This may be a very simple question to answer but oh well. I just separated a pair of my convicts into a 55 gallon tank... 3 inch pink female and 4 inch black male... how do I know what the likely hood of getting pink offspring is?
Convict Genetics?
- Thread starter catfishkemp
- Start date
this is what I was told last. but IDK
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Unless the black female carries the pink gene, all of the offspring will be black. If two of the black offspring later breed, one fourth of their fry will be pink and the rest black. One fourth of those will not carry any recessive gene for pink, but will be visually indistinguishable from the ones that do carry it.
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Unless the black female carries the pink gene, all of the offspring will be black. If two of the black offspring later breed, one fourth of their fry will be pink and the rest black. One fourth of those will not carry any recessive gene for pink, but will be visually indistinguishable from the ones that do carry it.
My Black Convict female doesn't have the Marbled gene but she created 50/50 Blacks/Marbles.
I bought them from a breeder on here, if your interested in some PM me. As far as getting then "Marbled" from square 1 I honestly have no idea.
So your pink convicts are homozygous recessive for the "pink" gene. We'll call that aa. Your "wild type" convicts can either be homozygous dominant (AA) or heterozygous (Aa), as both will look phenotypically wild type. If they're homozygous dominant any fry you get would turn out "wild type" but be carriers for the pink gene (Aa). If the "wild type" is heterozygous (Aa) you will get a mix, 50% pink (aa) and 50% heterozygous (Aa).
As far as JustinDavis's statement is concerned it is false, if his "wild type" female is homozygous dominant (AA) like he claims than any fry she has with a recessive type color morph will all turn out heterzygous (Aa) and hence "wild type". She must be a heterozygous carrier for the marbled phenotype in order to throw marbled fry. (Though the genetics behind marbled coloration isn't quite simple Mendelian genetics, it's more a codominant or incomplete type effect).
As far as JustinDavis's statement is concerned it is false, if his "wild type" female is homozygous dominant (AA) like he claims than any fry she has with a recessive type color morph will all turn out heterzygous (Aa) and hence "wild type". She must be a heterozygous carrier for the marbled phenotype in order to throw marbled fry. (Though the genetics behind marbled coloration isn't quite simple Mendelian genetics, it's more a codominant or incomplete type effect).
http://gregthecrazyfishguy.wordpress.com/
this site is about a guy who records his research of various breeding projects, you might find it interesting to read about the genetics, the jellybean convict article has a lot of info about the species.
this site is about a guy who records his research of various breeding projects, you might find it interesting to read about the genetics, the jellybean convict article has a lot of info about the species.
+1
I was a biology major and I don't know if I (or my genetics professor) could have said it any better. Great job!
-Cage
No that is NOT correct. No where does JustinDavis claim his 'wild type' is homozygous dominant (AA). He merely states it is not a carrier of the marbled gene. As one can see from reading through the link provied by Stratoquaris, that a 50/50 split of marbled/black is exactly the expected ratio if the 'wild type' parent is a carrier of the 'pink' gene and the other parent is homozygous marbled. When crossed with a marble, a 'wild type' parent does NOT have to be a carrier of the marbled gene, but rather only a carrier of the 'pink' gene in order to produce 50% marbled.