The data are generated from an initial parental cross. One parent displays the disease phenotype and one displays the wild-type (WT) phenotype. The WT parent always has a homozygous genotype. Your task is to perform a chi-square goodness of fit test on each of two F2 data sets, and make a decision, based on your statistical analyses as to which F2 data set provides greater evidence for indicating the correct mode of inheritance. In this problem, the true mode of inheritance is autosomal dominant, homozygous lethal. Evidence is measured in the following ways: the p-value is greater than 0.05, so we do not reject the null hypothesis, and the p-value is closer to 1.
The data are generated from an initial parental cross. One parent displays the disease phenotype and one displays the wild-type (WT) phenotype. The WT parent always has a homozygous genotype.
Your task is to perform a chi-square goodness of fit test on each of two F2 data sets, and make a decision, based on your statistical analyses as to which F2 data set provides greater
evidence for indicating the correct
In this problem, the true mode of inheritance is autosomal dominant, homozygous lethal.
Evidence is measured in the following ways: the p-value is greater than 0.05, so we do not reject the null hypothesis, and the p-value is closer to 1.
A few things of which to be mindful.
1. In the parental generation, the WT parent always has a homozygous genotype.
2. For the autosomal dominant mode of inheritance, the disease gene always homozygous for the disease allele in the parental generation.
3. For the autosomal dominant, homozygous lethal any person with the disease phenotype is heterozygous for their genotype.
4. The mode of inheritance is the same for both data sets, and so is the parental cross.
DATA SET 01
Parental cross: Mother with disease phenotype, Father with wild-type phenotype.
F1 COUNTS FOR DATA SET 01
| Gender | Phenotype | |
| Disease | Wild-type | |
| Male | 115 | 123 |
| Female | 131 | 128 |
F2 COUNTS FOR DATA SET 01
| Gender | Phenotype | |
| Disease | Wild-type | |
| Male | 112 | 128 |
| Female | 126 | 131 |
DATA SET 02
F1 COUNTS FOR DATA SET 02
| Gender | Phenotype | |
| Disease | Wild-type | |
| Male | 297 | 291 |
| Female | 294 | 298 |
F2 COUNTS FOR DATA SET 02
| Gender | Phenotype | |
| Disease | Wild-type | |
| Male | 276 | 311 |
| Female | 286 | 307 |
In this question, you are given two data sets that provide counts of F1 and F2 offspring with given genders and disease phenotypes.
The data are generated from an initial parental cross. One parent displays the disease phenotype and one displays the wild-type (WT) phenotype. The WT parent always has a homozygous genotype.
Your task is to perform a chi-square goodness of fit test on each of two F2 data sets, and make a decision, based on your statistical analyses as to which F2 data set provides greater
evidence for indicating the correct mode of inheritance.
In this problem, the true mode of inheritance is autosomal dominant, homozygous lethal.
Evidence is measured in the following ways: the p-value is greater than 0.05, so we do not reject the null hypothesis, and the p-value is closer to 1.
A few things of which to be mindful.
1. In the parental generation, the WT parent always has a homozygous genotype.
2. For the autosomal dominant mode of inheritance, the disease gene always homozygous for the disease allele in the parental generation.
3. For the autosomal dominant, homozygous lethal any person with the disease phenotype is heterozygous for their genotype.
4. The mode of inheritance is the same for both data sets, and so is the parental cross.
DATA SET 01
Parental cross: Mother with disease phenotype, Father with wild-type phenotype.
F1 COUNTS FOR DATA SET 01
| Gender | Phenotype | |
| Disease | Wild-type | |
| Male | 115 | 123 |
| Female | 131 | 128 |
F2 COUNTS FOR DATA SET 01
| Gender | Phenotype | |
| Disease | Wild-type | |
| Male | 112 | 128 |
| Female | 126 | 131 |
DATA SET 02
F1 COUNTS FOR DATA SET 02
| Gender | Phenotype | |
| Disease | Wild-type | |
| Male | 297 | 291 |
| Female | 294 | 298 |
F2 COUNTS FOR DATA SET 02
| Gender | Phenotype | |
| Disease | Wild-type | |
| Male | 276 | 311 |
| Female | 286 | 307 |
We do not reject either null hypothesis, since the chi-square goodness of fit test p-value is greater than 0.05 for each test.
However, Data Set 01's p-value is closer to 1.0, so we conclude that it provides more evidence for the mode of inheritance being that stated in the question.
We reject Data Set 01's null hypothesis, since the chi-square goodness of fit test p-value is less than 0.05 for that test.
We do not reject Data Set 02's null hypothesis, since the chi-square goodness of fit test p-value is greater than 0.05 for that test.
We conclude that Data Set 02 provides more evidence for the mode of inheritance being that stated in the question.
We reject Data Set 02's null hypothesis, since the chi-square goodness of fit test p-value is less than 0.05 for that test.
We do not reject Data Set 01's null hypothesis, since the chi-square goodness of fit test p-value is greater than 0.05 for that test.
We conclude that Data Set 01 provides more evidence for the mode of inheritance being that stated in the question.
We do not reject either null hypothesis, since the chi-square goodness of fit test p-value is greater than 0.05 for each test.
However, Data Set 02's p-value is closer to 1.0, so we conclude that Data Set 02 provides more evidence for the mode of inheritance being that stated in the question.
We reject Data Set 01's null hypothesis, since the chi-square goodness of fit test p-value is less than 0.05 for that test.
We do not reject Data Set 02's null hypothesis, since the chi-square goodness of fit test p-value is greater than 0.05 for that test.
We conclude that Data Set 01 provides more evidence for the mode of inheritance being that stated in the question.
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 2 images
