In an experiment, an ebony bodied, sepia eyed female was crossed with a homozygo
ID: 311523 • Letter: I
Question
In an experiment, an ebony bodied, sepia eyed female was crossed with a homozygous wild type male. The resulting progeny were allowed to mate with one another. Three repititions of the experiment were conducted. The following data were produced from the crosses. Test these data to determine if they are significantly different from the expected phenotypic ratio. Remember to use the 5% level of significance. Find X2 values for each experiement.
Exp. Wild Ebony Sepia Ebony body
No. Type Body Eye Sepia Eye
1 72 24 20 7
2 112 40 35 11
3 100 31 35 12
Find the X2 value for each experiment. Would pooling of the data be appropriate? If so please calculate and compare the pooled value. If not apporpriate please explain why.
Find Ho ratio, Degrees of freedom, critical value, X2 value and would you reject or not reject the Ho?
Explanation / Answer
As all results showing 9:3:3:1 ratio, we can pool up the data. It won't differ from exact values.
Null hypothesis The observed values are not devaiting from 9:3:3:1 ratio Categories Wild Type EbonyBody Sepia Eye Ebony body, Sepia Eye Total Observed values 72 24 20 7 123 Expected Ratio 9 3 3 1 16 Expected values 69.1875 23.0625 23.0625 7.6875 123 Deviation 2.8125 0.9375 -3.0625 -0.6875 D^2 7.91015625 0.87890625 9.37890625 0.47265625 D^2/E 0.114329268 0.038109756 0.406673442 0.06148374 0.620596206 X2 value 0.620596206 DF 4-1 3 Inference The calculated X^2 values is less than table values (7.84 @ 0.05 Probability and 3 DF), hence nullhypothesis accepted. Null hypothesis The observed values are not devaiting from 9:3:3:1 ratio Categories Wild Type EbonyBody Sepia Eye Ebony body, Sepia Eye Total Observed values 112 40 35 11 198 Expected Ratio 9 3 3 1 16 Expected values 111.375 37.125 37.125 12.375 198 Deviation 0.625 2.875 -2.125 -1.375 D^2 0.390625 8.265625 4.515625 1.890625 D^2/E 0.003507295 0.222643098 0.121632997 0.152777778 0.500561167 X2 value 0.500561167 DF 4-1 3 Inference The calculated X^2 values is less than table values (7.84 @ 0.05 Probability and 3 DF), hence nullhypothesis accepted. Null hypothesis The observed values are not devaiting from 9:3:3:1 ratio Categories Wild Type EbonyBody Sepia Eye Ebony body, Sepia Eye Total Observed values 100 31 35 12 178 Expected Ratio 9 3 3 1 16 Expected values 100.125 33.375 33.375 11.125 178 Deviation -0.125 -2.375 1.625 0.875 D^2 0.015625 5.640625 2.640625 0.765625 D^2/E 0.000156055 0.169007491 0.07911985 0.068820225 0.31710362 X2 value 0.31710362 DF 4-1 3 Inference The calculated X^2 values is less than table values (7.84 @ 0.05 Probability and 3 DF), hence nullhypothesis accepted. AFTER DATA POOLING Null hypothesis The observed values are not devaiting from 9:3:3:1 ratio Categories Wild Type EbonyBody Sepia Eye Ebony body, Sepia Eye Total Observed values 284 95 90 30 499 Expected Ratio 9 3 3 1 16 Expected values 280.6875 93.5625 93.5625 31.1875 499 Deviation 3.3125 1.4375 -3.5625 -1.1875 D^2 10.97265625 2.06640625 12.69140625 1.41015625 D^2/E 0.039092073 0.022085838 0.135646293 0.045215431 0.242039635 X2 value 0.242039635 DF 4-1 3 Inference The calculated X^2 values is less than table values (7.84 @ 0.05 Probability and 3 DF), hence nullhypothesis accepted.Related Questions
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