How are devices called that were on the market before FDA started reg.pdf

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How are devices called that were on the market before FDA started regulating medical devices in 1976? Predicate devices Pre-amendment devices Pre-approved devices None of the above What elements are part of the usual process of establishing safety and effectiveness for a PMA device? Human testing Bench testing Animal testing All of the above Solution Please find the answers below: Part 3: Choice b (Before the guidelines of FDA on 28th May 1976, the devices used to pateints were called as pre-amendment devices. A pre-amendment device is analysed by FDA for its safety, distribution and usability by patients by various means and at various levels of quarantine such as claims, technical, administrative and literature) Part 4: Choice d (FDA checks all the products for their safety of utilization by pateints. This includes all, the laboratory testing or bench testing, tests conducted in vitro and in vivo using cellular and animal models along with any data from human clinical trial data. This ensures the safety of utilization of any device by the patients.).

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Explain how substances get across membranes and into the cell. Solution The plasma membrane surrounding animal cells is where the exchange of substances inside and outside of cells takes place. Some substances need to move from the extracellular fluid outside cells to the inside of the cell, and some substances need to move from the inside of the cell to the extracellular fluid. the plasma membrane is choosy about what substances can pass through it, it is said to be selectively permeable. Selectively permeable or semipermeable means that only certain substances are able to pass through the membrane. Transporting substances across the plasma membrane can require that the cell use some of its energy. If energy is used, the transport is called active. If molecules can pass through the plasma membrane without using energy, the molecules are using passive transport. ACTIVE TRANSPORT Sometimes, the molecules are just too big to easily flow across the plasma membranes or dissolve in the water so that they can be filtered through the membrane. In these cases, the cells must put out a little energy to help get molecules in or out of the cell. Osmosis osmosis, the concentration of substances in the water is taken into consideration. If a solution is isotonic, that means the concentrations of the substances (solutes) and water (solvent) on both sides of the membrane are equal. If one solution is hypotonic, there is a lower concentration of substances (and more water) in it when compared to another solution. If a solution is hypertonic, there is a higher concentration of substances in it (and less water) when compared to another solution. For example, the blood in your body contains a certain amount of salt. The normal concentration is isotonic. If suddenly there is too high a concentration of salt, the blood becomes hypertonic (too many salt molecules). This excess of salt forces water out of the blood cells in an attempt to even things out. But the effect this action has is actually that of shrinking the blood cells.

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Does yeast have a post translation pathway for protein synthesis?If so how does it work? Please start explaining from nucleus transcription to translation in yeast with the promoters or genes that are specific to yeast. Solution Yes, yeast has post translational pathway for protein synthesis. The central dogma of molecular biology proposes that information flows from DNA to RNA through the process of transcription and from RNA to protein through translation. In eukaryotic cells like yeast cells, they posses phospholipid membranes allowing diverse biochemical and metabolic activities to be carried out simultaneously in distinct subcellular compartments.The membrane-bound organelles creates a requirement for the efficient targeting of proteins, which are primarily synthesised by ribosomes located in the cytosol, to the appropriate intracellular location.These proteins are first targeted to and translocated across the membrane of the endoplasmic reticulum (ER). The tail-anchor (TA) proteins are located at the C-terminus of the polypeptide.Therefore, by the time that the ER targeting signal is exposed to cytosolic factors the polypeptide will be released from the ribosome and TA proteins are post- translationally inserted into the ER membrane.Tail-anchored proteins are inserted by TRC40/Get3. A promoter is a DNA sequence that can recruit transcriptional machinery and lead to transcription of the downstream DNA sequence. All the yeast promoters depend on the availability of RNA polymerase holoenzyme, but is not affected by any transcriptional regulators..

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Determine whether the following statements about human ABO Wood groups are true or false._____ A child who is type o may have two type A parents _____A child who is type 0 may have one type AB parent _____A child who is type B may have two type AB parents _____A child who is type AB may have one type 0 parent Mary, who has blood type A, is not sure whether her child was fathered by John or Tom. The child has type B blood John has blood type O, and Tom has blood type AB. Can we determine which of these two men is more likely to be the father? If so, slow or explain how. Suppose Mary pursues further genetic testing, and discovers that her own genotype Is. If this is correct, why does it suggest that another mistake was made at some point? Use a Punnett square to support your answer. Solution 1. a) true b) false c) true d) false 2. Tom is likely to be the father. since, Mary has blood type A- genotype is AO or AA JOhn has blood type O- genotype is O O Tom has blood type AB- genotype is AB thus it is not possible for AO OR AA crossed with OO to have a child with blood type B while when crossed with AB it is possible to have a child with genotype BO which will be blood type B. 3. here the results show that mary is homozygous for the allele A thus her genotype would be AA thus it is likely that mary is not the mother of this child with blood group B. Mary\'s genotype cossed with tom would provide these results which does not include IBIO or IBIB which shloud be the genotype of the child.IAIA/IAIBIAIBIAIAIAIAIBIAIAIAIAIB.

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Describe a similarity transformation that maps trapezoid WXYZ to tr.pdf

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blackboard.ecu edu Take Test: Exam ll-Blou2300601201730 he map above shows the different DF mutants algned, with the boxes 1-g) showing tha locations of the dit QUESTION 2 s points Extra Credito save Answer During the course of experiments. Seymour Benzer discovered over 20.000 mutations in the T4 classified the mutations into two categories: 1) revertible mutants (which result from bacteriophage. He from mutations). point mutations), and nonrevertible mutants (which result Four riB partial deletion mutations were tested for their ability to complement five rlB point mutants. Use the data in this to place the point mutants in order on the rlig deletion gene maps below. (Note: A plus sign indicates complementation and a minus sign indicates no complementation.) Point Deletion Deletion Deletion Deletion utant Mutant Mutant Mutant Mutant D1 D2 D3 D1. D2 D3 D4. The map above shows the different DF mutants aligned, with the boxes (e1-9) showing the locations of the different mutations A thru E Enter the each DF mutant. This is an alignment, so the letters will be used more than once. On the correct letter in the n umbered blanks provided unde hown for DF1, point mutation A is shown n its correct location to get you started. Solution 1)D 2)B 3)A 4)C 5)B 6)A,C 7)E 8)D 9)C as these are the boxes which represent the mutants shown by the graphs in a given order..

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Bitcoin as an Ethical Dilemma closing case (Question below article) Bitcoin is an open-source, peer-to-peer digital currency introduced to the world on January 3, 2009, by developer Satoshi Nakamoto. The cryptocurrency is based on a protocol and software that allows instant peer-to-peer transactions and worldwide payments with minimal costs. In its few years of existence, bitcoin has seen unprecedented media coverage, a rollercoaster ride of epic spikes and epic plunges, and adopters from major retailers to lemon stands (e.g., Amazon, Target, Victoria\'s Secret, and Whole Foods). Bitcoin has also been covered by numerous major news organizations (e.g., ABC, CNBC, Forbes, Fox News, Reuters) as the most popular form of virtual currency. At the same time, ethical concerns exist with this new digital currency. The coupling of no regulations, virtually free movement of value, and a Ponzi scheme–like system have led renowned economist Paul Krugman to suggest that “bitcoin is evil.” At the basic level, Krugman says that “to be successful, money must be both a medium of exchange and a reasonably stable store of value.” He continues to say that “it remains completely unclear why bitcoin should be a stable store of value.” Joining in the discussion, Charlie Stross, the British writer of science fiction, says that “bitcoin looks like it was designed as a weapon intended to damage central banking and money issuing banks, with a Libertarian political agenda in mind—to damage states\' ability to collect tax and monitor their citizens\' financial transactions.” What is the difference between bitcoin and normal currency, such as the U.S. dollar? Bitcoin is an unregulated peer-to-peer digital currency that is not backed by any other commodity such as gold or silver. Bitcoins exist almost entirely in the digital, online world, although some bitcoins have actually been privately minted. The U.S. dollar, like many other stable currencies, are paper or coin currency issued by a national reserve–type bank (in the United States, it is the Federal Reserve Bank). This means that dollars are really Federal Reserve Notes that are printed or minted at the U.S. Bureau of Engraving and Printing. The dollar is so-called fiat money, which means that dollars derive their value from the U.S. government regulation or law. Interestingly, the United States decided in 2014 that bitcoins will be taxed as property, not currency, for International Revenue Services (IRS) purposes. The IRS defined bitcoin as a “convertible currency that can be used as a medium of exchange, a unit of account, and/or a store of value.” Technically, Bitcoin with a capitalized “B” refers to the technology and network associated with the currency, while bitcoin with a lower case “b” refers to the actual currency. The philosophy underlying the bitcoin is complete mistrust in authority or control—basically a perfectly stateless, market-based approach, with no country or region-level bank intervention. It is .

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2. This season\'s flu virus is spreading based on a logistic function. For a town of 50,000 people, they initially had 500 people infected. One month later, 1000 people were infected. a) Write the logistic function that models the spread of the flu. (6 pts) b) At what point will the spread of the virus begin to slow down (2 pts) c) How many people will be affected when the spread begins to slow down (2 pts) Solution.

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^^^ Q2. Discuss about Header Node? And also write a program for unordered single linked list and linked implementation of QUEUE? Solution Header node: Sometimes it is desire to keep an extra node at the front of a List. Such a node doesn’t represent an item in the List and is called a Header Node or a List Header. The Information field of such Header node might be unused but the Next field maintains the first node address. More often the information of such a node could be used to keep Global Information about the Entire List. One of the applications of Header node is the information portion of the Header node contains the number of nodes (not including Header) in the List. In this Structure the Header must be adjusted the total number of nodes when we apply addition or deletion operation on the List. We can directly obtained the total number of nodes without traversing the entire List. By using this application, we can develop Data Structure Stack easily. Since Header node information field maintains the number of elements that the Stack contains. Another application is we can simply develop the Data Structure Queue. Until now, two external pointers, Front & Rear, were necessary for a List to represent a Queue. However, now only a single external pointer is sufficient to maintain a Queue that is Head node. Because the Header node next field behave like a Front and information field of Header node maintains last node address so that it is behave like a Rear . Another possibility for the use of the information portion of a List Header is as a pointer to a Curr node in the List during a traversal process. This would be eliminate the node for an external pointer during traversal. #include #include #include #include struct node { char info; struct node *next; }; typedef struct node sl; void display(sl *); sl * create(sl *root) { sl *curr,*new; char val; printf(\"\ Enter $ to STOP --- Otherwise Continue : \"); fflush(stdin); scanf(\"%c\",&val); while(val != \'$\') { new=(sl *) malloc(sizeof(sl)); new->info = val; new->next = NULL; if(root == NULL) root = new; else { curr = root; while( curr->next != NULL ) curr = curr->next; curr->next = new; } printf(\" Enter $ to STOP --- Otherwise Continue : \"); fflush(stdin); scanf(\"%c\",&val); } printf(\"\ The Created Linked List Is \ \"); display(root); return root; } sl * delete(sl *root,char val) { sl *temp,*curr,*rear; curr = root; rear = NULL; while( curr != NULL && val != curr->info ) { rear = curr; curr = curr->next; } if(curr == NULL) { if( rear == NULL) { printf(\"\ Linked List is Empty\"); printf(\"\ Deletion is not Possible \"); } else { printf(\"\ deleted Node does not exist in List\"); printf(\"\ Deletion is not Possible \"); } } else { if( rear == NULL) { temp = root; root = root->next; } else { temp = curr; rear->next = curr->next; } printf(\"\ Node is deleted \"); printf(\"\ Deleted Node information is %c\",temp->info); free(temp); } display(root); return root; } sl * insert(sl *root,char va.

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Write the code segments to create a monitor that has the same behavior as a semaphore. Monitor semaphore { //initialization // synchronization condition variable // methods wait ( ) signal ( ) } Solution I assume the code is to be written in Java as no language is specified. The solution is below: public class Monitor_Semaphore { private int count; //initialize public Semaphore (int init) { count = init; } //signal synchronized public void S() { count++; notify(); } //wait synchronized public void W() throws InterruptedException { while (count == 0) wait(); count--; } public void Method1(){ W(); //critical section S(); } }.

Write the code segments to create a monitor that has the same behavi.pdf

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Which of the following is NOT a rationale for using mRNA intermediates in gene expression? to preserve its original, pristine copy of the gene DNA lacks the necessary information for protein production. to amplify the response to its changing needs to connect the information (DNA) with protein-producing organelles (ribosomes) Solution Answer b) DNA lacks the necessary information for protein production. Reason - This is not true because DNA contains all the necessary information required to code all the proteins. the same information is being transcribed on mRNA molecule during transcription process and the same information is translated into an amino acid sequence from mRNA into protein. in fact, DNA contains all the necessary information for protein production..

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Which elements are in series and which are in parallel in Figure P1.3.pdf

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What domain is horizontal gene transfer most common within? Multiple choice Horizontal gene transfer is common to all domains. Archaea Bacteria Bacteria and Eukarya Eukarya Solution Horizontal gene transfer is a mechanism by which short DNA fragments are transferred between organisms by direct uptake from environment (not via reproduction). It is most common in prokaryotes, although it occurs in other domains also. Horizontal gene transfer is a mechanism by which short DNA fragments are transferred between organisms by direct uptake from environment (not via reproduction). It is most common in prokaryotes, although it occurs in other domains also.Ans: c- bacteria.

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This problem is based on the \"B92\" method of quantum key distribution. See Physics World. May 1999. Photons are directed at a polaroid sheet held by Alice. The Transmission Axis (TA) of Alice\'s sheet is either vertical (V) or at 45degree. When Alice wants to send a \"1\" bit she uses the 45degree orientation and when she wants to send a \"0\" bit she uses the V orientation. Some distance away Bob has two polaroid sheets. His polaroid TA can be either horizontal (H) or at -45degree. When Bob uses his H sheet he assigns a \"1\" bit to any photon he observes. When Bob uses his -45degree sheet he assigns a \"0\" bit to any photon he observes. For each of the two types of photons that Alice can send make a table showing what Bob sees with each of his two polaroid orientations. For each case give the probability that Bob sees a photon. Bob and Alice keep the bits for which Bob saw a photon and they use these as the key. In this method they never discuss the polaroid orientations but just note the cases where Bob observed a photon. Show that for the kept subset of bits Bob and Alice agree on the bit values. Alice and Bob choose their polaroid sheets at random. What fraction of bits sent by Alice will be seen by Bob and therefore used? What happens if an evesdropper Eve, places a polaroid sheet (for example in the H direction) between Alice and Bob. What fraction of photons will Alice and Bob agree on now? Solution If Alice and Bob use the sheets at random, Bob would see 50% of bits sent by Alice on an average. if a polaroid with H direction is placed in between Alice and Bob, he would see a photon always. The phtons coming out of the intermdiate would be alinged along H, which can be consdiered as composed of +45 and -45. Thus Bob will always see a photon, whether he uses H sheet or -45 sheet.AliceAliceBobBobBob.Porb.V0H000-4511+451H111-4500.

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The _ operator can be used to determine a variable address The variable declared inside a structure declaration are called _. The _ operator allows us to access structure members Multiple Choices - Select the best answer and write it In the blank. Functions prototype are terminated with a semicolon. False True Many functions may have local variables with the same name. False True the execution of this program segment and write the outputs accordingly. Solution 1) The address operator can be used to determine a variable\'s address 2) The variables declared inside a structure declaration are called members 3) The dot operator allows you to access structure members. 1) Functions prototype are terminated with a semicolon True 2) Many functions may have local variables with the same name True.

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The main difference between passive and active transport is the specificity of the target molecule involved. the requirement for ATP. various gradient and osmotic pressures within the cell. the ability to maintain voltage across the membrane. diffusion versus membrane protein mediated transport. Co-transport occurs when active transport of a solute indirectly drives transport of another solute. 2 ions are transported in opposite directions across the membrane. a proton pump makes ATP. the voltage across the membrane suddenly changes. only during photosynthesis in plants. Match the following terms Phagocytosis Cell pooping. (_____) Pinocytosis Ingesting big particles. (_____) Exocytosis Cell sipping. (_____) Solution Answers : Question number 6 : The correct answer is B ( The requirement of ATP) Explanation : Passive transport is a type of transport in which ions or molecule move across a membrane down the concentration gradient. It doesnot uses any metabolic energy for this purpose. There are three types of passive transport which includes osmosis, simple diffusion and facilitated diffusion. In contrast to passive transport, active transport uses metabolic energy such as ATP to move an ion or a molecule against their concentration gradient (from a region of lower concentration to regions of higher concentration). Example of active transporter : Na+ K+ ATPase pump..

The main difference between passive and active transport is the spec.pdf

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The opposite movement of supination is ___. Pronation flexion extension opposition circumduction The lateral malleolus is part of which bone? Fibula Humerus Femur Patella Tibia The glenoid cavity is part of which bone? Humerus ulna scapule ilium clavicle Solution Ans. 1) Pronation (it is the opposite movement of supination). 2.) Fibula (Lateral malleolus is part of fibula). 3.) Humerus (Glenoid cavity is part of humerous)..

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The latter or final stages of an RNA based heriditary world were likely dominated by a. cellularized RNA heredity systems using ribosomes for the syntheis of proteins b. DNA based cells with nuclei c. RNA oligonucleotides experimenting with self replication d. RNA based cells with ribosomes but without nuclei e. cellularized RNA heredity systems experimenting with additional ribozymes for the synthesis of proteins Solution Finak stages of RNA based hereditary world were likely dominated by e. cellularized RNA heredity systems experimenting with additional ribozymes for the synthesis of proteins. The synthesis of new protreins at the final stage will lead to the inheritance of new traits..

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The following table is a summary of randomly chosen student evaluations of faculty at a university over a three-year period. The researcher is interested in whether the distribution of evaluations differs by faculty rank. Rank Evaluation Assistant Professor Associate Professor Professor Total Above Average 44 36 34 114 Below Average 31 29 51 111 Total 75 65 85 225 Which of the following correctly characterizes the p-value? a. p-value < .01 b. .01 < p-value < .025 c. .025 < p-value < .05 d. .05 < p-value < .10 e. p-value > .10 Do the above data provide evidence at the .025 level that evaluation and faculty rank are dependent? a. yes b. no c. cannot be determined d. maybe? e. Pie Rank Evaluation Assistant Professor Associate Professor Professor Total Above Average 44 36 34 114 Below Average 31 29 51 111 Total 75 65 85 225 Solution The following table is a summary of randomly chosen student evaluations of faculty at a university over a three-year period. The researcher is interested in whether the distribution of evaluations differs by faculty rank. Rank Evaluation Assistant Professor Associate Professor Professor Total Above Average 44 36 34 114 Below Average 31 29 51 111 Total 75 65 85 225 Chi-Square Test Observed Frequencies Column variable Calculations Row variable Assistant Professor Associate Professor Professor Total fo-fe Above Average 44 36 34 114 6 3.0667 -9.0667 Below Average 31 29 51 111 -6 -3.0667 9.0667 Total 75 65 85 225 Expected Frequencies Column variable Row variable Assistant Professor Associate Professor Professor Total (fo-fe)^2/fe Above Average 38 32.9333 43.0667 114 0.9474 0.2856 1.9088 Below Average 37 32.0667 41.9333 111 0.9730 0.2933 1.9604 Total 75 65 85 225 Data Level of Significance 0.025 Number of Rows 2 Number of Columns 3 Degrees of Freedom 2 Results Critical Value 7.3778 Chi-Square Test Statistic 6.3683 p-Value 0.0414 Do not Reject the null hypothesis Which of the following correctly characterizes the p-value? a. p-value < .01 b. .01 < p-value < .025 c. .025 < p-value < .05 d. .05 < p-value < .10 e. p-value > .10 Do the above data provide evidence at the .025 level that evaluation and faculty rank are dependent? a. yes b. no c. cannot be determined d. maybe? e. Pie Rank Evaluation Assistant Professor Associate Professor Professor Total Above Average 44 36 34 114 Below Average 31 29 51 111 Total 75 65 85 225.

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A female carrier of Tay-Sachs wants to have a child with a male carrier for Tay-Sachs. Tay- Sachs is inherited in an autosomal recessive manner. Using the letters \"T\" and T fill out the following table. Solution Autosomal recessive conditions are associated with non sex chromosomes and the disease allele must be present in homozygous condition to inflict the disease. Tay-Sachs is a sphingolipidoses associated with mutation in the HEXA gene situated on the chromosome 15. A defective copy of both the allele for HEXA gene will lead to Tay sachs. Lets say that the wild type HEXA gene is T and defective gene is t. genotype of Mother - Tt; phenotype of mother- carrier genotype of father-Tt; phenotype of father- carrier Cross between the two: Parents: Tt X Tt offsprings: TT; Tt; Tt; ttFemaleMaleGenotypeTT; Tt; Tt; ttTT; Tt; Tt; ttPhenotypeNormal; carrier; carrier; affectedNormal; carrier; carrier; affectedGenotypes of sperm/ovaSame as the individualsame as the individual.

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So I don\'t understand explicit equations. Whats the formula and how are they different than recursive? Solution in explicit formula we just need to know the term number to find any term of the sequence whereas in recursive formula each term is defined by its preceding term.so to find any term of the sequence we must know the preceding term lets understand by an example let a sequence be 1, 4 , 7 , 10 ... a1 = 1 , difference between 2 terms = 3 for explicit formula an = a1 + (n-1)*3 we just need to plug the value of n to get any term of the sequence for recursive formula an+1 = an + 3 to get any term of the sequence we need to know the preceding term for eaxample to find a3 we plug n = 2 a2+1 = a2 + 3 a3 = a2+ 3 tehrefore we need to know a2 term to get a3 term.

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