Drosophila Lab Report for Bios 221
Drosophila Lab Report for Bios 221 221
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Date Created: 03/17/16
The study of Mendelian heritance patterns on Drosophila Melanogaster 1 By Robin Cordero Introduction The common fruit fly, Drosophila Melanogaster, has been a well know model organism in genetic studies. With its small size, short life cycle, small genome, and inexpensive food requirement, Drosophila Melanogaster, is a perfect organism of study in todays experiment (UIC). Todays experiment will study the inheritance patterns of fruit flies. The traits that will be used are eye color and the presence or absences of wings. The wildtype phenotype will include winged and red eye fruit flies, and the mutant phenotype will include wingless (Apterous) and sepia (brown) eye color. To further understand the inheritance pattern of fruit flies a monohybrid and dihybrid crosses will be performed. In the first experiment, a monohybrid cross will be performed to study one particular trait of fruit flies. The F1 generation fruit flies will contain the phenotype wildtype form of wings however, the genotype form will be heterozygous for the mutant form of wings (Apterous or wingless). As denoted genotype, Ww x Ww will be the starting F1 generation. The F1 generation can be traced to its original parents which were homozygous dominant for the wildtype form of wings (WW) and homozygous recessive for the mutant form of wings (ww). The second experiment, dihybrid cross, will contain two particular traits: eye color and the presence or absence of wings. The F1 generation will contain the phenotype wildtype form of both traits (red eye and winged), but the genotype will contain the heterozygous mutant form of both traits, sepia (brown) and Apterous (wingless). Their genotype for the first generation can be denoted as SsWw x SsWw. The F1 generation may also be traced to their parental which were homozygous dominant for both traits SSWW and homozygous recessive for the mutant form of both traits ssww. 2 If the inheritance patterns were to followed median law of segregation and independent assortment than the F2 generation for monohybrid cross (Ww x Ww) should demonstrate the 3:1 phenotypic ration, with three containing the wildtype form of the wings and one containing the mutant form of wingless. On the other hand, the dihybrid cross (SsWw x SsWw) should demonstrate the phenotypic ration of 9:3:3:1, with nine contain the wildtype of both traits, three containing the wildtype form of wings and mutant form (sepia), three containing the mutant form of wings and wildtype (red eye) and one containing both the mutant from for both traits (Apterous and Sepia). Materials and Methods Medium: To prepare the medium culture place about 1 cup of agar (food material) on a small plastic bottle and then add about 1000 ml of distilled. Then make sure you that the agar and water are mix well. Next place a piece a cotton plug on each of the bottles containing the food material. Then add a small amount of dry yeast on the medium. The yeast will be used and served as a food source for the developing fly larva. Handling Flies: After F1 generation was placed into the bottles containing the medium, they were sterilized and counted for the F2 generation. First we began by placing a few drops of ether on the absorbent material of the etherizer. Then we inserted the etherizer inside the culture bottle so that the flies would sink to the bottom. Once inserted we placed the bottle on the side and shook the flies inside the bottle containing the etherizer for 30 seconds. Once all the flies were sterilized, the flies were transfer into a clean white cart. Then examine the flies using a dissecting microscope at 10X to 25X magnification. Using a soft brush move the flies on to the plate of the microscope and then examine your results. All the flies that are not needed will be discarded in 3 the fly morgue. The remaining etherized flies will be used for the further meeting and should be permitted to recover on the dry surface in the culture before they come in contact with the moist medium. Distinguishing Sex: In order to determine if the flies are male or female several characteristics can be distinguished. One characteristic that can be used is by examining the external genitalia under magnification. Male flies will exhibit dark color genitalia, which are visual on the ventral side of the abdomen. Males are also generally smaller than female flies. Males also contain sex comb, a characteristic that females lack. F1 Generation Cross: After determining the sex of the flies using the microscope, 8 female flies and 4 males containing the specific trait were tested and put back on to the bottle containing the culture media. For the dihybrid cross, exactly 13 females and 10 males along the two traits being tested were also put back on a separate bottle with the culture. Also assuring that the flies don’t die placed them in a lying position so that flies don’t get socked on their own culture medium. After the flies are awoken placed the bottle on an up right position. Scoring the F2 generation: After 7 days, remove the the F1 generation and remove any larvae that has developed within the bottle. For the F2 generation, mature flies were counted and categorize based on their trait and gender. For the monohybrid cross the trait that was taken in consideration was the phenotype wings (wing or wingless). For the dihybrid cross two traits were take into account containing the phenotypes for eye color and the presence and absence of wings. Once female and male flies are counted and separated base on their traits you can begin to analyze your data. 4 Results As expected, from looking at Table 1 we can see that the the overall monohybrid cross including both female and male flies had a chisquare of 0.00383, resulting in a pvalue of 0.9999. From this we can strongly agree that the observed ratio of fruit fly’s phenotype for the F2 generation was significant as the expected 3:1 ratio. More so, looking at table 2 we see that the overall dihydrid cross, which includes both male and female flies had a chisquare of 1.687, resulting in a pvalue of 0.9523. From the observed values we can strongly accept that the observed ratio of fruit flies for the F2 generation was significant as the expected phenotypic ratio 9:3:3:1. Table 1: Overall (Male and Female), Monohybrid Chisquare Analysis Phenotype Observed Expected (observed (observed (observed expected) expected)^ expected)^2/ 2 (expected) Apterous 22 21.75 0.25 0.0625 0.002873 Wildtype 65 65.25 0.25 0.0625 0.000957 Total 87 87 Chisquare:0.00383 Pvalues: 0.99 Df: 1 Accept/Reject: Accept Table 2: Overall (Male and Female), Dihybrid Chisquare Analysis Phenotype Observe Expected (observed (observed (observed d expected) expected)^2 expected) ^2/ (expected) WildType 35 36.5625 6.1875 38.285 1.0471 (winged) and Wild type (Red 5 eye) WildType 13 12.1875 1.9375 3.7539 0.3080 (winged) and Mutant (Sepia Eye) Mutant (wingless) 12 12.1875 1.9375 3.7539 0.3080 and Wildtype (Red eye) Mutant (wingless) 5 4.0625 0.3125 0.09765 0.024 and Mutant (Sepia Eye) Total 65 65 Chi square:1.68 71 Pvalues: 0.9523 Df: 3 Accept/Reje ct: Accept Discussion The null hypothesis assumes that any kind of difference from the observe results is solemnly due to chance. In order to determined if this is true a chisquare analysis have to be taken. If the pvalue falls below the p<0.0.5 than we can reject with confidence the null hypothesis and can therefore accept that it was no due to chance. Looking at the monohybrid cross results, the pvalue for the overall (males and female) were less than 0.05, so the null hypothesis can be strongly rejected and can therefore be said that the observed result was not due to chance. From what we can observed from the pvalue something other than chance may have caused the observed results. Similarly, the dihybrid cross experiment, had a pvalue of 1.687 and can therefore reject the null hypothesis and accept the alternative that is the observed ratio was not due to chance. Given from the results several factors could have contributed the outcome. One of the factors 6 could have been due to the environment. The environment in which the flies were put in could have been altered due to the temperature and could have increase the life cycle of the fruit flies. Furthermore, flies not being fully heterozygous for its traits could have affected the outcome of the results. For future experiments a bigger sample size should be used to accurately portray the inheritance patterns of fruit flies. Reference 1. Peedles, E. David, Sharon Whitmarsh, and Matthew R. "Basic Concepts in Drosophila Melanogaster Genetics." 2001: http://www.poultry.msstate.edu/pdf/courses/po3103/fly3.pdf 2. Reynolds, Kevin. “InheritancePatternsofDrosophilamelanogaster,theFruitFly”. Schoolof Natural Sciences at Ferrum College: http://kevinreynolds.weebly.com/uploads/1/0/6/4/10644034/fruit_fly_report.pdf 3. "Simple Mendelian Genetics in Drosophila.": http://math.hws.edu/javamath/ryan/Genetics1.html 4. Wiles, Spencer, and Kristian Hargadon. "Inheritance Patterns in Monohybrid and Dihybrid Crosses for Sepia Eye Color and Apterous (wingless) Mutations in Drosophila Melanogaster." HSC Journal of Sciences 2 (2013). Appendix (extra) Looking at tables 1and 2 can help us examine the inheritance of two particular traits: Eye color and Wing type. Wing type is controlled by an autosomal gene. An autosomal gene affects males and females equally. An organism that carrier WW or Ww would showcase flies with wings. On the other hand, flies that only contain w will be wingless. From looking at our data we see about and equal amount of both males and female possessing recessive and dominant traits for Wing type. 7 Eye color is controlled by a sexlinked gene. Male posse’s x and y gene, while female posses two XX chromosomes. For sexlinked genes males are more susceptible since males only have one x chromosome, meaning if the posses the recessive allele for a particular trait they would posses the recessive phenotype. In contrast female’s posses two x chromosomes, so it will take 2 recessive alleles two cause the recessive phenotype. In order to determined if males or female are prone to the dominant trait you can look at your F1 crosses and build a punnet square to help you determined the probability that their progeny (female or male) will posses’ dominant or recessive trait for the F2 generation. Tables 1: Monohybrid Cross (Males) Chisquare analysis Phenotyp Observe Expecte (observed (observed (observed e d d Expected)^ Expected)^2/(Expected) Expected) 2 Apterous 9 9.5 0.5 0.25 0.02631 Wildtype 29 28.5 0.5 0.25 0.008771 Total 38 38 Chisquare: 0.035081 Pvalues: 0.999 Df: 1 Accept/Reject: Accept Table 2: Monohybrid (Females) Chisquare analysis Phenotyp Observed Expected (observed (observed (observedexpected)^2/ e expected) expected)^2 (Expected) Apterous 13 12.25 0.75 0.5625 0.04591 Wildtype 36 36.75 0.75 0.5625 0.0153 Total 49 49 Chisquare: 0.06121 Pvalues: 0.99 Df: 1 Accept/Reject: Accept Table 3: Dihybrid (Male) Chisquare analysis 8 Phenotype Observed Expected (observed (observed (observed expected)^2 expected) ^2/ expected) (expected) WildType 35 36.5625 1.5625 2.441 0.06676 (winged) and Wild type (Red eye) WildType 13 12.1875 0.8125 0.6601 0.05416 (winged) and Mutant (Sepia Eye) Mutant (wingless) 12 12.1875 0.8125 0.6601 0.054 and Wildtype (Red eye) Mutant (wingless) 5 4.0625 0.9375 0.8789 0.2163 and Mutant (Sepia Eye) Total 65 Chi square:0.39122 Pvalues: 0.99 Df: 3 Accept/Reject: Accept Table 4: Dihybrid (Females) Chisquare analysis Phenotype Observed Expecte (observed (observed (observed d expected) expected)^ expected) 2 ^2/ (expected) WildType 28 31.5 3.5 12.25 0.3888 (winged) and Wild type (Red eye) WildType 12 10.5 1.5 2.25 0.21428 (winged) and Mutant (Sepia Eye) Mutant (wingless) 13 10.5 2.5 5.0625 0.48214 9 and Wildtype (Red eye) Mutant (wingless) 3 3.5 0.5 0.25 0.0714 and Mutant (Sepia Eye) Total 56 56 Chi square:1.1566 2 Pvalues: 0.99 Df: 3 Accept/Reject : Accept 10
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