Supplements of the water-soluble vitamins and _ are mos.pdf

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Supplements of the water-soluble vitamins ______ and _______ are most likely to cause toxicity symptoms. A.thiamin; riboflavin B.vitamin B-12; pantothenic acid C.vitamin B-6; vitamin C D.biotin; vitamin B-12 Solution Of the different water soluble vitamins, four i.e. folate, niacin, vitamin B6 and vitamin C have shown the potential for toxicity. Hence the correct answer is option C. – vitamin B6 and vitamin C..

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What are 4 different types of CVI systemsSolutionThe four typ.pdf

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Why is RNA more prone than DNA to forming secondary structures? Solution DNA has limited catalytic ability because it lacks functional groups that can participate in catalysis and has a regular structure that is not conducive to forming shapes required for catalysis.RNA molecules can catalyze some reactions because they have exposed hydroxyl functional groups and can fold into shapes that then can function in catalysis.The increased stability of RNA duplexes allows for highly diverse secondary and tertiary structures..

Why is RNA more prone than DNA to forming secondary structures.pdf

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What are the advantages and disadvantage of using the metric system of measurements? Solution Answer: Advantages: (1) Its based on decimal system instead of fractions which makes it a simpler system. (2) Has one unit for each quantity which eliminates the need of conversion. (3) It is universally accepted system. Disadvantages: (1) Difficult to make one acquainted with the metric system who is already using imperial units. (2) Difficult to use with fractions..

What are the advantages and disadvantage of using the metric system o.pdf

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Which of the following model organisms would be the best choice if you wanted to to and in vivo study of early animal cellular division? Xenopus Mouse E. coli Arabidopsis Drosophila Xenopus Mouse E. coli Arabidopsis Drosophila Solution Model organism is an organism selected for intensive scientific study based on features that make it easy to work with in the hope that findings will apply to other species . The correct answer is Arabidopsis. The aerial parts of the plants are generated by group of dividing cells called shoot apical meristem. To analyse, cell behaviour in these structures, we developed a technique to visualise living shoot apical meristems using confocal microscope. This method combined with fluorescent marker lines and vital stains, allows us to follow dynamics of cell proliferation and division..

Which of the following model organisms would be the best choice if y.pdf

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Write a C program that reads the words the user types at the command prompt (using the \'int argc, char * argv[] and store each unique letter in a Binary Search Tree. When a duplicate is encountered do not store the letter again and instead keep track of the count in the tree. Once the Binary Search tree has been created print out the tree both inorder and reverse order. Also print the highest and lowest alphabetically letter in the tree if any. Solution # include # include # include typedef struct BST { int data; struct BST *lchild, *rchild; } node; void insert(node *, node *); void preorder(node *); findMinimum(struct node* root) findMaximum(struct node* root) void reverseLevelOrder(struct node* root) void main(int argc,char argv) { char argv = \'N\'; int key; node *new_node, *root, *tmp, *parent; node *get_node(); root = NULL; clrscr(); printf(\"\ Program For Binary Search Tree \"); do { printf(\"\ 1.Create\"); printf(\"\ 2.Search\"); printf(\"\ 3.Recursive Traversals\"); printf(\"\ 4.Exit\"); printf(\"\ Enter your choice :\"); scanf(\"%d\", &argc); switch (argc) { case 1: do { new_node = get_node(); printf(\"\ Enter The Element \"); scanf(\"%d\", &new_node->data); if (root == NULL) /* Tree is not Created */ root = new_node; else insert(root, new_node); printf(\"\ Want To enter More Elements?(y/n)\"); argv= getch(); } while (argv == \'y\'); break; case 2: if (root == NULL) printf(\"Tree Is Not Created\"); else { printf(\"\ The Preorder display : \"); preorder(root); } break; } } while (argv != 4); } /* Get new Node */ node *get_node() { node *temp; temp = (node *) malloc(sizeof(node)); temp->lchild = NULL; temp->rchild = NULL; return temp; } /* This function is for creating a binary search tree */ void insert(node *root, node *new_node) { if (new_node->data < root->data) { if (root->lchild == NULL) root->lchild = new_node; return newNode(key); else insert(root->lchild, new_node); } if (new_node->data > root->data) { if (root->rchild == NULL) root->rchild = new_node; return newNode(key); else insert(root->rchild, new_node); } } if (key == node->key) { (node->count)++; return node; } /* This function displays the tree in preorder fashion */ void preorder(node *temp) { if (temp != NULL) { printf(\"%d\", temp->data); preorder(temp->lchild); preorder(temp->rchild); } } // Returns maximum value in a given Binary Tree int findMaximum(struct node* root) { // Base case if (root == NULL) return INT_MAXIMUM; // Return maximum of 3 values: // 1) Root\'s data 2) Max in Left Subtree // 3) Max in right subtree int res = root->data; int lres = findMaximum (root->lchild); int rres = findMaximum (root->rchild); if (lres > res) res = lres; if (rres > res) res = rres; return res; } // Returns minimum value in a given Binary Tree int findMinimum(struct node* root) { // Base case if (root == NULL) return INT_MINIMUM; // Return minimum of 3 values: // 1) Root\'s data 2) Max in Left Subtree // 3) Max in right subtree int res = root->data; int lres = findMinimum(r.

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What happens to jointly held stock between two siblings when one .pdf

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To write a program that implements the following C++ concepts 1. Data Encapsulation 2. Instantiate classes 3. Composition Class 4. Aggregation Class 5. Dynamic Memory 6. File Stream Make a program that reads a file and can generate reports. Each file will have phone calls records (see section 4). The program will process the data gathering all minutes and total amount for each report. The program should have a menu (see section 1): 1. New Report – this option will ask a file name (TE??????.tel). This option will instantiate an object of Report class, loading the whole file into memory. 2. Delete a report – this option will display all reports available in memory and ask which one you would like to delete. 3. Display a report – this option will display all reports available in memory and ask you which one you would like to generate a report. 4. Exit – Program ends. For option 1, you do not know how many reports are, meaning that you will create dynamically the reports. For long distance calls the formula will be (minutes call x $1.50). For local phone calls the formula will be (minutes call x $0.25) -------------------------------------------------------------------------------------------------------------------- ---------------------------- //Main.cpp *#define between(x,y,z) ((x>=y && x<=z)? true:false) #include #include #include #include \"Data.h\" #include \"Report.h\" #include \"Menu.h\" using namespace std; //uses class Report to create reports //uses class Menu to display a menu to choose int main() { int idx = 0; int count = idx; Menu menu(\"Reports\"); Data temp; string fileName; Report* phoneCompany; Report* newReport(Report*, int&); Report* deleteReport(Report*, int&, int); int displayReport(Report*, int); bool fileNameExists(string); menu.addMenuItem(\"New Report\"); menu.addMenuItem(\"Delete a Report\"); menu.addMenuItem(\"Display a Report\"); ofstream openFile; phoneCompany = NULL; phoneCompany = new Report[count]; do { cout << menu; cin >> menu; switch (menu.getChoice()) { case \'1\': cout << \"\ \\tEnter report name (TE??????): \"; cin >> fileName; openFile.open(fileName + \".tel\"); if (openFile.good()) { cout << \"\ \\tReport succesfully created.\ \"; phoneCompany[count].setFileName(fileName); count++; } else { cout << \"\ \\terror: Could not create report.\ \"; } break; case \'2\': if (count != 0) { do { system(\"cls\"); cout << \"\ \ \\tDelete a Report:\ \ \"; idx = displayReport(phoneCompany, count); if (between(idx, 0, count - 1)) { phoneCompany = deleteReport(phoneCompany, count, idx); idx = 0; } else cout << \"\ \\terror: Report could not be deleted.\ \"; } while (idx != count); } else { cout << \"\ \ \\tNo reports to delete.\ \ \"; system(\"pause\"); } break; case \'3\': if (count != 0) { do { system(\"cls\"); cout << \"Display a Report:\ \ \"; idx = displayReport(phoneCompany, count); system(\"cls\"); cout << \"\ \\t\" << phoneCompany[idx].getFileName(); cout << \"\ \ \\tWould you like to edit the report?\"; cout << \.

To write a program that implements the following C++ concepts 1. Dat.pdf

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The HER incentive programs incentive payments to eligible hospitals, and CAHs: to adopt, implement, upgrade, or demonstrate meaningful use of CEHRT to donate technology to understand and rural areas to participate in clinical trials Whish of the following entities is charged with managing the federal HER incentive programs? CMS ONCHIT congress Which of the following organizations is the principal federal entity charged with coordination of nationwide efforts to implement and use the most advanced HIT and the electronic of health information? CMD AHRQ ONCHIT The law that ended preexisting condition exclusions for children (health plans can no longer limit or deny benefits to children younger than 19 because of a preexisting condition)is: HIPAA HITECH PPACA The CLIA of 1988: eliminates the exception under the HIPAA privacy rule to an individual\'s right to access his or her protected health information when it id held by a CLIA-certified or CLIA-exempt laboratory. Created the exception under the security rule to an individual\'s right to access his or her protected health information when it id held by a CLIA-certified or CLIA-exempt laboratory. determined that patients may not access their protected health information when it is held by a CLIA-certified or CLIA-exempt laboratory. Project HealthDesign was a groundbreaking national program of Robert wood Johnson foundation\'s pioneer portfolio, designed to spark innovation in: Personal health technology EHR robotics Solution 8) CMS 7) a 9)c. ONCHIT 10)c. PPAC 11)a.e eliminated the exception 12)a. Personal health technology.

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The comet Shoemaker-Levy 9 broke up into a number of pieces called the ‘string of pearls’ as it approached close to Jupiter. The breakup happened because of the __________________________________ of Jupiter. (Fill in the blank) Solution Differential pull of the planet\'s enormous gravitational force on the near and far sides of the comet fragmented it into 21 or more large pieces and an enormous amount of smaller debris.

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The existence of universal values is still a matter of debate. But e.pdf

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\" The c program will implement the Caesar Cipher. Your program should take a command line argument, which will be the offset, while the input through standard in will be the part that will be rotated(encrypted/decrypted). To simplify this assignment you will only need to rotate characters that are alpha characters (letters). No other characters should be rotated. You can also assume that the input will only be a single line long. Your script will take a single command line argument, which will be the rotation argument to the Caesar c program you wrote above. Your script should then call the \"Program2_2.\" you wrote each of the .enc files. Your script should output a .dec file for each .enc file that exists. \" Solution #include #include #define MAXSIZE 1024 void encrypt(char*); void decrypt(char*); int menu(); int main(void) { char c, choice[2], s[MAXSIZE]; while(1) { menu(); gets(choice); if((choice[0]==\'e\')||(choice[0]==\'E\')) { puts(\"Input text to encrypt->\"); gets(s); encrypt(s); } else if((choice[0]==\'d\')||(choice[0]==\'D\')) { puts(\"Input text to decrypt->\"); gets(s); decrypt(s); } else break; } return 0; } void encrypt(char*str) { int n=0; char *p=str, q[MAXSIZE]; while(*p) { if(islower(*p)) { if((*p>=\'a\')&&(*p<\'x\')) q[n]=toupper(*p + (char)3); else if(*p==\'x\') q[n]=\'A\'; else if(*p==\'y\') q[n]=\'B\'; else q[n]=\'C\'; } else { q[n]=*p; } n++; p++; } q[n++]=\'\\0\'; puts(q); } void decrypt(char*str) { int n=0; char *p=str, q[MAXSIZE]; while(*p) { if(isupper(*p)) { if((*p>=\'D\')&&(*p<=\'Z\')) q[n]=tolower(*p - (char)3); else if(*p==\'A\') q[n]=\'x\'; else if(*p==\'B\') q[n]=\'y\'; else q[n]=\'z\'; } else { q[n]=*p; } n++; p++; } q[n++]=\'\\0\'; puts(q); } int menu() { puts(\"To encrypt, input e or E\ \"); puts(\"To decrypt, input d or D\ \"); puts(\"To exit, input any other letter\ \"); puts(\"Your choice:->\ \"); return 0; }.

The c program will implement the Caesar Cipher. Your program shou.pdf

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Amino acids are acids because they contain which functional group? Select one: a. carboxyl b. sulfhydryl c. carbonyl d. methyl e. amino Solution Amino acids possess acidic nature due to the presence of Carboxylic group I.e. CooH group. The CooH group has the capability to donate H+ or proton, so it act as an acid. According to Branstead lowry theory, compound that release H+ ion is called Acid. And due to the presence of CooH group, Amino acid act like acid.

Amino acids are acids because they contain which functional group S.pdf

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Suppose that theta is an acute angle of a right triangle and tan (the.pdf

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Legionellae, the genus of bacteria that causes Legionnaire\'s disease, are common in __. Hot water supplies water supplies Rivers and lakes above Drinking All of the What level of biosafety is needed for laboratory activities involving TB cultures at American Thoracic Society levels II and III? Solution 1. The genus Legionellae is a pathogenic group of gram-negative bacteria. It is common in many environment including soil and aquatic systems. The correct answer is All the above. 2. Biosafety level 3..

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Multiple Choice Question 1 What features are associated with the leaf epidermis? A= guard cells B= trichomes C= stomata D= all of the above are correct Question 2 The main function of root hairs is protection against herbivores? True or False Solution 1.A all of the above are correct. The epidermis of the leaf contains stomata which are the openings through which exchange of gases takes place and the guard cells control the opening and closing of the stomata. Trichomes are hair like appendages which prevent transpiration, and discourage herbivores to feed on the plant. 2.False. The main function of the root hairs is to collect water and minerals from the surroundings.

Multiple ChoiceQuestion 1What features are associated with the l.pdf

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In your own words, what is thw application of columbia CNA with 5% sheep blood agar? Solution Columbia CNA with 5% Sheep blood agar is a kind of selective as well as differential meadium. Mainly this is used for culture if Gram positive bacteria. In this media, differential growth of Entericocci, Staphylococci , hemolytic streptococci occurs while some clinical isolates like Klebsiella, Pseudomonas and Proteus are inhibited. Therefire this media is also used as differential medium. The differential selection occurs on the basis of hemolytic reactions..

In your own words, what is thw application of columbia CNA with 5 s.pdf

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List two properties of water and discuss how two substances in our body dissolve based on this. After listing properties, could you give 2-3 sentences per example please? Solution Water is a universal solvent and it has hydrogen bonding associated with highest heat capacity and is referred as “specific heat capacity”. This specific heat is predominantly due to intermolecular and intermolecular hydrogen bonding. In nature on land when heat is absorbed result in breaking of hydrogen bonds specifically followed by production of heat. This specific heat capacity of water is going to make the blood and lymph fluid to dissolve electrolytes within the cellular environment to maintain homeostasis by keeping blood pH -buffer as 7.4 Water is a universal solvent and it has both polar because it has both positive charge (H+ or H3O+) and negative charge (OH-). This polarity enables water to make substances to dissolve in it as charged species. This polar nature of water is going to enable \"bicarbonate substance to dissolve in blood\" in buffering system of the cellular respiration, finally involves an acid & base homeostatic mechanism further involving the equilibrium balance of carbonic acid(H2CO3), bicarbonate ion (HCO3-), and carbon dioxide (CO2) in order to maintain isotonic pH (7.4). On the other hand, water molecules possess a higher attractive power to each other so that these forces are referred as cohesive forces responsible for “surface tension” phenomenon” and these attractions are predominantly due to hydrogen bonding. 1). Due of the presence of hydrogen bonds and strong intermolecular forces, water has high specific heat index. Specific heat capacity of water is defined as; the amount of heat (energy in joules) required for raise of temperature of water per unit mass by 10C. Our body temperature is maintained higher than the surrounding temperature. The heat of vaporization of the water is higher (40.65 kJ/mol), this is due to the hydrogen bonds that connect water molecules. So, due to the temperature difference, water escapes out and carry the body heat (the specific heat capacity of the water is 5 times more than the sand). So, during the process of vaporization of water, water releases heat into the environment and cool the body. 2). Because of the hydrogen bonding, water exhibits high surface tension. So, water molecules on the surface are more strongly attracted to each other than to the air above. This is necessary for the functioning of alveoli. The layer of moisture over the alveoli is necessary for the gaseous exchange. 3). Due to the polar nature of water, water can dissolve inorganic salts that dissociate into ions in the presence of water (like dissolves like). So, the blood can carry many of the water soluble substances throughout the body. Water cannot dissolve nonpolar substances; they are dissolved in non-polar solvents. 4). Plasma contains approximately 91% of water; this maintains the viscosity of the blood..

List two properties of water and discuss how two substances in our b.pdf

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Interview Notes Darcy is 45 years old, single, and a U.S. citizen with a valid Social Security number. She had $45,000 in wages. During the interview with Darcy, you determine the following facts: Darcy’s son Chris, age 21, is unmarried and was a full-time student working on a degree in accounting during 2017. Chris’ income was $8,500 in wages, which he used to pay his tuition. He did not provide more than half his own support. Chris lived on campus during the school year, but came home on breaks and for the summer. Chris is in his third year of college. Chris has never had a felony drug conviction. Chris is a U.S. citizen with a valid Social Security number. Questions: 1. Who can claim Chris\' personal or dependency exemption? A. Chris can claim his personal exemption because he had earned income. B. Chris can claim his personal exemption if Darcy does not to claim him. C. Chris does not have a filing requirement so he cannot claim a personal exemption. D. Chris must file a tax return and claim zero personal exemptions because Darcy can claim him as a dependent on her tax return. 2. Darcy can claim the expenses Chris paid as qualifying expenses for the American opportunity credit if Darcy claims Chris as a dependent on her return. True False Solution Answer 1. To be dependent the following conditions should be met: *Chris is independent. because Chris lived on campus during the school year. A. Chris can claim his personal exemption because he had earned Answer 2. True Only qualifying educational expenses can be claimed on income taxes anyway. if Darcy claims Chris as a dependent on her return..

Interview NotesDarcy is 45 years old, single, and a U.S. citizen w.pdf

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Ligands bind reversibly to their receptors. Select the statements that describe what would happen if ligands bound irreversibly with their receptors. A) Pathway will be permanently active. B)The signal cannot be transduced in the cell. C)The cell cannot receive new signals. D)Cell metabolism cannot occur. Solution B) The signal cannot be transduced in the cell They work as antagonists or otherwise called as blockers. They have affinity but they do not have the efficacy for cognate receptors and hence the binding disrupts the interaction and inhibit the function of the agonist. Therefore, the signal remains untranslated into the cell and signal transduction does not occur..

Ligands bind reversibly to their receptors. Select the statements th.pdf

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In the rest of this lab, you will help build such a list class. The class definition is stored in intlist.h - please study it to know its parts: The public declarations show what a client program can do with an IntList object. Notice, however, that the list node structure definition is in the private area. Clients can\'t directly use or even refer to such nodes, but all that clients should care about are the values stored in these nodes anyway! Also notice that every list object will have one node pointer, to point at the first node in the list or at 0 if the list is empty. Now look at intlist.cpp. All but sum, contains, max, average and insertFirst are already done. You will implement those five methods in Step 4. The other parts are implemented at the bottom of the file - do not change what is done. An application file would create IntList objects and use them to solve problems. For this lab, our application is just a testing program - testlist.cpp - and you should look at that file next. Two IntList objects are created; then numbers read from the command line are appended to one of the lists; and finally the methods are tested for each list. You can use this Makefile to help you complile the program. If you store all four of these files in the same directory, then you just type \"make\" (without the quotes) to compile them and create the executable program. Step 4: Implement five linked list functions First: switch roles between pilot and navigator if you did not already do that. Always stay logged into the original pilot\'s account. You should be able to run the program now (assuming you compiled it in Step 3) - it requires you to enter a starter list of integers on the command line: Next time we\'ll run it properly. The contains method is checked for three values that ought to be in the list, plus one value that should not (the sum). Here is a sample run with initial values of 5, 7, 9 and 11: See that append, print and count all work. But the others need to be fixed. Use an editor (e.g., emacs) to make the following changes to intlist.cpp - do not change any of the other files. Fix the comment at the top to show your name(s) and the date. Implement the sum method. See the count method for guidance. Save, and then test your sum implementation - compile and execute testlist again. Verify sum is working before going on. Then implement the contains method, save again and test again. Continue with the other three functions in the same way: implement and test one at time. Here are correct results for the same sample data as above: FILES: Solution #include \"IntList.h\" #include using std::cout; // return sum of values in list int IntList::sum() const { int result = 0; Node *n = first; while (n) { result+= n->info; n = n->next; } return result; } // returns true if value is in the list; false if not bool IntList::contains(int value) const { Node *n = first; while (n) { if(n->info == value) return true; n = n->next; } return false; // REPLACE TH.

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