Show that f(x)=57^x+1 and f^-1(x)=7x-75 are inverses Choose the ap.pdf

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Show that f(x)=5/7^x+1 and f^-1(x)=7x-7/5 are inverses Choose the appropriate equality that will show that f(x) and f^-1(x) are inverses (f.f^-1)(x)= (f^-1.f)(x)=1 (f.f^-1)(x)= (f^-1.f)(x)=x (f.f^-1)(x)= (f^-1.f)(x)=1 (f.f^-1)(x)= (f^-1.f)(x)=x Solution Replace x by f^-1(x) in f(x) => 5/7(7x-7)/5+1 => (7x - 7)/7 + 1 = > x - 1 + 1 => x => f(f^-1(x)) = x.

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The DNA sequence below is a shortened version of the sequence on the previous page. This DNA sequence corresponds to the mRNA sequence only; two introns are not included. The numbers indicate the position of nucleotides within the sequence and indicates intervening nucleotides that are not shown. The gene is 626 nucleotides long. Recall that chromosomes are very long DNA molecules with millions of nucleotides, so this sequence is preceded and followed by seemingly endless sequences of nucleotides. IDNA is double-stranded. Write the sequence of the first 10 nucleotides of the complementary strand of DNA. Solution ans. DNA is a double stranded which consist of two strands and one strand is complemetary to the other.DNA contains complemetary base pairs in which adenine is linke d to tymine by two hydrogen bonds guanine to cytosine by three hydrogen bonds The seque of complementary strand of the DNA molecule is 5\' TGTAAACGAA.

The DNA sequence below is a shortened version of the sequence on the .pdf

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the conducting tissue found in vascular plants that functions in transportation of materials throughout the plant, but does not lend structure support to the stem is A. xylem B. phloem C.sporopollenin D.endosperm e. LYCOPHYLL Solution Answer : B. Phloem Phloem is responsible for the transporting sugars,proteins, and other organic molecules in plants. xylem is responsible for the transport of water and water soluble nutrients,taken up by the roots and also primary component of wood. Vascular plants are able to higher than other plants due to the rigidity of xylem cells, which support the plant..

the conducting tissue found in vascular plants that functions in tra.pdf

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Read the short article Horizontal Gene Transfer in E Coli and answer the following questions.(https://www.sciencedaily.com/releases/2015/05/150519105900.htm) Remember to use your own words; plagiarism of any kind will not be tolerated (see syllabus for further information on what constitutes plagiarism). (1) What is horizontal gene transfer (HGT)? (2) In the posted article, how many gene transfer elements were found for E. coli O104:H4? (3) What is the source of these gene transfer elements? (4) Should we be concerned about HGT in prokaryotes? If so, why? Solution (1) Horizontal gene transfer (HGT) is the lateral gene movement, wherein the movement of genetic material takes place between unicellular or multicellular organisms across species. This gene transfer is different from the vertical gene transfer in which the transmission of DNA occurs from parent to offspring. Often genetic material swaping occurs between bacteria of same as well as different species, through this means by aquiring one of the processes; transformation, transduction or conjugation. During HGT, plasmids, transposons or integrons act as vectors that transfer genes. (2) In the given article, the team from Madurai Kamaraj University in Madurai, Tamil Nadu, identified 38 gene transfer elements for E. coli O104:H4. (3) The research team identified 38 gene transfer prophage elements for E.coli O104:H4. They showed that the these elements act as genetic weapons protecting the bacteria from antibiotics. The source of these elements were viruses (the bacteriophages), that infect bacteria, and in the process bacteria aquired these elements from viruses. (4) In prokaryotes, horizontal gene transfer is a major concern. Aquisition of new genes through HGT is ability of Bacteria and Archaea to adapt to new changing environments. Druing HGT, genes that are responsible for antibiotic resistance in one species of bacteria can be transferred to another species of bacteria, and is the primary reason for the spread of antibiotic resistance. Besides antibiotic resistance, HGT also gives bacteria an ability to degrade novel compounds (such as different insecticides and pesticides). Thus the bacterial strains after HGT are much more dangerous as they are multi drug resistant and more virulent. These transfer elements when transferred to a pathogenic bacteia, can become a serious health hazard as they are resistant to antibiotics available with us and thus cannot be trated easily..

Read the short article Horizontal Gene Transfer in E Coli and answer.pdf

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Please select the best answer and click submit. The owners of fou.pdf

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One function of the spliceosome is to: (Select all that apply.) A. recognize the beginning and end of the introns. B. recognize the 5\' end of the primary transcript. C recognize the beginning and end of the exons. D recognize the 3\' end of the primary transcript. Solution The correct answer is :A. recognize the beginning and end of the introns. Reason: Spliceosome consists of snRNAs and snRNPs which together make a complex.This complex binds to the beginning and end of the introns and forms a loop which is later excised and the ends of exons are sealed..

One function of the spliceosome is to (Select all that apply.)A. .pdf

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Match the following organelles with their key functions in the cell nucleus rough endoplasmic reticulum Golgi apparatus lysosomes mitochondria chloroplasts peroxisomes cytosol carbon fixation site of protein synthesis ready for cellular distribution recycling of intracellular materials modification and packing of proteins for secreation site of protein synthesis that contains the majority of metabolic pathways containment of toxic reactions the site of the core genome oxidative phosphorylation Solution 1: Nucleus: The site of core genome;g 2: Rough ER: site of protein synthesis ready for cellular distribution; b 3: Golgi apparatus: Modification and packing of proteins for secretion; d 4: Lysosome: recylcing of intracellular material; c 5: Mitochondria: Oxidative phosphorylation;h 6: Chloroplast: Carbon fixation; a 7: Peroxisomes: Containment of toxic reaction; f 8: Cytosol: Site of protein synthesis that contains majority of metabolic pathways; e.

Match the following organelles with their key functions in the cell .pdf

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please choose the correct answerIn the ABO blood-type phenotype, w.pdf

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mo Press Submit until all questions are colored green and you have do.pdf

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matching: chose the stage of the light reactions which the statement occurs in. 1. _Produces ATP. 2. _Passes energized electrons down an electron transport chain to create ATP. 3. _Produces NADPH. 4. _Splits water to get electrons to replace energized electrons. 5. _Replaces energized electrons from electrons that have completed the 1st electron transport chain. a. water-splitting photosystem. b. NADPH-Producing photosystem. Solution During water splitting photosystem ,the electrons in water are removed and passed to the chlorophyll. The hydrogen ions extracted from the water molecule, and after two water molecules are splitted,the oxygen atoms takes electron and form molecular oxygen (O 2 ). In photosynthesis, water splitting donates electrons to the electron transport chain in photosystem II. So it splits water to get electrons to replace energized electrons. In the photosystem II (PSII) reaction center, energy from sunlight is used to extract electrons from water. The electrons travel via the chloroplast electron transport chain to photosystem I (PSI), which reduces NADP+ to NADPH. So it produces NADPH.

matching chose the stage of the light reactions which the statement.pdf

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Let E be the set of all even integers, and let O be the set of all odd integers. Explain why Z E O. A. Every integer is both even and odd. B. Some integers are both even and odd. C. Even integers and odd integers are integers. D. Every integer is either even or odd. E. No integer is both even and odd. Solution ans : D take any integer z, z is either odd or even so, for all z , Z is a subset of EUO.

Let E be the set of all even integers, and let O be the set of all o.pdf

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JavaIm not sure how to implement this code can someone help me .pdf

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Indexes prevent the database engine from having to scan every record in the database. True False Solution Answer: True Indexes prevent the database engine from having to scan every record in the database. Indexes provides the improvement of query performance by providing the database engine with fast way of locating a particular record based on the a column or columns provided in where condition..

Indexes prevent the database engine from having to scan every record .pdf

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Is this a probability Distribution? If so, find the Mean (mu) and the Standard deviation (a). If this is not a Probability Distribution, let the summation of P(x): (LP(x))= A, then find: A and let that he your missing row, Then find the Usual Values.! Solution The average population during 1994 was (204500 + 215000)/2 = 419500/2 = 209750. Incidence rate refers to the rate of occurrence of new cases of disease in a population over a specified period of time which also means the number of new cases per unit of population. Incidence rate of lyme disease in 1994 = Number of new cases of lyme disease in 1994/ average population during 1994 = 25/ 209750 = 0.011918951 % or, 0.12 % ( on rounding off to 2 decimal places).

Is this a probability Distribution If so, find the Mean (mu) and the.pdf

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in Java (ignore the last line thats hidden) Create a doubly linked list whose nodes contain Strings. Your list should include the following methods: Insert a node in the list in alphabetical order Find a node that matches a String Traverse the list forwards and print Traverse the list backwards and print Delete a node from the list Delete/destroy the list In addition to creating the class(es) for your linked list, you\'ll need to create a main method that thoroughly tests each function you build to show that it works. You should do these things in Solution / Java program to create , insert , traverse and delete a node from doubly linked list class LinkedList { static Node head = null; class Node { int data; Node next, prev; Node(int d) { data = d; next = prev = null; } } /* Let us create the doubly linked list 10<->8<->4<->2 */ list.push(head, 2); list.push(head, 4); list.push(head, 8); list.push(head, 10); System.out.println(\"Original Linked list \"); list.printList(head); /*Function to delete a node in a Doubly Linked List. head_ref --> pointer to head node pointer. del --> pointer to node to be deleted. */ void deleteNode(Node head_ref, Node del) { /* base case */ if (head == null || del == null) { return; } /* If node to be deleted is head node */ if (head == del) { head = del.next; } /* Change next only if node to be deleted is NOT the last node */ if (del.next != null) { del.next.prev = del.prev; } /* Change prev only if node to be deleted is NOT the first node */ if (del.prev != null) { del.prev.next = del.next; } /* Finally, free the memory occupied by del*/ return; } /* UTILITY FUNCTIONS */ /* Function to insert a node at the beginning of the Doubly Linked List */ void push(Node head_ref, int new_data) { /* allocate node */ Node new_node = new Node(new_data); /* since we are adding at the begining, prev is always NULL */ new_node.prev = null; /* link the old list off the new node */ new_node.next = (head); /* change prev of head node to new node */ if ((head) != null) { (head).prev = new_node; } /* move the head to point to the new node */ (head) = new_node; } /*Function to print nodes in a given doubly linked list This function is same as printList() of singly linked lsit */ void printList(Node node) { while (node != null) { System.out.print(node.data + \" \"); node = node.next; } } public static void main(String[] args) { LinkedList list = new LinkedList(); /* delete nodes from the doubly linked list */ list.deleteNode(head, head); /*delete first node*/ list.deleteNode(head, head.next); /*delete middle node*/ list.deleteNode(head, head.next); /*delete last node*/ System.out.println(\"\"); /* Modified linked list will be NULL<-8->NULL */ System.out.println(\"Modified Linked List\"); list.printList(head); } }.

in Java (ignore the last line thats hidden) Create a doubly linked l.pdf

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Imagine that you are looking at the amino acid sequences of the same protein from many different organisms. All of the organisms have a Cys at the same place in the protein sequence. What does that tell you about the role of that Cys in the protein? All of the organisms have a stretch of 3 amino acids that are Arg or Lys-Ser or Thr-Arg or Lys. What does that tell you about this part of the protein? Another part of the protein seems to have all of the possible amino acid sequences in the different organisms. What does that tell you about that part of the protein? Solution Answer: 1) Cysteine (Cys) and Methionine are the sulphur containng aminaocids. Cysteine presence in the teritary structure of a protein is a common thing due to formation of the disulphide linkage. Disulphide likages are formed by cysteine only but not methionine. The thiol group is oxidised to form disulphide linkage. 2) Arg or Lys-Ser, Thr-Arg or Lys From above sequence, it can be concluded that Basic group containing aminoacids such as Arginine (Arg) and Lysine (Lys) are forming bond with hydroxyl group containg aminoacids such as Serine (Ser) and Threonine(Thr). As basic aminoacids are insoluble in water, so they participate in protein structure by association with polar aminoacids (water loving) in the protein. 3) There are 20 aminacids with 64 codons which code for each aminoacid. Proteins seems to have all possible aminoacid sequences in different organisms. The aminoacids sequence of different protein vary with there functional properties, each protein function is based on the teritary arrangement of the polypeptide. Every gene has specific property and it codes for respective protein which is essential biological system, similarly, every enzyme has its owns aminoacid sequence to exhibit its catalytic activity..

Imagine that you are looking at the amino acid sequences of the same .pdf

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I need help implementing a Stack with this java programming assignment: Given a doubly linked list (code provided below) implement a Stack. You must use the doubly link list class and utilize it as an object in your implementation of the stack. A Stack is a Last in First Out (LiFO) structure which translates to the processes of outputting the last element which was inputted in the collection. The Stack is based on the process of putting things on top of one another and taking the item on top (think of a stack of papers). Operations to implement: push (E): Add an element to the start of the sequence pop: Remove an element from the start of the sequence Peek: Return the first element of the sequence without removing it atIndex(x): Return the element at the given index (x) Or throw an exception if it is out of bound (if you can control the user input then do that instead) Size: Return the size of the Stack isEmpty: Boolean, returns true if the Stack is empty Empty: Empty the Stack Code Part 1: public class DoublyLinkedList{ private DNode head; private DNode tail; private int size; public DoublyLinkedList(){ // construct an empty list this.tail = new DNode (null, null, this.head); this.head = new DNode (null, this.tail, null); this.size = 0; } public DoublyLinkedList(DNode next){ // constructs a list // out of a single node this.tail = new DNode (null, null, next); this.head = new DNode (null, next, null); next.changeNext(this.tail); next.changePrev(this.head); this.size = 1; } public DoublyLinkedList(Object [] objectArray){ // construct a list out of // an array this.tail = new DNode (null, null, this.head); this.head = new DNode (null, this.tail, null); DNode temp = this.head; for (Object e : objectArray) { //Anonomus function new DNode (e, temp.getNext(),temp).insertBetweenNodes(temp, temp.getNext()); temp = temp.getNext(); this.size += 1; } } public void addToFrontofList(Object toAdd){ // Appends the begining // of the list DNode temp = new DNode (toAdd, this.head.getNext(), this.head); this.head.getNext().changePrev(temp); this.head.changeNext(temp); this.size += 1; } public void addToendofList (Object toAdd) // appends the end of the list // with a node { DNode temp = new DNode (toAdd, this.tail, this.tail.getPrev()); this.tail.getPrev().changeNext(temp); this.tail.changePrev(temp); this.size += 1; } public void insertAfterNode(DNode current, Object input)// Inserts a new // a new node after // current node { current.insertAfterNode(input); this.size += 1; } public int getSize() // returns the size of the list { return this.size; } public String toString() { String result = \"\"; for (DNode temp = this.head.getNext(); temp.hasNext(); temp = temp.getNext()) { result += temp.getValue(); result += \" -> \"; } result += \"End of list\"; return result; } } and code part2: public class DNode{ private Object e; // place holder for the type of object the //collection is storing private DNode next; // reference to the next node in the sequence // of.

I need help implementing a Stack with this java programming assignme.pdf

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If a trait is sexually selected, the trait could appear the same in both males and females. Agree/Disagree? Explain: Solution No, all the sexually selected traits could not appear the same in both males and females. For example, in sex-linked diseases, the defective allele is present in X-chromosome. X- linked or sex-linked diseases could be developed in females only when they inherit two X-linked recessive alleles, one from the mother and another from the father. Whereas in males, only one defective allele will result in expression of disease phenotype. Similarly, Y-linked genes are not expressed in females as females do not contain the Y-chromosome..

If a trait is sexually selected, the trait could appear the same in .pdf

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html js node Create a function that reads from a file and returns a .pdf

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How does phospholipids structure relate to the selective permeability of the plasma membrane? critical feature of the plasma membrane is that it is selectively permeable This a lows the plasma membrane to regulate transport across cellular boundaries a function essential Drag the labels to fill in the table. Use only white targets, Pink tables for pink targets, and blue labels for blue targets. Solution 1. Hydrophobic can cross easily No transport protein required In lipid bilayer it has polar head and nonpolar tail. So, Non polar molecule can easily cross the membrane. It does not require any special mechanism for transport this molecules. 2. Polar molecules - Hydrophilic - Have difficulty to cross hydrophobic part - transport protein require for efficient transport. Hydrophobic molecules hates the hydrophilic so it does not allow the polar molecules easily cross in the lipid bilayers. 3. Ions - Hydrophilic - Have difficulty to cross hydrophobic part - transport protein require for efficient transport..

How does phospholipids structure relate to the selective permeability.pdf

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Geneticists use a shorthand notation to refer to the genotype of the different individuals with which they are working. For instance, the symbol denoting the recessive allele for the ability to roll your tongue might be written TRR. An individual that is heterozygous for this trait and also carries the wild-type allele might be denoted as ______. Note: Select all correct answers. +/TRR TR+/TRR TRR/TRR +/+ TR+/TR+ Solution In +/TRR the two alleles are \'+\' which denotes wild type, and TRR which is the recessive allele and since both the alleles are different it is heterozygous. In TR+/TRR the two alleles are \'TR+\' which has one wild type gene (+) and the other recessive (TR). The single wild type gene may confer the wild type trait depending on other factors such as dominance. An individual with these alleles are also heterozygous. TRR/TRR is homozygous recessive. +/+ is homozygous wild type. TR+/TR+ is homozygous..

Geneticists use a shorthand notation to refer to the genotype of the.pdf

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