The orange is which type of fruit Simple - Legume Simple - Drupe .pdf

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The orange is which type of fruit? Simple - Legume Simple - Drupe Simple - Berry Multiple Aggregate The peas is which type of fruit? Simple - Legume Simple - Drupe Simple - Berry Multiple Aggregate The pistachio is which type of fruit? Simple - Legume Simple - Drupe Simple - Berry Multiple Aggregate Solution Please find the answers below and reasons in parentheses: Answer 22: Chocie 3 (Botanically, orange fruit is considered as a multiple aggregate of berry and the juicy part or pulp is considered as succulent testa of the ovary. The peel of orange is considered as ovarian wall.) Answer 23: Choice 1 (A pea pod is considered as a simple legume by botanical nature. It can open across both walls which eliminates the seeds inside the pod) Answer 24: Choice2 (Pistachios belong to family of cashew or family Anacardiaceae botanically. The fruits of this family are simple drupes in nature. These fruits contain the ovarian wall as the hard covering of the fruit and the ovary itself modifies into edible seed).

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Give background information concerning EMT and include what is known about EMT in normal development as well as in neoplasms. What are strategies that could be used to overcome drug resistance in cancer cells by targeting EMT (minimum four paragraphs). Solution Answer: Cancer stem cell (CSC) hypothesis: This hypothesis elucidating that the CSC are the subpopulation and able to exert higher resistance associated with explicit capability to self-renew finally to differentiate into different offspring of cancer cells to generate a tumor. This hypothesis is based on the cellular mechanisms exclusively developed inside the CSC to exert resistance such pathways are mainly through Wnt signaling, Notch signaling. There are other targets to regulate proliferation and differentiation through ABC transporter system and receptor tyrosine kinases. These targets are the future stem cells anti-cancer therapies in regulating the proliferation & differentiation of cancer cells as CSCs have a potential relationship with epithelial-mesenchymal transition (EMT). Metastasis has many steps. The main cancer cell morphological features that support invasion, intravasation, and extravasation are epithelial-mesenchymal transition via cell -adhesion molecules. Invasion is mainly due to presence of cell adhesion molecules such as integrins on the cancer cells finally promote the movement of cancer cells to another region. This process is going to connect adjacent cells to tumor cells finally undergo epithelial-mesenchymal transition (EMT) result in formation of \"continuously dividing neoplasms\" with higher drug resistence to cure compared to normal cells Strategies that could be used to overcome drug resistance in cancer cells by targeting EMT: \"A combination therapy of neutraceuticals & chemotherapeutic agents\" are most promising strategies to overcome drug resistance in cancer cells by targeting EMTs, cancer stem cells (CSC) & Modulation of microRNAs do promote effective strategy to overcome drug resistance in cancer cells Oligonucletodie deliver: Another strategy could be used to overcome drug resistance in cancer cells by targeting EMT. In this method, synthesized microRNAs are going to make in the form of \"nanoparticle or microsomal\" formulation finally used to target microRNAs in cancer cells result in \"complete destruction of pre-mRNA\" finally there will no translation mechanism to synthesize \"cell survival proteins\" to proliferate.

Give background information concerning EMT and include what is known.pdf

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Explain how the current economic recession differs from the depression in the 1930\'s. Solution The Great depression was worldwide. It was due to the collapse of the international financial system It also resulted from the mutual adoption by many countries ( including USA) of high- tariff policies, which were intended to keep out foreign goods in order to protect domestic producers. The policies were called \"beggar thy neighbour\" strategies since they attempted to \"export\" unemployment by improving one country\'s trade position and hence demand for its goods at the expense of its trading partners And , of course, if each country keeps out foreign goods, the volume of world trade declines, providing a contractionary influence on the world economy Almost every country suffered a deep recession in the 1930s, but some countries did better than USA Sweden began an expansionary policy in the early 1930s and reduced its unemployment relatively quicker Britain\'s economy suffered high unemployment but in 1931 it went off the gold standard and the ensuing devaluation of the pound sterling set the stage for some improvement. Germany grew rapidly after Hitler came to power and expanded government spending. China escaped the recession until after 1931, because it had a floating exchange rate. In 1938, real GNP in US rose above its 1929 level for the first time in the decade, but it was not until 1942 , after the US formally entered world war II , that the unemployment rate finally fell below 5% The experience of the US in the early 1980s- the worst recession since the Great Depression, casts doubt on the optimistism of recovery. During the 1980s, the US experienced the largest sustained budget deficits in its peacetime history. Even so cutting spending or raising taxes was not politically popular. Gradually, in the 1990s, the deficit began to be brought under control, and toward the end of the decade the budget swung into surplus.This was due to raising the prices and government spending. By, Nishant Bhatt.

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Does yeast uses CopI and CopII formation for vesicle budding to grow asexually?Is CopII starts from the er then goes out the cell and CopI works for the molecules comes in the cell and goes to nucleus?Does copII carries proteins away to make the budding?Please explain in details. Solution Yes, yeasts reproduce asexually by an asymmetric division process known as budding. Initially, yeast forms a small protuberance which then grows to a full size and forms a bud. Then the nucleus splits and migrates into the daughter cell. After the movement of nucleus, the entire bud detaches from the parent cell by forming a constriction at the base. This process will repeat until a chain of bud cells are formed..

Does yeast uses CopI and CopII formation for vesicle budding to grow.pdf

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DNA molecules consist of chemically linked sequences of the bases adenine, guanine, cytosine and thymine, denoted A, G, C and T. A sequence of three bases is called a codon. A base may appear more than once in a codon. 1) Represent all possible base sequences using a tree diagram. 2) How many different codons are there? 3) The bases A and G are purines, while C and T are pyrimidines. How many codons are there whose first and third bases are purines and whose second base is a pyrimidine? 4) How many codons consist of three different bases? Solution.

DNA molecules consist of chemically linked sequences of the bases ad.pdf

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Data Structures PLEASE USING THIS C++ PROGRAM BELOW, I NEED HELP ON IMPLEMENTING BOTH QUEUE AND STACK TO PRINT \"I able\" REVERSE/BACKWARDS(PALINDROME) #include #include using namespace std; struct st { int year; st *next; }; class queue { public: int sz,i = 0; char chr[100]; queue() { //constructor for initializing front and rear to NULL front = NULL; rear = NULL; cout << \"enter size of queue: \"; cin >> sz; } // ~queue() // { // // } // in the enterqueue, create a new node, assign value and add it to queue // if F = R = NULL the ( ) else ( ) void enterqueue(int yr, char el) { //inserting into queue if queue not overflow if(i >= sz) { cout << \"queue over flow\"; return; } if (front == NULL ) { front = new st; front->next = NULL; front->year = yr; rear = front; chr[i] = el; i++; } else { st *temp = new st; temp->year = yr; rear->next = temp; rear = temp; chr[i] = el; i++; } } char deletequeue() { //delete front data from queue if data in the queue char tr; int j = 0; st *temp; temp = front; if( front == NULL) { cout << \"queue under flow\"; return \'a\'; } if(front == rear) { front = rear = NULL; return \'a\'; } front = front->next; delete temp; tr = chr[0]; i--; for(j = 0; j < i; j++) { chr[j] = chr[j+1]; } chr[j] = \'\\0\'; return tr; } bool isempty() { //it shows whether queue is empty if( front == NULL) { return true; } else return false; } bool isfull() { //it shows whether queue is full if(i == sz) return true; else return false; } void display() { //it displays queue data with characters of queue if invalid choice is entered st *tmp; int j = 0; cout << \"\ queue is:\ character \\tyear\"; for(tmp = front; tmp != rear; tmp = tmp->next, j++) { cout << \"\ \" << chr[j] << \" \" << tmp->year; } cout << \"\ \"<year; } private: st *front, *rear; }; int main() { //main method to call queue functions by choice queue s1; int ch,yr; char el; bool s; while(1) { cout << \"\ 1.ENTERQUEUE\ 2.DELETEQUEUE\ 3.ISEMPTY\ 4.ISFULL\ 5.EXIT\ enter your choice: \"; cin >> ch; switch(ch) { case 1: if(!s1.isfull()) { cout << \"\ enter a year : \"; cin >> yr; cout << \"\ enter a character : \"; cin >> el; s1.enterqueue(yr, el); } else cout << \"insertion not possible\"; break; case 2: s = s1.isempty(); if(!s) { el = s1.deletequeue(); cout << \"\ The character of queue \" << el <<\" is deleted\" << endl; } else cout << \"\ queue is empty.deletion not possible\"<< endl; break; case 3: s = s1.isempty(); if(s) cout << \"\ queue is empty\" << endl; else cout << \"\ queue is not empty\" << endl; break; case 4: s=s1.isfull(); if(s) cout << \"\ queue is full\" << endl; else cout << \"\ queue is not full\" << endl; break; case 5: exit(0); default : cout << \"invalid choice\" << endl; if(!s1.isempty()) s1.display(); } } return (0); } Solution #include #include using namespace std; struct st { int year; st *next; }; class queue { public: //int sz,i=0;//modified //code modified here// everything is fine with the code except this... queue is implemented correctly //you can .

Data StructuresPLEASE USING THIS C++ PROGRAM BELOW, I NEED HEL.pdf

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Count Red Nodes. Write a program that computes the percentage of red nodes in a given red- black BST. Test your program by running at least 100 trials of the experiment of inserting N random keys into an initially empty tree, for N = 10^4, 10^5, 10^6, and formulate an hypothesis. *It is experiment from textbook Algorithms, 3.3.42 Does anyone know the answer? Solution /* * Java Program to Implement Red Black Tree */ import java.util.Scanner; /* Class Node */ class RedBlackNode { RedBlackNode left, right; int element; int color; /* Constructor */ public RedBlackNode(int theElement) { this( theElement, null, null ); } /* Constructor */ public RedBlackNode(int theElement, RedBlackNode lt, RedBlackNode rt) { left = lt; right = rt; element = theElement; color = 1; } } /* Class RBTree */ class RBTree { private RedBlackNode current; private RedBlackNode parent; private RedBlackNode grand; private RedBlackNode great; private RedBlackNode header; private static RedBlackNode nullNode; /* static initializer for nullNode */ static { nullNode = new RedBlackNode(0); nullNode.left = nullNode; nullNode.right = nullNode; } /* Black - 1 RED - 0 */ static final int BLACK = 1; static final int RED = 0; /* Constructor */ public RBTree(int negInf) { header = new RedBlackNode(negInf); header.left = nullNode; header.right = nullNode; } /* Function to check if tree is empty */ public boolean isEmpty() { return header.right == nullNode; } /* Make the tree logically empty */ public void makeEmpty() { header.right = nullNode; } /* Function to insert item */ public void insert(int item ) { current = parent = grand = header; nullNode.element = item; while (current.element != item) { great = grand; grand = parent; parent = current; current = item < current.element ? current.left : current.right; // Check if two red children and fix if so if (current.left.color == RED && current.right.color == RED) handleReorient( item ); } // Insertion fails if already present if (current != nullNode) return; current = new RedBlackNode(item, nullNode, nullNode); // Attach to parent if (item < parent.element) parent.left = current; else parent.right = current; handleReorient( item ); } private void handleReorient(int item) { // Do the color flip current.color = RED; current.left.color = BLACK; current.right.color = BLACK; if (parent.color == RED) { // Have to rotate grand.color = RED; if (item < grand.element != item < parent.element) parent = rotate( item, grand ); // Start dbl rotate current = rotate(item, great ); current.color = BLACK; } // Make root black header.right.color = BLACK; } private RedBlackNode rotate(int item, RedBlackNode parent) { if(item < parent.element) return parent.left = item < parent.left.element ? rotateWithLeftChild(parent.left) : rotateWithRightChild(parent.left) ; else return parent.right = item < parent.right.element ? rotateWithLeftChild(parent.right) : rotateWithRightChild(parent.right); } /* Rotate binary tree node with left child */ private RedBlackNode rot.

Count Red Nodes. Write a program that computes the percentage of red.pdf

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apple 2008 historically what were Apples major competitive advantages and why did they fail to build on these advantages to lead the industry? Solution Apples major competitive advantages: Apple II and Macintosh were the machines that made Apple. They both had to following features. Brand loyalty among important segments Reasons for why did Apple fail to build on these advantages to lead the industry: Apple’s missteps: Entry of Rivals:.

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A partial results from a PERT analysis for a project with 50 activities shows that path A-K-Q Solution There are 50 activities shows the path A-K-Q. With the help of PERT (Program Evaluation and Review Technique), and CPM (Critical Path Method), the project manager can 1. Plan the project ahead of time and foresee possible sources of troubles and delays in completion. 2. Schedule the project activities at the appropriate times to conform with proper job sequence so that the project is completed as soon as possible. 3. Coordinate and control the projects activities so as to stay on schedule in completing the project. Thus PERT incorporates uncertainties in activity times in its analysis. It determines the probabilities of completing various stages of the project by specified deadlines. It also calculates the expected time to complete the project. An important and extremely useful by product of PERT analysis is its identification of various “bottlenecks” in a project. Application of PERT and CPM A partial list of applications of PERT and CPM techniques is 1. Construction projects (e.g., buildings, highways, houses, and bridges) 2. Preparation of bids and proposals for large projects. 3. Maintenance planning of oil refineries, ship repairs, and other large operations. 4. Development of new weapons systems and new manufactured products. 5. Manufacture and assembly of large items such as airplanes, ships, and computers. 6. Simple projects. Six steps : 1. Define the projects and all of its significant activities or tasks. 2. Develop the relationships among the activities. Decide which activities must precede and which must follow others. 3. Draw the network connecting all of the activities. 4. Assign time and/or cost estimates to each activity. 5. Compute the longest time path through the network; this is called the critical path. 6. Use the network to help plan, schedule, monitor and control the project..

A partial results from a PERT analysis for a project with 50 activit.pdf

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briefly write about the variability of natural systems and the significance of the rate of change and predictability to the stability and functioning of ecological systems Solution Answer:- The natural ecological systems are usually variable in physical and biological characteristics that are always in dynamic equilibrium with their biota. The ecological systems which diverse both in eveness and richness of species are usually stable and less prone to extinction and resist the disturbances created by any external agency. The keystone species play an important role in the ecosystem functioning and overall community stability. Thus variability and stability are essential for functioning of ecological systems..

briefly write about the variability of natural systems and the signi.pdf

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A graph is symmetric with respect to the vertical line corresponding.pdf

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X-Linked Recessive Write three rules to keep in mind when counseling someone about the likelihood of inheriting an X-linked recessive condition: About parents\' genotype. About parents\' phenotype. About the probability of the offspring showing the trait. If instead of having Hemophilia C (autosomal recessive), Lisa Keller\'s brother had Hemophilia A (X-linked recessive), what would be the probability that she was a carrier? 25% 50% 67% 100% More information is required to answer this question. Solution About genotype of parents- If the mother is a carrier i.e. carrying the recessive allele on one of her X chromosome and the father do not have the recessive allele in his genotype, then none of the daughters will suffer from the disease (since they will inherit only one recessive allele from the mother) but sons will have a 50% probability of suffering from the disease as they will inherit the recessive allele from the mother and the Y chromosome from the father, although there is a 50% probability of inheriting the wild type allele from the mother. Now if the mother is suffering from the disease i.e. has the recessive allele on both the X chromosome while the father is healthy having no allele for the disease, the daughters will always be a carrier but never suffer from the disease as they eill inherit the healthy wild type allele from the father as one X chromosome comes from the father and the other X chromosome from the mother. The sons will however all suffer from the disease as they inherit the Y chromosome from the father and the X chromosome having the recessive allele causing the disease will be inherited from the mother. Now suppose the mother is a carrier i.e. carries the recessive allele only on one X chromosome but the father suffers from the disease making his genotype XrY (Xr denoting recessive allele carrying X chromosome), then, 50%of their son will duffer from the disease while 50% of the daughters will also have a chance of suffering from the disease. If both the parents do not have a single recessive X linked disease gene on any of their X chromosome then none of the children will suffer from the disease. About phenotype of parents- If the mother is a carrier i.e. have the diseased allele on only one X Chromosome, then the phenotype of mother will be normal. But the father can never be a carrier for a X linked recessive disease as males have only one X chromosome. If the diseased allele in present on that X chromosome in males it will be expressed as there is no dominant healthy wild type allele to dominate the diseased gene. Thus if the father has the recessive diseased allele on X chromosome, the father\'s phenotype will be expresding the disease while if the father do not have the allele he will be healthy. On the other hand for the mother to express the disease the recessive allele has to be present on bith her X chromosome. Probability of offsprings showing the trait- If mother is a carrier and father is normal, then 5.

X-Linked Recessive Write three rules to keep in mind when counseling .pdf

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Write a Java Class to Implement a Generic Linked List Your list must be implemented as a singly-linked list of generic nodes, where each Node object has two instance variables: an object of the “type variable” class, and a pointer to the next node on the list. Your class will contain separate methods to handle each of the operations read from a data file (see II., below) Your class will also override toString() to return the objects on the list in the order in which they occur. Write a Test Class for Your LinkedList Class Your main method will read list operation instructions from a data file, until end-of-file and call the appropriate LinkedList method to execute each one. After each operation is executed, print out the operation and the updated list. The data file to be used is on the class web site and the operations are: APPEND X - Append object X to the end of the list ADD N X - Insert object X as the new Nth element in the list, increasing the size of the list by 1 E.g. Suppose the list is: head -> 1 -> 2 -> 3 -> 4->null After ADD 3 7 it would be: head -> 1 -> 2 -> 7 -> 3 -> 4->null DELETE N – Remove the Nth object from the list SWAP M N - Interchange the positions of the Mth and Nth objects on the list For credit, the two nodes must actually \"trade places\" in the list, and not merely swap their data values REVERSE - Reverse the order of the objects on the list This must be done by reversing the order of the nodes themselves, rather than by swapping the data stored To get credit for your reverse() method, it must use either one of these 2 algorithms: For each node on the list except the current “head” node, delete the node and insert it as the new head Use your swap() method 6. CLEAR – Clear the list (make it empty) No credit will be given for programs that use any additional data structures – either from the Java API or programmer defined -, other than your own LinkedList class txt file: Solution //Java Program to Implement Singly Linked List import java.io.BufferedReader; import java.io.FileNotFoundException; import java.io.FileReader; import java.io.IOException; /* Class Node */ class Nodes { protected Object data; protected Nodes next; /* Constructor */ public Nodes() { next = null; data = null; } /* Constructor */ public Nodes(Object d,Nodes n) { data = d; next = n; } /* Function to set link to next Node */ public void setLink(Nodes n) { next = n; } /* Function to set data to current Node */ public void setData(Object d) { data = d; } /* Function to get link to next node */ public Nodes getLink() { return next; } /* Function to get data from current Node */ public Object getData() { return data; } } /* Class linkedList */ class linkList { protected Nodes start; protected Nodes end ; public int size ; /* Constructor */ public linkList() { start = null; end = null; size = 0; } /* Function to check if list is empty */ public boolean isEmpty() { return start == null; } /* Function to get size of list */ public int getSize() { return .

Write a Java Class to Implement a Generic Linked ListYour list mus.pdf

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Write a class (BasketballTeam) encapsulating the concept of a team of basketball players. This method should have a single instance variable which is an array of basketball player objects. In addition to that class, you will need to design and a code a Playerclass to encapsulate the concept of a basketball player, assuming a basketball player has the following attributes: a name, a position, number of shots taken, and number of shots made. In your Player class you should write the following methods: A constructor taking a String for the name, a String for the position, an int for the number of shots taken, and an int for the number of shots made. Accessor, mutator, toString, and equals methods. A method returning the individual shooting percentage. Avg = Shots Made / Shots Taken. Hint: This method should return a double!!! In your BasketballTeam class you should write the following methods: A constructor taking an array of Player objects as its only parameter and assigning that array to the array data member of the class, its only instance variable. Accessor, mutator, toString, and equals methods. A method checking if all positions are different, returning true if they are, false if they are not. A method returning the shooting percentage of the team. Team Shooting Percentage = Total Shots Made / Total Shots Taken Hint: This should return a double!!! Also, this is not the average of the the averages. A method checking that we have a center (that is, the name of the position) on the team. If we do not have any return false; otherwise, it returns true; A method returning void that sorts the array of Player objects in ascending order using the number of shots taken as the sorting key. A method returning the array of Player objects sorted in ascending alphabetical order by name. Hint: Use compareTo method of the String class. A method returning the name of the player with the most shots made. A method returning the name of the player with the highest shooting average. In your BasketballTeamClient class, when you test all your methods, you can hard-code five basketball Player objects. This client should call all 12 methods including the constructor for full credit. When writing the Player class you should write a PlayerClient class that tests all its methods before integrating into BasketballTeam class. Solution public class Player { //Attributes private String name; private String position; private int shotsTaken; private int shotsMade; /** * Constructor * @param name * @param position * @param shotsTaken * @param shotsMade */ public Player(String name, String position, int shotsTaken, int shotsMade) { this.name = name; this.position = position; this.shotsTaken = shotsTaken; this.shotsMade = shotsMade; } /** * @return the name */ public String getName() { return name; } /** * @param name the name to set */ public void setName(String name) { this.name = name; } /** * @return the position */ public String getPosition() { return position; } /** * @p.

Write a class (BasketballTeam) encapsulating the concept of a tea.pdf

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Which of the following would you expect to happen in a plant that does not have enough water (multiple answers are correct) Without water, the plant cannot make the sugar it needs to grow the stem begins to dry out growth would be stunted because it is difficult to move nutrients around Without water, the plant would become limp. Solution Answer: All the choices given are correct! If Plant that does not have enough water; (a) Without water, plants cannot make the sugar it needs to grow. Plants make their own food by the process of photosynthesis.The amount of water, sunlight and carbon dioxide directly influences the process of food production for plant growth. Thus the process of photosynthesis slows down and even stops when water is absent, which causes internal food supplies to diminish. (b) the stem begins to dry out Scarcity of water leads to loss of turgor pressure. Stomata on the stem surface close as a result of less water (for water conservation) and as a result stem dries out. (c) the growth would be stunted because it is difficult to move nutrients around. As water moves above the plant system it carries vital minerals and nutrients to different areas of the plant. The stomas on the leaf and stem surface release waste products (oxygen) into the environment and takes CO2 in. This process of transpiration maintains the turgor pressure and also cools the plant in sunlight. In scarcity of water, transpiration process slows and, the plant show stunded growth due to lack of nutrients supply. (d) Without water the plant would become limp Turgor pressure holds the plant cells and keeps them erect. This pressure is lost without water and plants begin to collapse (wilting!) or give a limp appearance..

Which of the following would you expect to happen in a plant that doe.pdf

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What is the value of including personal information about yourself and your lifestyle when researchers are looking at your DNA? How does study different than many other similar sets of data gathered from GWAS and SNP Haplotypes?What do you think about the scope and purpose of this type of genetics data study. Solution Answer to including personal information and lifestyle lies in phenotype differences. Phenotype is influenced by genotype and environment. Hence although scientists and researchers look for DNA or genetic material of an individual but also take a look on the environmental conditions or life style which influences the phenotype of an organism. GWAS (Genome wide association study) is study of genetic variants which are associated with a trait like any major disease. By using SNPs (Short nuclear polymorphisms) short interspersed regions can be identified to locate the traits of interest. The environmental factors like sunlight can differ the response of an individual to it. One individual can be sensitiove to it but the other could be diseased on exposure to sunlight, thus having a knowledge about individual\'s lifestyle helps in tracking previous information. The scope of GWAS has wider implications. Disease dignosis using genetics data helps in developing drugs relating to it. It can be used to identify genes from different organisms. Ancestral hierarchy can be established..

What is the value of including personal information about yourself a.pdf

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What is a voluntary response sample? Solution Voluntary response sampling is heavily biased because it focuses on volunteered survey answers rather than a random sampling. What makes this biased is because the only people who volunteer are those who have a particular interest in the topic of the survey. They only include people who choose volunteer, whereas a random sample would need to include people whether or not they choose to volunteer. Often, voluntary response samples oversample people who have strong opinions and undersample people who don\'t care much about the topic of the survey. Thus inferences from a voluntary response sample are not as trustworthy as conclusions based on a random sample of the entire population under consideration..

What is a voluntary response sampleSolutionVoluntary response.pdf

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What adaptive benefit do these abilities give wolbachia? In other words, why does manipulating the host reproduction help its survival survival as parasite? What adaptive benefit do these abilities give wolbachia? In other words, why does manipulating the host reproduction help its survival survival as parasite? What adaptive benefit do these abilities give wolbachia? In other words, why does manipulating the host reproduction help its survival survival as parasite? Solution Manipulating the host reproduction help its survival as parasite.The improvement of complex lifestyles cycles can growth parasite health. Some parasites have, via host manipulation, further improved their reproductive success. Host manipulation in a few parasites, where they effectively can change the behavior or morphology of their intermediate host in a way that increases the parasite\'s trophic transmission It is a method of parasite transmission wherein the parasite manipulates the behavior of its host to growth transmission success to better trophic levels.

What adaptive benefit do these abilities give wolbachia In oth.pdf

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Think about autumn (Fall season) in places like New England (Pennsylvania, New York, Vermont, New Hampshire), where the leaves change color with the seasons. During the fall, which pigments are the first to be broken down in the leaves? How does that explain the fall colors we see? Why does the green color disappears during autumn (Fall season)? Please fully and clearly explain why the green color disappeares in trees located in New England during the autumn. Solution A number of factors play crucial role in changing the color of leaves in winter. Some are listed below: Similar is the reason with the New England trees..

Think about autumn (Fall season) in places like New England (Pennsyl.pdf

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The _________ is the protective chamber that houses the ovule and Later matures to become the fruit. A. Petal B. Ovary C. Carpel D. Stigma E. Sepals. What adaptation enhances bird\'s flying ability? A. hollow bones B. sternum. C. air sacs D. homeothermy E. all of the above. What plant characteristics are adaptations to land? A. internal fertilizations B. vascular tissues. C. seeds with waterproof coats. D. waxy leaves. E. all of the above. Which of which is characteristic of alternation of generation in land plants? A. haploid sporophytes make haploid spores. B. gametophytes produce spores that develop into gametes C. sporophytes and gametophytes are typically similar in appearance. D. meiosis in sporophytes produces haploid spores. E. either the gametophyte or the sporophyte is unicellular. The haploid stage in a plant life in a plant life cycle is called A. embryo B. zygote C. gametophyte D. sporophyte E. conspicuous Animals in which the blastopo Solution 31. ovary is the protective chamber that houses the ovule and later matures the fruit 32. tha hollow bones, air sacs , the sternum and homeothermy helps the birds to fly. the hollow bones helps in reducing the body weight which helps in flying. the air sacs are responsible for extracting more oxygen which hepls them to frly high. the sternum bones are present in the wings which are strong and flexible for the movement of the wings 33. all of the above are the plants characteristics for the plant to adapt land 34. 35. the haploid stage of plant life in plant life cycle is known as gametophyte. 36.in protostomes the blastopore becomes the mouth.

The _________ is the protective chamber that houses the ovule and Lat.pdf

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Suppose X follows a normal distribution with mean=1 and variance=4. .pdf

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