In an experiment an ebony bodied sepia eyed female was cross

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?

Solution

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.
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.
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.

Get Help Now

Submit a Take Down Notice

Tutor
Tutor: Dr Jack
Most rated tutor on our site