2. Do a Chi-square test of independence for the following pairs of genes on the

ID: 209691 • Letter: 2

Question

2. Do a Chi-square test of independence for the following pairs of genes on the two-point cross data (edited version attached). wings held out and cardinal pyes, short wings and ebony body (black body II), light eyes and black body_.black body and short wings, short wings and light eyes. For each one, include the contingency table, the chi-square calculation and a clear statement about whether there is clear evidence that these are not segregating independently The data below reflect the map distance tween various markers in Drosophila. These data have been generated by mating true-breeding double mutants with true-breeding wild-type flies. The wild-type F1 offspring were then mated in a test cross. Wild type Wings held out (ho) Cardinal eyes () 256 244 252 248 Wild type Wings held out (ho) Ebony body ho e 278 220 278 224 475 Wild type Cardinal eyes (c) Black hody (b) c h Wild type Short wings (sh) Wings held out (ho) 438 475 437 Wild type Short wings (sh) Ebony body (e) she 160 160 340 Wild type Black body (b) Short wings (sh) 253 249 252 246 Wild type Cardinal eyes c Curled wings (cu) c cu 371 129 129 371 Wild type Black body (b) Curled wings (cu) b cu 397 104 103 396 468 Wild type Ebony body (e) Light eyes (le) Wild type Short wings (sh) Light eyes (le) 310 190 190 310 467 le Wild type Light eyes (le) Black body (b) 255 245 256 244 428 Wild type Many wing veins (mv) Short wings (sb) mv 407 Wild type Wings held out (ho) Many wing veins (mv) ho mv 480 480

Explanation / Answer

In dihybrid test cross we get 1:1:1:1 phenotypic ratio in F2 generation.

For the first table we would do chi-square test by following way-

Therefore, all individuals would have equal numbers if there is no linkage, that is 250.

10004 = 250

Total progenies

Observed (O)

Expected (E)

O-E

(O-E)2

(O-E)2 / E

Wild type dominant

256

250

Wings held out (ho)

244

250

-6

Cardinal eyes (c)

252

250

Ho, c

248

250

-2

Total

1000

Calculate degree of freedom=

DF = (rows-1)(columns-1)

Here rows and columns are from the table of test cross which is

Number of rows would be = 4

Number of columns would be 2

Therefore DF =3 x 1= 3

Now according the table for DF 3 and X2 value 16, the significance value is 0.01, which is, lower than 0.05.

Therefore, we can reject the null hypothesis and choose the proposed hypothesis.