2.In Drosophila, pure breeding males with hairy legs were crossed to pure-breedi
ID: 187568 • Letter: 2
Question
2.In Drosophila, pure breeding males with hairy legs were crossed to pure-breeding dumpy- bodied females. All the F1 males were wild-type, and all the F1 females had hairy legs When intercrossed (males x females), the F1 flies produced an F2 distributed among four classes, with males and females having similar proportions of phenotypes. The F2 data are shown in the table below Observed numbers Phenotypes Wild-type Hairy legs Dumpy body Hairy, dumpy 138 140 60 62 400 (a) Give the genotypes of the parents and the F1. [ (b) What are the expected ratios in F2? (c) To the table below, add the expected numbers for each phenotypic class. Compare the expected numbers with the observed numbers by means of a chi square test at the 5% level of significance. Show your work and the items asked for below In the F2 the following types of flies were produced, distributed equally as to sex: Observed numbers Expected numbers, Calculations Phenotypes 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 (d) Do we reject the hypothesis given by the expected ratio at the 5% level of significance? How did you decide this?Explanation / Answer
Ans-
(Symbols:
Hairy legs - HH+
Dumpy body - dd+
Genotype of parents- Male- HY d+d+ (Hairy legs)
Female- H+H+dd (dumpy body)
F1 generatio- H+Y d+d (male ), H+H d+d (female)
(b) expected ratio in F2 generation- 3:3:1:1
`
(c)
Observed Expected (O – E)2 (O – E)2/E Chi square
Wild-type 138 150 (–12)2 144/150 = 0.96
Hairy 140 150 (–10) 2 100/150 = 0.67
Dumpy 60 50 (+10) 2 100/50 = 2.00
Hairy, dumpy 62 50 (+12) 2 144/50 = 2.88
400 400 6.51
Chi square test formula:
( Observed vakue – Expected value)2/Extected value
O; Observed value
E; Expected value
Formula: degree of freedom = total no. classes – 1
Number of degree of freedom for this exercise- 4 – 1 = 3
(d) Degree of freedom for this exercise is ‘3’, so we can find out the critical value of chi-square at the 5% level of significance given in the table-7.815
Since chi-square (6.51) is less than 7.816, the probability value is greater than 5%; thus, we can retain the hypothesis that the 3:3:1:1 ratio is possible.
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