Draw a contingency table for a chi square test of linkage. Assuming that you fin

ID: 205106 • Letter: D

Question

Draw a contingency table for a chi square test of linkage. Assuming that you find your chi square value to be 4.0 (with one degree of freedom), are your genes linked? If so, how far apart are they?

sn car 65 sn car+ 35 sn+ car 40 sn+ car+ 60 200 Table 5.3. Chi-square value Probability Degrees of Freedom Non Signi-Highly significant ficantsignificant 0.95 0.90 0.80 0.70 0.500.300.20 0.10 0.05 0.01 0.004 0.02 0.06 015 046107 642.73.846.64 0.10 0.2 045 0.739 2.41 3.22 4.605.999.2 0.35 058 .0 142237 3.664.646.2578211.34 0.71 106 65 2.20336 4.885.99 7.789.49 3.28 1.14 16 234 3.0046.067.29 9.24075.09 1.63 2.20 3.07 3.835.357.238.56 10.642.5916.81 2.17 2.83 3.82 4.67635 8.38 9.80 12.0214.0718.48 2.73 3.49 4.59 5.537349.52 11.03 13.365.520.09 3.32 4.17 538 639834 10.66 2.24 14.6816.9221.67 3.94 486 6.18 7.27 9.34178 13.44 15.99 18.31 23.21 10

Explanation / Answer

Null hypothesis: Ho: The genes are linked.

Alternate hypothesis: H1: The genes are not linked.

F1 cross was female “Xsn/car Xsn/car” crossed to male “Xsn+/car+ Y”

F1 result: Xsn/car Xsn+/car+ (wild type females-straight bristles, red eyes) and Xsn/car Y (sn car males-singed bristles, carnation eyes)

F2 cross: Xsn/car Xsn+/car+ crossed to Xsn/car Y

Results are given in the table:

Observed (O)

Expected (E)

(O-E)^2

2= (O-E)^2/E

sn car

225

4.5

sn car+

225

4.5

sn+car

100

sn+ car+

100

Total is 200

Chi square calculated value is 13.

Note that null hypothesis is accepted if calculated value is less than tabulated value.

Degrees of freedom = (rows-1) (columns-1) = (4-1) (4-1) = 9

Let the level of significance be 0.05

Tabulated value = 16.92

Now, because the calculated value is less than the tabulated value, so, null hypothesis is accepted. The genes ‘sn’ and ‘car’ are linked.

Suppose the chi-square value was 4 (as given), then, also the null hypothesis would be accepted and the genes would be linked (explanation given above).

Distance between the two genes = (number of recombinants / total) * 100 = (35 + 40) /200 *100 = 37.5mu