtml
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Old 10-23-2008, 09:27 AM   #6 (permalink)
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Default Re: ldominant/recessive coloring

okay, I am supposing here......the blue/tan has solids a few generations back on both side of the pedigree (blue, liver, black, white). But since the dog is blue/tan, that means it received the dilution from both sides. However, is there the possibility that they might also carry and pass on the solid black gene? Is the 'solid' gene a color gene or a pattern gene?
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Old 10-23-2008, 09:40 AM   #7 (permalink)
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Default Re: ldominant/recessive coloring

scarlett, thanx for the link. I have read it before, but I probably should read it again....and maybe again!!
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Old 10-23-2008, 10:29 AM   #8 (permalink)
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Default Re: ldominant/recessive coloring

I met my first blue last weekend - gosh he was cute but I bet Nikki is going to spend the next 10 or 15 years explaining that he isn't part chowchow.

Here's some more on blues
http://www.germanshepherds.com/forum...page=0&fpart=1

http://www.germanshepherds.com/forum...page=0&fpart=1
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Old 10-23-2008, 11:19 AM   #9 (permalink)
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Default Re: ldominant/recessive coloring

here's another question.....Ike is black sable with tar heels and penciling on the toes. I have read that the penciling might indicate carrying recessive black. Either it does or it doesn't.

Does it?
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Old 10-24-2008, 08:18 AM   #10 (permalink)
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Default Re: ldominant/recessive coloring

It seems most sables who do carry recessive black do have the toe penciling. But there are sables *with* toe penciling who do *not* carry recessive black (I own one right now, as does my mother). So it's no guarantee.
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