1. Tumors in Drosophila
This is a mutation linked to sex
The expression of this mutation is variable and is an example of incomplete penetrance
The mutation is temperature dependent.
The mutant phenotype is the presence of the tumor in the abdomen, torax and in some cases in the head or leggs.
The tumor develop during the larval stage.

5. Incidence of melanotic tumors between two lines of Drosophila melanogaster (tu(1)S2 ts and Oregon R) reared at 220 C and 290 C .
Objectives:
Analyze your 4 groups of Drosophila; two lines reared at different temperature (220C and 290 C).
Define your null hypotesis (H0)
Test H0 using the Ji- square of contingency.

6. TEST IF:
Is there difference in the presence of tumors between the male and female for the two lines of “moscas” reared at 220C and 290 C)?

7. b) Is there difference between “moscas” reared at 220C and 290 C for each of the two lines?.

8. c) Is there difference in the presence of tumors between the two lines reared at each of the different temperatures (220C and 290 C)?

9. Chi- square test for contingency 2x2 table
In many cases data are collected simultaneously for two variables and we want to test the hypothesis that the frequencies of occurrence in the various categories of one variable are independent of the frequencies in the second variable.
Data are organized in contingency tables (rows x colums).
The null hypothesis for contingency tables is that frequencies of observation found in the rows are independent of the frequencies of observation found in the column.
1 2 Total
A a b a+b
B c d c+d
Total a+c b+d N=a+b+c+d
χ2 = N(ad – bc )2
(a+b)(a+c)(c+d)(b+d)
Degree of freedom = (rows -1)(column-1)

10. (87/300)100
Degree of freedom = (rows -1)(column-1)

11. For contingency table 2 x 2 the formula can be simplified
H0: in the sampled population handedness is indipendent of sex
HA: in the sampled population handedness is not indipendent of sex
Boys girls Total
Left handed
Right handed
Total
1 2 Total
A a b a+b
B c d c+d
Total a+c b+d N=a+b+c+d
χ2 = N(ad – bc )2
(a+b)(a+c)(c+d)(b+d)
χ2 = (6x24 – 12x28)2= x34x52x36
χ2 = Σ (O – E )2 = [6-(34/70)18] 2 +[12-(36/70)18] 2 +[28-(34/70)52] 2 +[24-(36/70)52] 2 = 2.25
E (34/70) (36/70)18 34/70) (36/70)52
Degree of fredom = (2-1)(2-1)=1
Χ20.05_1=3.841
Therefore do not reject H0
Yates correction for continuity
χ2 = N[(ad – bc) -N/2 ]2 = abcd