Patterns of Inheritence in Drosophila Melanogaster

Patterns of Inheri 10ce in Drosophila MelanogasterThe Fruit locomote of MelanogasterIntroduction M whatever dewy- eyed patterns of heritage follow the laws of Mendel. Dominant traits forget always be expressed when present, and recessive allele exclusivelyelomorph traits provide sole(prenominal) be expressed when two recessive entirelyeles atomic number 18 present. When dawning a clear homozygous dominant trait with a pure homozygous recessive trait as the P gene proportionalityn, it is expected that all the effect in the F1 propagation forget express the dominant trait, since every issuing go forth receive mavin copy of the dominant allele from one heighten and one copy of the recessive allele from the other. In the F2 generation, the expected payoff lead be a 31 phenotypic ratio of dominant to recessive, and a 121 genotypic ratio of homozygous dominant to heterozygous to homozygous recessive (Campbell et al. 268).This simple he wilditary pattern pattern e xplains many of the inheritance phenomena exhibited in nature, merely some inheritance patterns go beyond Mendels laws of genetics. In incomplete dominance, n either allele is dominant over the other so the outcome is a decease of both traits. In codominance, both traits be expressed separately. In mitochondrial inheritance, all offspring will receive specific genes from the mother. In X- conjugate recessive traits, the alleles are dictated on the X chromosome, and these conditions frequently appear in males because they only have one copy of the X chromosome (Inheritance Patterns). When the exact inheritance pattern is unknown in a scratch, the ratios of each type of offspring help to determine if the inheritance pattern follows Mendels laws or if it is one of the above varieties.In output travel, the exit and brown genes for eye food color are located on autosomes. However, a mutation on the bloodless gene in fruit move on the X chromosome prevents any eye color from d eveloping at all (The Genetics of nerve center Color). The gene for flannel eye color is epistatic to the blushful and uninfected eye genes. This is how fruit flies are able to have ternion different eye colors when the purity mutation is not present, at that place will be a simple inheritance pattern betwixt red and mahogany tree look. When the mutation is present, the red or genus genus Sepia eyes will not be expressed because they will be dissemble behind the white mutation.Drosophila melanogaster were used in this procedure because they vomit up very quickly and are easily manageable. All their food and pissing needs are taken care of by the substance called media in the bottom of the vial. They are a convenient size because they are not as well big, but they are small enough to easily fall into place off traits under a microscope (The Fruit Fly and Genetics). The life pedal of the flies begins as eggs. From the eggs emerge the larvae, which look like petite worms. The larvae grow through tercet stages until they reach the pupal stage. The pupae mature and darken in color for three to four solar sidereal days until they break forrad from the pupal case to become adult flies (Development).In this experiment, three crosses were performed between different varieties of the fruit fly Drosophila Melanogaster. cross(a) 1 was between a burnt sienna eyed feminine and a wild type male, sink in 2 was between a white eyed female and a wild type male, and treat 3 was between a red eyed, vestigial winged female with a sepia eyed, normal winged male.In intersect 1, a simple pattern of Mendels laws is predicted to be expressed. Wild type flies with red eyes is the dominant phenotype over sepia dark-skinned eyes. Sepia nonreversible eyes are a result from a recessive gene, and only result when two sepia-eyed flies mate or when two heterozygous flies mate. Furthermore, sepia colored eyes is not dependent on the sex of the fly, so in the case of this cross all flies in the F1 generation should have red eyes, but be carriers for the sepia colored eye trait. In the F2 generation when the heterozygous flies mate, the predicted phenotypic ratio will be 31, where for every three red eyed flies there would be one sepia colored fly. The related genotypic ratio of homozygous dominant to heterozygous to homozygous recessive will be 121. Our surmise for cut across 1 is if there are no mutations and the cross follows Mendels laws of freelance assortment, thusly the ratio of red to sepia eyed flies will be 31 for the F2 generation.In cross 2, sex linked inheritance plays a role. The mutation for white colored eyes is X-linked recessive. When the white eyed female is cut through with a red eyed male, all the males in the F1 generation should exhibit the mutation, and all the females should have red eyes. This is because the males bunghole only accept a recessive allele from the mother and the Y chromosome from the father which does not carry the mutation for white eye color. The females will receive the red gene from the fathers X chromosome which will cover the white gene from the mother. The F2 generation produced by the white eyed male and heterozygous female will thus have a genotypic ratio of 1111. Therefore, our hypothesis for grouchy 2 for the F2 generation is that if the gene for the white eye mutation is located on the X chromosome, then the phenotypic ratio will be 1111 if sex is considered.In cross 3, the focus shifted from just face at eye colors to looking at eye colors and wing type. The fruit flies could either have normal wings or exhibit vestigial wings, which are shortened. Flies with vestigial wings have a defect in their vestigial gene located on the second chromosome. So, a dihybrid cross will be used to determine the predicted phenotypes and genotypes of the F1 and F2 generation. A dihybrid cross uses two traits with two alleles each, and so two different aspects of an organism ar e crossed. rudimentary wings are a recessive trait, so two recessive alleles must be inherited in order to express the trait. This is too the case with sepia colored eyes. So when a parent generation of a red eyed vestigial winged female is crossed with a sepia eyed normal winged male, all of the offspring in the F1 generation should have red eyes and normal wings. The F2 generation, however, are produced by heterozygotes and thus four phenotypes should be seen red eyed normal, sepia normal, sepia vestigial, and red eyed vestigial. Therefore, our hypothesis for the F2 generation in extend 3 is that if both traits follow Mendels laws of independent assortment for the dihybrid cross, then the predicted phenotypic ratio will be 9331.MethodologyMaterials used includeVialsMicroscopesFly nap and anesthesia billystickPaint brushesFly foodFly nettingCotton plugs orbit white index lineupProcedureFirst, prepare vials for the fruit flies to live in. Obtain three glass vials, and estimate a a couple of(prenominal) centimeters of Carolina Instant Drosophila Medium in each. After, put a a couple of(prenominal) drops of water in the culture and let it sit a a few(prenominal) minutes to soak in the medium. At this point, also put in a fly net. Obtain F1 flies from instructor for the three crosses. chit the vials for life. The flies need to be alive for active breeding purposes. However, plug that there are no F2 larvae yet, as this could be misleading for the results. return the food for moisture, and add water with a pipet if the food gets too dry.Anesthetize the fruit flies. Put the vials of flies upside down in the refrigerator, as this forces the flies into a state of inactivity. After approximately ten minutes, take the vials out and transport each of the three tubes (for the separate three crosses) into three different vials. Mark the three vials with tape for either cross 1, cross 2, or cross 3. Tap the flies into the new vials, and finish it with a cotton p lug with an anesthesia wand connected to it mean in Flynap. Wait a few minutes for the flies to drop by the wayside moving or flying to begin the next procedure.Shake the fruit flies onto a white index card and place the card under the dissecting microscope. Use the paintbrush to move the flies to the center of the viewing ambit in order to sex them and view them for the desired traits. Record the entropy in the information tables. Males have a solid black stomach and sex combs on their forelegs, while females have a striped stomach and no sex combs. Additionally, females are generally larger than males1.After arranging the data for the flies, place them in the morgue. Record F1 data for three days, or until F2 larvae are seen.Repeat the above procedure using the same vials, but this time using only F2 flies. Record data for three days. fate the extra flies free, and clean out all the vials thoroughly.ResultsCross 1Punnett square Cross between heterozygous male and heterozyg ous female for eye colorTable 1 Lab root word data for Cross 1Fasdfasdasdffasdfasdf twenty-four hour period 1Day 2Day 3TotalRed M325560Red F415863Sepia M0066Sepia F0044FasdfasdasdffasdfasdfDay 1Day 2Day 3TotalRed M2316Red F3205Sepia M0000Sepia F2002qi upstanding analysis for Lab classify selective information for Cross 1Expected TotalsRed x x/13 = 9.75 red(11-9.75)2/9.75 = 1.160Sepia x x/13 = 3.25 sepia(2-3.25)2/(3.25) = .481=1.641Degrees of Freedom 1.20 p .10Accept the null hypothesisTable 2 AP Bio 2015 Class Data for Cross 1FasdfasdasdffasdfasdfPd 8Pd 6TotalRed M395897Red F4664one hundred tenSepia M12719Sepia F11415F2sex1234567891011 kernel ruby-red MM1913835442168364822260 red-faced FF37121238513654365029310 burnt sienna MM4351210603512666 genus Sepia FF55316137211314988Chi Square compendium for AP Bio Group Data for Cross 1Expected TotalsRed x x/724 = 543(570-543)2/543 = 1.343Sepia x x/724 = 181(154-181)2/181 = 4.028= 5.371.01 p .001Reject the null hypothesisCross 2 Punnett Square Cross between heterozygous red eyed female and hemizygous white eyed maleTable 3 Lab Group Data for Cross 2 F1day 1day 2day 3 replete(p)RED M001515RED F2413257 livid M2431037WHITE F001313F2day 1day 2day 3totalRED M83011RED F96116WHITE M7108WHITE F63110Chi Square Analysis for Lab Group Data for Cross 2Expected TotalsRed M 1/4 x x/45 = 11.25(11-11.25)2/11.25 = .006Red F 1/4 x x/45 = 11.25(16-11.25)2/11.25 = 2.01White M 1/4 x x/45 = 11.25(8-11.25)2/11.25 = .939White F 1/4 x x/45 = 11.25(10-11.25)2/11.25 = .139= 3.094Degrees of granting immunity 30.50 p .30Accept the null hypothesisTable 4 AP Bio 2015 Class Data for Cross 2F112totalRED M601575RED F8758145WHITE M503585WHITE F101323F212367891011totalRED M2423228112338923211RED F25303871614421614231WHITE M1827209811131721162WHITE F202324111011121124176Chi Square Analysis for AP Bio Class Data for Cross 2Expected TotalsRed M 1/4 x x/780 = 195(211-195)2/195 = 1.312Red F 1/4 x x/780 = 195(231-195)2/195 = 6.646White M 1/4 x x /780 = 195(162-195)2/195 = 5.585White F 1/4 x x/780 = 195(176-195)2/195 = 1.851=15.394Degrees of freedom 3p .001Reject the null hypothesisCross 3Punnett Square Cross between two flies heterozygous for both red eyes and normal wingsTable 5 Lab Group Data for Cross 3F1sexday 2day 3totalRED / conventionM61640RED / NORMALF121762RED / VESTIGALM000RED / VESTIGALF000SEPIA / NORMALM055SEPIA / NORMALF011SEPIA / VESTIGALM000SEPIA / VESTIGALF000F2sexday 1day 2day 3RED / NORMALM1272RED / NORMALF131310RED / VESTIGALM120RED / VESTIGALF331SEPIA / NORMALM841SEPIA / NORMALF434SEPIA / VESTIGALM000SEPIA / VESTIGALF101Chi Square Analysis for Lab Group Data for Cross 3Expected TotalsRed normal 9/16 x x/93 = 52.313(57-52.313)2/52.313 = .420Sepia normal 3/16 x x/93 = 17.438(24-17.438)2/17.438 = 2.470Red vestigial 3/16 x x/93 = 17.438(10-17.438)2/17.438 = 3.172Sepia vestigial 1/16 x x/93 = 5.813(2-5.813)2/(5.813) = 2.501= 8.563Degrees of freedom 3.05 p .01Reject the null hypothesisTable 6 AP Bio 2015 Gr oup Data for Cross 3F1sex12totalRED / NORMALM283765RED / NORMALF425193RED / VESTIGALM505RED / VESTIGALF303SEPIA / NORMALM21113SEPIA / NORMALF459SEPIA / VESTIGALM000SEPIA / VESTIGALF000F2sex1234567891011totalRED / NORMALM71429301618212616710194RED / NORMALF211844341622363016116254RED / VESTIGALM438114930861167RED / VESTIGALF78993167041416120SEPIA / NORMALM548127