1) In peas, axillary flowers (A) is dominant to terminal flowers (a) and colored
ID: 268286 • Letter: 1
Question
1) In peas, axillary flowers (A) is dominant to terminal flowers (a) and colored flowers (R white (r). A true-breeding plant with white, terminal flowers is crossed with a true-h colored, axillary flowers. The Fi plants are then allowed to self-pollinate. The Fa generaion fha of: 299 plants with axillary, colored flowers 32 plants with terminal, white flowers 121 plants with axillary, white flowers 133 plants with terminal, colored flowers Using the chi-square test, determine if this outcome is consistent with the predicted ratios for ti n Provide the null and alternative hypotheses, chi-square statistic, P-value, and conclusion.(00 pExplanation / Answer
Null hypothesis : The observed values are not deviating from the mendilian dihybrid cross ratio i.e. 9:3:3:1.
Alternative hypothesis: The observed values are deviating from the mendilian dihybrid cross ratio i.e. 9:3:3:1.
Test static – Chisquare test:
Category
Axillary, colored
Axillary, white
Terminal, colored
terminal, white
Total
Observed values (O)
299
121
133
32
585
Exptected Ratio (ER)
9
3
3
1
16
Exprected Values (E)
329.0625
109.6875
109.6875
36.5625
Deviation (O-E)
-30.0625
11.3125
23.3125
-4.5625
D^2
903.7539
127.9727
543.4727
20.81641
D^2/E
2.746451
1.166702
4.954736
0.569338
9.437227
X^2
9.437227
Degrees of freedom
4-1 = 3
P –value: The p value at 0.05 probability and 3 DF is 7.82.
Conclusion: As the calculated chisquare value i.e. 9.43 is greater than the table value i.e. 7.82 at 3 DF and 0.05 probability, the null hypothesis is rejected and alternative hypothesis is accepted.
Category
Axillary, colored
Axillary, white
Terminal, colored
terminal, white
Total
Observed values (O)
299
121
133
32
585
Exptected Ratio (ER)
9
3
3
1
16
Exprected Values (E)
329.0625
109.6875
109.6875
36.5625
Deviation (O-E)
-30.0625
11.3125
23.3125
-4.5625
D^2
903.7539
127.9727
543.4727
20.81641
D^2/E
2.746451
1.166702
4.954736
0.569338
9.437227
X^2
9.437227
Degrees of freedom
4-1 = 3
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