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bronze from albino


Theplatymaster
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Both parents have to have the recessive gene that causes lack of pigment in albinism. If I breed all albino cory the fry are all albino. I do keep a tank with bronze and albino and when the bronze breed with the albino the fry are all bronze. I suppose it would be possible for a bronze to have one recessive gene and breed with an albino and some of the fry have both genes and be albino....

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Presuming that albinism in Corydoras aeneus follow Mendelian genetics, then yes, fry from Albino males and albino females should all be albino.

Here is how this works:

Albinism is a recessive (non-dominant) genetic trait.

So, let the gene for "Bronze" coloration be = B

And let the gene for "albino" coloration be = a

I've capitalized the "B" (for Bronze) because it is the dominant coloration, and let the "a" (for albinism) stay lower case because it is non-dominant.

Now, a Bronze-colored Corydoras aeneus can be either: B-a, where albinism is a present but recessive trait; or B-B, where the gene for albinism is not present at all.

But an albino Corydoras aeneus must possess both recessive genes: a-a

In theory at least, this means that albino Corydoras aeneus will _always_ pass on these albino genes to fry that will likewise carry the double-recessive colorations.

Now, it is possible that the perceived "albinism" is not true. Always look at the eyes to see if they are red.

The simple way to see this is to set up an old Punnett Square (H.S. Biology class):

One parental genetic marker trait is show across the top, and the other along the side...

     B     B

B

B

In this case, all four quarters of the fry will show the traits like this...

     B     B

B  BB  BB

B  BB  BB

Now, this the case of Albinism, this would look more like...

     a     a

a  aa  aa

a  aa  aa

So all of the fry should be albino.

Now, if one parent is heterozygous -- looks Bronze, but carries both genes, then the outcome looks like...

     B     a

a  Ba  aa

a  Ba  aa

In this scenario, 2 of the quarters of fry will be like the heterozygous parent -- looking Bronze but carrying both genes. The other 2 quarters will be albino

If both parents are heterozygous, then...

     B     a

B  BB  Ba

a  Ba  aa

Here, one quarter is true Bronze; 2 quarters are heterozygous (but will look Bronze, because that's dominant); and only one quarter will be albino

__________________

Having said all of that . . . I'm just arm-chairing this. I'd be glad to hear from a Biologist who can be more instructive. I have been told that certain species (e.g. Ancistrus) do not follow strict Mendelian genetics. But, anecdotally, I've seen it born out across several generations that _grand-fry_ from a color cross will split back out in Mendelian quarters, close to Punnett square predictions.

The way that I sometimes think about fish traits is to imagine a CUBE, where each side presents a different genetic markers...

806810864_ScreenShot2022-07-14at12_03_16PM.png.c5d2713a109800459bfa66eb841ec701.png

So the net result is ultimately a sort of "matrix" of genetic things, reflecting color, size, fin & body shape / length, etc.

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On 7/14/2022 at 11:04 AM, Fish Folk said:

Presuming that albinism in Corydoras aeneus follow Mendelian genetics, then yes, fry from Albino males and albino females should all be albino.

Here is how this works:

Albinism is a recessive (non-dominant) genetic trait.

So, let the gene for "Bronze" coloration be = B

And let the gene for "albino" coloration be = a

I've capitalized the "B" (for Bronze) because it is the dominant coloration, and let the "a" (for albinism) stay lower case because it is non-dominant.

Now, a Bronze-colored Corydoras aeneus can be either: B-a, where albinism is a present but recessive trait; or B-B, where the gene for albinism is not present at all.

But an albino Corydoras aeneus must possess both recessive genes: a-a

In theory at least, this means that albino Corydoras aeneus will _always_ pass on these albino genes to fry that will likewise carry the double-recessive colorations.

Now, it is possible that the perceived "albinism" is not true. Always look at the eyes to see if they are red.

The simple way to see this is to set up an old Punnett Square (H.S. Biology class):

One parental genetic marker trait is show across the top, and the other along the side...

     B     B

B

B

In this case, all four quarters of the fry will show the traits like this...

     B     B

B  BB  BB

B  BB  BB

Now, this the case of Albinism, this would look more like...

     a     a

a  aa  aa

a  aa  aa

So all of the fry should be albino.

Now, if one parent is heterozygous -- looks Bronze, but carries both genes, then the outcome looks like...

     B     a

a  Ba  aa

a  Ba  aa

In this scenario, 2 of the quarters of fry will be like the heterozygous parent -- looking Bronze but carrying both genes. The other 2 quarters will be albino

If both parents are heterozygous, then...

     B     a

B  BB  Ba

a  Ba  aa

Here, one quarter is true Bronze; 2 quarters are heterozygous (but will look Bronze, because that's dominant); and only one quarter will be albino

__________________

Having said all of that . . . I'm just arm-chairing this. I'd be glad to hear from a Biologist who can be more instructive. I have been told that certain species (e.g. Ancistrus) do not follow strict Mendelian genetics. But, anecdotally, I've seen it born out across several generations that _grand-fry_ from a color cross will split back out in Mendelian quarters, close to Punnett square predictions.

The way that I sometimes think about fish traits is to imagine a CUBE, where each side presents a different genetic markers...

806810864_ScreenShot2022-07-14at12_03_16PM.png.c5d2713a109800459bfa66eb841ec701.png

So the net result is ultimately a sort of "matrix" of genetic things, reflecting color, size, fin & body shape / length, etc.

You see, this is one of the reasons I like having you on the forum.  Great explanation

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