...The proteins are known as histones. The histones act as a scaffold organizing the structure of the DNA; a long string of the nucleotides. The histones are responsible for DNA folding and the overall formation of chromatin. Without the histones presence DNA itself would be extremely long in length, too long in length to fit inside the cell. The histones associate in octomers, 8 histones, to form beads on a string; Beads on a string describes the DNA wrapping around the octomer histone proteins. The term nucleoside describes the segments of DNA wrapped around histones. The Nucleosomes particles comprise of the complementary base pairings of DNA, which are wrapped around the histone proteins. The Nucleosomes are the fundamental repeating unit of a chromatin. It helps to package the large DNA genomes in the nucleus of a cell. It does this with the help of the histones, they in conjunction with one another, help from a tightly coiled structure. The Nucleosomes achieve this structure by compacting to form solenoid fibres. These fibres will fold or coil, to form complex loops, these are further condensed, which will later form...
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...ci3.1 BLOCK DIAGRAM Fig 3.1 block diagram of the vibration measurement system The above figure shows the block diagram of the vibration measurement system. The different blocks of the block diagram are the sensor which is used to measure the vibrations. The sensor we are using in this project is MMA8451Q accelerometer, which is able to measure the vibrations. The vibration data need to be measured from the acceleration data of the accelerometer by setting different thresholds. Then the data from the accelerometer need to be transferred to the microcontroller which is used to process the data. The microcontroller which we have selected to do the processing of the data is MKL25Z32VFM4 which is ARM® Cortex™-M0+ processor. This is best suited for the processing of the accelerometer data. There are different types of protocols that can be used to communicate with this controller. It supports SPI, I2C etc, the communication between the accelerometer and the microcontroller uses I2C protocol to transfer the data between two. For development we used MKL25Z128VLK4 controller. Even this is also ARM® Cortex™-M0+ processor. The reason for using this processor is this processor will be available on a FRDM KL25Z board from freescale and this board also has MMA8451Q accelerometer populated on the board. So till the completion of designing the PCB this board can be used for development. The processor which will be used in the final product and the processor which we are using now for development...
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...organ and tissue transplant: The Major histocompatibility complexes (MHC), think of them like trackers or identifications tags that are found on almost every cell that has a nucleus. They play an important role in activating the immune response and stopping further destruction. However, if MHC is not working properly, they can become antigenic. This means that the cell will not be able to recognize or distinguish self from non-self causing destruction of the cell. Thus, MHC plays an important role in organ and tissue transplantation success as well as can give rise to autoimmune diseases. In the case of organ and tissue transplant, it is crucial to match the MHC molecules of the donors and recipients to reduce risk of rejection of transplant. 2. Understanding the difference between humoral immunity versus cell-mediated immunity:...
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...TAQ 2 1 A voluntary muscle named the quadriceps has a structure that is divided into four separate portions. These include, the rectus femoris occupying the middle of the thigh and covering most of the other three quadriceps muscles, originating on the illium. The remaining three portions lay deep to rectus femoris and originate from the body of the femur that cover from thetrochanters to the condyles. Vastus lateralis is on the lateral side of the femur (outer side of the thigh). Vastus medialis is on the medial side of the femur (the inner part thigh). Vastus intermedius sits between vastus lateralis and vastus medialis on the front of the femur (on the top or front of the thigh), but deep to the rectus femoris. All four parts of the quadriceps muscle insert into the tibial tuberosity of thetibia. This process goes through the patella where the quadriceps tendon becomes the patellar ligament resulting in the attachment to the tibia (bbc, 2011)....
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...1. Antibodies are released in the body when antigens are present within its system and are responsible for neutralizing them. An Antigen is a foreign substance such as a virus that causes the immune system to release antibodies. An Antibody contains a set of antigen binding sites that it binds with Antigens when present. Antibodies bind with specific Antigens that were responsible for their production. Antibodies also contain light chains and heavy chains. The ends of these chains determine the specific antigens in which the antibody binds to and this part of the antibody is called the variable region. 2. The conjugation that occurs during the ELISA test is very important because it improves the specificity of the disease. Each well contains...
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...The objective of this lab is to become familiar with memory mapping for a HCS12 and address decoding. Results: What I found out after doing the lab is that I was able to determine what the CPU’s functions are and how the CPU handles the data. Also, I was able to determine the memory locations/address ranges as asked within our lab. Conclusions: I thought this lab was really awesome because I was able to observe, construct, and become familiarized with how memory mapping works, as well as address decoding. What I have implemented from this lab is that we can determine the memory size by converting hexadecimal to binary. Team: Sheree Elder ECET 330 Sheree Elder Name Program Signature Observations/Measurements: IV. 1: Memory...
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...2.151 Where pipes and cables enter buildings the entry point area is free of gaps. Observation: In the silo room the new pipe through the wall was not completely sealed. Facility External: 2.152 Facility grounds including the parking lots, are kept clean and well maintained. Observation: Pipe leading out of building allows material (suspect sugar) to backflow onto pavement on exterior of the building. Flying insects observed in this area. Decaying skunk along building perimeter. 2.153 Vegetation is kept short to discourage infestation and pest harborage. Grass is cut and weeds are kept to a minimum. 2.154 Grounds are graded to prevent standing water. Observation: Signs of erosion and past puddles within two foot perimeter of building. Mars representative on external site tour mentioned that water does pool in these areas. 2.155 Gutters and drains are clean and in good working order. Observation: Drain/screen in Clean Out of Place room was full of debri/product and showing evidence of flying insects. An open floor drain with no cover was observed in the Conche room. 2.156 A system of waste disposal is established. Trash and other refuse is disposed of in a safe and sanitary manner and emptied on a routine basis. 2.157 Building exterior is protected against rodent and insect entry, i.e. doors are tight fitting and window and door screens are in good condition. Truck seals / cushions are in place. Observation: The 4 roll-up doors in rail receiving don’t have brush seals...
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...the fractions obtained, the following equation was used: -ε = A / x l = C -A – Absorbance 0.045 ε- 0.0672 L - 1 So 0.045 / 0.0672 x 1 = 0.6696 rounded to 3sig fig = 0.670 and to 2sigfig for the 620 wavelength. The remaining concentrations of vitamin B12 fractions at wavelength 460 nm and 620nm were worked out in same way using the same equation. Converting absorbance figures The absorbance figures of vitamin B12 were multiplied by a 1000 to convert them into molarity E.g. 0.670 x 1000 = 670)mm) Fractions Absorbance at wavelength 460nm Concentration of vitamin B12 Sample = 0.0672 Vitamin B12 absorbance converted into Molarity (mM) Absorbance at wavelength 620nm Concentration of Vitamin B12 Sample = 0.0036 1 0.045 0.670 670 0.022 6.1 2 0.046 0.685 685 0.020 5.5 3 0.048 0.714 714 0.179 49.7 4 0.062 0.923 923 0.639 17.46 5 0.057 0.848 848 0.721 200 6 0.040 0.595 595 0.343 95.27 7 0.046 0.685 685 0.142 39.4 8 0.077 1.146 1146 0.052 14.4 9 0.126 1.876 1876 0.024 6.6 10 0.1910 1.880 1880 0.016 4.4 11 0.157 2.336 2336 0.010 2.7 12 0.151 2.247 2247 0.035 9.72 13 0.061 0.908 908 0.005 1.38 14 0.042 0.623 623 0.002 0.5 15 0.033 0.491 491 0.003 0.83 16 0.031 0.461 461 0.002 0.5 17 0.030 0.446 446 0.002 0.5 18 0.058 0.863 863 0.030 8.3 19 0.031 0.461 461 0.002 0.5 20 0.051 0.759 759 0.007 1.94 21 0.054 0.804 804 0.002 0.5 22 0.055 0.818 818 0.018 5 23 0.040 0.595 595 0.014 3.8 24 0.055 0.818 818 0.003 0.83 25 Q2. Wavelength 460nm ...
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...Julee Pettis Ch. 2 Reaction What did you learn? – Be sure to mention each section---What was confusing? – Only need 1 thing for the whole assignment What was most interesting? – Only need 1 thing for the whole assignment I learned that: 1. Quandaries: If the temperature of the earth increases by 5°F, the ocean will rise by one foot. (pg. 38) 2. Atoms and Molecules and the Meaning of Heat: I learned that you can see moving pieces of dust under a microscope. (pg. 40) 3. The Speed of Sound and the Speed of Light: I did not know that the speed of molecules is related to the speed of sound. (pg. 41) 4. The Enormous Energy in Heat: I learned that when an asteroid hit the Earth, the molecules were moving at 30km/s. (pg. 42) 5. Hiss and Snow: Electronic...
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...epidermidis didn’t coagulate, this is indicating a negative result. S. epidermidis doesn’t have the free coagulase enzyme; therefore, no coagulation was present. In conclusion, not only did I prove my assumption correctly, but I was also able to find out the exact name for my organism. After the gram stain procedure I was able to assume that my organism was gram positive bacteria. The next step for me was to use the agars and broth that was specifically used for gram positive species. Among all the agars and broths used for gram positive bacteria, I used mannitol salt agar, blood agar, DNA agar, and brain heart infusion broth. If you look back at tables 1.2-1.5 you’ll see how the unknown reacted to these agars. Going back to the previous lab reports I compared and observed the results. With the knowledge I’ve retained I was able to connect the dots with the results and figure out that my unknown is Staphylococcus epidermidis and is a gram positive bacteria. ...
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...Electrode 1 1. Utilizing the wire cutters, cut two pieces of copper wire to 6 inches in length. 2. Cut a 6 inch piece of plastic tube and thread the two wires through the tubing. 3. Abstract 1 inch of insulation from the wires that were threaded through the tub and wrap them around the cessation of the tube being conscientious that the wires do not physically contact each other. Secure the wires to the plastic tubing with sultry glue (figure 1). 4. Annex alligator clips to the antithesis terminus of the wires and consummate circuit as illustrated in figure 2. *electrode 1 engendered very low currents when analyzing beverages. Therefore, a second electrode was designed with a more preponderant surface area to engender higher currents. Electrode...
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...GI lab was a very interesting experience. Today, 2/2/18, I was able to observe several procedures including two EGDs (esophagogastroduodenoscopy), two colonoscopies, a bronchoscopy, and a PEG (percutaneous endoscopic gastrostomy) tube placement. For a few of my patients it was their first time undergoing a procedure like this and they seemed anxious and apprehensive. The staff in the room where kind and supportive to the patient. ‘The worst part is over’ is a phrase I heard over and over, in reference to the previous day’s bowel prep for the colonoscopy patients. Even brief conversation before the procedure seemed to calm the patient a bit. The sterile environment can be quite intimidating for patients, and I think that offering emotional support to a patient before the test is a very important part of patient care. The staff...
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...Exercise One Nt1310 Week 1 In: Computers and Technology Exercise One Nt1310 Week 1 NT1310:Week 1 Telecommunications By: Kenneth Martin You are an IT Network Specialist and are required to develop the design of the company’s telephone system for its new building which will begin construction in a few months. In order to get a background which will help when developing the new system, your supervisor asked you to research the current Plain Old Telephone Service (POTS) currently used by the company. You must research the system components and report back to your supervisor , the Telecommunications Manager, within a week. 1. Describe a local land line phone system based on the following Landline Telephone Components: a. Local Loop a local loop is the wired connection from a telephone company's central office in a locality to its customers' telephones at homes and businesses. This connection is usually on a pair of copper wires called twisted pair. The system was originally designed for voice transmission only using analog transmission technology on a single voice channel. Today, your computer's modem makes the conversion between analog signals and digital signals. With Integrated Services Digital Network (ISDN) or Digital Subscriber Line (DSL), the local loop can carry digital signals directly and at a much higher bandwidth than they do for voice only. b. Central Office In telephone communication in the United States, a central office (CO) is an office in a locality...
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...ITT Technical Institute IS3340 Windows Security Onsite Course SYLLABUS Credit hours: 4.5 Contact/Instructional hours: 60 (30 Theory Hours, 30 Lab Hours) Prerequisite(s) and/or Corequisite(s): Prerequisite: NT2580 Introduction to Information Security or equivalent Course Description: This course examines security implementations for a variety of Windows platforms and applications. Areas of study include analysis of the security architecture of Windows systems. Students will identify and examine security risks and apply tools and methods to address security issues in the Windows environment. Windows Security Syllabus Where Does This Course Belong? This course is required for the Bachelor of Science in Information Systems Security program. This program covers the following core areas: Foundational Courses Technical Courses BSISS Project The following diagram demonstrates how this course fits in the program: IS4799 NT2799 IS4670 ISC Capstone Project Capstone ProjectCybercrime Forensics NSA NT2580 NT2670 Introduction to Information Security IS4680 IS4560 NT2580 NT2670 Email and Web Services Hacking and Introduction to Security Auditing for Compliance Countermeasures Information Security Email and Web Services NT1230 NT1330 Client-Server Client-Server Networking I Networking II IS3230 IS3350 NT1230 NT1330 Issues Client-Server Client-Server SecurityContext in Legal Access Security Networking I Networking II NT1110...
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