Fly-CURE Lab Report Questions

Fly-CURE Lab Report Questions BIOL 118 – Genetics Due: December 4 th , 2023 Please provide short answer responses to the following questions. Most answers can be found from lab slides posted on Canvas and your lab worksheets. However, some questions will require you to consult literature published by Fly-CURE courses at other universities [example: 1] and use critical thinking to interpret results. Each question can be answered in one to two paragraphs and figures [2] can be substituted in place of text when appropriate (e.g., Q5). Please provide all citations in APA 7 th Edition; there is not a minimum or maximum number of citations required. Answers can be submitted via Canvas under the submission link on the Lab Resources module by 12/04 at 11:59PM. Each student should answer and submit questions independently, do not submit answers with your partner(s). [1] Vrailas-Mortimer et al., B.2.16 is a non-lethal modifier of the Dark82 mosaic eye phenotype in Drosophila melanogaster . MicroPubl Biol., 2021, doi: 10.17912/micropub.biology.000359 [2] https://www.internationalscienceediting.com/how-to-write-a-figure-caption/ 1. How could the identification of the location and function of M.2.2 meaningfully advance our understanding of human genetics? Provide an example of its application in cancer research. - The identification of the location and function of M.2.2 could be meaningful and advance understandings of human genetics. This could further be explained as around 75% of the genes in which are known to cause humans to be ill could also occur in flies. This makes for the model M.2.2 mutation identified in Drosophila to allow for further research on cancer. More than 90% of the genes that could cause cancer in humans also are possessed in Drosophila. In being able to identify it, it would be important to the pre-cancerous growths in which programmed cell death had already been lost. 2. How was the M.2.2 mutation generated and what does its inheritance pattern look like (dominant, recessive)? - The M.2.2 mutation is generated by a heterozygous cross with the deletion. The mutation lies within half of the strain. This could be inferred that if there were ten curly winged then there would be five straight winged. The inheritance pattern showcases recessive / lethal with M.2.2 and Df(2R), making them either straight winged or dead. 3. What are the three main functions of balancer chromosomes and how were balancer chromosomes used in this research? - Balancer chromosomes are critical as they allow for maintaining recessive lethal mutations. Specifically, it allows for the maintaining of mutations that are less fit, as well as balancing the fitness of organisms, and kills genotypes that don’t match the parents. Balancer chromosomes such as SM1 used in the research, had multiple overlapping inversions. This was made useful as it prevented recombination; the inversions served as a balancer when crossings of parents want to see a specific genotype come through to offspring. 4. Explain the layout of a deficiency stock and how they were used in this research. - A layout of a deficiency stock involves the categorization by the loss or deficiency of specific genes, allowing for the identification and characterization of gene interactions; creates for an organized and catalog like way to retrieve strains easily. In this research, the deficiency stock

allowed for the identification of the general area of the lethal. In doing so, there was a stock number, a deficiency name, a left break, right break, and the size (kb), as well as the number of genes affected to identify it in the area. 5. Explain the Punnett square for a cross between M.2.2//SM1 X Df(2R)//SM1. - The Punnett square for the cross between M.2.2//SM1 X Df(2R) // SM1 resulted in 1 straight, 2 curly, and 1 dead. This is due to the outcome of if the deletion does include the region where the M.2.2 reside, and the other if the deletion is in a different part of chromosome 2. The ratio would overall be 2:1, curly: straight or dependent on chance, it could 3:0. Though in doing this cross, it would not show differences between genders. 6. What were the outcomes of the crosses you and your partner screened? - The outcomes of my partner and I’s screening resulted in a 10:5 ratio; curly wings: straight wings. This could align with the Punnett square results as it would be a 2:1 ratio. This is highlighted in stock #9064, Df name: Df(2R) ED2426. This stock had a left break of 15.1 mb, right break of 15.6 mb. The size being about 500,000 mb and affecting about 72 genes. 7. Explain the process of mitotic recombination at the cellular level. Why was mitotic recombination needed to assess the function of M.2.2? Include a discussion of the role of homozygous Dark 82 alleles. - Occurring during mitosis, there is a genetic exchange between homologous chromosomes within a single somatic cell. There is an exchange of DNA segments between the chromosomes and results in the formation of new combinations of alleles on the chromosomes involved. This can lead to genetic mosaicism. The two cells are created without any changes to the DNA sequence because the alleles that have already existed are distributed irregularly as there is abnormal swapping of the chromosome arms. Mitotic recombination was needed to access the function of M.2.2 as it was seen that there were more patches of red facets seen in the controls. This being rather than individual red facets scattered all over the eye due to having two copies of M.2.2, making for the two copies of Dark82 to be overshadowed and grow less quickly. The mitotic combination could be inferred to have occurred before the fly matured. 8. Briefly describe your observations of mutant D. melanogaster eyes with and without M.2.2. What do these observations tell you about the phenotype of homozygous M.2.2? - The mutant D. melanogaster eyes with M.2.2 had circular eyes and no noticeable shape of cells, while the D. melanogaster eyes without the M.2.2 had oval eyes and grid shaped cells. These observations allow us to infer that the phenotype could vary in shape and color due to the amount or size the fly may have. It could mean that they also have limitations that can cause interruptions in the cells making it look a certain way, as well as the way a cell could glow can affect it. 9. Why did we extract genomic DNA from D. melanogaster and what was the genotype of the flies we used for the extraction?

4c724e396/QUANTIFICATION-OF-SINGLE-DROSOPHILA-FLY-GENOMIC-DNA-USING-UV- SPECTROPHOTOMETRY-NANODROP-AND-QUBIT-FLUOROMETRY.pdf. Accessed 30 Nov. 2023.