Extracted text: 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 F1 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. O E (O-E)? (O-E)/E Phenotype Disease, Male Disease, Female O-E 304 267 WT, Male WT, Female 285 301 Total 1157 DF p-value 1|Page
Extracted 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