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1 OBSERVATIONS: Table 1. Results of testcross of male heterozygous for three enzymes (Locus 1, 2 and 3). GENOTYPE # OF PROGENY LOCUS LAB SLOT _________ 1 2 3 TEST FOR EQUAL SEGREGATION OF ALLELES Table 2. Comparison of the number of individuals showing homozygous and heterozygous genotypes for each gene. ENZYME LOCUS 1 LOCUS 2 LOCUS 3 GENOTYPE NUMBER HYPOTHESIS: (ie., if the alleles of a gene are segregated equally, what should the results show in terms of numbers of each genotype(ie., homozygous and heterozygous)?): Table 3. Chi Square test for equal segregation of the alleles in the gene for enzyme LOCUS 1. GENOTYPE OBSERVED EXPECTED (O – E) (O – E)2 (O – E)2/E TOTAL X2 = Table 4. Chi Square test for equal segregation of the alleles in the gene for enzyme LOCUS 2. GENOTYPE OBSERVED EXPECTED (O – E) (O – E)2 (O – E)2/E TOTAL X2 = 2 Table 5. Chi Square test for equal segregation of the alleles in the gene for enzyme LOCUS 3. GENOTYPE OBSERVED EXPECTED (O – E) (O – E)2 (O – E)2/E TOTAL X2 = INTERPRETATION (one interpretation can include all 3 tests from Tables 3, 4 and 5): TEST FOR INDEPENDENT ASSORTMENT Table 6. Comparison of the numbers found for each genotype when considering each gene pair separately. GENOTYPE LOCUS 1 & 2 LOCUS 2 & 3 LOCUS 1 & 3 HYPOTHESIS ( ie., what ratio of the four possible genotypes should occur if the 2 genes are independently assorted?): Table 7. Chi Square test for independent assortment of genes of LOCUS 1 & 2. GENOTYPE OBSERVED EXPECTED (O – E) (O – E)2 (O – E)2/E TOTAL X2 = Table 8. Chi Square test for independent assortment of genes of LOCUS 2 & 3. GENOTYPE OBSERVED EXPECTED (O – E) (O – E)2 (O – E)2/E TOTAL X2 = 3 Table 9. Chi Square test for independent assortment of genes of LOCUS 1 & 3. GENOTYPE OBSERVED EXPECTED (O – E) (O – E)2 (O – E)2/E TOTAL X2 = INTERPRETATION (one interpretation can cover all 3 test from Tables 7,8 and 9): MAP DISTANCE If there is evidence of linkage, calculate percentage recombination to estimate map distance (show calculations ):