...Final Essay The chemistry 1100 class has been a very useful tool in helping me decide what I want to do with my life after college. Like most of the students in the biochemistry department, I had already firmly decided that I wanted to go to medical school after I graduate and eventually become a doctor and make money. What this class has helped me realize is that there are many more opportunities out there in the workforce that could be just as fulfilling and still be able to make enough income to live a comfortable lifestyle and help pay off student loans and provide for my family etc. Before taking this class I had a preconceived idea that research is boring and repetitive. I thought that all scientists did was stay confined in a lab all day working with test tubes and chemicals trying to find a solution problems. I was sure that I did not want to spend the rest of my life stuck in a lab trying to figure out an abstract idea. The reason why I wanted to become a doctor was because I wanted to do something that would make a difference in the lives of other people. What this class did for me was to open my eyes to a whole new opportunity to make a difference in the world and to prove that the stereotype that I had of research was wrong. I enjoyed that the biochemistry faculty took the time out of their busy lives to come and talk about their research and why they are doing what they are doing. From these speakers I was exposed to the vast opportunities there are for research...
Words: 458 - Pages: 2
...| UVA (mw/m2) | UVB (MW/m2) | | | IO | 59 | 263 | | | | | | UVA | UVB | Color | Thickness | | Average Attenuation Coefficient | Average Attenuation Coefficient | Green | 0.12 mm | 0.205 mm | 1661/m | -143. 9/m | Maroon | 0.13 mm | 0.25 mm | 1356/m | 1681.231 /m | Activity 1: Table 1:Initial Intensity In this activity, we measure first the initial intensity of UVA and UVB without the umbrella. We also measure for the thickness of two umbrellas, choosing the green and maroon colour of umbrella by getting the average of each colour. Then compute for the average attenuation coefficient of UVA and UVB of green and also for the maroon. Conclusion In this activity, we conclude that UVB has a greater intensity rather than UVA. Also the hue of the fabric of umbrellas has greater effect, the lighter the shade of the fabric the lower it protect against UV radiation. Activity 2: Experimental e/m ratio = 1.64 × 1011c/kg Accepted value of e/m ratio = 1.76 × 1011 c/kg % error = 6.18 % Table 2: Charge to mass (e/m) ratio Accelerating Potential (v) | 188 | Radius of the Helmhoitz coil (a) | 0.15m | No. of turns on each Helmhoitz coil (N) | 130 | Permeability constant (µO) | 4 π ×10-7 | Current through the Helmhoitz coils (I) | 1.66 | Radius of the electron beam path (r) | 0.037 | In this activity, we determine the charge to mass (e/m) ratio of an electron using the e/m apparatus. All units were given except for the radius of the electron...
Words: 439 - Pages: 2
...Biochemistry Task 2 Paul A. Lebeck 000490213 January 26, 2016 A. B. (Borges, 2014, Wolfe, 2015). C. (Wolfe, 2015). D. (Wolfe, 2015). E. The forces, bonds, and interactions by protein structures at the Tertiary level. There are Hydrophobic (nonpolar), Ionic bonds, Hydrogen (covalent) bonds, and Disulfide bonds, also called Disulfide Bridges. Hydrophobic are nonpolar bonds, meaning they cannot interact with water or aqueous solutions. Hydrophobic interactions will cause the protein to change shape to avoid making contact with such solutions. Considered weak bonds, but the proteins cluster tightly together on the interior of the protein, Van der Waals interaction take place between the proteins, again these are the weakest of the molecular bonds. Ionic Bonds are by definition bonds that are made up of charged particles. There are 20 Amino Acids, some with negatively charged terminals, some with positively charged terminals. This is a basic chemistry property that opposites attract. These are considered stronger bonds, but not the strongest. Next are hydrogen bonds. Considered stronger bonds than hydrophobic bonds, but weak compared to ionic and disulfide bonds. Hydrogen bonds are formed from Polar Covalent interactions. Two amino acids share a hydrogen electron and connect on the second amino acid oxygen atom. There must be a hydrogen donor on one amino acid, and a hydrogen acceptor on a second amino acid to complete the bond. The strongest...
Words: 776 - Pages: 4
...Study Guide: Biochemistry A. Hydrophilic vs Hydrophobic. Since biological chemistry occurs largely in an aqueous environment, the interaction of a biological molecule with water is very important. That interaction is influenced by two primary causes: size and polarity (charge). The smaller a molecule is, the more likely it is to be willing to associate with water (dissolve). Also, the more polar and/or charged a molecule is, the more likely it is to be willing to associate with water. Since biological molecules are often very large, it is common for the different parts of the molecule to interact differently in water. For instance, a protein, which is composed of many different amino acids which have a large variety of characters, may be hydrophobic in part of its sequence and hydrophilic in other parts. Hydrophilic (hydro=water; philios=love): Hydrophilic molecules or parts of molecules will dissolve in (interact with) water. Hydrophobic (hydro=water; phobio=fear): Hydrophobic molecules or parts of molecules will refuse to interact with water. If sufficiently hydrophobic, a molecule or part of a molecule will actively repel or exclude water. Hydrophilic/phobic characters are not an all-or-none phenomenon. Molecules fall along a scale, somewhere between extremely hydrophobic and extremely hydrophilic. Changing the parts of a molecule will often shift it more toward the hydrophobic or the hydrophilic end of the scale (depending upon the change)...
Words: 2812 - Pages: 12
...The Brad assay * is one of the most common forms of measuring absorbance through a spectrometer * uses a a dye (coomassie Brilliant Blue G-250) that binds to proteins and is absorbed at a wavelength of 465- 598 nm * Fast, therefore can meaure a 96 plate in 5 minutes * Absorbance shift allows the measure the protein at the 598 nm wavelength by using the spectrometer, can be justified through the beer-lambert law * The standard curve can be used to estimate the concentration of an unknown protein in solution by * Taking in account their known absorbance levels * Use line equation from curve * The unknown protein conc. in a sample with a known absorbance level can be determined Bradford Vs. Lowry * Lowry * Uses alkaline conditions- reduction cu 2+ to Cu+ * Folin phenol reagent- yellow to blue * Peptide bonds for monocovalent copper and radicals of try and trp react with folin * Concentrations 0.1-1.0 ug/ml of protein * Can be interfered with easily * Slow 40-60 * Colour varies with proteins * Criticial timing with procedure * * Bradford * Based on absorbance shift from brilliant blue when bound to proteins * 465- 598 (when bound) * Acidic conditions * Binds tightly to protein, and inhibits binding sites (little inference) * Limitations * Strong basic buffers Detergents interfere with binding of the cosmic blue, causing...
Words: 315 - Pages: 2
...BIOCHEMISTRY GRT1 Task 3 Breanna Jordan Alpha Beta Beta Alpha Oxygen Iron Atoms Heme Groups Oxygenated Hemoglobin * Formed via transportation of O2 to cells in tissue * O2 adheres to heme protein in Hgb * T (taut state) R (relaxed state) makes binding easier or releases De-oxygenated Hemoglobin * is not bound to oxygen molecule * Higher absorption * Blue-ish in color Bohr’s Effect CO2 produced through citric acid cycle, Hg carries oxygen from lungs to body's tissues. Hg releases oxygen for CO2 and affects pH levels. Normal pH range is 7.2 - 7.4 ↓ pH causes ↑ in the amount of oxygen being released in hemoglobin. An elevated pH will cause the oxygen to bind the hemoglobin proteins in RBCs. ("Hemoglobin," n.d.) RBCs are round cells that have concaved centers. They are flexible making it easy to move through blood vessels. Sickle cell RBC's are developed from mutations in DNA - mRNA transcriptions. They are crescent shape and become fibrous. This causes them to stick to one another. Once they begin to stick, the deoxygenated cells are unable to travel to the lungs to receive oxygen. As the CO2 builds up this causes lack of oxygen to the tissues causing pain. Sickle Cell cells also cause anemia due to the fact cells die faster than normal RBCs. Sickle cell disease is an inherited autosomal recessive pattern disorder. For a child to inherit the disease both parents must have the trait...
Words: 357 - Pages: 2
...Nancy Bosch Task 2 Student #000514178 Task 2: Protein Structure A. ("Amino Acids") B. (Rafael) (Wolfe) (Rafael) C. (Wolfe) D. (Wolfe) E. Explain the four forces (i.e., bonds or interactions) that stabilize a protein’s structure at the tertiary level. Hydrophobic interactions- R group in the amino acid is non-polar and therefore will avoid contact with water by joining together in the interior of the molecule, it will avoid contact with water. Van der Waals- this interaction occurs when the hydrophobic R groups that are packed together have a weak attraction that helps to reinforce the hydrophobic bond. Ionic bonding- The R group must have a charge. In this case the opposites attract (positive to negative/ negative to positive) forming a very strong bond.Disulfide bridges- These bonds are formed with the help of cysteine. One of the sulfur atoms from cysteine forms a single covalent bond with a second sulfur atom from another cycsteine located in the protein chain. (Wolfe) F. Explain how bovine spongiform encephalopathy (BSE) occurs at a molecular level by doing the following: 1. Explain the role of prions in BSE, including each of the following: ● how prions are formed A prion is a protein that causes infection. They are formed when a mutation occurs. A normal prion protein called PrPc is found in our neurons in our brains. This normal protein gets altered and becomes a PrPsc protein which then binds to another normal PrPc...
Words: 785 - Pages: 4
...DNA Mediated Charge Transport in p53 CHM 4390-U02 Rohit Vinod April 22, 2015 INTRODUCTION Human transcription factor p53 is considered to be an important and useful protein that plays a key role in cancer and tumor suppression. Many of the pathways in which p53 is utilized usually are apoptosis, cell deterioration also known as senescence, cell cycle arrest or DNA repair (Riley et. al, 2008). P53 also plays a role in glucose metabolism as a regulator of glycolytic and oxidative phosphorylation through positive and negative regulation of certain genes and the proteins that they encode for (Madan et. al, 2011). Usually if there are mutations in the genes that eventually translate into p53 then there are significant chance that tumors start to develop as seen in more than half of human cancers. In addition to such necessary roles, p53 also can sense oxidative stress and binds to redox-active DNA which often receives oxidative damage due to election migration through the DNA base stack. Many other sources of DNA oxidation can arise as a result of ionizing radiation, exogenous chemicals and even metabolic side products (Generaux et. al, 2010). DNA can be utilized in charge transport over distances of 100 base pairs or 34 nm in length (Augustyn et. al, 2013). The significance of the DNA oxidation and p53’s ability to sense oxidative stress is essential in understanding how DNA can be utilized in charge transport. In particular, redox activation in p53 occurs using DNA CT...
Words: 1302 - Pages: 6
...Western Governors University Biochemistry 10/16/2015 Task 1 Diagram 1: Showing the beginning of the DNA replication process Diagram 2: Showing the leading and lagging strand and the direction they replicate. Also showing Okazaki Fragments, and the 3” to 5” end of DNA. The arrows in the above diagram show the 3” to 5” direction that the DNA replicates; starting in the origin of replication and moving towards the 5 direction, using Okazaki fragments in the Lagging strand. Diagram 3: Showing a more detailed version of DNA replication. Showing the enzymes that assist in transcription. Diagram 4: Showing how mRNA assists in the transcription and translation process. Diagram 5: Showing the roles of tRNA and ribosomes during translation. Diagram 6: Showing how transcription takes place inside the nucleus and the translation takes place in the cytoplasm of the cell. Diagram 7: A detailed diagram showing protein translation. Showing how the DNA coding strand is matched by the mRNA. RNA is a multifaceted molecule. “RNA is an intermediary, carrying genetic information from the DNA to the machinery of protein synthesis” (Goodsell, 2003). The enzyme RNA polymerase creates different RNA molecules. RNA polymerase is made up of several proteins and they work together to surround DNA strands, unravel them and build an RNA strand based on the information inside the DNA. RNA polymerase needs to be accurate in its copying of genetic information and its production...
Words: 662 - Pages: 3
...Task 5: Lipids A. Fatty acids come from various sources within the body. The three primary fat sources are adipose tissue (which is where fat is stored), the liver from glycolysis and fats from the food we eat. Fats are stored in the form of lipids. This covers a multitude of compounds also including hormones and other substances that tend to be hydrophobic. Triglycerides are a type of lipid composed of three bonds between a glycerol and three fatty acids (these can be saturated or unsaturated). When triglycerides are broken down they are first simplified to the singular glycerol and three independent fatty acids. The fatty acids then enter beta-oxidation where they are further broken down into acetyl-CoA which consists of two carbon and one oxygen molecule, amongst other molecules. Acetyl-CoA then enters the citric acid cycle where it generates ATP. ATP production happens during electron transport phosphorylation where NADH and FADH2 from TCA pump hydrogen protons to the intermembrane space then they follow the proton gradient back into the mitochondrial matrix through ATP synthase generating ATP from ADP B. Two key differences between saturated and unsaturated fatty acids are the chemical structure as well as shape. All fatty acids maintain four bonds to all carbon atoms. Saturated fats are “saturated” with hydrogen atoms, these tend to come from animal sources but also include cheese and the occasional plant source such as coconut. Each carbon atom has a hydrogen or...
Words: 677 - Pages: 3
...Biochemistry Task 4 GRT1 208.5.4-01, 03-05, 5.5-02, 04-07 Western Governors University Biochemistry Task 4 GRT1 208.5.4-01, 03-05, 5.5-02, 04-07 A. Case 1: Hereditary Fructose Intolerance A1. Role of Enzymes in Processes Enzymes are proteins that carry out chemical reactions. They bind to substrates, which are basically substances that need to be broken down and changed into something else. When the enzyme and substrate bind, they form the enzyme-substrate complex. An enzyme will act in a specific way on the substrate that it is bound to in order to change it into a product, and at the end of the process, the enzyme is unchanged and ready to bind to the next substrate. An enzyme acts as a catalyst, something that lowers the energy required to complete a chemical reaction (activation energy) without itself being changed. (Hudon-Miller, 2012) In the case of fructose breakdown, an enzyme called fructokinase is responsible for splitting fructose into fructose 1-phosphate, a six-carbon fructose. Another enzyme called aldolase B splits fructose 1-phosphate into two three-carbon molecules, dihydroxyacetone phosphate (DHAP) and glyceraldehyde. These products are then able to enter the glycolysis pathway to be converted to pyruvate, which is essential for the citric acid cycle and the production of adenosine triphosphate (ATP) for cellular energy. A2. Deficiency in Aldolase B A hereditary deficiency in aldolase B could be caused by mutations in the ALDOB gene. An aldolase...
Words: 1387 - Pages: 6
...Cohort #2 A. Insert your original diagram, or series of diagrams, with clear labels, that demonstrates the process of DNA replication at the biochemical level. Click here to learn how to insert images into a Google Document. Check to see that you described the function of the enzymes (enzyme names end in “-ase”) as part of the labels for the diagram(s) and that you labeled the following: • DNA • replication fork • helicase- the helicase enzyme breaks up hydrogen bonds, which allows the DNA strand to unravel. • single-stranded binding proteins • leading strand • lagging strand • 5’ and 3’ ends of all DNA • primase- an enzyme that will create RNA Primer, creating a new strand able to bond to OH group. • RNA primer • DNA polymerase III- connects to one side of the exposed DNA chain reading from the 3’ to 5’ end. DNA polymarase III can only add from the 5’ to 3’ end of the newly developed strand of DNA. • Okazaki fragments (Wolfe, G, 2000). B. Insert your original diagram, or series of diagrams, with clear labels, that show the role of the ligase enzyme in the replication of DNA. Check to see that you described the function of the enzymes (enzyme names end in “-ase”) as part of the labels for the diagram(s) and that you labeled the following: • DNA • Okazaki fragments • DNA polymerase I- removes the...
Words: 692 - Pages: 3
...Biochemistry Task 5 Julie Rhonemus 000484390 July 29, 2015 Fat and Energy Production Fat molecules, known as triglycerides, are stored in adipose tissue in the body. Triglycerides are made up of glycerol and three fatty acid molecules. During beta oxidation the triglycerides are broken down into small units consisting of two carbon atoms and they are generated into acetyl CoA. NADH and FADH2 are created during beta oxidation and funneled into the electron transport chain. The acetyl CoA enters the citric acid cycle where NADH and FADH2 are also made and transported to the electron transport chain. The NADH and FADH2 from beta oxidation and the citric acid cycle are used to make ATP which is used for energy in the body’s cells. (O’Malley, 2014) Comparison of Fatty Acid Structures There are two types of fatty acids, saturated fatty acids and unsaturated fatty acids. Saturated fatty acids have a long carbon-hydrogen tail that forms a zigzag configuration. This zigzag configuration helps the saturated fatty acids fit together and that is why they are solid at room temperatures. Most saturated fatty acids come from animals, like butter and lard, but coconut oil is also a saturated fatty acid that is plant based. Unlike saturated fatty acids, unsaturated fatty acids do not have the long carbon-hydrogen tail. Instead the unsaturated fatty acid has a double bonded carbon somewhere along the chain that causes the chain...
Words: 762 - Pages: 4
...Biochemistry Task 2 Ashlyn Verrecchio 000502421 October 13, 2015 Requirements: A. B. C/D. E. F. Explain how bovine spongiform encephalopathy (BSE) occurs at a molecular level by doing the following: 1. Explain the role of prions in BSE, including each of the following: ● how prions are formed – Prions are an abnormal form of a normally harmless protein that is found in the brain and responsible for a variety of fatal neurodegenerative disease. ● the connection between misfolding and aggregation – When a protein begins misfolding it can lead to the protein aggregating, or better known as accumulating and clumping together, which can often be toxic. ● how prions lead to the disease – Prions can be found in the food in which cows eat (specifically sheep brain), when ingested, the harmful prion eventually can cause other normal proteins to begin misfolding and aggregating and than turn into harmful prions; this can take a long period of time, which leads to the build-up of these prions which eventually can lead to neurodegeneration. 2. Explain one possible role of a chaperone protein in BSE, including each of the following: ● how chaperones normally act in the cell – they assist proteins to correctly fold/assemble inside the cell ● how a chaperone protein can contribute to BSE – a defect in molecular chaperone interactions may lead to further progression of the disease instead of correction of the misfolding 3. Recommend two ways...
Words: 453 - Pages: 2
...Biochemistry Task 5 9/21/2015 A. Lipids in the form of triglycerides are broken down to produce ATP. Before oxidation, lipids are broken down into glycerol and fatty acids. The fatty acids then undergo beta oxidation. The fatty acid is activated by Coenzyme A which leads to the breakdown of the fatty acid into 2-carbon fragments called Acetyl-CoA. Acetyl-CoA moves on to the citric acid cycle. Electrons and hydrogens are removed from NADH and FADH-2 in the citric acid cycle and are carried to the electron transport system. NADH and FADH-2 also remove the electrons and hydrogens from the fatty acids and send them to the electron transport chain to help form ATP. After the hydrogens and electrons are carried to the electron transport system by NADH and FADH-2, they are used to make ATP from ADP and inorganic phosphate. The hydrogens in the electron transport system also combine with oxygen and form water. (O’mailley, 2014) B. Saturated fatty acids consists of single bonds and are therefore “saturated” with hydrogen. Within the chain, each carbon atom is bonded to two hydrogen atoms. Because of their regular structure, saturated fatty acids can stack easily. Since they can stack easily, they are solid at room temperature. Unsaturated fatty acids contain at least one double bond between carbon atoms. The double bond in their structure causes it to bend which makes it hard for unsaturated fatty acids to stack. For this reason, unsaturated fatty acids are liquid at...
Words: 696 - Pages: 3
