# Possible outcomes?



## Silver Moon (Sep 4, 2013)

So, been having a think about possible pairings, and think I've got one I'd like to try, but not 100% sure about the outcomes.

BEW Male; (cᵉ/c)









And black banded female; (a/a Wᵇᵈ/*)









Would they just all be black selfs carrying banded and beige?

Ideally I'd like to try and get some colour point beige into my lines if possible?


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## Fantasia Mousery (Jul 16, 2011)

Unless the doe carries some c-dilution, you will get 100% Black Banded babies, 50% of which will carry c^e and c.
To get CPB (c^e/c^h), you will need the Siamese-gene (c^h) into your lines. You can get that from pairing your Bone (BEW) buck to either a Siamese (c^h/c^h), Himalaya (c/c^h), or CPB.

About the Banded-gene... It's W^sh, not W^bd. And your doe only has one W^sh gene. You can read more here: http://hiiret.fi/eng/breeding/genetics/w-locus.html


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## Silver Moon (Sep 4, 2013)

Thanks for that reply.

My mistake on the code ( read the wrong one :lol: )

Unsure what the doe carries so could be some c-dilute ( or anything else for that matter ).

However, I think they will still get paired just to see what comes out of it and if there's anything interesting being carried


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## Frizzle (Oct 6, 2011)

Fantasia Mousery said:


> Unless the doe carries some c-dilution, you will get 100% Black Banded babies


I think you mean 50%, if she's only carrying one copy of banded.


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## Fantasia Mousery (Jul 16, 2011)

Frizzle said:


> Fantasia Mousery said:
> 
> 
> > Unless the doe carries some c-dilution, you will get 100% Black Banded babies
> ...


Yes. :shock: Sorry about that.


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## Silver Moon (Sep 4, 2013)

So 50% black banded, what would the other 50% be? Black selfs carrying banded?


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## Fantasia Mousery (Jul 16, 2011)

TRH said:


> So 50% black banded, what would the other 50% be? Black selfs carrying banded?


Black selfs _not_ carrying Banded.


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## Silver Moon (Sep 4, 2013)

I'm not quite sure I understand that diagram?


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## Frizzle (Oct 6, 2011)

It's something called a punnet square, & is used to diagram out what your possible out comes are as well as some basic chance. What it is depicting here is how banded works.

W^sh is the banded gene, taking place on an area called the W locus. Your mouse's W locus would be written as W^sh/W since she is displaying only one copy of banded. Therefore when she's crossed to a non banded mouse W/W, you end up with a mix of W^sh/W & W/W, about 50/50.


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## Fantasia Mousery (Jul 16, 2011)

W is variegated, so your mouse is W^sh/w, not W^sh/W.

But yes. Punnet squares are extremely helpfull when it comes to learning genetics and outcome.
What you do, is you put one mouse's gene code (you have to make a punnet square for each locus) on the two top boxes, and the other mouse's genes on the two boxes on the left side. Then you can fill in the four empty ones with the different outcomes, as the parents will give 1 gene each from the loci to their babies.


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## Frizzle (Oct 6, 2011)

Oops, yeah, FM's right.

As you get to know genetics more, you can actually make punnet squares that track several different genes at once. Say you want to know what the chance of getting satin and pied on the same mouse, from parents that are only both carriers for both. You plug it in, and you find that your chance is only 1:16. It's very fun. : )


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## Silver Moon (Sep 4, 2013)

Thanks for that help.

I've been having a play with these punnet squares, but still think I'm miles off :lol:

Example: BEW male to Champagne female;

All blacks. 
1:6 carrying B
1:6 carrying b ( chocolate )
1:6 carrying p ( pink eye dilution )
1:6 carrying P
1:6 carrying c
1:6 carrying C

or would everyone carry everything?

Maybe I'm way off the mark again :lol:


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## Fantasia Mousery (Jul 16, 2011)

Okay. Again, your BEW is a/a c^e/c. Champagne is a/a b^c/* d/d p/p - but ONLY if it's a proper Champagne. It can also be a/a b^c/* p/p (without the d-dilution) or a/a b/b p/p (still without the d-dilution).
b^c and b is almost the same thing, so let's just say it is.

First off, the genes presented as a capital letter (B, C, P and so on) cannot be carried, as they are dominant and the ones that are expressed. Small letters are the recessive genes, and they need two of those before you can see it. So, dominant genes cannot be carried, only recessive ones can.

But yes, from a BEW x Champagne pairing, all babies will be Black (a/a because both parents are a/a). Now, let's take one at a time, in alphabetical order.

b-locus: your Champagne is either b^c/b^c, b^c/b or b/b. We say it's b/b, to make it a bit simpler. She will give a b to all of her babies. But your BEW is B/* which means he can either be B/B or B/b. If he is B/B all the babies will be B/b (B from the father, b from the mother). If he is B/b, 50% of the babies will be B/b (B from father, b from mother), the other 50% will be b/b (a b from each parent). The BEW being B/B is most likely, but B/b is not impossible - this all depends on what is behind him, and you will never know for sure without a test like this pairing.

c-locus: your BEW is c^e/c, and will give c^e to 50% of his babies, and c to the other 50%. Your Champagne is C/* - any kind of c-dilution can be there, and again you can't know for sure without a test. Most likely she is C/C. 50% of the babies will be C/c^e (C from mother, c^e from father), the other 50% will be C/c (C from mother, c from father). But again, it is possible for your doe to carry some sort of c-dilution (c, c^e, c^h or even c^ch), but I won't cover that here, it will be too much right now, and it's not very likely.

d-locus: your Champagne is d/d (IF she is a proper Champagne), your BEW is D/* - most likely D/D. All babies will be D/d in that case. If your doe is D/d, the babies will be 50% D/d, 50% D/D. If your doe is D/D, all of the babies will also be D/D.

p-locus: your Champagne is p/p, your BEW is P/* - again, most likely P/P. All babies will be P/p.

So, as you can see: each baby gets, for each locus, a gene from the mother, and a gene from the father. The most likely outcome is therefore a/a B/b C/c or C/c^e D/d P/p (all Black babies carrying b, c or c^e, d, and p). Everyone would carry everything.
It is only if the two of them carries something in common that something other than Black will show up. For example: if half the babies are Black and the other half are Dove (a/a p/p) this means that the BEW carries p (is P/p).


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## Silver Moon (Sep 4, 2013)

Thanks ever so much for that reply


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## Laigaie (Mar 7, 2011)

In my experience, aabbddpp is not champagne as that term is described by NMC, UMC, AFRMA, FinnMouse. Champagne is aabbD*pp. Pink-eyed chocolate. Why are we using pink-eyed lilac (both chocolate and blue) for champagne? That would be too cool a color for champagne...


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