The Structure And Function Of Proteins Biology Essay

The Structure And Function Of Proteins Biology Essay Proteins are large macromolecules which consist of hydrogen, carbon and oxygen; proteins are polymeric chains that are built from monomers known as amino acids. Proteins have a major function in a living organism, for example, the replication of DNA, catalysing metabolic reactions (catalyst); stimulus response and also transporting molecules form one place to another. There are 20 different types of amino acids which synthesize proteins, however the function and different properties of each type of protein is due to the precise sequence and structure of the amino acids present. Each amino acid consists of a central carbon atom (C), which is attached to a hydrogen atom (H), an amino group (also known as NH2 group), a carboxyl group (- COOH, this gives up a proton hence why this is known as an acid) and also a unique side chain or R group. Amino acids are linked linearly via covalent peptide bonds, short chain amino acids are known as peptides whereas long chain formations of amino acids are called polypeptides, where the peptide bond is formed between the carboxyl group of one amino acid and the amino group on the neighbouring amino acid. This reaction occurs as a condensation reaction where there is a removal of a hydrogen atom from the amino group of one amino acid and the removal of a à ¢Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬OH group from the carboxyl acid from another amino acid forming a water molecule (Fig 1). http://ibhumanbiochemistry.wikispaces.com/file/view/CondensationReaction.jpg/31781961/CondensationReaction.jpg Fig 1: a condensation reaction between two amino acid molecules, there is a formation of a water molecule as a waste product. The unique side chain or R group is what disguises one amino acid from another; the overall structure and properties of the proteins are therefore dependent on sequence of the R group of each amino acid. Furthermore these variations of the R group and also the arrangements of the other amino acids would form a number of different polypeptides. Each protein consists of a different number of these polypeptide chains which are folded into complex three dimensional shapes therefore different proteins would have different shapes. There are four levels of protein organization found in polypeptides; these structures are known as: primary structure, secondary structure, tertiary structure and also quaternary structure. Primary structures is the basic structure of the levels of organization, the primary structure is the linear arrangements/sequence found of the amino acid in the protein, and also could be thought of as the covalent linkages found in the polypeptide chain or the protein, such as a disulphide bond. The secondary structure is the areas of folding found within the protein, where there is an ordered arrangement of the amino acids in some localized regions of the polypeptide molecule; hydrogen bonds play a vital role in stabilizing the folding patterns which are found in the protein molecule. Although the conformation of each protein molecule are considered unique, there are two main types of secondary structure, or folding patterns, that are often present; these are the alpha helix and the anti-parallel beta-pleated sheets, these two folding patterns are common due to the hydrogen bonding occurs between the N-H and C=O groups in the backbone of the polypeptide. However there are a number of other secondary structures but the alpha helix and the anti-parallel sheets are the most stable form of secondary structures found. Furthermore there may be a number of these two types of secondary structure found in a single polypeptide chain. An alpha helix is spiral structure where this could be either a right handed or left handed spiral, in which the peptide bonds are found to be Trans conformational and planar, it would also be found that the amino group of each of the peptide bonds is generally in the upward position where as the carboxyl group points in the downwards position. An alpha helix structure is generated when a single polypeptide chain has turned around itself to form a rigid cylinder where a hydrogen bond is formed between every fourth amino acid (fig 1.2), which links the C=O group of one peptide bond to the N-H group on another amino acid (fig 1.2). http://faculty.ccbcmd.edu/courses/bio141/lecguide/unit3/viruses/images/alphahelix.jpg Fig 1.2: shows the hydrogen bond formed between every fourth amino acid, also linking the N-H group and O=H group. There are two types of beta sheets; parallel and anti-parallel beta sheets. The Beta pleated sheets are extended polypeptide chains with another neighbouring polypeptide chain extending either parallel or anti-parallel to each other, this occurs due to the hydrogen bonds being formed between the segments of the polypeptide chain so are essentially place side by side. The parallel beta sheets is when the structure is shown to consist a polypeptide chain and neighbouring polypeptide chain that would run in the same direction (from the N-terminus to the C-terminus), is known as the parallel beta sheet (Fig 2.1), whereas when the polypeptide chain runs in the opposite direction of that of its neighbouring chain, it is known as an anti-parallel beta sheet (Fig 2.2). http://t2.gstatic.com/images?q=tbn:ANd9GcSXEJyNbzn7F6PlFREwMGrUg4oz5Ysk1Fho12R9GMWzGFSIQjfK9M9bVZ80 Fig 2.1: shows the parallel beta sheets, the dotted line represents hydrogen bonds. The polypeptide chains shown are placed side by side but run in the same direction so are parallel to each other. Fig 2.2: shows the anti-parallel beta sheets, the dotted line represents hydrogen bonds. The polypeptide chains shown are placed side by side but run in the opposite direction so are anti-parallel to each other. The beta sheet are stable structures that produces a very rigid, pleated structure; this is due to the beta sheet being stabilized by hydrogen bond being formed between the amino group on one polypeptide chain and the carboxyl group on the adjacent chain. The tertiary structure of a protein is the full three dimensional structure of the arrangements of atoms found within the polypeptide chain, this structure is the final geometric shape that protein assume and would be the highest level structure that a protein can attain, the structures include the alpha helix, beta sheets, random coils and also other structures such as loops and folds, which are formed between the N-terminus and the C-terminus. The tertiary structure is mainly stabilized by the formation of disulphide bonds, this is also known as a disulphide bridge because these bonds are formed by oxidation reaction of the side chains of cysteine, by oxidizing the two thiol groups (SH) which would form a disulphide bond (S-S) (fig 3). http://www.elmhurst.edu/~chm/vchembook/images/563cysdisulfide.gif Fig 3: Shows the equation of an oxidation reaction in the tertiary structure to form a disulphide bridge (S-S), where a molecule of water is formed. The quaternary structure of a protein is the arrangements of many different types of coiled and folded polypeptides to form a unique functional protein and is stabilized by several non-covalent bonding, where some of these types of bonding are also found in tertiary structures, for example; hydrogen bonding, Van Der Waals interactions, hydrophobic interactions and also ionic interactions. These can occur if there is more than one polypeptide chain present in a complex protein.

Luna Pen Case Writeup

Report on Luna Pen case PA major 0420830 Lee Joo-Hyun Q1. If you were in Erika’s position, what steps might give you the best chance of achieving your goals? First of all, I think it’s rather dangerous to prepare negotiation plan solely based on stereotype of specific race/nationality. But still it’s foolish to ignore cultural difference. Since negotiation counterpart Feng is not from mainland China but from Taiwan, I believe there would be little or no disadvantage of Erika being female.But in case DGG is planning a bigger picture (for example, sharing distribution system of Global Service in Asia region) than just resolving trademark right issue, I think it is better to consider the fact Taiwanese regard Guan-Xi (relationship) as important aspect in terms of long term, friendly relationship. Therefore it would probably be a good idea to prepare a male represent with similar position to Erika along with her. Secondly Erika needs to gather more information about the counterpart; Feng and Global Service.After all, negotiation process is to find compromise or creative option between my constraint conditions/needs and their constraint conditions/needs. I’ll discuss about it more in following Question #2. Thirdly, Erika needs to build up negotiation plan. She need to determine what DGG intend to/should earn and how far DGG can back down. After defining goals & constraints (or setting Maginot Line), assuming the worst case scenario is also important step.In this case, the worst case possible could be something like this; â€Å"Feng & Global Service refuses to negotiate, pointing out that DGG abandoned and stop making product under the name of Luna. Thus DGG is left with no other choice but to file an uncertain lawsuit which would be both time consuming and costly. † To prevent this from happening, it would be unwise to start the first contact with firm stand like writing Feng that his company must cease its unauthorized use of he L una name, and that DGG is prepared to file lawsuit if necessary. Instead, I think it’s better to mention that DGG is aware of the fact that Luna pen is selling well under the marketing and distribution power of Feng and Global Service and DGG is more than willing to negotiate about the usage of Luna name. Q2. What further information might you need before contacting Feng; and how might you realistically obtain it? First of all, DGG need to figure out brand value (or brand power) of Luna by doing brief market research.By doing so, DGG can tell how much the brand power affected the total sale of Feng’s Luna pen and use it as a sharable standard in negotiation. I believe this data can be obtained rather simply. All DGG have to do is to run series of survey to group of customers, asking their willingness to pay for given pen with/without brand name Luna. Secondly, it would be a good idea to find potential buyers that are willing to buy the trademark Luna. The information o f potential buyers could later be used as a mean to pressure Feng and Global Service if necessary.Since selling Luna brand would not harm DGG, by contacting company in pen making business, DGG can obtain information of potential buyers while spreading news that DGG is trying to sell Luna name as well. And as a last resort, DGG should investigate the legal position of DGG in this case. (Especially in terms of Taiwanese court) If DGG files a lawsuit, what are the odd of winning the case, how long will it take, and how much is the estimated cost of lawsuit?The answers to these questions are key factors to decide whether to file a lawsuit or not. If odds are against DGG, it will be more reasonable to approach this negotiation in more cooperative, soft stand. On the other hand, if the odds are in favor of DGG, DGG can use lawsuit both as leverage to pressure Feng and last resort to settle this matter. Such data can be obtained via corporate legal department.

Determination of a Chemical Formula

CHEM 1105 Experiment 4: Determination of a Chemical Formula Introduction When atoms of one element combine with those of another, the combining ratio is typically an integer or a simple fraction. The simplest formula of a compound expresses that atom ratio. When two or more elements are present in a compound, the formula still indicates the atom ratio. To find the formula of a compound we need to find the mass of each of the elements in a weighed sample of that compound.For example, if we resolved a sample of the compound NaOH weighing 40 grams into its elements, we would find that we obtained just about 23 grams of sodium, 16 grams of oxygen, and 1 gram of hydrogen. The sample of NaOH contains equal numbers of Na, O, and H atoms. Since this is the case, the atom ratio Na:O:H is 1:1:1, and so the simplest formula is NaOH. In terms of moles, we have one mole of Na, 23 grams, one mole of O, 16 grams, and one more of H, 1 gram. From this kind of argument we can conclude that the atom ra tio in a compound is equal to the mole ratio.We get the mole ratio from chemical analysis, and from that the formula of the compound. In this experiment, we will use these principles to find the formula of the compound with the general formula CuxCly zH2O, where x, y, and z are integers which establish the formula of the compound. The compound we will study is called copper chloride hydrate. We first drive out the water, which is called hydration. This occurs if we gently heat the sample to a little over 100*C. The compound formed is anhydrous (no water) copper chloride.If we subtract its mass from that hydrate, we can determine the mass of the water that was driven off, and using the molar mass of water, find the number of moles of H2O that were in the sample. Next, we need to the find either the mass of copper of chlorine in the anhydrous sample we have prepared. (It is easier to find one mass and find the other by difference. ) We do this by dissolving the anhydrous sample in wat er, which gives us a green solution. To that solution we add some aluminum metal wire, which will react to the ions, converting them to copper metal. As the reaction proceeds, copper metal will appear on the aluminum wire with typical red-orange color. When the reaction is complete, we remove the excess Al, separate the copper from the solution, and weigh the dried metal. From its mass we can calculate the number of moles of Cu in the sample. We find the mass of Cl by subtracting the mass of Cu from that of the anhydrous copper chloride, and from that value determine the number of moles of Cl. The more ratio for Cu:Cl:H2O gives us the formula of the compound. Experimental Weigh a clean, dry crucible, without a cover, accurately on the analytical balance. Place about 1 gram of the unknown hydrated copper chloride in the crucible.Then weigh the crucible and contents on the balance. Enter results on the Data page. Place the uncovered crucible on a clay triangle supported by an iron rin g. Light your Bunsen burner away from the crucible, and adjust the burner so that you have a small flame. Gentely heat the crucible as you more the burner back and forth. Do not overheat the sample. As the sample warms, the color will change from blue-green crystals to the anhydrous brown form. After all the crystals are brown, remove the burner, cover the crucible to minimize rehydration, and let cool for 15 minutes.Finally, weigh the cool uncovered crucible and contents. Transfer the brown crystals in the crucible to a 50-mL beaker. Rinse out the crucible with two 5-mL portions of distilled water, and add rinsings to the beaker. Swirl the beaker to dissolve crystals. The color will change to blue-green as the copper ions are rehydrated. Take about 20cm of 20-guage aluminum wire (~0. 25g) and form the wire into a loose spiral coil. Immerse the coil into the solution. As the copper ions are reduced, the color of the solution will fade. The reaction will take about 30 minutes to comp lete.The solution will be colorless and most of the copper metal that was produced will be on the Al wire. Add 5 drops of 6M HCl to dissolve any insoluble Al salts and clear up the solutions. Use your glass stirring rod to remove the copper from the wire. When finished, put the wire aside. In the beaker, you now have metallic copper in a solution containing aluminum salt. Next, we will use a Buchner funnel to separate the copper from the solution. Weight accurately a dry piece of filter paper that will fit the Buchner funnel, and record its mass. Put the paper on the funnel; apply light suction as you add a few mL of water to ensure a good seal.With suction on, decant the solution into the funnel. Wash the copper metal thoroughly with distilled water, then transfer the wash and all of the copper to the funnel. Rinse the copper of the paper once more and turn off suction. Add 5-mL of 95% ethanol to the funnel. After a minute or so, turn suction back on. Draw air through the funnel fo r about 5 minutes. With your spatula, lift the filter paper from the funnel. Dry the paper and copper under a heat lamp for 5 minutes. Allow it to cool to room temperature and then weigh it accurately. Results Atomic Masses:Copper (Cu)| 63. 55| Chlorine (Cl)| 35. 45| Hydrogen (H)| 1. 008| Oxygen (O)| 16. 00| Mass of crucible| 24. 374 g| Mass of crucible and hydrated sample| 24. 881 g| Mass of hydrated sample| 0. 507 g| Mass of crucible and dehydrated sample| 24. 763 g| Mass of dehydrated sample| 0. 389 g| Mass of filter paper| 0. 260 g| Mass of filter paper and copper| 0. 430 g| Mass of copper| 0. 170 g| No. mole of copper| 0. 003 moles| Mass of water evolved| 0. 118 g| No. moles of water| 0. 007 moles| Mass of chlorine in sample (by difference)| 0. 219 g| No. moles of chlorine| 0. 006 moles|Mole ratio, chlorine: copper in sample| 2:1| Mole ratio, water: copper in hydrated sample| 2:1| Formula of dehydrated sample| CuCl2| Formula of hydrated sample| CuCl2 ‘ 2H2O| Discussion Th e significance and relevance of the experimental results is that I was able to determine the chemical formula for the unknown compound, which was copper chloride hydrate. By finding out the mole ratio, I was able to find out the chemical formula. My results were precise and accurate. My results were expected, and gave me the answer CuCl2 (dehydrated sample) and CuCl2 2H2O (hydrated sample).Conclusion The experiment went as planned. During dehydration, the color changed from blue-green to brown (anhydrous), and returned back to blue-green when water was added. When the aluminum wire was added to the solution, the copper ions were reduced to the metal, and the wire was changed to a red-orange color. From the mass calculations of the samples, I was able to find the number of moles. With mole ratio of the hydrated and dehydrated, determining the chemical formula for each was easy. The conclusion is that it is possible to find the chemical formula of an unknown compound.

Street Gangs - 1329 Words

Gangs A gang is a group of recurrently associating individuals or close friends with identifiable leadership and internal organization, identifying with or claiming control over territory in a community, and engaging either individually or collectively in violent or other forms of illegal behavior. When a new member joins the gang he or she must go through an initiation, the most common initiation is â€Å"jumping in† or getting beaten by all the gang members and/or committing acts of theft or violence. In 2011 the National Gang Intelligence Center of the Federal Bureau of Investigation asserted that There are approximately 1.4 million active street, prison, and outlaw gang members comprising more than 33,500 gangs in the United States. In†¦show more content†¦This person will commit any crime or act of violence to further the goals and objectives of the gang. This person is usually in his/her late teens or early 20’s and on up into their 30’s. The †Å"gang leaders† are the high command in the gang’s structure. This gang member is probably the oldest in the posse and likely has the smallest criminal record and they often have the power to direct the gang’s activity, whether they are involved or not. There are many different factors that push the â€Å"at risk† to join gangs. Some common factors are the desire for power, money, protection, and respect. For many of the individuals these things only seem to be obtainable by joining a gang. Many of the people that join gangs are lacking social support and get an instant feeling of belonging and identity. Gangs are involved in all levels of street crime, from extortion, to robbery, to kidnapping, to drug trafficking. Cocaine is the primary drug of distribution by gangs in America, which have used the cities Chicago, Cape Town, and Rio de Janeiro to transport drugs internationally. When a gang has taken over a particular area or â€Å"turf† the gang w ill charge businesses or people payment for â€Å"protection†, usually from the gang itself. Gang violence refers to mostly those illegal and non-political acts of violence perpetrated by gangs against innocent people, property, or other gangs. Gangs could use violence to protect their â€Å"turf†Show MoreRelatedStreet Gang And Street Gangs950 Words   |  4 PagesApril 2015 Youth and Street Gangs There is no definite term for the word â€Å"gang†. State and other local government organizations tend to create their own definition. The depiction of â€Å"street gang† is consistently intertwined with â€Å"youth gang†. However, the term â€Å"street gang† can mean two particular meanings that raise it’s face value. 1st, it proposes a common quality of gangs: They usually contain a street presence. 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