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Is this right?

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  • 07-19-2008, 09:45 AM
    spix14
    Is this right?
    Some times I feel like I will never get all this genetic stuff straight in my head. Just to make sure I'm on the right track...

    Mojaves are co-dominant.
    BELs are the super form.
    Mojave x Normal= 50% Mojos, 50% Normals (het for nothing since there is no het mojo)
    Mojave x Mojave= 50% Mojo, 25% BEL, 25% Normal.
    BEL x Normal= 100% Mojo.
    BEL x BEL= not sure on this one. All BEL? All Mojos?

    Snake breeding is complicated business. :weirdface
  • 07-19-2008, 11:04 AM
    RandyRemington
    Re: Is this right?
    Pretty much right on. The only thing I would say differently is that ALL Mojaves ARE hets. Heterozygous doesn't mean normal looking gene carrier, it just works out that way with recessive morphs. Heterozygous really means having an unmatched pair of genes at whatever location you are talking about. With co-dominant morphs the hets are visible morphs. A mojave has one mojave mutant version of the lesser/mojave/phantom/Vin Russo/mocha whitesnake complex gene and one normal version of that same gene so it is truly a het. Because it doesn't look normal and the homozygous mojave looks different the mutation is classified as co-dominant.

    Also, as long as the BEL you are talking about was produced bye mojave X mojave (could be a combo with another mutant version of this same gene like lesser) then you are right on about BEL X normal producing 100% mojave.

    And a BEL bred to another BEL would produce 100% BEL because there would be no normal copies of this gene between the two parents so none of the offspring could get even one normal version to even be just mojave much less normal.
  • 07-19-2008, 11:16 AM
    spix14
    Re: Is this right?
    Awesome, thank you. So basically a Mojo is a visual het of BEL. I knew that in the back of my head but I keep thinking normal looking when I think het, I need to break myself of that, lol.

    What I really meant by "no het mojo" was that there are no hets for the mojo morph itself. Which is correct, right?

    Incidentally, are all normals produced by a co-dom to co-dom pairing just 100% normal?
  • 07-19-2008, 11:19 AM
    stangs13
    Re: Is this right?
    Quote:

    Originally Posted by spix14 View Post
    What I really meany by "no het mojo" was that there are no hets for the mojo morph itself. Which is correct, right?

    Incidentally, are all normals produced by a co-dom to co-dom pairing just 100% normal?

    Correct

    and

    Correct!:D
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