BRM's Bunny Thread! ❤️🐰❤️

[EmmaRainboe]

You can see it’s much nicer imagining both are homozygous in those locations.

 

[Kiki]

 

[EmmaRainboe]

Same. I’m just glad I found this site which I could just plug in letters, and didn’t actually have to write the whole thing manually.



https://scienceprimer.com/punnett-square-calculator

 

[BenlynFarm]

That is all soooooo very confusing and I have much to learn!!

Come on Mir! Have those babies!!!

 

[Kiki]

That is all soooooo very confusing and I have much to learn!!

Come on Mir! Have those babies!!!

I can't even try to learn the genetics stuff. It's tooo confusing. I still have trouble with understanding how my gray rabbit is called blue.

 

[BenlynFarm]

I can't even try to learn the genetics stuff. It's tooo confusing. I still have trouble with understanding how my gray rabbit is called blue.

Bahahahaha that’s so true. And yet my gray rabbit is called a Lilac

 

[EmmaRainboe]

Lilac and blue are caused by the same recessive dilute gene. Lilac is d/d on chocolate (b/b), and blue is d/d on black (B/b or B/B)

 

[EmmaRainboe]

Ooo! Emma, can you predict anything with a Broken Frosted Pearl and a Broken Orange??

If both parents are homozygous on all alleles where that is possible you have these three possibilities

50% AABBCcchdDDeeEnen (broken black agouti non-extention)

25% AABBCcchdDDeeenen (black agouti non-extention)

25% AABBCcchdDDeeEnEn (Charlie black agouti non-extention) -this one would be particularly interesting since Charlie's already have most of their pigment removed, plus the addition of the non-extention which will further remove dark pigmentation


The most complicated scenario leads to 156 unique allelic combinations.

This is the 5 trait cross with locus’ A-E, but then you have to multiple this Punnet Square by three (technically by four but two quarters will be broken so it is not unique allelic combinations) to account for the En locus.

 

[EmmaRainboe]

If both parents are homozygous on all alleles where that is possible you have these three possibilities

50% AABBCcchdDDeeEnen (broken black agouti non-extention)

25% AABBCcchdDDeeenen (black agouti non-extention)

25% AABBCcchdDDeeEnEn (Charlie black agouti non-extention) -this one would be particularly interesting since Charlie's already have most of their pigment removed, plus the addition of the non-extention which will further remove dark pigmentation


The most complicated scenario leads to 156 unique allelic combinations.

This is the 5 trait cross with locus’ A-E, but then you have to multiple this Punnet Square by three (technically by four but two quarters will be broken so it is not unique allelic combinations) to account for the En locus.

View attachment 3051586

Unfortunately this site only allows you to calculate five traits

 

[EmmaRainboe]

The most complicated scenario leads to 156 unique allelic combinations.

Which equivalates to 48 phenotypes, meaning there are 48 possibilities of what the babies will look like if both parents are heterozygous everywhere that is possible.