Drosophila Essay

1036 Words3 Pages

INTRODUCTION:
Drosophila is a small fruit fly, it is about 3mm long. This insect is a model organism most commonly used in developmental biology and genetics. The Drosophila fruit flies are especially suited in experiments because of their short life cycle which consist of two weeks; they easily reproduce many offspring, and are also cheap1. The drosophila contains four chromosomes that can easily be experimented on, which allows in-depth observation. In this experiment, Drosophila melanogaster were used to identify the properties of Mendelian inheritance. The Law of Segregation states that allele pairs separate during gamete formation and randomly unite during fertilization and is carried by every individual. The Law of Independent Assortment states that each parent randomly passes on alleles to their offspring. Although, the Law of Independent assortment does not take in account the patters of sex-linked inheritance.
The Drosophila live a distinct four-staged life cycle that requires approximately two weeks to reach complete maturity2. The stages are known as egg, larval, pupa, and adult. The egg is a small oval shape, and can barely be seen by the unaided eye, they are hatched the day after being laid. In order for the larva to molt and grow in size it consists of three stages: first instar, second instar, and third instar3; during this process the larvae is preparing itself for metamorphosis by shedding body parts and consuming excessive nutrients. Through metamorphosis the immature fruit fly attaches itself to an object and its outer shell hardens, it then begins the transformation process into an adult. Once the process is completed, the adult is then able to begin the sexual reproduction process within forty-eight hours....

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...Vestigial wings / ebony body 1 4.5 2.72
Vestigial wings / wild type body 11 13.5 .462
Wild type wings / ebony body 18 13.5 1.5
Wild type wings / wild type body 42 40.5 .055
Totals 72 4.73

DISCUSSION:
For both the monohybrid cross, and dihybrid cross chi-square tables were used to determine whether the deviation of the experiment was due to chance alone. The chi-square result for the monohybrid cross resulted in 6.53, ending up between .05 (X2= 5.991) and .01 (X2=9.210) with a degree of freedom of n=2 (3-1). This result leads to the rejection of the null hypothesis because there was only a 5% chance that the observations were due to chance alone. As for the dihybrid cross, the chi-square data resulted in 4.73 landing in between .20 (X2=4.642) and .05 (X2=7.815). This resulted in the null hypothesis being accepted since it is higher than .05.

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