Question 11
the following table shows the observed and expected crossover and non-crossover asci for the tan and gray spore genes of sordaria fimicola
from a cross between the two mutants. use the chi-squared test to determine if the gene - centromere estimated distance is significantly
different from the published map distances. [3 pt)
asci
crossover, observed
non-crossover, observed
crossover, expected
non-crossover, expected
tan spore gene
189
139
0.54 x 328) 177
(0.46 x 328) = 151
gray spore gene
204
124
0.66 x 328) = 216
10.34 x 328) = 112
only the gray spore gene is significantly different
only the tan spore gene is significantly different
both genes are significantly different
neither gene is significantly different

Option D

Neither gene is significantly different

Explanation:

Tan Spore gene

Cross-over

Observed=189

Expected=177

Observed-Expected=189-177=12

Standard deviation,

Non-cross-over

Cross-over

Observed=139

Expected=151

Observed-Expected=139-151=-12

Standard deviation,

Chi-square value=0.813559+0.953642=1.767202

Rounding off to three decimal places, chi-square value is 1.767

Gray Spore Gene

Cross-over

Observed=204

Expected=216

Observed-Expected=204-216=12

Standard deviation,

Non-cross-over

Cross-over

Observed=124

Expected=112

Observed-Expected=124-112=12

Standard deviation,

Chi-square value=0.666667+1.285714=1.952381

Rounding off to three decimal places, chi-square value is 1.952

Degree of freedom=n-1 where n is sample size, the sample size here are 2 hence degree of freedom is 2-1=1

From chi-square table, when degree of freedom is 1, chi-square value is 3.841

Since for both genes, the chi-square values are less than 3.841 then we should accept that data since it’s not significantly different