4.19.2007 - 4.19.2007 P F1 F2 cn cn (cinnabar eyes) x cn+...

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4.19.2007 P cn cn (cinnabar eyes) x cn + cn + (wild type eyes) F 1 cn + cn F 2 o May be specified how to get this generation in the problem, if you are not given specific instructions just interbreed the F 1 to get the F 2 in a MONOHYBRID CROSS (hybrid – heterozygocity and mono- one gene is in consideration) o Work the punnet square to get the answer cn+ cn+ cn+ cn+cn+ cn+cn cn cn+cn cncn o Punnet square is reenactment of Mendel’s first law and random fertilization o ½ of F2 will be heterozygous (wild) – cn+cn o ¼ will be homozygous cinnabar – cncn o ¼ will be homozygous wild type – cn+cn+ So ¾ wild type and ¼ cinnabar Dihybrid cross P ( Female ) cncn pb + pb + x cn + cn + pbpb ( male ) o Proboscis is the mouth part, looks like a trunk and pedia is foot so probosopedia is where you have a foot growing out of what should be the mouth part (male has this) o Female will always breed true to cn and pb+ o Male is true breeding with always having a cn and pb o Probosopedia gene also encodes an enzyme, so you already know F1 will have wild type eyes…what can you say about the proboscis? It is also there in these heterozygotes F1 o Heterozygous and wild type o cn + cn pb + pb F2 o We will interbreed the F1s o It depends on if the genes really do follow Mendel’s second law (if the genes are independent) we will work the two genes independently to get F2, but if not you must work it differently o What is the genotypic ratio in the F2 for a cross between two heterozygotes at the cinnabar locus? ¼, ½, ¼ as noted earlier cn ¼ cn + cn + ¼ pb + pb + , ½ pb + pb, ¼ pbpb o 1/16 cn + cn + pb + pb + o 1/8 cn + cn + pb + pb o 1/16 cn + cn + pbpb ½ cn + cn ¼ pb + pb + , ½ pb + pb, ¼ pbpb o 1/8 cn + cn pb + pb + o ¼ cn + cn pb + pb o 1/8 cn + cn pbpb ¼ cncn ¼ pb + pb + , ½ pb + pb, ¼ pbpb o 1/16 cncn pb + pb + o 1/8 cncn pb + pb o 1/16 cncn pbpb o Now do probosopedia – do a punnet square or you can know because its also a cross between two heterozygotes Pb ¼ pb + pb + ½ pb + pb
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¼ pbpb o What if we only look at F2s that are homozygous wild type at cinnabar? cn + cn + If ¼ of the F2 are homozygous wild type at proboscopedia…then ¼ of these homozygous wild types at cinnabar will be homozygous wild type at probosopedia…this is because the genes are independent of each other The two genotypes are independent because the two genes are independent o What this means is ¼ of this sub group (cn + cn + ) will be homozygous for proboscopedia Now you multiply, ¼ of the ¼ will have those two genotypes so you write that 1/16 of the F2
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4.19.2007 - 4.19.2007 P F1 F2 cn cn (cinnabar eyes) x cn+...

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