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.There are three possible modes of inheritance that underlie the generation of the data. All are monogenic. They are:(i) Autosomal Recessive.(ii) Homozygous Lethal Dominant.(iii) Autosomal Dominant. One's phenotype is determined by their genotype at the disease locus and the mode of inheritance, as we have seen with the Punnett Square.In this file you are provided with the true mode of inheritance.Your assignment 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.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.In the parental generation, the WT parent always has a homozygous genotype.For the autosomal dominant moi, the disease is always homozygous for the disease allele.For the autosomal dominant, homozygous lethal any person with the disease phenotype is heterozygous for their genotype.The mode of inheritance is the same for both data sets, and so is the parental cross. DATA SET 01The number of offspring in F1 and F2 generations: 510The true mode of inheritance is autosomal dominant.Parental cross: Father with disease phenotype, Mother with wild-type phenotype. F1 COUNTS FOR DATA SET 01PhenotypeGender. Disease Wild-Type Male 237 0Female 273 0 F2 COUNTS FOR DATA SET 01PhenotypeGender Disease Wild-Type Male 185 73Female 183 69 DATA SET 02The number of offspring in F1 and F2 generations: 899F1 COUNTS FOR DATA SET 02PhenotypeGender Disease Wild-Type Male 424 0Female 475 0 F2 COUNTS FOR DATA SET 02PhenotypeGender Disease Wild-Type Male 349 112Female 323 115Should be similar to this one The number of offspring in F1 and F2 generations: 854Punnett Square for Fi parents.Mother's allelesFather's allelesd.dDDdDddddddFrom intro, we know that WT must be homozygous, and because we are dealing with AutosomalDominant, Homozygous Lethal, other parent must be Dd. This table is identical to parental Punnettsquare.So, according to Punnett Square, F1 generation has 50% disease, 50% WT. Multiply by ½ to get male andfemale, or 25% for each.F1 COUNTS FOR DATA SET 02PhenotypeGender Disease Wild-TypeMale 209213Female 223209F2 COUNTS FOR DATA SET 022 |PagePhenotypeGender Disease Wild-TypeMale 219182Female 213240F2 CHI-SQUARE TEST ANALYSIS WITH TRUE MOITable for compute X^2 Goodness of Fit valuePhenotypeObsCounts(O) ExpPropExpCounts(E) (О-Е)213,5213,5213,5213,5(O-E)^2/E-31.55.526,5MaleWT1822194.64750.141690.25Male Disease0.25FemaleFemale DiseaseWT2400.253.28922130.25-0.50.001171X^2 Goodness-of-fit Statistic : 8.07962529Degrees of freedomP-value:3:0.0443944375P-value < 0.05. Reject null hypothesisfor your specified mode of inheritance.So, data set most consistent with MOI is Data Set 01, since we do not reject the null hypothesis, and wedo in Data Set 02. The number of offspring in F1 and F2 generations: 854Punnett Square for Fi parents.Mother's allelesFather's allelesd.dDDdDddddddFrom intro, we know that WT must be homozygous, and because we are dealing with AutosomalDominant, Homozygous Lethal, other parent must be Dd. This table is identical to parental Punnettsquare.So, according to Punnett Square, F1 generation has 50% disease, 50% WT. Multiply by ½ to get male andfemale, or 25% for each.F1 COUNTS FOR DATA SET 02PhenotypeGender Disease Wild-TypeMale 209213Female 223209F2 COUNTS FOR DATA SET 022 |PagePhenotypeGender Disease Wild-TypeMale 219182Female 213240F2 CHI-SQUARE TEST ANALYSIS WITH TRUE MOITable for compute X^2 Goodness of Fit valuePhenotypeObsCounts(O) ExpPropExpCounts(E) (О-Е)213,5213,5213,5213,5(O-E)^2/E-31.55.526,5MaleWT1822194.64750.141690.25Male Disease0.25FemaleFemale DiseaseWT2400.253.28922130.25-0.50.001171X^2 Goodness-of-fit Statistic : 8.07962529Degrees of freedomP-value:3:0.0443944375P-value < 0.05. Reject null hypothesisfor your specified mode of inheritance.So, data set most consistent with MOI is Data Set 01, since we do not reject the null hypothesis, and wedo in Data Set 02.

Name: You are given two data sets that provide counts of F1 and F2 offspring
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Description: 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) phen

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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. There are three possible modes of inheritance that underlie the generation of the data. All are monogenic. They are: (i) Autosomal Recessive. (ii) Homozygous Lethal Dominant. (iii) Autosomal Dominant. One's phenotype is determined by their genotype at the disease locus and the mode of inheritance, as we have seen with the Punnett Square. In this file you are provided with the true mode of inheritance. Your assignment 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. 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. In the parental generation, the WT parent always has a homozygous genotype. For the autosomal dominant moi, the disease is always homozygous for the disease allele. For the autosomal dominant, homozygous lethal any person with the disease phenotype is heterozygous for their genotype. The mode of inheritance is the same for both data sets, and so is the parental cross. DATA SET 01 The number of offspring in F1 and F2 generations: 510 The true mode of inheritance is autosomal dominant. Parental cross: Father with disease phenotype, Mother with wild-type phenotype. F1 COUNTS FOR DATA SET 01 Phenotype Gender. Disease Wild-Type Male 237 0 Female 273 0 F2 COUNTS FOR DATA SET 01 Phenotype Gender Disease Wild-Type Male 185 73 Female 183 69 DATA SET 02 The number of offspring in F1 and F2 generations: 899 F1 COUNTS FOR DATA SET 02 Phenotype Gender Disease Wild-Type Male 424 0 Female 475 0 F2 COUNTS FOR DATA SET 02 Phenotype Gender Disease Wild-Type Male 349 112 Female 323 115 Should be similar to this one