Week 4

Week 4 - Week 4, Lecture 1 Things that Mendel never saw...

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Week 4, Lecture 1 Things that Mendel never saw (cont’d) : 3) Lethal Allele - one of the allele combinations is lethal Aa x Aa -> 2:1 or 3 ratio A a A a AA Aa Aa aa If aa is lethal to the organism only the other phenotype(s) will show up in a 3 or 2:1 ratio or AA can be lethal to get a 2Aa: 1aa phenotypic and genotypic rqtio. The 1:2 ratio can also be 2Aa and one AA if the recessive allele is lethal (tail-less mice). 4) Gene Interaction – Epistasis (masking) flower color: 1. P Purple A x white A -> F1 purple self -> F2 3 purple : 1 white and 2. P Purple B x white B -> F1 purple self -> F2 3 purple : 1 white These two separate genes look exactly alike, producing similar results So 3. P white A x white B -> F1 purple self -> F2 9 purple : 7 white Out of two white flowers comes purple! How????? Lets look at the genotype: 1. AA__ x aa__ 2. __BB x __bb 3. aa_ x __bb So what are the missing genotypes? 1. AABB x aaBB 2. AA BB x AA bb 3. aaBB x AA bb Notice that white A (cross 1) and white B (cross 2) have opposite genotypes. Therefore when they are crossed together (cross 3) the F1 has a double heterozygous genotype, which shows that white A compliments white B resulting in a classic F2 ratio of 9:7 (or 9:3+3+1)
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In other words, it requires a dominant allele in both genes to produce a purple colored plant. If only one gene has a dominant allele or neither gene has a dominant allele the plant will be white. In the white1 x white2 cross complementation occurs. White1 complements white2 so that there is a dominant allele in each gene and the purple plant occurs. This is an example of Sequential Gene Action . This also makes sense metabolically. If enzyme A and enzyme B are both enzymes necessary for the purple colored plant then the lack of either enzyme would yield the white colored plant. If only the AA or Aa genotypes for gene A code for enzyme A and the same is true for gene B then either the aa or bb genotypes would block the production of one of the enzymes and the white colored plant would show. If there is no A gene then there is no A enzyme to convert A to B. If there is no B gene then there is no B enzyme to convert B to purple. A B purple. Comb shape in chickens: gene interaction only P Rose comb (RRpp ) x single (rrpp ) -> F1 Rose (Rrpp) sib mate -> 3 rose: 1 single P Pea comb (rr PP) x single (rr pp) -> F1 Pea (rrPp) sib mate -> 3 pea: 1 single P Rose (RRpp) x Pea (rrPP) -> F1 Walnut (RrPp) sib mate -> 9 walnut: 3 rose: 3 pea: 1 single What are the rules for Walnut Chicken? It is necessary for at least one of the dominant alleles for each gene to be present for the walnut color to show up (R_P_). In this example the genes are not interfering (or covering up the phenotype). If only one dominant allele showed up the corresponding color would show (R_pp gave a rose comb and rrP_ gave a pea comb). If no dominant alleles were present it would be single (rrpp). Epistasis
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Week 4 - Week 4, Lecture 1 Things that Mendel never saw...

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