Wow i'm a GEEK!

[JenC]

OK so I take it both the "Champagne" male and the White female were "het" for Brindle, and the "blue/champagne" gene was maybe co-dom? The male the Fawn/Brindle mother could have bred with may have been a "champagne/Blue"

Does that make sense?

 

[Baba-Lou]

Or she has champagne/blue in her. Without knowing who the father was makes it that much hardder unless you want to breed back from 3rd generation into the second to be sure.

 

[Serpwidgets]

I think brindle is dominant. Therefore, the white F1 female is het brindle (but it's masked) and the champagne F1 male is not carrying brindle.

Not sure what genetic makeup is in the champagnes. Is it possible to get a brindle champagne, or does brindle alter the color enough to make it appear non-champagne?

 

[Baba-Lou]

From what I've seen in my colonies my champagnes carry brindle but the brindle is not apparent.

 

[curiousL]

That is so funny! LOL!!! It looks like you have just as much fun breeding your mice as we do. Aren't they great. I have to admit when my boyfriend got me into the whole breeding snake thing I thought it was great, then when we started going through the whole mice breeding thing that was just fun because when they are so tiny and cute it just makes you smile, and of course trying to figure out their genetics is all too weird but interesting at the same time. How many mice do you have? I love the champagne color... :crazy02:

 

[Serpwidgets]

From what I've read, champagne is genotype aa-bb-pp. It's:
homozygous self at the agouti locus
homozygous brown at the brown (dilution) locus
homozygous for pink-eyed dilution at the P locus

As for brindle, apparently there are two brindles.

One is located at the agouti locus and is dominant to all but the "yellow" allele, including self. The male is not het for this, or he would be a brindle.

The other is sex-linked. It is lethal homozygous in females, and lethal hemizygous in males. So the only ones that express it are heterozygous females.

So, the F1 male is not the sex-linked variety, and this leaves the female as the carrier.

 

[Serpwidgets]

From what I've seen in my colonies my champagnes carry brindle but the brindle is not apparent.

What results were you getting where two non-brindles produced brindle offspring?

 

[JenC]

Wow..thanks everyone... :eek1:

CuriousL, right now i have 2.2 adults (the ones pictured and a chocolate male) and 7 fuzzies yet to be sexed, any males will be food, except from this litter.

From what I've read, champagne is genotype aa-bb-pp. It's:
homozygous self at the agouti locus
homozygous brown at the brown (dilution) locus
homozygous for pink-eyed dilution at the P locus

As for brindle, apparently there are two brindles.

One is located at the agouti locus and is dominant to all but the "yellow" allele, including self. The male is not het for this, or he would be a brindle.

The other is sex-linked. It is lethal homozygous in females, and lethal hemizygous in males. So the only ones that express it are heterozygous females.

So, the F1 male is not the sex-linked variety, and this leaves the female as the carrier.

Sure... :eek1: I'll pretend I understood that..

 

[Serpwidgets]

Sure... :eek1: I'll pretend I understood that..

LOL. Ok, I'm basing this on what I've read at other mouse sites (especially http://www.petrodents.com/BigFive.shtml) and assuming the IDs made are accurate... The following are all single-steps through piecing together the genotypes of the parents:

The male is a champagne, he's a non-agouti, brown, pink-eyed. Basically a triple recessive morph like a snow motley. Since he is homozygous, all of his offspring will get one non-agouti gene, one brown gene, and one pink-eyed gene from him. (He is "aa-bb-pp")

He didn't throw any albinos when bred to an albino, so we will assume he's not het. So he's aa-CC-bb-pp.

The female is carrying the brindle gene. It's dominant, so half of her offspring will be brindles. Since some of them are non-brindle, we know she is only het for it. A(vy) ?

Meanwhile, since she also produced a champagne (which must be homozygous non-agouti and must have inherited the non-agouti gene from her) that means she is het for non-agouti. ? a

Those two together add up to say her genotype at the agouti locus is heterozygous: brindle/non-agouti. A(vy) a

She's an albino, too, and it's recessive so all of her offspring are het for it. Her genotype there is "cc" so she is so far A(vy)a-cc.

Also, when crossed to a male carrying only b she threw Champagne offspring that are bb, and Blue offspring that are Bb. She is het at the B locus and is A(vy)a-cc-Bb.

Same thing goes for the P locus, so she's het there, too: A(vy)a-cc-Bb-Pp.

Are you having fun yet? Good, because we haven't covered the D locus. :grin01:

The genotype aa-dd is a blue mouse. So if that's what your blue fuzzy is, then he is aa-Cc-Bb-Pp-dd.

Since we know d came from the male, we can update his genotype to show he's also het for d: aa-CC-bb-pp-Dd

Since we know d came from the female she is either Dd or dd but we can't tell because she's an albino. So she's at least A(vy)a-cc-Bb-Pp-Dd. :santa:

So the cross you're doing is:
aa-CC-bb-pp-Dd X
A(vy)a-cc-Bb-Pp-Dd
Which comes out to:
50/50 brindle/non-agouti (black)
100% het albino
Half will express brown dilution
Half will express pink-eyed dilution
Half will express blue dilution

It gives you 16 possibilities which are:
Black
Black Brown (Chocolate)
Black Pink-eyed (Lilac)
Black Brown Pink-eyed (Champagne)
Black Blue (Blue)
Black Brown Blue (Dove)
Black Pink-eyed Blue
Black Brown Pink-eyed Blue
Brindle
Brindle Brown (reddish with brown stripes)
Brindle Pink-eyed
Brindle Brown Pink-eyed
Brindle Blue (Buff with blue stripes)
Brindle Brown Blue
Brindle Pink-eyed Blue
Brindle Brown Pink-eyed Blue

Much simpler, eh? :grin01: