Part II: Do a chi-square analysis on Mendel\'s data shown below (taken from Snud
ID: 184636 • Letter: P
Question
Part II: Do a chi-square analysis on Mendel's data shown below (taken from Snudstad and Simmons Genetics, 5th edition). TABLE 3.1 Results of Mendel's Monohybrid Crosses F2 Progeny Ratio Parental Strains Tall plants x dwarf plants Round seeds wrinkled seeds Yellow seeds X green seeds Violet flowers white flowers Inflated pods x constricted pods 882 inflated, 299 constricterd Green pods X yellow pods Axial flowers terminal flowers 787 tall, 277 dwarf 5474 round, 1850 wrinkled 6022 yellow, 2001 green 705 violet, 224 white 2.84:1 2.96:1 3.01:1 3.15:1 2.95:1 2.82:1 3.14:1 428 green, 152 yellow 651 axial, 207 terminalExplanation / Answer
Answer:
1)
Degree of freedom = number of categories - 1 = 2 - 1 = 1
Probability corresponding to Chi square value of 0.607 with 1 degree of freedom P = 0.4359. The result is not significant at p < 0.05 and Null Hypothesis is not rejected. Therefore, there is sufficient evidence that there is no difference between the results of the genetic cross and the predicted Mendelian ratio of 3:1.
2)
Degree of freedom = number of categories - 1 = 2 - 1 = 1
Probability corresponding to Chi square value of 0.263 with 1 degree of freedom P = 0.6081. The result is not significant at p < 0.05 and Null Hypothesis is not rejected. Therefore, there is sufficient evidence that there is no difference between the results of the genetic cross and the predicted Mendelian ratio of 3:1.
3)
Degree of freedom = number of categories - 1 = 2 - 1 = 1
Probability corresponding to Chi square value of 0.015 with 1 degree of freedom P = 0.9025. The result is not significant at p < 0.05 and Null Hypothesis is not rejected. Therefore, there is sufficient evidence that there is no difference between the results of the genetic cross and the predicted Mendelian ratio of 3:1.
4)
Degree of freedom = number of categories - 1 = 2 - 1 = 1
Probability corresponding to Chi square value of 0.391 with 1 degree of freedom P = 0.5318. The result is not significant at p < 0.05 and Null Hypothesis is not rejected. Therefore, there is sufficient evidence that there is no difference between the results of the genetic cross and the predicted Mendelian ratio of 3:1.
5)
Degree of freedom = number of categories - 1 = 2 - 1 = 1
Probability corresponding to Chi square value of 0.064 with 1 degree of freedom P = 0.8003. The result is not significant at p < 0.05 and Null Hypothesis is not rejected. Therefore, there is sufficient evidence that there is no difference between the results of the genetic cross and the predicted Mendelian ratio of 3:1.
6)
Degree of freedom = number of categories - 1 = 2 - 1 = 1
Probability corresponding to Chi square value of 0.451 with 1 degree of freedom P = 0.5019. The result is not significant at p < 0.05 and Null Hypothesis is not rejected. Therefore, there is sufficient evidence that there is no difference between the results of the genetic cross and the predicted Mendelian ratio of 3:1.
7)
Degree of freedom = number of categories - 1 = 2 - 1 = 1
Probability corresponding to Chi square value of 0.350 with 1 degree of freedom P = 0.5541. The result is not significant at p < 0.05 and Null Hypothesis is not rejected. Therefore, there is sufficient evidence that there is no difference between the results of the genetic cross and the predicted Mendelian ratio of 3:1.
Observed (O) Expected (3:1) (E) (O-E) (O-E)^2/E Tall Plants 787 798 -11 0.152 Dwarf Plants 277 266 11 0.455 Total 1064 1064 Chi Square (sum) 0.607