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# GENETICS EXPERIMENTS FLYLAB - Essay Example

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Fly Lab Experiment Name Subject Teacher Date               Fly Lab Experiment PART 1 MONOHYBRID CROSS a-d. Based on what I know about the principles of Mendelian genetics, the phenotypic ratio for crossing two parents that are homozygous for the genes is 1:1 or 100% because if the first gene were WW for wild type and the second were SS for sepia eyes, and one parent was WW while the other was SS, then all offspring would be W and S for both genes, naturally with wild type and sepia eye phenotypes…

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For the F2 generation, my expected phenotypic ratio was 1:1 still because the male and female flies involved were homozygous for the gene and since the phenotypes were wild type and sepia eyes, then both traits were dominant. When the F1 male and female flies were mated and this resulted in 3697 and 3717. With my expected phenotype ratio of 1:1, the Chi-Square result recommended rejecting my hypothesis. -------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #2 Phenotype Observed Hypothesis Expected Chi-Square Term Female: + 3697 1.0000 2447.7500 637.5756 Male: + 3717 1.0000 2447.7500 658.1536 Female: SE 1194 1.0000 2447.7500 642.1771 Male: SE 1183 1.0000 2447.7500 653.4951 Total 9791 4.0000 9791.0000 2591.4014 Chi-Squared Test Statistic = 2591.4014 Degrees of Freedom = 3 Level of Significance = 0.0000 Recommendation: Reject your hypothesis ----------------------------------------------------------------------- Another hypothesis was tested at 3:1, and this time, chi-square results favored the hypothesis. This means that in the F2 generation, there are 3 wild types for every 1 sepia-eyed fly. -------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #2 Phenotype Observed Hypothesis Expected Chi-Square Term Female: + 3697 3.0000 3671.6250 0.1754 Male: + 3717 3.0000 3671.6250 0.5608 Female: SE 1194 1.0000 1223.8750 0.7293 Male: SE 1183 1.0000 1223.8750 1.3651 Total 9791 8.0000 9791.0000 2.8305 Chi-Squared Test Statistic = 2.8305 Degrees of Freedom = 3 Level of Significance = 0.4185 Recommendation: Do not reject your hypothesis -------------------------------------------------------------------------- PART 1 MONOHYBRID CROSSES For the cross of the wild-type male and the ebony-bodied female fly, the expected phenotypic ratio is 1:1 for the F1 generation and 3:1 for the F2 generation, based on the results of previous experiments. The results were the same with the actual ratios for the F1 and F2 generation experiments. For the F1 generation, there were equal number of wild types and ebony bodies. For the F2 generation, there are 3 wild types for every ebony bodies. -------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #3 (F1 Generation) Phenotype Observed Hypothesis Expected Chi-Square Term Female: + 501 1.0000 496.5000 0.0408 Male: + 492 1.0000 496.5000 0.0408 Total 993 2.0000 993.0000 0.0816 Chi-Squared Test Statistic = 0.0816 Degrees of Freedom = 1 Level of Significance = 0.7752 Recommendation: Do not reject your hypothesis (1:1) -------------------------------------------------------------------------- Chi Square Hypothesis Using Cross #4 (F2 GENERATION) Phenotype Observed Hypothesis Expected Chi-Square Term Female: + 383 3.0000 376.1250 0.1257 Male: + 387 3.0000 376.1250 0.3144 Female: E 109 1.0000 125.3750 2.1387 Male: E 124 1.0000 125.3750 0.0151 Total 1003 Read More
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