Patterns of Inheritence in Drosophila Melanogaster

Patterns of Inheritence in Drosophila Melanogaster The Fruit Flies of Melanogaster Introduction:    Many simple patterns of inheritance follow the laws of Mendel. Dominant traits will always be expressed when present, and recessive traits will only be expressed when two recessive alleles are present. When crossing a pure homozygous dominant trait with a pure homozygous recessive trait as the P generation, it is expected that all the offspring in the F1 generation will express the dominant trait, since every offspring will receive one copy of the dominant allele from one parent and one copy of the recessive allele from the other. In the F2 generation, the expected outcome will be a 3:1 phenotypic ratio of dominant to recessive, and a 1:2:1 genotypic ratio of homozygous dominant to heterozygous to homozygous recessive (Campbell et al. 268). This simple inheritance pattern explains many of the inheritance phenomena exhibited in nature, but some inheritance patterns go beyond Mendel’s laws of genetics. In incomplete dominance, neither allele is dominant over the other so the outcome is a blend of both traits. In codominance, both traits are expressed separately. In mitochondrial inheritance, all offspring will receive specific genes from the mother. In X-linked recessive traits, the alleles are located on the X chromosome, and these conditions frequently appear in males because they only have one copy of the X chromosome (â€Å"Inheritance Patterns†). When the exact inheritance pattern is unknown in a cross, the ratios of each type of offspring help to determine if the inheritance pattern follows Mendel’s laws or if it is one of the above varieties. In fruit flies, the red and brown genes for eye color are located on autosomes. However, a mutation on the white gene in fruit flies on the X chromosome prevents any eye color from developing at all (â€Å"The Genetics of Eye Color†). The gene for white eye color is epistatic to the red and white eye genes. This is how fruit flies are able to have three different eye colors when the white mutation is not present, there will be a simple inheritance pattern between red and sepia eyes. When the mutation is present, the red or sepia eyes will not be expressed because they will be masked behind the white mutation. Drosophila melanogaster were used in this procedure because they reproduce very quickly and are easily manageable. All their food and water 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 too big, but they are small enough to easily distinguish traits under a microscope (â€Å"The Fruit Fly and Genetics†). The life cycle of the flies begins as eggs. From the eggs emerge the larvae, which look like tiny worms. The larvae grow through three stages until they reach the pupal stage. The pupae mature and darken in color for three to four days until they break forth from the pupal case to become adult flies (â€Å"Development†). In this experiment, three crosses were performed between different varieties of the fruit fly Drosophila Melanogaster. Cross 1 was between a sepia eyed female and a wild type male, Cross 2 was between a white eyed female and a wild type male, and Cross 3 was between a red eyed, vestigial winged female with a sepia eyed, normal winged male. In Cross 1, a simple pattern of Mendel’s laws is predicted to be expressed. Wild type flies with red eyes is the dominant phenotype over sepia colored eyes. Sepia colored 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 3:1, 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 1:2:1. Our hypothesis for Cross 1 is if there are no mutations and the cross follows Mendel’s laws of independent assortment, then the ratio of red to sepia eyed flies will be 3:1 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 crossed 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 can 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 father’s 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 1:1:1:1. Therefore, our hypothesis for Cross 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 1:1:1:1 if sex is considered. In cross 3, the focus shifted from just looking 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 are crossed. Vestigial wings are a recessive trait, so two recessive alleles must be inherited in order to express the trait. This is also 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, s epia normal, sepia vestigial, and red eyed vestigial. Therefore, our hypothesis for the F2 generation in Cross 3 is that if both traits follow Mendel’s laws of independent assortment for the dihybrid cross, then the predicted phenotypic ratio will be 9:3:3:1. Methodology: Materials used include: Vials Microscopes Fly nap and anesthesia wand Paint brushes Fly food Fly netting Cotton plugs Plain white index card Procedure: First, prepare vials for the fruit flies to live in. Obtain three glass vials, and estimate a few centimeters of Carolina Instant Drosophila Medium in each. After, put a few drops of water in the culture and let it sit a few minutes to soak in the medium. At this point, also put in a fly net. Obtain F1 flies from instructor for the three crosses. Check the vials for life. The flies need to be alive for active breeding purposes. However, ensure that there are no F2 larvae yet, as this could be misleading for the results. Check 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 close it with a cotton plug with an anesthesia wand connected to it soaked in Flynap. Wait a few minutes for the flies to stop 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 field in order to sex them and view them for the desired traits. Record the data in the data tables. Males have a solid black abdomen and sex combs on their forelegs, while females have a striped abdomen and no sex combs. Additionally, females are generally larger than males[1]. After recording 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. Set the extra flies free, and clean out all the vials thoroughly. Results: Cross 1 Punnett square: Cross between heterozygous male and heterozygous female for eye color Table 1: Lab Group data for Cross 1 Fasdfasdasdffasdfasdf Day 1 Day 2 Day 3 Total Red M 3 2 55 60 Red F 4 1 58 63 Sepia M 0 0 6 6 Sepia F 0 0 4 4 Fasdfasdasdffasdfasdf Day 1 Day 2 Day 3 Total Red M 2 3 1 6 Red F 3 2 0 5 Sepia M 0 0 0 0 Sepia F 2 0 0 2 Chi Square Analysis for Lab Group Data for Cross 1 Expected Totals: Red:  ¾ x x/13 = 9.75 red (11-9.75)^2/9.75 = 1.160 Sepia:  ¼ x x/13 = 3.25 sepia (2-3.25)^2/(3.25) = .481 ∑=1.641 Degrees of Freedom: 1 .20 > p > .10 Accept the null hypothesis Table 2: AP Bio 2015 Class Data for Cross 1 Fasdfasdasdffasdfasdf Pd 8 Pd 6 Total Red M 39 58 97 Red F 46 64 110 Sepia M 12 7 19 Sepia F 11 4 15 F2 sex 1 2 3 4 5 6 7 8 9 10 11 total RED M M 19 13 8 35 44 21 6 8 36 48 22 260 RED F F 37 12 12 38 51 36 5 4 36 50 29 310 SEPIA M M 4 3 5 12 10 6 0 3 5 12 6 66 SEPIA F F 5 5 3 16 13 7 2 1 13 14 9 88 Chi Square Analysis for AP Bio Group Data for Cross 1: Expected Totals: Red:  ¾ x x/724 = 543 (570-543)^2/543 = 1.343 Sepia:  ¼ x x/724 = 181 (154-181)^2/181 = 4.028 ∑= 5.371 .01 > p > .001 Reject the null hypothesis Cross 2: Punnett Square: Cross between heterozygous red eyed female and hemizygous white eyed male Table 3: Lab Group Data for Cross 2      F1 day 1 day 2 day 3 total RED M 0 0 15 15 RED F 24 1 32 57 WHITE M 24 3 10 37 WHITE F 0 0 13 13 F2 day 1 day 2 day 3 total RED M 8 3 0 11 RED F 9 6 1 16 WHITE M 7 1 0 8 WHITE F 6 3 1 10 Chi Square Analysis for Lab Group Data for Cross 2 Expected Totals: Red M: 1/4 x x/45 = 11.25 (11-11.25)^2/11.25 = .006 Red F: 1/4 x x/45 = 11.25 (16-11.25)^2/11.25 = 2.01 White M: 1/4 x x/45 = 11.25 (8-11.25)^2/11.25 = .939 White F: 1/4 x x/45 = 11.25 (10-11.25)^2/11.25 = .139 ∑= 3.094 Degrees of freedom: 3 0.50 > p > .30 Accept the null hypothesis Table 4: AP Bio 2015 Class Data for Cross 2 F1 1 2 total RED M 60 15 75 RED F 87 58 145 WHITE M 50 35 85 WHITE F 10 13 23 F2 1 2 3 6 7 8 9 10 11 total RED M 24 23 22 8 11 23 38 9 23 211 RED F 25 30 38 7 16 14 42 16 14 231 WHITE M 18 27 20 9 8 11 13 17 21 162 WHITE F 20 23 24 11 10 11 12 11 24 176 Chi Square Analysis for AP Bio Class Data for Cross 2: Expected Totals: Red M: 1/4 x x/780 = 195 (211-195)^2/195 = 1.312 Red F: 1/4 x x/780 = 195 (231-195)^2/195 = 6.646 White M: 1/4 x x/780 = 195 (162-195)^2/195 = 5.585 White F: 1/4 x x/780 = 195 (176-195)^2/195 = 1.851 ∑=15.394 Degrees of freedom: 3 p > .001 Reject the null hypothesis Cross 3: Punnett Square: Cross between two flies heterozygous for both red eyes and normal wings Table 5: Lab Group Data for Cross 3 F1 sex day 2 day 3 total RED / NORMAL M 6 16 40 RED / NORMAL F 12 17 62 RED / VESTIGAL M 0 0 0 RED / VESTIGAL F 0 0 0 SEPIA / NORMAL M 0 5 5 SEPIA / NORMAL F 0 1 1 SEPIA / VESTIGAL M 0 0 0 SEPIA / VESTIGAL F 0 0 0 F2 sex day 1 day 2 day 3 RED / NORMAL M 12 7 2 RED / NORMAL F 13 13 10 RED / VESTIGAL M 1 2 0 RED / VESTIGAL F 3 3 1 SEPIA / NORMAL M 8 4 1 SEPIA / NORMAL F 4 3 4 SEPIA / VESTIGAL M 0 0 0 SEPIA / VESTIGAL F 1 0 1 Chi Square Analysis for Lab Group Data for Cross 3 Expected Totals: Red normal: 9/16 x x/93 = 52.313 (57-52.313)^2/52.313 = .420 Sepia normal: 3/16 x x/93 = 17.438 (24-17.438)^2/17.438 = 2.470 Red vestigial: 3/16 x x/93 = 17.438 (10-17.438)^2/17.438 = 3.172 Sepia vestigial: 1/16 x x/93 = 5.813 (2-5.813)^2/(5.813) = 2.501 ∑= 8.563 Degrees of freedom: 3 .05 > p >.01 Reject the null hypothesis Table 6: AP Bio 2015 Group Data for Cross 3 F1 sex 1 2 total RED / NORMAL M 28 37 65 RED / NORMAL F 42 51 93 RED / VESTIGAL M 5 0 5 RED / VESTIGAL F 3 0 3 SEPIA / NORMAL M 2 11 13 SEPIA / NORMAL F 4 5 9 SEPIA / VESTIGAL M 0 0 0 SEPIA / VESTIGAL F 0 0 0 F2 sex 1 2 3 4 5 6 7 8 9 10 11 total RED / NORMAL M 7 14 29 30 16 18 21 26 16 7 10 194 RED / NORMAL F 21 18 44 34 16 22 36 30 16 11 6 254 RED / VESTIGAL M 4 3 8 11 4 9 3 0 8 6 11 67 RED / VESTIGAL F 7 8 9 9 3 16 7 0 41 4 16 120 SEPIA / NORMAL M 5 4 8 12 7

“Naming of parts” by Henry Reed, and “War is Kind” by Stephen Crane Essay

â€Å"War†¦ouhh†¦.What is it good for†¦absolutely nothing!† sang Edwin Starr in 1965. He felt the same vibe that both Henry Reed and Stephen Crane felt in their poems, â€Å"Naming of Parts† and â€Å"War is Kind.† Although these authors may not have said it as straightforward as Starr did in his hit single â€Å"War,† they still had just as much hatred of war. Both Reed and Crane have developed their perspectives on war through their writing styles, their usage of figurative language, and their attitudes toward war in general. Henry Reed and Stephen Crane both have very different writing styles. Reed’s style in â€Å"Naming of Parts† is built upon juxtaposition. Guns and gardens, soldiers and bees: the poem relates the unrelated in order to draw a clear line between the horrors of war and the fruits of nature. However, the poem goes further than just contrasting opposites. The structure and language of the poem combine to show how one should become the other in hopes that the harmonious image of this Eden transforms the unnatural feat of war. His overall structure also serves to make nature better. Each stanza is split between the dry, unimaginative language of the first speaker, probably the drill sergeant, and the poetic language used by the second speaker to describe nature. In every stanza, the gentle and peaceful language of the second speaker is quite dominant over the monotone voice of the drill instructor. This shows that war disturbs the balance of nature. Stephen Crane, in â€Å"War is Kind,† develops his style by using vivid imagery and irony. Through doing this, he leads the reader directly to his perspective of war. He feels war is a horrible way to solve problems and uses irony to tell us that war is blatantly stupid. Nothing good has ever come from it and nothing ever will. The American flag, â€Å"The unexplained glory, flies above them† to symbolize that the glory they were fighting for was not earned righteously, it was stolen by â€Å"these little men†¦born to drill and die.† Both authors also heavily use figurative language to help create a picture of what they saw in their minds as they wrote these poems. Henry Reed’s entire poem is entangled in figurative language. He shows us a perfect balance of the world of nature in the sections of the poem that describe the garden. The garden is a symbol of life and beauty: a magical place, â€Å"silent† and â€Å"eloquent.† In the garden, we see the personification of branches which  Ã¢â‚¬Å"hold in the gardens their silent, eloquent gestures.† We are told of blossoms that â€Å"are fragile and motionless, never letting anyone see / any of them using their finger.† We also witness bees â€Å"assaulting and fumbling the flowers.† These examples of figurative language create a picture in our mind to which Reed can build his theme upon. Crane uses more subtle figurative language to get his point across in â€Å"War is Kind.† The excellent use of irony draws us to his imagery and metaphors. The metaphor in the middle of the second stanza helps point in the direction of the true meaning of the poem. â€Å"And his Kingdom – a field where a thousand corpses lie,† proves that all of the dead bodies after the war do not belong there. This world is God’s creation and war was not part of His plan. In â€Å"War is Kind† and â€Å"Naming of the Parts,† both authors’ attitudes toward war are similar but only on the basic level. They both believe war is a tremendous waste of time for it solves nothing at all. For Stephen Crane, this deeply ironic poem is not only an attack on war and all of its horror but also a statement against violence of any kind. This includes the violence that we observe daily, mans inhumanity to man, and the rage and fury within our own hearts, which are equally as destructive. The poem comments on â€Å"those little souls who thirst for fight†, who find virtue in something as horrifying as slaughter and excellence in a field of a thousand corpses. In â€Å"Naming of the Parts,† Reed tells of the instructor that insists that the men â€Å"not let [him] / see anyone using his finger†. At the end of the same stanza, the blossoms are seen â€Å"never letting anyone see / any one of them using their finger†. Alth ough not directly stated in the poem, perhaps the soldiers should take a cue from the blossoms, and in turn nature, not to use their fingers, especially on the trigger. This contributes to his negative view of war. He structures nature to be more powerful than these soldiers and in turn society. Nature’s â€Å"silent† and â€Å"eloquent† state of being show that war should never be used as a solution to a problem. Although both of these poems were in some way about the topic of war, each author has developed their own way of conveying how they feels through their unique writing styles, how they use figurative language, and their own  attitudes about war. Imagine what the world would be like if only we stopped and actually thought what we were doing. Maybe then we would realize, like Edwin Starr and these two authors did, that war is not the solution to the problem, but instead the root.

Low Molecular Weight Heparin

Low Molecular Weight Heparin Paper Low molecular weight heparin is typically used for patients who need to be treated for deep vein thrombosis. Deep vein thrombosis (DVT) is a blood clot (thrombus) in a deep vein usually in the legs. These clots are dangerous because they can break loose, travel through the bloodstream to the lungs, and block blood flow in the lungs (pulmonary embolism). There are many reasons why clots form in a patient.Mainly when a patient is inactive or bedridden for long periods of time, surgery can damage a blood vessel so a clot can form, or even cancer can cause DVT to form. Treatments for DVT are drugs called anticoagulants that can prevent the blood from clotting thus preventing the adverse effects from a clot. Low molecular weight heparin works by binding to a substance called antithrombin III (which is the major inhibitor of thrombin in the blood). The overall effect of heparin is that it turns off the coagulation pathway and prevents clots from forming.I t can be used as a subcutaneous injection which can be given in an outpatient setting with no increased risk of recurrent thromboembolism or bleeding complications. Since most patients with DVT require one or more diagnostic tests, treatment with intravenous heparin and a three to seven day hospital stay thus making low molecular weight heparin a better alternative. (aafp. org1999) However, with low molecular weight heparin, being a subcutaneous injection makes the process easier for the patient since they do not have to spend all that extra time in the hospital.The ultimate consequence of a blood clot can be stroke or heart attack so prevention of these events is the consequence of this drug. Unfortunately, just like many other drugs there are some serious side effects to taking LMWHs. They are contraindicated with patients with an indwelling epidural catheter; they can be given two hours after the epidural is removed. If it is given before the epidural is taken out then they have found it to be associated with epidural hematoma. Bleeding is the main concern when taking anticoagulation therapy.Some of the other common adverse effects to heparin are hematoma, nausea, anemia, thrombocytopenia, fever, and edema. There is a low chance for side effects with monitoring and patient awareness. When evaluating a patient on anticoagulants the nurse needs to ensure patients know the side effects to be aware of and arrange follow up care. Cranberry juice should be avoided since it can affect the INR results. Patients should seek emergency medical care for injuries, particularly a head injury, due to the hemorrhage risk.As a nurse you need to monitor your patient while on these drugs because of the bleeding factor. (nursingtimes. net2012) References Gee, Emma. (2011) How to look after a patient on anticoagulant therapy. January 22, 2011 Retreived from www. nursingnet. net on July 2012 Lilley, Rainforth Collins, Harrington, Snyder. (2011) Pharmacology and the Nursing Proce ss Copyright 2011 Mosby Inc. Rydberg, J Eric MD. (1999) Low Molecular Weight Heparin in Prevention and Treating DVT Retrieved from www. aafp. org

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