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Discussion Starter · #1 ·
Ok so I'm in the process of learning budgie mutations and I thought I Would do some punnet squares and post the results here to see if I'm on the right track. SO here are my results from my example budgies.

male- yellow base, single dark factor, double violet factor
female- yellow base split to white base, double dark factor, single grey factor, single violet, opaline

so would the results be
males
100% yellow based
50% double dark
50% single dark
100% single grey
100% double violet
100%split to opaline

Females
100% split to whiteface
50% ouble dark
50%single dark
100% no grey
100% single violet
50% normal -not opaline
50% opaline


If I'm wrong please explain why I am wrong.
 

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~Crazy Bird Lady~
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Your on the right track,but there are a few mistakes I can see. Colors are not sex-linked. So it wouldn't be that males are split for a white base and females aren't. It would just be half the chicks would be split and half not, gender doesn't matter. Same with the grey and violet genes, gender doesn't matter.

Opaline on the other hand is Sex-linked, so in order to get visual males, the both parents need to carry the mutation. In your example the males would all be Split for Opaline and the females would not get the gene.
 

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Discussion Starter · #3 ·
Well, I did a basic punnet square with four squares and I got the top row was yellow and I thought the top was males and the bottom I got B( yellow) b( white) in each of the squares. I thought that was for females. Is that wrong?
 

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Your results should be:

Males
100% split for opaline
50% single factor violet, 50% double factor violet
50% homozygous yellow base, 50% yellow base split for white base
50% single factor grey
50% single dark factor, 50% two dark factors

Females
100% normal (no opalines)
50% single factor violet, 50% double factor violet
50% homozygous yellow base, 50% yellow base split for white base
50% single factor grey
50% single dark factor, 50% two dark factors

A sex-linked gene like opaline must be present in the male for it to be visable in any of the chicks

Any double factor mated to a single factor will produce 50% each of single & double.
Male = factor/factor (double factor)
Female = factor/normal (single factor)
Taking 1 from each parent:
1st: factor from male + factor from female = factor/factor
2nd: factor from male + normal from female = factor/normal
3rd: factor from male + factor from female = factor/factor
4th: factor from male + normal from female = factor/normal
Result 50% each
 
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