crossed to pure-breeding dumpy maee wild-type, and all the F1 females had hairy
ID: 3165211 • Letter: C
Question
crossed to pure-breeding dumpy maee wild-type, and all the F1 females had hairy legs emales having similar proportions of phenotypes. The F2 2.in Drosophila, pure breeding males with hairy legs were When intercro four classes, with males and data are shown in the table below. Intercrossed (m sed (males x females), the Fi fies produced an F2 distributed among similar pr Observed numbers Wild-type Hairy legs Dumpy body Hairy, dumpy 138 140 60 62 400 (a) Give the genotypes of the parents and the F1. 12 pts (b) What are the expected ratios in F2? [1 pt] (c) To the table below, add the expected numbers for each phenotypic class. Compare the expected numbe 5% level of significance. Show your work and the items asked for below. [4 pts) rs with bserved numbers by means of a chi square test at the the o In the F2 the following types of flies were produced, distributed equally as to sex: Observed Expected numbers numbers Calculations Wild-type Hairy legs Dumpy body Hairy, dumpy 138 140 60 62 400 Chi square test formula: Value of chi square: Formula for Degrees of freedom: Number of degrees of freedom for this exercise: ?H | Do we reject the hypothesis given by the expected ratio at the 5% level of sign did you decide this?Explanation / Answer
a)Genotype of the parent is HH+(hairy leged male),dd+ for dumpy body .
PARENTS-HH+ X dd+--------------> F1 (Wild type male).HY;d+d+ and H+H+/d/d(Hairy leg female).
b)Expected ratio in F2
138/400-0.3
140/400-0.3
60/400-0.15
62/400-0.15
Therfore the expecte ratio is 3:3:1:1
c)Expected number of phenotyp-Wild type-3/4x400=300/2=150
Hairy leg-3/4x400=300/2=150
Dumpy body-1/4x400=100/2=50
Hairy dumpy-1/4x400=100/2=50
(O-E)2/E
Using Chi square test formula- O-E=-12 0.96
-10 0.67
+10 2.00
+12 2.88
----------------
6.51
Formula for degree of freedom-no of class-1
4-1=3
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