Name: Samantha Pipher Section: 03
1. Describe the sex and phenotype of the mutant fly. Describe the phenotype as it compares to the wild type.
You can obtain this information by clicking on the fly and looking at the Properties window.
The sex of the grounded mutant fly is female and the phenotype as compared to the wildtype is wingless, but otherwise normal for other traits besides wings. This would be directly compared to the normal (winged) wildtype phenotype that is represented by an added (+).
2. You want to determine the genotype of grounded. You are unsure whether the mutant allele confers a phenotype that is dominant or recessive to wild type. You are also not sure if grounded is true-breeding or not. To determine its
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5. Now you have determined some facts about the grounded allele and the trait that it causes. Given what you know, do you expect the mutant F1 flies to be homozygous or heterozygous for the allele that causes the grounded trait? According to your reasoning, if you mated two mutant F1 flies, what percentage of flies would you expect to be wild type versus mutant in the F2 progeny? Draw a Punnett square of this cross to justify your answer.
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+).
An expected F2 cross would be as follows: (The expected ratio would be 3:1 in this case – wildtype : mutant).
ap+ ap ap+ ap+ ap+ ap ap+ ap ap+ ap ap ap
6. Now mate a mutant F1 female fly with a mutant F1 male fly. Out of the 50 F2 progeny, what percentage of flies are wild type and what percentage are mutant
You can collect individual flies that you have generated (for use in future experiments) by dragging them
parent carried the b allele. The F1 offspring of such a cross would be Bb, and
Calculate the ratios of the genotypes and phenotypes of the offspring in the F1 generation.
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
We started out with three populations; B, D, and G. In order for us to properly create controlled genetic crosses, we had to ensure that all the female flies were “virgins”.
5. Again, set the number of offspring to the maximum of 6. Then, click the Cross button repeatedly until these parents have produced about 100 F1 offspring.
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
we said goodbye and placed them in the fly morgue. We allowed the F2 larval
This lab had 2 exercises. Exercise 9.1 involved observing pictures of 60 F2 offspring and recording the phenotypes for 6 different traits. Exercise 9.2 required us to perform the “chi-square test” to determine whether the data we collected matches the standard Mendelian ratio.
In cross B, a red-eyed female fly was placed into a vial with a white-eyed male fly and a red- eyed male fly for a breeding period of two weeks. At least eight red-eyed females were produced from this cross. These eight females were used to conduct a polymerase chain reaction. Comparing to the control groups, the results show that seven out of eight flies has the Xw+ Xw genotype because all of them had both bands appear on the gel when both alleles were present. This indicates that all of them are heterozygous. They are all red-eye females. Fly 7, however either of the band appear. This means that an error occurred, and one possible error could be not enough DNA were extracted therefore the gel could
You are also provided with a heterozygous female, and a homozygous recessive male for a genetic cross. In this particular female, all the dominant alleles are on one chromosome, and the recessive counterparts are on the other homologous chromosome. Due to a chromosomal condition, in the female no recombination occurs between the M and N loci. Normal recombination occurs between the L and M loci. Diagram this cross, and show the genotypes and frequencies of all offspring expected from this cross.
This experiment looks at the relationship between genes, generations of a population and if genes are carried from one generation to another. By studying Drosophila melanogaster, starting with a parent group we crossed a variety of flies and observe the characteristics of the F1 generation. We then concluded that sex-linked genes and autosomal genes could indeed be traced through from the parent generation to the F1 generation.
The main purpose for this conducting this experiment was to further knowledge on Mendelian Genetics and how traits are inherited from generation to generation. Something that we attempted to solve was which traits were considered dominant and which were considered to be recessive. Drosophila melanogaster also known as fruit flies were used in this experiment, Dihybrid crosses were done to gather information on how characteristics are linked from generation to generation. Our crosses consisted of female wild type with male sepia eye/ ebony body and female ebony with male vestigial. It is shown that some inheritance patterns are due to unlinked genes and linked genes.
Intro: In this laboratory experiment we cross bred fruit flies (Drosophila melanogaster) in order to identify which traits were passed onto their offspring. Since the 20th century the common fruit fly has been a useful and one of the best organisms for the study of genetics. Thomas Hunt Morgan was responsible for the work that demonstrated the huge advantages of fruit flies and led to them becoming a common model for scientific research and testing. Its relatively short generation time (~2 weeks) produces a high rate of reproduction and since they are small and easy to grow, they make for a cheap and simple experiment to do in the laboratory. They also only have four pairs of chromosomes which allow being able to detect which traits pass on to their offspring quite easy.
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.