You are given a stock of female Drosophila that appears wild type. You cross them with...
You are given a stock of female Drosophila that appears wild type. You cross them with wild-type males and collect progeny, recording the following results: 1,000 wild-type females, 450 wild-type males, and 50 white-eyed males. (Note the male to female ratio.)
How many genes are assorting? What phenotypes do they control? Is linkage indicated? If so, what is the map distance?
Homework Answers
Representation:
Considering trait of eye color:
Wildtype: W (dominant)
- Male= XWY
- Female= XW XW (Homozygous), XW Xw (Heterozygous)
White eye: w (recessive)
- Male= XwY
- Female= Xw Xw
Cross: Wildtype (heterozygous) Female with wildtype male:
- XW Xw x XWY
|
F1 |
Female |
||
|
Male |
XW |
Xw |
F2 |
|
XW |
XW XW |
XW Xw |
Female |
|
Y |
XWY |
XwY |
Male |
- XW XW = Wildtype female (Homozygous) = 25% =1/4
- XW Xw = Wildtype female (Heterozygous)= 25%=1/4
- XWY = Wildtype male = 25%=1/4
- XwY = White eyed males = 25%=1/4
Phenotype ratio of wild type female: Wildtype male: White eyed males
= 2:1:1
Wildtype: mutant = 3:1
Thus,
- There is one gene with two alleles (dominant for wildtype eye color and recessive for white color eye) which are assorting.
- There independent assortment and no linkage.
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