I really need help with dihybrid crosses plz!

Genetics as it applies to evolution, molecular biology, and medical aspects.

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Enzyme
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Post by Enzyme » Fri Jan 06, 2006 4:50 pm

sdekivit wrote:
Enzyme wrote:
xnabizkox wrote:GgTt x GgTt

______[GT]___[Gt]___[gT]___[gt]
[GT] _ GGTT _ GGTt _ GgTT _ GgTt

[Gt] _GGTt___ GGtt__ GgTt__ Ggtt

[gT] _ GgTT _GgTt _ ggTT __ ggTt

[gt] __GgTt___ Ggtt__ ggTt __ggtt


If you want to find the chances for having heterozygous for both treats, the answer would be 4/16 which we can simplify to 1/4. YES, IT IS CORRECT :wink:.


with my 'probability-tree': Gg = 1/2 and Tt = 1/2

--> GgTt = 1/2 * 1/2 = 1/4 = 4/16


He he, your 'probability tree' is very nice :D.
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xnabizkox
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Post by xnabizkox » Fri Jan 06, 2006 6:19 pm

Thank you. It was actually a lot harder to make then I first imagined. But it's all in the name of science :D
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sdekivit
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Post by sdekivit » Fri Jan 06, 2006 7:42 pm

ty enzyme :!: :idea: :D

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Enzyme
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Post by Enzyme » Sat Jan 21, 2006 10:29 pm

He he, you're welcome ;).
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