5 We are given phenotype counts of F2 individual genotypes

Last updated: 6/16/2023

5 We are given phenotype counts of F2 individual genotypes from a F cross The gene is on the X Chromosome and the Po cross consists of an affected father and an unaffected mother who is homozygous for the wild type allele at the gene Specifying Phenotype Male Affected Male Unaffected Female Affected Female Unaffected 0 35 The statistic cannot be calculated 1 013 0 64 0 02 Count 156 139 0 300 2 points Given the following table of observed data for an autosomal disease Observed Number Phenotype Disease Wild type 47 109 What appears to be the most reasonable conclusion The most likely expected proportions for the phenotypes are 1 2 for Disease phenotype 1 2 for Wild type phenotype based on the Chi Square Goodness of Fit Test The disease does not appear to follow an AD AR or ADHL mode of inheritance based on the Chi Square Goodness of Fit The most likely expected proportions for the phenotypes are 1 4 for Disease phenotype 3 4 for Wild type phenotype based on the Chi Square Goodness of Fit Test The most likely expected proportions for the phenotypes are 3 4 for Disease phenotype 1 4 for Wild type phenotype based on the Chi Square Goodness of Fit Test hogod on the Chi Square Goodness of Fit Test