Consider a situation where you have a parental cross with the mother and father phenotypes listed below. Remember that the genotype for the wild-type parent is always homozygous. The counts of the F1offspring are listed in Table 1. Two randomly selected individuals are selected and mated to produce a set of F2individuals. If you test the F2counts to determine whether they are consistent with an autosomal dominant mode of inheritance, what is your decision? Use a 0.05 significance level. PARENTAL CROSS Parental cross: Mother with disease phenotype, Father with wild-type phenotype. Table 1. F1 DATA Gender Phenotype Disease Wild-type Male 0 23 Female 0 34 Table 2. F2 DATA Gender Phenotype Disease Wild-type Male 7 25 Female 0 25 A. Do not reject the null hypothesis that the F2 data are consistent with an autosomal dominant mode of inheritance; chi-square goodness of fit test p-value is greater than 0.05. B. Reject the null hypothesis that the F2 data are consistent with an autosomal dominant mode of inheritance; chi-square goodness of fit test p-value is 1e-06. C. Reject the null hypothesis that the F2 data are consistent with an autosomal dominant mode of inheritance; the p-value for the chi-square goodness of fit test is less than 0.05. D. Reject the null hypothesis that the F2 data are consistent with an autosomal dominant mode of inheritance; the p-value for the chi-square goodness of fit test is greater than 0.05. E. Do not reject the null hypothesis that the F2 data are consistent with an autosomal dominant mode of inheritance; chi-square goodness of fit test statistic value is between 0 and 1.
Consider a situation where you have a parental cross with the mother and father
The counts of the F1offspring are listed in Table 1. Two randomly selected individuals are selected and mated to produce a set of F2individuals. If you test the F2counts to determine whether they are consistent with an autosomal dominant mode of inheritance, what is your decision? Use a 0.05 significance level.
PARENTAL CROSS
Parental cross: Mother with disease phenotype, Father with wild-type phenotype.
Table 1. F1 DATA
| Gender | Phenotype | |
| Disease | Wild-type | |
| Male | 0 | 23 |
| Female | 0 | 34 |
Table 2. F2 DATA
| Gender | Phenotype | |
| Disease | Wild-type | |
| Male | 7 | 25 |
| Female | 0 | 25 |
A. Do not reject the null hypothesis that the F2 data are consistent with an autosomal dominant mode of inheritance; chi-square goodness of fit test p-value is greater than 0.05.
B. Reject the null hypothesis that the F2 data are consistent with an autosomal dominant mode of inheritance; chi-square goodness of fit test p-value is 1e-06.
C. Reject the null hypothesis that the F2 data are consistent with an autosomal dominant mode of inheritance; the p-value for the chi-square goodness of fit test is less than 0.05.
D. Reject the null hypothesis that the F2 data are consistent with an autosomal dominant mode of inheritance; the p-value for the chi-square goodness of fit test is greater than 0.05.
E. Do not reject the null hypothesis that the F2 data are consistent with an autosomal dominant mode of inheritance; chi-square goodness of fit test statistic value is between 0 and 1.
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