2. In corn, colored seed coat (R) is dominant to colorless (r), and green plant
ID: 200827 • Letter: 2
Question
2. In corn, colored seed coat (R) is dominant to colorless (r), and green plant color (G) is dominant to yellow (g). Two F1 plants, each heterozygous for both characteristics, are testcrossed to homozygous recessives, and their progeny are combined to produce the following totals Colored green Colored yellow Colorless green Colorless yellow 100 97 103 100 Use chi-square analysis to test these data for independent assortment of the two characteristics a. b. When the progeny of each of the two heterozygous plants were scored separatel following results were obtained Plant 1 Phenotypes Colored green Colored yellow Colorless green Colorless yellow Plant 2 12 85 95 12 92 Use chi-square analysis to test each data set for independent assortment. Explain the results of these three chi square analysis. (hint, what does it tell you about combining data in experiments?) C.Explanation / Answer
Test static –Chisquare test:
Category
Colored green
Colored yellow
Colorless green
Colorless yellow
Total
Observed values (O)
100
97
103
100
400
Exptected Ratio (ER)
1
1
1
1
4
Exprected Values (E)
100
100
100
100
Deviation (O-E)
0
-3
3
0
D^2
0
9
9
0
D^2/E
0
0.09
0.09
0
0.18
X^2
0.18
Degrees of freedom
4 -1 =3
Inference: The calculated chisquare value i.e. 0.18 is less than the table value i.e. 7.92 at 3 DF and 0.05 probability. Hence the null hypothesis is accepted.
Plant 1:
Null hypothesis : The observed values are not deviating from the 1:1:1:1 ratio.
Test static –Chisquare test:
Category
Colored green
Colored yellow
Colorless green
Colorless yellow
Total
Observed values (O)
88
12
8
92
200
Exptected Ratio (ER)
1
1
1
1
4
Exprected Values (E)
50
50
50
50
Deviation (O-E)
38
-38
-42
42
D^2
1444
1444
1764
1764
D^2/E
28.88
28.88
35.28
35.28
128.32
X^2
128.32
Degrees of freedom
4-1=3
Inference: The calculated chisquare value i.e. 128.32 is greater than the table value i.e. 7.92 at 3 DF and 0.05 probability. Hence the null hypothesis is rejected. Which indicates that these genes are linked.
Plant 2:
Null hypothesis : The observed values are not deviating from the 1:1:1:1 ratio.
Test static –Chisquare test:
Category
Colored green
Colored yellow
Colorless green
Colorless yellow
Total
Observed values (O)
12
85
95
8
200
Exptected Ratio (ER)
1
1
1
1
4
Exprected Values (E)
50
50
50
50
Deviation (O-E)
-38
35
45
-42
D^2
1444
1225
2025
1764
D^2/E
28.88
24.5
40.5
35.28
129.16
X^2
129.16
Degrees of freedom
4-1=3
Inference: The calculated chisquare value i.e. 129.16 is greater than the table value i.e. 7.92 at 3 DF and 0.05 probability. Hence the null hypothesis is rejected. Which indicates that these genes are linked.
c. The plant 1 results indicates that those genes are arranged in Cis pattern, plant 2 results indicates that those genes are arranged in Transmanner. The combined results indictes null results.
Category
Colored green
Colored yellow
Colorless green
Colorless yellow
Total
Observed values (O)
100
97
103
100
400
Exptected Ratio (ER)
1
1
1
1
4
Exprected Values (E)
100
100
100
100
Deviation (O-E)
0
-3
3
0
D^2
0
9
9
0
D^2/E
0
0.09
0.09
0
0.18
X^2
0.18
Degrees of freedom
4 -1 =3
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