*2. You are given a true breeding strain of Drosophilia with normal wings and no
ID: 141773 • Letter: #
Question
*2. You are given a true breeding strain of Drosophilia with normal wings and no eyes and another strain of Drosophilia with miniature wings and eyes. You cross a minature winged and eyed female with a normal winged, no eyes male. The in the F1 generation all females have normal wings and eyes and all the males have miniature wings and eyes. You cross these F1 progeny and get the following phenotypic distribution of 500 progeny F2 generation 170 normal wings, eyes (85 males and females) 80 normal wings, eyeless (40 males and females) 190 minature wings, eyes (95 males and females) 60 minature wings, eyeless (30 males and females) Use all the provided information to answer the questions below.
A. For each character, which traits are dominant and which traits are recessive?
B. Are any of the traits sex-linked? If yes, which ones.
C. Give the genotypes for the parents and the FI males and females.
D. Assuming independent assortment, what are the expected numbers of progeny for each of the phenotypes in the F2 generation.
E. You redo the F2 cross and examine 5000 F2 flies and preform a chi-square goodness of fit test. You get a p-vale of 0.04. How would you interpret your results in relation to your hypothesis in part A concerning inheritance of eye type and wing type?
Explanation / Answer
As per data provided
Answer A Let the wings be represented by N (normal wings), n (miniature wings) where normal are dominant over miniature wings
The condition of having eyes be represented by E (Eyed), e (eyeless) where having eyes is dominant over being eyeless
Answer B Wing type is found to be sex linked, having eyes is autosomal
Answer C Genotypes of parents:-
Female = XnXnEE (miniature wings, eyed), gametes: XnE
Male = XNYee (Normal wings, eyeless), gametes : XNe and Ye
F1 generation : XNXnEe (Normal wings, eyed female) and XnYEe (miniature wings, eyed male)
Answer D F2 generation:
XnE
Xne
YE
Ye
XNE
XNXnEE
Normal wings eyed female
XNXnEe
Normal wings eyed female
XNYEE
Normal wings eyed male
XNYEe
Normal wings eyed male
XNe
XNXnEe
Normal wings eyed female
XNXnee
Normal wings eyeless female
XNYEe
Normal wings eyed male
XNYee
Normal wings eyeless male
XnE
XnXnEE
Miniature wings eyed female
XnXnEe
Miniature wings eyed female
XnYEE
Miniature wings eyed male
XnYEe
Miniature wings eyed male
Xne
XnXnEe
Miniature wings eyed female
XnXnEE
Miniature wings eyed female
XnYEe
Miniature wings eyed male
XnYee
Miniature wings eyeless female
Ratio of phenotypes irresepective of sex:
Normal wing, eyed: Normal wing, eyeless: Miniature wing, eyed: Miniature wing, eyeless = 6:2:6:2
Expected progeny, If total flies are 500,
Normal wing, eyed and Miniature wing, eyed = (6/16) x 500 = 187.5 or 188 (rounded up) each with 94 females and 94 males of each type
Normal wing, eyeless and Miniature wing, eyeless = (2/16) x 500 = 62.5 or 62 (round down) each with 31 females and 31 males of each type
Answer E
p-value of 0.04 indicates that null hypothesis is rejected and there is significant difference in expected phenotypes and observed phenotype of given cross.
XnE
Xne
YE
Ye
XNE
XNXnEE
Normal wings eyed female
XNXnEe
Normal wings eyed female
XNYEE
Normal wings eyed male
XNYEe
Normal wings eyed male
XNe
XNXnEe
Normal wings eyed female
XNXnee
Normal wings eyeless female
XNYEe
Normal wings eyed male
XNYee
Normal wings eyeless male
XnE
XnXnEE
Miniature wings eyed female
XnXnEe
Miniature wings eyed female
XnYEE
Miniature wings eyed male
XnYEe
Miniature wings eyed male
Xne
XnXnEe
Miniature wings eyed female
XnXnEE
Miniature wings eyed female
XnYEe
Miniature wings eyed male
XnYee
Miniature wings eyeless female
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