L - a paralyzed fly In this case a true breeding paralyzed...

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L From the last lecture, we followed gene segregation in a cross of a true breeding shibire fly with a wild type fly. Shibire x wild type F 1 : all not paralyzed F 2 : 3 not paralyzed : 1 paralyzed This is the segregation pattern expected for a single gene. But in an actual experiment how do we know that the phenotypic ratio is really 3 : 1 ? There is no logical way to prove that we have a 3 :1 ratio. Nevertheless, we can think of an alternative hypothesis then show that the alternative hypothesis does not fit the data. Usually, we then adopt the simplest hypothesis that still fits the data. A possible alternative hypothesis is that recessive mutations in two different genes are needed to get
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Unformatted text preview: a paralyzed fly. In this case a true breeding paralyzed fly would have genotype: aaaaa / aaaaa , bbbbb / b Whereas wild type would have genotype: AAAAA / AAAAA , BBBBB / B F1: AAAAA / aaaaaBBBBB / b not paralyzed F2: p( aaaaa / aaaaa and bbbbb / bbbbb ) = ( 1/ 4 ) 2 = 1 /16 p( aaaaa / aaaaa and BBBBB / ––––– ) = 1 /4 x 3 /4 = 3 /16 p( bbbbb / bbbbb and AAAAA / ––––– ) = 3 /16 p( AAAAA / ––––– and BBBBB / ––––– ) = the rest = 9 /16 This is the classic ratio for two gene segregation 9 : 3 : 3 : 1 paralyzed...
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