Drosophila Melanogaster Lab Experiment

Suppose the feather color of a bird is controlled by two alleles, D and d. The D allele results in dark feathers, while the d allele results in lighter feathers.

Jack J, DeLotto Y. Effect of Wing Scalloping Mutations on Cut Expression and Sense Organ Differentiation in the Drosophila Wing Margin. Genetics. 1992;131(2):353-363.

Apply your understanding of how alleles assort and combine during reproduction to evaluate a scenario involving a monohybrid cross.

It was decided that there would be 80 vestigial flies and 20 wild type flies to total to an initial population of 100 drosophila. Next, the flies were anesthetized flies using Fly Nap. The flies were counted out to reach desired ratio, sexing the flies making sure there are equal amounts of males and females to be sure there is ample individuals to allow successful mating. The fly’s food was prepared by taking a frozen rotten banana, cutting it in half, mashing up the banana meat, and mixing yeast into it. The

Based on your results for the female offspring, predict whether color blindness is a dominant or recessive trait. Explain your reasoning.

Butterflies have many genes which are expressed into ways that are either dominant, or recessive. For example to have blue eyes the dominant allele would be (B) and the recessive allele would be (b).

METHODS: In this experiment, the instructor provided us with 30 ebony individuals and 20 wild type individuals. In order to get an exact amount of each type, we anesthetized the flies and counted them off by gently using a fine point paint brush. Then all 50 Drosophila were put into a population cage which had a lid that had six holes for the centrifuge tubes. Two food tubes and four clean, empty tubes were added on the first day. Each food tube consisted of half a cup full of food mixed with 6-7 milliliters of water. This was the fly medium. The food should turn blue once the water is added. Each tube was labeled with a number and with the date. Every two to three days we added one more food tube until all 6 tubes contained the fly medium. After all 6 tubes were filled, the following days after we exchanged the first food tube with a new food tube. At the end of the experiment, we fed the flies with a total of 8 food tubes. Then the flies were anesthetized, again. At the end of this four week lab, the number of living ebony and wild

Mendel’s law of independent assortment deals with dihybrid crosses meaning that independent assortment dealt with the crosses in Group 2 (ap+/ap; se+/se x ap+/ap; se+/se) and Group 4 (vg+/vg; se+/se x vg+/vg; se+/se). This is also the law of independent assortment as the cross deals with the production of haploid cells to the offspring (Gen.: Analysis & Principles, p28). Independent assortment is observed in these two crosses as there are to different traits within the

but homozygous recessive for seed color (Ttyy). If 80 offspring are produced, how many are expected to be

We set up a vial agar; it had two drosophila f1female and five drosophila f1 males. After one week pupas were visible and the parental were removed. A week after this the developed f2 drosophilas were counted after being analyzed.

It would be expected that the mutant F1 flies would be heterozygous for the allele responsible for the grounded trait. If two F1 flies were mated, the percentage of flies that would be expected to be wildtype in the F2 generation would be 25% mutants given that the mutant allele (ap) is predicted to be recessive and, leaving 75% to be wildtype (ap+).

we said goodbye and placed them in the fly morgue. We allowed the F2 larval

The mutation of these flies were unknown. This vial served as the parental generation. The parents were left to mate and then transferred to a new a vial. Virgin females were then collected and kept in a separate vial. The female virgins were then crossed with balancer chromosome male flies.

11. The progeny of a Drosophila female (heterozygous at three loci: y, ct, and w) crossed to a wild type male are listed below:

For our first generation (F1) of flies we chose to cross apterous (+) females and white-eye (w) males. We predicted that the mutation would be sex linked recessive. So if the female was the sex with the mutation then all females would be wild type heterozygous. Heterozygous is a term used when the two genes for a trait are opposite. The males would all be white eye since they only have one X chromosome. If the males were the sex that had the mutation then all the flies would be wild type but the females would be heterozygous.