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The degrees of freedom is critical to computing the p-value for the chi-square goodness of fit. We will discuss the p-value in the next lab. The degrees of freedom for the Chi-Square goodness of fit is equal to (the number of non-zero rows in the table) - 1. The number of non-zero rows is the number of expected values that are non-zero. For example: Phenotype ... Expected ... Disease ... 234 ... Wild-type ... 6743 ...

The degrees of freedom is critical to computing the p-value for the chi-square goodness of fit. We will discuss the p-value in the next lab. The degrees of freedom for the Chi-Square goodness of fit is equal to (the number of non-zero rows in the table) - 1. The number of non-zero rows is the number of expected values that are non-zero. For example: Phenotype ... Expected ... Disease ... 234 ... Wild-type ... 6743 ...

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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The degrees of freedom is critical to computing the p-value for the chi-square goodness of fit. We will discuss the p-value in the next lab. The degrees of freedom for the Chi-Square goodness of fit is equal to (the number of non-zero rows in the table) - 1. The number of non-zero rows is the number of expected values that are non-zero. For example:

Phenotype ... Expected ...
Disease ... 234 ...
Wild-type ... 6743 ...

has two rows where the Expected value is non-zero. So the degrees of freedom is 2-1 = 1.

Another example:

Phenotype ... Expected ...
Disease ... 2374 ...
Wild-type ... 0 ...

has one row where the Expected value is non-zero. So the degrees of freedom is 1-1 = 0. With 0 degrees of freedom, the chi-square test cannot be calculated.

A third example (sex-linked):

Phenotype ... Expected ...
Disease-Male ... 2374 ...
Disease-Female ... 9456  
Wild-type-Male ... 1001 ...
Wild-type-Female ... 235 ...

has four rows where the Expected values are non-zero. So the degrees of freedom is 4-1 = 3

Assignment on determining whether specified MOI is consistent with observed data In this assignment, I would like you to compute the chi-square goodness of fit test for an observed data set, using two different MOIS. A critically important point in our work is that MOI just means the proportions that correspond to the different phenotypes. For this exercise, the two MOIS are Autosomal Dominant (AD) and Sex Linked Recessive (SLR). The data in the table below is the counts of F2 offspring from an F1 cross. We use the following notation for the alleles at the gene of interest: Notation: D = disease allele at gene d = wild-type allele at gene Let's provide some information about each MOI. AD: Each of the Fi parents is heterozygous for the D allele. SLR: The observed F2 data are the offspring from an Fi cross where the father is affected (disease) and the mother is an unaffected carrier (wild-type, or WT). That is, the mother's genotype has one disease allele and one WT allele in it. In Table 1, I provide the observed data counts (in dark blue). Notice that I stratify by gender. Table 1. (0-E)² (O-E)/E Phenotype Disease, Male Disease, Female WT, Male E O-E 304 267 285 WT, Female 301 Total 1157 DF p-value 1| Page
Transcribed Image Text:Assignment on determining whether specified MOI is consistent with observed data In this assignment, I would like you to compute the chi-square goodness of fit test for an observed data set, using two different MOIS. A critically important point in our work is that MOI just means the proportions that correspond to the different phenotypes. For this exercise, the two MOIS are Autosomal Dominant (AD) and Sex Linked Recessive (SLR). The data in the table below is the counts of F2 offspring from an F1 cross. We use the following notation for the alleles at the gene of interest: Notation: D = disease allele at gene d = wild-type allele at gene Let's provide some information about each MOI. AD: Each of the Fi parents is heterozygous for the D allele. SLR: The observed F2 data are the offspring from an Fi cross where the father is affected (disease) and the mother is an unaffected carrier (wild-type, or WT). That is, the mother's genotype has one disease allele and one WT allele in it. In Table 1, I provide the observed data counts (in dark blue). Notice that I stratify by gender. Table 1. (0-E)² (O-E)/E Phenotype Disease, Male Disease, Female WT, Male E O-E 304 267 285 WT, Female 301 Total 1157 DF p-value 1| Page
In Table 2, provide expected proportions for two different MOIS: AD and SLR. The values in this table are computed using information from the Punnett Square and the specified MOI. Once the proportions are determined, we can fill in the values in Table 3, E(xpected) column. Table 2. AD proportions SLR proportions Phenotype Disease, Male Disease, Female WT, Male WT, Female Your assignment is to fill in Table 1 twice (make two copies of Table 1), once using the AD proportions in Table 2 to compute the E(xpected) column, and once using the SLR proportions in Table 2 to compute the E(xpected) column. Then, follow through, decide whether either/both/neither of the specified MOIS are consistent with the observed data, and report your results. I'll get you started with the AD MOI (Table 3): Table 3. (O-E)? (0-E)'/E Phenotype Disease, Male Disease, Female WT, Male E О-Е 304 =1157 x 0.375 (433.875) 285 267 WT, Female 301 =1157 x 0.125 (144.625) Total 1157 DF p-value You must fill in the rest of the cells, compute the Chi-Square statistic, determine the degrees of freedom (DF), compute the p-value, and determine whether you reject or do not reject the null hypothesis that you specified the correct MOI. Remember, you are doing this work for two MOIS, and remember to show all work. 2I Page
Transcribed Image Text:In Table 2, provide expected proportions for two different MOIS: AD and SLR. The values in this table are computed using information from the Punnett Square and the specified MOI. Once the proportions are determined, we can fill in the values in Table 3, E(xpected) column. Table 2. AD proportions SLR proportions Phenotype Disease, Male Disease, Female WT, Male WT, Female Your assignment is to fill in Table 1 twice (make two copies of Table 1), once using the AD proportions in Table 2 to compute the E(xpected) column, and once using the SLR proportions in Table 2 to compute the E(xpected) column. Then, follow through, decide whether either/both/neither of the specified MOIS are consistent with the observed data, and report your results. I'll get you started with the AD MOI (Table 3): Table 3. (O-E)? (0-E)'/E Phenotype Disease, Male Disease, Female WT, Male E О-Е 304 =1157 x 0.375 (433.875) 285 267 WT, Female 301 =1157 x 0.125 (144.625) Total 1157 DF p-value You must fill in the rest of the cells, compute the Chi-Square statistic, determine the degrees of freedom (DF), compute the p-value, and determine whether you reject or do not reject the null hypothesis that you specified the correct MOI. Remember, you are doing this work for two MOIS, and remember to show all work. 2I Page
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