BIS 2B Lecture Notes - Lecture 8: Warfarin, Gamete, Selective Breeding

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BIS 2B Full Course Notes
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BIS 2B Full Course Notes
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More genetics study questions answers: f(aa) = p2 =. 2x. 2=. 04 f(aa) = 2pq = . 2 x . 8 x 2 = . 32 f(aa) = q2 =. 8x. 8=. 64. The observed frequency of aa homozygotes is much higher than predicted by a. When combined with information from the lectures and reading, these results imply that the frequency of the a allele is increasing, that the frequency of gray morphs is increasing. What color the moth happens to be is related to probability of being eaten. We might discuss this example in lecture if we have time (you can also google pepper moths and you will get the full story): frequency of alleles in the parental generation: blue birds = yy. In this case, the green birds are all homozygotes, yy. Number of y alleles in yy birds: y = . 4 + (0. 5) (0) = . 4 (there are no heterozygotes so this is why we multiply by 0)

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