Lecture 4 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 F1: all not paralyzed F2:3 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 a paralyzed fly In this case a true breeding paralyzed fly would have genotype: d/a, b/b Whereas wild type would have genotype: A/A,B/B F1: A/a B/b not paralyzed F2: p(@/a and b/b)=(14)2=1/6 p(a/andB-)=1/4×3/4=3/16 pb/band4/-)=3/16 P(A/-and B/-)=the rest:9/16 This is the classic ratio for two gene segregation 9: 3: 3: 1 aralyzed For our hypothesis we should see a phenotypic ratio of 15 not paralyzed 1 paralyzedLecture 4 Lecture 4Lecture 4 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 ↓ F1: all not paralyzed ↓ F2: 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 a paralyzed fly. In this case a true breeding paralyzed fly would have genotype: a/a , b/b Whereas wild type would have genotype: A/A , B/B F1: A/a B/b not paralyzed F2: p(a/a and b/b) = (1/4 )2 = 1/16 p(a/a and B/–) = 1/4 x 3/4 = 3/16 p(b/b and A/–) = 3/16 p(A/– and B/–) = the rest = 9/16 This is the classic ratio for two gene segregation 9 : 3 : 3 : 1 paralyzed For our hypothesis we should see a phenotypic ratio of 15 not paralyzed : 1 paralyzed