How can a topologically associating domain (TAD) mutation effect mor.pdf

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How can a topologically associating domain (TAD) mutation effect more that one gene? Explain why a TAD mutation shouldn\'t have any effect on the normal shape of a protein? And what seems to be one of the normal functions of a TAD.? Solution Topological associated domain are genome region whose DNA sequence prefrentially contact each other. TAD is a fundamental structural unit that regulates elements to their cognate promoters.Function: TAD are critical for accurate gene expression Any aleration can lead to disease.

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Inhaling and exhaling Label the figure that shows \"INHALATION\" and the figure that shows \"EXHALATION\" in targets (a) and (b). Then drag the other labels to the appropriate locations on the figures. Solution a)Inhalation b)Exhalation c)External intercoastal muscles contract d)Air enters body e)Air leaves body f)Internal intercoastal muscles contract g)Diaphragm contracts(moves down) h)Diaphram relaxes(moves up).

Inhaling and exhaling Label the figure that shows INHALATION and.pdf

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In accordance to the PLSS land description system, locate the section of land in the township; locate the parcel land in the section; and determine the acreage of the land parcel. Given the following: W 1/2, NW 1/4, SE 1/4, SE 1/4, Section 28, R2W, T7S, New Mexico Mexico Meridian Area of Land parcel acres Calculate the end area (meters^2) of the land cross section below using the alternative coordinate method. Solution.

In accordance to the PLSS land description system, locate the section.pdf

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Hello, I need help writing a code in C. It wants to do things with files. It gives us two files. The code must first ask which file they want, then open the file. Once opened, the file will have three colums: the names alphabetically, the rank of the name (a name with a rank of 232 is the 232nd most popular name in 2014), and the total number of recorded instances. The program then prompts the user to input a name, when inputed, If the name is found in the database, display the name rank, and number of the name in sequence with the prior two names and the following two names. If the name is not shown, show only the two entries before and after the position where the name would be. My major is mechanical so i have never been very good at coding. Any help would be greatly appreciated. Thank you for your time. Solution #include int main() { int c; char str[30]; printf(\"Enter your choice \ 1.First file\ 2.Second file\ \"); scanf(\"%d\",&c); switch(c) { case 1: FILE *fptr=fopen(\"First file\",rb); break; case 2:FILE *fptr=fopen(\"Second file\",rb); break; default: printf(\"Enter the correct option\ \"); return 1; } struct emp { char name[20]; int pos; int num; }e1; printf(\"Enter the name for search the name in that file:\ \"); scanf(\"%s\",str); while(1) { fread(fptr,1,&e1); if(strcmp(e1->name,str)==1) { printf(\"%s\\t%d\\t%d\ \",e1->name,e1->pos,e1->num); break; } } fclose(fptr); return 0; }.

Hello, I need help writing a code in C. It wants to do things with f.pdf

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Fix the following C program in linux. When it is compiled it gives the following error: warning cast to pointer from integer of different size. Solution #ifndef endLocation #define endLocation #include \"myLinkedlist.h\" #include \"myString.h\" class endLocation can use atoi int travelCost; // can use atoi public: endLocation(); endLocation(String, int, int); endLocation(endLocation& rhs); void setName(String); void setTime(int); void setCost(int); String getName(); int getTime(); int getCost(); }; #endif // endLocation //.cpp implementation endLocation::endLocation() endLocation::endLocation(endLocation& rhs) void endLocation::setName(String inputName) void endLocation::setTime(int inputTime) void endLocation::setCost(int inputCost) String endLocation::getName() come locationName; } int endLocation::getTime() come travelTime; } int endLocation::getCost() come travelCost; } Here is my main int main( int argc, char* const argv[] ) range of lines here for(int i = 0; i < num_lines; ++i)dissect and store all strings size = findIndex(line, \'|\'); strncpy(name, line, size); index = size + 1; size = findIndex(line + index, \'|\'); strncpy(time, line + index, size); index += size + 1; size = findIndex(line + index, \'|\'); strncpy(cost, line + index, size); } in.close(); } else.

Fix the following C program in linux. When it is compiled it gives t.pdf

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7. Wet ear wax (W) is dominant over dry ear wax (w) W=wet ear wax w=dry ear wax a. A 3:1 phenotypic ratio of F1 progeny indicates that the parents are of what genotype? b. A 1:1 phenotypic ratio of F1 progeny indicates that the parents are of what genotype? Solution Ans: This is the result of a problem common in trying to determine the nature of inheritance patterns in humans. The results are combined from offspring from several families. Thus the sticky x sticky cross includes families in which parents may be (AA x AA; Aa X AA). aa x aa..

7. Wet ear wax (W) is dominant over dry ear wax (w)W=wet ear wax.pdf

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Examine your opinion on the death penalty. Defend why you are for the death penalty or against it. Be sure to be specific in your defense on your stance. Solution Death penalty or execution is govt. sanctioned punishment by death.Iam in favour of death penalty as:- 1.) Detterent to crime:- Death penalty is given to criminals who do first degree crimes ike murder , terrorism and espionage.It produces a strong detterent effect on crime and criminals. 2) Fear of death:- Though death is inevitable we all still fear it. Imposing death punishment for brutal crimes like rapes and murders will definately make the criminal think twice before committing a crime. 3) No escape:- A criminal murderer or serial killer can escape imprisonment in the long run leading to more crime and threat to society.. Death penalty puts an end to crime. 4) Saves lifes:- Death penalty produces a strong detterent effect on crime and criminals and hence saves lifes..

Examine your opinion on the death penalty. Defend why you are for th.pdf

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Do you think that knowing about aggression between groups of chimpanzees is useful in understanding conflicts between human societies? Why or why not? Solution In many primates where the range is small , contact with one or more groups of the same species is a very common feature. However the nature of these encounters varies from species to species. Primate groups remain permanently in a specific locality which is described as there home range. Members of the same group exist in this range however it is not necessary that the individuals of the group should remain in the same range but may change over to other ranges and join a new community. This home range consists of a central core area with highest concentration of the predictable resources and represent the region where the group is more likely to be found.It also represents the territory of the group which is defended against intrusion. There may be overlapping of the home range between the groups but not the core area. Territoriality is a behavior exhibited by many primates to protect their home range. This is seen extensively where the home ranges are small and permits patrolling and protection of the territory. Biological continuum refers to the fact that the organisms are related through common ancestry and the behavioral traits seen in one organisms are seen in others also but in varying degrees. The man is considered to be a highly evolved social organism with a huge brain and thinking capacity. He is more related to chimpanzees and the main difference between us and the other primates is that the difference in behavior is merely quantitative and not qualitative. Interactions between groups of chimpanzees include aggression, display chasing and sometimes violent fighters also. Male chimpanzees are highly intolerable to unfamiliar chimpanzees, specially males and aggressively defend their territories and resources. Lethal aggression is a common feature among the chimpanzees. It is common in groups of male chimpanzees which do patrolling around their territory. Sometimes they are also accompanied by one or two females. During this patrolling, they exhibit silent movement, stop frequently, sniff around and climb tall trees and sit on them for hours together to survey the area. They appear to be tense during such situations and any sudden sound causes them to exhibit nervous behavior like touching each other for reassurance or embrace each other. When they encounter members of another community, They exhibit some form of aggression till the other group retreats. Chimpanzees are also known to exhibit cruelty, and aggression combined with compassion, tenderness and altruism which is comparable to human behavior, but the main difference is that chimpanzees are not aware of the suffering they inflict on others while humans do. Thus, the behavior of chimpanzees can be more related to human behavior, but the difference lies in quantitative than qualitative.

Do you think that knowing about aggression between groups of chimpan.pdf

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Diana’s Death Unraveled Hemophilia causes a lack of blood clotting, but Diana died from overactive blood clotting. Factor V Leiden is the most common hereditary hypercoagulability disorder amongst Eurasians. About 5% of Caucasians in North America are affected, and it is inherited as an autosomal dominant mutation that causes high levels of clotting Factor V. It can cause deep vein thrombosis (blood clot in the vein), pain, life-threatening strokes, and heart attacks. 1. Based on the types of mutations we’ve discussed so far in class, which does Factor V Leiden fall into? (1pt) 2. Kevin Keller’s aunt’s husband and his girl cousin have died of heart attacks, but his mother, sister, and brother are fine. Lisa Keller’s uncle is fine, but her aunt had multiple miscarriages before she died. Her mother suffered from deep vein thrombosis, and her grandfather died of a heart attack. Recall that Lisa has a brother with hemophilia C and an unaffected sister. Create a pedigree diagramming these relationships. (3 pts) b. Who do you think is more likely to carry the Factor V Leiden mutation? (1 pt) a. Kevin Keller b. Lisa Keller c. Both c. What do you think happened with the Morrison’s and Keller’s when Olivia and Diana were young? (2 pt) 3. There is enough research evidence to support phenotypic changes in hemophiliacs with Factor V Leiden mutations. Based on what we’ve discussed in chapter 4, what are some suggestions as to why this might happen? Be sure to describe the normal symptoms of hemophilia and how protein levels are affected to support your answer. (3pts) Solution 1. Missense substitution mutation leads to Factor V Leiden . The gene which codes to the factor is F5. A mutation of this gene where Guanine is substituted by Adenine which results in the changing of amino acid Arginine to Glutamine. I am not sure of the answers for the remaining questions. Please post those questions again for other experts to answer. Thank you..

Diana’s Death UnraveledHemophilia causes a lack of blood clotting,.pdf

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Create a new java class called ListNode. Implement ListNode as a generic version of IntNode. public class ListNode { } Create a generic class called Queue, which is a linked list implementation of a queue data type. Queue must have the following methods: • public void enqueue(T value){} This method should add a new node to the back of the queue. • public T dequeue(){} This method should remove a node from the front of the queue and return its value • public T peek(){} This method should return the value of the front node without removing it from the queue Create a new generic class called Stack, which is a linked list implementation of a stack data type. Stack must have the following methods: • public void push(T value){} This method should add a new node to the top of the stack. • public T pop(){} This method should remove a node from the top of the stack and return its integer value • public T peek(){} This method should return the integer value of the top node without removing it from the stack Solution import java.util.Iterator; public class LinkedList extends AbstractList { private ListNode front; // refers to first node in list (null if empty) public LinkedList() { front = null; // null front means empty } public void add(int index, E value) { if (index == 0) { // insert at the front front = new ListNode(value, front); } else { ListNode current = goTo(index - 1); ListNode newNode = new ListNode(value, current.next); current.next = newNode; } } public E get(int index) { ListNode current = goTo(index); return current.data; } public int indexOf(E value) { int index = 0; ListNode current = front; while (current != null) { if (current.data == value) { return index; } index++; current = current.next; } return -1; } public Iterator iterator() { return new LinkedListIterator(); } public void remove(int index) { if (index == 0) { // removing from the front front = front.next; } else { ListNode current = goTo(index - 1); current.next = current.next.next; // deletes the node } } public void set(int index, E value) { ListNode current = goTo(index); current.data = value; } public int size() { int count = 0; ListNode current = front; while (current != null) { current = current.next; count++; } return count; } public String toString() { if (front == null) { return \"[]\"; } else { String result = \"[\" + front.data; ListNode current = front.next; while (current != null) { result += \", \" + current.data; current = current.next; } result += \"]\"; return result; } } private ListNode goTo(int index) { ListNode current = front; for (int i = 0; i < index; i++) { current = current.next; } return current; } public class ListNode { public E data; public ListNode next; public ListNode(E data) { this.data = data; this.next = null; } public ListNode(E data, ListNode next) { this.data = data; this.next = next; } } private class LinkedListIterator implements Iterator { private ListNode current; // current position in list public LinkedListIterator() { current = f.

Create a new java class called ListNode. Implement ListNode as a gen.pdf

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Complete the following short answer questions: 1. How do the viruses propagate between computers? 2. What is social engineering? 3. How will an SPI firewall handle a packet containing a TCP segment which is an acknowledgement? 4. Revise the access control list (ACL) in Figure 3-23 (page 119 in the textbook) to permit access to an FTP server with IP address 10.32.67.112. 5. How will the ACL in Figure 3-23 (page 119 in the textbook) handle a packet that attempts to open a connection to an FTP server? Explain. 6. For each of the following passwords, first state the kind of attack that would be necessary to crack it. Justify your answer. Then say whether or not it is an adequate password, again giving specific reasons. a) password b) Winter1 c) SpringBreaK d) 2!T*d (00-10) e) 9g&8tY7#?s+445=232+ Solution Answers: 1)How do the viruses propagate between computers? -> The browsers gain more features and functions more methods for virus propagation appear directly to a computer across the network. Web pages : ->Either via a download, web code, or a cross-site scripting (CSS) attack, web pages can be a source of viral infection. As browsers gain more features and functions more methods for virus propagation appear. Messages: -> E-mail messages as well as IM or chat messages can be a vector for viral infection. -> Whether the virus is spread through links, embedded code, or an attachment depends on the type of message and the client software. Infected programs or Trojans : -> Running an infected program that you copied from a CD or other media (or downloaded) is a means of infection. Sometimes the virus is a Trojan, where it appears to be a useful program, but is in fact a virus. 2. What is social engineering? -> Social engineering is an attack vector that relies heavily on human interaction and often involves tricking people into breaking normal security procedures. -> Social engineering is the art of manipulating people so they give up confidential information. -> A social engineer runs what used to be called a \"con game. Popular types of social engineering attacks : 1)Phishing: -> Phishing is when a malicious party sends a fraudulent email disguised as a legitimate email, often purporting to be from a trusted source. 2) Pretexting: ->Pretexting is another form of social engineering where attackers focus on creating a good pretext, or a fabricated scenario, that they can use to try and steal their victims’ personal information. 3. How will an SPI firewall handle a packet containing a TCP segment which is an acknowledgement? -->Process it through the Access Control List (ACL) -> Stateful Packet Inspection: Connections have distinct states or stages. ->Different states are subject to different attacks. Stateful firewalls use different filtering rules for different states. Different because it uses filter rules in order to accept or deny traffic. ->An example would be windows firewall. 4. Revise the access control list (ACL) in Figure 3-23 (page 119 in.

Complete the following short answer questions1. How do the virus.pdf

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Briefly describe the two competing theories regarding the origin of Homo sapiens and its distributional expansion from its origin in Africa. What does the genetic data say about these two hypothesis, which genes are most useful for these analyses and why? Given a large section of a genome, say a complete chromosome, how could you locate genes? Given a large section of a gene, say a complete exon, how could you determine reading frame? Solution 1. There are many competing therapies for the origin of Homo sapiens. Two such theories are Multi-regional hypothesis and Out of Africa 2 model. Multi-regional Hypothesis: The multi-regional hypothesis argues that our early hominid ancestors, including Homo ergaster and Homo heidelbergensis, migrated out of Africa and thus the evolution of modern humans took place in different parts of the world, a process termed regional continuity. This theory places great emphasis on the notion of steady evolutionary alterations or changes that happen in different regions and produce diverse variations of the species. Evolution of this kind is kept at a regular rate due to an amalgamation of cultural progress and ‘gene flow’ or interbreeding, thus keeping all lineages evolving at the same time. This process is characterized as parallel evolution, which suggests a subtle morphological resemblance between populations of species who are geographically separated. Out of Africa 2 model: It is widely recognized among both archaeological and anthropological academics that archaic Homo populations did leave Africa in an initial phase of globalisation, called the Out of Africa 1 model. Following this, the population replacement hypothesis maintains that modern humans evolved in Africa from the ancestral hominids that did not travel out of this continent in the first stage of global colonisation. It is then argued in this model that once evolved as anatomically modern humans, Homo sapiens travelled out of Africa to explore, colonise and rapidly replace the archaic Homo population. This movement suggests a second wave of hominid species emerging from Africa, thus the title: the Out of Africa 2 model..

Briefly describe the two competing theories regarding the origin of H.pdf

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Atlas Advertising Atlas Advertising is a regional advertising agency with oces in Boston, New York, Providence, Washington, D.C., and Philadelphia. 1. Describe the types of networks you think they would have (e.g., LANs, BNs, WANs) and where they are likely to be located. 2. What types of standard proto-cols and technologies do you think they are using at each layer (e.g., see Figure 1-5)? Solution 1. Within their offices,internally they will use local area network (LAN) to communicate among themselves.In the same city ,if they have another office,they can use metropolitian area network(MAN),which is used to communicate in a city or within a cities.For their offices in other cities,they will use WAN(Wide area network)which is used for communication worldwide . 2. The standard protocols and technologies used are TCP-IP for the wired transmission and UDP for wireless Transmission.

Atlas Advertising Atlas Advertising is a regional advertising agency.pdf

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4. Show that Prefix(L)-x: for some y, xy E L is regular if L is regular Solution As L is a regular language it will have a DFA. The new language prefix(L) should accept x if xy (for some y) belongs to L. As we are given that y can be anything, this can be achieved by taking all states which have a path from start and will reach a final state for some string, we make all those states as final states. This DFA will accept a substring of any string accepted by L. Now as prefix(L) has a DFA, it is a regular language..

4. Show that Prefix(L)-x for some y, xy E L is regular if L is regul.pdf

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26) Which is a function of an operating system? O Browse the Internet O Edit images O Protect against malware O Provide user interface Next Question Solution 26. Provide user interface All other mentioned functionality can be achieved by other software 27. Defender 28. APIs 29. device driver Without device drivers you can use your hardware functionality optimally 30. Application Software: User can not directly interact with hardware directly, they need some kind of application software to interact with them.

26) Which is a function of an operating system O Browse the Internet.pdf

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1.Declare a student structure that contains : - Student\'s first and last name -Student ID 2. Create the following functions: -getStudentInfo(void) Declares a single student, uses printf()/scanf() to get input Returns the single student back - printStudentInfo(student_t *st_ptr) Takes the pointer to a student (to avoid making a copy) -Prints out all of the student information 3. Declare an array of five students 4. Using a for loop and getStudentInfo() function, get input of all the students 5. Using a for loop and printStudentInfo() function, print all the output of all students. Solution getStudentInfo(void) { printStudentInfo(student_t *st_ptr) { str_ptr.getStudentInfo(); display(stud); return 0; } int main() { printStudentInfo(str_ptr); } return 0; }.

1.Declare a student structure that contains - Students first an.pdf

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1. Describe how plant vasculature is optimized for water transport in the tallest trees 2. What property of water molecules allow water to stay together while being transported to extreme heights? And how does temperature affect this? 3.Water and minerals enter the plant through _______, and are then transported through plant\'s vasculature, via the _________. This flow is maintained through constant ________, where water exists through __________ in the leaves. The sugars that are made during photosynthesis in the ____________ are transported downwards through the vasculature via the _______. 4. Give an example of ways plant have evolved to cope with drought, Solution Water transport in tree occurs by cappilary action .Water is trasported by root to leaves through xylem . xylem is made up by tube like cells tracheids and vessel . These join together from root to leaves . Two forces are usful to flow water upward 1. root pressure and 2. transpiration . Root pressure occure when water flow into root due to osmotic pressure . Transpiration occures due to evoporation from leaf surface . The adhesive nature of water molecules to inner cappilary wall help to keep water column against gravitational force . 2 Adhesive nature of water to inner cappilary wall . Temprature increases transpiration and low osmatic pressure in leaves . 3. water and mineral entered the plant through ROOT and are then trasported through plant vasculature , via the XYLEM These flow is maintained through constant TRANSPIRATION where water exist through STOMETA in the leaves . Thesuger that are made during photosynthesis in the leaf are transported downward through the vasculature via PHLOEM. 4 Plant have evolved to cope with drought as xerophytic plant have well develop root system , fleshy leaves and stometa sunk in fleshy leave to avoid tranpiration ..

1. Describe how plant vasculature is optimized for water transport i.pdf

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A19730 Which, ofthe foliowrgconsidered a tusa.a.for restructuring ro.pdf

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