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Fruit Fly Lab

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Introduction

Alice Wang Genetics of Organisms Question: Is an unknown Drosophila mutation dominant or recessive, and is it X-linked or autosomal? Hypothesis: The offspring of a pair of reciprocal Drosophila crosses is determined by which the mutation is dominant or recessive and whether it is x-linked or autosomal. Materials: * Drosophila cultures * Dry active yeast * Fly nap wands * Ether * Fruit fly media * Small paint brushes * Dissecting microscopes * Glass vials * Foam stoppers Outline: 1. First week: Immobilize and remove the adult flies of the original vial. Observe them carefully under the dissecting microscope. Separate the males from the females and look for the mutation(s). Note whether the mutation(s) is/are associated with the males or females. 2. Place the parents in the morgue( jar containing alcohol). Label the vial containing the eggs or larvae with the symbols for the mating. Also label the vial with your name and date. Place the vial in a warm location. 3. Second week: Begin by observing the F1 flies. Immobilize and examine all the flies. Record their sex and characteristics. Consider the conclusion s that can be drawn from these data. Place five or six pairs of F1 flies in a fresh culture bottle and the rest of the flies in the morgue. 4. Third week: Remove the F1 flies from the vials and place them into the morgue. ...read more.

Middle

apterous 8 8 Red x wild 46 29 Sepia x wild 4 9 Cross 4 Results: F2 Generation Phenotype # males # females Red x apterous 2 5 Red x wild 2 16 Sepia x wild 2 1 Cross of Apterous from F2 Generation Cross # Generation Phenotypes male Phenotypes female # males # females Offspring (F3) 5 F2 Red x Apterous, Sepia x Apterous Red x Apterous. Sepia x Apterous 4 1 4 2 sepia x apterous, red x apterous, sepia x wild, red x wild Cross 5 Results: F3 Generation Phenotype # males # females Sepia x wild 3 2 Red x apterous 6 7 Sepia x apterous 4 5 Red x wild 8 11 The data that I have collected is not very reliable. My results indicate that apterous is a recessive trait while the wild type is dominant according to offspring. However, my group made a huge mistake in the F1 generation. For cross three and cross four of the F1 generation, the offspring phenotype should not have contained any wild type fruit flies because apterous winged is a recessive phenotype. Instead of crossing apterous fruit flies, my group mistakenly crossed a different vial of flies. Instead of waiting for the original vial of (parent) fruit flies to lay offspring for the F1 generation, my group crossed the parent fruit flies. ...read more.

Conclusion

Thus, incorrectly identifying the characteristics of the flies could have also greatly affected the results received. Also, instead of waiting for the original vial of (parent) fruit flies to lay offspring for the F1 generation, my group crossed the parent fruit flies. Hence, improper calculation of numbers could have also caused inaccurate results. Finally, some flies could have gotten stuck in the medium and could have been identified. Improvements can be made if the investigation crossed more fruit flies, thus giving a more veritable number of fruit flies for each phenotype and hence a more reliable data. Finally, crossing the F1 generation instead of the parent generation would be more veritable for the data collected. Apterous x sepia X Apterous x wild (ww x rr) x (Ww x Rr) wr wr wr wr wr WR WwRr WwRr WwRr WwRr WwRr Wr Wwrr Wwrr Wwrr Wwrr Wwrr wR wwRr wwRr wwRr wwRr wwRr wr wwrr wwrr wwrr wwrr wwrr Red x Apterous = 7/28 Wild x Wild = 18/28 Sepia x wild = (3/18) The Chi square (?2) test is used for discontinuous values. ?2 = ? ((O-E)2/E) Where E = expected values O = observed values ?= sum of Observed Expected (O - E) (O - E)2 (O - E)2/ E Wild type (18/28) (14/28) 2 Wild type (18/28) (14/28) 2 Table to show Chi square analysis and probability of F1 generation for monohybrid cross Degrees of freedom = number genotypes - 1 2 genotypes - 1 = 1 degree of freedom Critical value = 3.84 ...read more.

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