n = N 2^(rt)n = no of people at a given timeN = initial populat.pdf

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n = N *2^(rt) n = no of people at a given time N = initial population r = rate of population growth t = time so 59 = 35 * 2^ (r*50) so r = 0.015 so n = 35 * 2^ (20r) = 43.13 million Solution n = N *2^(rt) n = no of people at a given time N = initial population r = rate of population growth t = time so 59 = 35 * 2^ (r*50) so r = 0.015 so n = 35 * 2^ (20r) = 43.13 million.

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public class Storm { //Attributes private String stormName; private int stormYear; private String stormStart; private String stormEnd; private int stormMag; /** * Constructor * * @param stormName * @param stormYear * @param stormStart * @param stormEnd * @param stormMag */ public Storm(String stormName, int stormYear, String stormStart, String stormEnd, int stormMag) { this.stormName = stormName; this.stormYear = stormYear; this.stormStart = stormStart; this.stormEnd = stormEnd; this.stormMag = stormMag; } /** * @return the stormName */ public String getStormName() { return stormName; } /** * @param stormName the stormName to set */ public void setStormName(String stormName) { this.stormName = stormName; } /** * @return the stormYear */ public int getStormYear() { return stormYear; } /** * @param stormYear the stormYear to set */ public void setStormYear(int stormYear) { this.stormYear = stormYear; } /** * @return the stormStart */ public String getStormStart() { return stormStart; } /** * @param stormStart the stormStart to set */ public void setStormStart(String stormStart) { this.stormStart = stormStart; } /** * @return the stormEnd */ public String getStormEnd() { return stormEnd; } /** * @param stormEnd the stormEnd to set */ public void setStormEnd(String stormEnd) { this.stormEnd = stormEnd; } /** * @return the stormMag */ public int getStormMag() { return stormMag; } /** * @param stormMag the stormMag to set */ public void setStormMag(int stormMag) { this.stormMag = stormMag; } @Override public String toString() { return \"\ \" + getStormYear() + \": \" + getStormName() + \" \" + ((getStormMag() == -1) ? \"(no info)\" : ((getStormMag() == 0) ? \"(tropical storm)\" : \"(hurricane level \" + getStormMag() + \")\")) + \": \" + ((getStormStart().equals(\"\")) ? \"(no start)\" : getStormStart().substring(0, 2) + \"/\" + getStormStart().substring(2)) + \" - \" + ((getStormEnd().equals(\"\")) ? \"(no end)\" : getStormEnd().substring(0, 2) + \"/\" + getStormEnd().substring(2)); } } import java.io.File; import java.io.FileNotFoundException; import java.util.Scanner; public class Database { private static final int MAX_SIZE = 50; //Attributes private Storm[] stormArr; private int count; /** * Constructor * Accepts a file and attempts to read it * Fills the storm array with the data */ public Database(File fileName) { //Initialize array this.stormArr = new Storm[MAX_SIZE]; this.count = 0; //Scanner to read from the file Scanner in = null; try { in = new Scanner(fileName); //Read data from the file while(in.hasNextLine()) { //Year of storm/ Name of storm/ mmdd storm started/ mmdd storm ended/ magnitude of storm String line = in.nextLine(); String[] data = line.replaceAll(\"/\", \"/ \").split(\"/\"); if(data.length < 5) System.out.println(\"Database entry not in the correct format: \" + line); //Add data to array this.stormArr[this.count] = new Storm(data[1].trim(), Integer.parseInt(data[0].trim()), data[2].trim(), data[3].trim(), (data[4].trim()..

public class Storm { Attributes private String stormName;.pdf

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picture is missingSolutionpicture is missing.pdf

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package employeeType.employee; public class Employee { private String firstName; private String lastName; private char middleInitial; private boolean fulltime; private char gender; private int employeeNum; public Employee(String firstName, String lastName, char middleInitial,char gender,int employeeNum,boolean fulltime ) { setFirstName(firstName); setLastName(lastName); setMiddleInitial(middleInitial); setFulltime(fulltime); setGender(gender); setEmployeeNum(employeeNum); } public String getFirstName() { return firstName; } public void setFirstName(String firstName) { this.firstName = firstName; } public String getLastName() { return lastName; } public void setLastName(String lastName) { this.lastName = lastName; } public char getMiddleInitial() { return middleInitial; } public void setMiddleInitial(char middleInitial) { this.middleInitial = middleInitial; } public boolean isFulltime() { return fulltime; } public void setFulltime(boolean fulltime) { this.fulltime = fulltime; } public char getGender() { return gender; } public void setGender(char gender) { if(gender==\'M\' || gender==\'F\') { this.gender = gender; } else { this.gender=\'F\'; } } public int getEmployeeNum() { return employeeNum; } public void setEmployeeNum(int employeeNum) { while(true) { if(employeeNum>=10000 && employeeNum<=99999) { this.employeeNum = employeeNum; } else { new IllegalArgumentException(\"Invalid Employee Num.should be between 10000 and 99999(inclusive) \"); continue; } } } Override(Have to keep at the rate symbol before the override where ever it is in this project) public boolean equals(Object obj) { if (this == obj) return true; if (obj == null) return false; if (getClass() != obj.getClass()) return false; Employee other = (Employee) obj; if (employeeNum != other.employeeNum) return false; if (firstName == null) { if (other.firstName != null) return false; } else if (!firstName.equals(other.firstName)) return false; if (fulltime != other.fulltime) return false; if (gender != other.gender) return false; if (lastName == null) { if (other.lastName != null) return false; } else if (!lastName.equals(other.lastName)) return false; if (middleInitial != other.middleInitial) return false; return true; } Override public String toString() { System.out.println(getEmployeeNum()); System.out.println(getFirstName()+\", \"+getLastName()+\" \"+getMiddleInitial()); System.out.println(\"Gender: \"+getGender()); if(isFulltime()) { System.out.println(\"Status: Full Time\"); } else { System.out.println(\"Status: Part Time\"); } return \" \"; } } _____________________________________________________________________ package employeeType.subTypes; import employeeType.employee.Employee; public class HourlyEmployee extends Employee { private double wage; private double hoursWorked; public HourlyEmployee(String firstName, String lastName, char middleInitial, char gender,int employeeNum ,boolean fulltime ,double wage) { super(firstName, lastName, middleInitial, gender, employeeNum, ful.

package employeeType.employee;public class Employee { private.pdf

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OSI (Open Systems Interconnection) is reference model for how applications can communicate over a network. A reference model is a conceptual framework for understanding relationships.In the Open Systems Interconnection (OSI) communications model, the Network layer knows the address of the neighboring nodes in the network, packages output with the correct network address information, selects routes and quality of service, and recognizes and forwards to the Transport layer incoming messages for local host domains.It is the third layer, situated in detween transport layerand data-link layer.The network layer is the third level of the Open Systems Interconnection Model (OSI Model) and the layer that provides data routing paths for network communication. Data is transferred in the form of packets via logical network paths in an ordered format controlled by the network layer. The main aim of this layer is to deliver packets from source to destination across multiple links (networks). If two computers (system) are connected on the same link then there is no need for a network layer. It routes the signal through different channels to the other end and acts as a network controller.It also divides the outgoing messages into packets and to assemble incoming packets into messages for higher levels.The Network Layer is responsible for end-to-end (source to destination) packet delivery including routing through intermediate hosts, whereas the Data Link Layer is responsible for node-to-node (hop-to-hop) frame delivery on the same link. Internet Protocol, or IP, is the method that governs how computers share data across the Internet. When one computer sends data, such as an email or a web form, its message gets parsed into small packets that contain the sending computer\'s Internet address, the receiving computer\'s address, and part of the message. Internet Protocol serves several basic functions. The Internet Protocol (IP) is the method or protocol by which data is sent from one computer to another on the Internet. Each computer (known as a host) on the Internet has at least one IP address that uniquely identifies it from all other computers on the Internet. .IP packet headers contain addresses that identify the sending computer and the receiving computer. Routers use this information to guide each packet across communication networks and connect the sending and receiving computers.TCP/IP stands for Transmission Control Protocol / Internet Protocol. It is the communication protocol used for Internet and similar networks such as Internet and Extranet. It controls and manages the data transmission over the Internet. It also defines a mechanism through which every computer on the Internet is identified separately. Every computer on the Internet must have this protocol. Solution OSI (Open Systems Interconnection) is reference model for how applications can communicate over a network. A reference model is a conceptual framework for understanding relationships.I.

OSI (Open Systems Interconnection) is reference model for how applic.pdf

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Note: Modified code code: #include #include #include #include #include using namespace std; #pragma warning(disable: 4996) typedef enum { male = 0, female = 1 } gender; // enumeration type gender struct dog { char name[30]; gender genderValue; char breed[30]; int age; float weight; }; int count = 0; // the amount of dogs currently stored in the list (initialized at 0) struct dog list[30]; // initialize list of dogs // forward declaration of functions void flush(); void branching(char); void helper(char); int add(char*, char*, char*, int, float, struct dog*); // 30 points char* search(char*, int, struct dog*); // 10 points void display(); void save(char* fileName); void load(char* fileName); // 10 points int main() { load(\"Dog_List.txt\"); // load list of dogs from file (if it exists) char ch = \'i\'; printf(\"Assignment 5: Array of Structs and Enum Types\ \ \"); printf(\"Dog Adoption Center\ \ \"); do { printf(\"Please enter your selection:\ \"); printf(\"\\ta: add a new dog to the list\ \"); printf(\"\\ts: search for a dog on the list\ \"); printf(\"\\td: display list of dogs\ \"); printf(\"\\tq: quit and save your list\ \"); ch = tolower(getchar()); flush(); branching(ch); } while (ch != \'q\'); save(\"Dog_List.txt\"); // save list of dogs to file (overwrite if it exists) return 0; } // consume leftover \'\ \' characters void flush() { int c; do c = getchar(); while (c != \'\ \' && c != EOF); } // branch to different tasks void branching(char c) { switch (c) { case \'a\': case \'s\': helper(c); break; case \'d\': display(); break; case \'q\': break; default: printf(\"Invalid input!\ \"); } } // The helper function is used to determine how much information is needed and which function to send that information to. // It uses values that are returned from some functions to produce the correct ouput. // There is no implementation needed here, but you should study this function and know how it works. // It is always helpful to understand how the code works before implementing new features. // Do not change anything in this function or you risk failing the automated test cases. void helper(char c) { char input[100]; if (c == \'a\') { printf(\"\ Please enter the dog\'s information in the following format:\ \"); printf(\"\\tname:gender:breed:age:weight\ \"); fgets(input, sizeof(input), stdin); // discard \'\ \' chars attached to input input[strlen(input) - 1] = \'\\0\'; char* name = strtok(input, \":\"); // strtok used to parse string char* genderValueString = strtok(NULL, \":\"); char* breed = strtok(NULL, \":\"); int age = atoi(strtok(NULL, \":\")); // atoi used to convert string to int float weight = atof(strtok(NULL, \":\")); // atof used to convert string to float int result = add(name, genderValueString, breed, age, weight, list); if (result == 0) printf(\"\ That dog is already on the list\ \ \"); else printf(\"\ Dog added to list successfully\ \ \"); } else // c = \'s\' { printf(\"\ Please enter the dog\'s information in the following forma.

Note Modified codecode#includeiostream #include stdio.h.pdf

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Oligotrophic area Usually, an olidgotropic organism is a one that can survive in an environment which has low levels of nutrients. So, an oligotropic area would be an environment which has less nutrient to support life. These environments are usually of air, ice, soil or water with low levels of nutrients. Eutropicarea In contrast to oligotropic, eutropic areas are nutrient rich areas. Here, due to excess nutrients/organic matter, a large number of algae and other plants would grow excessively, depleting the oxygen content of the water body. Due to depletion of oxygen, the other aquatic organisms might be affected. Upwelling area This occurs in oceanic ecosystems. Usually, the top or the upper surface of the ocean will be warmer and nutrient-depleted. Due to the wind, the nutrient-rich water will be moved to the upper surface, replacing the water present earlier (nutrient-depleted). This area is nothing but a nutrient-rich area, that was once nutrient-depleted. So, the difference between the three is in terms of the nutrient availability. Oligotropic is a one which has low amounts of nutrients. Eutropic is a one which has high amounts of nutrients and Upwelling area is a one which is nutrient-rich but was once nutrient-depleted. The second question cannot be answered because, stratification is a very common term used to refer to layers. Here, I do not know which stratification you are talking about. Solution Oligotrophic area Usually, an olidgotropic organism is a one that can survive in an environment which has low levels of nutrients. So, an oligotropic area would be an environment which has less nutrient to support life. These environments are usually of air, ice, soil or water with low levels of nutrients. Eutropicarea In contrast to oligotropic, eutropic areas are nutrient rich areas. Here, due to excess nutrients/organic matter, a large number of algae and other plants would grow excessively, depleting the oxygen content of the water body. Due to depletion of oxygen, the other aquatic organisms might be affected. Upwelling area This occurs in oceanic ecosystems. Usually, the top or the upper surface of the ocean will be warmer and nutrient-depleted. Due to the wind, the nutrient-rich water will be moved to the upper surface, replacing the water present earlier (nutrient-depleted). This area is nothing but a nutrient-rich area, that was once nutrient-depleted. So, the difference between the three is in terms of the nutrient availability. Oligotropic is a one which has low amounts of nutrients. Eutropic is a one which has high amounts of nutrients and Upwelling area is a one which is nutrient-rich but was once nutrient-depleted. The second question cannot be answered because, stratification is a very common term used to refer to layers. Here, I do not know which stratification you are talking about..

Oligotrophic area Usually, an olidgotropic organism is a one that .pdf

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Maroochy Shire Sewage Spill Case Study Student’s Name Institution Affiliation The Maroochy Shire Sewage Spill case study is as story of an Australian man who was sent to prison for two years. He was found guilty of hacking into the Maroochy Shire, Queensland computerized waste management system and caused millions of liters of raw sewage to spill out into local parks, rivers and even the grounds of a Hyatt Regency hotel. The Maroochy dore District Court heard that 49-year-old Vitek Boden had conducted a series of electronic attacks on the Maroochy Shire sewage control system after a job application he had made was rejected by the area\'s Council. At the time he was employed by the company that had installed the system. Boden made at least 46 attempts to take control of the sewage system during March and April 2000. On 23 April, the date of Boden\'s last hacking attempt, police who pulled over his car found radio and computer equipment. Later investigations found Boden\'s laptop had been used at the time of the attacks and his hard drive contained software for accessing and controlling the sewage management system. Vitek should have appealed to the board if he felt that the application was wrongly rejected. The situation should have been handled before Vitek hacking the system. The reason for this is because he affected the larger populace and environment by his act. The sewage spill was significant. It polluted over 500 metres of open drain in a residential area and flowed into a tidal canal. Cleaning up the spill and its effects took days and required the deployment of considerable resources. \"Marine life died, the creek water turned black and the stench was unbearable for residents,\" said Janelle Bryant, investigations manager for the Australian Environmental Protection Agency. Solution Maroochy Shire Sewage Spill Case Study Student’s Name Institution Affiliation The Maroochy Shire Sewage Spill case study is as story of an Australian man who was sent to prison for two years. He was found guilty of hacking into the Maroochy Shire, Queensland computerized waste management system and caused millions of liters of raw sewage to spill out into local parks, rivers and even the grounds of a Hyatt Regency hotel. The Maroochy dore District Court heard that 49-year-old Vitek Boden had conducted a series of electronic attacks on the Maroochy Shire sewage control system after a job application he had made was rejected by the area\'s Council. At the time he was employed by the company that had installed the system. Boden made at least 46 attempts to take control of the sewage system during March and April 2000. On 23 April, the date of Boden\'s last hacking attempt, police who pulled over his car found radio and computer equipment. Later investigations found Boden\'s laptop had been used at the time of the attacks and his hard drive contained software for accessing and controlling the sewage management system. Vitek should have appealed to the board if.

Maroochy Shire Sewage Spill Case Study Student’s Name Institution Af.pdf

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It lacks the origin of replication which is most important for the i.pdf

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In developmental biology, an embryo is divided into two hemispheres: the animal pole and the vegetal pole within a blastula. The animal pole consists of small cells that divide rapidly, in contrast with the vegetal pole below it. In some cases, the animal pole is thought to differentiate into the later embryo itself, forming the three primary germ layers and participating in gastrulation. The vegetal pole contains large yolky cells that divide very slowly, in contrast with the animal pole above it. In some cases, the vegetal pole is thought to differentiate into the extraembryonic membranes that protect and nourish the developing embryo, such as the placentain mammals and the chorion in birds. The development of the animal-vegetal axis occurs prior to fertilisation. Sperm entry can occur anywhere in the animal hemisphere. The point of sperm entry defines the dorso-ventral axis - cells opposite the region of sperm entry will eventually form the dorsal portion of the body a. Males release so many sperm that the egg is covered by them. b. The egg has a plasma membrane, a vitelline envelope, and a jelly coat. c. Acrosome enzymes digest away the zona pellucida around the egg as it extrudes a filament that attaches to a receptor on the vitelline jelly layer envelope. d. This interaction between filament and receptor is a lock-and-key reaction that is species- specific. e. The egg plasma membrane and the sperm nuclear membrane fuse, allowing the nucleus to enter. f. Fusion takes place and the zygote begins development. g. As soon as the plasma membranes of sperm and egg fuse, the plasma membrane and the vitelline envelope undergo changes that prevent entrance of any other sperm. h. The vitelline envelope now becomes the fertilization envelope. Early Developmental Stages 1. Development includes events and processes that occur as a single cell becomes a complex organism. 2. All chordate embryos go through same early developmental stages: zygote, morula, blastula, early and late gastrula. 3. The presence of yolk, dense nutrient material, affects how the embryonic cells complete the first three stages. 4. Following fertilization, a zygote undergoes cleavage, cell division without growth. a. DNA replication and mitosis occur repeatedly; the cells get smaller each division. b. As deuterostomes, lancelets have a radial and indeterminate pattern of cleavage. 1) In radial cleavage, any plane passing through will divide the embryo into symmetrical halves. 2) In indeterminate cleavage, cells have not differentiated; if separated, each one develops a complete organism. 5. Because the lancelet has little yolk, the cell divisions are equal in the resulting morula. 6. A cavity called the blastocoel develops forming the hollow ball called the blastula. 7. Gastrulation is invagination of some cells of the blastocyst into blastocoel to form three primary germ layers. a. The outer layer of cells becomes the ectoderm. b. The inner layer of cells becomes the endoderm. c.

In developmental biology, an embryo is divided into two hemispheres.pdf

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Bryophytes- Development of primitive vasculature for water transport, though water absorption mechanism not present. Pteridophytes- Specialised vascular tissue for water transport, ability to collect water from moist soil. Gymnosperms- development of roots and vascular tissue for water absorption and conduction. Development of naked seeds that are wind pollinated. Sporophyte is dominant over the gametophyte. Angiosperms- Well developed vascular system, seed enclosed in seed coat. The earlier groups faced challenges in their transition from an aquatic to a terrestrial mode of life as they faced the problem of desciccation. Hence, they dried out quickly. Most earlier land plants were restricted to moist environments as a result. Minimizing the gametophyte and magnifying the sporophyte has led to the development of seed habit. Further, with the development of heterospory, the megaspore remains in the parent sporophyte and the microgametophyte travels to reach the ovule. Seed plants can reproduce independently of water. Solution Bryophytes- Development of primitive vasculature for water transport, though water absorption mechanism not present. Pteridophytes- Specialised vascular tissue for water transport, ability to collect water from moist soil. Gymnosperms- development of roots and vascular tissue for water absorption and conduction. Development of naked seeds that are wind pollinated. Sporophyte is dominant over the gametophyte. Angiosperms- Well developed vascular system, seed enclosed in seed coat. The earlier groups faced challenges in their transition from an aquatic to a terrestrial mode of life as they faced the problem of desciccation. Hence, they dried out quickly. Most earlier land plants were restricted to moist environments as a result. Minimizing the gametophyte and magnifying the sporophyte has led to the development of seed habit. Further, with the development of heterospory, the megaspore remains in the parent sporophyte and the microgametophyte travels to reach the ovule. Seed plants can reproduce independently of water..

Bryophytes- Development of primitive vasculature for water transport.pdf

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givenSolutiongiven.pdf

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Ethernet II framing (also known as DIX Ethernet, named after DEC, Intel and Xerox, the major participants in its design), defines the two-octet EtherType field in an Ethernet frame, preceded by destination and source MAC addresses, that identifies an upper layer protocol encapsulating the frame data. The Length/EtherType field is the only one which differs between 802.3 and Ethernet II. In 802.3 it indicates the number of bytes of data in the frames payload, and can be anything from 0 to 1500 bytes. Frames must be at least 64 bytes long, not including the preamble, so, if the data field is shorter than 46 bytes, it must be compensated by the Pad field. The reason for specifying a minimum length lies with the collision-detect mechanism. In CSMA/CD a station must never be allowed to believe it has transmitted a frame successfully if that frame has, in fact, experienced a collision. In the worst case it takes twice the maximum propagation delay across the network before a station can be sure that a transmission has been successful. If a station sends a really short frame, it may actually finish sending and release the Ether without realising that a collision has occurred. The 802.3 design rules specify an upper limit on the maximum propagation delay in any Ethernet installation, and the minimum frame size is set to be more than twice this figure. In Ethernet II, on the other hand, this field is used to indicate the type of payload carried by the frame. Solution Ethernet II framing (also known as DIX Ethernet, named after DEC, Intel and Xerox, the major participants in its design), defines the two-octet EtherType field in an Ethernet frame, preceded by destination and source MAC addresses, that identifies an upper layer protocol encapsulating the frame data. The Length/EtherType field is the only one which differs between 802.3 and Ethernet II. In 802.3 it indicates the number of bytes of data in the frames payload, and can be anything from 0 to 1500 bytes. Frames must be at least 64 bytes long, not including the preamble, so, if the data field is shorter than 46 bytes, it must be compensated by the Pad field. The reason for specifying a minimum length lies with the collision-detect mechanism. In CSMA/CD a station must never be allowed to believe it has transmitted a frame successfully if that frame has, in fact, experienced a collision. In the worst case it takes twice the maximum propagation delay across the network before a station can be sure that a transmission has been successful. If a station sends a really short frame, it may actually finish sending and release the Ether without realising that a collision has occurred. The 802.3 design rules specify an upper limit on the maximum propagation delay in any Ethernet installation, and the minimum frame size is set to be more than twice this figure. In Ethernet II, on the other hand, this field is used to indicate the type of payload carried by the frame..

Ethernet II framing (also known as DIX Ethernet, named after DEC, In.pdf

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content analysis refers to an analytical process for measuring the s.pdf

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Because acetone is soluble in water.Since both are miscible They c.pdf

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Answer: Dichlorodiphenyltrichloroethane (DDT) is an organochlorine insecticide. DDT causes the sodium channels in insect neurons to open which causing them to fire uncontrollably (repeatedly generate an impulse). This accounts for the repetitive body tremors, spasms and eventual death of the insect. Solution Answer: Dichlorodiphenyltrichloroethane (DDT) is an organochlorine insecticide. DDT causes the sodium channels in insect neurons to open which causing them to fire uncontrollably (repeatedly generate an impulse). This accounts for the repetitive body tremors, spasms and eventual death of the insect..

AnswerDichlorodiphenyltrichloroethane (DDT) is an organochlorine .pdf

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AnswerNote Entire skeleton of the code is provided. The below co.pdf

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ANS: D (ventral root) Ventral root: The spinal cord grows into ventral and dorsal sides. The Posterior median sulcus groove is called dorsal side and anterior median sulcus groove is called ventral side The ventral roots contain efferent fibers and dorsal root contains afferent fibers. The dorsal roots contain afferent fibers and the ventral roots contain efferent fibers. Solution ANS: D (ventral root) Ventral root: The spinal cord grows into ventral and dorsal sides. The Posterior median sulcus groove is called dorsal side and anterior median sulcus groove is called ventral side The ventral roots contain efferent fibers and dorsal root contains afferent fibers. The dorsal roots contain afferent fibers and the ventral roots contain efferent fibers..

ANS D (ventral root)Ventral rootThe spinal cord grows into ven.pdf

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a). In both unicellular and multicellular organisms the purpose of cell division is growth, maintenance and reproduction. in unicellular organisms like bacteria, the cell division takes place by mitosis to produce two idential daughter cells. In multicellular organisms the all the cells except the gametes undergo mitosis to produce daughter cells that look like parent cells. The cells undergoing mitosis also maintain same number of chromosomes similar to parent cells. The gemetes undergo meiosis or reduction division to produce half number of chromosomes from each parent cell and when both male and female gametes fuse, they produce complete set of chromosomes. b). Cells becoming old or cells undergoing damage require replacement of cells by cell division. The age of cell and any changes causing damage in cells determine the occurrance of cell division in both the prokaryotes and eukaryotes. c). The prokaryotes have small genomes and may contain less than 500 to more than 5000 genes. The bacteria have a small circular chromosome and show specific variation with other bacterial species. The bacteria contain extra chromosomal circular DNA called a plasmid. The plasmid is capable of replicating on its own and often carries the non essential genes that can be transferred from one cell to another. The bacterial genes are organized in to clusters known as operons that can be transferred at a time. The genes of the operons contain protein encoding gene as well as regulatory sequences. The prokaryotes are single celled organism whereas the eukaryotes are multi cellular. The eukaryotic genomes are large in size than the prokaryotes and may range 10 MB to over 100,000MB. Prokaryotes contain a single chromosome and eukaryotes contain several. The eukaryotic genes exhibit special characteristics like presence of non-coding or intervening sequences or introns and repetitive sequences that are responsible for large genome size in eukaryotes. d). In unicellular organisms like bacteria, the cell division takes place by mitosis to produce two idential daughter cells. In multicellular organisms the all the cells except the gametes undergo mitosis to produce daughter cells that look like parent cells. The cells undergoing mitosis also maintain same number of chromosomes similar to parent cells. The gemetes undergo meiosis or reduction division to produce half number of chromosomes from each parent cell and when both male and female gametes fuse, they produce complete set of chromosomes. Solution a). In both unicellular and multicellular organisms the purpose of cell division is growth, maintenance and reproduction. in unicellular organisms like bacteria, the cell division takes place by mitosis to produce two idential daughter cells. In multicellular organisms the all the cells except the gametes undergo mitosis to produce daughter cells that look like parent cells. The cells undergoing mitosis also maintain same number of chromosomes similar to parent cells. Th.

a). In both unicellular and multicellular organisms the purpose of c.pdf

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A chemical property describes a substance and its ability to change into a new substance with different properties. Chemical properties of matter are : FLAMMABILITY: Flammability is the ability to burn. example:wood can be burned to create smoke and ash REACTIVITY: Reactivity is when 2 substances get together,something can happen like bubbling,fizzing or color change. example:Iron reacts with Oxygen to form rust Solution A chemical property describes a substance and its ability to change into a new substance with different properties. Chemical properties of matter are : FLAMMABILITY: Flammability is the ability to burn. example:wood can be burned to create smoke and ash REACTIVITY: Reactivity is when 2 substances get together,something can happen like bubbling,fizzing or color change. example:Iron reacts with Oxygen to form rust.

A chemical property describes a substance and its ability to change .pdf

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9. (b ) Myelinated axons are called nerves in the PNS. Nerves can be myelinated and unmyelinated axons present in PNS. This myelin forms insulation and helps in proper functioning of the nervous tissue. 10. (a) MAO breaks down Norepinephrine intracelluraly in the presynaptic nerve terminal. It is located inside the mitochondria. COMT breaks NE extracellularly. NE is broken down by these two enzymes. Solution 9. (b ) Myelinated axons are called nerves in the PNS. Nerves can be myelinated and unmyelinated axons present in PNS. This myelin forms insulation and helps in proper functioning of the nervous tissue. 10. (a) MAO breaks down Norepinephrine intracelluraly in the presynaptic nerve terminal. It is located inside the mitochondria. COMT breaks NE extracellularly. NE is broken down by these two enzymes..

9. (b ) Myelinated axons are called nerves in the PNS.Nerves can b.pdf

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