Apoptosis And Inhibitor of apoptosis proteins Family: A

Apoptosis And Inhibitor of apoptosis proteins Family: A   Background Apoptosis is an orchestrated biological cellular process that occurs in physiological and pathological conditions(1). It is essential for regulating development, homeostasis, and immune-system function in organisms(2). In mammalian cells, apoptosis is mediated by a family of cysteine proteases named caspases which are initially expressed in cells as inactive procaspase precursors and are activated by two pathways, the extrinsic ( or death receptor) and intrinsic (or mitochondrial) apoptotic pathways(1). The extrinsic pathway is activated by the binding of ligands such as Fas ligand (FasL) and tumour necrosis factor (TNF) to death receptors on the cell surface, FAS and the TNF receptor (TNFR), respectively, which leads to the formation of the death-induced signalling complex (DISC)(3)(4). DISC recruits caspase-8 and promotes the cascade of procaspase activation that follows(5). The intrinsic pathway is triggered by extracellular and intracellular stresses, such as high cytosolic [ca+2 ], hypoxia, severe oxidative stress, DNA damage(5), which results in the permeabilization of the outer mitochondrial membrane, the release of pro-apoptotic molecules such as cytochrome C and others into the cytoplasm(6), the formation of the apoptosome- a large protein complex that is made up of cytochrome C, apoptotic protease activating factor 1 (APAF1) and caspase-9 and caspase activation(7). On the other hand, cell death is also modified by other mitochondrial proteins such as apoptosis-inducing factor(AIF), second mitochondria- derived activator of caspase (Smac), direct IAP Binding protein with low PI (DIABLO)   and Omi/high temperature requirement protein A (Htr A2)(7). Smac/ DIABLO or Omi/HtrA2 induces cell death independently of caspase activation by counteracting inhibitor of apoptosis (IAP)- mediated caspase inhibition(7)(8)( Fig. 1). The upstream caspase for the intrinsic pathway is caspase 9, while that of the extrinsic pathway is caspase 8. The intrinsic and extrinsic pathways cleave the precursor forms of effector caspases, such ascaspase-3, caspase-6 and caspase-7(9). Activated effector caspases cleave many vital cellular proteins such as protein kinases, cytoskeletal proteins, DNA repair proteins and inhibitory subunits of endonucleases family and break up the nuclear scaffold and cytoskeleton(9). They also activate DNAase, that further degrade nuclear DNA(10), which together contribute to the typical morphological changes in apoptosis. Dysregulation of apoptosis has been implicated in numerous pathological conditions, including cancer(1). Besides, targeting the apoptotic pathways for cancer treatment is supported by several findings emphasizing the role of aberrant apoptosis in tumorigenesis and also resistance to anticancer treatment. Evasion from apoptosis is critical for tumor growth and a hallmark of cancer(11). One of the mechanisms by which evasion of apoptosis occurs is   disrupted balance of pro-apoptotic and anti-apoptotic proteins(1). A delicate balance between pro-apoptotic and anti-apoptotic mechanisms determines whether a cell death signal can activate the apoptotic program. It is not the absolute quantity but rather the ratio of these pro-and anti-apoptotic proteins that controls the regulation of cell death. In this balance, pro-apoptotic proteins activate apoptosis and anti-apoptotic proteins inhibit apoptosis(12)(13). Inhibitors of apoptosis protein (IAPs)   are important members of the anti-ap optotic family of proteins that can inhibit   caspase activation and play a key role in regulating of apoptosis in many species(1). Inhibitor of apoptosis proteins (IAPs): The inhibitor of apoptosis proteins are a group of structurally and functionally similar proteins that regulate programmed cell death, cytokinesis and signal transduction(14). The   IAP gene is 1.6 kb in size encoding a 31 kDa protein with a zinc finger-like motif. Many IAP family members have been identified in almost all species from viruses to mammals(15). They are characterized by the baculovirus IAP repeats (BIR) domain at the N- terminus, the name of which derives from the original discovery of these apoptosis suppressors in the genome of baculoviruses(16). The BIR domain contains approximately 70 amino acids. Although the number of BIR domains varies among IAP members, each BIR domain is made up of cysteine and histidine residues in a well-defined pattern (CX2CX16HX6C)(15). IAP acts as endogenous inhibitor of caspases by binding of their conserved BIR domains to the active sites of caspases in vitro and   vivo. IAPs inhibit caspases by promoting the degradation of active caspases, or by sequestering the caspases away from their substrates(17). When IAP family members are overexpressed, cancer cells no longer proceed to apoptosis and become increasingly resistant to standard chemo- and radiation therapies(18)(19). Many studies have established a circumstantial association between IAPs and cancer. Pathological overexpression of several IAP family members has been detected in several classes of human cancers(20)(21)(22). The eight IAPs identified in humans are cIAP1, cIAP2, NAIP, Survivin, XIAP, apollon, ILP-2 and livin(23). Interestingly, many data have shown that c-IAP1, c-IAP2 and XIAP are   broadly expressed in normal cells(24)(22). In normal tissues, IAPs could have some potential physiological roles, such as the regulation of the immune system(25), the response to cell damage(25), cell survival and differentiation(26). On the other hand, it has been proven in many studies that survivin, unlike other IAPs, is prominently expressed in vast majority of neoplasms but not in differentiated normal tissues(27). Survivin has been reported to be overexpressed in various cancers including breast and lung cancer, prostate, gastric, colon, bladder and esophageal carcinomas, osteosarcomas and lymphomas(28)(29). Overexpression of survivin was also found to be significantly associated with poor prognosis and decreased survivial rates in many cancers(30)(31). Survivin: Survivin (also Called IAP 4) is a protein with a crucial role in regulating both cell division and apoptosis. It is the smallest member of the IAP family(29). Survivin, a 16.5 kDa intracellular   protein of 142 amino acid, was discovered in 1997 by Ambrosini and colleagues(32). Structurally, survivin contains a single BIR domain. This domain is essential for its anti-apoptotic activity(33). However, instead of a ring finger domain (RING) near the C-terminus shared by others members of the IAPs, survivin contains a C-terminus alpha-helical coiled-coil (CC) domain which is thought to be important for its interaction with microtubules, hence its roles in cell cycle(34)(35) In normal tissues, survivin shows cell -cycle dependent expression during cell division. Its expression increases in G2/M phase and decreases rapidly in G1(29). The regulation of survivn expression and function is complex and can occur at various levels, including transcriptional regulation, post-translational modification, and protein stability regulation(27). it is regulated by a number of factors such as: NF-nB(36), insulin-like growth factor I/mTOR(37), Ras oncogene family(38), E2F, Sp1, TCF, and heat shock protein (Hsp) 90(39)(40).   Survivin is also regulated by p53 wild type. Additionally, post-transcriptional phosphorylation has been proven to play a   regulatory role in survivin activation(41). Biologic function of survivin Survivin as an inhibitor of apoptosis The mechanism by which survivin inhibits apoptosis is still controversial. Initially, survivin and other IAPs were postulated to inhibit apoptosis directly by interfering with the function of caspase-3, caspase- 7, and caspase-9(42).   In support of this model, it was shown that survivin can interact with Smac/DIABLO physically, thus placing survivin in a central position in the dynamic balance of proapoptotic and antiapoptotic factors(43). However, Structural analyses of survivin indicated later that any effect on caspase should be indirect, as it lacks the amino acid sequence that is essential in other IAPs for caspase binding. Also, the survivin gene is highly conserved in a wide range of organisms, and all of its orthologues are involved in mitotic regulation but not in cytoprotection(44).   Studies of cells from survivin-knockout mice have cast further doubt on the existence of a direct link between survivin and apoptosis(45). Later experiments indicated that Survivin inhibits active caspase-9 but not active caspase-3 and caspase-7. And, survivin mediated inhibition of caspase-9 requires interaction and cooperation with other molecules such as   HBXIP (hepatitis B X-interacting protein)(46) and   XIAP (X-linked inhibitor of apoptosis protein) which also known as inhibitor of apoptosis protein 3 (IAP3)(47) (Fig. 3). Survivin also provides cytoprotection to cells through the inhibition of the AIF pathway, which is known to induce caspase-independent DNA fragmentation(48). Survivin as a promotor of mitosis The cell- cycle dependent expression of survivin in normal tissues   supports strongly its role in cell division. During mitosis, survivin acts in a narrow time window at metaphase and anaphase. It is acting as an interphase between the centromere/central spindle and the chromosomal passenger complex (CPC)(49). CPC is a hetero-tetrameric complex which localizes to different sites at different times during mitosis, and   is composed of four components:   Aurora-B Kinase (enzymatic component), Borealin/Dasra, Survivin and inner centromere protein (INCENP)(50)(51). CPC is essential for proper chromosome segregation and cytokinesis(52). Inactivation of mammalian survivin -or its orthologues in lower organisms results in cytokinesis abnormalities, particularly spindle defects(53)(54) (Fig. 3)(55). Survivin facilitating angiogenesis In addition to its roles in apoptosis and mitosis, survivin   promotes angiogenesis. it is strongly expressed in endothelial cells (EC) during   the proliferative phase of angiogenesis(56)(57) and the antisense-mediated suppression of survivin   during angiogenesis stimulates vascular regression in vitro(58). Besides, exposure of cultured vascular EC to angiogenic factors such as VEGF and bFGF result in increasing survivin expression (both mRNA and protein)(59)(60). Survivin expression In normal physiological conditions, survivin is usually expressed in   embryonic lung and fetal organs in the developmental Stages(61). The protein is also detected in mature tissues with high proliferation potential such as thymus, placenta, CD34+ stem cells and basal colonic epithelial cells(61)(62)(63). However survivin seems to be selectively expressed in transformed cells and in most human cancers. Many studies have shown that survivin, unlike other IAPs, is prominently expressed in the vast majority of neoplasms but not in the differentiated normal tissue(27). Based on detection of protein by immunohistochemistry and mRNA by polymerase chain reaction techniques, overexpression of survivin has been reported in various human malignancies including lung cancer(64), breast cancer(65)(66); stomach(67)(68), esophagus(69), liver(70)(71), ovary cancers(72), brain(73) and hematological cancers(74). Additionally, the immunological responses which detected against survivin supports its specific up-regulation in malignant cells(75)(76). Survivin protein has also been shown to induce cytotoxic T-lymphocytes (CTL) response in   breast cancer, melanoma and chronic lymphatic leukemia patients(76). Survivin expression can be deregulated in cancer by several mechanisms, including amplification of the survivin locus on chromosome 17q25 (77), demethylation of survivin exons(78), increased promoter activity(79), and increased upstream signaling in the phosphatidylinositol 3-kinase or mitogen activated protein kinase pathways(80). Overall, increased survivin expression in several malignancies is associated with cancer survival or disease recurrence, and resistance to chemotherapy or radiotherapy. In a study of 275 patients with breast cancer demonstrated that survivin was a significant prognostic factor and predicted the outcome independent of patients age, tumor size and histologic grade(81). In the case of ovarian cancers, survivin expression was correlated with poor prognostic factors such as: high histologic  grade, mutant p53, and poor histologic type(81)(82). Also, previous studies demonstrated that survivin was expressed in benign brain and pituitary tumors. Although survivin   was also present in normal pituitary tissue,   the level of the gene expression was 6-fold higher in tumors than in normal pituitary tissue(83). In a study of 222   patients who underwent radical cystectomy, survivin was expressed in 64% of bladder tumors and 94% of malignant lymph nodes, but not in normal bladder specime ns and its expression correlated with disease recurrence and disease-specific mortality(84).   Also, increased survivin expression has been associated with an unfavorable survival or disease recurrence in colorectal cancer(85), particularly in stage II disease in esophageal cancer(86), hepatocellular carcinoma(87), lung cancer(88), glioma(89), leukemia(90), and other cancer types. A   study in oral cancer demonstrated that the extent of survivin expression was negatively correlated with the degree of differentiation(91). Additionally, survivin overexpression may be a predictive factor to determine response to chemotherapy and radiotherapy in patients with bladder cancer(92), breast cancer(93), multiple myeloma(94), lung cancer(95) and lymphoma(96)(97). On other hand, patients with lower survivin expression were more responsive to preoperative chemotherapy with 5-flourouracil and cisplatin in esophageal cancer(98). It is also reported that patients with lower survivin expression in pretreatment biopsies were more responsive to radiotherapies in rectal cancer(99). While Overexpression of survivin was associated with   resistance to a taxol-based   therapy for ovarian carcinomas(100). In addition to full-length transcript (survivin (wild type)), five splice variants, which result from splicing of survivin BIRC5 gene pre-messenger RNA (mRNA), have been described: survivin-ΔEx3, survivin-3B, survivin-2ß, survivin2ÃŽ ± and survivin 3ÃŽ ± with different structure and function(101)(102)(103). Previous studies showed that an imbalance in the alternative transcript ratios may affect the cell to be resistant or sensitive to apoptosis(104). This alternative splicing of Survivin has been shown to have correlation with disease activity in various patient studies. For example, studies showed that Survivin-ΔEx3 and survivin-3B were found to be highest in tumors with advanced histological grade and were associated with poor prognosis(105)(106). On other hand, the expression of survivin-2ß was significantly higher in small tumor size and was inversely associated with axillary node positive carcinomas(106). Besides different splicing forms, immunohistochemical studies have demonstrated that survivin also localized   in distinct nuclear and cytoplasmic subcellular pools. Cytosolic Survivin is believed to act as apoptotic suppressor while nuclear Survivin is postulated to regulate cell division(29). There are conflicting data of pathological significance of nuclear Survivin.   Some Splicing studies showed that nuclear staining of survivin is associated with favorable prognosis(107), while others showed Its expression in the nuclei of tumor cells appears to be associated with unfavorable clinical outcomes(108)(109). Also, the cellular localization of Survivin isoforms   differs. while survivin-2ß   and Survivin 2a are localized in both nuclear and cytoplasmic compartments, survivin-ΔEx3 is localized in both mitochondria and nucleus(110). Additionally, Methylation and Phosphorylation are critical requirements for survivin function. Several observations show that survivin is unmethylated in cancer but may be selectively methylate  d in normal tissues with individual variations(111)(112). Methylation may play an important role in the p53 mediated suppression of survivin(113). Another critical requirement for survivin function is the phosphorylation on Thr34(114) Treatment approaches: Due to important role of Survivin in tumor cell division, apoptosis, chemo resistance and survival, survivin represents a unique target for biologic therapy in many human malignancies. Several novel experimental therapeutic strategies have been developed to block the expression or function of Survivin in tumour cells. These include immunotherapeutic approaches to induce immune response against Survivin, small molecule inhibitors/antagonists of   survivin function, and nucleic acid based approaches which interfere with Survivin gene expression(115)   such as antisense oligonucleotides (ASOs), ribozymes and small interfering RNAs (siRNAs)(116). Also, Vaccine approaches such as dendritic cell based (DC) vaccines, DNA vaccines(117), peptide vaccines for Survivin have also been evaluated in preclinical or clinical studies. Survivin ASOs were first used against malignant melanoma cell lines. Transfection with the ASOs triggered spontaneous apoptosis linked to decreased endogenous survivin expression(118) . Treatment with LY2181308, a specific inhibitor of Survivin mRNA which has already entered the phase 1 trial(119). YM-155 is a novel small-molecule survivin suppressant which inhibits survivin mRNA transcription and protein expression in p53-deficient cancer cells in vitro(120). YM155 has also shown to be effective in vivo models of prostate, pancreatic, and lung cancer(120)(121). Ribozyme mediated approaches have also been evaluated for inhibition of Survivin expression. Down-regulation of human Survivin gene expression and increased apoptosis was achieved by using two hammerhead ribozymes (RZ-1, RZ-2) targeting human Survivin mRNA (122) PIQL: Success-Tolerant Query Processing in the Cloud PIQL: Success-Tolerant Query Processing in the Cloud Advanced Topics in Foundations of Databases PIQL: Success-Tolerant Query Processing in the Cloud Stavros Anastasios Iakovou Introduction In our days it is widely know that modern web applications are directly linked with databases. In addition, the number of the users is highly increas- ing through the time and as result the related databases start overloading. Furthermore, despite the fact that data indepence would be ideal for im- plemeting lithe applications developers abandoned this idea in order to avoid expensive queries. Hence, Michael Armbrust et al.[1] implemented a new declarative language called PIQL, a scale independent language. A large number of frameworks have already appeared in order to assist developers to create modern web applications. However, this plethora of websites with millions of users led to database failures due to lack of request managing. As a result, there was a demand on implementation of a new system that will control all these requests and provide efficient results to users. A few methods have introduced and one of the most popular is NoSQL. Despite the fact that NoSQL provided a high level interface, data indepen- dence created scalability problems since a large number of queries took a lot of time. This led to to several issues like performance failing and user disatisfaction as well. In order to avoid this bad situation scientists hand coded key/value implementations. On the one hand, this provided the de- sirable scalability but, on the other hand is was not easy enough for the developers to write that kind of code to parallelize their queries so as to fi achieve high scalability. Another significant issue is time consuming functions rewrites. Now, once we talked about several problems occured by queries in the next section we will discuss about PIQL. More specifically, we will present this method and give a brief summary of the implementation. In the rest of the document we will discuss about the performance of the previous imple- mentation. What is PIQL? In this section we will discuss and analyze the PIQL (PerformanceInsightful Query Language) model. One important advantage of PIQL is that intro- duced the notion of scale independency. More specifically, the model pre- serves the logical data independence. The most significant about data this technique is that performace maintains not only on small datasets but also in large as well. For this reason this is called success-tolerant since the success is for every large dataset. But why PIQL is successful? The answer is on the limitation on key/value store operations. As we previously mentioned, one goal of PIQL is to avoid issues when the database gets larger. PIQL uses static analysis in order to fi the correct number of operation in every step of the execution. Before we move to the next step of the analysis of the methodology we should mention the four queries classes. The fi one is called constant since the processing time is constant. The second one is the bounded class. More specifically this class refers to bounded data when the site becomes more popular. For instance, in case of Facebook every user has a limit of 5000 friends. The third class is called sub-linear or linear and is referring to queries that become more successfull when the data increase linearly. The last one is Super-linear where intermediate calculations are necessary for the queries. Now, once we mentioned all the necessary theoritical parts of PIQL we will discuss on its structure. Every server is directly connected with a Distributed Key/Value Store. Hence, this methodology maintains the scalability and the response time is now predictable. A significant drawback of this technique is that a specific key/value store is required so as to maintain data locality. On the other hand, this method is non-blocking and according to Chen et al.[2] can reduce memory latency. Another important benefit of PIQL is that extends the cardinality con- straint of regular direction to diff ent directions as well. More specifi , these cardinalities provide several information on its relationships. For in- stance, a Facebook user should have no more than 5000 friends. This is a very significant information since selecting the wrong number for limita- tions can lead back to the previous problems. Thinking again the Facebook limitations for the maximum number of friends on Facebook, according to Brandtzg et al.[3] a significant issue that occurs is the lack of privacy. Hence, the limitations are not only important for the performance but also for the user protection as well. In addition, the same person can create a new profit for free and add his new friends there. As a result, 5000 friends is not actually a limitation for a user and is provided in terms of privacy and performance. According to Michael Armbrust et al.[1] their algorithm for scale inde- pendent optim ization contains two phases. The fi one is reffering to stop operator insertion. In order to maintain scalability, the algorithm starts by fi a linear join ordering on the query parser. Depsite the fact that stop operator is already contained due to LIMIT which in contained in the reg- ular query, scientists have introduced data-stop operators which are pushed in lower levels in order to preserve the initial rules without the demand of  restart the whole system. Next, after fiphase 1 the second step which is called remote op- erator matching. As we previously mentioned we should ensure scalabiliy. Hence, the intermediate results are bounded. But how all these logical op- erators are mapped on remote operators? For Index Scan, that means that maximum one attribute can be affected by predicates. As for Index Foreign Key join the number of tuples after the join is less than or than the tuples of the initial plan. References [1] Armbrust, Michael, et al. PIQL: Success-tolerant query processing in the cloud. Proceedings of the VLDB Endowment 5.3 (2011): 181-192. [2] Chen, Tien-Fu, and Jean-Loup Baer. Reducing memory latency via non- blocking and prefetching caches. Vol. 27. No. 9. ACM, 1992. [3] Brandtzg, Petter Bae, Marika Lders, and Jan Hvard Skjetne. Too many Facebook friends? Content sharing and sociability versus the need for pri- vacy in social network sites. Intl. Journal of HumanComputer Interaction 26.11-12 (2010): 1006-1030.

Breaking Family Ties

Thai Ngo Barbara Estermann English 96 February 25, 2013 â€Å"Breaking Family Ties† Norman Rockwell’s â€Å"Breaking Family Ties† gives us a look into the change of the post Great Depression and World War II generation. How America itself had changed so much in the passed 25 years from the greatest economic depression to being the greatest country on earth. It also shows the heartbreaking moment of a boy preparing to leave his father and dog and be on his own for the first time.The father, tired from a life of hard works, sacrificed everything so that his son can go to college; the young man, representing the post Great Depression and World War II generation, is making a better life than what his father had by getting higher education. In Norman Rockwell’s painting â€Å"Breaking Family Ties†, a boy sits with his father and dog preparing to leave for college. The young man and his father sit on a board of the family farm truck. At the bottom corner of the painting, shows a single rail. Suggesting that they are waiting for a train. On the ground is the son’s suitcase with a â€Å"State U† sticker.Books are stacked on top of the suitcase. The young man has his tie and socks perfectly matched, and is wearing white trouser and matching jacket. Sitting with his hand folded, the young man looked eagerly toward the train track, ready for the next chapter in his life. His father; however, sits slumped with his and his son’s hats in his hand as if he didn’t want the son to leave. The father looked at the opposite direction of the rail; as if he didn’t want to see the train come and take his son away. Although the father and son are looking in opposite direction, the sense of family bond is still strong.The father, probably in his late fifty, has been through the Great Depression and World War II; values his family more. The generation coming out of the Depression and World War II has been through some of the greatest challenges this country had ever faced. They worked hard their whole life, and now all they want is for their children to have a better life than they. Norman Rockwell perfectly describe the generation shift in â€Å"Breaking Family Ties† as the father, old beyond his years, sacrificed everything so that his son can go to college.He does not want to see his son leave, because he is everything to him. Still, he knows that his son must leave to grow up and be independent. The son, treated like royalty by the father, is ready to move on to the next chapter in his life and be away from the protection of his father. The painting gives mix emotion because of the two different emotions of the main characters. Norman Rockwell shows the differences between the two-generation through the father and son. The experience of leaving you’re parents and being on your own is universal to everyone.The universal message in â€Å"Breaking Family Ties† is being inde pendent for the first time. I was seventeen when I left home by myself to go to college. Preparing to leave home, I was like the boy in â€Å"Breaking Family Ties† eager to move on to the next chapter in my life and to be independent. I was only thinking of myself, and not how leaving would affect my parents. My parents were sad to see me leave, but they knew that I was making a better life for myself by going to college. Just like the Father in â€Å"Breaking Family Ties†. Leaving home for me was the beginning of a new chapter in my life.While for my parents, it was seeing me grow up too fast. Norman Rockwell’s perfectly showed the heartbreaking moment and a son leaving to be on his own for the first time. He also reminded America of how much our country had changed with the two generations. Like all of Norman Rockwell’s works, everything in â€Å"Breaking Family Ties† has a meaning behind it. Which is I can relate to it even if it was painted yea rs before I was born. Norman Rockwell is one of America greatest artist because of the universal meanings behind his works.

Diffusion and Osmosis

The Effects of Osmosis and Diffusion The experimentation of last week’s lab was in order to test the many effects of diffusion and osmosis amongst four experiments. One such experiment was testing the effects of molecular weight on diffusion in relation to the use of Agar. The methods performed included the use of two acids, HCl and acetic acid. Both acids were placed into an Agar-filled dish and, over increments of 15 minutes, data collection was taken based off the diffusion rate and the diameter length of both the HCl and the Acetic Acid.The resulting factor was the HCl exhibited a greater rate of diffusion, directly resulting in a lager diameter. This implies that the HCl ultimately has a smaller molecular weight. The next experiment was based off osmosis of an animal cell; a chicken egg. After submerging two different chicken eggs in distilled water and 10% salt water, once again intervals of 15 minute data collection was taken for a total of one hour. After each interval the weight in grams was taken and then the eggs were placed back into the solution for further analysis.Ultimately, the egg in distilled water exhibited an increase in weight while the egg in salt water was the opposite; a decrease in weight. This conclusion proves that water diffusion occurs from a hypotonic solution to a hypertonic solution. Osmosis in a plant cell was tested by comparing an Elodea cell in pond, distilled, and salt water. After obtaining samples of the Elodea cell and preparing a wet mount of each leaf using all three types of water, observations of the cells in a compound microscope was the next step.From there, comparisons of all three types of solutions in order to determine the apparent differences in osmosis were needed. When examined, the cell in pond water was not as defined; this result implied that water left the hypotonic cytoplasm of the cells causing it to wither in a way. Introduction In order to conduct the experiments of this lab, a hypothesis is n o doubt necessary. In reference to the effects of molecular weight on diffusion a person is lead to believe that since the atomic mass unit of Acetic Acid is greater than that of HCl, the rate of diffusion of Acetic Acid will be slower and therefore produce a smaller diameter.As stated by Watson (2011), â€Å"larger molecules diffuse more slowly because of resistance from molecules of the medium. † This â€Å"medium† is the means of passing through the spaces in between a molecule. This was as well stated by (Watson 2011). Reiterating what was described, unlike smaller molecules, which can fit through a medium more easily, in turn allowing for a faster and more sufficient means of diffusion, a larger molecule has the resistance from a specific medium, which in a way is pulling back molecules therefore causing a prolonged time of diffusion.This resistance is a direct correlation and explanation as to why the diffusion rate of a relatively larger molecule exhibits a long er rate of diffusion, as with the comparison of hydrochloric acid and acetic acid, and ultimately the purpose of this experiment. Based on the background information acquired on osmosis of an animal cell, it is safe to assume that after each interval of fifteen minutes, the weight of the animal cell in distilled water will continually grow, while the egg in salt water will decrease in weight.Derived from information provided by (Fisher, Williams, & Lineback 2011), an animal cell, which is hypertonic, placed into a hypotonic solution of distilled water will cause water to diffuse into the hypertonic cell, seeing as diffusion occurs from hypotonic to a hypertonic solution. With any type of diffusion process, the particles that are being diffused tend to travel from a concentration that is greater to one that is smaller; moving down in the concentration gradient. This is the direct result of the increase in weight of the animal cell in the experiment.In relation to a chicken egg, the l argest living cell, it is predicted that the containing molecules will be too large to pass the membrane and water will flow into the egg (Reece 2011). The matter of the animal egg being placed into a solution of 10% salt is the directly opposite of the above stated. Osmosis within a plant cell placed in pond water will show a wilted cell wall based on the continual impeding force of the water on the wall. Aquatic plants tend to be hypertonic in their natural environment causing the plant to exhibit a â€Å"swollen† or turgid structure.Materials and Methods In order to accurately and sufficiently test the hypothesis of the effects of molecular weight on diffusion, agar was one substance that was used. Agar in the presence of acids turns from a yellowish color to a more violet color. This same dish contained to holes with which two acids could be placed-HCl and acetic acid. From basic chemistry knowledge one knows that the molecular weight of HCl in comparison to Acetic Acid i s smaller in size; that information was given from Watson (2011).This is significant because it will later give way to the rate of diffusion of the two different acids. Constant observations, recordings, and measurements were required for this experiment, only in the intervals of 15 minutes. Over a period of one hour it was noticeable that the HCl exhibited a greater rate of diffusion and a great length in diameter, in comparison to acetic acid. The most important factor when dealing with this diffusion experiment, was the methods taken to prove that HCl had a greater rate of diffusion than acetic acid.Initially, soaking a chicken egg in a small solution of acetic acid and 2 parts tap water will allow for better experimentation of the rate of osmosis of an animal cell. The overall scope of this particular experiment was to weigh two eggs using a triple beam balance in order to get an initial weight of the eggs before beginning the process of the lab. After doing so, the eggs were pl aced into two solutions, one being distilled water and the other 10% salt. Proceeding these steps were the 15 minute intervals of time, and after, a recording of the weight of the egg.This process was done until a total of 60 minutes was reached for both the distilled water solution and the 10% salt solution. After acquiring all results and data, a conclusion could be based. Once acquiring three samples of Elodea leaves, preparing three different wet mounts was the following step. From there, after ten minutes an observation of all the samples under a compound microscope was the following method needed in order to determine the characteristics of the leaves. The leaf in the pond water demonstrated the leaf cell in â€Å"normal† conditions, while the distilled water and NaCl were not â€Å"normal† conditions.Results The findings of the effects of molecular weight diffusion conclude that ultimately the molecular weight of a molecule affects the rate of diffusion directly . The greater the weight, the slower the diffusion process will be; that was the case for acetic acid, and it was in part due to the diffusion of particles through the medium. In addition to that, the measurement of the diameter of both acids also was directly affected by the molecular rate. All the comparisons in the diameter readings of the two acids can be found in table 2.All readings for both acids were taken over an increment of 15 minutes for an hour. In total, HCl produced a larger diameter due to its smaller amu. See table 2. In comparing the affects of distilled water to 10% salt water and the rate of osmosis of an animal cell, the rate of osmosis proved most sufficient in distilled water, rather than in the salt water, with an apparent increasing weight distribution in the distilled water, and a decrease in weight in the salt water. These changes in weight loss and gain are exhibited in Table 1.Even though it is obvious that both eggs exhibited either weight loss or gain, both eggs also showed a sudden spike it the gain or loss around the time frame of 15 minutes and 45 minutes, yet again illustrated in Table 1. Discussion After conducting the diffusion experiment using agar and examining the results, it is apparent what the outcome of diffusion is when comparing HCl and acetic acid atomic weights. It is as well safe to assume the resulting outcomes of future comparisons of two molecules of with different atomic mass units.The use of agar in this specific experiment is much useful due to the properties and characteristics of the extract. The agar, in the presence of an acid, turns from a yellowish color to one that is pink; because of this characteristic, it was possible to measure the distance from the center outward of the agar when placed into a dish of HCl and acetic acid (Watson 2011). As explained before, these measurements allowed for sufficient data in determining the rate off diffusion for both acids. Table 2 will provide a visual for the d ata that was collected from the experiment.In the end, a conclusion was established that the rate of diffusion was most prominent in HCl, the acid with the smallest amu. Simply the definition of diffusion itself will aid in understanding why molecules of a higher molecular weight will diffuse slower in comparison to one of a smaller weight. Any substance will diffuse down its concentration gradient, the region along which the density of a chemical substance decreases (Reece 2011). It is understood that the molecular weight is how much mass a substance has, and mass can be determined by how tightly packed particles are-density.A molecule with a high mass, ultimately a high density, will illustrate a slower rate of diffusion. With regards to the cell that is the egg, the rate of osmosis proved to be greater in the distilled water as compared to that of the 10% salt. This is in part due to the size of the particles that make up the egg as well as surround the egg. If there is a higher concentration of nonpenetrating solutes in the surrounding solution, then water will tend to leave the cell (Reece 2011). This definition provides an understanding of what is happening to the egg when it is submerged into the 10% salt solution.Comparing the egg to the salt solution, there is a higher concentration of nonpenetrating solute in the salt solution, nonpenetrating being the particles that cannot cross the membrane, and this in return allows water to leave the egg which ultimately causes dehydration for the egg, resulting in weight loss recorded in Table 1. The complete opposite is the case for the distilled water which would result in weight gain for the egg. Literature Cited Fisher, K. , Williams, K. , & Lineback, J. (2011). Osmosis and diffusion conceptual assessment. CBE Life Sciences Education, 10(4), 418-429. doi: 10. 187/cbe. 11-04-0038 Reece, J. B. 2011. Campbell Biology. 9th ed. San Francisco (CA): Pearson Education Inc. 125-139 p. Watson, C. M. (2011). Diffusion and osmosis. In Biology 1441 Laboratory: Cellular and Molecular Biology (pp. 76-91). Boston: Pearson Learning Solutions. Tables and Figures Figure 1 percentage change in wait of eggs between 15 minute intervals [pic] |Weight of Egg (grams) | |Time Water 10% Salt | |0 75. 60 91. 65 | |15 76. 00 91. 46 | |30 76. 10 91. 39 | |45 76. 10 91. 5 | |60 76. 10 91. 23 | Table 1 A comparison in weight and change of each egg in DI water and a 10% salt solution. |Start time |HCl |Acetic Acid | | |15 min |16 mm |16 mm | | |30 min |18 mm |19 mm | | |45 min |23 mm |22mm | | Table 2 ———————– 60 min26mm23 mm Diffusion and Osmosis Kristen Demaline Bio 1113, Lab 3: Diffusion and Osmosis Osmolarity of Plant Cells In this class, we learned about hypertonic, hypotonic, and isotonic solutions. Hypertonic solutions have a higher concentration of solutes outside of the membrane, hypotonic solutions have a lower concentration of solutes outside the membrane, and isotonic solutions have an equal amount of solutes inside and outside of the membrane (Morgan & Carter, 66). When the solute concentration is not equal, the water concentration is not equal, so water will move from a higher concentration to a lower concentration in a process called osmosis.In this experiment, we cut 4 pieces of potato, weighed them, and let each soak in a different sucrose solution for about an hour and a half. Our solutions consisted of distilled water (. 0 sucrose molarity), . 1 sucrose molarity, . 3 sucrose molarity, and . 6 sucrose molarity. Our question was â€Å"which solutions are hypertonic, which are hypotonic, and which are isotonic ? †. This can all be determined through weight change. We hypothesized that distilled water would be a hypotonic solution, the . 1M would be a hypotonic solution, the . 3M would be an isotonic solution, and the . 6M would be a hypertonic solution. We thought that . M would be the isotonic solution because its molarity is in the middle. If . 3M is in fact an isotonic solution, then the water concentration is the same inside and outside of the membrane and there should be no water movement resulting in no weight change. If distilled water and . 1M are hypotonic solutions, then the concentration of water is higher on the outside, so water will move into the potato where water concentration is lower, causing a weight gain. Finally if . 6M is hypertonic, then water concentration is lower on the outside, so water will move from the inside of the potato to the solution, causing the potato to lose weight.After about an hour and a half we took the potato pieces out of the solutions the y were soaking in, patted the water off of them, and weighed them for a second time. The initial weight and final weight was recorded, which can be seen in Table 1. The potato piece that was soaking in the distilled water had a 3. 1% weight gain, and the potato piece that was soaking in . 1M sucrose had a 2. 1% weight gain. The potato piece had no weight change in the . 3M sucrose solution. And the potato piece that was soaking in . 6M sucrose solution had a 5. 7% weight loss.The weight changes can be easily seen in Graph 1. Table 1: Change in Weight |Sucrose Molarity: |0M |0. 1M |0. 3M |0. 6M | |final weight (g) |16. 4 |14. 7 |17. 7 |13. 2 | |initial weight (g) |15. 9 |14. 4 |17. 7 |14 | |weight change (g) |0. 5 |0. 3 |0 |0. 8 | |%change in weight |3. 10% |2. 0% |0% |5. 70% | Graph 1: [pic] As you can see, the results supported our hypothesis. Distilled water is a hypotonic solution, which makes sense because there is no concentration of solute in it. The water moved to the potato because the potato has more sucrose concentration, meaning a lower water concentration. The potato that was soaking in . 1M sucrose solution also gained weight as an effect of having a lower water concentration inside, but its weight gain percentage was lower because the solution had more solute than the distilled water. The potato soaking in . M sucrose solution had no change because the concentration of sucrose was the same in the potato as it was in the solution, as we predicted. The potato lost weight in the . 6M sucrose solution because the amount of sucrose inside the potato was less than the solution causing water movement from the potato to the solution. These results clearly demonstrate the process of osmosis. The water moved from a region where concentration is higher to a region where concentration is lower in every case, just like it would in a cell. Of course there is always a possibility of human error in weighing, labeling, and so on.One mistake our group made was tha t we forgot to look at the time when we put the potatoes in the solution, so we took them out a couple minute after the group next to us took theirs out, since we started at about the same time. When our results were compared to the results of other groups, they still seemed to match up. Repeating the experiment multiple times would give even clearer results. Diffusion of Starch, Salt, and Glucose Diffusion is when molecules move from an area where they are high in concentration to an area where they are low in concentration (Morgan & Carter, 66).In this experiment, we tested the ability of certain substances to pass through a semi-permeable membrane in the process of diffusion. Our semi-permeable membrane was dialysis tubing that was presoaked in water. We tied one end of the tubing with string, filled it with a solution that contained starch, salt, and glucose, and then we tied the other end. We weighed it, so we could later weigh it to discover if there was any weight change. We then placed the dialysis tubing into a beaker of distilled water.Our question was â€Å"which of these substances would be able to pass through the dialysis tubing, or semi-permeable membrane? †. After we let the tubing soak for 30 minutes, we could test for the presence of starch, salt, and glucose using 3 tests (iodine test for starch, silver nitrate test for salt, and Benedict’s reagent for glucose). Our hypothesis was that we would find the presence of all three substances in the distilled water. We thought this because we knew that molecules naturally diffuse when surrounded with an area with less concentration, but we didn’t know how much the semi-permeable membrane would interfere.Our other hypothesis was that water would enter the tubing as substances escaped it. We thought that due to osmosis, the water would move from the area of higher concentration (outside the tubing) to the area of lower concentration (inside the tubing). If our hypothesis was corr ect and all substances made it through the membrane, then we would expect to see the tubing gain weight and the original distilled water test positive for each substance, using our 3 tests, after the 30 minutes.To carry out the tests we had a positive control for each substance. The positive controls allowed us to see the results of the tests when we knew the solution contained the substances being tested for. We filled 3 test tubes with the starch/salt/glucose solution (positive controls) and 3 test tubes with the distilled water that the dialysis tubing had been soaking in. We put three drops of iodine in a positive control test tube, and three drops into a distilled water test tube to test for starch.Then we put five drops of silver nitrate into a positive control test tube, and five drops into a distilled water test tube to test for salt. Lastly, we put five drops of Benedict’s reagent into a positive control test tube, five drops into a distilled water test tube, and pla ced them both into boiling water to test for glucose. We recorded the color of each, which can be found in Table 2. We also weighed the tubing after it had soaked for 30 minutes and recorded it with the initial weight, which can be found in Graph 2. Table 2: Results of Diffusion Tests Test tube |Initial color |Final color | |starch pos control |cloudy, white |dark purple | |starch experiment |clear |yellow | |salt pos control |cloudy, white |cloudy, white | |salt experiment |clear |cloudy, white | |glucose pos control |cloudy, white |orange | |glucose experiment |clear |orange |Graph 2: [pic] If we look at Table 2 we see that we got the same color in the distilled water as we got in the positive control for the salt test and the glucose test, meaning that the distilled water tested positive for those substances. For the starch test, the positive control turned dark purple, but the distilled water turned yellow, meaning that it tested negative. If these results are correct, then star ch was unable to pass through the semi-permeable membrane. This made our hypothesis false, but not completely. We were still correct about the salt and the glucose making it throught the membrane.Our other hypothesis was correct. Graph 2 displays a weight gain showing that osmosis occured, like predicted. Just like with every experiment, there is room for human error. In this experiment, a mistake that could easily be made is with tying the ends of the tubing and making sure there is no leaks. That mistake could even go unnoticed leading to false results, because it makes it look like the substances made it through the membrane when in actuallity the substances accidently spilled into the distilled water. I think these experiments were successful in demonstrating diffusion and osmosis.The diffusion experiment clearly showed that substances move down a concentration gradient until concentration is equal everywhere, unless something is holding the substances back, like a membrane. The osmosis experiment showed that water always moves down its concentration gradient also. They both showed a search for balance, or equilibrium, on a level that is hard to see without investigation. References Morgan, J. G. and M. E. B. Carter. 2013. Energy Transfer and Development Lab Manual. Pearson Learning Solutions, Boston, MA.    |Points |Self-Assessment |Total Earned | |Introduction |2 |  2 |   | |Results |2 |  2 |   | |Figures/Tables |3 |  3 |   | |Discussion |3 |  3 |   | |Total |10 |  10 |   | Diffusion and Osmosis Kristen Demaline Bio 1113, Lab 3: Diffusion and Osmosis Osmolarity of Plant Cells In this class, we learned about hypertonic, hypotonic, and isotonic solutions. Hypertonic solutions have a higher concentration of solutes outside of the membrane, hypotonic solutions have a lower concentration of solutes outside the membrane, and isotonic solutions have an equal amount of solutes inside and outside of the membrane (Morgan & Carter, 66). When the solute concentration is not equal, the water concentration is not equal, so water will move from a higher concentration to a lower concentration in a process called osmosis.In this experiment, we cut 4 pieces of potato, weighed them, and let each soak in a different sucrose solution for about an hour and a half. Our solutions consisted of distilled water (. 0 sucrose molarity), . 1 sucrose molarity, . 3 sucrose molarity, and . 6 sucrose molarity. Our question was â€Å"which solutions are hypertonic, which are hypotonic, and which are isotonic ? †. This can all be determined through weight change. We hypothesized that distilled water would be a hypotonic solution, the . 1M would be a hypotonic solution, the . 3M would be an isotonic solution, and the . 6M would be a hypertonic solution. We thought that . M would be the isotonic solution because its molarity is in the middle. If . 3M is in fact an isotonic solution, then the water concentration is the same inside and outside of the membrane and there should be no water movement resulting in no weight change. If distilled water and . 1M are hypotonic solutions, then the concentration of water is higher on the outside, so water will move into the potato where water concentration is lower, causing a weight gain. Finally if . 6M is hypertonic, then water concentration is lower on the outside, so water will move from the inside of the potato to the solution, causing the potato to lose weight.After about an hour and a half we took the potato pieces out of the solutions the y were soaking in, patted the water off of them, and weighed them for a second time. The initial weight and final weight was recorded, which can be seen in Table 1. The potato piece that was soaking in the distilled water had a 3. 1% weight gain, and the potato piece that was soaking in . 1M sucrose had a 2. 1% weight gain. The potato piece had no weight change in the . 3M sucrose solution. And the potato piece that was soaking in . 6M sucrose solution had a 5. 7% weight loss.The weight changes can be easily seen in Graph 1. Table 1: Change in Weight |Sucrose Molarity: |0M |0. 1M |0. 3M |0. 6M | |final weight (g) |16. 4 |14. 7 |17. 7 |13. 2 | |initial weight (g) |15. 9 |14. 4 |17. 7 |14 | |weight change (g) |0. 5 |0. 3 |0 |0. 8 | |%change in weight |3. 10% |2. 0% |0% |5. 70% | Graph 1: [pic] As you can see, the results supported our hypothesis. Distilled water is a hypotonic solution, which makes sense because there is no concentration of solute in it. The water moved to the potato because the potato has more sucrose concentration, meaning a lower water concentration. The potato that was soaking in . 1M sucrose solution also gained weight as an effect of having a lower water concentration inside, but its weight gain percentage was lower because the solution had more solute than the distilled water. The potato soaking in . M sucrose solution had no change because the concentration of sucrose was the same in the potato as it was in the solution, as we predicted. The potato lost weight in the . 6M sucrose solution because the amount of sucrose inside the potato was less than the solution causing water movement from the potato to the solution. These results clearly demonstrate the process of osmosis. The water moved from a region where concentration is higher to a region where concentration is lower in every case, just like it would in a cell. Of course there is always a possibility of human error in weighing, labeling, and so on.One mistake our group made was tha t we forgot to look at the time when we put the potatoes in the solution, so we took them out a couple minute after the group next to us took theirs out, since we started at about the same time. When our results were compared to the results of other groups, they still seemed to match up. Repeating the experiment multiple times would give even clearer results. Diffusion of Starch, Salt, and Glucose Diffusion is when molecules move from an area where they are high in concentration to an area where they are low in concentration (Morgan & Carter, 66).In this experiment, we tested the ability of certain substances to pass through a semi-permeable membrane in the process of diffusion. Our semi-permeable membrane was dialysis tubing that was presoaked in water. We tied one end of the tubing with string, filled it with a solution that contained starch, salt, and glucose, and then we tied the other end. We weighed it, so we could later weigh it to discover if there was any weight change. We then placed the dialysis tubing into a beaker of distilled water.Our question was â€Å"which of these substances would be able to pass through the dialysis tubing, or semi-permeable membrane? †. After we let the tubing soak for 30 minutes, we could test for the presence of starch, salt, and glucose using 3 tests (iodine test for starch, silver nitrate test for salt, and Benedict’s reagent for glucose). Our hypothesis was that we would find the presence of all three substances in the distilled water. We thought this because we knew that molecules naturally diffuse when surrounded with an area with less concentration, but we didn’t know how much the semi-permeable membrane would interfere.Our other hypothesis was that water would enter the tubing as substances escaped it. We thought that due to osmosis, the water would move from the area of higher concentration (outside the tubing) to the area of lower concentration (inside the tubing). If our hypothesis was corr ect and all substances made it through the membrane, then we would expect to see the tubing gain weight and the original distilled water test positive for each substance, using our 3 tests, after the 30 minutes.To carry out the tests we had a positive control for each substance. The positive controls allowed us to see the results of the tests when we knew the solution contained the substances being tested for. We filled 3 test tubes with the starch/salt/glucose solution (positive controls) and 3 test tubes with the distilled water that the dialysis tubing had been soaking in. We put three drops of iodine in a positive control test tube, and three drops into a distilled water test tube to test for starch.Then we put five drops of silver nitrate into a positive control test tube, and five drops into a distilled water test tube to test for salt. Lastly, we put five drops of Benedict’s reagent into a positive control test tube, five drops into a distilled water test tube, and pla ced them both into boiling water to test for glucose. We recorded the color of each, which can be found in Table 2. We also weighed the tubing after it had soaked for 30 minutes and recorded it with the initial weight, which can be found in Graph 2. Table 2: Results of Diffusion Tests Test tube |Initial color |Final color | |starch pos control |cloudy, white |dark purple | |starch experiment |clear |yellow | |salt pos control |cloudy, white |cloudy, white | |salt experiment |clear |cloudy, white | |glucose pos control |cloudy, white |orange | |glucose experiment |clear |orange |Graph 2: [pic] If we look at Table 2 we see that we got the same color in the distilled water as we got in the positive control for the salt test and the glucose test, meaning that the distilled water tested positive for those substances. For the starch test, the positive control turned dark purple, but the distilled water turned yellow, meaning that it tested negative. If these results are correct, then star ch was unable to pass through the semi-permeable membrane. This made our hypothesis false, but not completely. We were still correct about the salt and the glucose making it throught the membrane.Our other hypothesis was correct. Graph 2 displays a weight gain showing that osmosis occured, like predicted. Just like with every experiment, there is room for human error. In this experiment, a mistake that could easily be made is with tying the ends of the tubing and making sure there is no leaks. That mistake could even go unnoticed leading to false results, because it makes it look like the substances made it through the membrane when in actuallity the substances accidently spilled into the distilled water. I think these experiments were successful in demonstrating diffusion and osmosis.The diffusion experiment clearly showed that substances move down a concentration gradient until concentration is equal everywhere, unless something is holding the substances back, like a membrane. The osmosis experiment showed that water always moves down its concentration gradient also. They both showed a search for balance, or equilibrium, on a level that is hard to see without investigation. References Morgan, J. G. and M. E. B. Carter. 2013. Energy Transfer and Development Lab Manual. Pearson Learning Solutions, Boston, MA.    |Points |Self-Assessment |Total Earned | |Introduction |2 |  2 |   | |Results |2 |  2 |   | |Figures/Tables |3 |  3 |   | |Discussion |3 |  3 |   | |Total |10 |  10 |   |

Why Burning Driftwood Makes Colored (Toxic) Fire

Did you know you can burn driftwood, especially from the ocean, to get a fire with blue and lavender flames? The colored fire comes from excitation of the metal salts that have soaked into the wood.   While the flames are pretty, the smoke given off of the fire is toxic. Specifically, driftwood releases a lot of dioxin from combustion of salt-soaked wood. Dioxins are carginogenic, so burning driftwood from beaches is not recommended. Some coastal communities have considered burn bans on driftwood to reduce the levels of pollution from the smoke. All smoke contains particulates which can cause health problems when the smoke is inhaled, but you may have been unaware of the additional issue with burning driftwood.