Case Study Using The Fruit Fly, Drosophila Melanogaster
Abstract
The basic principle of Mendelian inheritance was studied using the fruit fly, Drosophila Melanogaster. A Cinnabar Brown female was crossed with Wild Type male. The flies were mated in a jar containing a nutrient medium and then placed in an incubator. The experiment was conducted over a period of 5 weeks. Cinnabar (cn) and brown (bw) are two loci in Drosophila with recessive eye colour mutations. Mutations in cn cause a bright red eye and mutations in bw cause a brown eye. Double mutants have white eyes. As predicted, all the F1 generation of flies were wild type. The F1 generation flies were crossed, and their progeny (F2 generation) produced 4 different types of phenotypes (wild type, cinnabar, brown and white) in the phenotypic …show more content…
The F2 generation (the offspring) are in their larval stage of development. The P2 generation (the parent flies) must be discarded to insure that no interference with the offspring will occur. The P2 generation was transferred from the medium bottle to an empty bottle, and then etherised with ether and discarded. The medium bottle containing the F2 generation was placed back in the fly incubator room at 25°C.
Week 5: Analysis of the F2 generation
All the larvae of the F2 generation have developed into flies. All the flies were etherized, counted and classified into their specific phenotype classes. The flies were analysed under a dissecting microscope in order to classify their phenotype classes. To determine whether the obtained results differed from the expected ratio, the Chi-square test was used.
Results
Cinnabar(cn) and brown (bw) are two loci in Drosophila with recessive eye colour mutations. Mutations in cn cause a bright red eye and mutations in bw cause a brown eye. Double mutants have white eyes. In crossing the Cinnabar Brown female fly (cn/cn ; bw/bw) and the Wild Type male fly (cn+/cn+ ; bw+/bw+), the cross was as follows:
P1: (♀) cn/cn ; bw/bw x (♂) cn+/cn+ ; bw+/bw+ cn/bw cn+/bw+ …show more content…
This P value is greater than 0.05 and therefore there is a high probability that the difference between observed and expected results is due to chance. The null hypothesis is valid for the individual Chi-square test.
Figure 1: A Table showing the Probability of results when compared to the degree of freedom and the Chi-square value
Table 5: Chi-square test to compare obtained phenotypic ratio to expected ratio (combined group results)
Phenotype Observed Expected (O-E)² (O-E)²
E
Wild Type 1903 9/16 x 3345=
1882 (1903-1882)² =
441 411/1882 = 0.219
Cinnabar 608 3/16 x 3345=
627 (608-627)² =
361 361/627 = 0.576
Brown 536 3/16 x 3345=
627 (536-627)² =
8281 8281/627 = 13.2
White 298 1/16 x 3345=
209 (298-209)² =
7921 7921/209 =37.9 Total: 3345 Total: 3345 X²=∑▒((O-E)²)/E
=
The basic principle of Mendelian inheritance was studied using the fruit fly, Drosophila Melanogaster. A Cinnabar Brown female was crossed with Wild Type male. The flies were mated in a jar containing a nutrient medium and then placed in an incubator. The experiment was conducted over a period of 5 weeks. Cinnabar (cn) and brown (bw) are two loci in Drosophila with recessive eye colour mutations. Mutations in cn cause a bright red eye and mutations in bw cause a brown eye. Double mutants have white eyes. As predicted, all the F1 generation of flies were wild type. The F1 generation flies were crossed, and their progeny (F2 generation) produced 4 different types of phenotypes (wild type, cinnabar, brown and white) in the phenotypic …show more content…
The F2 generation (the offspring) are in their larval stage of development. The P2 generation (the parent flies) must be discarded to insure that no interference with the offspring will occur. The P2 generation was transferred from the medium bottle to an empty bottle, and then etherised with ether and discarded. The medium bottle containing the F2 generation was placed back in the fly incubator room at 25°C.
Week 5: Analysis of the F2 generation
All the larvae of the F2 generation have developed into flies. All the flies were etherized, counted and classified into their specific phenotype classes. The flies were analysed under a dissecting microscope in order to classify their phenotype classes. To determine whether the obtained results differed from the expected ratio, the Chi-square test was used.
Results
Cinnabar(cn) and brown (bw) are two loci in Drosophila with recessive eye colour mutations. Mutations in cn cause a bright red eye and mutations in bw cause a brown eye. Double mutants have white eyes. In crossing the Cinnabar Brown female fly (cn/cn ; bw/bw) and the Wild Type male fly (cn+/cn+ ; bw+/bw+), the cross was as follows:
P1: (♀) cn/cn ; bw/bw x (♂) cn+/cn+ ; bw+/bw+ cn/bw cn+/bw+ …show more content…
This P value is greater than 0.05 and therefore there is a high probability that the difference between observed and expected results is due to chance. The null hypothesis is valid for the individual Chi-square test.
Figure 1: A Table showing the Probability of results when compared to the degree of freedom and the Chi-square value
Table 5: Chi-square test to compare obtained phenotypic ratio to expected ratio (combined group results)
Phenotype Observed Expected (O-E)² (O-E)²
E
Wild Type 1903 9/16 x 3345=
1882 (1903-1882)² =
441 411/1882 = 0.219
Cinnabar 608 3/16 x 3345=
627 (608-627)² =
361 361/627 = 0.576
Brown 536 3/16 x 3345=
627 (536-627)² =
8281 8281/627 = 13.2
White 298 1/16 x 3345=
209 (298-209)² =
7921 7921/209 =37.9 Total: 3345 Total: 3345 X²=∑▒((O-E)²)/E
=