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Different Cell Death Mechanisms between Ischemic Core and Penumbra Regions after Cerebral Ischemia

Different Cell Death Mechanisms between Ischemic Core and Penumbra Regions after Cerebral IschemiaIschemic cell death involves a complex cascade of events in cerebral ischemia. This phenomenon arises from permanent or prolonged occlusion of a cerebral artery (5). When a focal ischemic stroke occurs, the core of brain tissue exposed to the dramatic blood flow reduction. This eventually causes to necrotic cell death. Around this necrotic core, there is a region of hypoperfused, electrically silent tissue receives enough blood flow to keep neurons alive. This zone, which is affected less severely by an ischemic stroke, is called “ischemic penumbra”. This is associated with a delayed death of neurons.Ischemic cell death is initiated by an inhibition of oxidative phosphorylation, and creates an early maelstrom of activity (3). The causes of ischemic cell death include decreased pH, decreased ATP, and initiation of free radical production by the mitochondrial chain. Moreover, rapid depletion of energy after cerebral ischemia leads to a loss of membrane potential and depolarization of nerves (1). The secondary changes in ion and chemical concentration would be followed up. The ischemic insult in the brain has numerous pathophysiological consequences including excitotoxicity, DNA damage, and production of free radical.The process of cell death has at least three major stages:An early development of ionic and chemical changesResult of activation of perpetratorsSubsequent change in critical functions and structures that denote or least to cell death by one of three different routes, necrosis, apoptosis or autophagyMajor Features of Ischemic Cell DeathThe sufficient availability of ATP and an intact mitochondrial function determine shifting apoptosis-doomed neurons away from necrosis (6). When energy is plentiful, the apoptotic route will predominate. In contrast, necrosis is the end result of a bioenergetic result from ATP depletion. Ischemia/hypoxia may be a powerful stimulus for autophagic lysosomal cell death in brain. The combined hypoxia/ischemia injury-induced pyramidal neuron death by both caspase-dependent and –independent pathways and the damaged neurons exhibit hallmarks of autophagy (5). Moreover, the ischemic neuronal death was rescued by an Atg7 deficiency.In both proximal and distal middle cerebral artery (MCA) occlusion, (1, 3, 5)Core ischemic regionsPenumbra regions1. Cell DeathNecrosisApoptosis and Autophagy2. Ratio(Apoptosis : Necrosis)1:19:13. Time periodTen to twentyminutes produces scattered dead neurons in the core (2)Undergo apoptosis after several hours or days4. Blood FlowReduced to

의/약학| 2012.07.09| 7페이지| 2,000원| 조회(91)

Brain, apoptosis, Core, necrosis, ischemia, a..

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Unknown bacteria 평가A+최고예요

METHODSAll methods were to figure out what unknown bacteria (#149) it was. The procedures were followed by Benson’s Microbiological Applications by Alfred E. Brown. The reserve stock and working stock cultures of unknown bacteria were prepared by using an aseptic technique. The working stock cultures would be used for the rest of further protocols.To figure out which temperature the unknown bacteria would grow better, two plates were prepared after inoculating unknown bacteria. The plates were put into different temperature environments. The clear and abundant colonies would show the optimal temperature for the unknown bacteria. To figure out the morphology of the unknown bacteria, gram staining was processed, and EMB and PEA test plates were also used. The morphology was decided by shapes and colors in the gram staining, and motility in SIM medium.Cultural characteristics of an organism were decided by amount of growth, colors of colonies, opacity, elevation and form on a reserve stocermentation of glucose and lactose was done by Kligler’s iron agar. To detect the production of hydrogen sulfide and indole, SIM medium was used.RESULTSThe identification of the unknown bacteria was determined by Bergey’s Manual of Determinative Bacteriology, 9th Ed. The unknown bacteria (# 149) were Pseudomonas aeruginosa according to the results from the tests. All observed results matched with expected results except optimal temperature test, gelatin test, Voges-Proskauer test, and H2S production test in SIM media.The analysis of structure of the bacteria was done by morphological tests. The optimal temperature of the unknown bacteria was expected to be 37°C, but the result was 25°C.Table 1. Morphological tests for unknown bacteriaTests of morphological testsObserved resultsExpected resultsOptimal Temperature25°C37°C 3EMB plateGrowthGrowth2PEA plateNo growthNo growth2Gram stainingPink rod shapePink rod shape 1MMMotileMotile11Brown, 2Holt et al., 3TodarTo differentiate the unknown baex.php/Pseudomonas, 6http://bac.hs.med.kyoto-u.ac.jpHydrolytic and degradative tests revealed that there were almost no exoenzymes, which degrade large macromolecules into smaller units, in the unknown bacteria.Table 4. Hydrolytic and degradative tests for unknown bacteriaTests of hydrolytic and degradativeObserved resultsExpected resultsStarch hydrolysisBlueBlue2Casein hydrolysisNo clear zoneNFTryptophan hydrolysisNo organic layerNo organic layer2Urea hydrolysisPinkNFPhenylalanine deaminationNo color changeNo color change52Holt et al., 5ZhaoNF = information not foundFermentation of glucose was confirmed by Kligler’s iron agar test, and motility and non-indole production were confirmed by SIM media. However, production of hydrogen sulfate did not agreed with an expected result. Proteus mirabilis in SIM media, which was used as a positive control, showed black precipitate because of production of hydrogen sulfate. The unknown bacteria were expected to not have a black precipitate, but a optimal temperature was 37°C, the colonies were seen well on both 25°C and 37°C plates. The translucent colonies were very small and distributed well on both plates. That is because the temperature does not affect much on the growth of bacteria unless temperature is too low, around 4°C, or too high.P. aeruginosa did not produce proteases, which means the bacteria were not able to grow in gelatin, while the expectation was producing proteases1. Because of technical error, the bacteria might not transfer into the gelatin tube, so it made an error in this test.The color change in VP test means the organism ferments glucose to produce 2,3-butanediol. By Brown1, if an organism has a positive result in MP test, it usually has a negative result in VP test. Since, P. aeruginosa has a negative result; no color change, the VP test would show the positive result; red color change. The expected result might show different results because of having different environments; such as a different tempe pathogen which means it would not infect easily on healthy individuals. P. aeruginosa carries a 40-60% mortality rate10. It is always listed as one of the top three most frequent Gram-negative pathogens and is linked to the worst visual diseases10. Moreover, it has a low antibiotic susceptibility and acquires resistance easily8. Thus, it is a hard pathogen to eliminate.The multitest systems, which are Enterotube II, Oxi/Ferm Tube II, and API Staph systems, can be used to determine unknown bacteria. The Enterotube II is for identification of Enterobacteriacease. Gram-negative rod or coccobacillus is appeared in one of the Enterobacteriaceae. The Oxi/Ferm Tube II system is used for the gram negative with oxidase-positive bacteria. It consists with 12 media and 14 physiological tests, and identifies nonfastidious oxidative-fermentative gram-negative rods1. The API Staph system identifies gram-positive cocci and staphylococci. Staphylococcus, Kocuria and Micorococcus can be determined by 9

자연과학| 2010.03.26| 9페이지| 2,000원| 조회(220)

lab, Microbiology, Method, Result, discuss..

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Characterization of Organic Solids and Liquids by Elemental Analysis, Melting Point, Boiling Point, and Infrared Spectroscopy

Methods and BackgroundThe purpose of this lab was to find out the identity of one unknown solid sample and one unknown liquid sample by using different analytical technique and calculation. First, the empirical formula was found by calculating molecular weight, with the percentage of composition of each element in the molecule, and through calculating hydrogen index deficiency number. Boiling point and melting point and IR spectra was also obtained. Figuring out the identity of an unknown compound is possible by obtaining its invariant physical constants. Physical constants or properties, such as boiling point, melting point, index of refracting, density, solubility and spectroscopy can be obtained by experimental methods. However, identification of an unknown compound is possible only with the previously known compounds. A melting point of a substance is defined as the temperature at which the liquid and solid phases exist in equilibrium withclusionThe mass spectrometry, melting point, boiling point, and infrared spectroscopy helped us determine the structure and the molecular formula of unknown solid I and unknown liquid J. Infrared spectroscopy measures the molecular vibrations due to bond stretching, bending, and rotation. IR graphs of a compound give the peaks and intensities of each of these vibrations. Specific bond vibrations occur at only certain areas; thus, giving us the functional groups that are present in an unknown compound. This graph can help us determine possible structures resembling the functional groups. Mass spectrometry determines the molecular weight and the elemental composition of a compound. This information is used to determine the empirical and the mo

자연과학| 2010.03.26| 5페이지| 1,500원| 조회(285)

lab, Melting point, IR Spectroscopy, Chem..

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Preparation of alkyl chlorideby sn1

In this lab we observed the conversion of a tertiary alcohol, (CH3)3COH, into a tertiary alkyl chloride, (CH3)3CCl, using hydrochloric acid. This reaction occurs via nucleophilic substitution of one group for another group. In this case, -Cl group was substated for an -OH group. We tested the presence of this compound by performing sodium iodide and silver nitrate tests. There are two different kinds of nucleophilic substitutions: Sn1 and Sn2. Which mechanism a compound might go through depends largely on the structure of the alkyl group bearing the leaving group. Leaving groups are usually basic groups like Cl- or H2O and the nuleophile usually has a pKa value less than 5. Sn1 undergoes one order of kinetics. This means that the reactant takes part in the rate determining step, or the slowest step, of the reaction. During Sn1 mechanism, the reactant protinates itself and forms a stable leaving group. The leaving group is kicked out dConclusionTert-butylchloride Tert-butanolIn this experiment, we obtained 3.4 grams of organic layer, tert-butylchloride, after simple distillation. This experimental mass was only 35.1% of the original product. Multiple factors like improper separation of the organic and the aqueous layer, and boiling off the organic layer might have resulted in this low yield of the sample. When we separated the organic layer from the aqueous layer we might have poured in some of the organic layer into the beaker with the aqueous layer. Thus, some of the organic layer was not distilled via simple distillation and that difference in mass was not taken into consideration when we measured the experimental mass of tert-butylchloride. Furthermore, tert-butanol has a boiling point of 83oC, so it some of the tert-butanol was still mixed with

자연과학| 2010.03.26| 6페이지| 1,500원| 조회(137)

lab, SN1, Preparation, Chemistry, organic, alkyl chloride

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base extraction

Physical State: white crystal Physical State: white crystalIn this experiment, we separated a binary mixture of acetanilide and benzoic acid by liquid-liquid extraction and recrystallized the extracts. We then purified each isolated solid by recrystillizing it from boiling water. We measured the melting points and compared to reported values. Finally, we calculated the % yield obtained for each solid based on the mass of starting mixture.Liquid-liquid extraction is used widely to isolate and purify molecules from most chemical reaction. During this process, a solute is distributed between two immiscible liquids, extracting phase and original phase, when it comes to contact with them. How effective the extracting acid or base is depends on the Keq of the reaction. For an effective extracting acid or base the Keq must be greater than 1. In this case, most of the solute will be in the extracting ConclusionRecrystallization is used to purify solid or liquids. It occurs in two processes: extraction and filtration. During extraction, solutes are transferred from one solvent into another. The two solvents used in this lab were NaOH and CH2Cl2. Since dichloromethane is nonpolar only organic molecules like acetanilide will adhere to it. However, benzoic acid will adhere to the polar NaOH since like soluets dissolves in like solvents. These two solvents separate the solution into two distinct layers: organic layer and the aqueous layer. During extraction choosing the right solvents can make the process go faster. The best solvents are when the they are immiscible with the original solution, they do not react and form an equilibrium with the solutes in the mixture, they must remove the preferred component from the solution, and they must be easily separated from the solutes. After extraction comes filtration. This is when

자연과학| 2010.03.26| 5페이지| 1,500원| 조회(287)

Extraction, lab, Base, Chemistry, organic

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