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 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 A. 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. B. 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. C. 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. D. 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. E. 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

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 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 A. 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. B. 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. C. 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. D. 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. E. 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

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Question

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 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

A. 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.

B. 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.

C. 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.

D. 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.

E. 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.

Definition Definition Measure of central tendency that is the value that occurs most frequently in a data set. A data set may have more than one mode if multiple categories repeat an equal number of times. For example, in a data set with five item—3, 5, 5, 29, 473—the mode is 5 because it occurs twice and no other value occurs more than once. On a histogram or bar chart, the element with the highest bar represents the mode. Therefore, the mode is sometimes considered the most popular option. The mode is useful for nominal or categorical data (e.g., the most common color car that users purchase), but it is problematic for continuous data because it is more likely not to have any value that is more frequent than the other.

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 5 steps

SEE SOLUTION Check out a sample Q&A here